Graduate Texts in Physics

Jean-Louis Basdevant

Lectures on Quantum MechanicsWith Problems, Exercises and Their Solutions

Second Edition

Graduate Texts in Physics

Series editors

Kurt H. Becker, Polytechnic School of Engineering, Brooklyn, USAJean-Marc Di Meglio, Université Paris Diderot, Paris, FranceSadri Hassani, Illinois State University, Normal, USABill Munro, NTT Basic Research Laboratories, Atsugi, JapanRichard Needs, University of Cambridge, Cambridge, UKWilliam T. Rhodes, Florida Atlantic University, Boca Raton, USASusan Scott, Australian National University, Acton, AustraliaH. Eugene Stanley, Boston University, Boston, USAMartin Stutzmann, TU München, Garching, GermanyAndreas Wipf, Friedrich-Schiller-Universität Jena, Jena, Germany

Graduate Texts in Physics

Graduate Texts in Physics publishes core learning/teachingmaterial for graduate- andadvanced-level undergraduate courses on topics of current and emerging fields withinphysics, both pure and applied. These textbooks serve students at the MS- orPhD-level and their instructors as comprehensive sources of principles, definitions,derivations, experiments and applications (as relevant) for their mastery and teaching,respectively. International in scope and relevance, the textbooks correspond to coursesyllabi sufficiently to serve as required reading. Their didactic style, comprehensive-ness and coverage of fundamental material also make them suitable as introductionsor references for scientists entering, or requiring timely knowledge of, a research field.

More information about this series at http://www.springer.com/series/8431

Jean-Louis Basdevant

Lectures on QuantumMechanicsWith Problems, Exercises andTheir Solutions

Second Edition

123

Jean-Louis BasdevantDépartement de PhysiqueEcole PolytechniquePalaiseau CedexFrance

ISSN 1868-4513 ISSN 1868-4521 (electronic)Graduate Texts in PhysicsISBN 978-3-319-43478-0 ISBN 978-3-319-43479-7 (eBook)DOI 10.1007/978-3-319-43479-7

Library of Congress Control Number: 2016947037

© Springer International Publishing Switzerland 2007, 2016This work is subject to copyright. All rights are reserved by the Publisher, whether the whole or partof the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations,recitation, broadcasting, reproduction on microfilms or in any other physical way, and transmissionor information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilarmethodology now known or hereafter developed.The use of general descriptive names, registered names, trademarks, service marks, etc. in thispublication does not imply, even in the absence of a specific statement, that such names are exempt fromthe relevant protective laws and regulations and therefore free for general use.The publisher, the authors and the editors are safe to assume that the advice and information in thisbook are believed to be true and accurate at the date of publication. Neither the publisher nor theauthors or the editors give a warranty, express or implied, with respect to the material contained herein orfor any errors or omissions that may have been made.

Printed on acid-free paper

This Springer imprint is published by Springer NatureThe registered company is Springer International Publishing AG Switzerland

Preface

This book is a revised and extended version of my introductory lectures on quantummechanics that I delivered for many years at the Ecole Polytechnique. It is not atextbook. I was dragged into writing it by friends, among whom are many formerstudents of mine.

I have published two books in the same Springer collection with my colleagueand friend Jean Dalibard.

One is a textbook: Quantum Mechanics, J.-L. Basdevant and Jean Dalibard,Heidelberg: Springer-Verlag, 2002 (revised in 2005).

The other one is a collection of problems and their solutions: The QuantumMechanics Solver, J.-L. Basdevant and Jean Dalibard, Heidelberg: Springer-Verlag,2000 (completely revised in 2005). All of these problems concern contemporaryexperimental or theoretical developments, some of which had appeared in thespecialized literature a year or so before we gave them as written examinations.Needless to say that if the second of these books is somewhat unusual, there aredozens of excellent textbooks on quantum mechanics, among which some mas-terpieces which I often consult and refer to.

The remarks that eventually convinced me to write the present text are twofold.The textbooks I had written on the subject, both in French and in English, wereterribly deprived of life, action, thoughts, and questioning which I always liked toput in the narrative account of the ideas and applications of the subject, during mylectures. The human aspect of the experimental investigations and of the ensuingdiscovery of basic principles made the lectures lively (besides the fact that theminds need to rest for few minutes after following a difficult argument). I alwaysthought that teaching science is incomplete if it does not incorporate the humandimension, be it of the lecturer, of the audience and of the topic to which it isdevoted.

The second is that my original publication was totally deprived of any exerciseor problem. Very many remarks were that the book helped a lot to understandquantum mechanics, but it did not help much to work out applications. Therefore ithas been one of my major goals to fill that void. The present book contains ten

v

problems, placed when the tools to solve them have been treated, and followed bytheir solutions and possible comments. Many of them refer to quite modern physics(there is no overlap with The Quantum Mechanics Solver). It also contains sixty orso shorter exercises whose solutions are given at the end of the book.

I must say a few words about the content of this book. First, of course, mylectures evolved quite a lot in 25 years. Actually they were never the same from oneyear to the next. Minds evolve; student’s minds as well as mine. Science evolves:during that period there appeared numerous crucial experimental and technologicalsteps forward.

So each lecture itself must be considered as a superposition of texts and topics,which I could not have covered completely in about an hour and a half. I used tomake selections according to my mood, to latest experimental results, to the evo-lution of the student’s minds in mathematics, in physics, and in regard to the worldthey were facing. The first lecture always consisted of a general description ofcontemporary physics and of the various courses that students were offered in theircurriculum. I have reproduced an example in Chap. 1.

Another point which is amusing and quite characteristic of the French highereducational system (devised more than two centuries ago) is that the students of theEcole Polytechnique, who were all selected after a stiff entrance examination, andwhose ambitions in life were diverse—in science, in industry, in business, and inhigh public office—all had to follow this introductory physics course. The officialreason put forward was that whatever they were going to work on later, QuantumMechanics and Fundamental physics would be indispensable in their occupation, aswould Pure and Applied Mathematics. The famous mathematician LaurentSchwartz, the man I admired most, who was my colleague, liked to be asked thequestion: what’s the use of doing mathematics? “It’s very simple,” he said.

Mathematicians study Lp spaces, negligible sets, and representable functors. One mustcertainly do mathematics. Because mathematics allows to do physics. Physics allows tomake refrigerators. Refrigerators allow to keep Lobsters, and Lobsters are useful formathematicians who can eat them and therefore be in a good mood to do Lp spaces,negligible sets, and representable functors. It’s obviously useful to do mathematics.

I also tried to attract students to physics. I must admit that when LaurentSchwartz retired, we discussed that point, and we came to the conclusion thatobviously it was impossible to convince our students that what we had taught themwould be indispensable, even if they were to manage a large company: “I do notmanage to persuade them because they know very well that I am objectively wrong[…] on that issue I failed completely,” he said. And I fully agreed with him.

Nevertheless, quantum mechanics is an ideal subject because one can be inter-ested in it for a variety of reasons such as the physics itself, the mathematicalstructure of the theory, its technological spinoffs, as well as its philosophical orcultural aspects. And the task was basically to think about the pedagogical aspects,in order to satisfy audiences that went up to 500 students during the last ten years.I do think it is a part of their indisputable personal culture. It seems difficult to graspthe concepts and the functioning of quantum mechanics past a certain age.

vi Preface

I thank Jean Dalibard, who is now my successor, and Philippe Grangier for theirconstant help during the last 10–15 years. They are in particular responsible for partof the text on quantum entanglement and Bell’s inequalities, of which they areworldwide known specialists. I am deeply grateful to James Rich and to AlfredVidal-Madjar. Both of them contributed immensely on all of this book. Discussingwith them was an everlasting pleasure, and they taught me a lot of physics.

I want to thank Jean-Michel Bony and to pay a tribute to the memory of LaurentSchwartz. Both had the patience to explain to me with an incredible profoundnessand clarity the mathematical subtleties of quantum mechanics. This enabled me toeliminate most of the unnecessary mathematical complications at this stage, and stillto be able to answer the questions of my more mathematically minded students.Indeed, if quantum mechanics has been a rich field of investigation for mathe-maticians, it is really the physics that is subtle in it.

Paris, France Jean-Louis Basdevant

Preface vii

Contents

1 The Appeal of Physics. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.1 The Interplay of the Eye and the Mind. . . . . . . . . . . . . . . . . . . 11.2 Advanced Technologies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.3 The Pillars of Contemporary Physics . . . . . . . . . . . . . . . . . . . . 6

1.3.1 Mysteries of Light . . . . . . . . . . . . . . . . . . . . . . . . . . . 61.3.2 Fundamental Structure of Matter . . . . . . . . . . . . . . . . 8

1.4 The Infinitely Complex . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91.5 The Universe . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121.6 Physical Constants . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

2 A Quantum Phenomenon . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152.1 Wave Behavior of Particles. . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

2.1.1 Interferences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182.1.2 Wave Behavior of Matter . . . . . . . . . . . . . . . . . . . . . 192.1.3 Analysis of the Phenomenon . . . . . . . . . . . . . . . . . . . 21

2.2 Probabilistic Nature of Quantum Phenomena . . . . . . . . . . . . . . 222.2.1 Random Behavior of Particles . . . . . . . . . . . . . . . . . . 222.2.2 A Nonclassical Probabilistic Phenomenon . . . . . . . . . 22

2.3 Conclusions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 232.4 Appendix: Notions on Probabilities . . . . . . . . . . . . . . . . . . . . . 272.5 Exercises. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

3 Wave Function, Schrödinger Equation . . . . . . . . . . . . . . . . . . . . . . . 353.1 Terminology and Methodology. . . . . . . . . . . . . . . . . . . . . . . . . 353.2 Principles of Wave Mechanics . . . . . . . . . . . . . . . . . . . . . . . . . 37

3.2.1 The Wave Function . . . . . . . . . . . . . . . . . . . . . . . . . . 373.2.2 Schrödinger Equation. . . . . . . . . . . . . . . . . . . . . . . . . 38

3.3 Superposition Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 403.4 Wave Packets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

3.4.1 Free Wave Packets . . . . . . . . . . . . . . . . . . . . . . . . . . 413.4.2 Fourier Transforms . . . . . . . . . . . . . . . . . . . . . . . . . . 413.4.3 Shape of Wave Packets . . . . . . . . . . . . . . . . . . . . . . . 43

ix

3.5 Historical Landmarks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 443.6 Momentum Probability Law . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

3.6.1 Free Particle. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 453.6.2 General Case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46

3.7 Heisenberg Uncertainty Relations . . . . . . . . . . . . . . . . . . . . . . . 463.8 Controversies and Paradoxes. . . . . . . . . . . . . . . . . . . . . . . . . . . 503.9 Exercises. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 523.10 Appendix: Dirac δ “Function”, Distributions. . . . . . . . . . . . . . . 533.11 Appendix: Fourier Transformation . . . . . . . . . . . . . . . . . . . . . . 58

3.11.1 Uncertainty Relation . . . . . . . . . . . . . . . . . . . . . . . . . 613.12 Exercises. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62

4 Physical Quantities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 634.1 Statement of the Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64

4.1.1 Physical Quantities . . . . . . . . . . . . . . . . . . . . . . . . . . 644.1.2 Position and Momentum . . . . . . . . . . . . . . . . . . . . . . 65

4.2 Observables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 664.2.1 Position Observable . . . . . . . . . . . . . . . . . . . . . . . . . . 674.2.2 Momentum Observable . . . . . . . . . . . . . . . . . . . . . . . 674.2.3 Correspondence Principle . . . . . . . . . . . . . . . . . . . . . 684.2.4 Historical Landmarks. . . . . . . . . . . . . . . . . . . . . . . . . 69

4.3 A Counterexample of Einstein and Its Consequences . . . . . . . . 694.3.1 What Do We Know After a Measurement? . . . . . . . . 714.3.2 Eigenstates and Eigenvalues of an Observable. . . . . . 724.3.3 Wave Packet Reduction . . . . . . . . . . . . . . . . . . . . . . . 73

4.4 The Specific Role of Energy. . . . . . . . . . . . . . . . . . . . . . . . . . . 744.4.1 The Hamiltonian . . . . . . . . . . . . . . . . . . . . . . . . . . . . 744.4.2 The Schrödinger Equation, Time and Energy . . . . . . 754.4.3 Stationary States . . . . . . . . . . . . . . . . . . . . . . . . . . . . 764.4.4 Motion: Interference of Stationary States . . . . . . . . . . 77

4.5 Schrödinger’s Cat . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 784.6 Exercises. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82

5 Energy Quantization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 835.1 Methodology. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83

5.1.1 Bound States and Scattering States . . . . . . . . . . . . . . 845.1.2 One-Dimensional Problems . . . . . . . . . . . . . . . . . . . . 85

5.2 The Harmonic Oscillator. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 855.3 Square Well Potentials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 875.4 Double Well, the Ammonia Molecule. . . . . . . . . . . . . . . . . . . . 92

5.4.1 The Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 925.4.2 Stationary States, the Tunnel Effect . . . . . . . . . . . . . . 935.4.3 Energy Levels . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 955.4.4 Wave Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 965.4.5 Inversion of the Molecule . . . . . . . . . . . . . . . . . . . . . 97

x Contents

5.5 Illustrations and Applications of the Tunnel Effect . . . . . . . . . . 995.6 Tunneling Microscopy, Nanotechnologies. . . . . . . . . . . . . . . . . 1025.7 Exercises. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1045.8 Problem. The Ramsauer Effect . . . . . . . . . . . . . . . . . . . . . . . . . 106

5.8.1 Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1075.9 Problem. Colored Centers in Ionic Cristals . . . . . . . . . . . . . . . . 108

5.9.1 Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112

6 Principles of Quantum Mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . . 1196.1 Hilbert Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1206.2 Dirac Formalism . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124

6.2.1 Notations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1246.2.2 Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1266.2.3 Syntax Rules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1276.2.4 Projectors; Decomposition of the Identity . . . . . . . . . 128

6.3 Measurement Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1286.3.1 Eigenvectors and Eigenvalues of an Observable . . . . 1296.3.2 Results of the Measurement

of a Physical Quantity . . . . . . . . . . . . . . . . . . . . . . . . 1306.3.3 Probabilities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1306.3.4 The Riesz Spectral Theorem . . . . . . . . . . . . . . . . . . . 1316.3.5 Physical Meaning of Various Representations . . . . . . 133

6.4 Principles of Quantum Mechanics. . . . . . . . . . . . . . . . . . . . . . . 1336.5 Heisenberg’s Matrices. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1366.6 The Polarization of Light, Quantum “Logic” . . . . . . . . . . . . . . 1406.7 Exercises. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 144

7 Two-State Systems. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1477.1 The NH3 Molecule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1487.2 “Two-State” System . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1487.3 Matrix Quantum Mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . . 1517.4 NH3 in an Electric Field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 154

7.4.1 Uniform Constant Field . . . . . . . . . . . . . . . . . . . . . . . 1557.4.2 Weak and Strong Field Regimes . . . . . . . . . . . . . . . . 1567.4.3 Other Two-State Systems. . . . . . . . . . . . . . . . . . . . . . 157

7.5 Motion of Ammonia Molecule in an Inhomogeneous Field . . . 1587.5.1 Force on the Molecule in an Inhomogeneous

Field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1587.5.2 Population Inversion . . . . . . . . . . . . . . . . . . . . . . . . . 160

7.6 Reaction to an Oscillating Field, the Maser . . . . . . . . . . . . . . . 1607.7 Principle and Applications of the Maser . . . . . . . . . . . . . . . . . . 163

7.7.1 Amplifiers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1647.7.2 Oscillators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1647.7.3 Atomic Clocks and the GPS . . . . . . . . . . . . . . . . . . . 1657.7.4 Tests of Relativity . . . . . . . . . . . . . . . . . . . . . . . . . . . 166

Contents xi

7.8 Exercises. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1697.9 Problem: Neutrino Oscillations . . . . . . . . . . . . . . . . . . . . . . . . . 170

7.9.1 Mechanism of the Oscillations;Reactor Neutrinos . . . . . . . . . . . . . . . . . . . . . . . . . . . 172

7.9.2 Oscillations of Three Species; AtmosphericNeutrinos . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 174

7.9.3 Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1767.9.4 Comments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 181

8 Algebra of Observables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1838.1 Commutation of Observables . . . . . . . . . . . . . . . . . . . . . . . . . . 183

8.1.1 Fundamental Commutation Relation . . . . . . . . . . . . . 1848.1.2 Other Commutation Relations . . . . . . . . . . . . . . . . . . 1848.1.3 Dirac in the Summer of 1925 . . . . . . . . . . . . . . . . . . 185

8.2 Uncertainty Relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1868.3 Evolution of Physical Quantities . . . . . . . . . . . . . . . . . . . . . . . . 187

8.3.1 Evolution of an Expectation Value . . . . . . . . . . . . . . 1878.3.2 Particle in a Potential, Classical Limit . . . . . . . . . . . . 1888.3.3 Conservation Laws . . . . . . . . . . . . . . . . . . . . . . . . . . 190

8.4 Algebraic Resolution of the Harmonic Oscillator . . . . . . . . . . . 1918.5 Commuting Observables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 194

8.5.1 Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1958.5.2 Example. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1968.5.3 Tensor Structure of Quantum Mechanics . . . . . . . . . . 1968.5.4 Complete Set of Commuting

Observables (CSCO) . . . . . . . . . . . . . . . . . . . . . . . . . 1978.5.5 Completely Prepared Quantum State . . . . . . . . . . . . . 198

8.6 Sunday September 20, 1925 . . . . . . . . . . . . . . . . . . . . . . . . . . . 1998.7 Exercices. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2018.8 Problem. Quasi-Classical States of the Harmonic

Oscillator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2048.8.1 Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 205

8.9 Problem. Benzene and C8 Molecules . . . . . . . . . . . . . . . . . . . . 2088.9.1 Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 209

8.10 Problem. Conductibility of Crystals; Band Theory . . . . . . . . . . 2108.10.1 Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 215

9 Approximation Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2199.1 Perturbation Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 219

9.1.1 Definition of the Problem . . . . . . . . . . . . . . . . . . . . . 2199.1.2 First Order Perturbation Theory . . . . . . . . . . . . . . . . . 2219.1.3 Second Order Perturbation to the Energy Levels . . . . 223

9.2 The Variational Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2249.3 Exercises. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 227

xii Contents

10 Angular Momentum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22910.1 Fundamental Commutation Relation . . . . . . . . . . . . . . . . . . . . . 230

10.1.1 Classical Angular Momentum . . . . . . . . . . . . . . . . . . 23010.1.2 Definition of an Angular Momentum Observable . . . 23010.1.3 Results of the Quantization . . . . . . . . . . . . . . . . . . . . 231

10.2 Proof of the Quantization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23110.2.1 Statement of the Problem. . . . . . . . . . . . . . . . . . . . . . 23110.2.2 Vectors jj;m[ and Eigenvalues j and m . . . . . . . . . 23310.2.3 Operators J� ¼ Jx � iJy . . . . . . . . . . . . . . . . . . . . . . . 23310.2.4 Quantization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 235

10.3 Orbital Angular Momenta . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23610.3.1 Formulae in Spherical Coordinates . . . . . . . . . . . . . . 23610.3.2 Integer Values of m and ‘ . . . . . . . . . . . . . . . . . . . . . 23610.3.3 Spherical Harmonics . . . . . . . . . . . . . . . . . . . . . . . . . 237

10.4 Rotation Energy of a Diatomic Molecule . . . . . . . . . . . . . . . . . 23910.5 Interstellar Molecules, the Origin of Life . . . . . . . . . . . . . . . . . 24110.6 Angular Momentum and Magnetic Moment . . . . . . . . . . . . . . . 245

10.6.1 Classical Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24610.6.2 Quantum Transposition . . . . . . . . . . . . . . . . . . . . . . . 24710.6.3 Experimental Consequences. . . . . . . . . . . . . . . . . . . . 24810.6.4 Larmor Precession . . . . . . . . . . . . . . . . . . . . . . . . . . . 24910.6.5 What About Half-Integer Values of j and m? . . . . . . 250

10.7 Exercises. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 250

11 The Hydrogen Atom . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25311.1 Two-Body Problem; Relative Motion . . . . . . . . . . . . . . . . . . . . 25411.2 Motion in a Central Potential . . . . . . . . . . . . . . . . . . . . . . . . . . 25611.3 The Hydrogen Atom. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 260

11.3.1 Atomic Units; Fine Structure Constant . . . . . . . . . . . 26111.3.2 The Dimensionless Radial Equation . . . . . . . . . . . . . 26211.3.3 Spectrum of Hydrogen. . . . . . . . . . . . . . . . . . . . . . . . 26411.3.4 Stationary States of the Hydrogen Atom . . . . . . . . . . 26511.3.5 Dimensions and Orders of Magnitude . . . . . . . . . . . . 26711.3.6 Historical Landmarks. . . . . . . . . . . . . . . . . . . . . . . . . 268

11.4 Muonic Atoms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26911.5 Exercises. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27211.6 Problem. Decay of a Tritium Atom . . . . . . . . . . . . . . . . . . . . . 275

11.6.1 Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 276

12 Spin 1/2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27912.1 Experimental Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27912.2 Spin 1/2 Formalism . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28112.3 Complete Description of a Spin 1/2 Particle . . . . . . . . . . . . . . . 283

12.3.1 Observables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28412.4 Physical Spin Effects . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 285

Contents xiii

12.5 Spin Magnetic Moment. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28512.6 The Stern–Gerlach Experiment . . . . . . . . . . . . . . . . . . . . . . . . . 28612.7 Principle of the Experiment . . . . . . . . . . . . . . . . . . . . . . . . . . . 287

12.7.1 Semi-classical Analysis . . . . . . . . . . . . . . . . . . . . . . . 28712.7.2 Experimental Results . . . . . . . . . . . . . . . . . . . . . . . . . 28812.7.3 Explanation of the Stern–Gerlach Experiment . . . . . . 28912.7.4 Successive Stern–Gerlach Setups . . . . . . . . . . . . . . . . 29112.7.5 Measurement Along an Arbitrary Axis . . . . . . . . . . . 293

12.8 The Discovery of Spin . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29412.8.1 The Hidden Sides of the Stern–Gerlach

Experiment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29412.8.2 Einstein and Ehrenfest’s Objections. . . . . . . . . . . . . . 29612.8.3 Anomalous Zeeman Effect . . . . . . . . . . . . . . . . . . . . . 29712.8.4 Bohr's Challenge to Pauli . . . . . . . . . . . . . . . . . . . . . 29812.8.5 The Spin Hypothesis . . . . . . . . . . . . . . . . . . . . . . . . . 29812.8.6 The Fine Structure of Atomic Lines . . . . . . . . . . . . . 299

12.9 Magnetism, Magnetic Resonance . . . . . . . . . . . . . . . . . . . . . . . 30012.9.1 Spin Effects, Larmor Precession . . . . . . . . . . . . . . . . 30112.9.2 Larmor Precession in a Fixed Magnetic Field . . . . . . 30112.9.3 Rabi's Calculation and Experiment . . . . . . . . . . . . . . 30212.9.4 Nuclear Magnetic Resonance. . . . . . . . . . . . . . . . . . . 30612.9.5 Magnetic Moments of Elementary Particles . . . . . . . . 307

12.10 Entertainment: Rotation by 2π of a Spin 1/2 . . . . . . . . . . . . . . 30812.11 Exercises. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 310

13 Addition of Angular Momenta . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31313.1 Addition of Angular Momenta . . . . . . . . . . . . . . . . . . . . . . . . . 313

13.1.1 A Simple Case: The Addition of Two Spins 1/2 . . . . 31513.1.2 Addition of Two Arbitrary Angular Momenta . . . . . . 318

13.2 One-Electron Atoms, Spectroscopic Notations . . . . . . . . . . . . . 32213.2.1 Fine Structure of Monovalent Atoms. . . . . . . . . . . . . 323

13.3 Hyperfine Structure; The 21 cm Line of Hydrogen. . . . . . . . . . 32513.4 Radioastronomy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33013.5 The 21-cm Line of Hydrogen . . . . . . . . . . . . . . . . . . . . . . . . . . 33313.6 The Intergalactic Medium; Star Wars . . . . . . . . . . . . . . . . . . . . 33613.7 Exercises. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 341

14 Identical Particles, the Pauli Principle . . . . . . . . . . . . . . . . . . . . . . . . 34314.1 Indistinguishability of Two Identical Particles . . . . . . . . . . . . . 344

14.1.1 Identical Particles in Classical Physics. . . . . . . . . . . . 34414.1.2 The Quantum Problem. . . . . . . . . . . . . . . . . . . . . . . . 34514.1.3 Example of Ambiguities . . . . . . . . . . . . . . . . . . . . . . 345

14.2 Two-Particle System; The Exchange Operator . . . . . . . . . . . . . 34614.2.1 The Hilbert Space for the Two-Particle System . . . . . 346

xiv Contents

14.2.2 The Exchange Operator BetweenTwo Identical Particles . . . . . . . . . . . . . . . . . . . . . . . 346

14.2.3 Symmetry of the States . . . . . . . . . . . . . . . . . . . . . . . 34814.3 The Pauli Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 349

14.3.1 The Case of Two Particles. . . . . . . . . . . . . . . . . . . . . 34914.3.2 Independent Fermions and Exclusion Principle . . . . . 35014.3.3 The Case of N Identical Particles . . . . . . . . . . . . . . . 350

14.4 Physical Consequences of the Pauli Principle . . . . . . . . . . . . . . 35214.4.1 Exchange Force Between Two Fermions . . . . . . . . . . 35214.4.2 The Ground State of N Identical Independent

Particles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35314.4.3 Behavior of Fermion and Boson Systems

at Low Temperatures . . . . . . . . . . . . . . . . . . . . . . . . . 35414.4.4 Stimulated Emission and the Laser Effect . . . . . . . . . 35714.4.5 Uncertainty Relations for a System

of N Fermions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35814.5 Problem: Discovery of the Pauli Principle . . . . . . . . . . . . . . . . 361

14.5.1 Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36514.6 Problem. Heisenberg Relations for Fermions.

The Way to Macroscopic Systems . . . . . . . . . . . . . . . . . . . . . . 36914.6.1 Uncertainty Relations for N Fermions . . . . . . . . . . . . 36914.6.2 White Dwarfs and the Chandrasekhar Mass . . . . . . . 37114.6.3 Neutron stars . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37314.6.4 Mini-boson Stars . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37514.6.5 Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 377

15 Lagrangian and Hamiltonian, Lorentz Force in QuantumMechanics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38515.1 Lagrangian Formalism and the Least Action Principle . . . . . . . 38615.2 Canonical Formalism of Hamilton and Jacobi. . . . . . . . . . . . . . 38915.3 Analytical Mechanics and Quantum Mechanics . . . . . . . . . . . . 39115.4 Classical Charged Particle in an Electromagnetic Field. . . . . . . 39315.5 Lorentz Force in Quantum Mechanics . . . . . . . . . . . . . . . . . . . 394

15.5.1 Hamiltonian . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39415.5.2 Gauge Invariance . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39415.5.3 The Hydrogen Atom Without Spin in a Uniform

Magnetic Field. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39615.5.4 Spin 1/2 Particle in an Electromagnetic Field . . . . . . 397

15.6 Exercises. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39715.7 Problem. Landau Levels . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 399

15.7.1 Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 400

16 The Evolution of Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40316.1 Time-Dependent Perturbation Theory . . . . . . . . . . . . . . . . . . . . 40316.2 Interaction of an Atom with an Electromagnetic Wave. . . . . . . 407

Contents xv

16.2.1 The Electric Dipole Approximation . . . . . . . . . . . . . . 40816.2.2 Justification of the Electric Dipole Interaction . . . . . . 40816.2.3 Absorption of Energy by an Atom . . . . . . . . . . . . . . 40916.2.4 Selection Rules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41116.2.5 Spontaneous Emission . . . . . . . . . . . . . . . . . . . . . . . . 411

16.3 Decay of a System . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41316.3.1 The Radioactivity of 57Fe . . . . . . . . . . . . . . . . . . . . . 41316.3.2 The Fermi Golden Rule . . . . . . . . . . . . . . . . . . . . . . . 41516.3.3 Orders of Magnitude . . . . . . . . . . . . . . . . . . . . . . . . . 41616.3.4 Behavior for Long Times . . . . . . . . . . . . . . . . . . . . . 417

16.4 The Time-Energy Uncertainty Relation. . . . . . . . . . . . . . . . . . . 42016.4.1 Isolated Systems and Intrinsic Interpretations . . . . . . 42116.4.2 Interpretation of Landau and Peierls . . . . . . . . . . . . . 42216.4.3 The Einstein–Bohr Controversy . . . . . . . . . . . . . . . . . 422

16.5 Exercises. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42316.6 Problem. Molecular Lasers . . . . . . . . . . . . . . . . . . . . . . . . . . . . 425

16.6.1 Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 427

17 Entangled States. The Way of Paradoxes . . . . . . . . . . . . . . . . . . . . . 43317.1 The EPR Paradox . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43417.2 The Version of David Bohm. . . . . . . . . . . . . . . . . . . . . . . . . . . 435

17.2.1 Bell's Inequality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43717.2.2 Experimental Tests . . . . . . . . . . . . . . . . . . . . . . . . . . 441

17.3 The GHZ Experiment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44317.4 Quantum Cryptography; How to Take Advantage

of an Embarrassment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44617.5 The Quantum Computer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45117.6 Quantum Teleportation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45417.7 Concluding Remarks. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 456

18 Solutions to the Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 459

Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 495

xvi Contents

Chapter 1The Appeal of Physics

1.1 The Interplay of the Eye and the Mind

Physics is a fascinating adventure between the eye and the mind, between the worldof phenomena and the world of ideas. Physicists look at Nature and ask questions towhich they try and imagine answers. And the interplay, often the quarrel, betweenwhat we observe and what we construct “logically” is always a source of amazement.A discovery is often the seed of creation of new and unthought ideas to explore.

For instance: why do stars shine? It’s important. The sun is an ordinary star, similarto 80 % of the 200 billion stars of our galaxy. But it is unique and incomparable,because it is our star. In mass, the sun is made of 75 % hydrogen and 25 % helium(actually a plasma of electrons and nuclei). Its parameters are

radius R = 700,000 km, mass M = 2 1030 kg,

power (luminosity) L = 4 1023 kW, surface temperature T = 6000 K.

One mustn’t overestimate the power of the sun. We are much more efficient. Ifyou calculate the power-to-mass ratio, the sun has a score of 0.2 mW/kg which isvery small. We consume on the average 2000 kilo-calories per day; that is, 100 W,25 % of which is used by the brain. Our brain has a power of 25 W! It is consequently10,000 times more powerful than the sun for a given mass! One always tells kidsthey are brilliant, but without explaining why, and where that can lead them.

So, why does the sun shine? One usually thinks that it shines because of thepowerful thermonuclear reactions that take place inside it. But that is not really true!I want to show you that, contrary to common prejudices, it is gravitation that makesstars shine and that thermonuclear reactions cool them permanently.

• Stars shine because they are hot and any hot body radiates energy.• They are hot because of gravity. Stars are huge masses of gas, mainly hydrogen,

which are strongly compressed by the inward pressure of their own weight. This

© Springer International Publishing Switzerland 2016J.-L. Basdevant, Lectures on Quantum Mechanics,Graduate Texts in Physics, DOI 10.1007/978-3-319-43479-7_1

1

2 1 The Appeal of Physics

brings them to high temperatures. Stars are self-gravitating systems in equilibriumunder their own weight.

• OK, but you may object that a hot compressed gas loses energy by radiating. If itloses energy, then it contracts and it cools down.

• Well, the amazing thing is that, on the contrary, a self-gravitating gas does contractwhen it loses energy, but its temperature increases!

This is understandable. If the size of a self-gravitating system decreases, thegravitational inward pressure increases and, in order to maintain equilibrium, thethermal outward pressure increases, (the mean energy of the constituents of the gasmust increase).

Because temperature reflects motion (i.e., kinetic energy of the constituents ofa gas), if the particles move faster, the temperature increases. Therefore, if a self-gravitating system contracts, its temperature increases.

This is quite easy to formulate. A star such as the sun can be represented by anideal gas of N ∼ 1057 particles, at some average temperature 〈T 〉. The temperatureitself varies from 15 million degrees in the center to 6000 degrees on the surface.

• The gravitational potential energy of a sphere of mass M and radius R is propor-tional to Newton’s constant, to the square of the mass, and to the inverse of theradius EG = −γGM2/R, where γ is a dimensionless constant of order 1 (γ = 3/5if the mass distribution is uniform in the sphere). The potential energy is negativebecause one must give energy to the system in order to dissociate it.

• In a self-gravitating system, the total kinetic energy Ekin of the orbiting particles,that is, the internal energy U of the gas, is equal to half of the absolute value of thepotential energy: U = Ekin = 1/2|EG |. This is obvious for circular orbits arounda massive center, and it can be generalized with no great difficulty.

• Therefore, the total energy of the star is E = Ekin + EG = −(1/2)γGM2/R .Again, it is negative because the system is bound. One must bring energy in orderto dissociate it.

• On the other hand, the average temperature 〈T 〉 of the star is related to the meankinetic energy of the constituents by Boltzmann’s relation (3/2)Nk〈T 〉 = Ekin =(1/2)γGM2/R, with N ∼ 1057 .

• When the star radiates, it loses energy. Its energy decreases and becomes more neg-ative, therefore its radius decreases, it is compressed, and its temperature increases.When the star loses energy it radiates more and more strongly.

Therefore, stars shine because of gravitation. Since it was formed in a molecularcloud, the sun, whose present mass and radius we know, has lost a gravitationalenergy of ΔE � 1041 J; its average temperature is of the order of 〈T 〉 � 3 milliondegrees, which is quite acceptable.

Now, we must think! Paleontologists teach us the following.

• The “blue-green algae” or cyanobacteria, who are responsible for the birth oflife on earth because they manufactured the oxygen in the atmosphere, existed3.5 billion years ago.

1.1 The Interplay of the Eye and the Mind 3

• Our cousin, the Kenyapithecus, lived 15 million years ago, and our ancestors Lucy,an Ethiopian Australopithecus afarensis, as well as her cousin Abel in Chad, lived3.5 million years ago. Orrorin (the ancestor of the millennium) lived 6 million yearsago in Kenya, and the present champion, Toumaï lived in Chad 7 million years ago.

• Dinosaurs lived 200 million years ago; they consumed a lot of greenery.

Therefore, the sun must have been stable during all that time. It must have hadapproximately the same power (the weather stayed roughly the same) and the sameexternal temperature (the sun radiates in the visible part of the spectrum wherephotosynthesis takes place, allowing vegetables to grow).

Now, if the sun was stable, we can evaluate roughly when it started to shine. If ithad the same power L for a long time, we can evaluate the time that it took to get ridof the energy ΔE ∼ 1041 J, that is, a time t = ΔE/L ∼ 10 million years. The sunstarted shining 10 million years ago. Therefore, the sun is only 10 million years old!

Consequently, we have just proven scientifically that dinosaurs never existed; theywere simply invented to make Jurassic Park. The Kenyapithecus was just inventedto give us a superiority complex. Because the sun did not shine at that time!

Something is wrong in our above reasoning. Of course it may be that, in order tosave energy, the Creator turned on the sun once in a while, just when it was necessaryfor archaeologic discoveries.

Since it is not my purpose here to inspire religious vocations, we must find some-thing else. Actually, the answer is right in front of us. Suppose there is a sourceof energy in the sun, and that, at some temperature, something ignites in the gas.The combustion releases energy. It increases the energy of the gas, which becomesless negative. Therefore the gas expands, which is understandable. But if its radiusincreases, then its temperature decreases!

If a self-gravitating gas loses energy, its temperature increases; if it gains energy,its temperature decreases. It has a negative specific heat. And that’s great. A com-bustion stabilizes the star’s temperature. An excess of combustion cools the gasand slows down combustion. Conversely, an insufficient combustion rate heats thegas and revives combustion. The combustion energy is peacefully evacuated at con-stant temperature. The system is self-regulated. The energy we receive is indeed dueto thermonuclear reactions, but as long as the combustion lasts, the star evacuatesthat energy in stable conditions. That is exactly what we need for blue-green algae,dinosaurs, Lucy, Orrorin and their companions!

So, our star is stable, but for how long? As long as the fuel is not exhausted. Withthe mass and power of the sun, one can check that if the combustion were chemical,for instance,

2H → H2 + 4, 5 eV, that is, 2 eV per proton,

the available energy would be 1038 J; the lifetime of the sun would be at most 30,000years, which is much too short. On the contrary, nuclear fusion reactions such as

4 p →4 He + 27 MeV, that is, � 7,000,000 eV per proton,

4 1 The Appeal of Physics

is a million times more energetic, which leads us to roughly ten billion years. Andwe have made in three pages a theory of the sun which is not bad at all in firstapproximation!

The conclusion is that stars shine because of gravitation, which compresses themand heats them. Nuclear reactions, which should make them explode, simply allowthem to react against gravitational collapse. They cool down the stars permanentlyand give them a long lifetime. The sun has been shining for 4.5 billion years and willcontinue to do so for another 5 billion years.

That is an example of the confrontation of a physicist’s ideas with the observedworld. And that’s what is interesting in physics. If the ideas we have do not correspondto what we see, we must find other ideas. One cannot change Nature with speeches.

In physics, one can make mistakes but one cannot cheat.

One can do lots of things with speeches. In 1894, Edwin J. Goodwin, a countrydoctor in Indiana, found that the number π was too complicated. So he decided thatfrom then on, π should be equal to 3 (π version 3.0),1 which is much simpler foreveryone. Well, that doesn’t work so well! One can observe that if π were equal to3, four inches of tires would be missing on bicycles, which would be uncomfortable,and five inches of stripes would be missing on a French colonel’s hat, which wouldbe inelegant.

However, Goodwin convinced a state representative who introduced a bill “TheHouse bill No. 246, Indiana State Legislature 1897” which decided that from thenon the number π should be equal to 3. The full House passed the bill by a vote of 67to 0. At the Indiana Senate, the bill was nearly passed, but one senator observed thatthe General Assembly lacked the power to define mathematical truth. He added thathe thought consideration of such a proposition was not dignified or worthy of theSenate. He moved the indefinite postponement of the bill, and the motion carried.

Politicians learn to make speeches and scientists learn to use their intelligence.It is a radically different way of thinking. The two methods happen to be usefulin practice: there exist scientists who can explain their findings, and there existintelligent politicians.

1.2 Advanced Technologies

There are many other reasons to learn physics, of course. Our world is filled withadvanced technologies such as the Internet, GPS, optoelectronics, nanotechnologies,and so on. Many of these new technologies come from the results of fundamentalresearch obtained in the last 10 or 20 years, sometimes in very recent years. Verymany of them are tightly linked with quantum mechanical techniques, and we willcross a few of them in the course of this book.

1Actually the figure he proposed was 3.2, but it’s less fun.

1.2 Advanced Technologies 5

Fig. 1.1 Left Forest of microlasers, each of which is a pile of pancakes of alternating slices ofGaAs and GaAl semiconductors. The diameter of each element is 0.5µm, the height is 7µm.(Courtesy Emmanuel Rosencher) Right Scanning electron microscope (SEM) image of quantumdots fabricated through electron beam lithography on a bidimensional GaAl layer. These structuresare used to study the behavior of electrons, which are confined into tiny spaces approximately 10electrons per dot. The diameter of each quantum dot is 200 nm (Image: C.P. Garcia, V. Pellegrini,NEST (INFM), Pisa.)

Figure 1.1 shows two examples of nanotechnology components. On the left, thedetails of a sample of microelectronics. This was made in the 1990s. It consists of aforest of microlasers each of which is a pile of 10 nm pancakes of alternating slicesof gallium–arsenide and gallium–aluminum semiconductors. We come back to suchdevices. These components have numerous applications in infrared technologies.Infrared sensors are used as temperature sensors for night vision, on automobiles tosee pedestrians at night, in rescuing operations in the ocean, to measure the temper-ature of the earth and of the ocean from satellites, in telecommunications with fiberoptics, and so on.

On the right, there is a much more recent arrangement of quantum dots fabricatedin 2010 through electron beam lithography and subsequent dry-chemical etching ona quasi bidimensional layer (GaAl heterostructure) by C.P. Garcia, V. Pellegrini inPisa. These structures are used to study the behavior of electrons, which are confinedinto tiny spaces approximately 10 electrons per dot. The diameter of each quantumdot is 200 nm (which means that a billion of these structures easily fit on the tip ofyour finger).

What is really amazing is the size. The size of each individual elements, of theorder of 10 nm, is that of a virus, the smallest living object. In order to imagine theorder of magnitude, if instead of making lasers or dots one made letters (which isquite possible, even though it may seem ridiculous) one could write and read on1 mm2 of silicon, the complete works of Sigmund Freud, Carl von Clausewitz, KarlMarx, Shakespeare, Snoopy, Charlie Brown, Anaïs Nin and Leo Tolstoy etc. (whichmay be useful during a boring lecture).

These components are called quantum components because in order to conceive,to manufacture, and to use such components, one cannot bypass quantum physics.

6 1 The Appeal of Physics

Micro- and nanotechnologies are undergoing tremendous progress at present. Inelectronics, one of the present world records consists of a transistor that is 18 nm long,a hundred times smaller than the smallest present transistors. One could put threebillion such transistors on a dime. It is the physical lower limit due to the Heisenberginequalities. One builds automated microsystems that possess the three functions ofbeing sensors, of processing information, and of activating a reaction or a response.Such systems are found in all sectors of technology, from electronic equipment ofcars up to medicine, including telecommunications, computers, and space technolo-gies. Medical researchers are developing customized nanoparticles the size of bloodcells that can deliver drugs directly to diseased cells in the body. In the early 2000susage for cancer detection and treatment has become a systematic and incredibly suc-cessful direction. Nanotechnology increases the capabilities of electronics deviceswhile they reduce their weight and power consumption. Nanotechnological solarcells can be manufactured at significantly lower cost than conventional solar cells.Nanotechnologies may make space-flight more practical. They are used as chemicalsensors for a variety of applications, to improve the quality of air and water.2

One can multiply the number of such examples. Here again, whatever one’s ownperspectives are, one must be familiar with such developments, be it only to positiononeself in front of them. One must be capable of inventing and acting.

1.3 The Pillars of Contemporary Physics

In order to understand contemporary physics, three fundamental links are necessary:quantum mechanics, statistical mechanics, and relativity.

Quantum mechanics, which is dealt with in this book, is the complete and funda-mental theory of structures and processes at the microscopic scale, that is, atomic,molecular, or nuclear scales. It is the fundamental and inescapable field. All physicsis quantum physics.

The first success of quantum mechanics is to explain the structure of matter, atomsand molecules. But it is in the interaction of atoms and molecules with radiation thatone finds the greatest progress, both fundamental and technological, in recent years.

1.3.1 Mysteries of Light

Light has always been considered as a physical phenomenon of its own, the greatmystery. It is our first tool to explore the world, to probe the cosmos as well as theinfinitely small. In physics, light delivered simultaneously the two great discoveries ofthe 20th century: relativity with Michelson, Einstein, Lorentz, and Poincaré in 1905,

2See for instance http://www.understandingnano.com/nanotech-applications.html.

1.3 The Pillars of Contemporary Physics 7

and quantum physics with the black-body theory of Planck in 1900 and Einstein’sphoton in 1905.

The nature of light has always been a mysterious and fundamental question. Thefirst theory of light originated from the importance given to light rays. Just look atFig. 1.2.

This drawing seems quite ordinary, not at all scientific. Fifty percent of childrendraw the sun that way. But 50 % is extraordinary, because you have never seen thoselight rays attached to the sun. This child cannot explain why she drew them, but foreverybody their presence is perfectly natural.

In nature, one can see light rays only under special circumstances, when light ispartially screened by clouds or trees. And the fact that light rays are straight, andthat they materialize the perfect straight lines of geometry was always considered asfundamental.

For thousands of years, a sacred character was attributed to light rays, as one cansee in Fig. 1.3. In Egyptian as well as in Christian culture, light rays are a mediumthrough which the beyond becomes accessible to humans.

In the 18th century, Newton decided that light was made of corpuscles, becauseonly particles can travel along straight lines. However, since the end of the 17th cen-tury, interference and diffraction phenomena were known and the 19th century sawthe triumph of wave optics. Nobody could imagine the incredible answer of quantumtheory. Einstein understood in 1905 that light was both wavelike and corpusclelike.Quantum optics, that is, the quantum description of electromagnetic radiation, alsoplays a decisive role in modern science and technology. The interaction between radi-ation and matter has produced laser physics. Lasers beams are the modern legendarylight rays.

The manipulation of cold atoms with laser beams is one of the highlights ofpresent fundamental research. There are numerous practical applications: inertialcontrolled fusion, optoelectronics, gyrolasers, and others. Intensive work is carriedout on optical computers.

Fig. 1.2 Child’s drawing

8 1 The Appeal of Physics

Fig. 1.3 Left Stele of Taperet (around 900–800 B.C.) Taperet worships the sun god Horakhty whoserays are materialized by lily flowers of all colors (Le Louvre Museum, Paris.) Right Il Sodoma,Saint Sebastian (1526) (Galleria Pitti, Florence.)

1.3.2 Fundamental Structure of Matter

Elementary particle physics started a bit more than one century ago with the discoveryof the electron by J.J. Thomson in 1897. It tries to answer two questions:

• What is the world made of?• How does the world work?

In one century, one has found a nearly complete answer. At present, we possess asimple theory of the Universe, called the Standard model, in which a small numberof elementary constituents of matter, quarks and leptons, interact through a simpleset of forces. And that theory explains all natural phenomena!

In October 1989, a measurement, done in the CERN LEP collider in Geneva,allowed us to count the number of different constituents of matter. There are 24 ofthem.

The validity of the Standard model is constantly verified experimentally more andmore accurately. The next to last elements, the top quark and the τ neutrino, wereobserved respectively in 1995 and 2001. The discovery of the Higgs boson at theCERN Large Hadron Collider facility in June 2012 was a celebrated event.

Many physicists consider the Standard model to be very close to the end of thestory in the infinitely small structure of matter, and, for the moment, there is no exper-imental evidence against that. It is a problem of esthetics and a semi-metaphysicalproblem, namely the whereabouts of the big bang.

Matter is made of atoms. In 1910, Rutherford discovered that atoms are made oftiny but heavy nuclei bound to electrons by electromagnetic forces. In the 1930s, peo-ple showed that nuclei also have an internal structure. They are systems of nucleons(protons and neutrons), bound by nuclear forces of small range and large intensity.Then, in the 1960s, people understood that nucleons are not elementary either. They

1.3 The Pillars of Contemporary Physics 9

have an internal structure: they are systems of three quarks. There are two sorts ofquarks, the u (up) quark of charge +2/3 and the d (down) quark of charge −1/3.The proton is a (uud) system, and the neutron a (udd) system. Quarks are imprisonedagainst each other by “gluons”.

What is amazing in the Standard model is that apparently quarks and leptons(electrons, neutrinos, etc.) are experimentally pointlike. “After” them, there is nothingelse. Electrons and quarks are elementary down to 10−18 m. They are the true elementsof matter.

Actually, this end of the story is a problem. The model works too well! Pointlikeobjects are not consistent with what we know from quantum field theory or fromgeneral relativity. At very short distances, it seems that the notion of particles mustbe replaced by some other concept: superstrings, which are extended objects. Thisis one of the major problems of fundamental physics. This problem is related tosomething we have not yet mastered, unifying general relativity, which is primarily ageometrical theory, with quantum mechanics which is basically nongeometrical. Inthis problem, we might find the answer to fascinating questions such as: why is thedimensionality of space equal to three? The answer is probably that actually thereare more dimensions, ten altogether, but that the additional ones cannot be seen withthe naked eye. Like a bug on a straw, it seems that the bug moves up and down on aone-dimensional space, the straw, but the bug itself knows that it can also turn aroundalong the surface of the straw, and its world is two-dimensional.

Nuclear physics (i.e., the physics of atomic nuclei) is a beautiful and complexfundamental field of research, but it is also an engineering science that plays a con-siderable role in our societies.

It has many aspects. In medicine, nuclear magnetic resonance imaging, as wellas the various applications of radioactivity, and proton and heavy ion therapy, arerevolutions in medical diagnosis and therapy. It is needless to emphasize the problemsof energy in the world. It is a fact that in order to dismantle a nuclear plant, it takes 50years, and in order to launch a new nuclear option (in fusion or in fission) it will take30 or 40 years. In any case, we are concerned with that question for many reasons,in particular because of safety and the disposal of nuclear waste.

1.4 The Infinitely Complex

Now, it is very nice to know the laws of physics at the microscopic scale, but we mustsome day turn back to the physical world at our scale, namely macroscopic physics.When we eat a pound of strawberry pie, we don’t think we’re eating half a pound ofprotons, half a pound of neutrons, and a little overweight of electrons. It’s perfectlytrue, but it’s silly, it’s perverse, and it’s disgusting.

Statistical physics studies the global and collective behavior of large numbers ofparticles or systems whose individual properties are known. It is a great discoveryof the last decades that one cannot reconstruct everything from the very beginning,that is, microscopic laws. As soon as one deals with large numbers of constituents,

10 1 The Appeal of Physics

there appear new phenomena, new singularities or regularities that are specificallymacroscopic. These are related to the number of constituents rather than to theirspecific nature. Examples are:

• Collective effects, phase transitions• Shapes, ordered structures• Irreversibility, life and death

This kind of problem (i.e., physics of the infinitely complex world) is one of themost fascinating fields of physics at present. To understand it, to dominate it, willhave a considerable impact not only in physics, but in biology where reproducibleordered structures are fundamental, to some extent in economics, and maybe someday in sociology. The most fascinating system is the brain itself.

At this point, there appears a much simpler and more relevant answer to thequestion of what is the use of doing physics. Physics is fun; it’s amusing.

Take a simple example. The fact that water freezes at 0 ◦Celsius is a very ancientscientific observation. Everyone knows that. At school, that property is used to definewater: “Water is a colorless tasteless liquid, it is used to wash, some people even drinkit, and it freezes at zero degrees Celsius!”

But, one day, we learn physics. We learn that water is a liquid made of H2Omolecules that wander around at random. Ice is a crystal where the same moleculesH2O are well organized in a periodic structure.

That’s really an amazing phenomenon! Why on earth do those molecules decideat 0◦ to settle down in an ordered structure? It is a mystery! We all know how difficultit is, after a break, to put in an ordered state a number of children at a playground orscientists whose natural tendency is to be dispersed.

Therefore, because we have learned some physics, we discover a very deep aspectin a very familiar fact: the freezing of water. And that’s when we make progress.

But, in order to do that, one must learn to observe and ask oneself questions aboutreality. Creativity is much more important than knowledge or equations, and it isfundamental to develop it and to preserve it. Physics, and in particular experimentalphysics, is an excellent field for that operation.

Materials

Physics of condensed matter, as opposed to corpuscular physics, is a broad domaincommon to physics, to mechanics, to chemistry, and to biology.

Materials have perhaps the most important role in the evolution of scienceand technology, including semiconductors, steels, concretes, composite materials,glasses, polymers, paints, and so on. Practically all the important breakthroughs ofthe progress of mankind are associated with the discovery and the use of new materi-als: think of stones, flint, bronze 10,000 years ago, iron, more recently aluminum andaeronautics, silicon, electronics and computer science, carbon nanotubes and theirderivatives.

Up to the 1970s, it was customary to differentiate between the mechanical prop-erties of solids, that is metallurgy, and electrical properties. But thanks to quantum

1.4 The Infinitely Complex 11

physics and statistical physics, materials science has become a unified theory, becausewe can understand it from its microscopic aspect.

Solids are aggregates of atoms or molecules that are bound by the electrons ofcrystalline bonds. These electrons form a more or less hard cement that determinesthe mechanical properties, resistance, hardness, and plasticity. And it is, in turn, thephysics of these electrons that determines the electrical and thermal properties. Allthese properties are intimately connected.

At first, it is difficult to appreciate the importance and the depth of such a globalsynthetic understanding. Metallurgy was for a long time purely empirical. By manip-ulating such and such a mixture, one used to obtain such and such a result; knowledgewas transmitted by word of mouth. Sometimes it was great, such as in Syria in the13th century. There was a problem in the weapons industry for making swords. Ironis a resistant material, but it is soft and iron swords got bent easily. On the otherhand, carbide is hard, but it breaks easily. Damascus steel consisted of alternatingsheets of iron and carbide. This allowed them to make swords that were both hardand resistant (sometimes physics isn’t that funny; it would have been much morefun if the result had been soft and fragile). It was a revolution in weaponry, and it isvery clever from the modern point of view; in fact this is an example of compositematerials.

The best composite materials that people try to imitate are biological compositessuch as bones or shells. These associate the hardness of limestone apatite, which isfragile but hard, with the resistance of biological collagen.

For modern purposes, one must conceive a material directly in view of the func-tion it should have, namely the desired mechanical, electrical, chemical, and opticalproperties. And this is done more and more systematically.

We have already mentioned nanotechnologies. Let’s add that in recent years therehas been a technological breakthrough with what one calls as smart materials, forinstance, materials with shape memory. A piece of material can have some shape(think of a metal wire) that we can change. The surprise is that a smart materialrecovers its initial shape if it is heated. This does not occur with just any material. Thealloys with shape memory are, for instance, metal alloys (such as nickel–titanium)that undergo a phase transition between two crystalline structures, martensite andaustenite, called a martensitic transition at some temperature. One can give a materialthe shape one wants above the transition temperature. It holds that shape below thetransition point, but one can change this shape by a plastic deformation. If afterthat change, one heats the material, it recovers its original shape because there aredomains that “remember” the initial shape and convey the structure to the entirematerial.

The industrial issues are huge. The applications of such materials are found inmany different domains such as opening up satellite antennas, bone or tooth pros-thesis, and heart and blood vessel surgery. One can crumple pieces of smart materialat usual temperature (20 to 25 ◦C) and insert them in a blood vessel. Once theyhave reached their destination they open up and take their functional shape at thetemperature of the human body, 37 ◦C.

12 1 The Appeal of Physics

There exist in addition hysteresis phenomena. One can “educate” such materialsand construct artificial “muscles” that can transform heat into work. Again, industrialissues are huge.

1.5 The Universe

To end this brief panorama of physics, one should mention astrophysics. The threebasic fields—quantum mechanics, statistical physics, and relativity—are deeply con-nected in astrophysics, and in cosmology—the history of the Universe—together withGeneral Relativity. This is a fascinating subject, one of the most exciting perhapsat present. New observations appear constantly. On September 14, 2015, LIGO, forthe first time, physically sensed distortions in spacetime caused by passing gravi-tational waves generated by two colliding black holes nearly 1.3 billion light yearsaway! This discovery is undoubtedly one of the greatest scientific achievements ofour times.3

The reader will understand that this subject would take us too far in this introduc-tion, together with questions such as: are we alone in the Universe? More and moreextrasolar planets are being discovered, around other stars, so are there other beingswho think like we do in the Universe?

1.6 Physical Constants

Units:Angström 1 Å = 10−10 m (∼size of an atom)

Fermi 1 fm = 10−15 m (∼size of a nucleus)

Electron-volt 1 eV = 1.60218 10−19 J.

Fundamental Constants:Planck’s constant h = 6.6261 10−34 J s,

� = h/2π = 1.05457 10−34 J s= 6.5821 10−22 MeV s

Velocity of light c = 299 792 458 m s−1

�c = 197.327 MeV fm � 1973 eV Å

Vacuum permeability μ0 = 4π10−7 H m−1, ε0μ0c2 = 1

3See for instance https://www.ligo.caltech.edu/page/what-are-gw.

1.6 Physical Constants 13

Boltzmann’s constant kB = 1.38066 10−23 J K−1 = 8.6174 10−5 eV K−1

Avogadro’s number NA = 6.0221 1023

Electron charge qe = −q = −1.60218 10−19 C and e2 = q2/(4πε0)Electron mass me = 9.1094 10−31 kg, mec2 = 0.51100 MeV

Proton mass mp = 1.67262 10−27 kg, mpc2 = 938.27 MeV,mp/me = 1836.15

Neutron mass mn = 1.67493 10−27 kg, mnc2 = 939.57 MeV

Fine structure constant (dimensionless) α = e2/(�c) = 1/137.036Classical radius of the electron re = e2/(mec2) = 2.818 10−15 m

Compton wavelength of the electron λc = h/(mec) = 2.426 10−12 m

Bohr radius a1 = �2/(mee2) = 0.52918 10−10 m

Ionisation energy of Hydrogen EI = mee4/(2�2) = α2mec2/2 = 13.6057 eV

Rydberg’s constant R∞ = EI /(hc) = 1.09737 107 m−1

Bohr magneton μB = qe�/(2me) = −9.2740 10−24 J T−1

= −5.7884 10−5 eV T−1

Nuclear magneton μN = q�/(2mp) = 5.0508 10−27 J T−1

= 3.1525 10−8 eV T−1.

Updated values can be found at http://wulff.mit.edu/constants.html

Chapter 2A Quantum Phenomenon

If, at a party, you ask someone to state a physics formula, the odds are that the answerwill be E = mc2. Nevertheless, the formula E = hν, which was written in the sameyear 1905 by the same Albert Einstein concerns their daily life considerably more.

In fact, among the three great scientific events of the beginning of the 20th cen-tury, 1905 with the special relativity of Einstein, Lorentz, and Poincaré, 1915 withEinstein’s general relativity, an extraordinary reflection on gravitation, space, andtime, and 1925 with the elaboration of quantum mechanics, it is certainly the lastthat has had the most profound impact on science and technology.

The first Nobel prize for relativity was awarded in 1993 to Taylor and Hulse forthe double pulsar. Nobel prizes for quantummechanics can hardly be counted (of theorder of 120) including Einstein’s for the photon in 1921. That reflects discoverieswhich have had important consequences. About 30% of the gross internal productof the United States comes from byproducts of quantum mechanics.

Quantum mechanics is inescapable. All physics is quantum physics, from ele-mentary particles to the big bang, semiconductors, and solar energy cells.

It is undoubtedly one of the greatest intellectual achievements of the history ofmankind, probably the greatest of those that will remain from the 20th century, beforepsychoanalysis, computer science, or genome decoding.

This theory exists. It is expressed in a simple set of axioms that we discuss inChap.6. Above all, this theory works. For a physicist, it even works too well, insome sense. One cannot determine its limits, except that during 10−43 s just after thebig bang, we don’t knowwhat replaced it. But afterwards, that is, nowadays, it seemsunbeatable.

However, this theory is subtle. One can only express it in mathematical language,which is quite frustrating for philosophers. Knowing mathematics is the entrancefee to the group of the happy few who can understand it, even though, as we show,the core of these mathematics is quite simple. It is the physics that is subtle. Moreimportant perhaps, we show how and why quantum mechanics is still a subject ofdebate as to its interpretation and its intellectual content. In some sense, mankind

© Springer International Publishing Switzerland 2016J.-L. Basdevant, Lectures on Quantum Mechanics,Graduate Texts in Physics, DOI 10.1007/978-3-319-43479-7_2

15

16 2 A Quantum Phenomenon

has made a beautiful and successful intellectual construction that escapes humanunderstanding to some extent. As Richard Feynman put it: “I think I can safely saythat nobody understands quantum mechanics”.1

The discovery of quantummechanics could have happened by analyzing a varietyof physical facts at the end of the 19th century. The notion of quanta was proposedin 1900 by Max Planck. Planck had found semi-empirically a remarkable formula toexplain a problem that fascinated people, the spectrum of black-body radiation. Thefrequency distribution of radiation inside an oven at temperature T depends only onthe temperature, not on the nature or shape of the oven. It is a universal law. Planckobtained the good result

u(ν) = 8πh

c3ν3

e(hν/kT ) − 1, (2.1)

whereν is the frequency, T the temperature, and k Boltzmann’s constant, by assumingthat radiation of frequency ν can exchange energy with the inner surface of the ovenonly by discrete quantities that are integermultiples of an elementary energy quantumhν,

ΔE = nhν. (2.2)

Planck understood that the constant h in the above formula, which now bears hisname and whose value is

h ≈ 6, 62 10−34 j.s,

is a fundamental constant of nature, as the velocity of light c in relativity andNewton’sconstantG in gravitation. For technical simplicity, wemainly use the reduced Planckconstant

� ≡ h

2π≈ 1, 05 10−34j.s.

Planck’s formula works remarkably well. The direct verification would requireus to be inside an oven. We have the great luck to live inside the cosmic backgroundradiation of the big bang, which cooled down as the Universe expanded. The temper-ature of that radiation is at present 3K. Its observation and its more and more precisemeasurement (Fig. 2.1) is perhaps the best observational evidence in favor of the bigbang theory, as well as of Planck’s formula.

Planck’s quanta were somewhat mysterious, and it was Einstein who made adecisive step forward in 1905, the sameyear as he did forBrownianmotion theory andfor special relativity. By performing a critique of Planck’s ideas, and for reasons dueto equilibrium considerations (i.e., entropy) Einstein understood that the quantizedaspect is not limited to the energy exchanges between radiation and matter, but thatit must be present in the electromagnetic field itself. Light, which was known to bea wave propagation phenomenon since the beginning of the 19th century, must also

1The Character of Physical Law, MIT Press, Cambridge, MA 1965.

2 A Quantum Phenomenon 17

Fig. 2.1 Wave-numberdistribution of the cosmicbackground radiationmeasured in 1992 by theCOBE satellite. Theagreement between Planck’sformula at a temperatureT = 2.728 K lies within theline (Photo credit: Mather etal., Astrophys. J., 420, 439,(1994). http://lambda.gsfc.nasa.gov/product/cobe/firas_image.cfm.)

exhibit a particlelike behavior. Light of frequency ν is carried by particles, photonsas the chemist Gilbert called them in 1926, of energy

E = hν, (2.3)

and momentum p = �k, where k is the wave vector k = 2π/λ, as was proven exper-imentally by Compton in 1921.

In that respect, Einstein understood an essential feature of quantum theory, theso-called “dual” manifestations of the properties of light, which appear to be bothwavelike and particlelike. In the course of his work, Einstein found the explanationof the photoelectric effect, which was one of the first experimental confirmationsof his ideas. Such ideas were considered revolutionary or even iconoclastic becausethey seemed to contradict Maxwell’s equations which were a great triumph of the19th century.

At the same time, atomic spectroscopy was one of the great enigmas of physics.The third breakthrough, which derives in some respect from Einstein’s ideas, camein 1913 from Niels Bohr.

There are three parts in Bohr’s ideas and results.

• He postulated that matter is also quantized and that there exist discrete energylevels for atoms, which was verified experimentally by Franck and Hertz in 1914.

• He postulated that spectral lines which had been abundantly observed during the19th century, came from transitions between these energy levels. When atomsabsorb or emit radiation, the positions of spectral lines are given by the difference

νnm = |En − Em |h

. (2.4)

• Finally, Bohr constructed an empirical model of the hydrogen atom that worksremarkably well and gives the energy levels En of this atom as

18 2 A Quantum Phenomenon

En = − mqe4

2(4πε0)2�2n2, (2.5)

where n is a positive integer. With that formula, where all physical constants areknown from different experiments, the wavelengths λ = c/νnm of spectral linescoincide with experiment to one part in a thousand.Bohr’s formula (2.5) expresses the famous “Rydberg constant” of spectroscopistsin terms of fundamental constants, which impressed people, in particular,Einstein.2

So we are facing three similar formulae, E = hν. The first (2.2) is an assumptionabout the interaction of radiation andmatter, the second (2.3) has to do with radiationitself, and the third (2.4) is a property of atoms, namely matter.

Bohr’s success was fantastic, but it was too easy. Actually one realized later onthat it was a piece of luck due to the fact that the hydrogen atom is a simple two-body system. This easy result generated an obscure prequantum era, where peopleaccumulated recipes for more complicated atoms, with fluctuating results deprivedof any global coherence.

2.1 Wave Behavior of Particles

The synthetic and coherent formulation of quantummechanicswas performed around1925. It is due to an incredible collective work of talented people such as Louisde Broglie, Schrödinger, Heisenberg, Max Born, Dirac, Pauli, and Hilbert, amongothers. Never before, in physics, had one seen such a collective effort to find ideascapable of explaining physical phenomena.

We are now going to discover some of the main features on a simple concreteexperiment that shows the wavelike behavior of particles. This is symmetric in somerespect to the particlelike behavior of light. We show that the behavior of matterat atomic scales does not follow what we expect from daily “common sense.” It isimpossible to explain it with our immediate conceptions.

In order to understand quantum mechanics, one must get rid of prejudices andideas that seem obvious, and one must adopt a critical intellectual attitude in frontof experimental facts.

2.1.1 Interferences

Let us recall interference phenomena in wave physics, optics, or acoustics, in thesimple case of Young slit fringes.

2The 1/n2 behavior was known since 1886 and Balmer’s empirical discovery.

2.1 Wave Behavior of Particles 19

Fig. 2.2 Sketch of a Young two-slit interference experiment

One sends a light beam on a screen pierced with two slits, and one observes thevariation of the intensity of light on another screen as a function of the distance x tothe center (Fig. 2.2).

The two slits act as secondary sources in phase, and the amplitude of the wave ata point C of the screen is the algebraic sum of the amplitudes issued from each ofthem.

If the two waves are in phase, the amplitude is twice as large. If they are out ofphase by π the amplitude vanishes; there is no luminous energy at that point. Andthere exist all intermediate cases.

In other words, the amplitude at some point is the sum of amplitudes reachingthat point,

Amplitude at C : AC = A1 + A2 , Intensity : I (x) = |AC |2. (2.6)

The amplitudes emitted by the two slits add up, the intensity is the square of thatsum and it presents a periodic variation, the distance of fringes being x0 = λD/a.

2.1.2 Wave Behavior of Matter

We turn to the wave behavior of matter. In 1923, Louis de Broglie made the bold butremarkable assumption that any particle of mass m and of velocity v possesses an“associated” wave of wavelength

λ = h

p; (2.7)

where p = mv is the momentum of the particle and p its norm.

20 2 A Quantum Phenomenon

Fig. 2.3 Double slit Young interference experiment performed with neon atoms cooled down to amilliKelvin (left part). Each point of the figure (right part) corresponds to the impact of an atomon the detector. Interference fringes are clearly visible

Louis de Broglie had many reasons to propose this. In particular he had in mindthat the discrete energy levels of Bohr might come from a stationary wave phenom-enon. This aspect struck the minds of people, in particular that of Einstein, who wasenthusiastic.

In order to verify such an assumption, it is natural to perform interference anddiffraction experiments. The first experimental confirmation is due to Davisson andGermer in 1927. It is a diffraction experiment of an electron beam on a nickel crystal.

It is more difficult to perform a Young double-slit interference experiment withelectrons. However, a group of Japanese physicists from Nippon Electronics (NEC)performed in 1994 a beautiful interference experiment of cold atoms in Youngslits. Neon atoms are initially trapped in stationary laser waves (so-called opticalmolasses). They are then released and undergo free fall across a two-slit device. Theslits are 2µm large, they are 6µm apart. The scale in Fig. 2.3 is distorted.

What do we observe in Fig. 2.3? The distribution of impacts of atoms on thedetecting plate is similar to the optical intensity in the same device. The fringes areat the same positions provided Louis de Broglie’s relation is satisfied λ = h/p. (Ofcourse, one must take care of the uniform acceleration in this particular setup.)

The same phenomenon can be observed with other particles: neutrons, heliumatoms, hydrogen molecules, the same relation holds between the wavelength and themomentum. The present record is to perform interferences with large molecules suchas fullerenes, that is, C60 molecules.3

Therefore matter particles exhibit a wave behavior with a wavelength given by deBroglie’s formula.

3O. Nairz, M. Arndt, A. Zeilinger, American Journal of Physics, Vol. 71, 319 (2003), and referencestherein.

2.1 Wave Behavior of Particles 21

2.1.3 Analysis of the Phenomenon

Now, a number of questions are in order.

What is this wave? And why is this result so extraordinary?

It is extraordinary because atoms are known to be particles. An atom has a size ofthe order of an angström (0.1nm) and it is pointlike at the scales of interest (µm ormm).With a counter, one canmeasurewhether an atomhas arrived at some pointwithan accuracy as fine as one wishes. When an atom is detected, it has a well-definedposition; it does not break up into pieces; it is point-like.

But a wave fills all space. A wave, on the surface of water, is the whole set ofdeformations on all points of that surface.

So, what is a particle? Is it a pointlike object or is it spread out in the entire space?A simple glance at Fig. 2.4 shows that we are facing a conceptual contradiction.

How can we escape this contradiction? Actually, the phenomenon is much richerthan a simple wave phenomenon; we must observe experimental facts and use ourcritical minds.

Fig. 2.4 Top two sourceinterferences on the surfaceof water; the radial lines arenodes of interferences.Bottom tracks of particles inthe Aleph detector of LEP atCERN

22 2 A Quantum Phenomenon

Since atoms are particles, we can send them individually, one at a time, and all inthe same way.

This proposition is perfectly decidable; it is feasible experimentally.Wecan triggerthe device so that it releases atoms one after the other and that they are all releasedin the same way.

2.2 Probabilistic Nature of Quantum Phenomena

2.2.1 Random Behavior of Particles

What do we observe? Actually, we can guess that from Fig. 2.3.

• Each atom has a well-defined impact; an atom does not break into pieces.• But the positions of the impacts are distributed at random. In other words, to thesame initial conditions, there correspond different impacts.

In other words, atoms, or particles in general, have a random behavior. Each atomarrives where it wants, but the whole lot is distributed with a probability law similarto the intensity observed in optics or acoustics:

P(z) ∝ I(optical)(λ = h/p).

Therefore, there is a second difference with classical physics: to identical initialconditions, there correspond different final conditions. The impact of a single particleis unpredictable, the whole set of impacts has a well defined probability distribution.

But, one can object that random, or probabilistic, phenomena exist in classicalphysics, such as playing dice, or heads and tails, and so on.

True, but the big problem is that this is by no means a classical probabilisticphenomenon, as in usual probability theory. Why is that?

2.2.2 A Nonclassical Probabilistic Phenomenon

If we block one of the slits, the atoms will pass through the other one and theirdistribution on the detector shows no sign of any interference. If we block the otherslit, the distribution is approximately the same, up to a small global shift (1µm/1mm)Now let’s make a logical reasoning and perform the critique of what we say.

1. We send the atoms one by one. These are independent phenomena; atoms don’tbother each other; they do not act on each other’s trajectory.

2. Each atom has certainly gone through one of the slits.3. We can measure which slit each atom went through. There exist techniques for

this; send light on the slits, put counters, and so on. It is feasible.

2.2 Probabilistic Nature of Quantum Phenomena 23

4. If we perform this measurement, we can separate the outgoing atoms in twosamples, those that have passed through the first slit, and those that have passedthrough the second one. And we know where each atom arrived.

5. For those that passed through the first slit, everything is as if the second slit wereblocked, and vice versa. Each sample shows no interference.

Now, we have two independent samples, and we can bring them together. Clas-sically, the result we would obtain by opening the two slits should be the sum, thesuperposition of the two distributions such as (2.5). But not at all!

It’s even worse! Opening a second slit (i.e., giving an extra possibility for theatoms to reach the detector) has prevented the atoms from arriving at certain points.That’s really incredible to be able to stop some people from entering your house byopening another door!

We must admit that the usual logical ideas of probability theory do not apply. Wecannot explain the phenomenon in classical terms. It is a non-classical probabilisticphenomenon.

2.3 Conclusions

At this point, it seems we are at a logical dead end. How can we find our way?Our argument, however logical it may seem, leads to wrong conclusions. There issomething we haven’t thought about. Because physics is consistent. The answer isexperimental. What actually happens is the following.

1. If we measure by which slit each atom passed, we can indeed make the separationand indeed we observe the sum of two distributions such as in Fig. 2.5. Thereforewe no longer observe interferences; they disappear.It is another experiment!

Fig. 2.5 Same experiment as in Fig. 2.3 but opening only one slit. The interference fringes disappearand one observes a diffraction pattern (this figure is not experimental)

24 2 A Quantum Phenomenon

2. Conversely, if we do observe interferences, it is not possible to know throughwhich slit each atom passed. We can talk about it, but we can’t do anythingwith it.Knowing by which slit an atom has passed in an interference experiment is aproposition that has no physical meaning; it is undecidable. It is perfectly correctto say that an atom passed through both holes at the same time, which seemsparadoxical or absurd classically.

Whatwaswrongwas to assume implicitly that, at the same time,we couldmeasureby which slit each atom passed and observe interferences. We assumed that withoutchecking it.

We can draw the following conclusions.

• First, a measurement perturbs the system. If we do not measure by which slitthey pass, the atoms are capable of interfering. After we perform this measurement,they are in another state where they are no longer capable of interfering. They havebeen perturbed by the measurement.

• Secondly and consequently, there is no trajectory in the classical sense. If weobserve an atom in an interference experiment, we know when and where it wasemitted and where and when it was detected, but we cannot say where it was in themeantime.

However, these two ideas seemed obvious in classical physics. The fact that wecan make a measurement as accurate as we wish without affecting the system is anold belief of physics. Physicists used to say that they just needed to improve themeasuring apparatus. Quantum physics tells us that there is a absolute lower boundto the perturbation that a measurement produces.

The notion of a trajectory, namely that there exists a set of points by which wecan claim and verify that a particle has passed at each moment, is as old as mankind.Cavemen knew that intuitively when they went hunting. It took centuries to constructa theory of trajectories, to predict a trajectory in terms of initial conditions. Newton’sclassical mechanics, celestial mechanics, ballistics, rests entirely on that notion, butits starting point is beaten up by the simple quantum phenomenon we just examined.

Classically, we understand the motion of a particle by assuming that, at eachmoment one can measure the position of a projectile, that the collection of the resultsconsists of a trajectory, and that we can draw a reproducible conclusion independentof the fact that we measure the positions at any moments. We learn these ideas asif they were obvious, but they are wrong. More precisely: in order to penetrate thequantum world, one must get rid of such ideas. Figure2.6, or analogous ones, iscompletely wrong in quantum mechanics.

Of course, one mustn’t go too far. These are very good approximations in theclassical world. If a policeman stops you on a freeway saying you were driving at80miles an hour, the good attitude is to claim, “Not at all! I was driving peacefullyat 35mph on the little road under the bridge, and your radar perturbed me!” Unfor-tunately, he won’t believe you even if he knows some physics. Because it is Planck’sconstant � that governs such effects. However, in quantum driving one must changethe rules. Changing the rules consists of constructing the theory of all that.

2.3 Conclusions 25

Fig. 2.6 Stroboscopicpicture of the free fall of anapple which then bounces onthe floor. This is a goodexample of the a priorirepresentation of an intuitivephenomenon that cannotshow up in quantummechanics (WilliamMcLaughlin, “The resolutionof Zeno’s paradoxes,” Sci.Amer., 1994)

Phenomenological Description

The interference phenomenon would be very complicated to explain if we did nothave the luck that it so closely resembles usual interference, with, in addition, asimple formula for the wavelength λ = h/p.

So, let’s try and use the analogy with wave physics in order to formalize Louisde Broglie’s idea. Here, we should be able to explain the interference experiment inthe following way.

• The behavior of an atom of velocity v and momentum p = mv in the incomingbeam corresponds to that of a monochromatic plane wave

ψincident = e−i(ωt−p·r/�), k = p/�, λ = 2π/k = h/p, (2.8)

which has the good wave vector k = p/� and the good wavelength.

26 2 A Quantum Phenomenon

• After the two slits, the behavior is that of the sum of two waves each of which hasbeen diffracted by a slit

ψoutgoing(x) = ψ1 + ψ2, (2.9)

which would describe, respectively, the behavior of the atom if it passed throughone of the slits, the other one being blocked. We can calculate the phase shift ofthese waves at any point because we know the wavelength.

• Finally, the probability for an atom to reach some point C of the detector is simplythe modulus squared of that sum

P(C) = |ψC |2. (2.10)

We just follow the same argument as for usual interferences.We now have an answer to one of our questions above; what is the physical

meaning of these waves?In usual wave physics, one manipulates electromagnetic or acoustic wave ampli-

tudes which add up and whose modulus squared gives intensities, that is, energydensities.

Our quantum waves are probability amplitudes. They add up and the modulussquared of the sum gives us probabilities, or probability densities.

One does not work directly with probabilities but with these intermediate tools,these probability amplitudes that add up.

The interference experiment gives us the wavelength, but not the frequency ω ofthe waves. Louis de Broglie made a good choice by assuming that this frequency isrelated to the energy of the particles in the same way as for Einstein’s photons

ω = E/�, that is, ν = E/h, (2.11)

where E = p2/2m is the kinetic energy of the atoms. This leads to the completestructure of de Broglie waves:

ψincident = e−(i/�)(Et−p·r), where E = p2/2m , (2.12)

which is the probability amplitude for the presence of a particle at point r and timet of a particle of momentum p = mv.

Remark Notice that because the kinetic energy and the momentum are related byE = p2/2m, one can find with this expression a wave equation, which is satisfiedwhatever the value of the momentum p. Indeed, if we take the time derivative on onehand, and the Laplacian on the other, we obtain

∂

∂tψincident = − i E

�ψincident, and Δψincident = − p2

�2ψincident,

2.3 Conclusions 27

therefore, because E = p2/2m, we have the wave equation

i�∂ψ

∂t= − �

2

2mΔψ, (2.13)

which is nothing but the Schrödinger equation for a free particle.4

Of course, we are not completely finished. For instance, atoms have a particlelikebehavior that is obscure in all that. But we’re getting closer.

2.4 Appendix: Notions on Probabilities

Probabilistic Phenomena

Consider a set of phenomena of the same nature on which we repeatedly make thesame observation or measurement. For instance, play dice, measure a temperature oran economic parameter etc. Each observation belongs to some set Ω of outcomes.This set can be discrete, continuous, or a more complicated object such as a set offunctions.

The set Ω is the set of a priori possible outcomes of the experiment. One alsospeaks of events: “the roulette number is even”, “the observed temperature is betweenT0 and T1”, etc.

Suppose we repeat an experiment a large number of times N , Ω being the set ofpossible outcomes. Consider a specific event α and Nα the number of times, amongN , where α occurs. The observed number Nα depends on the specific sequence ofexperiments. One calls the empirical frequency of the event α in this sequence ofexperiments, the ratio:

fα(N ) = Nα/N .

The fundamental empirical observation is that when N becomes large, if the succes-sive repetitions of the experiment are done independently (the result of an experimenthas no a priori influence on the conditions in which the other experiments are done),the frequencies fα(N ) tend, for each event α, to a well defined limit. To each eventα there corresponds a number P(α) called the probability of event α, related to theempirical frequency by the relation:

P(α) = limN→∞ fα(N ).

4It is surprising that de Broglie didn’t think of writing this equation, or its relativistic equivalent—since he used the relativistic energy-momentum relation E2 = (p2c2 + m2c4).

28 2 A Quantum Phenomenon

Clearly, one has P(α) ≥ 0, P(Ω) = 1, P(∅) = 0, and if (Ai )i∈I is a finite family ofdisjoint events:

P(⋃

i∈IAi ) =

∑

i∈IP(Ai ).

Examples of Probability Laws

Discrete Laws

The Simple Alternative

In this case, there are only two possible outcomes, α = 1 or 2 (example: heads ortails). We note p the probability of the outcome 1 and q that of the outcome 2. Weobviously have p + q = 1.

The Generalized Alternative

There are n outcomes α = 1, 2 . . . n. For instance one can place in an urn m1 ballsmarked with the sign 1,m2 balls marked 2, . . .. If the draw does not distinguish theballs, the probability law consists in the set of numbers p1, p2, . . ., pn such that:

pα = mα∑nβ=1 mβ

withn∑

α=1

pα = 1.

Probability Laws on R or Rn

A probability law P on R (resp. Rn) is said to be of density p, p being a positiveintegrable function such that

∫ +∞−∞ p(x) dx = 1 (resp.

∫Rn p(x) dnx = 1), if, for any

interval (resp. any volume) I :

P(I ) =∫

Ip(x) dx .

It is useful to treat the discrete and continuous cases in the same formalismbyworkingwith the distribution function:

F(t) = P(] − ∞, t]).

Examples

1. Exponential law:

p(x) ={

λe−λx if x ≥ 0 (λ > 0)0 if x < 0.

,

which yields:

F(t) =∫ t

−∞p(x)dx =

{0 if t < 01 − e−λt if t > 0

.

2.4 Appendix: Notions on Probabilities 29

Fig. 2.7 The Gaussian probability law for μ = 0 and σ = 1

2. Gauss’s law of parameters μ,σ (Fig. 2.7):

p(x) = 1

σ√2π

exp− (x − μ)2

2σ2with μ ∈ R,σ ∈ R∗. (2.14)

Random Variables

Definition

Consider the example of the game with n outcomes α1, . . ., αn of respective prob-abilities p1, . . ., pn . If, in this game, we win some amount of money xα when theoutcome is α, the number xα which is a function of the (random) outcome of theexperiment is called a random variable.

In the above example, the set of the {xα} is discrete. One calls a discrete randomvariable x a set of numbers xα (positive, negative, complex) each of which is asso-ciated to an outcome of a discrete random event. The couples {xα, pα} define theprobability law of the random variable x .

In the sameway, one can consider continuous randomvariables. Let x be a randomvariable which takes its values in an interval [a, b]. The probability density p(x)(positive or zero) defines the probability law of this random variable if the probabilityto find, in an experiment, a value between x and x + dx is p(x) dx . We obviouslyhave

∫ ba p(x) dx = 1. The generalization toRn is straightforward.

Conditional Probabilities

Consider two types of events [A] and [B]. We are led to defining the conditionalprobability of the event B knowing A, noted P(B/A) by

P(B/A) = P(B ∩ A)

P(A)as long as P(A) > 0.

If X is a discrete random variable, one can define the conditional probabilityP(B|X = x) of the event B when X = x , i.e. knowing the event {X = x}.

30 2 A Quantum Phenomenon

Example: The Exponential Decay Law

When a radioactive particle exists at time t , its probability to decay in the time interval]t, t + Δt] is independent of its past history. Therefore the conditional probabilitythat the time X at which the particle decays is between t and t + Δt , knowing that{X > t}, is independent of t and equal to P{0 < X ≤ Δt}:

P{0 < X ≤ Δt} = P{t < X ≤ t + Δt}P{X > t} .

If we call F the distribution function of X we obtain the functional relation:

F(Δt) = F(t + Δt) − F(t)

1 − F(t).

The function F therefore satisfies the differential equation:

F ′(t) = F ′(0) (1 − F(t)).

Therefore, setting λ = p(0) = F ′(0) (λ is a decay rate), we get F(t) = 1 − e−λt .The density of the law of X is then:

p(x) ={

λe−λx for x ≥ 00 for x < 0

.

Since λ has the dimension of the inverse of a time, we can note it 1/τ where τ is thelifetime (or mean life). This exponential law is met in many practical applications(physics, pharmacology, reliability, etc.).

Independent Random Variables

Consider two discrete random variables X and Y with values in E1 and E2 respec-tively. One says that X and Y are two independent variables if the observation of Xdoes not give any information on Y , and vice-versa. In other words, the conditionalprobability to find x if one knows y is independent of y (and vice versa).

This can be expressed in a symmetric form in x and y by:

P({X = x, Y = y}) = P({X = x}) P({Y = y}).

The variables X and Y are independent if and only if the law of the couple (X, Y ) isthe product of the laws of X and of Y .

Binomial Law and the Gaussian Approximation

Consider an experiment consisting in repeating N consecutive times and indepen-dently an experiment with two outcomes (for instance heads or tails). The firstoutcome, noted 1, has a probability p to happen, and the second, noted 0, has the prob-

2.4 Appendix: Notions on Probabilities 31

ability q = 1 − p to happen. Such a sequence of experiments is called a Bernoullisequence.

Since the successive partial experiments are assumed to be independent, the prob-ability for a given sequence (x1, . . . , xN ) is given by:

P(x1, . . . , xn) = P[X1 = x1] . . . P[XN = xN ] = pk qN−k,

where k is the number of 1 in the sequence (x1 . . . xN ). We now consider the randomvariable X = X1 + · · · + XN representing the number of times 1 appears in the Nsuccessive draws:

P[X = k] =(Nk

)pkqN−k ≡ b(k; N , p).

This law b(k; N , p) is called the binomial law of parameters N and p.

Normal Approximation of the Binomial Law

Using Stirling’s formula: n! ∼ √2πn nn e−n , we obtain for n � 1:

b(k; n, p) ∼√

1

2πnpqexp− (k − np)2

2npq,

i.e. a Gaussian law for k, with μ = np and σ = √npq .

Moments of a Probability Distribution

Mean Value or Expectation Value

Consider a function ϕ(x) of the random variable x (ϕ(x) is a new random variable).We define its mean value or equivalently expectation value 〈ϕ〉 as

〈ϕ〉 ={∑

α ϕ(xα) pα discrete case∫ ba ϕ(x) p(x) dx continuous case (a < x < b)

We note 〈x〉 the mean value of the variable x itself:

〈x〉 =∫

x p(x) dx .

This quantity is equivalently called the mathematical expectation, or expectationvalue: if we gain the amount xα when the result is α, then we expect to gain on theaverage 〈x〉.Expectation Values of Usual Laws

1. Variable of the simple alternative: 〈X〉 = p.2. Binomial law b(k; n, p): 〈k〉 = np.

32 2 A Quantum Phenomenon

3. Geometric law P{X = k} = (1 − p)pk (k ≥ 0): 〈X〉 = p/(1 − p).4. Poisson law P{X = k} = e−λ λk/k! (k ≥ 0): 〈X〉 = λ.

Example

In the exponential decay above, the mean time that the particle spends before itdecays, or the expectation value of its lifetime, is:

〈t〉 =∫ ∞

0

t

τe−t/τdt = τ .

Variance and Mean Square Deviation

Consider a real random variable x whose expectation value is 〈x〉 = m. The meansquare deviation of x , noted σ or Δx , is defined by:

(Δx)2 = σ2 = 〈(x − 〈x〉)2〉,

σ2 is also called the variance of the probability law.One readily checks, by expandingthe square term, that:

σ2 = 〈x2〉 − 2〈x〉〈x〉 + 〈x〉2 = 〈x2〉 − 〈x〉2.

The smaller σ is, the more probable it is to find a value of x close to the mean value.The quantity σ measures the deviation from the mean value.

Variance of Usual Laws

1. The simple alternative: σ2 = p(1 − p).2. Binomial law: σ2 = np(1 − p). Note that the relative dispersion σ/〈X〉 tends to

zero as n−1/2 when n → ∞.3. Gaussian law: the variance coincides with the parameter σ2 of (2.14).4. Geometric law: σ2 = p/(1 − p)2.5. Poisson law of parameter λ : σ2 = λ.

Bienaymé–Tchebycheff Inequality

Note m the mean value and σ2 the variance of the discrete real variable X . One canshow that:

P({|X − m| ≥ τσ}) ≤ 1/τ 2, (2.15)

which proves that for a small variance, there is a small probability to find X far fromits expectation value.

Error Function

In the particular case of the Gaussian law (m,σ), one calls error function Φ(τ ) thequantity P({|X − m| ≤ τσ}). One has:

2.4 Appendix: Notions on Probabilities 33

Φ(τ ) =∫ +τσ

τσ

1

σ√2π

e−x2/2σ2dx .

Some values of Φ(τ ) are the following

τ 1 2 3Φ(τ ) 0.68 0.95 0.99

2.5 Exercises

1. Distribution of impacts

We observe the impacts on a target in the xy plane. The observable is assumedto obey a probability law of density p(x, y) = (2πσ2)−1 exp(−ρ2/(2σ2)) whereρ = (x2 + y2)1/2 is the distance from the origin of the impact point. What is theprobability law of ρ?

2. Is this a fair game?

Suppose that one offers you the following game: Bet one euro and throw three dice.If number 6 (or any number you choose in advance) does not show up, you lose yourbet; you get paid 2 euros if it shows up on one dice, 3 euros if it shows up on two,and 6 euros if it shows up on the three of them. Calculate the expectation value ofwhat you gain (which is negative if you lose) and see if it is reasonable to play.

3. Spatial distribution of the molecules in a gas

Consider in a volume V (22.4 l for instance) N molecules (6 × 1023 for instance).Consider an enclosed volume v (10−3 cm3). How many molecules are there on theaverage in v? What are the fluctuations of this number?

Chapter 3Wave Function, Schrödinger Equation

In the first chapter, we described an interference experiment of atoms which, as wehave understood, is both a wave and a probabilistic phenomenon.

We now want to construct the theory of this experiment. More generally, wewant to find the quantum theory of the simplest problem of classical mechanics; thenonrelativistic motion of a particle of mass m in a field of force.

This is called wave mechanics. It is due to de Broglie and to Schrödinger. Wegeneralize it later on.

We do not want to say what the nature of an atom or an electron is; we simplywant to determine their behavior in a field of force. In celestial mechanics, one doesnot worry about the nature of planets. They are considered as points whose motionwe can calculate.

3.1 Terminology and Methodology

Terminology

Before we start, we must agree on the meaning of words and on the methodology.We cannot avoid using ordinary language. Words are necessary. But words can alsobe traps when discussing phenomena that are so new and unusual. We constantly usethe following words: physical system, state, physical quantities.

The foundation of physics is experimental observation and the measurementprocess that consists of characterizing aspects of reality that we observe, by num-bers. These aspects of reality are elaborated into concepts of physical quantities (forinstance, velocity, energy, electric intensity, etc.).

In given circumstances, we say that a physical system (i.e., an object pertaining toreality) is in a certain state. The state of the system is “the way the object is” (i.e., theparticular form in which its reality can manifest itself). That is what we are interested

© Springer International Publishing Switzerland 2016J.-L. Basdevant, Lectures on Quantum Mechanics,Graduate Texts in Physics, DOI 10.1007/978-3-319-43479-7_3

35

36 3 Wave Function, Schrödinger Equation

in. We want to know the state of an atom in space, not its internal structure, whichwe study later. We possess some more or less complete knowledge of a state if weperform some set of measurements of physical quantities on the system.

We give a name to the system.We can call it a particle, an atom, or an electron. Buta particle is simply for the moment an object with a well-defined massm and electriccharge q, and which preserves any internal structure it may have in the experimentsof interest. We are not interested for the moment in the possible internal degrees offreedom of the particle.

Methodology

The construction of our theory has the following elements that we describe in theclassical case of Newtonian mechanics of a massive particle in a field of force thatwe assume to derive from a potential energy V (r).

1. We must first describe the state of the system. This means associating with thisstate a mathematical representation that defines it from an operational point ofview. In Newton’s theory, the state of a massive particle in space is described attime t by six numbers; its position r and velocity v, or its momentum p = mv.

2. Then, we must know the law that governs the time evolution of the state ofthe system when it is placed in given conditions; that is, we must be able topredict the state at time t , given the state at time t = 0. In Newton’s theory, itis the fundamental law of dynamics d p/dt = f , that allows us to calculate thetrajectory.

3. Next, we must know the laws that enable us to calculate the results of measure-ments of physical quantities, laws that transform the mathematical representationof the state of a system into measurable numbers. In Newton’s theory, physicalquantities are functions of the state variables r and p.

4. Finally, we must address a question that is absent in Newtonian theory. In whatdoes the measurement process result? What do we know after a measurement?

In this chapter, we study the first two questions. We examine the two others in thenext chapter.

In quantum mechanics, there is no direct intuitive link between physical conceptsand theirmathematical representation aswecanfind in classical physics. For instance,any child knows that instantaneous velocity and acceleration exist because in a carthere is an object, the speedometer, whose indications correspond to the variousphysical feelings of the child. When later on, the child learns the mathematicalnotion of derivatives, it is quite natural to associate that notion with the physicalconcept that has become familiar.

So, what we do is to place ourselves in the position of de Broglie, Schrödinger,Einstein, and Born, after they have investigated all alternative possibilities.

• We give the principles of the theory.• We check that they are consistent, and that they account for observed phenomena;we will then understand how the theory works.

3.1 Terminology and Methodology 37

• Finally, we say a fewwords about how such ideas came to theminds of people. Thisis useful for understanding the theoretical scheme and approaching it in a humanway. Principles were not dictated by any superior being. It is human scientists whofought with reality in order to elaborate them.

Later on, we shall be capable of understanding that the mathematical structurewhich is perhaps the most important in quantum mechanics is the first of the fourelementary operations, addition. But it takes some time to fully appreciate that.

3.2 Principles of Wave Mechanics

The Interference Experiment

In the previous chapter, we have described the experimental aspects of quantuminterferences of particles on the specific case of atoms.

A beam of particles of given momentum p is sent on a diffracting setup. Thisgives rise to an interference pattern similar to light interferences. The impact pointof a given atom can be anywhere with a probability law proportional to the intensityof light fringes.

We attempted to make an empirical description using de Broglie waves, whichare probability amplitudes that add and whose modulus squared gives a probability

ψincident = e−(i/�)(Et−p·r), where E = p2/2m. (3.1)

These waves obey the free particle Schrödinger equation

i�∂ψ

∂t= − �

2

2mΔψ. (3.2)

We now go back to a deductive presentation.

3.2.1 The Wave Function

• DescriptionSchrödinger and Born understood in 1926 that the complete description of a par-ticle in space at time t is performed with a complex wave function ψ(r, t), whosephysical interpretation is that the probability dP(r) to find the particle in a vicinityd3r of the point r is given by

dP(r) = |ψ(r, t)|2 d3r. (3.3)

38 3 Wave Function, Schrödinger Equation

This is indeed a probabilistic description, but it is a nonclassical one. One can-not describe a quantum system with probabilities only. The wave function is aprobability amplitude (it is necessarily complex). Its modulus squared gives theprobability density of finding the particle at point r at time t and, quite naturally,the integral of this quantity over all space is equal to one:

∫|ψ(r, t)|2 d3r = 1, (3.4)

which is obvious but fundamental (the particle must surely be somewhere).

• Probabilistic interpretationThe meaning of this description is the following. We prepare N atoms indepen-dently, in the same state, so that, when each of them ismeasured, they are describedby strictly the same wave function. Then the result of a position measurement isfor each of them as accurate as we wish (limited by the accuracy of the measuringapparatus) but is not the same for all.The set of impacts is distributed in space with the probability density |ψ(r, t)|2.One can plot the histogram of the distribution. The set of N measurements ischaracterized by an expectation value 〈x〉 and a root mean square dispersion (orsimply dispersion, for short) Δx ,

〈x〉 =∫

x |ψ(r, t)|2 d3r; (3.5)

similarly, the square of the dispersion (Δx)2 is by definition

(Δx)2 = 〈x2〉 − (〈x〉)2 = 〈(x − 〈x〉)2〉. (3.6)

It is a theorem of probability theory that the probability of finding a result within afew times Δx of the value 〈x〉 is close to one. If the accuracy δx of the measuringapparatus is not as fine as the dispersion Δx we can say the particle has a well-defined position.A probabilistic description can perfectly well accommodate the description of a“pointlike” object (Fig. 3.1).

3.2.2 Schrödinger Equation

In 1926, Schrödinger discovered that when the particle is placed in a field of forcethat derives from a potential V (r), the time evolution of the wave function, thereforeof the state of the particle, is given by the partial differential equation

3.2 Principles of Wave Mechanics 39

Fig. 3.1 Simple example of a histogram of the distribution of a position measurement for particlesprepared all in the same state ψ

i�∂

∂tψ(r, t) = − �

2

2mΔψ(r, t) + V (r)ψ(r, t). (3.7)

It is the same as previously (3.2) for a free particle. The incorporation of the forceslies in the second term on the right hand side.

1. The Schrödinger equation is the wave equation. It replaces the fundamental lawof dynamics (mγ = f ).

2. It is a partial differential equation of first order in time. Therefore, if the stateψ(r, 0) is known at some initial time t = 0, the equation determines the stateψ(r, t) at any further time t .

3. One cannot prove this equation (any more than one can prove mγ = f ). Itsjustification is that it works: when one calculates, it gives the correct results.When he tried to apply de Broglie’s idea to the hydrogen atom, Schrödinger wroteseveral other equations. These were more clever, in his mind, because they hada relativistic structure. But, if they gave the correct 1/n2 term, they did not givethe proper relativistic corrections, which were small but known. His wife saidthat for three weeks, he had been in a terrible mood. Then, he surrendered andabandoned relativity, saying that “God decided that things were so.” Actually,when he did this, he did not know about the electron spin, which contributes inan important way to these corrections. That is a piece of luck. Most of quantumphysics is nonrelativistic, therefore comparatively simple.

4. Nearly all quantum mechanical problems consist of solving that equation. Wecome back to it constantly.

At this point, our real problem is to become familiar with this new concept ofa wave function. We want to understand its structure, its properties. We want tounderstand why we need a whole function in order to describe the state of a particle,whereas for Newton six numbers were sufficient.

40 3 Wave Function, Schrödinger Equation

3.3 Superposition Principle

Consider our empirical analysis of atom interferences and de Broglie waves. Every-where the atom is free, and the potential vanishes (except on the screen which atomscannot cross).

Starting from Louis de Broglie’s choice for the frequency ω, we had deducedthe Schrödinger equation in the absence of forces, with V = 0. Conversely, thede Broglie waves are particular solutions of the Schrödinger equation for V = 0.These two principles are the mathematical formulation of de Broglie’s idea.Schrödinger’s real contribution lies in the incorporation of forces.

The previous analysis seems correct; it will account for experimental observationprovided one condition is satisfied. This condition is that after the slits the wavefunction at a point C is indeed the sum

ψC = ψA + ψB (3.8)

of the wave function which would describe the atom if it went through A (B beingblocked) and of the wave function which would describe the atom if it went throughB (A being blocked), each of which would give diffraction distributions showing nointerference fringes.

Here, we have in front of us the most fundamental thing of all this chapter. Wavefunctions have the right to add up. That is the fundamental property ofwave functions.

We promised addition; there it is!More generally, let ψ1 and ψ2 be two wave functions, then the combination

Ψ = αψ1 + βψ2, (3.9)

where α and β are complex numbers (one can produce a phase shift or an attenuationof a wave) belong to the family of wave functions. This sum is a possible wavefunction, and this first principle is called the superposition principle.

The superposition property is completely connected with the interference phe-nomenon. It is much more important than the formula λ = h/p; any other formulawould not change the fundamentals of the experiment.

In other words, two states have the right to add up in order to construct a thirdone.

That is the fundamental notion. What is amazing with mathematicians is thatwhen they see simple things such as that, they understand immediately underlyingstructures. They tell us that the set of wave functions {ψ(r, t)} is a complex vectorspace. If one imposes the normalization condition (3.4),

∫ |ψ|2 = 1, it is what onecalls a Hilbert space, the space of square integrable functions. This property, whichexpresses interferences, is much more important than the concept of a wave functionitself.

Does this theory account for Young slit interferences?

3.3 Superposition Principle 41

Yes, because the Schrödinger equation is linear and any linear combination ofsolutions is a solution. If one formulates the problem of sending a plane wave fromthe left of the slits, one can prove that in the vicinity of the axis and, at large enoughdistances, the usual interference formulae apply. It is a complicated mathematicalproblem, but it has a well-defined solution. (Note that the problem of interferenceson the surface of water presented in the previous chapter, Fig. 2.4, is much simplermathematically, because the two sources are independent. Here, one must take intoaccount that it is the same plane wave that is incident on the slits.)

3.4 Wave Packets

However, de Broglie waves are not wave functions. In fact they are not normalizable.This is not a very difficult problem. It is frequently encountered in wave physics: a

planewave does not exist in practice. It is a useful idealization thatmakes calculationseasy, but physically it would fill all space at all times. A physical wave is alwayslocalized in space at a given moment, and localized in time at a given place. A waveis never exactly monochromatic; there is always some dispersion in frequency andin wavelength.

3.4.1 Free Wave Packets

The representation of a realistic physical situation is a linear superposition of mono-chromatic plane waves, each of which is a particular solution of the Schrödingerequation, of the form

ψ(r, t) =∫

ϕ( p) e−(i/�)(Et−p·r) d3 p

(2π�)3/2, where E = p2

2m, (3.10)

such that all these waves interfere destructively outside some region of space. Thisis called a wave packet. In this formula, we have introduced the constant (2π�)3/2

for convenience, and the complex function ϕ( p) is arbitrary (it determines ψ) exceptthat the expression exists and we want it to be properly normalized (we want that∫ |ψ|2 = 1).

This expression satisfies the Schrödinger for a free particle. One proves in math-ematics that it is the general solution of the free Schrödinger equation.

3.4.2 Fourier Transforms

Note: some further mathematical considerations are treated in Sect.3.11.

However, we want the resulting wave function to be normalizable. How can wedo this without having to check each time?

42 3 Wave Function, Schrödinger Equation

The tool that answers this question is the Fourier transformation. The Fouriertransformation is one of the most important mathematical structures discovered inthe 19th century. It has numerous applications in mathematics, in electronics, inphysics, in chemistry, and so on.

For our purpose, we only need to have in mind one definition and three conse-quences.

One morning in 1812, Fourier found out that he could solve many problems bytransforming a function g of a real variable k into another function f of a real variablex with the formula

f (x) = 1√2π

∫eikx g(k) dk. (3.11)

• The inverse transformation, which allows us to get back g knowing f , is givenby

g(k) = 1√2π

∫e−ikx f (x) dx . (3.12)

The similarity between the two expressions (3.11) and (3.12) is such that we can saythat f and g are Fourier transforms of each other.

• The second property is that the Fourier transformation is what is called anisometry. If f1(x) and f2(x) are, respectively, Fourier transforms of g1(k) and g2(k),then we have the Parseval–Plancherel theorem:

∫f ∗1 (x) f2(x) dx =

∫g∗1(k) g2(k) dk. (3.13)

Of course in all this, we assume that all expressions exist and behave properly.• The third property is that the more the support of |g(k)|2 is concentrated (around

some value k0), the larger is the support of | f (x)|2 (and vice versa). If we normalizef and g to one, so that

∫ | f |2dx = 1 and∫ |g|2dk = 1, we can consider |g(k)|2 and

| f (x)|2 as probability laws for the variables k and x , respectively, and if we considerthe resulting expectation values and dispersions

〈k〉 =∫

k |g(k)|2 dk ; (Δk)2 = 〈k2〉 − 〈k〉2, (3.14)

and similarly for 〈x〉 and Δx in terms of f , the product of the dispersions Δx andΔk is constrained by the inequality

Δx Δk ≥ 1/2. (3.15)

3.4 Wave Packets 43

3.4.3 Shape of Wave Packets

Coming back to wave packets (3.10), the constant � is present in order to havedimensionless quantities because p.r has the dimension of an action.

We see that ψ is the Fourier transform of ϕ( p) e−i Et/� which implies that

∫|ψ(r, t)|2 d3r =

∫|ϕ( p)|2 d3 p. (3.16)

Therefore,∫ |ψ|2 = 1 if and only if

∫ |ϕ|2 = 1. In the choice of ϕ above, we onlyneed to have

∫ |ϕ|2 = 1; ϕ is otherwise arbitrary.We see that:

1. If ϕ is very concentrated around some value p0 the wave function will be close toa monochromatic plane wave in a large region of space. It is therefore a realisticwave function for a beam of atoms. A simple case is represented in Fig. 3.2. Ifwe look at the wave function closely, it resembles a plane wave; if we look at itfrom far away, it may seem to be a concentrated distribution.

2. Conversely, it suffices to exchange the two functions in order to obtain the wavefunction of an atom whose position is well localized in space; that is, ψ is con-centrated near r0, as can be seen in the symmetric Fig. 3.3.

Fig. 3.2 Wave packet obtained with a square form of φ(p) localized in the vicinity of a value p0(top). The two figures at the bottom show the probability distribution (actually the square of the realpart) with two scales of the variable which differ by a factor of 10

44 3 Wave Function, Schrödinger Equation

Fig. 3.3 Real part of thewave function ψ(x)corresponding to a localizedfunction φ(p) in the vicinityof p0 and vice versa, in twocases

3.5 Historical Landmarks

Now, how and when did these ideas emerge?In 1923, Louis deBroglie,who initially studied history, proposed hiswave hypoth-

esis. His thesis is a beautiful text with five chapters. The Science Faculty of theSorbonne was a little bit embarrassed. Was that brilliant or foolish?

• Paul Langevin gave de Broglie’s thesis to Einstein in April 1924, and Einstein wasvery enthusiastic for several reasons.

• First of all, the wave hypothesis could lead to energy quantization as a stationarywave problem, provided one used proper mathematics. This quantization could beachieved without abandoning the continuity of physical laws. Bohr kept on sayingthat discontinuity was a fundamental property of matter, Einstein was shocked bysuch an idea.

• In addition, Einstein managed to explain problems of statistical physics with thewave assumption. He advertised the idea, saying that it was much more than ananalogy with photons; it was very deep.

• At the beginning, Schrödinger found that the idea was elegant, but he was skepticalbecause of relativity. It was Einsteinwho urged Schrödinger towork on the subject.Actually, Schrödinger called that the wave theory of Einstein and de Broglie. Heacknowledged that without Einstein he would not have done anything.

• It is Schrödinger who introduced the Greek letter ψ for the wave function, whichhas become a tradition. However, he made a mistake in its interpretation.

• It isMaxBorn (amathematician,who had the chair of theoretical physics inGöttin-gen) who gave the probabilistic interpretation at the end of 1926. What happenedis that, using the Schrödinger equation, Born calculated the scattering of electrons

3.5 Historical Landmarks 45

on nuclei. He noticed that the square of the wave function |ψ|2 gave the valueof the measured intensity. At the same time, the development of Geiger–Müllercounters provided an alternative measurement technique. Instead of measuringelectric current intensities, one could count numbers of electrons; in other words,one did statistics. Hence the interpretation of |ψ|2 as a probability density.

This probabilistic interpretation achieved the synthesis of two “complementary”aspects of the behavior of particles (electrons or atoms). In the same experiment, anatom behaves as a wave when it interacts with the Young slits, and as a particle whenit is detected (a detector and a diffracting system do not act in the same way).

That is the real problem. An atom is not schizophrenic; it is neither a wave nora particle in the classical sense. It is a well-defined quantum object. In given exper-imental situations, this quantum object may behave as a Newtonian particle or as awave, which seems contradictory to those who possess only classical concepts.

3.6 Momentum Probability Law

So, everything seems to work well up to now. But one can go much further.

3.6.1 Free Particle

We said that the wave functionψ describes completely the state of the particle at timet . Therefore, it must contain the information on the particle’s velocity or momentum.But we do not see that information at the moment.

So, let’s invent. The Fourier transformation is fantastic because it enables us toinvent. Now, we use mathematics just as in the case of instantaneous velocity andderivatives.

In fact, the properties of the Fourier transformation suggest that the probabilitylaw of momentum is given by |ϕ( p)|2; that is,

dP( p) = |ϕ( p)|2 d3 p. (3.17)

This can be proven1; we won’t do this here. We have an idea and we wish to see ifit is plausible and consistent, and if it works.

1The proof can be found in R.P. Feynman’s Thesis Ph.D. Princeton University, 1942; See R.P.Feynman and A.R. Hibbs, Quantum Mechanics and Path Integrals McGraw-Hill, New York,p. 96–100; There is a direct derivation in J.-L. Basdevant and J. Dalibard, Quantum Mechanics,Chap.2, Sect. 6, 2005.

46 3 Wave Function, Schrödinger Equation

1. |ϕ|2 is nonnegative, and, because |ψ|2 is normalized to one, then |ϕ|2 is alsonormalized to one, its integral is equal to one. Therefore it has all the propertiesof a probability law.

2. There is a one-to-one correspondence between ϕ and ψ, which satisfies ourrequirement that the information on the momentum is contained in the wavefunction. If our idea is true, the Fourier transformation extracts the informationon the momentum of the particle from the wave function.

3. De Broglie’s assumption says that the wave function of a free particle of well-defined momentum p0 should be a monochromatic plane wave. In order to getclose to that, one must find a function ϕ that is large in the vicinity of p0 andsmall elsewhere as in Fig. 3.2.

Now, we see that we can use amathematical notion in order to represent a physicalidea. The Fourier transform of the wave function is the probability amplitude for themomentum of the particle (again, this can be proven).

That’s a fantastic physical result! If the particle is localized in momentum p,then it is wavelike in position x . If it is localized in position, then it is wavelikein momentum, as one can see in Fig. 3.3. The space of positions and the space ofmomenta are reciprocal spaces, conjugate of each other. In order to describe the stateof the particle, we can use either of them, in a symmetrical way.

3.6.2 General Case

In fact, not only is this true, but it is general. For a particle placed in an arbitrarypotential, if we consider the Fourier transform of the wave function,

ϕ( p, t) =∫

ψ(r, t) e−(i/�)(p·r) d3r

(2π�)3/2, (3.18)

then ϕ( p, t) is the probability amplitude for the momentum,

dP( p) = |ϕ( p, t)|2 d3 p. (3.19)

The free particle case is a particular case for the behavior in time.Now,we can calculate various expectation values of themomentumor of functions

of the momentum.

3.7 Heisenberg Uncertainty Relations

Here, we face an extraordinary discovery, a central result, and, at the same time, atragedy.

Indeed, it is a consequence of Fourier analysis and of Eq. (3.15) that, whatever thewave function is, whatever the state of the system, the dispersions on measurementsof positions and momenta along the same axis always satisfy the inequalities

3.7 Heisenberg Uncertainty Relations 47

Δx Δpx ≥ �

2, Δy Δpy ≥ �

2, Δz Δpz ≥ �

2, (3.20)

which are called the Heisenberg uncertainty relations.Here is a true physical constraint on a wave packet. If we compress it in one

variable, it expands in the other! If it is compressed in position (i.e., localized) thenit must be spread out in momentum. If it is compressed in momentum, it is spreadout in space.

What is the physical meaning of Heisenberg’s inequalities?

• Once more, suppose we prepare N systems in the same state. For half of them,we measure their positions x ; for the other half, we measure their momenta px .Whatever way we prepare the state of these systems, the dispersions obey theseinequalities.

• These are intrinsic properties of the quantum description of the state of a particle.• Heisenberg uncertainty relations have nothing to do with the accuracy of measure-ments. Each measurement is done with as great an accuracy as one wishes. Theyhave nothing to do with the perturbation that a measurement causes to a system,inasmuch as each particle is measured only once.

• In other words, the position and momentum of a particle are defined numericallyonly within limits that obey these inequalities. There exists some “fuzziness” inthe numerical definition of these two physical quantities. If we prepare particles allat the same point, they will have very different velocities. If we prepare particleswith a well-defined velocity, then they will be spread out in a large region of space.

• Newton’s starting point must be abandoned. One cannot speak simultaneously ofx and p. The starting point of classical mechanics is destroyed.

Some comments are in order.A plane wave corresponds to the limit Δp = 0. Then Δx is infinite. In an inter-

ference experiment, the beam, which is well defined in momentum, is spread out inposition. The atoms pass through both slits at the same time. We cannot “aim” at oneof the slits and observe interferences.

The classical limit (i.e. how does this relate to classical physics) can be seen in avariety of ways that are more or less equivalent. One possibility is that the orders ofmagnitude of x and p are so large that �/2 is not a realistic constraint. This is the casefor macroscopic systems. Another possibility is that the accuracy of the measuringdevices is such that one cannot detect the quantum dispersionsΔx andΔp. We showlater on how one recovers quantitatively the laws of classical mechanics themselves.

Size and Energy of a Quantum System

Our third comment is that now we can do a lot of physics. In fact, uncertaintyrelations enable us to estimate orders of magnitude of various effects without solvingcomplicated equations.

In the lowest energy state of a quantum system, its “ground state,” that is, whenit is no longer excited, the product Δx Δp is of the order of �,

48 3 Wave Function, Schrödinger Equation

Δx Δpx ∼ �. (3.21)

This allows us to estimate quickly orders of magnitude. Of course we will alwaysmiss a numerical factor of order 1. This factor can only be obtained by solvingequations, but we will know the orders of magnitude.

For instance, there is a relation between the size and the energy of a quantumsystem.

Consider a particle that is bound to a fixed center, for instance an electron arounda proton in an atom, or nucleons (neutrons and protons) in a nucleus. We assume thecenter is fixed at position r0; that is, we neglect quantum effects of the center whichis very massive. The particle has some wave function. In the center of mass system,we can choose the origin of coordinates so that we have by assumption 〈r〉 = 0,〈 p〉 = 0.

The dispersion Δx is of the order of r0, the size of the system. The square of themomentum dispersion Δp is equal to 2m times the kinetic energy

Δp2 = 〈p2〉 = 2m 〈Ekin〉.

There is therefore a relation between the kinetic energy and the size of the system

〈Ekin〉 � �2

2mr20.

The smaller the system is, the larger its kinetic energy.For an external electron in an atom, the size is of the order of an Angström, and

we obtain a kinetic energy of a few eV.In a nucleus, we lose a factor of 2000 because of the mass, but the size, one fermi,

gives a factor of 1010 larger; we therefore obtain energies of the order of tens ofMeV.We do obtain the correct orders of magnitude, because the kinetic energy, bindingenergy, and potential energy are of the same order of magnitude unless there existsa pathology.

This also explains why, in order to probe matter at short distances, one must uselarge energies. It is necessary to use powerful particle accelerators.

Stability of Matter

More important is that uncertainty relations allow us to prove the stability of matter,which is one of the greatest contradictions of classical physics. In theworld ofNewtonand Maxwell, matter should be unstable and the world should collapse.

This is an inevitable consequence of the theories of Newton and Maxwell. Con-sider the very simple case of a hydrogen atom; one electron orbiting around a protonin the Coulomb potential V (r) = −q2

e /4πε0r . Suppose the orbit is circular, for sim-plicity, of radius r . Mechanical equilibrium implies mv2/r = q2

e /4πε0r2, and theenergy of the electron is therefore:

3.7 Heisenberg Uncertainty Relations 49

Fig. 3.4 Effectiveinteraction potential due tothe interplay betweenCoulomb attraction andquantum repulsion owing touncertainty relations in anatom

E = p2

2me+ V (r) = −1

2

q2

4πε0r.

This energy is not bounded from below; the more the radius decreases the morethe energy decreases. Now, in its circular motion, the electron is accelerated. Theconsequence of Maxwell’s equations is that it must then radiate and lose energy.Therefore, from the classical point of view, matter is unstable. The electron shouldradiate continuously and it should collapse on the nucleus (r → 0) by radiating aninfinite amount of energy.

It is perhaps the most serious problem of classical physics, although people didnot realize that because the electron and the nucleus were found rather late.

Uncertainty relations preserve us from this catastrophic fate and suppress thisinconsistency. Let 〈r〉 be the average distance of the electron and the proton, whichwe consider fixed (the recoil is negligible). The Coulomb energy is of the order ofq2e /4πε0〈r〉. If we use the order of magnitude (3.21), the kinetic energy is Ek ≥

�2/2me〈r〉2. Therefore, the total energy is of the order of

E ≥ �2

2me〈r〉2 − q2

4πε0〈r〉 . (3.22)

This quantity is bounded from below. Its minimum is attained for 〈r〉 = 4πε0�2/

(meq2) ∼ 0.53 10−10 m, which is bounded from below, and leads to

Emin ∼ − me

2�2

(q2

4πε0

)2

= −13.6 eV.

This is a fundamental result. The uncertainty relations put a lower bound on theaverage distance of the electron and the proton, as well as on their potential energy,and their total energy. This explains the stability of matter (Fig. 3.4).

50 3 Wave Function, Schrödinger Equation

The above argument is not rigorous; it can be made rigorous because one canprove other forms of the uncertainty relations, in particular that for any system wehave 〈p2〉 ≥ �

2〈1/r〉2. If we apply this to 〈E〉 = 〈p2〉/2m − (q2e /4πε0)〈1/r〉 the

argument becomes rigorous: 〈1/r〉 is bounded from above. The uncertainty relationscreate a “Heisenberg pressure,” which reacts against the fact that the electron andproton come too close to each other. Consequently, this results in an equilibriumsituation and the quantum impossibility that matter collapses.

This term has the behavior of a centrifugal force, but it does not come from angularmomentum. We show later on that, in its quantum ground state, the electron of thehydrogen atom has zero angular momentum and does not radiate.

3.8 Controversies and Paradoxes

Heisenberg was a young assistant of Max Born and, as early as 1924, he had elab-orated his own version of quantum mechanics which seems at first very differentfrom what we are doing (we come back to this point). It was only in 1927 that hestated his uncertainty principle (the proof was given later). It is one of the mostfundamental elements of quantum mechanics. It is compulsory for the consistencyof quantum mechanics. If an experiment contradicts Heisenberg’s relations, all ofquantum mechanics is destroyed.

The 1927 Solvay Congress

The uncertainty principle set off terrible debates. Some peoplewere astonished; someothers were enthusiastic. And there was the tragedy we mentioned above. The year1927 was a breaking off for one of the founders of quantum mechanics, and notthe least, Albert Einstein, one of the greatest physicists in history. In 1905 he hadoriginated the theory of Brownian motion, he had invented special relativity, andhe had invented the photon, for which he was awarded the Nobel prize in 1921. In1915, he had constructed general relativity, a most beautiful theory. He had madetremendous contributions to statistical physics and to quantum physics. In 1917, heunderstood the existence of stimulated emission, a keypoint for laser physics. At thefifth SolvayCongress, inBrussels inOctober 1927, he rose up and revolted against theprobabilistic aspect of quantum mechanics, and against uncertainty relations whichhe disliked profoundly. Einstein did not understand what was going on.

Einstein’s revolt concerned two points. One is the notion of a complete descriptionof reality. He thought that a complete description is possible in principle, but that theprobabilistic description is simply quicker to handle and more convenient. The otherreason is the notion of determinism: same causes produce same effects.

Einstein said the famous: “God does not throw dice!” Actually, Einstein’s wordsare in a letter he wrote to Max Born on December 4th, 1926,2 “The theory produces

2“Die Theorie liefert viel, aber dem Geheimnis des Alten bringt sie uns kaum näder. Jedenfalls binich überzeugt, dass der nicht würfelt.”

3.8 Controversies and Paradoxes 51

a good deal but hardly brings us closer to the secret of the Old One. I am at allevents convinced that He does not play dice.” Because the theory works and givesgood results, it must be some intermediate step: there must exist a more completeunderlying theory, involving, for instance, “hidden variables”3 to which we do notyet have any access and over which we average things.

And, at the Solvay meeting, Einstein refused and reacted abruptly. This was duepartly to the dogmatic attitude of Niels Bohr. So Einstein attacked; in particular heattacked uncertainty relations. Uncertainty relations are, in that respect, similar tothe Carnot principle and perpetual motion in thermodynamics. Einstein accumulatedcounterexamples. But, of course, his counterexamples were far from being obvious.One had to work hard to disprove them.

The EPR Paradox

In 1935, Einstein proposed a famous paradox: the EPR—Einstein, Podolsky, andRosen—paradox which we shall describe in Chap.17.

In that “gedanken experiment,” by considering a couple of correlated particles,he showed how because of momentum conservation, one could in principle know asaccurately as one wished both the position and the momentum of one of the particles,which “beats” uncertainty relations.

Hidden Variables, Bell’s Inequalities

In another version, proposed 20 years later by David Bohm, with spin, it is a genuinepuzzle. This goes so far that physicists addressed the question as to whether thereexists a more complete theory involving “hidden variables” of which we would havesome ignorance.

After all, when we play cards, we are blind to the identity of each card that is dealt.But if one used sophisticated enough devices, one could certainly tell the differencebetween the ace of spades and the seven of hearts. In order to play end enjoy it, weforbid ourselves to have a more complete knowledge which exists in principle.

The amazing thing is that John Bell, in 1965, was able to show that this assump-tion leads to quantitative measurable consequences that are in opposition to thepredictions of quantum mechanics.4 We come back to this in Chap.17.

The Experimental Test

This attracted the attention of the general public and the press. There are philosophicalconsiderations attached to that problem. In 1979, at the Cordoba Colloquium, therewere many discussions on parapsychology, levitation, oriental philosophy, and otherbold considerations. Thiswas reported by theFrenchnewspaperLeMondeofOctober24, 1979 (considered as very serious intellectually). All of that was based on theEinstein–Podolsky–Rosen paradox. But exactly three years later, on December 15,

3Actually Einstein never used that word.4J.S. Bell, Physics 1, 195 (1964).

52 3 Wave Function, Schrödinger Equation

1982, in the same newspaper, the phraseology changed. One can see the words:“Experiment might ruin Einstein’s hopes,” “God probably plays dice,” and so on.

Why did that change of phraseology occur? Because, since the late 1970s, aseries of experiments have been performed, among which are those of Clauser andFreedman5 and of Alain Aspect and his group6 in Orsay, to try to see whether onecould falsify quantum mechanics.

In fact the results were negative! Quantum mechanics holds, even though it hurtssome “common sense.” We must face experimental results and revise our way ofthinking. These are extraordinary experiments which we discuss in Chap.17. We seewhat kind of thoughts Feynman had when he said in 1965 “I think I can safely saythat nobody understands quantum mechanics”.7

3.9 Exercises

1. Spreading of the wave packet of a free particle

a. Consider a free particle moving along the axis x . Show that the time derivativeof 〈x2〉t can be written:

d〈x2〉tdt

= A(t) with A(t) = i�

m

∫x

(ψ

∂ψ∗

∂x− ψ∗ ∂ψ

∂x

)dx .

b. Calculate the time derivative of A(t) and show that:

d A

dt= B(t) with B(t) = 2�

2

m2

∫∂ψ

∂x

∂ψ∗

∂xdx .

c. Show that B(t) is constant.d. By setting:

v21 = �

2

m2

∫∂ψ

∂x

∂ψ∗

∂xdx

and ξ0 = A(0), show that 〈x2〉t = 〈x2〉0 + ξ0t + v21 t

2.e. Show that (Δxt )2 = (Δx0)2 + ξ1t + (Δv)2t2 holds, with:

ξ1 = i�

m

∫x

(ψ0

∂ψ∗0

∂x− ψ∗

0∂ψ0

∂x

)dx − 2x0v0,

5S.J. Freedman, J.F. Clauser, Experimental test of local hidden-variable theories, Phys. Rev. Lett.28.938, 1972.6Experimental Realization of Einstein–Podolsky–Rosen–Bohm Gedankenexperiment: A New Vio-lation of Bell’s Inequalities, A. Aspect, P. Grangier, and G. Roger, Physical Review Letters, 49,91 (1982); Experimental Test of Bell’s Inequalities Using Time-Varying Analyzers, A. Aspect, J.Dalibard and G. Roger, Physical Review Letters, Vol. 49, 1804 (1982).7The Character of Physical Law, MIT Press, Cambridge, MA 1965.

3.9 Exercises 53

where ψ0 ≡ ψ(r, 0). The coefficient ξ1 can be interpreted physically using theresults of the next chapter, as the correlation at time 0 between position andvelocity: ξ1/2 = 〈xv〉0 − x0v0.One can verify that the constraint on ξ1 resultingfrom the fact that (Δxt )2 > 0 is equivalent to the condition that Δxt Δpt ≥ �/2at each time t .

2. The Gaussian wave packet

Consider the wave packet given by:

ϕ(p) = (πσ2�2)−1/4 exp

(− (p − p0)2

2σ2�2

)(3.23)

a. For t = 0 show that Δx Δp = �/2.b. Show that the spatial width of the wave packet at time t is given by:

Δx2(t) = 1

2

(1

σ2+ t2σ2

�2

m2

). (3.24)

3. Characteristic size and energy in a linear or quadratic potential

Using an argument similar to that of the stability of matter, evaluate the characteristicsize and energy of a particle with mass m moving in (i) a one-dimensional harmonicpotential V (x) = mω2x2/2; (ii) a one-dimensional linear potential V (x) = α|x |.

3.10 Appendix: Dirac δ “Function”, Distributions

Definition of δ(x)

We often refer to point-like objects in physics. The mass density ρ(r) (or the chargedensity) of such an object is not a function in the usual sense, since it is everywherezero except at point r0, but its “integral” is finite:

∫ρ(r) d3r = m.

The δ “function”, introduced by Paul Dirac, can describe such a density. Its math-ematical definition was elaborated by the mathematician Laurent Schwartz in theframework of distribution theory, which we shall briefly describe in the next section.

In this section we present the (mathematically improper) names and formalismused by physicists. For a real variable x , the “function” δ(x) has the followingproperties:

δ(x) = 0 for x �= 0 and∫ +∞

−∞δ(x) dx = 1. (3.25)

54 3 Wave Function, Schrödinger Equation

For any function F(x) regular at x = 0, we have by definition:

∫F(x) δ(x) dx = F(0). (3.26)

By a change of variables, one can define the function δ(x − x0) for which:

∫F(x) δ(x − x0) dx = F(x0). (3.27)

The generalization to several dimensions is straightforward. Consider for instancer = (x, y, z), we will have:

δ(r − r0) = δ(x − x0) δ(y − y0) δ(z − z0), (3.28)

that is to say: ∫F(r) δ(r − r0) d3r = F(r0).

Examples of Functions Which Tend to δ(x)

One can construct distributions which are nearly point-like, using functions concen-trated in the vicinity of a point x0 (Fig. 3.5). In order to do so, we consider sequencesof functions depending on a parameter which determines their width (yε(x), gσ(x) inthe first two following examples). Although these functions have no limit in the usualsense when their width goes to zero, the integral of their product with any functionF regular at x = x0 remains well defined and tends to the limit F(x0).

Fig. 3.5 Examples of functions concentrated in the vicinity of a point, whose limit, in the senseof distributions, is equal to δ(x)

3.10 Appendix: Dirac δ “Function”, Distributions 55

1. Consider the sequence of functions yε(x) (Fig. 3.5a) defined by:

yε ={1/ε for |x | ≤ ε/20 |x | > ε/2

. (3.29)

Then

∫ +∞

−∞F(x) yε(x) dx = 1

ε

∫ ε/2

−ε/2F(x) dx = F(θε/2) with − 1 ≤ θ ≤ 1.

In the limit ε → 0, yε(x) “tends” to δ(x).

2. Gaussian function (Fig. 3.5b): gσ(x) = 1√2πσ

exp(−x2/2σ2).

By the change of variables y = x/σ,

∫ +∞

−∞F(x) gσ(x) dx = 1√

2π

∫ +∞

−∞e−y2/2 F(σy) dy.

In the limit σ → 0, the above integral remains well defined and gives the resultF(0). As σ → 0, gσ(x) “tends” to δ(x).

3. Square of a “sine cardinal” (Fig. 3.5c): sin2(xY )/(πx2Y ) with Y → ∞. Since∫ +∞

−∞sin2 x

x2dx = π, we have

∫ +∞

−∞sin2 xY

πx2Ydx = 1.

4. “Sine cardinal” (Fig. 3.5d): sin(xY )/(πx) with Y → ∞.

Since∫ +∞

−∞sin x

xdx = π, we have for all Y :

∫ +∞

−∞sin xY

πxdx = 1.

The last case differs from the previous examples in the sense that for x �= 0, thefunction sin(xY )/(πx) does not tend to zero in the sense of functions when Y → ∞.On the contrary, it oscillates more and more rapidly. It is only “on the average” thatit vanishes.

Properties of δ(x)

1. δ(x) is an even function: δ(x − x0) = δ(x0 − x) (justmake the changeof variablesin (3.27)).

2. We have δ(ax) = 1

|a|δ(x) (a real). Indeed we find for a > 0:

∫ +∞

−∞F(x) δ(ax) dx =

∫ +∞

−∞F(u/a) δ(u)

du

a= 1

aF(0).

For a < 0, we use the fact that δ is even.

Distributions

The previous notions can be formalized rigorously using distribution theory. Wesketch this theory in order to extract some useful results.

56 3 Wave Function, Schrödinger Equation

The Space S

We shall consider the vector space S whose elements are complex valued functionsϕ(x) of one (or several) real variable x and which satisfy the following conditions:the functions ϕ(x) are indefinitely differentiable and, as x tends to infinity, they tendto zero, as well as all their derivatives, more rapidly than any power of 1/|x |. Forexample, the functions e−x2 , xne−x2 are elements of S.

Linear Functionals

A continuous linear functional f on space S is a mapping of S onto the complexnumbers ( f : S → C), such that to each ϕ in S, there corresponds the complexnumber noted ( f,ϕ). This mapping satisfies the following properties:

1. Linearity:( f,α1ϕ1 + α2ϕ2) = α1( f,ϕ1) + α2( f,ϕ2), (3.30)

whatever the complex numbers α1 and α2 and the functions ϕ1 and ϕ2 belongingto S.

2. Continuity: if the sequence of functions ϕ1, ϕ2, . . ., ϕn tends to zero in S, thesequence of numbers ( f,ϕ1), ( f,ϕ2), . . ., ( f,ϕn) tends to zero.N.B. We say that the sequence ϕn tends to zero if xk(d/dx)k

′ϕn tends to zero

uniformly in x whatever the integers k and k ′ positive or zero.

These functionals are called tempered distributions and their set is named S′.

Example

1. Let f (x) be a locally integrable function which remains bounded by a powerof |x | as |x | → ∞. One can associate to it a functional, also noted f , with theformula:

( f,ϕ) =∫

f (x)ϕ(x) dx for any ϕ in S. (3.31)

2. Dirac δ distribution. It is the functional which to any functionϕ(x) in S associatesthe number ϕ(0). This is written as:

(δ,ϕ) = ϕ(0). (3.32)

It is convenient for physicists (but improper) to write:

∫δ(x)ϕ(x) dx = ϕ(0).

In the examples we showed in the previous section, the statement yε or gσ tendto δ is incorrect. However, the statement:

limσ→0

(gσ,ϕ) = (δ,ϕ) for any ϕ in S (3.33)

3.10 Appendix: Dirac δ “Function”, Distributions 57

is perfectly correct. One says that gσ (or yε, . . .) tends to δ in the sense of distri-butions.

Derivation of a Distribution

When the distribution is associated to a differentiable function f (x), as in (3.31),one can write, all operations being legitimate:

( f ′,ϕ) =∫ +∞

−∞d f (x)

dxϕ(x) dx = −

∫ +∞

−∞f (x)

dϕ

dxdx = −( f,ϕ′).

One defines the derivative d f/dx or f ′ of an arbitrary linear functional f in the setof tempered distributions by the relation:

(d f

dx,ϕ

)= −

(f,dϕ

dx

). (3.34)

Example

1. Derivative δ′ of δ:(δ′,ϕ) = −(δ,ϕ′) = −ϕ′(0), (3.35)

which the physicists will write as∫

δ′(x)ϕ(x) dx = −ϕ′(0).2. Consider the step function (Heaviside’s function) defined by:

Θ(x) ={0 x < 01 x ≥ 0

. (3.36)

It is locally integrable, andΘ belongs to the space of tempered distributions. Wecan calculate its derivative:

(Θ ′,ϕ) = −(Θ,ϕ′) = −∫ +∞

−∞Θ(x)ϕ′(x) dx = −

∫ +∞

0ϕ′(x) dx = ϕ(0),

hence the remarkable equality (in the sense of distributions):

dΘ(x)

dx= δ(x). (3.37)

3. In three dimensions, one has (cf. Exercise 1 at the end of this chapter):

Δ(1/r) = −4πδ(r). (3.38)

58 3 Wave Function, Schrödinger Equation

3.11 Appendix: Fourier Transformation

Definition

Consider the vector space S above, whose elements are complex valued functionsϕ(x) of one (or several) real variable x .

Let f (k) be a function of the above space S. We call Fourier transform of f (k)the function g(x) defined by the integral:

g(x) = 1√2π

∫ +∞

−∞f (k) eikx dk. (3.39)

From the definition of S, this integral exists and it is infinitely differentiable in x . Thegeneralization to several variables: x = (x1, . . . , xn), k = (k1, . . . , kn) is obvious:

g(x) = 1

(√2π)n

∫f (k) eik·x dnk. (3.40)

Fourier Transform of a Gaussian

Consider the particular case of a Gaussian function:

f (k) = e−k2/(2σ2)

σ√2π

⇒ g(x) = 1√2π

∫ +∞

−∞e−k2/(2σ2)

σ√2π

eikx dk. (3.41)

This integral can be calculated in several ways, for instance by noting that g(x)satisfies the differential equation:

g′(x) + xσ2g(x) = 0 and that g(0) = 1/√2π,

therefore:

g(x) = e−x2σ2/2

√2π

. (3.42)

The Fourier transform of a Gaussian is also a Gaussian. We notice that:

1√2π

∫e−ikx g(x) dx = 1√

2π

∫e−ikx e

−x2σ2/2

√2π

dx = e−k2/(2σ2)

σ√2π

= f (k).

(3.43)

Inversion of the Fourier Transformation

The Fourier transformation f (k) → g(x) is a mapping of S(k) in S(x). This trans-formation can be inverted, and we have:

3.11 Appendix: Fourier Transformation 59

f (k) = 1√2π

∫ +∞

−∞e−ikx g(x) dx . (3.44)

In order to show this result, we consider the integral:

hσ(k) = 1√2π

∫ +∞

−∞e−ikx e−x2σ2/2 g(x) dx

= 1√2π

∫ +∞

−∞e−ikx e

−x2σ2/2

√2π

(∫ +∞

−∞eixk

′f (k ′) dk ′

)dx .

In the limit σ → 0 we recover the right hand side of (3.44); furthermore, in thedouble integral above, for σ �= 0, all integrals are absolutely convergent and we caninterchange the integrations. After integrating on x , and using the result (3.43), weobtain

hσ(k) = 1

σ√2π

∫ +∞

−∞e−(k ′−k)2/2σ2

f (k ′) dk ′.

By a change of variable y = (k ′ − k)/σ,

hσ(k) = 1√2π

∫ +∞

−∞e−y2/2 f (k + σy) dy.

In the limit σ → 0, we obtain the wanted result

limσ→0

hσ(k) = 1√2π

∫ +∞

−∞e−y2/2 f (k) dy = f (k).

Therefore the Fourier transformation is a reciprocal transformation:

f (k) = 1√2π

∫ +∞

−∞e−ikx g(x) dx ←→ g(x) = 1√

2π

∫ +∞

−∞eikx f (k) dk.

(3.45)

This is why we shall say indiscriminately that f (k) and g(x) are Fourier transformsof one another.

Parseval–Plancherel Theorem

Consider two functions f1(k) and f2(k) of S, and their Fourier transforms g1(x)and g2(x). A fundamental property of the Fourier transformation is the Parseval–Plancherel theorem:

∫f ∗1 (k) f2(k) dk =

∫g∗1(x) g2(x) dx . (3.46)

60 3 Wave Function, Schrödinger Equation

Indeed, using the definition of g2(x), we have:

∫g∗1(x) g2(x) dx = 1√

2π

∫g∗1(x)

(∫eikx f2(k) dk

)dx

= 1√2π

∫∫eikx g∗

1(x) f2(k) dk dx .

On the other hand, from the definition of f1(k) (and therefore of f ∗1 which is its

complex conjugate), we find:

∫f ∗1 (k) f2(k) dk = 1√

2π

∫ (∫eikx g∗

1(x) dx

)f2(k) dk

= 1√2π

∫∫eikx g∗

1(x) f2(k) dk dx,

which proves the result.We can introduce in the space S a scalar product defined by:

〈 f1, f2〉 =∫ +∞

−∞f ∗1 (k) f2(k) dk, (3.47)

to which we associate the norm ‖ f ‖ = √〈 f, f 〉. This scalar product is invariantunder the Fourier transformation (〈 f1, f2〉 = 〈g1, g2〉) which is called an isometrictransformation.

Example Some one-dimensional Fourier transforms are given in the Table3.1.

Table 3.1 Fourier transformsof some usual functions

Function or distribution f (k) Fourier transform g(x)( ddk

)nf (k) (−i x)ng(x)

kn f (k)(−i d

dx

)ng(x)

f (a k) 1|a|g(x/a)

eix0k f (k) g(x + x0)

f (k + k0) e−ik0xg(x)

δ(k) 1/√2π

1√2π δ(x)

e−k2/(2σ2) σ e−x2σ2/2

3.11 Appendix: Fourier Transformation 61

3.11.1 Uncertainty Relation

Consider a function f (k) and assume that | f (k)|2 is the probability law for therandom variable k. It is therefore normalized as:

∫| f (k)|2 dk = 1. (3.48)

With this probability law, one can define the expectation value 〈k〉 of k. By a changeof variable k = 〈k〉 + q , one can shift to the centered variable q of zero expectationvalue. We assume this has been done and we continue to call the centered variable k.The mean square deviation Δk of k, is therefore:

(Δk)2 =∫

k2 | f (k)|2 dk. (3.49)

Because of the isometry, the Fourier transform g(x) of f (k) verifies∫ |g(x)|2 dx = 1. The function |g(x)|2 can be considered as the probability law of therandom variable x . This variable will be centered if 〈x〉 = ∫

x |g(x)|2 dx = 0. Oth-erwise, we can, by a change of variables, switch to a centered variable (a translationin x does not affect | f (k)|2). The mean square deviation Δx of x is:

(Δx)2 =∫

x2 |g(x)|2 dx . (3.50)

One has the following theorem:

Whatever the function f , one has the inequality:

Δx Δk ≥ 1/2. (3.51)

The equality occurs only for Gaussian functions.

Proof Consider the integral I (λ) = ∫ |k f (k) + λ d fdk |2 dk where λ is a real number.

One has:

I (λ) =∫

k2 | f (k)|2 dk + λ

∫k

(f ∗ d f

dk+ d f ∗

dkf

)dk + λ2

∫ ∣∣∣∣d f

dk

∣∣∣∣2

dk.

The first term is equal to (Δk)2, the second term, after integrating by parts, gives:

∫ +∞

−∞kd| f |2dk

dk = −∫ +∞

−∞| f |2 dk = −1.

As for the third, we notice that d f /dk is the Fourier transform of −i xg(x). Owingto (3.46), we therefore have:

62 3 Wave Function, Schrödinger Equation

∫ ∣∣∣∣d f

dk

∣∣∣∣2

dk =∫

x2|g(x)|2 dx = (Δx)2 .

Weobtain finally I (λ) = (Δk)2 − λ + λ2(Δx)2. But I (λ) is positive for allλ, whichis possible only if 1 − 4(Δx)2(Δk)2 ≤ 0. This is nothing but the inequality (3.51).

The inequality (3.51), called an uncertainty relation, shows that when the supportof a function is concentrated, the support of its Fourier transform is spread. One canshow that in the case of a Gaussian, and only in that case, we obtain a saturation ofthe inequality: Δx Δk = 1/2.

3.12 Exercises

4. Laplacian operator in three dimensions

By applying (3.34), show that one has in three dimensions Δ(1/r) = −4πδ(r).

5. Fourier transform and complex conjugation

We choose a function f (k) in the space S and we denote its Fourier transform g(x).Determine the Fourier transform of f ∗(k). Characterize the properties of g(x) whenf (k) is real and symmetric ( f (k) = f (−k)).

Chapter 4Physical Quantities

What we have done up to now on wave mechanics seems complicated at first sight.Indeed there are new words, complicated formulae, but the number of importantthings to remember is not that large.

Actually, everything can be summarized as follows.

1. The superposition principle says that the state of a particle in space at time tis described by a wave function ψ(r, t). This wave function has a probabilisticinterpretation. We have said, but not used yet, that the wave function belongs to avector space. This fundamental property expresses interference phenomena. Weshow in a moment that it is not trivial at all.

2. The Schrödinger equation gives the time evolution of the wave function. Thistheoretical scheme explains interference phenomena of particles through the formof de Broglie waves.

3. By studying wave packets and the properties of the Fourier transformation, wehave learned a lot. A wave packet can be very close locally to a monochromaticplane wave, but still be normalizable. If one looks at it from a distance, it can bewell localized in space.It also has good velocity. The group velocity of a wave packet is equal to thevelocity of a classical particle. This was one of the successes of Louis de Broglie.More generally, by making use of the Fourier transformation one can show thatthe time derivative of the expectation value of a wave packet is equal to theexpectation value of the momentum divided by the mass: d〈x〉/dt = 〈p〉/m.

4. The probability amplitude for themomentumof the particle is given by the Fouriertransformof thewave function. In otherwords, the Fourier transformation extractsin the wave function the information on the momentum of the particle.

5. Finally, using Fourier analysis, we have ended upwith the Heisenberg uncertaintyrelations, which are a cornerstone of quantum mechanics.

© Springer International Publishing Switzerland 2016J.-L. Basdevant, Lectures on Quantum Mechanics,Graduate Texts in Physics, DOI 10.1007/978-3-319-43479-7_4

63

64 4 Physical Quantities

4.1 Statement of the Problem

We now finish our program: we examine the question of physical quantities and theirmeasurement.

We discuss the following questions simultaneously. The first is: what does onefind by measuring a physical quantity on a system whose state ψ is known?

The second is: what does the measurement process consist of? We discuss thisquestion by using one of Einstein’s counter examples. That is the point on which thephilosophical considerations we mentioned, and Einstein’s objections to quantummechanics, bear.

Next, we examine a physical quantity that plays a central role, the energy. Andwe finish this chapter by describing the celebrated Schrödinger cat problem whichshows the kind of debate one can have on the interpretation measurement in quantummechanics.

Here we are dealing with a difficult subject. Things are far from obvious, but theirmanipulation is simple and they become quickly familiar when one manipulates thetheory. We really face the problem that the relation between physical concepts andtheir mathematical representation is not obvious in quantum mechanics.

What we are going to say on physical quantities would be terribly abstract andpuzzling if we did not have the Fourier transformation at our disposal. It is difficultfor us, but it wasn’t for the people who were constructing quantum mechanics. Wewill understand why.

4.1.1 Physical Quantities

Physical quantities are things corresponding to aspects of the reality of a system thatcan be measured, that is, they can be characterized by numbers. In classical physics,considering a particle, the result of the measurement of a quantity A is a number aand this single number at time t .

In quantum mechanics, what we said concerning position and momentum holdsfor any physical quantity. If one uses the same apparatus as in classical physics, andif ψ is the state of the system, quite generally, the result of a measurement on onesystem is some value ai . But if we measure that quantity on a large number N ofidentical systems, all in the same stateψ, we obtain N values that are not all the samein general. There is some probability law for this quantity.

Consequently, the complete outcome of the measurement of a quantity A on thestate ψ is the whole set of results ai and associated probabilities pi , or rather theprobability amplitudes α(ai ). This set may be a continuous set, as for the positionof a particle; it may also be a discrete set, as for the energy levels of an atom. Thetheory must therefore be able to define laws that allow us to calculate these numbersstarting with the wave function ψ:

4.1 Statement of the Problem 65

ψ(r, t) −→ numbers {ai ,α(ai ), p(ai ) = |α(ai )|2}.

Here, we are interested in only one combination of these numbers, the expectationvalue of the results

〈a〉 =∑

i

ai p(ai ), in the discrete case, or 〈a〉 =∫

a p(a) da. (4.1)

We say a few words on the possible results of a measurement, but all this, results andamplitudes, becomes simpler when we have Hilbert space analysis at our disposal.

Notice something rather obvious intuitively: it is an experimental fact that

• The various results only depend on the nature of the system, not on its state. Forinstance, the energy levels are different for an electron in an atom and for a protonin a nucleus.

• It is the total outcome, i.e. the probability amplitudes that depend on the state ofthe system.We must therefore transform a function into numbers.

4.1.2 Position and Momentum

However, one can object that we already know a lot. We know the probability lawsfor the position and for the momentum. Therefore we can calculate a whole lotof expectation values. For instance, we can calculate the expectation value of thepotential energy of the particle. At each point, the potential energy of the particlewould be V (r). The expectation value, averaged over the probability law, is givenby the integral

〈V 〉 =∫

V (r)|ψ(r, t)|2d3r. (4.2)

Similarly for any function of the momentum, for instance, the kinetic energy

〈Ek〉 =∫

p2

2m|ϕ(r, t)|2d3 p. (4.3)

True, but there are two objections.

• First of all, for what concerns the momentum, thanks to the beautiful properties ofthe Fourier transformation, one can calculate the expectation value of any functionof the momentum. But it is slightly frustrating. We must take the wave function,calculate an integral,

ϕ( p, t) =∫

ψ(r, t) e−(i/�)p·r d3r

(2π�)3/2, (4.4)

66 4 Physical Quantities

and then start over again, and calculate another integral. Altogether, this meanscalculating a triple integral.

• And, more seriously, there exist physical quantities that are functions of both theposition r and the momentum p. For instance, the total energy is E = p2/2m +V (r) and, even worse, the angular momentum is L = r × p.

There, we don’t know what to do. Even if we can calculate the expectation valueof the energy by adding the two expressions (4.2) and (4.3),

〈E〉 = 〈Ek〉 + 〈V 〉, (4.5)

we don’t know the probability law for the energy. It’s even worse for the angularmomentum L. What are its probability laws and its expectation value?

Do we need to invent a new type of Fourier transform each time and be amazed?One would never believe it. Even though, in the end, it is indeed what will happen;the structure of the Fourier transformation is quite general.

Wemust have a general principle.Wemust be able to calculate everything directlyin terms of the wave function ψ that contains all the physical information on thesystem.

4.2 Observables

The principle that we have to understand and learn can be stated in a single sentence.But it is a sentence that seems terribly abstract and far away from physics.

To each physical quantity A, there corresponds an observable A, a mathematicalobject that is a linear Hermitian operator acting on the space of wave functions,such that the expectation value of the result of a measurement of the quantity Aon the system in a state ψ is given by the number:

〈a〉 =∫

ψ∗(r, t) [ Aψ(r, t)] d3r. (4.6)

For themoment, we don’t understandmuch, except that a linear operator is a linearmapping of the space onto itself. For instance we can multiply the function by x ,take its derivative with respect to x , or take its Laplacian (provided these operationsare legal).

That observables are Hermitian is inessential for the moment. We show later onthat it comes from the fact that measured quantities are real numbers.

Therefore, our principle says that there exist linear Hermitian operators (whichremain to be defined) that perform the operation of transforming a function intonumbers using the above prescription.

4.2 Observables 67

The observable A is the tool that extracts from the wave function the informationabout the quantity A. It transforms the wave function into another function; wemultiply by the complex conjugate of the wave function, integrate, and there lies, in(4.6), the desired expectation value.

4.2.1 Position Observable

For what concerns space properties, this is completely trivial. It simply consists ofdeveloping the modulus square and in writing

< x >=∫

ψ∗(r, t) x ψ(r, t)d3r , < V >=∫

ψ∗(r, t)V (r)ψ(r, t)d3r, (4.7)

which indeed have good form. The position and potential energy observables are themultiplication of the wave function by x, y, and z, or by V (r).

4.2.2 Momentum Observable

But for momentum it is less obvious, except that the form

< px >=∫

ϕ∗(r, t) px ϕ(r, t)d3 p (4.8)

has good structure.In fact, if we remember the definition of the Fourier transform, its inversion, and

the Plancherel Theorem (3.13), we see that we simply have to find a function whoseFourier transform is pxϕ. But that is very simple; we only need to take the derivative

∂

∂xψ(r, t) =

∫i px�

ϕ( p, t) e(i/�)p·r d3 p

(2π�)3/2, (4.9)

or�

i

∂

∂xψ(r, t) =

∫px ϕ( p, t) e(i/�)p·r d3 p

(2π�)3/2. (4.10)

Therefore (�/ i)∂ψ/∂x is the Fourier transform of pxϕ

�

i

∂

∂xψ ←→ pxϕ,

and thanks to Plancherel, this is becoming really interesting and nontrivial.

68 4 Physical Quantities

We obtain

< px >=∫

ϕ∗(r, t) px ϕ(r, t)d3 p =∫

ψ∗(r, t)(

�

i

∂

∂xψ(r, t)

)d3r. (4.11)

In otherwords, in our principle, the observable px associatedwith the x componentof the momentum is

px = �

i

∂

∂x. (4.12)

This holds for other components, and the vector observable p is simply

p = �

i∇. (4.13)

And there we win! We don’t have to calculate the Fourier transform. We know howto express 〈px 〉 or the expectation value of any function of the momentum directlyin terms of the wave function.

For instance, the expectation value of the kinetic energy, Ek = p2/2m in Eq. (4.3)is

〈Ek〉 =∫

ψ∗(r, t) (−�

2

2m)Δψ(r, t) d3r. (4.14)

4.2.3 Correspondence Principle

Now, we are nearly home. We just have to fix the form of other observables. In orderto do this, we simply remember classical physics and use a principle that ensures, aswe show later on, that in the classical limit one recovers the classical equations. Thisis called the correspondence principle.

In classical mechanics, physical quantities are functions A(r, p) of the positionand momentum variables. The correspondence principle consists of choosing inquantum mechanics the same functions of the position and momentum observables.

To the quantity A(r, p) there corresponds the observable A = A(r, p) .

For instance

Ek = −�2

2mΔ, L = �

ir × ∇ , that is, L z = �

i(x

∂

∂y− y

∂

∂x).

4.2 Observables 69

4.2.4 Historical Landmarks

Why is this principle of physical quantities so difficult to accept for us, whereas itcaused no difficulty for the people who built quantum mechanics? Because we aremissing a link (or some mathematics) and because history is not a logical sequencebut an arborescence.

In 1926, there were actually two versions of quantummechanics. There was wavemechanics (in 1923 de Broglie’s hypothesis, in 1926 Schrödinger’s equation). Butthere was also matrix mechanics, which had been constructed since 1924 in a mostinspired way by Heisenberg, and had been developed by Heisenberg, Born, andJordan.

Furthermore, both versions gave good results. So people quarreled. Are electronsor atoms waves or matrices? That was a terrible question. Physicists discussed andargued about fundamental questions, but not this question of observables, which isbasically technical.

Born haddeclaredSchrödinger’sworks, namelyfinding energy levels as stationarywave problems, as being of “unsurpassed greatness in theoretical physics.” In 1926,Schrödinger had given a series of talks in Copenhagen. There, he, who was a manof great distinction and class, had been literally attacked by Niels Bohr, rigid anddogmatic, to such an extent that he said, “If one must stick to that idea of quantumjumps, I really regret getting involved in all this business!”

Niels Bohr had answered him: “But, Herr Schrödinger, you shouldn’t, given all thepublicity that your works have given to our ideas”. One can imagine the atmosphere.

But Schrödinger and Dirac, the little Mozart of quantum mechanics, were wisepeople and they brought back peace, as we show in Chap.6, because they knew somemathematics. Independently, at the end of 1926, they showed that the two approacheswere equivalent, and they saw this principle emerge in a natural way. There was noproblem. They had both the physical concepts and the mathematical tools.

Wave mechanics was interested primarily in the state of a system; matrix mechan-ics concentrated primarily on physical quantities. For our information, the importanceof the Fourier transform, which has been very useful for us, was pointed out quitelate, in 1927, by C.G. Darwin.

4.3 A Counterexample of Einstein and Its Consequences

At this stage, in order to progress, it is useful to analyze one of Einstein’s counterex-amples. This will allow us to address the question, “OK, God plays dice, but is it afull time job?” In other words, are there situations where indeterminism disappearsand God ceases to throw dice?

Einstein did not like the probabilistic aspect of the theory. So he attacked the rootof the problem. Indeterminism is inscribed in the uncertainty relations, so why notbreak them to pieces, once and for all.

70 4 Physical Quantities

Fig. 4.1 Gedanken experiment of Einstein in order to beat Heisenberg’s uncertainty relations

Fig. 4.2 Velocity distribution of the particles after they pass through the second diaphragm, owingto the uncertainty relations

Fig. 4.3 Diffraction ofa wave by a small hole

The first counterexample, “gedanken experiment,” of Einstein is the follow-ing. Between two diaphragms, one puts a device made of two cogwheels that letpass particles of a well-defined velocity (as in Fizeau’s measurement of the veloc-ity of light). The second diaphragm has a dimension δz as small as one wishes(Figs. 4.1, 4.2 and 4.3).

The cogwheel device ensures that particles which pass through the seconddiaphragm have a well-defined longitudinal velocity vlong . But, in order for the par-ticles to pass through the diaphragm, it is necessary for them to have a transversevelocity vt less than vt ≤ vlongδz/L .

4.3 A Counterexample of Einstein and Its Consequences 71

We can make the size of the diaphragm δz as small as we wish. Therefore, imme-diately after the particles have passed the diaphragm, we are sure of their positionsalong z, up to δz, and we know their velocity vt up to vlongδz/L .

We therefore beat Heisenberg!• “Nein, nein, nein! Not at all!” said Heisenberg to Einstein.• “Warum nicht? Why not?” asked Einstein.• “Look at the result of the experiment. If you measure the velocity vz after

particles have passed through the diaphragm, for a large number of particles you willnot find only one value up to vlongδz/L , but a whole lot of possible values.”

• “Why?” asked Einstein.• “Because of the uncertainty relations! These particles are emitted by a small

hole and have a small spreading δz in position along z. Therefore, they have a largespreading in momentum Δpz ≥ �/2δz (Fig. 4.2).”

• “You’re making fun of me,” said Einstein. “If you use uncertainty relations toprove uncertainty relations, something’s going wrong in your mind!”

• “OK, you prefer de Broglie waves,” said Heisenberg, “but it’s exactly the sameresult. The wave diffracted by a small hole is more and more dispersed in the wavevector as the hole gets smaller (Fig. 4.3). Uncertainty relations also exist in usualwave physics. They express globally one aspect of diffraction. It is only becausewe apply them to particles on which we have some a priori knowledge that we areshocked.”

The only thing we know with certainty is that after passing through the seconddiaphragm, the particles have passed through a hole of size δz, therefore there is adispersion in momentum Δpz which is larger than �/2δz in the statistical sense (i.e.,if we measure a large number of particles).

This is what Heisenberg explained to Einstein: the longitudinal and transversevelocities belong to the past; talking about them is history or philosophy. One cannotdo anything with that. And Heisenberg added, “It is a matter of personal belief,without any operational content, to decide whether the past history of a particle hasa physical reality. It is something one cannot measure.” Heisenberg came close toconvincing Einstein just before the 1927 Solvay meeting.

4.3.1 What Do We Know After a Measurement?

But we have learned a lot with this counterexample.We know something for sure after the particle exits. It has passed through the

hole. Therefore, its position is well-defined (up to δz).But, placing a hole and checking that a particle has passed through it is simply

one of the many ways to measure the position of that particle. We could use anymeasuring apparatus and measure any physical quantity; the situation would be thesame.

1. Therefore, we know something for sure after a measurement on a single system.Wehave found somewell-definedmeasured value (up to δz in this particular case).

72 4 Physical Quantities

Therefore, if we redo the measurement immediately afterwards (immediatelymeans that the wave function hasn’t had time to evolve appreciably) we will findthe same result z0 up to δz, and only that result. Therefore, in performing thatmeasurement, we have obtained information on the state of the particle after themeasurement process.Ameasurement on a single system provides us with some information on the stateof the system after the measurement. This can also be considered as preparingthe state of the system.

2. However, can we check that Einstein was partially right and that the velocity waswell-defined by his device?Yes, of course; we must put a speedometer before the diaphragm. For N particles,the velocity is well-defined, but then a position measurement will show that theparticles are delocalized; that is, the position dispersionΔz is very large before thediaphragm. Andwe can alsomove the hole around in all possible positions, whichwill give us the position probability before particles pass through the hole. Afterthat, by applying the Fourier transformation (which in full rigor means makingan assumption on the phase) we will also get a measurement of the velocity.It is by a measurement on N identical systems that we can acquire informationon the state before the measurement.

3. We can imagine the origin of many philosophical questions at the time, and stillnow: what is physical information? What is reality? Does reality exist indepen-dently from the fact that one observes andmeasures? Does a tree falling in a forestmake noise even if no one is there to listen to it?

4.3.2 Eigenstates and Eigenvalues of an Observable

Now, obviously, this analysis implies much more. Let’s pursue this argument.

1. Just after a position measurement, there is no longer any determinism on theposition. A further position measurement will give us a known value; the valueof the position is sure and well-defined. The state is such that God has stoppedplaying dice!

2. The completely general consequence is that, for any physical quantity A, theremust exist particular states such that the result of a measurement is sure andwell-defined.

3. What are the corresponding particular wave functions? This is formalized usingthe notions of eigenfunctions and eigenvalues of the observable A.

A functionψα is an eigenfunction of A if the application of A on this function givesthe same function multiplied by a number aα which is the corresponding eigenvalue

Aψα(r, t) = aαψα(r, t). (4.15)

With this definition, the theorem is very simple.

4.3 A Counterexample of Einstein and Its Consequences 73

Theorem 1 If ψ is an eigenfunction of A, then the result of a measurement of A iscertain and equal to the corresponding eigenvalue.

Proof. The expectation value of A is aα, and the expectation value of A2 is a2α,therefore the dispersion vanishes, and there is no uncertainty. Indeed, we have

〈a〉 =∫

ψ∗α Aψα d

3r = aα

∫|ψα|2 d3r = aα, (4.16)

but 〈a2〉 =∫

ψ∗α A

2ψα d3r = a2α

∫|ψα|2 d3r = a2α,

therefore (Δa)2 = 〈a2〉 − 〈a〉2 = 0 QED.

Later on, we show that the converse is true. Therefore the result of a measurementis well-defined if and only if the wave function ψ is an eigenfunction of A. We canalso say that “the value of A is well-defined.”

Careful: all of this is easier to formulate and to understand with discrete proba-bilities than with continuous probabilities.

Hence the answer to the question, “What are the possible results of a measure-ment?” Obviously these are the eigenvalues of the corresponding observable. If weare sure that in a further measurement, we will find the same value ai that we havealready found, that means that the number ai found previously is one of the eigen-values of A.

4.3.3 Wave Packet Reduction

Now, the last consequence and not the least.The measurement, which we can decide to perform at some instant, is an irre-

versible act that changes the wave function completely. Before the measurement ofa given quantity, the particle had some wave function that led to a variety of possibleresults in a measurement of that quantity. After the measurement, it has another wavefunction for which the result is unique.

This is completely irreversible! Once we have made the measurement, we cannotdecide that we haven’t made it and come back to the previous wave function. Inother words, a measurement, which we are free to perform or not, modifies thewave function instantaneously in all space. Because of the very intuitive examplepresented above, this is called the reduction of the wave packet. Before, it was spreadout in space; just after the measurement it is very concentrated. And this wave packetreduction postulate is essential in the formalization of quantum mechanics.

It is understandable that there lies the source of serious problems which stilllast in the interpretation of quantum mechanics. That seems to be an instantaneousphenomenon at a distance. Isn’t that in contradiction to relativity?

74 4 Physical Quantities

In fact this is the point on which Einstein constructed his EPR paradox a fewyears later. No signal can travel faster than light; it is not possible to change thewave function instantaneously at a distance. However, this is what experiments onBell’s inequalities have shown to occur. Something does happen instantaneously at adistance. Experiment says that the predictions of quantummechanics are right. But noinformation can travel faster than light because, in order to observe the phenomenon(the correlation of the values of two quantities at a distance), one must phone otherpeople at other places to know what happened there.

We now have all the elements of the theory. Notice that the language is gettingcloser and closer to that of linear algebra: we have spoken of linear mappings, weused the words Hermitian, eigenvalues, and indeed, in a while, quantum mechanicsbecomes more and more similar to matrix calculus.

4.4 The Specific Role of Energy

In the following, we study various systems, and various physical quantities. But oneof these quantities has a special role and we encounter it constantly because it hasconsiderable importance. That is energy.

4.4.1 The Hamiltonian

How did Schrödinger test his equations?

• He was looking for wave equations. The problem was to find energy levels as theresult of a stationary wave problem. Therefore, Schrödinger looked for stationarysolutions of his equations (he actually wrote several before he ended up with thegood one), solutions of the form

ψ(r, t) = φ(r)e−iωt . (4.17)

Inserting this into the Schrödinger equation, the term e−iωt factorizes andwe obtaina time-independent equation

−�2

2mΔφ(r) + V (r)φ(r) = �ωφ(r). (4.18)

• Schrödinger knew a lot of mathematics. He knew that the physically acceptablesolutions (i.e., for us square integrable functions) of such a problem form a discreteset {φn(r), �ωn}. Therefore, the frequencies and the energy levels are quantized

En = �ωn.

4.4 The Specific Role of Energy 75

• But all of this must be consistent. If we are seeking energy levels of atoms, we arelooking for particular solutions such that the energy is well-defined.

Now, let’s consider a particle in a potential V (r). Classically its total energy isE = p2/2m + V , and the corresponding observable is therefore

H = p2

2m+ V = − �

2

2mΔ + V . (4.19)

The equation

− �2

2mΔψ(r, t) + Vψ(r, t) = Eψ(r, t) , that is, Hψ(r, t) = Eψ(r, t), (4.20)

is the eigenvalue equation for the energy. It is the same as (4.18), since the timevariable does not play any role. We can suppress it as in (4.18) which is called thetime-independent Schrödinger equation.

The energy observable H (4.19) is called the Hamiltonian of the system.How is it that Hamilton, who lived from 1805 to 1865, is involved in quantum

mechanics, although he existed one century before its discovery? Well, he deservesit! The story is fascinating. Although he did not know quantummechanics, Hamiltonhad understood that the structure of classical mechanics and quantum mechanics isthe same.

He had first shown that geometrical optics was a limit of wave optics for smallwavelengths. Hewas fascinated by variational principles, in particular by the similar-ity between Fermat’s principle in optics and the least action principle of Maupertuis.

In 1828, Hamilton wrote the following extraordinary observation, “Newtonianmechanics is only one of the possible mechanics. It corresponds to the same limit asgeometrical optics compared to wave optics, whereas geometrical optics is only anapproximation.” Nobody noticed, even Hamilton himself. The great mathematicianFelix Klein pointed that out with some regrets in 1891. Of course there was noexperiment at that time which showed that this idea could have any application.Planck’s constant appeared nowhere.

4.4.2 The Schrödinger Equation, Time and Energy

We observe a most remarkable property. The Schrödinger can be written in the form

i�∂

∂tψ = Hψ. (4.21)

This is the true Schrödinger equation. It is “simpler” than the other one. Naturallyit boils down to the same thing in the present context. But it is more general, as weshow later. It is valid for any system and, when we generalize quantum mechanics,

76 4 Physical Quantities

the wave function ψ is replaced by another object, a vector in Hilbert space, theHamiltonian H is an operator in that space, and the Schrödinger equation is simply(4.21).

If we look at it from that point of view, it tells us some thing remarkable. Whatdetermines the time evolution of the state of a system is the energy observable!

Caution! The information on the energy of a system is contained in the wavefunction, and H is the tool that enables us to extract it. Nevertheless, there exists arelation between two fundamental but equally mysterious physical concepts: energyand time.

We must be careful. Quite often, people think that scientists are partially idioticand that they know only complicated things such as mathematics. In order to makeyou feel comfortable at parties, they will ask you about Fermat’s theorem, blackholes, or what was there before the big bang. But take care, if they ask you whatare time or energy, it’s far better to shift the conversation to football, the globaltemperature increase, or tornadoes.

Theword energy is everywhere. Time is one of themost difficult physical conceptsto define. I’ve looked it up in the Oxford Concise Dictionary, in order to see simpledefinitions.

• Energy is “a body’s power of doing work by virtue of stresses resulting from itsreaction to other bodies.”

• Time is “The progress of continued existence viewed as affecting persons orthings.”

That’s superb, but we haven’t really made much progress.

We don’t know what time is. Does time simply exist?

• The past no longer exists.• The future does not yet exist.• The present instant is of measure zero in the mathematical sense; as soon as it’s

arrived, it’s already gone.

In the Confessions, Saint Augustine wrote: “What is time? If no one asks me, Iknow. As soon as someone asks the question, and I want to explain it, I no longerknow.”

It’s true that we don’t knowwhat time is, no more than what energy is (as opposedto other concepts such as velocity, flux, wavelength, and so on).What is extraordinaryis to see that, if we don’t know what time and energy are, there is a well-definedrelation between the two concepts. It is before us in quantum mechanics. It alsoexists in Hamilton’s analytical mechanics.

4.4.3 Stationary States

In order to see that in a more concrete way, we use the notions introduced above.Consider an isolated system, that is, a system whose potential energy does not

depend on time.

4.4 The Specific Role of Energy 77

We have just seen that states with well-defined energies, eigenstates of energy(energy levels of atoms as for Schrödinger) for which ΔE = 0, have no uncertaintyin the energy.

They have a particularly simple time dependence

ψn(r, t) = φn(r)e−i En t/�. (4.22)

It is then remarkable that in such cases |ψ|2 does not depend on time.The position probability law, the expectation of the position, does not depend on

time. The system does not move! No physical quantity changes!Such states are called stationary states. The functions φn(r) satisfy the eigenvalue

equation (4.18), that is,Hφn(r) = Enφn(r),

in the space variables only, the time-independent Schrödinger equation.Therefore, if a system has a well-defined energy, it cannot evolve. For that system,

time does not exist. No evolution, no motion can occur.If energy is well-defined, there are no oscillations for a pendulum, no Kepler

motion along an orbit, and so on. The planetary model of atoms is wrong. Electronsdon’t orbit around the nucleus.

If the energy is well-defined, the system is frozen in time. That’s a rather fright-ening observation.

4.4.4 Motion: Interference of Stationary States

What causes motion? (This seems to be a question for Greek philosophers in antiq-uity.)

In order for motion to appear, the wave function must be a linear superposition ofstationary states in interference. For instance, one can easily check that a superposi-tion of two stationary states of different energies E1 and E2

ψ = λϕ1e−i E1t/� + μϕ2e

−i E2t/� (4.23)

has a position probability |ψ|2 in which the crossed term depends on time. However,for such a system ΔE = 0, the energy is not well-defined.

Motion occurs, and time exists, only if energy is not well-defined, and if Godplays dice with energy. Then, there are given nonvanishing probabilities of findingE1, E2, and so on. In order for motion to appear, the system must be an interferencebetween states of different energies. That’s a rather fascinating relation between timeand energy.

Actually, we have just written a fundamental technical result, useful for all thatfollows. Suppose that at t = 0 the wave function is a given superposition of stateswith well-defined energies

78 4 Physical Quantities

ψ(r, t = 0) =∑

n

cn φn(r), (4.24)

where the φn(r) are the energy eigenfunctions. Then its time evolution can be writtendirectly, without solving any equation:

ψ(r, t) =∑

n

cn φn(r)e−i En t/�. (4.25)

One can readily check that this expression satisfies the Schrödinger equation (4.21)with the boundary condition (4.24). The evolution is known; motion is known. Weshow later on that it is a fundamental theorem of Hilbert space analysis that any wavefunction can be written in that way.

Therefore the solution of the evolution problem in quantummechanics, for an iso-lated system, always involves finding its energy eigenfunctions and eigenvalues, thatis, solving the time-independent Schrödinger equation or, equivalently, the eigen-value problem of the Hamiltonian. The time evolution follows immediately.

4.5 Schrödinger’s Cat

To end this Chapter, we show one of the most famous paradoxes on the problemof measurement in quantum mechanics, in particular on the superposition of statesand wave packet reduction. It’s necessary for our culture; people also like to talkabout it at parties. It’s not that important if one does not understand everything at thebeginning.

That’s something that always makes me sad. Schrödinger was really an extraor-dinary man.

The photograph (Fig. 4.4) shows that he was somewhat fancy, but a very pleasantand friendly man.

But he had an enormous defect: he didn’t like cats. I like cats very much.

The Dreadful Idea

In 1935, Schrödinger had the following monstrous idea. Suppose one puts a catin a steel chamber (Fig. 4.5) with a diabolic device consisting of a single atom ofradioactive chlorine 39 which decays with a mean life of 60min into argon 39. Thismeans that the probability for the atom to decay within an hour is 50%. If it decays,it emits an electron that is detected by a Geiger–Müller counter, which itself operatesan apparatus, a hammer that falls on a capsule of cyanide, which kills the cat instantly.It’s simply revolting.

Now, one observes the cat after an hour; the horrible questions are the following:

1. What killed the cat?2. When did the cat die? (Of course, if the cat is still alive, we redo the experiment

an hour later, which is horrendous.)

4.5 Schrödinger’s Cat 79

Fig. 4.4 Erwin Schrödinger in 1956. (All rights reserved)

Fig. 4.5 Diabolic device imagined by Schrödinger

The atom, more exactly its nucleus, is a quantum object. Its state is not describedby a wave function of the type we have seen, but by another mathematical objectwith the same name ψ. The atom has a 50% probability of having decayed after onehour, therefore its quantum state after an hour is

ψ(atom) = 1√2(ψ(nondecayed atom) + ψ(decayed atom)). (4.26)

80 4 Physical Quantities

Fig. 4.6 Siné: Schrödinger’scat (© Siné)

It is an equalweight quantum superposition.As long as one has notmeasuredwhetherit has decayed, the atom is in a superposition of the two states, nondecayed atom anddecayed atom.

But, because of the diabolic device, the entire system, including the cat, hasbecome a quantum system. And, the cat is alive if the atom has not decayed; it isdead if the atom has decayed.

To describe a cat requires a very complicated wave function, with some 1027

variables. But that’s inessential. The state of the cat can be inferred directly from thestate of the atom. It is a linear superposition of the states {live cat} and {dead cat},both at the same time, in interference

ψ(cat) = 1√2(ψ(live cat) + ψ(dead cat)). (4.27)

The cat is both alive and dead. It is an abominable state, quite inconceivable and verydifficult to represent, as one can see in Fig. 4.6!

So, coming back to our outrageous questions: what killed the cat? When did thecat die?

• Cyanide? No, it’s simplistic; it was in a capsule that must be broken. Similarly,the hammer must be released.

• The decay? Maybe, but the atom is a quantum system. Its state is a superposition;the superposition must cease; that is, there must be a wave packet reduction inorder to know whether the atom has decayed. Then it will be possible to accusethe atom.

• Therefore, we must have a measuring instrument in order to know whether theatom has decayed.

• The horror is that the porthole and the cat form an instrument that enables us tosee whether the atom has decayed.

• Therefore, onemust observe the cat in order to reduce the wave packet of the atom!• It is by observing the cat that we reduce the wave packet of the cat.• It is when we observe the cat that we commit the atrocity of killing it, destroyingany hope for the poor animal.

• Before we looked at it, the cat was in a much more profitable state (4.27)!

4.5 Schrödinger’s Cat 81

Fig. 4.7 Consequence of theobservation of the cat

Wigner had suggested that observation is a transcendental act of consciencedestroys the superposition. He thus addressed the question of the role of the observerin quantum physics. But one can reply that one can put a camera with shape recogni-tion that typeswhether the cat is dead or alive and puts the answer in an envelope. Onereads the letter a year after. If Wigner was right, the transcendental act of consciencewould go backwards in time, because we exert it a year later (Fig. 4.7).

Actually, in this disgusting example, what is shocking is not so much the wavepacket reduction but the superposition. The quantum superposition of states, whichseems natural when we apply it to objects deprived of souls, such as atoms or elec-trons, becomes very disconcerting if we apply it to familiar objects, as in Fig.4.8.

The Classical World

In order to settle thematter, I gave that as a written examination tomy students. Otherstudents are lucky to escape such torture. The answer is that it is quite conceivableto manufacture paradoxical states such as the superposition {live and dead cat}, thatis, paradoxical macroscopic states.

But such states are extremely vulnerable and fragile. They imply a coherence,or a conspiracy of the 1027 particles, which gets destroyed in an incredibly shorttime because of the (thermal) interaction with the environment. This is called“decoherence” theory. The Nobel Prize in physics 2012 was awarded to jointly toSerge Haroche and David J. Wineland “for ground-breaking experimental methodsthat enable measuring and manipulation of individual quantum systems”. In 1996,

Fig. 4.8 The superpositionprinciple applied to familiarobjects. (Courtesy of Siné.)

82 4 Physical Quantities

S. Haroche and his collaborators1 managed to study the behavior of Schrödinger“kittens”, i.e. systems involving a small number of atoms, and to verify their pre-dicted decoherence. They were able to make a movie of the time evolution of acat-state to a classical state by decoherence.2 Consequently, one must point out thatthe so-called “macroscopic world”, the world of large objects, is not identical to theworld of objects that follow classical physical laws. The classical world is the worldof large objects or systems that are furthermore stable under quantum fluctuationsin their interactions with the external world.

I tried to explain that to my own cat, but either he didn’t understand or he justwasn’t interested.

4.6 Exercises

1. Expectation values and variances

Consider the one-dimensional wave functionψ(x) = √2/a sin(πx/a) if 0 ≤ x ≤ a,

and ψ(x) = 0 otherwise. Calculate 〈x〉,Δx, 〈p〉,Δp and the product Δx Δp.

2. The mean kinetic energy is positive

Verify that for any wave function ψ(x), the expectation value 〈p2〉 is positive.3. Real wave functions

Consider a real one-dimensional wave function ψ(x). Show that 〈p〉 = 0.

4. Translation in momentum space

Consider a one-dimensional wave function ψ(x) such that 〈p〉 = q and Δp = σ.What are the values of 〈p〉 and Δp for the wave function ψ(x)eip0x/�?

5. The first Hermite function

Show that the wave function ψ(x) = e−x2/2 is an eigenfunction of the operator(x2 − ∂2/∂x2) with eigenvalue 1.

1M. Brune, E. Hagley, J. Dreyer, X.Matre, A.Maali, C.Wunderlich, J.M. Raimond and S. Haroche,Phys. Rev. Lett. 77, 4887 (1996).2S. Delglise, I. Dotsenko, C. Sayrin, J. Bernu, M. Brune, J.M. Raimond and S. Haroche, Nature455, 510 (2008).

Chapter 5Energy Quantization

We now solve quantum mechanical problems in order to see how the theory works.We we consider the motion of particles in simple potentials and the quantization ofenergy.

We are going to do three things.

• First explain the position of the problem.• Next, see in two simple cases the origin of quantization of energy levels.• And finally, we study an example that is basically as simple as these two but muchmore subtle in its consequences. This is a model of the ammoniamolecule NH3, onwhich true quantum mechanics effects appear. We will discover the tunnel effect,one of the most important quantum effects. By generalizing our results, this leadsus to modern applications and nanotechnologies.

5.1 Methodology

The solution of such a problem consists of solving the Schrödinger equation. TheHamiltonian of a particle in a potential V is

H = − �2

2mΔ + V (r). (5.1)

We study solutions of given energy (energy is conserved; it is a constant of themotion). These are stationary states which have a very simple time-dependence

ψn(r, t) = φn(r) e−i En t/�. (5.2)

© Springer International Publishing Switzerland 2016J.-L. Basdevant, Lectures on Quantum Mechanics,Graduate Texts in Physics, DOI 10.1007/978-3-319-43479-7_5

83

84 5 Energy Quantization

The functions φn are the eigenfunctions of the Hamiltonian H and En are the corre-sponding eigenvalues

Hφn(r) = Enφn(r). (5.3)

One must impose boundary conditions on the wave function.

5.1.1 Bound States and Scattering States

In classical physics, there is a difference between two regimes in the motion of aparticle in a potential according to the value of the energy.

If it is greater than the value of the potential at infinity, E ≥ V (∞), the system isin a scattering state. The particle goes to infinity as t tends to infinity, and extractsitself from the field of force.

If the energy is smaller than the value of the potential at infinity, E < V (∞), it isa bound state. The particle is on an orbit. At any time its position remains confinedin a finite region of space.

The same distinction exists in quantum mechanics, but going from one regime tothe other is not as simple as in classical physics, both physically and technically. Inclassical physics, we are always interested in the trajectory. In quantum mechanics,the physical quantities of interest are not the same in the two cases.

The general problemof three-dimensional scattering, which is of great importanceparticularly in atomic, nuclear and particle physics, is technically complicated.

Here, we are only interested in bound states.A bound state of well-defined energy is defined by the fact that its wave func-

tion satisfies the eigenvalue equation of the Hamiltonian (5.3) and that it is squareintegrable. In other words it is a “good” wave function

∫|ψn(r)|2 d3r = 1. (5.4)

This condition is essential and corresponds to the fact that classically the particleis confined in a finite region of space.

Therefore, we are interested in finding the set {φn, En} of eigenfunctions andcorresponding eigenvalues of the Hamiltonian, with this boundary condition (5.4).

In mathematics, one can prove that it is a discrete set. This is the origin of energyquantization: the En are the energy levels of the system.

The most general bound state is a linear superposition of stationary bound states.It evolves in a nontrivial way: a wave packet moves around in the potential. In thelimit of large energy values, this motion can become the classical motion.

If we relax the normalization condition (5.4), there exist solutions of the equation

Hψ(r) = Eψ(r), (5.5)

5.1 Methodology 85

for a continuous set of values of the energy E . Such solutions must be properlyinterpreted, and they correspond to scattering states. Their linear superpositions arewave packets that move in space.

5.1.2 One-Dimensional Problems

Here, we concentrate on simple one-dimensional problems. For instance, a marbleon a rail. This is simpler technically, we treat three-dimensional problems later. Theeigenvalue equation is then an ordinary second-order differential equation

− �2

2mψ′′n (x) + V (x)ψn(x) = Enψn(x). (5.6)

We wish to determine the functions {ψn(x)} and the numbers {En} with the normal-ization condition ∫

|ψn(x)|2 dx = 1. (5.7)

In addition, we consider simple potentials for which analytical solutions exist, inorder to become familiar with the physics.

5.2 The Harmonic Oscillator

A first example is the harmonic potential

V (x) = 1

2mω2x2. (5.8)

This corresponds classically to the sinusoidal motion, of frequency ω, of a particleelastically bound to a center (here x0 = 0). It is called a harmonic oscillator. It onlyhas bound states (V (∞) = ∞).

The eigenvalue equation is

− �2

2m

d2ψ(x)

dx2+ 1

2mω2x2 ψ(x) = E ψ(x). (5.9)

The harmonic oscillator has numerous applications. For motions of small amplitudearound its equilibrium position, any system has harmonic oscillations.

Eigenvalues, Eigenfunctions

This equation is a classic of 19th century mathematics. We turn to dimensionlessquantities

86 5 Energy Quantization

ε = E

�ω, y = x√

a, with a = �

mω, (5.10)

and the eigenvalue equation becomes

1

2

(y2 − d2

dy2

)φ(y) = ε φ(y). (5.11)

The square integrable solutions ψ(x) = √a φ(x/

√a), were found by Charles

Hermite:

φn(y) = cn ey2/2 dn

dyn

(e−y2

)= cn e

−y2/2 Hn(y), (5.12)

where Hn(y) is a polynomial of degree n, called a Hermite polynomial. It containsonly even (resp., odd) powers of y if n is even (resp., odd), and the normalizationconstant is cn = π−1/4 2−n/2 (n!)−1/2. The corresponding eigenvalues are

εn = n + 1

2, n nonnegative integer. (5.13)

The differential equation (5.11) has solutions for all positive values of ε. However,in general, these solutions increase at infinity as � e+y2/2. It is only for the set ofvalues (5.13) of ε that the solutions are square integrable.

Energy Levels, Eigenfunctions

The energy levels of the one-dimension harmonic oscillator are therefore equal-spaced

En = (n + 1

2) �ω = (n + 1

2) hν. (5.14)

Planck had correctly guessed the term nhν. Energy exchanges of the oscillator intransitions between levels occur for integer multiples of �ω = hν,

ΔEnn′ = (n − n′)hν.

But Planck had no means (and no reason) to guess the presence of the constant �ω/2,which is called the zero point energy, and which can be measured. This constant isessential in order to satisfy uncertainty relations. In fact the classical equilibrium stateof the oscillator consists of being at rest at the minimum of the potential. In otherwords, its velocity is zero and its position is at the minimum of the potential. Thisis contrary to Heisenberg’s uncertainty relations. The zero-point energy is inevitableand measurable.1

This result, which can be extended easily to three dimensions, has numerous appli-cations. It allows us to understand the vibration spectra of molecules, and the specific

1This energymaymanifest itself inwhat is called the “vacuumenergy” of theUniverse in cosmology.

5.2 The Harmonic Oscillator 87

Fig. 5.1 The first four Hermite functions (abscissa: xa1/2); ψ0(x) is a Gaussian; ψ1(x) is thisGaussian multiplied by x(2a)1/2, and so on.

heats of solids (molecules in a crystal vibrate around their equilibrium position). Itis a basic tool in field quantization and in relativistic quantum physics.

The eigenfunctions are real and orthogonal

∫ψ∗n(x)ψn′(x) dx = δn,n′ . (5.15)

From the definition of Hermite functions (5.12), one obtains the action of the oper-ators x and d/dx on ψn(x):

√2

ax ψn(x) = √

n + 1 ψn+1(x) + √n ψn−1(x) (5.16)

√2a

d

dxψn(x) = √

n ψn−1(x) − √n + 1 ψn+1(x). (5.17)

The first four Hermite polynomials are (Fig. 5.1):

H0(x) = 1, H1(x) = 2x, H2(x) = 4x2 − 1, H3(x) = 8x3 − 12x .

5.3 Square Well Potentials

Square Potentials

Even simpler models consist of what are known as “square potentials”, which arepiecewise constant.

The solutions are, piecewise, exponential or sinusoidal according to the sign ofV − E (E is a number we want to determine).

1. In regions where V = V0 and E − V0 > 0, the wave functions have a sinusoidalbehavior ψ(x) ∝ e±ikx with k = √

2m(E − V0)/�2.2. When E−V0 < 0, thewave functions have an exponential behaviorψ(x) ∝ e±Kx

with K = √2m(V0 − E)/�2.

88 5 Energy Quantization

Fig. 5.2 Sandwich of AlGaAs–GaAs–AlGaAs. The central slice of GaAs has a width of 6nm. Onthe vertical axis, the aluminum concentration is indicated. The shape corresponds to the variationof the potential as “seen” by a conduction electron (i.e., the electrostatic potential averaged overone period of the crystal lattice) (Photograph due to Abbas Ourmazd, ATT Bell Labs.)

In order to obtain the bound states of such systems, we require as usual thatthe wave functions be square integrable. But in addition, the wave functions mustbe continuous as well as their first derivative at the points of discontinuity of thepotential. This prescription is a consequence of the Schrödinger equation. It can beproven easily. One can understand it by considering a step in the potential as the limitof a continuous function (for instance, V ∼ limλ→0 tanh(x/λ)) and by reductio adabsurdum.

This procedure allows us to determine the values of the energy E , as we will see.In modern microelectronic technologies, such potentials have many applications.In “sandwiches”of alternating thin layers of semiconductors (GaAsandGaAlAs),

shown in Fig. 5.2, one can manufacture 2–5nm wide quantum wells (we showedan example of this in the first chapter). The quantum confinement of electrons insuch domains has opened a new era in electronics, computer components, energy,and medicine.2 These are used in optoelectronics, because the corresponding tran-sitions between electron levels (ΔE ∼ 50–200 meV) are in the infrared part of thespectrum.

At the end of this chapter, we show how one can obtain such a picture withquantum tunneling microscopes.

What is important in such models is to understand the physics, that is, the quali-tative results. Exact calculations can then be given safely to computers.

Symmetric Square Well

Energy quantization is a theorem, but let’s see how it occurs. We consider the energylevels of a particle in a symmetric potential well of depth V0 and width 2a, centeredat x = 0.

2Tagged: quantumdots,MITTechnologyReviewCambridgeMass.Nov.2015;M.A.Reed, QuantumDots, Sci. Amer., Jan. 93 ; L.L. Chanz and L. Esaki, Semiconductor Quantum Heterostructures,Phys. Today, 45, p. 36 (1992).

5.3 Square Well Potentials 89

We choose the origin of energies at the bottom of the potential well, so that theenergy E , which we want to determine, is the kinetic energy of the particle insidethe well.

We are only interested in bound states, states for which 0 ≤ E < V0, which inclassical physics correspond to a particle that is confined inside the well. We assumethe energy is not sufficient for the particle to jump out of the well. We set

k =√2mE

�2, and K =

√2m(V0 − E)

�2, (5.18)

and we have

k2 + K 2 = 2mV0

�2. (5.19)

It is straightforward to solve the Schrödinger equation. The wave functions are expo-nentials on the left and on the right (regions I ′ and I ) and sinusoids in the middle(region I I ).

An important simplification comes from the symmetry of the problem V (x) =V (−x). One can indeed classify the solutions in two categories: symmetric (or even)solutions, and antisymmetric (or odd) solutions. In fact, because the Hamiltonian issymmetric H(x) = H(−x), if we change x into−x , we obtain for any solutionψ(x)

H(x)ψ(x) = Eψ(x), and H(x)ψ(−x) = Eψ(−x).

In other words, if ψ(x) is a solution of the Schrödinger equation, then ψ(−x) is alsoa solution for the same eigenvalue of the energy. Therefore, ψ(x) ± ψ(−x) is eithera solution for the same value of the energy, or it is identically zero. This is a particularcase of an important feature in quantum mechanics. To the invariance laws of theHamiltonian, there correspond symmetry properties of the solutions. (This can alsobe obtained directly here by performing the calculation; we could have made thesame remark on the harmonic oscillator.)

The symmetric (ψS) and antisymmetric (ψA) solutions have the forms

ψS : (I ′) ψ(x) = B eKx , (I I ) ψ(x) = A cos kx, (I ) ψ(x) = B e−Kx ,

ψA : (I ′) ψ(x) = −D eKx , (I I ) ψ(x) = C sin kx , (I ) ψ(x) = D e−Kx ,

(5.20)

where the constants A, B,C , and D are determined by the continuity of ψ and ψ′ atx = ±a. Notice that in (5.20) we have omitted terms that increase exponentially atinfinity because of the normalizability condition.

We must join these expressions at x = ±a. When V is continuous, the equationhandles the problem by itself. If not, as is the case here, we must impose that ψ andψ′ are continuous at x = ±a.

90 5 Energy Quantization

We obtain

ψS : A cos ka = Be−Ka, and k A sin ka = KCe−Ka

ψA : C sin ka = De−Ka, and kC cos ka = −K De−Ka, (5.21)

and, if we take the ratios,

k tan ka = K , for (ψS), and − k cot ka = K , for (ψA). (5.22)

These two relations between k and K must be completed by the definition (5.19)which can be written as

k2a2 + K 2a2 = 2mV0a2

�2. (5.23)

In the (ka, Ka) plane, this is the equation of a circle. We must therefore find theintersections of this circle with the curves Ka = ka tan ka and Ka = −ka cot ka asrepresented in Fig. 5.3b. Suppose the width 2a of the well is given, these intersectionskn form a finite set, and they correspond alternatively to even and odd solutions. Thenumber of solutions, or bound states, increases with V0 (deeper well). There is onlyone bound state if V0 is less than

a√2mV0

�<

π

2or V0 <

π2�2

8ma2. (5.24)

The energy levels En = �2k2n/2m are quantized. This quantization is not a conse-

quence of the continuity conditions, which simply enable us to calculate the eigen-values En , but of the normalizability that eliminates exponentially increasing termsin (5.20).

Fig. 5.3 Square well potential: a shape of the potential; b graphical solution giving the energylevels; c limit of an infinitely deep well.

5.3 Square Well Potentials 91

Fig. 5.4 Temperaturedistribution in a breath(artificial colors). Thebackground is black, thebreath comes out of themouth at body temperatureand it expands at roomtemperature (Courtesy ofEmmanuel Rosencher.)

The solutions can be classified by increasing values of the energy E , according tothe number of nodes of the wave functions (Sturm–Liouville theorem). The lowestenergy state is called the ground state.

Notice an essential difference from classical mechanics: the particle has a nonva-nishing probability to be in the classically forbidden regions, (I ) and (I ′), where itskinetic energy would be negative and is therefore energetically illegal.

However, it does not propagate in those regions; it can penetrate them, but itbounces off the edges x = ±a. The wave function decreases exponentially with amean penetration distance of 1/K . This is analogous to the skin effect in electromag-netism. We can see how the “classical limit” appears: 1/K → 0, if we make � → 0or if the mass is large m → ∞.

As we said, quantum wells such as the one represented in Fig. 5.2 have alloweddecisive improvements in advanced technologies, in particular infrared technologiessuch as shown in Fig. 5.4.

Infinite Well, Particle in a Box

Another simple, but interesting, limiting case is that of an infinitely deep wellV0 = ∞. The particle is confined inside a region that we place for conveniencebetween x = 0 and x = L as in Fig. 5.3c.

In this limit of the previous case, thewave function vanishes,ψ(x) = 0, outside theinterval [0, L]. The continuity conditions are then different: only the wave functionis continuous (this can be seen by taking the limit V0 → ∞ on the above solutions).

The normalized eigenfunctions of the Hamiltonian are then:

ψn(x) =√

2

Lsin(

nπx

L) n integer > 0. (5.25)

92 5 Energy Quantization

The corresponding energy levels are

En = n2π2

�2

2mL2n integer > 0. (5.26)

Quantization appears here as a simple stationary wave phenomenon.This calculation can easily be generalized to three dimensions, that is, the case of

a particle in a box. If we consider a rectangular box of sides (a, b, c), the solutionscan be factorized:

Ψn1,n2,n3(r) =√8√

abcsin(

n1πx

a) sin(

n2πy

b) sin(

n3πz

c) , (5.27)

E = En1,n2,n3 = �2π2

2m

(n21a2

+ n22b2

+ n23c2

). (5.28)

This very simple result has numerous applications. Many systems can be approxi-mated by infinite wells: molecules in a gas, neutrons inside a neutron star, conductionelectrons in a metal, and others.

5.4 Double Well, the Ammonia Molecule

Consider nowa similar problem,wherewe really do quantummechanics anddiscoverunexpected results. This is the case of a symmetric double potential well. At thebeginning it is similar to the infinite potential well, but with a potential barrier inthe middle: in other words, an infinite well containing two wells of width a centeredrespectively at ±b, and separated by a barrier of height V0 and width Δ = 2b − a.

5.4.1 The Model

Consider a concrete example, the ammonia molecule NH3.In its lowest energy states, this molecule has the shape of a pyramid with the

nitrogen atom at the top and the three hydrogen atoms at the base, on an equilateraltriangle. It is a complex object made of 14 particles (10 electrons and 4 nuclei) andthere are many possible motions of this system. However, the lowest energy motionscorrespond to the global displacement of the triangle of hydrogen atoms, which wecall collectively a “particle of mass m”, with respect to the nitrogen atom along thesymmetry axis of the molecule.

When the abscissa x of this plane varies along x > 0, the potential energy ofthe system has a minimum that corresponds to a classical equilibrium configurationat x = b. However, the molecule can invert itself, as an umbrella, and there existsanother symmetric stable configuration for a negative value of x = −b. The potentialenergy is symmetric with two minima and a maximum in between. This maximumcorresponds to an unstable configuration where the four atoms are in the same plane.

5.4 Double Well, the Ammonia Molecule 93

Fig. 5.5 The ammoniamolecule:a the two classical configurations;b the actualmolecular potentialenergy (full line) and the simplified model (dashed line) that describes the reversal of the molecule

These two configurations of classical equilibrium are not physically equivalentbecause the molecule possesses an intrinsic angular momentum, and one can definethe “right” and the “left”, as in an umbrella.

This model can be solved numerically, but it suffices to study the square wellmodel V (x) drawn in a dashed line in Fig. 5.5b. This potential consists of two wellsof width a centered at b and −b, respectively, and separated by a barrier of heightV0 and width Δ = 2b − a.

We ask the following question. For E > V0 there are periodic oscillations fromleft to right. But when E < V0, for a given value of the energy E , there are alwaystwo possible classical configurations of same energy, one in the left-hand side well,the other in the right-hand side well.

In particular, there are two positions of equilibrium, one on the left, the other onthe right; both have the same energy.

What is the quantum situation?

5.4.2 Stationary States, the Tunnel Effect

From the calculational point of view, the problem is strictly analogous to what wejust did. We consider the energy levels of a particle of mass m such that classicallythe particle cannot cross the potential barrier: E < V0.

As previously, we define

k =√2mE

�2and K =

√2m(V0 − E)

�2. (5.29)

94 5 Energy Quantization

Fig. 5.6 Symmetric solution (a), and antisymmetric solution (b) of lowest energy in the doublesquare well potential which is a model of the ammonia molecule

The problemhas the symmetry (x ↔ −x).We can classify the solutions accordingto their parity or symmetry.

The solutions are sinusoidal in the regions L and R and exponential in the middleregion M . The wave functions must vanish for x = ±(b + a/2), and the eigenfunc-tions of the Hamiltonian are of the form

ψ(x) = ±λ sin k(b + a/2 + x) L region

= λ sin k(b + a/2 − x) R region

ψ(x) = μ cosh Kx Symmetric solution M region

ψ(x) = μ sinh Kx Antisymmetric solution (5.30)

The two lowest energy solutions are represented in Fig. 5.6. The ground state issymmetric; the first excited state is antisymmetric.

We observe that the wave functions exist in the classically illegal middle regionE < V0, which is not surprising. That comes from the exponential of the simplesymmetric well which is cut off at a finite distance. This results in a symmetrizationor antisymmetrization of the wave functions.

Therefore, in all stationary states, the particle has the same probability to be onthe right and on the left.

So it is a two-well problem, but, because the particle has a non-vanishing proba-bility to be in the classically forbidden middle region, these two wells are coupled bythe tunnel effect. Things resemble a classical situation where one would have drilleda narrow quantum tunnel in the intermediate potential barrier, in order to allow theparticle to communicate between the two wells.

5.4 Double Well, the Ammonia Molecule 95

5.4.3 Energy Levels

The continuity of the function and of its derivative at points x = ±(b − a/2) yieldsthe conditions:

tan ka = − k

Kcoth K (b − a/2) for a symmetric solution ψS ,

tan ka = − k

Ktanh K (b − a/2) for an antisymmetric solution ψA.

This, together with the condition k2+K 2 = 2mV0/�2 gives transcendental equations

that can be solved numerically.However, in order to understand the physics of the problem in a simplemanner, we

assume the orders of magnitude are such that: V0 E , that is, K � √2mV0/�2 =

constant and KΔ 1, which is quite reasonable in the specific case of the NH3

molecule. Under such conditions, we end up with

tan ka � − k

K

(1 ± 2e−KΔ

), (5.31)

where the + sign corresponds to ψS , and the − sign to ψA. With this equation, wecan calculate the quantized values of ka. These values appear on the graph in Fig. 5.7as the positions of the intersections of the successive branches of y = tan ka withthe two straight lines y = −εAka and y = −εSka. These intersections are locatedin the vicinity of ka ∼ π. The two constants εA and εS are:

εA = 1

Ka

(1 − 2e−KΔ

), εS = 1

Ka

(1 + 2e−KΔ

). (5.32)

They are close to each other, and such that εA < εS � 1, because Ka ka ∼ π.

Fig. 5.7 a Graphical determination of the energy levels in the double well; b the two first levels arelower than the ground state of a single well centered at L or R (E0 → E ′

0), and there is a splittingby tunneling between these two levels (E ′

0 → EA and ES)

96 5 Energy Quantization

For K infinite, V0 infinite, we recover two independent wells. The particle is inone of them. The energies are En = n2π2

�2/2ma2. The order of magnitude of ES

and EA is E0 = π2�2/2ma2.

For finite K (V0 finite), the two levels are shifted downwards to E ′0, which is

intuitively understandable. The two wells communicate by quantum tunneling andthe particle “feels” an effective well broader than a. This lowering is accompaniedby a splitting in two sublevels as can be seen in Fig. 5.7. The symmetric state is moretightly bound.

In our approximation (K k, Ka 1), we obtain

kS ∼ π

a(1 + εS), kA ∼ π

a(1 + εA), (5.33)

with εS and εA � 1. Using (5.31) and (5.33), we obtain a mean energy E ′0 =

(EA + ES)/2

E ′0 � �

2π2

2ma2

(1 − 2

Ka

). (5.34)

The splitting EA − ES of the two levels by the tunnel effect is

EA − ES ≡ 2A � �2π2

2ma2

[1

(1 + εA)2− 1

(1 + εS)2

]� �

2π2

2ma28e−KΔ

Ka, (5.35)

or

A ≡ EA − ES

2� �

2π2

2ma24

Kae−KΔ. (5.36)

Because K � √2mV0/� in this approximation, the splitting tends to zero exponen-

tially when the width Δ, or the height V0, of the potential barrier increase.In fact, in all its consequences, namely the splitting EA − ES and the value of the

wave function in the middle region, the tunnel effect is proportional to e−KΔ whereΔ = 2b − a and K ∼ √

2m(V0)/�2 (or the square of this for probabilities).We also notice that A → 0 extremely rapidly in the limit � → 0. This is in contrast

with the polynomial level spacing, in �, of the harmonic oscillator or potential welllevels. Here, the splitting is exponential in −1/�.

We therefore observe a first differencewith the classical case. There are indeed twolowest lying energy states, but classically they have the same energy (E0), whereashere they are split. This qualitative feature is general, for any symmetric double wellof arbitrary shape.

5.4.4 Wave Functions

The two corresponding wave functions, which are shown in Fig. 5.6, are such that theprobability of finding the particle on the left or on the right is the same. If the particle

5.4 Double Well, the Ammonia Molecule 97

is in a well-defined energy state, it has the same probability of being on either side.It is both on the right and on the left.

That result is really contrary to classical observations. Classically, in its two lowestenergy states, the particle is either on the right or on the left. Here it is both on theright and on the left at the same time.

Is that a real difference? That’s not really convincing for the moment, because ofthe statistical interpretation. If, classically, we fix the condition that the energy beminimum, then it is natural to find half of the particles on the right and the other halfon the left. We must find something else.

Something else means doing true quantum physics, taking into account the timeparameter.

5.4.5 Inversion of the Molecule

How can we put the particle on the right- or left-hand side in quantum mechanics?We must prepare it in a state where its wave function is concentrated on the right oron the left. But we know how to do that! We just need to look at the wave functions inFig. 5.6. We just have to take the sum and the difference of these two eigenfunctions,according to the superposition principle,

ψR = (ψS + ψA)/√2 , and ψL = (ψS − ψA)/

√2. (5.37)

The resulting wave functions describe states where almost all the probability isconcentrated on one side only, on the left for ψL and on the right for ψR . (Actuallythe residual probability is of the order of ≈ e−2KΔ, which is very small.) Thesetwo states correspond to the “classical” configurations, molecule on the right andmolecule on the left (Fig. 5.8).

This observation is interesting. We can phrase it in the following way. ψR is alinear superposition of states with a well-defined energy that interferes destructively

Fig. 5.8 Classical configurations of the ammonia molecule

98 5 Energy Quantization

on the left. Similarly,ψL is a linear superposition of states with a well-defined energythat interferes destructively on the right. In this way, interferences are really simple;there is nothing more from the mathematical point of view.

We can invert these relations and say that states of well-defined energies, whichare on both sides, are interfering superpositions of “contradictory” classical states.Classically, one is either on the right or on the left. Nobody is both on the right andon the left: well nearly nobody, some people once in a while, but very few.

Here, it seems perfectly natural. But that is what seemed shocking withSchrödinger’s cat! What was inconceivable for a cat seems perfectly natural here.Notice that, at that time, we had forgotten a possible state ψ−(cat) = (ψ(live cat) −ψ(dead cat))

√2. Frankly, the second elementary operation, subtraction, is just as

noble as the first one. But mind you: the probabilities to be on either side (or deador alive) are the same although the two states are definitely not the same! Puzzling!

Now we can apply what we know on time evolution. The state ψR is not a station-ary state; the system does not have a well-defined energy: p(Es) = p(Ea) = 1/2.Therefore it must evolve in a non-trivial way with time. Its time evolution is:

ψ(x, t) = 1√2

(ψS(x) e

−i ES t/� + ψA(x) e−i EAt/�

)

= e−i ES t/�

√2

(ψS(x) + ψA(x)e

−iωt), (5.38)

where we have introduced the Bohr frequency �ω = EA − ES = 2A of the system.Now, we face something truly astonishing and unpredictable. At time T = π/ω

the particle is on the left! Its wave function is ψL , up to a phase factor,

ψ(x, T ) = e−i EST/�

√2

(ψS(x) − ψA(x)) .

This phenomenon is really contrary to classical mechanics where, if at t = 0 theparticle is at rest in one of the wells, it stays there!

In quantum mechanics, there is a permanent oscillation between the two wellsat the frequency ω = 2A/�. The wave function and the probability density “flow”periodically from one well to the other. The particle keeps on shifting from one sideto the other. The cat oscillates permanently between life and death. That is trulynon-classical.

In the specific case of NH3, this phenomenon is called the “inversion of theammonia molecule.” If, at t = 0, we prepare it in a classical configuration, it reversesitself periodically at the Bohr frequency.

The inversion frequency can be measured very accurately. In the lowest energystate, the splitting is 2A ≈ 10−4 eV, hence a frequency ν = 24 GHz, a wavelengthλ =1.25 cm, and a period τ = 4.2 10−11 s.

In fact, NH3 possesses an electric dipole moment. The center of gravity of positivecharges is different from the center of gravity of negative charges (the nitrogen atom

5.4 Double Well, the Ammonia Molecule 99

attracts electrons more strongly). This electric dipole moment D is inverted whenthe molecule reverts. This produces the emission or absorption of a radio wave atthat frequency. (Actually, at the microscopic level, a molecule can absorb or emit aphoton of energy hν ≈ 10−4 eV in a transition between the two states. It is only on amacroscopic sample that the electric dipole argument is valid.) That is a characteristicline of NH3, a fingerprint of that molecule which is used in radioastronomy to detectammonia in the interstellar medium.We come back later to an important application,the ammonia maser.

It was totally impossible to predict the order of magnitude of that frequencyqualitatively, by dimensional analysis, before we understood the mechanism of thetunnel effect and before we found the exponential (5.36).

5.5 Illustrations and Applications of the Tunnel Effect

This brings us to a series of remarks on the tunnel effect, of great importance inquantum mechanics.

Sensitivity to the Parameters

The exponential dependence of the splitting is fundamental. An exponential variesvery rapidly. This same effect, this same mechanism explains phenomena whoseorders of magnitude are incredibly different.

Consider for instance NH3 and similar molecules ND3, PH3, AsH3. This wastreated in detail by Townes and Schawlow3 who gave the form of realistic potentialsin this kind of physics. It is instructive to go from NH3 to AsH3:

NH3 : V0 = 0, 25 eV , b = 0, 4 Å : ν0 = 2, 4 1010 Hz ,

ND3 : V0 = 0, 25 md = 2mp , b = 0, 4 Å : ν0 = 1 600 Hz ,

PH3 : V0 = 0, 75 eV , b = 1 Å : ν0 = 2, 4 1010 Hz ,

AsH3 : V0 = 1, 50 eV , b = 2 Å : ν0 = 1, 6 10−8 Hz .

A change by a factor of 6 in V0 and 5 in the size b produces a spectacular decreaseof 18 orders of magnitude in the inversion frequency between NH3 and AsH3. ForAsH3, the frequency is one inversion in two years, which is not measurable. One canonly calculate it theoretically. In other words, AsH3, which seems quite similar toNH3 from a chemical point of view, behaves as a classical structure from the point ofview studied here, simply because the arsenic atom is 5 times larger than nitrogen.

There are a variety of similar situations where, by quantum tunneling, a processoccurs through a nonclassical transition across a potential barrier. The first physicalphenomenon that was explained in this way was the alpha decay of nuclei. This was

3C.H. Townes and A.L. Schawlow, Microwave Spectroscopy, Chap.12. New York: McGraw-Hill,(1955).

100 5 Energy Quantization

Fig. 5.9 Double potential well as seen by an electron when two atomsare separated by a distanceΔ

understood byGamow in 1928.Many other phenomena have the same origin, such ascatalysis, the formation of interstellar molecules on interstellar dust, nuclear fusionand fission, and so on.

Molecular Structure

Valence Electrons

Similar examples are provided by electrons in molecules.Consider, for instance, the case of two identical atoms located at a distance

Δ of each other. An external electron “sees” a double well as shown in Fig. 5.9(for simplicity, we assume the atoms are at a given distance that we can vary). Wechoose the origin of energies such that V → 0 for x → ∞. If Δ is sufficiently large,one can safely consider that V ∼ 0 halfway between the atoms.

In order to get from one atom to the other, an electron in an energy level E0 < 0must cross a potential barrier of height −E0 and width Δ. We can calculate thetypical time T it takes to get from one atom to the other.

Suppose the kinetic energy Ek of the electron in one well is of the order ofthe binding energy |E0| (for the hydrogen atom, this is exact, owing to the virialtheorem). In the exponential of the tunnel effect, we have K = √

2m|E0|/�. For anelectron bound in an atom, one has in good approximation Ka ∼ 1. The exponentialdependence of the oscillation frequency in terms of the parameter KΔ remains true.We can use (5.36) in the form A ∼ Eke−KΔ ∼ |E0|e−KΔ, up to a numerical factorof order one.

In a molecule or in a crystal, the distances of atoms are of the order of 0.1 nm. Ina gas at usual temperatures and pressures, they are roughly ten times larger (∼3 nm).The binding energies of valence electrons in an atom are of the order of a few eV.One then finds:

Molecule: Δ = 0.2 nm |E0| = 4 eV A = 1 eV T = 10−15 s ,

Gas: Δ = 3.0 nm |E0| = 4 eV A = 10−12 eV T = 10−3 s .

The time to get from one atom to the other is very small for valence electrons in amolecule or in a solid. These electrons are completely delocalized in the molecular

5.5 Illustrations and Applications of the Tunnel Effect 101

structure. Conversely, this phenomenon is completely negligible in gases. In fact,owing to thermal motion, two molecules in a gas remain at a distance of, say, 3 nmfor a length of time smaller than 10−10 s. Quantum tunneling oscillations have aperiod of 10−3 s and they cannot occur appreciably on such a small time scale.Therefore, the idea that in a gas at room temprature each electron belongs to a givenmolecule is quite acceptable.

Molecular Binding

Similarly, in this result, one can find a starting idea for the explanation of molecularbinding, which cannot be understood classically.

Consider the simplestmolecule, theH+2 ion,made of two protons and one electron.

The decrease E0 → E ′0 explains that it ismore favorable energetically for the electron

to have an equal probability to be on both protons rather that being bound to one ofthem and let the other live its own life. Classically, the moon belongs to the earth.It could belong to the planet Mars, but it has chosen the earth and it sticks to thatchoice.

However, quantummechanically, the fact that the electron belongs to both protonsat the same time stabilizes it. This effect increases as the distance between protonsdecreases. However, a compromise must be found, owing to the Coulomb repulsionof protons. An equilibrium situation results. After doing the calculations, one canprove that the H+

2 ion is bound.If, in addition, one takes into account spin and the Pauli principle, the splitting

and the fact that the symmetric state is more tightly bound ES < EA accounts for thechemical covalent bond. In other words, quantum tunneling is responsible for ourexistence!

Potential Barriers

One can find exercises of the following type. A particle of energy E hits a potentialbarrier of width Δ and height V0 > E . Classically, the particle should bounce back.What is the tunneling probability?

For an electron and with atomic orders of magnitude E = 1eV, V0 = 2eV,Δ ≈ 0,1nm, we obtain a probability of p = 80% that the particle crosses the barrierwhich is completely anticlassical.

In the same conditions, a proton mp ∼ 2000me has a probability of crossing thebarrier of p ≈ 10−19 because of the mass effect. In other words, saying that nuclei,protons and neutrons, have well-defined positions in an atom or a molecule makessense.

But at nuclear scales V0 = 2E ≈ 10 MeV and Δ ≈ 1 fm, the probability is 80%,and a proton is delocalized in a nucleus.

102 5 Energy Quantization

5.6 Tunneling Microscopy, Nanotechnologies

An important practical application is the construction by Binnig and Rohrer in the1980s of the scanning tunneling microscope (STM, 1986 Nobel prize).

A conducting tip is moved along a surface at a very short distance Δ ≈ 10Å. Apotential difference is applied, and the electrons pass from the surface to the tip bythe tunnel effect. The current is extremely sensitive to the distance (actually to theelectrostatic potential V0). That way, one can detect incredibly fine details≈0.01 nm,and one can make a map of the surface. An example is shown in Fig. 5.10.

Nanotechnologies

With the tunnel effect one gets near to science fiction by inventing nanotechnologies,that is, creating operation techniques at the nanometer scale, at distances comparableto the size of the smallest living systems, viruses.4

On September 29, 1989, D. Eigler, research engineer at IBM, was able to manip-ulate individual atoms on a metal surface. He picked them with a tip and put themon another site of the surface (the device is similar to that shown in Fig. 5.10).

Eigler first managed to write the letters IBM with 35 xenon atoms on a nickelsubstrate. One year later, he was able to construct an electronic switch whose movingpart was made of a single atom (5 nm high)!

In Fig. 5.11, one can see the progressive construction of a “coral reef” of 48 ironatoms on a nickel substrate by atom manipulation. Figure5.12 shows a side view of

Fig. 5.10 a Principle of a tunneling microscope. A thin tip is moved in the vicinity of a solidsurface with piezoelectric transductors. One adjusts the distance of the tip to the surface in such away that electric current due to tunneling between the surface and the tip is constant. This providesa mapmaking of the electron density distribution (actually the electrostatic potential) at the surfaceof the crystal. An example is shown in bwhere one can see a surface of In Sb. The Sb atoms appearto be raised. The actual size of the sample in the figure is ∼3nm (After Y. Liang et al., J. Vac. Sci.Technol. B9, 730, (1991).)

4See for instance Zooming into the nanoworld http://www.nano.geo.uni-muenchen.de/SW/images/zoom.html..

5.6 Tunneling Microscopy, Nanotechnologies 103

Fig. 5.11 Progressive construction of a “coral reef” made of 48 Fe atoms on an nickel crystallinesurface by atommanipulation. M.F. Crommie, C.P. Lutz, D.M. Eigler, E.J. Heller.Waves on a metalsurface and quantum corrals. Surface Review and Letters 2 (1), 127–137 (1995)

Fig. 5.12 Side view of theelectron “lake” inside thecoral reef of the previouspicture. The waves arestationary de Broglie waves.STM Image Gallery-IBM

104 5 Energy Quantization

the corral reef. The “waves” correspond to the surface density of electrons trappedinside the structure. In other words we see the ground state of de Broglie waves in acircular two-dimensional well directly (this is a consequence of the Pauli principle;we see the probability densities of higher excitations because the lower ones arealready occupied). We could have considered this problem in Sect. 5.3.

Classical Limit

These developments are truly impressive. However, what is even more striking per-haps is what one cannot do with the tunnel effect.

Indeed, quantum mechanics opens the possibility of fantastic dreams. One canget to another galaxy by quantum tunneling and see extraterrestrials. One can wina bike race by crossing mountains by quantum tunneling without any fear of beingdope-tested, or, more seriously, see a piece of dust cross a tulle curtain by quantumtunneling. In principle, quantum mechanics gives us the possibility to do all that.Unfortunately, the probability is very small. One can, as an exercise, calculate theprobabilities and one ends up with the most extravagantly small numbers one canimagine p ∼ exp(−10(30 to 65)). Such numbers are impossible to write in the binarysystem! In order to win the Nobel prize, it is undoubtedly much more promising tobuy a cat. The probability that a cat will type the unified theory of the Universe, orShakespeare’s works by walking at random on the keyboard of a computer withoutmaking any mistake is enormously larger: p ∼ exp(−10(4 to 6)).

All of this is meaningless for simple reasons. There exist other terms in theHamiltonian of a piece of dust that allow it to cross a curtain: someone can dothe cleaning; there can be a small hole in the curtain, and so on. But it is fascinatingto see that an effect which is so important for our existence cannot manifest itselfopenly at our scale.

5.7 Exercises

1. Uncertainty relation for the harmonic oscillator

Using the recursion relations satisfied by the Hermite functions (5.16), show that, ina state of energy En given by (5.13) one has 〈x〉 = 〈p〉 = 0. Calculate 〈x2〉 and 〈p2〉and show that the zero point energy is essential in order to preserve the uncertaintyrelations.

2. Time evolution of a one-dimensional harmonic oscillator

Consider a harmonic oscillator of Hamiltonian H = p2/2m +mω2 x2/2 and its firsttwo normalized eigenfunctions φ0(x) and φ1(x).Consider a system which at time t = 0 has the wave function:

ψ(x, t = 0) = cos θ φ0(x) + sin θ φ1(x) with 0 ≤ θ < π .

5.7 Exercises 105

a. What is the wave function ψ(x, t) at time t?b. Calculate the expectation values 〈E〉, 〈E2〉 and ΔE2 = 〈E2〉 − 〈E〉2. Explain

their time-dependence.c. Calculate the time evolution of 〈x〉, 〈x2〉 and Δx .

3. Three-dimensional harmonic oscillator

Consider in three dimensions a particle ofmassm and theHamiltonian H = p2/2m+mω2 r2/2 where r2 = x2 + y2 + z2.

a. What are the energy levels and their degeneracies?b. How do these results change in the case of an anisotropic potential:

V = m(ω21x

2 + ω22 y

2 + ω23z

2)/2 ?

4. One-dimensional infinite potential well

Consider an infinite potential well of width a: V (x) = 0 for 0 < x < a and V = ∞otherwise.

a. Show that in the energy eigenstate ψn(x), one has 〈x〉 = a/2 and Δx2 = a2(1−6/n2π2)/12.

b. Consider the wave-function ψ(x) = Ax(a − x).

(i) What is the probability pn to find the particle in the n-th excited state?(ii) From this set of probabilities, calculate the expectation values 〈E〉 and 〈E2〉

for that wave function.

We recall that∑∞

k=0(2k+1)−2n = π2/8 for n = 1, π4/96 for n = 2, and π6/960for n = 3.

c. Check that if one applies blindly the correspondence principle, i.e. if one usesH 2 = (�2/2m)2d4/dx4 in the definition of 〈E2〉, one obtains the absurd resultΔE2 < 0. What is the reason for this?

5. Isotropic states of the hydrogen atom

The energy levels of spherically symmetric states of the hydrogen atom can beobtained in the following one-dimensional calculation. Consider an electron of massm in the potential V (x) such that V = ∞ if x ≤ 0 and V = −A/x if x > 0, whereA = q2/4πε0, and q is the elementary charge. We set α = q2/(4πε0�c) � 1/137(dimensionless constant) where c is the velocity of light.

a. Show that the wave function ψ(x) = C x e−x/a for x ≥ 0 and ψ(x) = 0 forx < 0, is an eigenfunction of the Hamiltonian with energy E for a given valueof a. Express E and a in terms of m, α, � and c.

b. Calculate the numerical values of E and a. One can use mc2 = 5.11 105 eV and�c = 197 eV nm.

106 5 Energy Quantization

c. Determine the normalization constant C in terms of a.d. Calculate the expectation value of 1/x in the state |ψ〉 and deduce from that the

expectation value of the kinetic energy.What is the relation, valid also in classicalmechanics, between these two quantities?

6. δ-function potentials

a. Consider a particle of mass m in the one-dimensional potential V (x) = αδ(x),α < 0. We are interested in bound states (E < 0).

(i) Assuming that the wave function ψ(x) is continuous at x = 0 (which can beproven), find the relation between the discontinuity of its derivative and ψ(0)by integrating the Schrödinger equation between x = −ε and x = +ε.

(ii) How many bound states are there? With what energies? We set K =√−2mE/� and λ0 = −�2/mα.

b. Consider the double δ-function potential:

V (x) = α (δ(x + d/2) + δ(x − d/2)) .

(i) Write the general formof bound statewave functions.What is the quantizationcondition?

(ii) Discuss the number of bound states as a function of the distance d betweenthe two wells.

5.8 Problem. The Ramsauer Effect

In 1921, Ramsauer noticed that for some particular values of the incident energy,rare gases such as helium, argon or neon were transparent to low-energy electronbeams. This can be explained in the following one-dimensional model. Consider astationary solution of the Schrödinger equation of positive energy E , for a particleof mass m in the following one-dimensional potential (V0 > 0):

V (x) = 0 for |x | > a , V (x) = −V0 for |x | ≤ a.

We set q2 = 2m(V0 + E)/�2, k2 = 2mE/�

2 and we are interested in a solutionof the form

ψ(x) = eikx + A e−ikx x ≤ −a,

ψ(x) = B eiqx + C e−iqx − a < x ≤ a,

ψ(x) = D eikx x > a .

5.8 Problem. The Ramsauer Effect 107

1. Write the continuity relations at x = −a and x = a.2. Setting Δ = (q + k)2 − e4iqa(q − k)2, calculate the transmission probability

T = |D|2. Calculate the reflection probability R = |A|2. Check that R + T = 1.3. Show that T = 1 for some values of the energy. Interpret this result and the

Ramsauer effect.4. In Helium, the lowest energy at which the phenomenon occurs is E = 0.7 eV.

Assuming that the radius of the atom is a = 0.1 nm, calculate the depth V0 of thepotential well inside the atom in this model.

5. How does the reflection coefficient behave as the energy E tends to zero? Whenone sends very slow hydrogen atoms on a liquid helium surface, these atomsbounce back elastically instead of being adsorbed. Explain this phenomenonqualitatively.

5.8.1 Solution

1. The continuity equations are at x = −a:

e−ika + Ae+ika = Be−iqa + Ceiqa

ik(e−ika − Aeika) = iq(Be−iqa − Ceiqa),

and at x = +a:

Beiqa + Ce−iqa = Deika and iq(Beiqa − Ce−iqa) = ik Deika .

2. Setting Δ = (q + k)2 − e4iqa(q − k)2, one obtains:

D = 4kq

Δe−2i(k−q)a A = (k2 − q2)

Δe−2ika (1 − e4iqa) .

We have |Δ|2 = 16k2q2 + 4(k2 − q2)2 sin2 2qa and:

R = |A|2 = 4(k2 − q2)2

|Δ|2 sin2 2qa T = |D|2 = 16k2q2

|Δ|2

where R + T = 1.3. For all values of q such that sin 2qa = 0, i.e. qa = nπ/2, the transmission

probability is equal to 1, and there is no reflection, T = 1 , R = 0. Thishappens when the size of the well 2a is a multiple of λ/2, where λ = 2π/q is thede Broglie wavelength of the particle inside the potential well. All the reflectedwaves interfere destructively and the well becomes transparent to the incidentwave (more precisely, the wave reflected in x = −a, which does not enter the

108 5 Energy Quantization

well, interferes destructively in the backward direction with the sum of all thewaves undergoing multiple reflections in the well).

4. The corresponding energies are:

En = n2π2�2

8ma2− V0,

Choosing n = 1 and E = 0.7 eV, we obtain V0 = 9.4 − 0.7 = 8.7 eV.5. When E tends to 0, k also tends to 0 and the transmission probability vanishes. The

incident particle is reflected by the potential well. The sticking of the hydrogenatoms on the liquid helium surface occurs when the hydrogen atoms enters thepotential well at the vicinity of the surface. In this well, the hydrogen atom mayloose energy via the emission of a wave propagating at the surface of the liquid(ripplon): after such a process, the energy of the hydrogen atom is too low to exitthe well, and the atom is trapped at the surface of the liquid. At very low incidentenergy, incoming hydrogen atoms have a vanishing probability to enter the well,and hence the absorption probability tends to zero.

5.9 Problem. Colored Centers in Ionic Cristals

Consider the diatomic crystal NaCl. It is called an ionic crystal because, when thecrystal forms, the outer electron of a sodium atom is transfered to a chlorine atom.Hence, in the crystal, the electronic configuration is (Na+, Cl−), and the electrostaticinteraction between the Na+ and Cl− ions is responsible for the binding of thestructure. The crystal is face centered cubic for both ions. Schematically, the cristal,as seen parallel to one face of the cube can be represented as in Fig. 5.13. Such a

Fig. 5.13 Structure of the ionic Na+ Cl−

5.9 Problem. Colored Centers in Ionic Cristals 109

Fig. 5.14 Structure of anFcenter in a NaCl crystal

structure, called the NaCl structure, is very frequently encountered. It is in particularthe structure of all alkali halides.

These crystals are transparent if they are sufficiently pure. However, if they areirradiated by energetic photons (X or γ rays) alkali halides become coloured. Thereason for this is the following. A photon can eject an anion from its site, creatingan unoccupied site called a vacancy. This anion vacancy, surrounded by positivelycharged ions, can trap an electron and restore the local electrical neutrality of thecrystal. The trapped electron has a series of energy levels. It can absorb light andjump from the ground state to an excited state. This process is responsible for thecolour of the crystal. The electron traped in the vacancy is called a coloured centre,or F center (from the german Farbezentrum). The structure of an Fcenter is shownon Fig. 5.14.

The Mollwo-Ivey Law

Let a be the lattice spacing, i.e. the distance between two neighbouring ions Na+ andCl−. Measurements of the wavelengths λ or energies ε of absorption lines on variousalkali halides have been performed by Mollwo and Ivey. The results are displayedon Fig. 5.15. They show that the variation of the absorption energy with the latticespacing a follows a simple law.

1. Express the empirical law that emerges from these measurements as

ε = K an (5.39)

where ε is in eV and a is in Å. This is called the Mollwo-Ivey law.Since, in good approximation, the absorption energy ε depends only on the latticespacing a and not on its particular nature, one may assume that the shapes ofFcenters are the same for all of these crystals and that they only differ by theirsizes.The simplest model one can build consists in assuming that the Z positive ionsnearest neighbours to the Fcenter form a cubic square well potential inside which

110 5 Energy Quantization

Fig. 5.15 Energy of theabsorption peaks of variousalkali halides, versus thelattice spacing a

the electron is trapped. In first approximation, we shall assume it is an infinitelydeep potential well:

V = 0 for 0 < x < a, 0 < y < a, and 0 < z < aV = ∞ for x < 0, y < 0, z < 0, or x > a, y > a, z > a.

2. What is the number Z of positive ions nearest neighbours to an F center?3. Give the energy levels E1 and E2 of the ground state and of the first excited state in

the potential well, and the corresponding wave functions. What is the degeneracyof each level?

4. Assuming that the absorption of light is due to the transition of the electrons fromE1 to E2, express the absorption energy in terms of the lattice spacing a. Sincethis model leads to an expression of the form (5.39), compare the experimentaland the theoretical values of the exponent n and of the constant K .

5. Clearly, the previous simple model accounts quite successfully for the exponentn but not for the constant K .In order to cure this defect, we remark that the size of the square well is ratherarbitrary. By introducing an effective size a0 = αa, choose α in order to fit theexperimental date. Give a brief physical comment on the effective size a0 of thewell. Plot the theoretical curve on Fig. 5.15.

The Jahn–Teller Effect

When a state of a non linear molecule is degenerate, one can show that a distortionof the molecule lifts the degeneracy and stabilizes the molecule. This general effectis called the Jahn–Teller effect.5

Here, the Fcenter and the surrounding ions can be considered as a pseudomoleculewhich can undergo a Jahn–Teller distortion, as we shall now see.

5H.A.Jahn and E.Teller, Proc.Royal Soc. A161, 220, 1937.

5.9 Problem. Colored Centers in Ionic Cristals 111

Fig. 5.16 Distortion of anF center

1. Let us distort the potential well of the vacancy into a parallelepiped as shown onFig. 5.16. The lengths along the x and y axes are equal, we note them c, and thelength along the z axis is b. It is reasonable to assume that, owing to the rigidity ofthe crystal, this distortion occurs at constant volume, i.e. a30 = bc2. The distortionis characterized by the parameter η = b/c.Show that this distortion lifts the degeneracy of the excited level E2.Calculate the dependence of the excited levels on the parameter η. Show that, forone of the excited states (specify which one), the energy has a minimum E0

2 fora certain value η0 of the distortion. Is the Fcenter stretched along the z axis orflattened against the (x, y) plane?

2. Calculate the variation with respect to η of the ground state energy E1. Calculatethe value E0

1 = E1(η0).3. Plot the variations of E1 and E2 as a function of η.

The Stokes Shift

We can now give a simple account of the absorption and emission of light by anFcenter. In section 1, we have described the absorption of light by an Fcenter. Aftera time of the order of 10−6 s, the electron makes a transition to the ground state andemits radiation, called “luminescence”.

Experiment shows that the emission lines are systematically shifted towards longerwavelengths—or equivalently smaller energies—than the corresponding absorptionlines. This shift, an example of which is shown on Fig. 5.17 is called the Stokes shift.

1. Let us first assume that most lines are shifted to the infrared part of the spectrum,which is not visible. Under this assumption, by what simple mechanism do theFcenters colour a crystal when the crystal is placed in visible light?

2. What are, respectively, the colours of the crystals KI, KCl, and NaCl after theyhave been exposed to X-rays ?

We recall that the colours of the spectrum of visible light are, for increasing val-ues of the energy, red (from 1.65 to 2.0 eV), orange (from 2.0 to 2.1 eV), yellow(from 2.1 to 2.3 eV), green (from 2.3 to 2.55 eV), blue (from 2.55 to 2.65 eV)and violet (from 2.65 to 3.1 eV).

112 5 Energy Quantization

Fig. 5.17 Absorption andemission spectra of anFcenter in KBr at lowtemperature. The maxima ofthe two lines are respectivelyat 2.06 and 0.92 eV. Data aretaken from W. Gebbart andA. Kuhnert, Physica StatusSolidi, Vol. 14, p. 157 (1966)

We also recall that “complementary colours” are colours which, when associated,give back white light. The main couples of complementary colours are yellow-violet, red-green and blue-orange. Hence when blue is absorbed by a substancein natural white light, the substance appears to be yellow.

3. We shall now attempt to give a simple description of the Stokes shift. We shallassume that the elecronic excitation or de-excitation times are negligible com-pared to typical times for local distortions of the crystal, these being, in turn,much shorter than the lifetimes of the excited states (of the order of 10−6 s).Under these assumptions, give a simple description of the absorption and emissionof light by an Fcenter, using the results of Sect. 2.

4. More quantitatively, show that the results of section 2 give a good account of theexperimental result shown on Fig. 5.17.

5. Justify the assumption made in question3.1 by showing that for most crystals ofFig. 5.15, the emission line is in the infrared part of the spectrum. Specify forwhich crystals this occurs.

5.9.1 Solution

Section1

1. The experimental points lie on a straight line in a log-log plot. The experimentallaw is of the form ε = Kan with K � 68 and n � −1.85.

2. There are Z = 6 positive ions at a distance a/2 of the Fcenter.3. Choosing the origin at a vertex of the cube,

(a) the ground state, with energy E1 = 3�2π2/(2ma2), is not degenerate; itswave function is

ψ = (2/a)3/2 sin(πx/a) sin(πy/a) sin(πz/a);

5.9 Problem. Colored Centers in Ionic Cristals 113

Fig. 5.18 Absorption linesof Fcenters in various alkalihalides; comparison of thedata and the modeldeveloped in question 1.5

(b) the first excited state has a three-fold degeneracy ψ2x , ψ2y , ψ2z with, forinstance,

ψ2z = (2/a)3/2 sin(πx/a) sin(πy/a) sin(2πz/a)

corresponding to an energy E2 = 6�2π2/(2ma2).

4. The transition E1 → E2 corresponds to the absorption of an energy ε = E2 −E1 = 3�

2π2/(2ma2) where a is the lattice spacing. This expression is of thetype (5.39) with K = 112 and n = −2. The value of n is close to what isexperimentally observed (−1.85). The constant K is quite overestimated.

5. If the effective extension of the potential is a0 = αa, the theoretical formulebecomes ε = 3�

2π2/(2ma2α2). Using the value α = 1.13 corresponding toK = 87 (and n = −2), one obtains a good fit to the data as shown on Fig. 5.18.

The effective size of the cube is 13% greater than the lattice size. This is notsurprising since the six neighbouring positive ions each attract the electron of theFcenter. In a more realistic potential model of the Fcenter, the probability for theelectron to be outside the vacancy should be non-zero.

Section2

1. Consider the state

ψ2z = (2/a0)3/2 sin(πx/a0) sin(πy/a0) sin(2πz/a0) .

Under the distortion, it becomes:

ψ′2z = (2/c)(2/b)1/2 sin(πx/c) sin(πy/c) sin(2πz/b) ,

114 5 Energy Quantization

and the corresponding energy E2 = 6π2�2/(2ma20) becomes

E ′2z = �

2π2

2m(2

c2+ 4

b2) .

Setting η = b/c, and imposing that the distortion occurs at constant volume,a30 = c2b, one has c = a0η−1/3 and b = a0η2/3, hence

E ′2z = �

2π2

2ma20(2η2/3 + 4η−4/3) .

Similarly, one finds that

E ′2x = E ′

2y = �2π2

2ma20(5η2/3 + η−4/3) .

Clearly, E ′2x = E ′

2y on one hand, and E ′2z on the other are different from E2, and

different from one another. The distortion lifts partially the degeneracy.If we study the variation of E ′

2z and E ′2x with respect to η, we find that both

energies have minimum values:

• E ′2z is minimum for η = 2, where it reaches the value

E ′2z(2) = 4.76

�2π2

2ma20;

• E ′2x is minimum for η = √

(2/5) � 0.63, where it reaches the value

E ′2x (

√(2/5)) = 5.52

�2π2

2ma20.

Both values are smaller than E2. The first is the absolute minimum. Hence theenergy of the first excited state has the minimum value E0

2 = 4.76 (�2π2)/

(2ma20), for a value η = 2 of the distortion parameter. Since η > 1 this cor-responds to an Fcenter stretched along the z axis.

2. When the Fcenter is distorted, the ground state energy is

E ′1 = �

2π2

2m(2

c2+ 1

b2) = �

2π2

2ma20(2η2/3 + η−4/3) .

This function is minimum for η = 1, i.e. an undistorted center. Any distortionwill increase the energy of the ground state. We have, in particular at η = 2 wherethe excited state energy is minimum, E ′

1(η0) = E ′1(2) = 3.57 �

2π2/(2ma20) .

3. The variations of the energy levels with the distortion are shown on Fig. 5.19.

5.9 Problem. Colored Centers in Ionic Cristals 115

Fig. 5.19 Variation of theenergy levels, in units�2π2/2ma20 , with the

distortion parameter η

Section3

1. If the emission of light is in the infrared part of the spectrum, it will not produce acolouring of the crystal. The colour is only due to absorption. The observed colouris the complementary colour to that of the absorbed radiation (of energy ε).

2. Among the crystalsmentioned, NaCl absorbs violet light (ε � 2.75 eV), its colouris therefore yellow. Similarly, KI is green, and KCl is violet.The first time this problemwas given to students, it was actully accompanied withthree plastic bags containing respectively yellow, green and light violet crystals.The question was to determine what type of alkali halide was contained in eachof them. The crystals had been irradiated overnight in a Van de Graaf accelerator.

3. If an Fcenter distorts itself after absorbing energy, i.e. when it is in the excitedstate, its energy will decrease down to E0

2 . If it de-excites, it will emit a photon ofsmaller energy than the energy of the absorbed photon. Hence the Stokes shift.Using the Franck–Condon principle, one may represent the successive steps ofthe absorption-emission process as follows (Fig. 5.20).

• 0: Fcenter in its ground state;• 1: Absorption of a photon of energy ε = E2 − E1, instantaneous transition tothe degenerate state ψ2;

• 2: Distortion of the Fcenter. The electronic energy decreases down to E02 .

The corresponding energy difference E2 − E02 is transfered to the thermal

vibrations (phonons) of the crystal.

Fig. 5.20 Schematic timedescription of theabsorption-emission processby an Fcenter

116 5 Energy Quantization

• 3: De-excitation. This process is instantaneous, and occurs on a distortedcenter. The emitted photon has energy E0

2 − E ′1(η0).• 4: The Fcenter recovers its original symmetry. The corresponding energy

E ′1(η0) − E1, is again released in the crystal thermal vibrations.

4. From the previous considerations, the energy of the emission line is, within ourmodel,

ε′ = E02 − E ′

1(η0) = �2π2

2ma20(4.76 − 3.57) = 1.19

�2π2

2ma20.

This emission energy is smaller than the absorption energy; the ratio is ε′/ε ∼ 0.4.The experimental result for KBr is ∼0.44 (see Fig. 5.17). The agreement of themodel with experiment is quite acceptable.

5. The ratio ε′/ε calculated above does not depend on a0, and therefore should notdepend on the nature of the cristal. For an absorption energy near the upper part ofthe visible spectrum, i.e. ∼3.1 eV, the calculated emission energy is of the orderof 3.1 × 0.4 = 1.14 eV which lies in the infrared region. We therefore concludethat if the absorbed light is in the visible part of the spectrum, (crystals rangingfrom RbI to KF on Fig. 5.15) then the emission lines lie outside of the visiblespectrum. We assumed that in Sect. 3.1.

Further Comments on Fcenters

1. The mechanism by which the Fcenters form is yet unclear. There are severalproposals (the most plausible beeing due to Pooley) which are based on theassumption that the X-ray photons can ionize the anions once (A− → A) ortwice (A → A+). The resulting species, either electrically neutral or positivelycharged, is then in a very unstable situation in the middle of all the positive ions.It is then ejected from its site, leaving behind a vacancy (Fcenter) and reachingan interstitial position.The colour can also be obtained by adding impurities (such as a few Ca++ ionsin NaCl) to the crystal. This the reason why many minerals with a marked ioniccharacter are found coloured in nature, while they are transparent if they are pure(like quartz). They were contaminated by other ions when they crystallized.

2. The model of Sect. 1 accounts for the Mollwo-Ivey law quite reasonably. It is, ofcourse, very simplistic. The actual potential is by no means infinitely deep. Byelectron spin resonance experiments, one can show that the wave function extendsup to the eighth ionic shell surrounding the Fcenter, i.e. much further than a0/2.

3. The Fcenters can move around. A nearby anion can jump in the F center, whichtherefore moves in the reverse direction. This process involves the crossing of apotential barrier, and is favored by an increase in temperature. The mobility of anFcenter increases with the temperature.

5.9 Problem. Colored Centers in Ionic Cristals 117

Owing to this mobility, the Fcenters tend to disappear, for instance when theyreach the surface of the crystal. One can see the colour disappear if the crystalsare heated.The colour can also disappear progressively if the crystals are exposed to naturallight. In fact, the Fcenters can then be ionized by ultra-violet photons, which caneject the electron from its vacancy.

Chapter 6Principles of Quantum Mechanics

During the years 1925–1927, quantum mechanics took shape and accumulated suc-cesses. But three persons, and not the least, addressed the question of its structure. InZürich there was Erwin Schrödinger, in Göttingen David Hilbert, and in CambridgePaul Adrien Maurice Dirac (Fig. 6.1).

Hilbert was 65; he was considered the greatest living mathematician since thedeath of Poincaré. Schrödinger was 40. Dirac was a young 23-year-old student inCambridge, simply brilliant.

Their thoughts, which we can follow, generated the basic principles of quantummechanics. Here, we do the following.

• First we write a more concise and general formulation of what we have done byusing the formalism of Hilbert space and the notations invented by Dirac.

Fig. 6.1 Schrödinger, Hilbert, and Dirac at the end of the 1920s (All rights reserved)

© Springer International Publishing Switzerland 2016J.-L. Basdevant, Lectures on Quantum Mechanics,Graduate Texts in Physics, DOI 10.1007/978-3-319-43479-7_6

119

120 6 Principles of Quantum Mechanics

• Then, we prove in a simple manner some results we have already guessed onobservables, in order to be able to state the general principles of quantummechan-ics, valid for any system.

• This allows us to discover at lastHeisenberg’smatrices and the “matrixmechanics”that had been elaborated by Heisenberg, Born, Jordan, and Pauli between 1924and 1925.

• In the course of this, we understand quite simply how Schrödinger and Diracshowed, in 1926, the equivalence of the two approaches: wave mechanics andmatrix mechanics.

• Finally, we illustrate these principles on a quantum phenomenon that is directlyvisible and familiar. It is the only one: the polarization of light.

From the point of view of Hilbert, Schrödinger, and Dirac, what were the prob-lems? First there were two versions of quantum mechanics and one had to imposesome order.

But even without that, our theory is nice and appealing, but it is not very pleasantesthetically. First, it is restricted to the motion of a particle in space, and it must begeneralized. Also, it is somewhat ambiguous. In wave mechanics, one describes thestate of a particle in space at time t by a wave function ψ(r, t). However, that thedescription is not unique. The Fourier transform of the wave function ϕ(p, t) is acompletely equivalent description of the state of the particle. We can perfectly wellstate our principles using ϕ(p, t). For instance, the position observable x is then

x = i�∂

∂ px.

To get a clear insight of this consists of doing theoretical physics. We now speakabout mathematics. We do not want to do mathematics in the sense of taking careof rigor and convergence questions, but we want to use mathematics and to see howthis allows to understand the physics better.

Actually, the only difficulty is the language. After one has become familiar withthe language, things becomemuch simpler; life is easier. But learning a new languageis always difficult at the beginning.

6.1 Hilbert Space

Consider two wave functions and their Fourier transforms. Because of Plancherel’stheorem, the two following integrals are equal,

∫ψ∗1(r, t)ψ2(r, t) d3r =

∫ϕ∗1( p, t)ϕ2( p, t) d3 p. (6.1)

Whenmathematicians see such properties, they understand the underlying structures.Indeed such integrals can be viewed as scalar products. The extraordinary idea ofpeople such as Banach, Hilbert, and Fréchet was to consider functions as vectors or

6.1 Hilbert Space 121

points in vector spaces, and to use a geometric language in order to solve problemsof analysis.

What happens here is quite similar to what happens in ordinary geometry. Agiven vector can be represented by different sets of coordinates in different referencesystems. But the lengths, the angles, namely scalar products, are independent of thespecific reference system. We show that there are, in fact, many representations ofthe state of a system. Each one has its own advantages.

Two-Dimensional Space

Before entering the core of the subject, we can recall some simple notions aboutHermitian spaces, that is, complex vector spaces of finite dimension. These notionsare useful in what follows.

In two dimensions, for simplicity, one can represent a vector u by the columnmatrix of its components. The conjugate vector is the line matrix u where we trans-pose and take the complex conjugate of each coordinate. We denote 〈v|u〉 as theHermitian scalar product of u and v:

u =(u1u2

), u = (

u∗1, u

∗2

), 〈v|u〉 = v∗

1u1 + v∗2u2. (6.2)

This scalar product is positive definite and the norm ‖u‖ of a vector is defined by

‖u‖2 = 〈u|u〉. (6.3)

The Hermitian conjugate M† of a matrix M is obtained by transposing and takingcomplex conjugates of numbers M†

i j = (Mji )∗.

A matrix is said to be Hermitian if it is equal to its Hermitian conjugate M = M†.The eigenvalues of a Hermitian matrix are real. The corresponding eigenvectors,normalized to one, form an orthonormal basis, or a Hermitian basis of the space.

Square Integrable Functions

Let us now consider square integrable functions that are of interest in quantummechanics.A complex function f (x)of a real variable x is said to be square integrableif it satisfies ∫ ∞

−∞| f (x)|2 dx < ∞. (6.4)

In mathematics, one calls this set of functions L2(R), so we write f ∈ L2(R).Square integrable functions form a complex vector space (any linear combination

of square integrable functions is square integrable). The extension to three dimensions(or three variables) g(x, y, z) is straightforward; the corresponding space is denotedL2(R3).

At this point, we consider the results of Charles Hermite in 1860. In fact, consid-ering complex functions, Hermite defined anHermitian scalar product of two square

122 6 Principles of Quantum Mechanics

integrable functions f and g by

〈g| f 〉 =∫

g∗(x) f (x) dx . (6.5)

This is linear in f and antilinear in g, and it possesses the Hermitian symmetry

〈g| f 〉 = 〈 f |g〉∗. (6.6)

This allows us to define the norm ‖ f ‖ of the function f by

‖ f ‖2 =∫

| f (x)|2 dx . (6.7)

Algebraically, it is exactly the same as in finite-dimensional spaces considered above.It is convergence, that is, topological properties, that is different.

Now, Hermite made a very remarkable discovery. Without being aware of that,he studied the quantum harmonic oscillator eigenvalue problem

(x2 − d2

dx2)ϕn(x) = εnϕn(x)

or

hϕn(x) = εnϕn(x) with h = (x2 − d2

dx2), (6.8)

and he found all the square integrable solutions {ϕn(x), εn},

ϕn(x) = γn ex2/2 dn

dxn

(e−x2

), εn = 2n + 1, n integer ≥ 0. (6.9)

These functions are normalized to one (‖ϕn‖ = 1) if

γn = π−1/4 2−n/2 (n!)−1/2. (6.10)

The Hermite functions ϕn(x) are orthonormal (i.e., orthogonal and normalized toone) as one can check; they form a free orthonormal set.

But Hermite found a most remarkable property. All square integrable functionscan be expanded on the set of Hermite functions,

∀ f ∈ L2(R) , f (x) =∞∑

n=0

Cnϕn(x), (6.11)

where the components Cn of f are

Cn = 〈ϕn| f 〉, (6.12)

6.1 Hilbert Space 123

because the ϕn(x) are orthogonal and normalized. In other words the Hermite func-tions are a complete set, called a Hilbert basis of L2(R).

That is the great discovery! Square integrable functions form a Hilbert spacewhich has the three properties of being a complex vector space, where a Hermitianpositive definite scalar product is defined, and that contains Hilbert bases (the ϕn(x)are an example of such a basis for the space of square integrable functions in onevariable L2(R)).

This space is infinite-dimensional. For mathematicians it is more interesting tostudy than a two-dimensional space, but psychologically, for us it is essentiallysimilar. We are not concerned with topological properties (even though we mentionsome of them, and they play an important role inmore elaborate quantummechanicalproblems). The algebraic rules are the same as in finite-dimensional spaces.

Therefore, in a Hilbert basis, f is entirely determined by the set of its components

f (x) ⇐⇒ {Cn}.

Actually, one can “forget” about the elements of the basisϕn(x)which are simplykept in a catalogue, andwork directly with the components {Cn} that define the vectorf ∈ L2(R).For instance, consider a function g, whose expansion is

g(x) =∞∑

n=0

Bnϕn(x). (6.13)

The scalar product 〈g| f 〉 is expressed simply in terms of the components of f andg, Cn = 〈ϕn| f 〉 and Bn = 〈ϕn|g〉, as

〈g| f 〉 =∑

B∗nCn, (6.14)

and the norm of f is given by

‖ f ‖2 =∑

|Cn|2, (6.15)

which is simply the extension of the Pythagorean theorem. (In infinite dimensions,mathematicians call that the Bessel–Parseval theorem.) If f is normalized to 1, wehave ∑

‖Cn‖2 = 1.

One thing that one does not fully appreciate at first is that the geometrical proper-ties of a Hilbert space are very similar to those of a Euclidian space: the Pythagoreantheorem and the triangle inequality hold in both.

The result of the application of h (Eq. (6.8)) to f is

124 6 Principles of Quantum Mechanics

h f (x) =∑

Cnεnϕn(x). (6.16)

Hence the scalar product of h f and f (or, more generally, any function g) is

〈 f |h f 〉 =∑

εn|Cn|2. (6.17)

We can therefore use a geometrical language in order to speak about problems ofanalysis.

6.2 Dirac Formalism

Coming back to quantum mechanics, Hilbert and Dirac understood in 1927 that thewave function ψ and its Fourier transform ϕ are simply two representations of thesame unique mathematical object, a vector in Hilbert space. And they were able togive a clear formulation of the theory with these ideas. Dirac invented notations thathave been adopted by mathematicians.

From now on, we say that the state of a system is described a any time t by a statevector

|Ψ (t)〉, which belongs to a Hilbert space EH , (6.18)

and which Dirac calls a “ket”. The functions ψ and ϕ are simply particular represen-tations of this vector. For a particle in three-dimensional space R3, the Hilbert spaceis the space of square integrable functions in three variables L2(R3), but this can begeneralized to any system.

6.2.1 Notations

Vectors

1. The vectors are denoted by the symbol: |name〉.In (6.18) we have indicated the presence of the time variable. Of course, thereexist fixed vectors, such as the elements of a basis that can be denoted |ϕn〉 or |n〉.

2. The Hermitian scalar product of ψ1 and ψ2 is denoted

〈ψ2|ψ1〉 = 〈ψ1|ψ2〉∗. (6.19)

It is not commutative, it has Hermitian symmetry, and it is linear on the right andantilinear on the left.

3. This scalar product is positive definite which allows us to define the norm ‖ψ‖ ofa vector by

6.2 Dirac Formalism 125

‖ψ‖2 = 〈ψ|ψ〉. (6.20)

The norm of a vector ψ vanishes if and only if ψ is the null vector. The norm ofa state vector is always equal to one.

4. At the beginning, it is useful to have a little dictionary, where one can see thatDirac’s notations are obviously quicker to use (Fig. 6.2).

5. The elements of the dual space are denoted

〈ψ(t)| ∈ E∗H (6.21)

and called bras (this is justified owing to a theorem of F. Riesz). This notationcomes from the scalar product. A rule in Dirac’s formalism is the contraction ofproducts. When a bra 〈u| is on the left of a ket |v〉 the expression contracts in a“bracket”, that is, the number λ = 〈u|v〉. In other words, the bra “eats” the ketand gives this number.We encounter other examples of this contraction rule, which comes from thetensor structure of quantum mechanics.

6. Hilbert basisA Hilbert basis is a free, orthonormal and complete set of vectors

{|n〉} : 〈n|m〉 = δnm .

Fig. 6.2 Dirac versus wavefunctions dictionary

Dictionary

Dirac formalism Wave functions

|ϕ〉 ϕ(r)

|ψ(t)〉 ψ(r, t)

〈ψ2|ψ1〉∫

ψ∗2(r)ψ1(r) d3r

‖ψ‖2 = 〈ψ|ψ〉 ∫ |ψ(r)|2 d3r

〈ψ2|A|ψ1〉∫

ψ∗2(r)Aψ1(r) d3r

〈a〉 = 〈ψ|A|ψ〉 ∫ψ∗(r)Aψ(r) d3r

126 6 Principles of Quantum Mechanics

Any vector |ψ〉 can be expanded on this basis as

|ψ〉 =∑

n

Cn|n〉, with Cn = 〈n|ψ〉. (6.22)

Cn is the component of |ψ〉 along |n〉. All Hilbert spaces possess Hilbert bases.

6.2.2 Operators

Consider now linear operators, that is, linear mappings of the space onto itself. Wewill keep the same notation A as before:

|χ〉 = A|ψ〉 , (ψ ,χ) ∈ EH . (6.23)

1. We are interested in the following numbers which are the scalar products of(6.23) with some other vector:

〈ψ2|χ1〉 = 〈ψ2|( A|ψ1〉). (6.24)

One can show that this product is associative, that is, one can define the action ofA in the dual space

〈ψ2|χ1〉 = 〈ψ2|( A|ψ1〉) = (〈ψ2| A)|ψ1〉 = 〈ψ2| A|ψ1〉. (6.25)

We call this scalar product the matrix element of A between ψ1 and ψ2.This is the same as in finite-dimensional spaces. In ordinary matrix calculus, thisexpression would be of the type vMu, where M is a matrix.In order to be rigorous, mathematicians have another way of writing: 〈ψ2| Aψ1〉,which does not show the useful associativity. Of course, they are right to do soin their work. One cannot bypass rigor. In infinite-dimensional spaces there existoperators whose “domain” (i.e., the set of functions on which they act safelyand produce square integrable results) is not the entire space. For instance, ifwe multiply a square integrable function by x , the result is not always squareintegrable. Here, we do not worry about such a question.Again, the basic theme in all this is that we act as if we were in finite-dimensionalspaces. To a large extent the topological questions can be put aside for whatconcerns us at the present level.

2. Adjoint operators. The adjoint operator A† of an operator A can be defined bythe relation

〈ψ2| A|ψ1〉 = 〈ψ1| A|ψ2〉∗ , ∀ψ1,ψ2 ∈ EH . (6.26)

6.2 Dirac Formalism 127

It is the same definition as for matrices: we transpose and take the complexconjugate.

3. Self-adjoint operators. An operator A is said to be self-adjoint or Hermitian ifA = A†, or, equivalently

〈ψ| A|ψ〉 is real for all vectors ψ of EH . (6.27)

But we already know this expression. If we look at that above dictionary, (6.27)is the expectation value of the quantity A.Nowwe remark that the results of measurements, in particular expectation values,are real numbers. Therefore, observables are Hermitian operators A = A† as weannounced in (4.6).

Theorem 2 Observables are Hermitian operators.

4. Commutation of observables.We can see easily with the correspondence principle that observables do notcommute in general. The product of A and B is not the same as the product of Band A. One defines the commutator [ A, B] of two operators as

[ A, B] ≡ A B − B A. (6.28)

One can check the fundamental commutation relation

[x, px ] = i� I , (6.29)

where I is the identity operator.We show later on that commutation relations between observables play a funda-mental role. They allow us, in particular, to derive uncertainty relations for anycouple of physical quantities.

6.2.3 Syntax Rules

Before we come back to observables and measurement results, we make two obser-vations on syntax rules in Dirac’s formalism.

1. Contraction of products. In this formalism, expressions are products of terms.Such products can be contracted. We have seen this for the scalar products; it istrue for any expression (as we have said, this comes from the tensor structure ofthe theory).With this rule, we discover a special kind of operators of the form

|u〉〈v|. (6.30)

128 6 Principles of Quantum Mechanics

This object is a linear operator because if we apply it to a ket |ψ〉, the bra 〈v| eatsthe ket |ψ〉, which gives the number λ = 〈v|ψ〉, and one ends up with the vectorλ|u〉.

2. Hermitian conjugate of an expression. Just as in finite-dimensional spaces, onemust transpose and take the conjugate of each term. In other words, we reverse theorder of the factors, and we change the kets into bras and vice versa, the operatorsin their adjoints and the numbers in their complex conjugates. The Hermitianconjugate of λ|ϕ〉〈ψ| A† B is λ∗ B† A|ψ〉〈ϕ|.

6.2.4 Projectors; Decomposition of the Identity

Projector. Consider a Hilbert basis {|n〉}. The operator

Pn = |n〉〈n| (6.31)

is the projector on the basis vector |n〉 (this remark applies to any vector that isnormalized to one). Indeed we have

P2n = Pn and Pn|ψ〉 = Cn|ψ〉,

where Cn is the component (6.22) of |ψ〉 on the vector |n〉.One can define a projector Pν on a subspace n ∈ {ν} by

Pν =∑

n∈{ν}|n〉〈n|. (6.32)

If we extend this to the entire space, we obtain the important closure relation, alsocalled the decomposition of the identity

∑

n

|n〉〈n| = I . (6.33)

6.3 Measurement Results

We can now understand that, in a measurement of a quantity A,

• The possible results of themeasurement are the eigenvalues an of the observable A,• And the probability of finding the result an is the modulus square of the scalarproduct of the state vector with the corresponding eigenvector:

p(an) = |〈ϕn|ψ(t)〉|2.

6.3 Measurement Results 129

6.3.1 Eigenvectors and Eigenvalues of an Observable

We denote |ϕn〉 the eigenvectors of A and an the corresponding eigenvalues, that is,by definition

A|ϕn〉 = an|ϕn〉. (6.34)

We constantly use the following theorems, well known in finite-dimensionalspaces.

Theorem 3 The eigenvalues an of an Hermitian operator are real.

The proof is straightforward. If we multiply (6.34) on the left by 〈ϕn|, we obtain

〈ϕn| A|ϕn〉 = an〈ϕn|ϕn〉.

The left-hand side is real because A is Hermitian, and 〈ϕn|ϕn〉 is real and positive.Therefore an is a real number.

Theorem 4 The eigenvectors corresponding to different eigenvalues are orthogonal.

We multiply (6.34) on the left by another eigenvector 〈ϕm |; we obtain

〈ϕm | A|ϕn〉 = an〈ϕm |ϕn〉 = am〈ϕm |ϕn〉.

In the last expression, A acts on the left on 〈ϕm |. Therefore, we obtain

(an − am)〈ϕm |ϕn〉 = 0, (6.35)

so that either an = am , or, if an = am , 〈ϕm |ϕn〉 = 0.We can therefore choose a set of eigenvectors {ϕn} that are orthonormal.Let us mention a minor and inessential technical difficulty in this context. It

can happen that to the same eigenvalue an there correspond several independenteigenvectors |ϕn,k〉

A|ϕn,r 〉 = an|ϕn,r 〉 , r = 1, . . . , k. (6.36)

In such a case, one says that the eigenvalue an is degenerate with a degeneracy oforder k. Notice that the projector on the eigensubspace of an , of dimension k, is

Pn =k∑

r=1

|ϕn,r 〉〈ϕn,r |, (6.37)

where we assume we have chosen the |ϕn,k〉 to be orthonormal

〈ϕn′,k ′ |ϕn,k〉 = δnn′δkk ′ .

130 6 Principles of Quantum Mechanics

6.3.2 Results of the Measurement of a Physical Quantity

Theorem 5 The result of themeasurement of a quantity A on a system is certain (i.e.,with probability one) if and only if the system is in an eigenstate of the observableA.

We have already seen in Chap. 4 that if ψ is an eigenvector of A then the result ofthe measurement is certain since the dispersion vanishes, Δa = 0. The converse iseasy to prove. Consider the norm of the vector |χ〉 = ( A − 〈a〉 I )|ψ〉, where I is theidentity. We obtain

‖( A − 〈a〉 I )|ψ〉‖2 = 〈ψ|( A − 〈a〉 I )2|ψ〉 = 〈ψ| A2|ψ〉 − 〈a〉2 = Δa2. (6.38)

If the dispersion is zero Δa2 = 0, the norm of the vector ( A − 〈a〉 I )|ψ〉 vanishes.This vector is therefore the null vector and A|ψ〉 = 〈a〉 |ψ〉. Therefore, if Δa = 0,|ψ〉 is necessarily an eigenvector of A with eigenvalue 〈a〉.Theorem 6 The result of a measurement of a quantity A on a single system is oneof the eigenvalues an of the observable A.

In Chap.4, we made the following remarks.

• For a given system and a given quantity, nothing restricts the accuracy of themeasurement. One finds some value with the accuracy allowed by the measuringinstruments.

• By consistency, if, immediately after the measurement, we perform another mea-surement of the same quantity on the system that has already been measured,we will find the same answer with probability one. (This is a postulate on theconsistency of physics; no experiment has ever contradicted this.)

• Therefore, the measurement on a system is actually a means to prepare this systemin a new state for which we know the value of A exactly (Δa = 0).

• Owing to Theorem 5, the system is then necessarily in an eigenstate of A, andtherefore the value found previously is an eigenvalue of A.

• Obviously, after themeasurement, the state vector of the system is in the eigenspacecorresponding to the eigenvalue that has been obtained.

6.3.3 Probabilities

What is the probability of finding the result an by measuring A on a system whosestate is ψ? There, the answer comes from geometry.

In fact, the question is to know “how much” of the eigenvector |ψn〉 does the statevector contain. By the superposition principle, we can understand that if it is 100%|ψn〉 the probability is equal to 1, and if it does not have any component along |ψn〉,if it doesn’t contain it at all, the probability is zero; one will never find the value an .

6.3 Measurement Results 131

How can we evaluate this probability? The answer is that the probability is themodulus squared of the component of the state vector |ψ〉 along the normalizedeigenvector |ϕn〉, or, equivalently, of the scalar product of |ψ〉 and |ϕn〉,

p(an) = |〈ϕn|ψ(t)〉|2. (6.39)

The scalar product 〈ϕn|ψ(t)〉 is the probability amplitude α(an) to find an .Do we really find that by some magic inspiration? No; it is practically written

above in (6.17) in a particular case.Let’s go back to Sect. 6.1, to the Eqs. (6.15) and (6.17), and to the case of the

harmonic oscillator. The {ϕn} are the eigenstates of the energy: hϕn(x) = Enϕn(x).The components Cn = 〈ϕn|ψ〉 of the state vector on this basis are these scalarproducts. The normalization of the state vector (i.e., the probabilistic interpretation)and the Pythagorean theorem tell us that

∞∑

n=0

|Cn|2 = 1, (6.40)

as should be satisfied by a probability law. The expectation value of the energy (6.17)is indeed given in terms of the possible issues En by

〈E〉 =∞∑

n=0

En|Cn|2 = 1. (6.41)

This shows that in this case the |Cn|2 = |〈ϕn|ψ〉|2 are the probabilities of findingEn . Indeed, it is a theorem, which is not difficult to prove for a finite set, that if weknow the outcomes of a probability law and its moments (i.e., 〈Ek〉 for all values ofthe integer k) we know the probabilities (obviously we can calculate the expectationvalue of any power of h and obtain the value of 〈Ek〉).

And that is completely general! Above, we never referred to the specific form ofh, of ϕn , or of the values En . We simply used the fact that the {ϕn} form a Hilbertbasis, and that they are the eigenvectors of h.

6.3.4 The Riesz Spectral Theorem

This property relies on the spectral theorem of Frederic Riesz, which is a fundamentaltheorem of Hilbert space analysis.

Theorem 7 Spectral theorem. The set {|ϕn〉} of eigenvectors of an Hermitian oper-ator A forms a Hilbert basis of the space.

This is well known in finite-dimensional spaces and matrix calculus.

132 6 Principles of Quantum Mechanics

We have deliberately stated it in a mathematically incorrect way for an infinitedimensional space. The true statement is: To any self-adjoint operator, there corre-sponds a decomposition of the identity and a spectral decomposition.

In other words, if we forget about possible degeneracies for simplicity,

1. Any vector |ψ〉 can be decomposed on the basis{|ϕn〉} of the eigenvectors of A,

∀ |ψ〉, |ψ〉 =∑

n

Cn|ϕn〉 , with Cn = 〈ϕn|ψ〉. (6.42)

If |ψ〉 is a normalized state vector, then

〈ψ|ψ〉 =∑

n

|Cn|2 = 1. (6.43)

2. The decomposition of the identity, or closure relation, is

I =∑

n

|ϕn〉〈ϕn|. (6.44)

3. The operator A has a spectral decomposition; that is,

A =∑

n

an|ϕn〉〈ϕn|. (6.45)

4. ThereforeA|ψ〉 =

∑

n

Cn an|ϕn〉 , and 〈a〉 =∑

n

an |Cn|2. (6.46)

We can read in (6.43) and (6.46) that the numbers {|Cn|2} are the probabilities offinding the results an , as previously.

As stated in Theorem 7, this “theorem” isn’t quite true; why? It is true in spirit,but not in form; it lacks rigor. In fact, there exist pathologies. For instance, thereexist operators which, when they are applied to some vectors, “push them out” ofthe Hilbert space, such as x or d/dx . Their eigenvectors do not belong to the Hilbertspace, but to another space called the space of eigendistributions.

The eigenfunctions of px are not square integrable, because they are planewaves ∝ eip0x/�. But the amazing fact is that one can nevertheless expand anysquare integrable function on this set, this is simply the Fourier transformationf (x) = (2π)−1/2

∫g(p)eipx/� dp. One can expand a square integrable function on

a continuous set of functions that does not belong to the Hilbert space, a continuousbasis that belongs to another space. Of course, it is quite feasible to write quantummechanics in a rigorous way, but it is tedious and too complicated at our stage, andit does not bring anything new physically. It suffices to be aware of that.

6.3 Measurement Results 133

6.3.5 Physical Meaning of Various Representations

At this point, things are becoming physically interesting.In a given Hilbert basis, it is obvious that the state vector |ψ〉 is completely

determined by the set of its components {Cn},

|ψ〉 ↔ {Cn = 〈ϕn|ψ〉}

which we can write as a column vector, the corresponding bra being the conjugateline vector.

This representation of the state vector is completely equivalent to the wave func-tion ψ(r, t).

Therefore, there are not only two, but an infinite number of equivalent represen-tations of the state of the system. What is their physical meaning?

In the basis of the eigenstates of the Hamiltonian, the interpretation of this rep-resentation is simple and crystal clear: the Cn’s are the amplitudes to find En in anenergy measurement. Therefore,

• The representation ψ(r, t) is more convenient if we are interested in the propertiesof the particle in space,

• Its Fourier transform ϕ(p, t) is more convenient if we are interested in its momen-tum properties,

• And the components {Cn} in the basis of energy eigenstates are more convenientif we are interested in the energy of the particle.

But, owing to Riesz’s theorem, this can be done with any physical quantity, forinstance, the angular momentum, which we examine later on and which also hasdiscrete eigenvalues. This can be thought of as a “generalization” of the propertiesof the Fourier transform.

6.4 Principles of Quantum Mechanics

We are now able to state the general principles of quantum mechanics. Up to atechnical detail, which we discuss below, these are the following three principles.

The Principles

I. Superposition principleWith each physical system is associated aHilbert space EH . The state of the system

is defined at any instant by a vector |ψ(t)〉 of EH normalized to one.

134 6 Principles of Quantum Mechanics

Comment. This means that any linear superposition of state vectors |ψ〉 =∑Ci |ψi 〉, with Ci complex such that |ψ〉 is normalized, is a possible state vec-

tor. Notice that the convention ‖ψ‖ = 1 leaves an indetermination. A state vector isdefined up to an arbitrary phase factor eiδ . However, it is an overall phase factor: therelative phases of different states of the system are essential. If |ψ′

1〉 = eiδ1 |ψ1〉 and|ψ′

2〉 = eiδ2 |ψ2〉, the superposition of states C1|ψ′1〉 + C2|ψ′

2〉 is different from thesuperposition C1|ψ1〉 + C2|ψ2〉.II. Physical quantities

1. To each physical quantity A there corresponds a linear Hermitian operator A inEH : A is the observable corresponding to the quantity A.

2. Let |ψ〉 be the state of the system when a measurement of A is performed. What-ever the state |ψ〉 is, the only possible results of the measurement are the eigen-values an of the observable A.

3. Denoting by Pn the projector of the subspace associated with the eigenvalue an ,the probability of finding the value an in a measurement of A is:

P(an) = ‖ψn‖2 where |ψn〉 = Pn|ψ〉. (6.47)

This reduces to (6.39) in the absence of degeneracies, but takes into account allcases in a geometrical way.

4. Immediately after a measurement of A that has given the value an , the system isin a new state |ψ′〉:

|ψ′〉 = |ψn〉‖ψn‖ . (6.48)

Comments. Relation (6.47) can be written in the equivalent forms:

P(an) = 〈ψ|Pn|ψ〉 = |〈ψ|ψn〉|2. (6.49)

Principle II.2 is called the principle of quantization and II.3 is the principle of spectraldecomposition. Principle II.4 is the principle of wave packet reduction. It is thequantitative form of the fact that a measurement perturbs the system.

III. Evolution in timeLet |ψ(t)〉 be the state of a system at time t . As long as no measurement is performedon the system, its evolution in time is given by the Schrödinger equation:

i�d

dt|ψ(t)〉 = H |ψ(t)〉, (6.50)

where H is the observable energy, or the Hamiltonian of the system.The state vector ψ(t) depends on time and evolves in Hilbert space according to

this first-order ordinary differential equation.

6.4 Principles of Quantum Mechanics 135

The Case of a Continuous Spectrum

This is the case for the position and momentum variables. In such a case, the onlyprediction that makes sense is to find a result inside some range [a, a + da[. Thediscrete probability law (6.47) is replaced by a continuous law. In the case of theposition observable x , the law is

P(x) dx = |ψ(x)|2 dx, (6.51)

where ψ(x) is the wave function of Chap. 3, and similarly for the momentum vari-able p.

Interest of This Synthetic Formulation

This formulation has many advantages. It is general. It is precise mathematicallyspeaking. It exhibits the important features of the theory.

It is instructive to compare what we just did with the history of Maxwell’s equa-tions. On October 27, 1864, Maxwell presented his memoir on the unification ofelectricity and magnetism to the Royal Society.

Maxwell used 283 symbols to write his equations. Here is a sample:

Magnetic Force (α, β, γ)

⎧⎪⎨

⎪⎩

dγdy − dβ

dz = 4π p′dαdz − dγ

dx = 4πq ′dβdx − dα

dy = 4πr ′.

“In these equations for the electromagnetic field, we have introduced twentyvariable quantities,” said Maxwell, who added “Between these twenty quantities, wehave found twenty equations. Therefore these equations are sufficient to determineall the quantities involved provided we know the conditions of the problem.”

Of course, we canwrite the equations in a simpler way by using vectors and vectoranalysis which exhibits the rotation invariance of the equations.

∇ · B = 0 , ∇ · E = ρ

ε0, ∇ × E = −∂B

∂t, c2∇ × B = j

ε0+ ∂E

∂t.

This way of writing uses only 59 symbols and it leads easily to results such as thepropagation equation in vacuum

(∂2

c2∂t2− Δ

)E = 0.

However, relativistic invariance is the fundamental property that underliesMaxwell’s equations and when a theorist writes them, they appear as

∂μFμν = jν, (6.52)

where relativistic invariance is explicit, and one uses only 8 symbols.

136 6 Principles of Quantum Mechanics

Of course, when it comes to constructing the antenna of a satellite, we must comeback to more concrete quantities and recall that in (6.52) there exist a number ofimplicit conventions (ε0 = c = 1, Aμ = (ϕ, A), Fμν = ∂μAν − ∂ν Aμ, jμ = (ρ, j),and so on). But the proof of intermediate results is much simpler.

6.5 Heisenberg’s Matrices

We can now understand what Heisenberg’s matrices are.

Matrix Representation of Operators

A vector of the Hilbert space can be represented in a Hilbert basis {|ϕn〉} by a columnvector made of its components Cn .

It is therefore natural to expect that any linear operator A acting on this space hasa matrix representation in the same basis {|ϕn〉}.

In fact, consider a vector |ψ〉 and its expansion

|ψ〉 =∑

n

Cn|ϕn〉. (6.53)

If we apply a linear operator A on this, we obtain another vector |χ〉with components{Bn} in the basis {|ϕn〉}

A|ψ〉 = |χ〉 =∑

n

Bn|ϕn〉. (6.54)

If we multiply (6.54) on the left by the basis vector |ϕn〉 we obtain

Bn = 〈ϕn| A|ψ〉 =∑

m

〈ϕn| A|ϕm〉Cm, (6.55)

where we have inserted the expansion (6.53) of |ψ〉.This is simply the matrix relation between the coefficients {Bn} and {Cn},

Bn =∑

m

An,mCm, (6.56)

where the matrix elements An,m are given by

An,m = 〈ϕn| A|ϕm〉. (6.57)

This is the expected result: an observable A is represented in this basis by thematrix (An,m), which acts on the line or column vectors above.

6.5 Heisenberg’s Matrices 137

Matrices X and P

It is interesting to write the matrix representations of the operators x and px inthe basis of the harmonic oscillator eigenstates. In order to do this, we can use therecursion relations (5.16) of the Hermite functions (Chap. 5).

We obtain:

x ⇒√

�

2mω

⎛

⎜⎜⎜⎜⎜⎝

0√1 0 0 . . .√

1 0√2 0 . . .

0√2 0

√3 . . .

0 0√3 0 . . .

......

......

⎞

⎟⎟⎟⎟⎟⎠, (6.58)

p ⇒ −i

√mω�

2

⎛

⎜⎜⎜⎜⎜⎝

0√1 0 0 . . .

−√1 0

√2 0 . . .

0 −√2 0

√3 . . .

0 0 −√3 0 . . .

......

......

⎞

⎟⎟⎟⎟⎟⎠. (6.59)

In this basis, the matrix of the Hamiltonian is of course diagonal.

These are examples of Heisenberg’s matrices ! But how did he end up there?

Heisenberg’s Thoughts

In 1924–1925, Heisenberg was a young assistant of Max Born in Göttingen (he was23). He did things that his contemporaries found difficult to comprehend.

Heisenbergwas an amazing person. It is difficult to understand hiswayof thinking.He was both very simple and infused with Northern philosophy.

He admiredNielsBohrwhowas a propagandist of positivism and ofKierkegaard’sprinciple, “Any new experimental field can only be analyzedwith concepts of its own.It is not possible to use concepts and principles used previously in other contexts.”

And Göttingen was a fabulous place. The philosopher Husserl had been there.He was the founder of phenomenology, “One must come back from discourses andopinions to facts,” that is, describe what one observes in the simplest possible lan-guage before trying to interpret it. Husserl had left his student Wittgenstein, a greatspecialist of language, who said that “It is the language which must adapt to factsand not the reverse. If one attempts to adapt the interpretation of a phenomenon witha language which is already formed and filled with a priori’s, one is bound to drawwrong conclusions on the nature of things.”

So, Heisenberg said to himself, “I cannot talk about the position x and velocity v

of an electron in an atom. I can only talk about what I am able to see, the positionsand the intensities of spectral lines.”

In the classical theory of radiation, a fundamental consequence of Maxwell’sequations and of relativity is that if a charged particle is accelerated, it radiates energy.The simplest example is a charged dipole with a sinusoidal motion x = a cos(ωt). It

138 6 Principles of Quantum Mechanics

radiates and loses energy. At large distances, the field decreases not as 1/r2 as for afixed charge, but as 1/r . Therefore, the energy radiated through a sphere of radius Ris independent of R; its flux is conserved. The radiated amplitude is proportional tothe acceleration E ∝ x ′′ ∝ aω2, and the total radiated power is therefore proportionalto P ∼ a2ω4. An arbitrary periodic motion of period T = 2π/ω can be expanded ina Fourier series:

x =∑

n

aneinωt ,

and the resulting radiation occurs at frequencies νn = nω/2π with intensities pro-portional to In ∝ n4ω4a2n .

Now, said Heisenberg, in atoms there are two indices. The atomic frequenciesfollow the rule

νnm = (En − Em)/h.

So, Heisenberg introduced “quantum amplitudes” or “quantum quantities” with twoindices, and a standard time behavior

A =⇒ Anme−i(En−Em )/�.

(There is a technical difference between the Schrödinger picture, where the statevector evolves and the observables are fixed, and Heisenberg’s where the state vectoris fixed, but the observables evolve in time.) The positions of spectral lines areνnm = (En − Em)/h and their intensities are proportional to |Anm |2.

Using such ideas,Heisenberg developed a theory thatworked quitewell. However,he realized that in order for the product of two quantum quantities to have a propertime-dependence and to be expressed in terms of the two quantities, he needed toinvent a “symbolic multiplication,” for instance,

A2 →∑

k

Anke−i(En−Ek )/�Akme

−i(Ek−Em )/� = (∑

k

Ank Akm) e−i(En−Em )/�

in other words,(A2)nm =

∑

k

Ank Akm .

Max Born was intrigued. He was a mathematician, and that reminded him of hisyouth. Those are just the rules of matrix calculus! Born was very excited.

Thematrices ofHeisenberg are there in front of us in (6.58).What seems incredibleis that in 1925 mathematicians knew of the existence of matrices, as examples ofnoncommutative algebras, but they considered them as very formal objects, and theyweren’t used to working with matrices in practice.

So, Born was very excited: “We must absolutely go further in Heisenberg’s ideas!In his works, there is an underlying structure to be explored!” He publicized Heisen-berg’s work widely.

6.5 Heisenberg’s Matrices 139

On July 17, 1925, Born took the train from Göttingen to Hanover and, in thesame compartment, he met Pauli. He was excited and said to Pauli, “You know,Heisenberg’s quantities are matrices! Do you want to collaborate?”

Pauli, who was very young but very famous because of his monumental treatiseon relativity, which he had written in 1922 when he was 22, was a tough character.Most of the time, he found other people’s ideas either stupid or obvious. So he repliedto Born, “You’re going to spoil Heisenberg’s beautiful physical ideas with your futilemathematics.” They quarreled.

In the same compartment, there was a young and shy mathematics assistant,Pascual Jordan, who said to Born, once the train had arrived and Pauli had left,“I have worked on matrices. Maybe I can help you.” On the evening of the nextday, Born and Jordan established the fundamental commutation relation between thematrices (X) and (P)

(X)(P) − (P)(X) = h

2πi I, (6.60)

where I is the unit matrix, that Born called the fundamental equation of the quantenmechanik,1 that is, the mechanics specific to quanta. This led to three fundamentalarticles of Born, Heisenberg, and Jordan.

One must say that in the meantime, Pauli, who was pragmatic, had reconsideredhis opinion onmatrices, and he had calculated the energy levels of the hydrogen atomwith matrices in 1925 before Schrödinger’s calculation.2 Pauli managed to calculateall eigenvalues of an infinite matrix, and some effects that were not measured untilthe 1980s!

Now, Born, Heisenberg, and Jordan went to see Hilbert, who was the great mathe-matics professor in Göttingen, and questioned him about matrices. “It’s very formal,”said Hilbert. “The only case when I observed it was useful in practice is in eigen-value problems related to differential equations with boundary conditions.” Born,Heisenberg and Jordan retired politely, thinking that poor Hilbert did not understandthe issue. Six months later, Hilbert had fun saying, “If these arrogant youngsters hadlistened to me, they would have found the Schrödinger equation six months beforehim.” Indeed the Schrödinger equation was greeted as the great step forward becauseit enabled the calculation of energy levels as a stationary wave problem, in oppositionto quantum restrictions on classical trajectories, as advocated by Bohr.

We come back to Dirac in Chap.8. He appeared in the same summer of 1925;he was just 23. We have done, backwards, the work of unification performed bySchrödinger at the end of 1926 and independently by Dirac at the beginning of 1927.Schrödinger knew a lot of mathematics; he had worked on eigenvalue problems.Dirac was simply a young genius.

This brought to an end the quarrels between the pros of wave mechanics and thepros of matrix mechanics. On the other hand, this opened a new field of research to

1The formula pq − qp = ih/(2π) is carved on the gravestone of Born and his wife in Göttingen.2In order to do this, Pauli used the SU (2) × SU (2) symmetry of the hydrogen atom.

140 6 Principles of Quantum Mechanics

mathematicians. In 1927Hilbert andvonNeumann laid themathematical foundationsof quantummechanics where one can find the first steps of the theory of distributionsdeveloped by Laurent Schwartz in 1946.

6.6 The Polarization of Light, Quantum “Logic”

In order to illustrate all this and to show oncemore that themost importantmathemat-ical structure of quantummechanics is addition, we examine a quantum phenomenonthat is the only one directly visible.

Light waves are transverse and they possess a polarization that describes thebehavior of the electric field in the plane transverse to the electric field.

There are several types of polarizations. The light coming out of a projector isnonpolarized; it is in a statistical mixture of polarization states. A polarizer, forinstance a polaroid, filters linear polarization along its optical axis (it can be ananisotropic medium that absorbs light whose polarization is perpendicular to theaxis). In Fig. 6.3, left, this axis is assumed to be horizontal.

If we place another polarizer, also called an analyzer, at an angle θ with the firstone, the transmitted intensity is proportional to I ∝ cos2 θ (Fig. 6.3, middle). Atθ = 45◦ the intensity is half of that which came out of the first polarizer. If the axisof the analyzer is vertical, at θ = 90◦, no light comes out (Fig. 6.3, right).

Classically, this is how Fresnel understood and explained the phenomenon. Itseems to be elementary geometry. Notice that it happens in the same manner inde-pendently of the wavelength.

But actually, this is purely a quantum phenomenon; it can and it must be describedexactly by this quantum state formalism. (This can be done in classical optics asStokes understood in the 19th century.) In fact:

1. Light is composed of photons.2. Photons are elementary particles; they cannot be broken into pieces.

Fig. 6.3 Outgoing light from a horizontal polarizer (left). Intensity across an analyzer at an angleθ (middle); extinction if the analyzer is at 90◦ of the first one (right)

6.6 The Polarization of Light, Quantum “Logic” 141

Fig. 6.4 Reappearance oflight issuing from twocrossed polarizers if a thirdpolarizer at some angle isinserted between them

So, let’s discuss the matter in terms of photons.3 When a photon impinges on apolaroid, either it goes through it or it doesn’t; there is no other choice for it.

Of course, with a macroscopic source, a lot of photons are produced. A lightbeam of 1 watt carries∼1018 photons per second. At an angle θ, it is a fraction cos2 θof these photons that comes out of the analyzer. In other words, each photon has aprobability of cos2 θ to get through.

This is even clearer if we cross the polarizer and analyzer at a right angle θ = 90◦.Nothing gets through. States of orthogonal polarizations are incompatible; there is azero probability that a photon in the horizontal polarization state can be found in thevertical polarization state. Nowwe can observe an amazing phenomenon. If we inserta third polarizer at some non-zero angle, say 45◦, between the crossed polarizers,the light reappears (Fig. 6.4) although we have inserted an absorbing object that canonly reject all photons polarized perpendicular to its axis! (Actually, is it really theonly thing it can do? No! It’s not a triviality to say that it is also able to let photonspass if their polarization is parallel to its axis.)

We know the solution. We must describe polarization states of photons in a two-dimensional Hilbert space. In this space, we can choose as basis states the states ofhorizontal and vertical linear polarization, which we denote

| →〉 and | ↑〉. (6.61)

If the photon is in the state | →〉 it passes through the horizontal polarizer with prob-ability 1. If it is in the state | ↑〉, it is absorbed by this polarizer, but it passes througha vertical polarizer with probability one. By definition, these states are orthogonal〈↑ | →〉 = 0 (Fig. 6.3).

We denote |θ〉 as the state of a photon polarized linearly along a direction at anangle θ with the horizontal axis (0 ≤ θ < π). This state is a linear superposition of

3Polarization of light comes from the fact that the photon is a “spin one” particle. It is a pointlikemassless particle that carries an intrinsic angular momentum whose projection on the direction ofpropagation is either +� or −�. The reader will understand that such odd properties are outside thescope of this book.

142 6 Principles of Quantum Mechanics

the basis states (6.61), as is the orthogonal state |θ + π/2〉:

|θ〉 = cos θ| →〉 + sin θ| ↑〉, |θ + π/2〉 = − sin θ| →〉 + cos θ| ↑〉. (6.62)

For the particular value θ = π/4, we have (this does not restrict the generality ofour argument)

| ↗〉 = 1√2(| →〉 + | ↑〉), | ↖〉 = 1√

2(−| →〉 + | ↑〉); (6.63)

this relation can be inverted:

| →〉 = 1√2(| ↗〉 + | ↖〉), | ↑〉 = 1√

2(−| ↗〉 + | ↖〉). (6.64)

The explanation of the observations (Fig. 6.3) is that the probability for a hori-zontally polarized photon to get through a polarizer at an angle θ is

p(→, θ) = |〈θ| →〉|2 = cos2 θ (6.65)

as announced before; that is, p(→, 45◦) = 1/2.We now come to the observation of Fig. 6.4. After crossing the polaroid success-

fully at 45◦, by the principle of wave packet reduction, the photon is necessarily inthe state | ↗〉, which can be decomposed according to (6.63). In this new state, it isin both states | →〉 and | ↑〉. It is therefore natural to observe that it can cross thevertical polarizer with probability 1/2, whereas this was forbidden in the absence ofthe intermediate polarizer. If this latter polarizer is at an angle θ, the probability offinding that a photon crosses the entire setup is

p(→, θ,↑) = cos2 θ sin2 θ. (6.66)

Notice that, although all we have done here is ordinary Euclidian geometry in twodimensions, it is necessary for the space to be complex, i.e. Hermitian, in order todescribe all the pure polarization states. There exist states with complex componentssuch as

|ΨL ,R〉 = 1√2

(| →〉 ± i | ↑〉) . (6.67)

One can check that such states keep the same form under an arbitrary rotation ofthe linear polarization basis states. These states correspond to left and right circularpolarized states (more generally, elliptic polarization states).

Quantum “Logic”

This allows us to understand better the difference between classical and quantumlogics, namely the difference between “or"and “and.”

6.6 The Polarization of Light, Quantum “Logic” 143

The polaroids are filters that let photons pass if their polarization is along theiraxis and eject them otherwise.

Let’s use a metaphor (after all, at the end of such a Chapter, we deserve it!). Let’suse other words. Instead of horizontal and vertical, let’s say ladies and gentlemen. Ahorizontal polaroid is a filter that only allows ladies to pass. Similarly, a vertical oneonly lets gentlemen pass. Of course, if one places the two filters one after the other,no one passes. And, if we want to know how many gentlemen are in a population, itsuffices to put a vertical polarizer and to measure the outgoing intensity.

Let’s say that a polaroid at 45◦ allows smokers to get through, and a polaroid at135◦ does so for the nonsmokers. No smoker is a nonsmoker.

Now assume we are quantum kids trying to understand set theory. We first makea selection of ladies with a horizontal polarizer. Then, in this new sample, we selectthose who smoke with a polarizer at 45◦, as in Fig. 6.5 middle.

In classical logic, the succession of the two filters allows the passage of peoplewho are both ladies and smokers, that is, the intersection of the sets {ladies} and{smokers}. We can in fact check that none of them is a nonsmoker.

But see what happens if we look at who they are! We observe with great horrorthat, by placing a vertically oriented polarizer, half of the people we had selectedinitially get through (Fig. 6.5 right). In other words half of these ladies are gentlemen!In quantum mechanics there is no way out but to conclude that half of the ladies whosmoke are in fact gentlemen!

Therefore there appears to be a very different way of conceiving logic for quantumchildren when they play. If they play with cubes and spheres each of which canbe either blue or red, it’s a hard task, if not an impossible one, to try to find theintersections of sets.

Whatever you do in quantum mechanics, smokers are always both ladies andgentlemen, ladies are always both smokers and nonsmokers.

All that is simply the mechanism of the superposition principle and the reductionof the wave packet.

Fig. 6.5 Difference between the quantum and and the classical or: outgoing light after crossing aseries of polarizers. Left, only one polarizer,middle, a second polarizer at 45 ◦, right a third polarizeron top of the second, perpendicular to the first one

144 6 Principles of Quantum Mechanics

6.7 Exercises

1. Translation and Rotation Operators

a. Consider a one-dimensional problem and a wave function ψ(x) which can beexpanded in a Taylor series. Show that the operator T (x0) = e−i x0 p/�, where x0is a length and p is the momentum operator, is such that:

T (x0)ψ(x) = ψ(x − x0).

N.B. The expansion eiu = ∑∞n=0 (i u)

n/n! is mathematically legitimate.

b. We now consider a two-dimensional problem in an xy plane and we define the zcomponent of the angular momentum operator by:

L z = x py − y px = −i�

(x

∂

∂y− y

∂

∂x

)= −i�

∂

∂θ,

where the polar coordinates r, θ are defined by r = (x2 + y2

)1/2and θ =

arctan(y/x). Show that the operator R(ϕ) = e−iϕL z/�, where ϕ is dimension-less, is such that :

R(ϕ)ψ(r, θ) = ψ(r, θ − ϕ).

2. The Evolution Operator

Consider a system whose Hamiltonian does not depend on time (isolated system).Show that the state vector at time t , denoted |ψ(t)〉, can be deduced from the statevector |ψ(t0)〉 at initial time using:

|ψ(t)〉 = U (t − t0) |ψ(t0)〉 with U (τ ) = e−i Hτ/�. (6.68)

Show that U (τ ) is unitary, i.e., U † = U−1.

3. Heisenberg Representation

Consider an isolated system whose Hamiltonian is H . We denote |ψ(0)〉 the statevector of the system at time t = 0. We want to calculate the expectation value a(t)of the results of the measurement of an observable A at time t .

a. Express a(t) in terms of |ψ(0)〉, A and the evolution operator U (t) defined in theprevious exercise.

b. Show that a(t) can be written as the expectation value of an operator A(t) for thestate |ψ(0)〉. Show that A(t) is determined from:

i�d A(t)

dt= [ A(t), H ] and A(0) = A. (6.69)

6.7 Exercises 145

This approach is called Heisenberg representation (or Heisenberg picture): the statevector is time independent, and the operators obey the Heisenberg equation (6.69).

4. Dirac Formalism with a Two-State Problem

Consider twonormalized eigenstates |ψ1〉 and |ψ2〉of aHamiltonian H correspondingto different eigenvalues E1 and E2 (one can set E1 − E2 = �ω)

a. Show that |ψ1〉 and |ψ2〉 are orthogonal.b. Consider the state |ψ−〉 = {|ψ1〉− |ψ2〉}/

√2, calculate the expectation value 〈E〉

of the energy and the dispersion ΔE in this state.c. Assume that at t = 0 the system is in the state |ψ(t = 0)〉 = |ψ−〉. What is the

state of the system |ψ(t)〉 at time t?d. Consider an observable A defined by A|ψ1〉 = |ψ2〉, and A|ψ2〉 = |ψ1〉. What are

the eigenvalues a of A in the subspace generated by |ψ1〉 and |ψ2〉?e. Construct the corresponding combinations of |ψ1〉 and |ψ2〉, which are eigenvec-

tors of A.f. Assume that at t = 0 the system is in the state |ψ−〉 corresponding to the eigen-

value a = −1. What is the probability to find a = −1 in a measurement of A ata later time t?

Chapter 7Two-State Systems

In the previous chapter, we saw how Heisenberg’s matrix mechanics arose in1924–1925. What we want to do here is to come back to the problem of the NH3

molecule, seen in Chap.5 and to do matrix mechanics on this particular case, in asimilar way to what we did in wave mechanics by considering the simple problemof the motion of a particle in space.

Thiswill allow us to become familiar withDirac’s formalism and to treat problemsin amuch simpler way, mathematically speaking, thanwith wave functions.We showthat a great deal of quantummechanics can be performedwith very littlemathematics,namely two-dimensional matrices.

We have seen a first example of this with the polarization of the photon in theprevious chapter. In other cases, the link with physics might have been obscure if wehad started quantum mechanics in this way. We might have had problems in relatingsimple calculations with actual physical phenomena if we hadn’t already treated theproblem of the NH3 molecule as in Chap. 5.

This leads us to a series of applications. Here, we consider the behavior of theammonia molecule in an electric field. This enables us to understand the mechanismof masers and to see various applications such as atomic clocks, the GPS system,and the tests of the predictions of relativity on time.

The problem we present at the end of this chapter is devoted to a remarkable phe-nomenon, which is impossible to imagine with classical concepts: neutrino oscilla-tions. It is only after having understood quantumoscillations in two-state systems thatone can understand this amazing discovery where an elementary particle in vacuumcan change into another one periodically. The experimental results on this questionhave been awarded many awards, among which the 2002 and 2015 Nobel Prizes forPhysics.

© Springer International Publishing Switzerland 2016J.-L. Basdevant, Lectures on Quantum Mechanics,Graduate Texts in Physics, DOI 10.1007/978-3-319-43479-7_7

147

148 7 Two-State Systems

7.1 The NH3 Molecule

We recall what we found on NH3. We had made a model of the inversion motionof this molecule by a symmetric double well whose minima correspond to the twoclassical equilibrium configurations. We saw that the lowest energy level is actuallysplit in two sublevels by the tunnel effect:

• ES = E0 − A0, corresponds to a symmetric wave function ψS , and• EA = E0 + A0, corresponds to an antisymmetric wave function ψA.

We constructed states corresponding to the “left and right” classical configurationsas linear superpositions of these stationary states. And we understood the inversionmotion of the molecule between these states.

In the transitions between the two energy levels, the molecule emits or absorbsradiation at the Bohr frequency ν = 24GHz, related to the splitting. This emissioncan easily bedetected andmeasured since themolecule has an electric dipolemoment.

Therefore one can measure directly the splitting ES − EA. It is even measured soaccurately that it is nonsense to try to calculate it theoretically from first principles.

7.2 “Two-State” System

This is where we follow Heisenberg.

We study physical processes that involve states of the NH3 molecule which aresuperpositions of only the two lowest energy states

|ψ〉 = a|ψS〉 + b|ψA〉.

In other words, we are interested in physical situations where the state vector |ψ〉 ofthe molecule remains in a two-dimensional subspace of the Hilbert space, which isinfinite-dimensional. This type of situation is called a two-state system or a two-levelsystem, that is, d(EH ) = 2. There are an infinite number of states, but all are linearcombinations of two of them. We are going to do quantum mechanics in the case ofa two-dimensional Hilbert space where the mathematics are simple.

Of course, this is conceivable mathematically.

• Can it be achieved physically?• Does it have any physical interest?

Yes! It is indeed Heisenberg’s starting point! In NH3, one knows everything inprinciple, but it’s an awfully complicated problem. It is a system of 14 particles, 10electrons, and 4 nuclei, with pairwise Coulomb interactions. The Hamiltonian is

7.2 “Two-State” System 149

H =4∑

n=1

p2n2Mn

+10∑

i=1

p2i2me

−4∑

n=1

10∑

i=1

qeqn4πε0|rn − r i |

+ q2e

2

10∑

i=1

10∑

j=1

1

4πε0|r i − r j | + 1

2

4∑

n=1

4∑

m=1

qnqm4πε0|rn − rm | , (7.1)

where (r i , pi ) are the positions and momenta of electrons, and (qn, Mn), (rn, pn)are the charges, masses, positions, and momenta of the nuclei.

Our potential model allowed us to understand qualitatively the tunnel effect, with-out which the radiowave of ν = 24GHz would be a mystery, because the energysplitting is comparatively very small.

But the general problem of finding these wave functions in 42 variables is muchtoo complicated. And, above all, it is completely uninteresting. The description ofthe positions of these 14 particles is of no interest.

However, we know for sure, owing to the spectral theorem,

1. That there exists a set of energy levels En and corresponding eigenfunctionsψn . (It is an imaginary catalogue. We cannot write these functions, but we only needto know they exist; we can talk about them even though we do not know them.)

2. And we know the {En} experimentally.

For instance, there exists a whole part of the spectrum due to the inversion motiondiscussed in Chap. 5. It is a series of levels whose energies are higher and higherand where the splitting increases with the energy until it fades away, as depicted inFig. 7.1.

The following orders of magnitude are of interest. The lowest energy splitting is2A0 ≈ 10−4 eV, the splitting between this doublet and the following one is E1−E0 ≈0.12eV, and the splitting of the latter doublet (E1) is 2A1 ≈ 4 × 10−3 eV.

If we denote this imaginary catalogue {|ψn〉}, we can represent the most generalstate of the NH3 molecule as

Fig. 7.1 Energy levels andcorresponding wavefunctions in a double well.The abscissa axis is taken,for a given wave function, atthe value of thecorresponding energy level(Courtesy of Denis Gratias)

150 7 Two-State Systems

|ψ〉 =∑

n

an|ψn〉, (7.2)

or by an infinite column vector whose components an are the probability amplitudesof finding the system in each energy level

|ψ〉 :

⎛

⎜⎜⎜⎜⎜⎜⎜⎜⎝

a1a2a3...

an...

⎞

⎟⎟⎟⎟⎟⎟⎟⎟⎠

with probabilities P(En) = |an|2

by definition of the {an}.This is what Heisenberg does, and it’s more realistic than a description in space

by a wave function of the 14 particles because1. One can measures the values En ,2. And one can control much more easily the energy of such a complex system

than all these positions.It is not deeper than a wave function representation, but it is not less deep for the

moment.Now, is it possible to construct physically states that are only superpositions of

the two lowest energy states?Yes, of course, becausewe control the energy. For instance, in a gas at temperature

T , we know that the ratio of populations of molecules in energy states E1 and E2

is given by Boltzmann’s factor n(E2)/n(E1) = e−(E2−E1)/kT . Therefore, insertingthe above values,

• At 100K: NA ≈ NS , n(E1)/n(E0) ≈ 6 × 10−7 which is a small probability, onemakes the approximation to consider it to be zero;

• If that’s not enough, we can go down to 50K, p ≈ 3×10−13, or to 25K, p ≈ 10−25

(not a single molecule in 22.4L).

In order for the first excited level E1 to play a role and for p(E1) to be nonnegli-gible, one must reach temperatures of T ≈ 1300K.

Temperature gives us a cutoff on the finite number of significant components ofthe state vector.

We know this type of situation. Gravity holds us on the ground. If we navigate,we live in a effective two-dimensional world in first approximation. If we climbmountains or fly in a spacecraft, then we must worry about the third dimension.

7.3 Matrix Quantum Mechanics 151

7.3 Matrix Quantum Mechanics

Vectors

Consider a two-dimensional subspace of EH generated by the set {|ψS〉, |ψA〉}whichforms a basis of the subspace. We can represent an arbitrary state |ψ〉, by a two-component vector,

|ψ〉 = a|ψS〉 + b|ψA〉, with P(ES) = |a|2, P(EA) = |b2|. (7.3)

In a matrix representation, this is written as

|ψS〉 :(10

), |ψA〉 :

(01

), |ψ〉 :

(ab

),

for instance, |ψR/L〉 : 1√2

(1

±1

).

Hamiltonian

What is the expression of the Hamiltonian? Of course, not the full Hamiltonian ofthe molecule, but its restriction to the subspace of interest. The basis states are bydefinition energy eigenstates of H which is a diagonal matrix, and, in the subspace,it is the 2 × 2 matrix:

H =(E0 − A 0

0 E0 + A

). (7.4)

The Schrödinger equation is

i�d

dt|ψ(t)〉 = H |ψ(t)〉, with |ψ(t)〉 =

(α(t)β(t)

). (7.5)

This gives us two uncoupled equations whose solutions are

|ψ(t)〉 = e−i(E0t/�)

(a eiωt/2

b e−iωt/2

), (7.6)

where we have introduced the Bohr frequency ω = 2A/�.

Observables

In fact, in this subspace, any linear operator, any observable, is a 2 × 2 matrix. The“restriction” of an observable to the subspace is the first 2 × 2 block of the infinitematrix in the basis of our imaginary catalogue.

Now, how are we going to do calculations? In wave mechanics, we have a differ-ential equation and a potential. What should we do here?

152 7 Two-State Systems

Simplymakemodels of observables (i.e., 2×2matrices) instead ofmakingmodelsof potentials as in wave mechanics.

Eigenvalues and Eigenvectors of a Hermitian 2 × 2 Matrix

Before we continue, it is useful to recall a few formulae on 2 × 2 matrices. AnyHermitian 2 × 2 matrix can be written as

A =(

a c eiφ

c e−iφ b

), (7.7)

where a, b, c are real and φ is a phase.The eigenvalues of A are

λ± = 1

2

(a + b ±

√(a − b)2 + 4c2

), (7.8)

corresponding to eigenvectors

|ψ+〉 =(

cos θsin θ e−iφ

), |ψ−〉 =

( − sin θcos θ e−iφ

), with tan 2θ = 2c

(a − b). (7.9)

Position Observable

The following observable plays a central role here.We can define a “position” observ-able by the matrix

X = x0

(0 11 0

), (7.10)

where x0 is a length parameter, roughly speaking equal to the positions of the minimaof the double well. The observable X has eigenvalues ±x0 and eigenvectors

|ψ±〉 = 1√2

(1

±1

)= |ψR/L〉, (7.11)

that is, the “classical” configurations of Chap. 5.This is indeed the restriction of the position observable x in the previous sense. It

has the same structure as the first 2× 2 block of Eq. (6.58) in the harmonic oscillatorbasis.

Strictly speaking, it is not the position observable but rather the disposition withrespect to the center. Our assumptions forbid us to talk about the position with greateraccuracy than the half-width of each well. This is the consequence of the Heisenbergrelations. If we prepare a wave function whose dispersion in x is much smaller thanΔx = a/2 (i.e., the half width of one of the wells) then one cannot remain in the two-dimensional subspace. One must attain the levels E1. We need more eigenfunctionsin order to localize the particle better as one can understand in Fig. 7.1.

7.3 Matrix Quantum Mechanics 153

Examples

We can do a series of simple exercises. Consider the state

|ψ(t)〉 =(

a eiωt/2

b e−iωt/2

), with �ω = 2A. (7.12)

The expectation value of X in this state is

〈X〉 = 〈ψ(t)|X |ψ(t)〉 = 2x0Re(a∗b e−iωt ); (7.13)

that is, for |ψ(0)〉 = |ψR〉 with a = b = 1/√2,

〈X〉 = |〈ψR|X |ψ(t)〉|2 = x0 cosωt, (7.14)

and a probability

Pt (X = +x0) = |〈ψR|ψ(t)〉|2 = cos2(ωt

2), (7.15)

which shows simply the inversion of the NH3 molecule.For the moment, there is nothing really new, but calculations are very simple. We

have got rid of the potential and of wave functions. All the interesting physics is inthe value of the parameter A which is given by experiment, and in the parameter x0which defines the size of the molecule.

This is a model. In order to improve it, we must increase the size of the Hilbertsubspace, namely take more terms into account in the expansion (7.2).

Basis of Classical Configurations

It is interesting to make a change of basis and in particular to express vectors andoperators in the alternative basis of classical equilibrium configurations {|ψR〉, |ψL〉}.We then express quantum effects in terms of classical situations.

The Hamiltonian is not diagonal in this basis, as opposed to X :

H =(

E0 −A−A E0

), X = x0

(1 00 1

). (7.16)

The off-diagonal terms are transition terms that allow transitions L ↔ R, the inver-sion of themolecule. If A = 0, the two classical configurations have the same energy.

The energy levels are the eigenvalues of H . The calculation is straightforward:we indeed obtain the eigenvalues and eigenvectors we know

E− = E0 − A, |ψS〉 = (|ψR〉 + |ψL〉)/√2,

E+ = E0 + A, |ψA〉 = (|ψR〉 − |ψL〉)/√2.

154 7 Two-State Systems

Interference and Measurement

At this point, it is instructive to apply the principles. Suppose we start with an energyeigenstate, say |ψS〉,

|ψS〉 = 1√2(|ψR〉 + |ψL〉).

If we measure X , we can find ±x0 with probabilities 1/2 = (1/√2)2.

Suppose themeasurement has given the result+x0; the state after themeasurementis

|ψR〉 = 1√2(|ψS〉 + |ψA〉). (7.17)

If we measure X again immediately afterwards, before the oscillation is appreciable,we find +x0 with probability 1; the state after the measurement is |ψR〉.

Now, suppose that, on this new state |ψR〉, we measure not X but the energy Ewhich we are sure was E = ES when we started. One can read in (7.17) that wedo not always find ES but the two possibilities ES and EA, each with a probabilityof 1/2.

We see in this case how the measurement has perturbed the system. At the begin-ning, the state was |ψS〉; at the end it is a mixture of |ψS〉 and |ψA〉 in interference,for which 〈E〉 = (ES + EA)/2.

All of this results from the superposition principle on one hand and the filteringof which a measurement consists.

We remark that, consequently, a positionmeasurement implies aminimum energyexchange with the system. Here, on the average, the exchange of energy is equalto A.

Needless to say that all of this is quite similar to what we saw with the photonpolarization at the end of Chap. 6.

7.4 NH3 in an Electric Field

Let us comeback toNH3. In order tomake those energy or positionmeasurements,wemust control the energy in a more refined way than thermal motion. The temperaturehas allowed us to be in a subspace; now we want to manufacture individual states ofa given energy.

We have seen that the NH3 molecule has an electric dipole moment (Fig. 7.2). Wecan make this electric dipole interact with an electric field. How can we describe thisproblem?

Classically, if a system with an electric dipole moment D is placed in an electricfield E it acquires a potential energy W = −E · D. If we assume that E and D areparallel, this becomes

W = −E D. (7.18)

7.4 NH3 in an Electric Field 155

Fig. 7.2 The two classicalconfigurations of themolecule NH3 and theassociated electric dipole

If we place the NH3 molecule in an electric field E , what is the quantum potentialenergy W , which is related to the observable D, according to the correspondenceprinciple, by

W = −E D? (7.19)

The answer is simple. In the particular configurations of interest here, D is simplyproportional to X , that is,

D = q0 X =(

0 d0d0 0

), W =

(0 −d0E

−d0E 0

)(7.20)

where q0 is an effective charge and d0 is an electric dipole moment that is measuredexperimentally: d0 ≈ 3 × 10−11 (eV)/(V/m) = 5 × 10−30 C.m.

In other words, if we measure X and we find ±x0 with some probabilities, ameasurement of D will give ±d0 with the same probabilities. And W is the productof D times the number E , the value of the electric field.

The only difficulty, here, is to accept that it is a good model for the observable D,within our assumptions, to be proportional to X .

The potential energy observable W is simply the product of the observable D bythe numerical value of the applied electric field.

The only real justification for this choice is that it works very well.From then on, things are quite simple. The Hamiltonian of the molecule in a field

is the sum of the free Hamiltonian (7.4) and the potential energy (7.20).

7.4.1 Uniform Constant Field

In a uniform constant field, if we set η = d0E , the Hamiltonian is

H =(E0 − A −η−η E0 + A

). (7.21)

Finding the energy levels and the corresponding eigenstates amounts to diagonalizingthis matrix.

156 7 Two-State Systems

Fig. 7.3 Energy levels of aNH3 molecule in an electricfield

Turning to (7.8) and (7.9), the eigenvalues and eigenvectors of H are

E− = E0 −√A2 + η2 |ψ−〉 =

(cos θsin θ

), (7.22)

E+ = E0 +√A2 + η2 |ψ+〉 =

(− sin θcos θ

)with tan 2θ = η/A. (7.23)

The variation of these energy levels with the applied electric field E is representedin Fig. 7.3. The validity of the result rests on the condition that E must not be largeenough to reach the levels E1. Otherwise one would have to take care of higher levels(4-, 6-, etc. level systems). The value of d0 leaves a lot of space.

7.4.2 Weak and Strong Field Regimes

It is interesting to consider two limits: the weak field and the strong field limits.The borderline between the two domains, weak and strong fields, is grosso modoE ∼ A/d0 ≈ 1.7× 106 V/m for NH3. The quantity d2

0/A is called the polarizabilityof the molecule. It is large for NH3 in its ground state because the splitting A0 issmall.

We notice that if we consider excited levels E1, A1 ≈ 40 A0, the polarizabilityis much smaller. The borderline is then around 7 × 107 V/m. For usual fields of106 V/m, the polarization of the states E1 is completely frozen, these states do notparticipate in energy exchanges through an electric field; our starting assumption isagain consistent.

Weak Field

In the weak field regime, such that E d0/A or θ 1, the levels and eigenstatesare to lowest order in E :

7.4 NH3 in an Electric Field 157

E∓ � E0 ∓(A + d2

0E2

2A

), (7.24)

|ψ−〉 � |ψS〉 + d0E2A

|ψA〉, |ψ+〉 � |ψA〉 − d0E2A

|ψS〉. (7.25)

This is understandable. In the absence of a field, the molecule has a symmetricprobability and 〈D〉 = 0. The effect of the field is to polarize the molecule, whichacquires a mean electric dipole moment proportional to the field 〈D〉 ≈ ±d2

0E/A,hence an energy which is quadratic in E .

Strong Field

For strong fields, E � d0/A or θ � π/4, the effect of the field dominates over thetunnel effect; the molecule is completely polarized. The eigenstates are close to theclassical configurations |ψR/L〉 with 〈D〉 = ±d0 and the energies are E ± d0E ; theresponse to the field is linear as for a classical dipole.

There is a competition between two effects:

• The tunnel effect tends to symmetrize the molecule, which results in a vanishingdipole moment ⇒ 〈D〉 = 0.

• The field pulls the molecule toward the classical configurations |ψR/L〉, where ithas a dipole moment 〈D〉 = ±d0.

7.4.3 Other Two-State Systems

We have made progress on the ammonia molecule. All this could have been donewith wave functions, but it would soon have been complicated. Here, the calculationof energy levels amounts to diagonalizing a matrix.

We come back below to the ammonia maser.We insist that this is simply an example, because the same mathematics, finite

dimensional matrix calculus, applies to many effects. Some of them are exactlyfinite-dimensional such as

• The spin 1/2 of particles such as electrons, protons, quarks, and so on• The polarization of the photon• The physics of “strange” neutral mesons K0 � K0, and “beautiful” mesons B0 �

B0,• The universality of weak interactions and the mixing matrix of d, s, b quarks• The quantum oscillations of neutrinos, which we show below.

It can also be an approximate model as for lasers, the chemical bond, and nuclearmagnetic resonance.

158 7 Two-State Systems

7.5 Motion of Ammonia Molecule in an InhomogeneousField

We come back to the case of the ammonia molecule. With the tools we have devel-oped, we can understand the principle of the maser, which has been a revolution inthe physics of radiowaves, in telecommunications, and in astrophysics. We do notprove technical results here. They are intuitive and we come back to them in thesimilar, but simpler, case of the electron spin in Chap.12.

7.5.1 Force on the Molecule in an Inhomogeneous Field

How can we separate energy eigenstates of the molecule? How can we prepare asample of NH3 in only one energy state?

Consider for definiteness the weak field case. The eigenstates are close to |ψS〉and |ψA〉 with energies

E∓ = E0 ∓√A2 + d2

0E2 � E0 ∓ A ∓ d20E2

2A. (7.26)

The last term is simply the potential energy V∓ of the molecule in the field. We seethat it is different according to the internal quantum state of the molecule.

We prepare a molecular beam. Suppose, first that all molecules are in the state|ψS〉. If these molecules which are “big” classical objects cross a region where thereis an inhomogeneous field, i.e. ∇E2 �= 0, their energy depends on the point wherethey are, therefore a force will act on them

F− = −∇ V− = + d20

2A∇E2. (7.27)

Similarly, if they are all in the state |ψA〉 the force acting on them is

F+ = −∇ V+ = − d20

2A∇E2. (7.28)

The force is different according to the internal state of the molecule. The two forcesactually have opposite signs.

Therefore, the molecular beam will follow a different path according to whetherthe internal state of the molecules is |ψS〉 or |ψA〉. What is quantum mechanical isthe internal state of the molecule, not the motion of its center of gravity.

What happens if themolecule is in a superposition of the states |ψS〉 and |ψA〉?Thisproblem is slightly more complicated. The state of a molecule must be representedby a vector such as we have done above, but the components of this vector dependon the position R of the center of gravity of the molecule,

7.5 Motion of Ammonia Molecule in an Inhomogeneous Field 159

Fig. 7.4 Stabilization of the beam |ψ+〉 and divergence of the beam |ψ−〉 in an electric quadrupolefield (E2 ∝ y2 + z2)

|ψ〉 =(

Ψ1(R, t)Ψ2(R, t)

).

The probabilistic interpretation is that |Ψ1|2 is the probability density of being atpoint R in the internal state |ψS〉, and similarly |Ψ2|2 is the probability density ofbeing at point R in the state |ψA〉.

This can be proven. A similar problem arises in the simpler, case of the electronwith its spin in the Stern–Gerlach experiment, Chap.12. What is quite interesting isthat the respective expectation values of the positions of the wave packets Ψ1(R, t)and Ψ2(R, t) each evolve according to Newton’s laws with, respectively, the poten-tials (7.27) and (7.28). All happens as if different forces acted on these components.

Therefore an inhomogeneous field allows to perform the following operations(Fig. 7.4).

1. One observes a spatial separation of the molecules according to their internalstate. This device is a filter that selects the states |ψS〉 and |ψA〉. Notice also thatin a well chosen field one can select any real linear superposition of |ψS〉 and|ψA〉.

2. Here we face an incredible phenomenon. There are only two quantum trajectorieswhereas classically, if the electric dipole moments were oriented at random thereshould be a continuous set of impacts on a screen.The first time physicists saw such a behavior was in the experiment of Stern andGerlach (Chap.12). They thought it was an experimental proof of the quantizationof trajectories advocated by Bohr.

3. But if we can select the states |ψS〉 and |ψA〉 in space, that means we measuretheir energies. If a molecule arrives above, it has an internal quantum energy ES;if a molecule arrives below, it has energy EA. The respective numbers of themgive the probabilities that in the incoming beam the molecules are in the states|ψS〉 and |ψA〉; we obtain information on the initial state.

4. This apparatus is a concrete example of a quantum mechanical measuring appa-ratus. It transfers internal quantum degrees of freedom into classical space–timeproperties.

160 7 Two-State Systems

5. It is also a device to prepare the molecules in the states |ψS〉 or |ψA〉, or in linearsuperpositions of them (it is sufficient to vary the value of the field E in the device).

Here we can at last speak of the measurement question in quantum mechanics. Wesee that in order tomake the separation, onemust have, for a given velocity, aminimallength, therefore a minimal time. Ameasurement is never instantaneous or pointlike.A measurement always has a finite extension in space and time.

A little calculation (again with spin 1/2) shows that the condition for the mea-surement to be possible is universal. One must have the inequality

ΔE Δt > �, (7.29)

where ΔE is the transverse kinetic energy acquired by the molecules and Δt thetime they spend in the inhomogeneous field zone. This is called the “time-energyuncertainty relation” by analogy with the previous ones. Actually, there is a bigdifference because time t is not an intrinsic observable of a system (even though itcan be measured).

7.5.2 Population Inversion

Actually, in the case of NH3, if one uses a molecular beam and an electric quadrupolefield, the beam |ψA〉 is stable and can be focused, whereas the beam |ψS〉 is unstableand gets dispersed. What is interesting is that by this technique (a molecular beam,a diaphragm, an inhomogeneous field), one can perform what is called a populationinversion. One can select all molecules in the state ΨA. One breaks the (A) − (S)

thermal equilibrium there was in the initial beam.This is only one example of a technique of population inversion. There are many

others.We notice that in order for this population inversion to occur, it is not necessary

to align the axis of a molecule along the field. Whatever the value of E , the ψA beamis focused and ψS is dispersed.

7.6 Reaction to an Oscillating Field, the Maser

In order to make a maser, we will force the molecules in the state |ψA〉 to give backtheir energy 2A by making a transition to the state |ψS〉. We must force them becausespontaneously, these molecules do fall back in the state |ψS〉, but they do it veryslowly. The mean time to do this transition is of the order of one month, which ismuch too long for our purpose.

In quantum physics, a system can absorb a photon of energy hν and reach anexcited state; it can fall back in the initial level by emitting a photon spontaneously.

7.6 Reaction to an Oscillating Field, the Maser 161

However, there exists a third mechanism which was understood by Einstein as soonas 1917, called stimulated emission. If the excited system is placed in an electro-magnetic field properly tuned to the Bohr frequency, it can undergo a transition tothe lower state very rapidly. This is done by a resonance mechanism.

Technically, this exercise is not much more difficult than the previous one, but itis new. We describe the position of the problem and the result. Again the calculationitself is done later on with spin 1/2 and magnetic resonance.

We place the molecule in an oscillating field

E = E0 cosωt,

and we set η = d0E0. The Hamiltonian is

H =(

E0 − A −η cosωt−η cosωt E0 + A

). (7.30)

We see the difference with previous problems. The Hamiltonian H now dependsexplicitly on time. The system is not isolated, one cannot speak of stationary states.We must solve the Schrödinger equation. If we write |ψ(t)〉 = a(t)|ψS〉 + b(t)|ψA〉,we must solve the equation i�(d/dt)|ψ(t)〉 = H |ψ(t)〉 in order to determine theevolution of the system.

If we write the state vector of a molecule as

|ψ(t)〉 =(a(t)b(t)

), (7.31)

the Schrödinger equation is a first-order differential system:

i�a = (E0 − A)a − ηb cosωt (7.32)

i�b = (E0 + A)b − ηa cosωt. (7.33)

Setting a(t) = e−i(E0−A)t/�α(t) and b(t) = e−i(E0+A)t/�β(t), we obtain:

2i α = −ω1 β(ei(ω−ω0)t + e−i(ω+ω0)t

), (7.34)

2i β = −ω1 α(e−i(ω−ω0)t + ei(ω+ω0)t

). (7.35)

This set of coupled equations involves three frequencies:

ω, ω0 = 2A

�, and ω1 = η

�= d0E0

�. (7.36)

Starting with state |ψA〉 at t = 0, a(0) = 0 and b(0) = 1, we want to calculatethe probability |a(t)|2 of finding the system in state |ψS〉 at time t .

162 7 Two-State Systems

Physically, this differential system corresponds to forced oscillations with a reso-nance phenomenon at ω = ω0.1 There is no analytic solution, however, one obtainsa good approximation in the vicinity of the resonance ω ∼ ω0, if we neglect termsthat oscillate rapidly in time e±i(ω+ω0)t . This leads to an exactly soluble problem, forwhich we can give the solution.

Near the resonance, the transition probability PA→S(t) that at time t themoleculesundergo a transition to the state |ψS〉 under the influence of the oscillating field, andthat they release their energy 2A = EA − ES is given by

PA→S(t) � ω21

(ω − ω0)2 + ω21

sin2(√

(ω − ω0)2 + ω21

t

2

). (7.37)

This formula is due toRabi. As can be seen in Figure 7.5a, the probability PA→S(t)oscillates in time between 0 and a maximum value Pmax given by

Pmax = ω21

(ω − ω0)2 + ω21

.

When the frequency ω of the applied field is varied (Fig. 7.5b), the maximum proba-bility Pmax has a characteristic resonant behavior, with a maximum equal to 1 at theresonance, that is, for ω = ω0. The width at half maximum of the resonance curveis ω1.

If the frequency of the external field is tuned in the vicinity of the resonance,|ω − ω0| ω1, practically all the molecules will release their energy 2A at the timeT = π/ω1. This energy emission occurs in the form of an electromagnetic radiationof frequency ν = ω0/2π = 24GHz. This is called stimulated emission. The smallerω1, the narrower the resonance curve of Fig. 7.5b is, and the greater the time to obtainthat emission.

Fig. 7.5 Rabi oscillations:a Probability of finding themolecule in the state |ψS〉 asa function of time.b Resonance curve showingthe maximum transitionprobability as a function ofthe external field frequency ω

1Actually, there are two resonances at ω = ω0 and ω = −ω0, but the two values are equivalent forwhat concerns us here.

7.6 Reaction to an Oscillating Field, the Maser 163

Therefore, if spontaneously the molecules decay in ≈1 month, with this phenom-enon one can force them to do it quickly, and to beat the natural time constant ofthe process. This happens whatever the intensity of the exciting field, provided it isproperly tuned.

7.7 Principle and Applications of the Maser

In practice, a NH3 maser (microwave amplification by stimulated emission of radia-tion),whichwas invented byTownes in 1951,works in the followingway, representedin Fig. 7.6. One starts with a molecular beam of velocity v coming from an oven ata temperature of 100K. One then separates the molecules in the state |ψA〉 by anelectric quadrupole field. The beam then enters a high-frequency cavity where thereis a field E0 cosωt andwhose length L is adjusted2 so that L/v = T = (2n+1)π/ω1.The outgoing molecules are in the state |ψS〉 and they have released their energy 2Ain the cavity in the form of an electromagnetic radiation of frequency ω0. In the setupmade by Townes in 1951, a beam of 1014 molecules s−1 gave a power of 10−9W ona single frequency 24GHz with a width of 3000Hz, a quality factor of Q = 107.

This maser effect, which is based on two elements, population inversion (whichbreaks the thermodynamic equilibrium) and stimulated emission which generatesthe quick and coherent transition, was discovered by Townes in 1951 (1964 Nobelprize). It has numerous applications.

Lasers (light) work in a different frequency range, with a more sophisticatedtechnique of population inversion, optical pumping. They are treated as three- orfour-state systems. There is no basic difference from what we have seen otherwise.

There are essentially three types of applications of such devices.

Fig. 7.6 Sketch of a NH3 maser device

2It is not necessary that L be exactly adjusted to the correct value; the transition probability isappreciable if one doesn’t have the bad luck of falling on the unfavorable values T = 2nπ/ω1. Inpractice a feedback device adjusts the length of the cavity so that the signal is maximum.

164 7 Two-State Systems

7.7.1 Amplifiers

One can amplify in a selective way, and without any background noise, a very weaksignal. Hence there are very important applications in radioastronomy in order tostudy the interstellar medium, as we show in Chap. 13.

Initially, Townes used a molecular beam of 1014 molecules per second. Nowadaysone uses solid-state masers, such as a crystal of ruby (a crystal of Al2O3 with Cr+3

ions of a concentration of ∼0.05%). This allows gains of 36dB. Ruby masers wereused in 1965 by A. Penzias and R.W. Wilson when they discovered the cosmicbackground radiation at 3K, one of the major observational proofs in favor of thebig bang.

The mechanism of the amplification is simple if one visualizes it in terms ofphotons. The population inversion is achieved on a macroscopic number of atoms,such as in a crystal of ruby.As shown in Fig. 7.7, a photon coming from the interstellarmediumwill induce the transition of a first atom by stimulated emission. This resultsin two tuned photons both of which can in turn generate stimulated emission on otheratoms, leading to four photons, and so on. The chain reaction yields the amplification.

The device presented in Fig. 7.6 is not an amplifier but an emitter or an oscillator,because the outgoing intensity is independent of the incoming intensity. In order to seethe system act as an amplifier, one must calculate its response to an incoherent signalspread out in frequency. Our calculation only concerns a monochromatic coherentfield.

7.7.2 Oscillators

In Fig. 7.6, we actually see an oscillator. A field as small as one wishes can be self-produced in the cavity; one evacuates the electromagneticwave of frequency 24GHz,which results in a very monochromatic output wave.

Fig. 7.7 Chain reaction ofa tuned photon on a set ofexcited atoms

7.7 Principle and Applications of the Maser 165

7.7.3 Atomic Clocks and the GPS

As oscillators, masers have allowed the construction of atomic clocks, which arethe present standard time-keeping devices. Such devices use, for instance, a jet ofcesium atoms (isotope 133Cs). The ground state of cesium is split by the hyperfineinteraction, which we describe later on. The physical origin of this effect is magneticinstead of being electric. It results from the magnetic interaction of the nuclear spinand the spin of the valence electron, and a splitting in two levels |g1〉 and |g2〉, ofenergies E1 and E2. Since 1967, the definition of the hertz is given by the fact thatthe 133Cs hyperfine frequency ν12 = (E2 − E1)/h is equal to 9,192,631,770Hz. Itis a considerable improvement compared to astronomical definitions.

In order to transform the above device into an atomic clock, we prepare a jet ofcesium atoms in the state |g1〉. These atoms cross a cavity inside which one injectsan electromagnetic wave of frequency ν, and one adjusts ν in order to maximizethe number of outgoing atoms in the state |g2〉. The frequency ν is thus locked atν12. One can measure a time interval by simply counting the number of oscillationsduring that time interval. The principle is shown in Fig. 7.8. Atomic clocks werefirst developed in the 1950s by Zacharias at the National Institute of Standards andTechnology (NIST). The clock NIST-7 together with ten others in the world thusserves as a “time keeper.”

Present clocks have a relative accuracy of 10−15, so that the time standard isthe most accurate of all standards. This explains that, in 1983, the definition of themeter was changed by fixing the value of the velocity of light to an integer value:c = 299,792,458ms−1.

Such accuracy is essential both in applied physics and in a variety of tech-nical devices, such as positioning and navigation, be it terrestrial with the GPSsystem (Fig. 7.9), or on satellites. They are commonly used to guide ships, they

Fig. 7.8 Sketch of an atomic clock (Courtesy of Patrick Bouchareine)

166 7 Two-State Systems

Fig. 7.9 Satellite of the GPSNavstar system. There are 24of them orbiting around theEarth

equip airplanes. In 1990, they appeared in automobiles where they are commonlyused by now: almost all common automobiles have a GPS guiding system for dailypurposes. They exist in cell phones, inmountain climbing safety devices, in elaboratewrist watches etc.

Perhaps more amazing is their use in space navigation. The recent mission of theCassini spacecraft, launched in 2000, saw, on January 14 2005, the Huygens probeland on the planet Titan. Titan is a satellite of Saturn that is similar to the Earthbecause of its atmosphere, though it is much colder (in some sense it is an Earth ina refrigerator). This has yielded an impressive number of results and will continueto do so. In the final phase, the probe had to guide itself by its own means and landproperly (the time of propagation of a signal to the earth is one hour). (See http://www.esa.int/SPECIALS/Cassini-Huygens/index.html.)

7.7.4 Tests of Relativity

As soon as atomic clocks appeared, people thought of directly verifying the predic-tions of relativity on time, in particular the twin paradox. In this “paradox” the twinwho has traveled comes back younger. His proper time is shorter than the time of theother twin who stayed on Earth. His clock must therefore be late compared to theclocks that remained on Earth,

t = t0√1 − v2

c2

, Δt = t − t0 � t0v2

2c2, (7.38)

7.7 Principle and Applications of the Maser 167

where t is the time measured by the twin who stayed on Earth, and t0 the time of thetwin who traveled.

In 1975, Alley3 sent 4 atomic cesium clocks for a 15-h flight on an airplane andcompared them with 11 clocks that stayed on Earth. Those clocks should have beenlate. Not at all; they they had gained 47 × 10−9 s!

In fact, there is an effect of the opposite sign due to gravitation and predicted bygeneral relativity:

(Δt

t

)

GR

= ΔΦ

c2,

(Δt

t

)

SR

= − v2

2c2,

where Φ is the gravitational potential. In the conditions of flight of Alley’s clocks,the result is the sum of

(Δt

t

)

GR

= 53 × 10−9 s,

(Δt

t

)

SR

= −6 × 10−9 s.

The predictions of relativity are verified here with an accuracy of 1%.These effects were measured in 1979 by R.F.C. Vessot and collaborators.4 A

hydrogen maser was sent to an altitude of 10,000km by a Scout rocket, and thevariation in time of its frequency was made as the gravitational potential increased(algebraically). There are many corrections, in particular due to the Doppler effectof the spacecraft and to the Earth’s rotation. It was possible to test the predictionsof general relativity on the variation of the pace of a clock as a function of thegravitational field with a relative accuracy of 7×10−5. This was done by comparisonwith atomic clocks, or masers, on Earth. Up to now, it has been one of the bestverifications of general relativity. The recording of the beats between the embarkedmaser and a test maser on Earth is shown in Fig. 7.10. (These are actually beatsbetween signals that are first recorded and then treated in order to take into accountall physical corrections.) A simple calculation shows that with this accuracy, onemust specify the exact altitude of a clock up to one meter or so, in order to knowone’s time. Time is longer at the top of a conference hall than at the bottom. Studentsknow this very well.

3C. Alley, “Proper Time Experiments in Gravitational Fields with Atomic Clocks, Aircraft, andLaser Light Pulses,” in Quantum Optics, Experimental Gravity, and Measurement Theory, eds.Pierre Meystre and Marlan O. Scully, Proceedings Conf. BadWindsheim 1981, 1983 Plenum PressNew York, ISBN 0-306-41354-X, pg 363427.4R.F.C. Vessot, M.W. Levine, E.M. Mattison, E.L. Blomberg, T.E. Hoffman, G.U. Nystrom, B.F.Farrel, R. Decher, P.B. Eby, C.R. Baugher, J.W. Watts, D.L. Teuber, and F.D. Wills, “Test of Rela-tivistic Gravitation with a Space-Borne Hydrogen Maser”, Phys. Rev. Lett. 45, 2081, (1980).

168 7 Two-State Systems

(a)

(b)

(c)

(d)

(e)

Fig. 7.10 Beats between a maser onboard the spacecraft launched by a Scout rocket and a maseron Earth at various instants in GMT. a Signal of the dipole antenna; the pointer shows the delicatemoment when the spacecraft separated from the rocket (it was important that the maser onboard hadnot been damaged by vibrations during takeoff). During this first phase, the special relativity effectdue to the velocity is dominant. b Time interval of “zero beat” during ascent when the velocityeffect and the gravitational effect, of opposite signs, cancel each other. c Beat at the apogee, entirelydue to the gravitational effect of general relativity. Its frequency is 0.9Hz. d Zero beat at descent.e End of the experiment. The spacecraft enters the atmosphere and the maser onboard ceases towork (Courtesy of R.F.C. Vessot)

7.8 Exercises 169

7.8 Exercises

1. Linear three-atom molecule

We consider the states of an electron in a linear three-atom molecule (such as N3 orC3) with equally spaced atoms L ,C, R at a distance d from one another.

Let |ψL〉, |ψC 〉 and |ψR〉 be the eigenstates of an observable B, corresponding toan electron localized respectively in the vicinity of the atoms L , C and R:

B|ψL〉 = −d|ψL〉; B|ψC 〉 = 0; B|ψR〉 = +d|ψR〉.

In the basis {|ψL〉, |ψC 〉, |ψR〉}, the Hamiltonian of the system is represented by thematrix:

H =⎛

⎝E0 −a 0−a E0 −a0 −a E0

⎞

⎠ a > 0.

a. Calculate the energy levels and eigenstates of H .b. Consider the ground state; what are the probabilities to find the electron in the

vicinity of L , C and R?c. Suppose the electron is in the state |ψL〉, and we measure its energy. What values

can we find, with what probabilities? Calculate 〈E〉 and ΔE in this state.

2. Crystallized Violet and Malachite Green

The active principle of the dye 42555 (called “Crystallized Violet”) is the organicmonovalent cation C[C6H4N(CH3)2]+3 . The skeleton of this ion ismade of three iden-tical branches (Fig. 7.11). The electronic deficit responsible for the positive chargecan be taken from either of these three branches. One can treat the electronic stateof this ion as a three-state system. The Hamiltonian H is not diagonal in the basis{|1〉, |2〉, |3〉} (which we assume orthonormal) because of tunnelling between theseclassical configurations.

Fig. 7.11 The three possible configurations of the molecule

170 7 Two-State Systems

a. We work in the basis {|1〉, |2〉, |3〉} corresponding to “classical configurations”.We choose the origin of energies in order to have 〈1|H |1〉 = 〈2|H |2〉 =〈3|H |3〉 = 0. We set 〈1|H |2〉 = 〈2|H |3〉 = 〈3|H |1〉 = −A where A is realpositive (A > 0).Write the matrix H in this basis. Comparing with the case of the ammonia mole-cule NH3, justify briefly the choice of this matrix.

b. Consider the states |φ1〉 = (|1〉 + |2〉 + |3〉)/√3 and |φ2〉 = (|2〉 − |3〉)/√2.Calculate the expectation value 〈E〉 and the dispersionΔE in each of these states.Interpret the result.

c. Determine the energy levels of the system. Give a corresponding orthonormaleigenbasis. Is this basis unique?

d. The value of A is A ≈ 0.75eV. Why is this ion violet?

We recall that the colors of the spectrum of natural light are, for increasingenergies (E = hc/λ), red (from ≈1.65 to 2.0eV); orange (from ≈ 2.0 to2.1eV); yellow (from ≈2.1 to 2.3eV); green (from ≈2.3 to 2.55eV); blue(from ≈2.55 to 2.65eV); violet (from ≈ 2.65 to 3.1eV). The main couples of“complementary colors” which produce white light when they are associated,are yellow-violet, red-green and blue-orange.

e. One replaces the N(CH3)2 group of the upper branch by a hydrogen atom. Weassume that the sole effect of this substitution is to increase 〈1|H |1〉 by an amountΔ > 0, and that it leaves the other matrix elements of H unchanged.

(i) Show that A is still an eigenvalue of the Hamiltonian. What are the otherenergy levels of this new system?

(ii) How do they behave in the limits Δ A and Δ � A?

f. This modified ion (coloring 42,000 “malachite green”) absorbs light of twowave-lengths: 620 and 450nm. Calculate Δ and comment on the agreement betweentheory and experiment.One can use hc ≈ 1240eV.nm.

7.9 Problem: Neutrino Oscillations

We consider here a quantum oscillation effect that is completely contrary to commonintuition. It is an oscillation between two or more types of pointlike elementaryparticles. The effect consists of the fact that if we prepare a particle in the vacuum, itcan transform spontaneously and periodically into one or several types of differentelementary particles. In other words, we face a phenomenon of “successive periodichermaphroditism” between elementary particles.5

5Successive hermaphroditism exists in zoology, for instance, in several fish species.

7.9 Problem: Neutrino Oscillations 171

Lepton Families

In β decay or, more generally, inWeak interactions, the electron is always associatedwith a neutral particle, the neutrino νe. There exists in nature another particle, theμ lepton, or muon, whose physical properties seem completely analogous to thoseof the electron, except for its mass mμ � 200 me. The muon has the same Weakinteractions as the electron, but it is associated to a different neutrino, the νμ.

A neutrino beam produced in an accelerator can interact with a neutron (n) in anucleus and give rise to the reactions

νe + n → p + e and νμ + n → p + μ, (7.39)

whereas the reactions νe + n → p + μ or νμ + n → p + e are never observed. Thereactions (7.39) are used in practice in order to detect neutrinos.

Similarly, a π− meson can decay via the modes

π− → μ + νμ (dominant mode) and π− → e + νe, (7.40)

whereas π− → μ + νe or π− → e + νμ are never observed. This is how one canproduce neutrinos abundantly (it is easy to produce π mesons). In (7.40) we haveintroduced the antiparticles νμ et νe. There is a (quasi) strict symmetry between par-ticles and their antiparticles, so that, in the sameway as the electron is associatedwiththe neutrino νe, the antielectron, or positron, e+ is associated with the antineutrinoνe. One observes the “charge-conjugate” reactions of (7.39) and (7.40)

νe + p → n + e+ , νμ + p → n + μ+ and π+ → μ+ + νμ. (7.41)

In all what follows, what we will say about neutrinos holds symmetrically for anti-neutrinos.

The experimental discovery of the electron antineutrino is due to Cowan andReines in 1956. They operated near the Savannah River nuclear reactor, and theyobserved the reaction

νe + p → n + e+, and not νe + n → p + e. (7.42)

The νe antineutrinos came from the many beta decays of the type n → p + e + νin the reactor. (The second reaction in (7.42) does not occur because the electron isassociated with its neutrino and not with the antineutrino νe which is the partner ofthe antielectron e+.)

In 1975, a third lepton, the τ , was discovered. It is much more massive, mμ �3500 me, it is associated with its own neutrino ντ , and it obeys the same physicallaws as the two lighter leptons, except for mass effects. Since the 1990s, the experi-mental measurements at the LEP colliding ring in CERN have shown that these threeneutrinos νe, νμ, ντ (and their antiparticles) are the only ones of their kinds (at leastfor masses less that 100GeV/c2).

172 7 Two-State Systems

For a long time, physicists believed that neutrinos were zero-mass particles, asis the photon. In any case, their masses (multiplied by c2) are considerably smallerthan the energies involved in experiments where they are observed. Therefore, manyexperimental limits on these masses are consistent with zero. However, both theo-retical and cosmological arguments suggested that this might not be the case. Theproof that neutrino masses are not all zero is a great discovery of the 1990s.

For this discovery, the 2002 Nobel prize was awarded to Raymond Davis Jr. apioneer of this physics, and toMasatoshi Koshiba who led experiments performed inJapan with detectors particularly well adapted to that kind of physics. The Japaneseexperiments have gathered an impressive amount of results.

The 2015 Nobel prize for physics was awarded to Takaaki Kajita (Super-Kamiokande) and Arthur B. McDonald (SNO) “for the discovery of neutrino oscil-lations, which shows that neutrinos have mass”.

In the present study, we show how the mass differences of neutrinos can bemeasured by a quantum oscillation effect. The idea is that the “flavor” neutrinosνe, νμ and ντ , which are produced or detected experimentally are not eigenstates ofthe mass, but rather linear combinations of mass eigenstates ν1, ν2, ν3, with massesm1,m2,m3.

The neutrinos observed on earth have various origins. They can be produced inaccelerators, in nuclear reactors, and also in the atmosphere by cosmic rays, or inthermonuclear reaction inside stars, in particular the core of the sun, and in supernovaeexplosions.

7.9.1 Mechanism of the Oscillations; Reactor Neutrinos

In this first part, we consider oscillations between two types of neutrinos, the νe andthe νμ. This simple case will allow us to understand the underlying physics of thegeneral case. We will analyze the data obtained with nuclear reactors. The averageenergy of the (anti-)neutrinos produced in reactors is E = 4MeV, with a dispersionof the same order.

In all what follows, we will assume that if m is the neutrino mass and p and E itsmomentum and energy, the mass is so small that the energy of a neutrino of mass mand momentum p is

E =√p2c2 + m2c4 � pc + m2c4

2pc, (7.43)

and that the neutrino propagates to very good approximation at the velocity oflight c.

Let H be the Hamiltonian of a free neutrino of momentum p, which we assumeto be well defined. We note |ν1〉 and |ν2〉 the two eigenstates of H :

7.9 Problem: Neutrino Oscillations 173

H |ν j 〉 = E j |ν j 〉, E j = pc + m2j c

4

2pc, j = 1, 2.

m1 and m2 are the respective masses of the states |ν1〉 and |ν2〉, and we assumem1 �= m2.

The oscillations of freely propagating neutrinos come from the following quantumeffect. If the physical states of the neutrinos which are produced (reactions (7.40))or detected (reactions (7.39)) are not |ν1〉 and |ν2〉, but linear combinations of these:

|νe〉 = |ν1〉 cos θ + |ν2〉 sin θ, |νμ〉 = −|ν1〉 sin θ + |ν2〉 cos θ (7.44)

where θ is a mixing angle to be determined, these linear combination of energyeigenstates oscillate in time and this leads to measurable effects.

7.9.1 At time t = 0, one produces a neutrino of momentum p in the state |νe〉.Calculate the state |ν(t)〉 at time t in terms of |ν1〉 and |ν2〉.7.9.2 What is the probability Pe for this neutrino to be detected in the state |νe〉at time t? The result will be expressed in terms of the mixing angle θ and of theoscillation length L

L = 4π�p

|Δm2| c2 , Δm2 = m21 − m2

2. (7.45)

7.9.3 Calculate the oscillation length L for an energy E � pc = 4MeV and a massdifference Δm2c4 = 10−4 eV2.

7.9.4 One measures the neutrino fluxes with a detector located at a distance � fromthe production area. Express the probability Pe as a function of the distance � = ct .

7.9.5 The mass of the muon satisfies mμc2 = 106MeV. Conclude that in such anexperiment one cannot detect muon neutrinos νμ with the reaction (7.39). We recallthat mpc2 = 938.27MeV and mnc2 = 939.57MeV.

7.9.6 The detectors measure neutrino fluxes with an accuracy of ∼10%.

(a) Assuming Δm2c4 = 10−4 eV2, determine the minimal distance �min where toput a detector in order to detect an oscillation effect. For this calculation, assumethe mixing in (7.44) is maximum, i.e. θ = π/4.

(b) How does �min change if the mixing is not maximum?

7.9.7 Several experiments on neutrinos produced by nuclear energy plants havebeen performed in Chooz and in Bugey in France. The most recent data comes fromthe KamLAND collaboration, in Japan. The results are given on Fig. 7.12.

(a) Explain the results of Fig. 7.12, except that of KamLAND.

174 7 Two-State Systems

Fig. 7.12 Ratio between the numbers of observed electron neutrinos and those expected in theabsence of oscillations as a function of the distance � to the reactor

(b) TheKamLAND experiment, which started operating in 2002,6 consisted inmea-suring the neutrinos coming from all the (numerous) reactors in Japan and neigh-boring countries, which amounts to taking an average distance of � = 180km.Putting together that data and the results of numerous experiments performedon solar neutrinos, the physicists of Kamland come to the following results:

|Δm2| c4 = 7.1 (± 0.4) × 10−5 eV2, tan2 θ = 0.45 (± 0.02). (7.46)

Show that these values are consistent with the result Pe = 0.61 (± 0.10) ofFig. 7.12.

7.9.2 Oscillations of Three Species; Atmospheric Neutrinos

We now consider the general formalism with three neutrino species. We denote|να〉, α = e,μ, τ the “flavor” neutrinos and |νi 〉, i = 1, 2, 3 the mass eigenstates.These two bases are related to one another by the Maki–Nagawaka–Sakata (MNS)matrix U ,

|να〉 =3∑

i=1

Uαi |νi 〉, U =⎛

⎝Ue1 Ue2 Ue3

Uμ1 Uμ2 Uμ3

Uτ1 Uτ2 Uτ3

⎞

⎠ (7.47)

This matrix is unitary (∑

i U∗βiUαi = δαβ) and it can be written as:

6See for instance Atsuto Suzuki, Proc. of Nobel Symposium 129, Physica Scripta T121 (2005).

7.9 Problem: Neutrino Oscillations 175

U =⎛

⎝1 0 00 c23 s230 −s23 c23

⎞

⎠

⎛

⎝c13 0 s13 e−iδ

0 1 0−s13 eiδ 0 c13

⎞

⎠

⎛

⎝c12 s12 0

−s12 c12 00 0 1

⎞

⎠

where ci j = cos θi j and si j = sin θi j . The complete experimental solution of theproblem would consist in measuring the three mixing angles θ12, θ23, θ13, the phaseδ, and the three masses m1, m2, m3. We consider situations such that (7.43) is stillvalid.

7.9.1 At time t = 0 a neutrino is produced with momentum p in the state |ν(0)〉 =|να〉. Express, in terms of the matrix elements Uαi , its state at a later time t . Writethe probability Pα→β(t) to observe a neutrino of flavor β at time t .

7.9.2 We define the oscillation lengths at an energy E � pc by:

Li j = 4π�p

|Δm2i j |c2

, Δm2i j = m2

i − m2j . (7.48)

Notice that there are only two independent oscillation lengths sinceΔm212 +Δm2

23 +Δm2

31 = 0. For neutrinos of energy E = 4GeV, calculate the oscillation lengths L12

and L23. We will choose for |Δm212| the result given in (7.46), and we will choose

|Δm223| c4 = 2.5 × 10−3 eV2, a value which will be justified later on.

7.9.3 The neutrino counters have an accuracy of the order of 10% and the energyis E = 4GeV. Above which distances �12 and �23 of the production point of theneutrinos can one hope to detect oscillations coming from the superpositions 1 ↔ 2and 2 ↔ 3?

7.9.4 The Super-Kamiokande experiment, performed in 1998, consists in detecting“atmospheric” neutrinos. Such neutrinos are produced in the collision of high energycosmic rays with nuclei in the atmosphere at high altitudes. In a series of reactions,π± mesons are produced abundantly, and they decay through the chain:

π− → μ− + νμ followed by μ− → e− + νe + νμ, (7.49)

and an analogous chain for π+ mesons. The neutrino fluxes are detected in an under-ground detector by the reactions (7.39) and (7.41).

To simplify things, we assume that all muons decay before reaching the surfaceof the Earth. Deduce that, in the absence of neutrino oscillations, the expected ratiobetween electron and muon neutrinos

Rμ/e = N (νμ) + N (νμ)

N (νe) + N (νe)

would be equal to 2.

176 7 Two-State Systems

7.9.5 The corrections to the ratio Rμ/e due to the fact that part of the muons reachthe ground can be calculated accurately. Once this correction is made, one finds, bycomparing the measured and calculated values for Rμ/e

(Rμ/e)measured

(Rμ/e)calculated= 0.64 (± 0.05).

In order to explain this relative decrease of the number of νμ’s, one can think ofoscillations of the types νμ � νe and νμ � ντ . The Super-Kamiokande experimentconsists in varying the time of flight of the neutrinos by measuring selectively thedirection where they come from, as indicated on Fig. 7.13. The neutrinos comingfrom above (cosα ∼ 1) have traveled a distance equal to the atmospheric height plusthe depth of the detector, while those coming from the bottom (cosα ∼ −1) havecrossed the diameter of the Earth (13,400km). Given the weakness of the interactionof neutrinos with matter, one can consider that the neutrinos propagate freely on ameasurable distance between a few tens of km and 13,400km.

The neutrino energies are typically 4GeV in this experiment. Can one observe aνe � νμ oscillation of the type studied in the first part?

7.9.6 The angular distributions of the νe and the νμ are represented on Fig. 7.13,together with the distributions one would observe in the absence of oscillations.Explain why this data is compatible with the fact that one observes a νμ � ντ

oscillation, no νe � ντ oscillation, and no νe � νμ oscillation.

7.9.7 In view of the above results, we assume that there is only a two-neutrinooscillation phenomenon: νμ � ντ in such an observation. We therefore use the sameformalism as in the first part, except that we change the names of particles.

By comparing the muon neutrino flux coming from above and from below, givean estimate of the mixing angle θ23. In order to take into account the large energydispersion of cosmic rays, and therefore of atmospheric neutrinos, we replace theoscillating factor sin2(π�/L23) by its mean value 1/2 if � � L23.

The complete results published by the Super-Kamiokande experiment are

|Δm223| c4 = 2.5 × 10−3 eV2, θ23 = π/4, θ13 = 0.

Do they agree with the above considerations?

7.9.3 Solution

Mechanism of the Oscillations: Reactor Neutrinos

7.3.1 Initially, the neutrino state is |ν(0)〉 = |νe〉 = |ν1〉 cos θ+|ν2〉 sin θ. Therefore,we have at time t

7.9 Problem: Neutrino Oscillations 177

Fig. 7.13 Left production of atmospheric neutrinos in collisions of cosmic rays with terrestrialatmospheric nuclei. The underground detector measures the flux of electron and muon neutrinos asa function of the zenithal angle α. Right number of atmospheric neutrinos detected in the Super-Kamiokande experiment as a function of the zenithal angle (this picture is drawn after K. Tanyaka,XXII Physics in Collisions Conference, Stanford 2002)

|ν(t)〉 = |ν1〉 cos θ e−iE1t/� + |ν2〉 sin θ e−iE2t/�.

7.3.2 The probability to find this neutrino in the state |νe〉 at time t is

Pe(t) = |〈νe|ν(t)〉|2 = ∣∣cos2 θ e−iE1t/� + sin2 θ e−iE2t/�∣∣2 ,

which gives, after a simple calculation:

Pe(t) = 1 − sin2(2θ) sin2(

(E1 − E2)t

2�

).

We have E1 − E2 = (m21 − m2

2)c4/(2pc). Defining the oscillation length by

L = 4π�p/(|Δm2| c2), we obtain

Pe(t) = 1 − sin2(2θ) sin2(

πct

L

).

178 7 Two-State Systems

7.3.3 For an energy E = pc = 4MeV and a mass difference Δm2c4 = 10−4 eV2,we obtain an oscillation length L = 100km.

7.3.4 The time of flight is t = �/c. The probability Pe(�) is therefore

Pe(�) = 1 − sin2(2θ) sin2(

π�

L

). (7.50)

7.3.5 A νμ energy of only 4MeV is below the threshold of the reaction νμ + n →p+μ. Therefore this reaction does not occur with reactor neutrinos, and one cannotmeasure the νμ flux.

7.3.6 In order to detect a significant decrease in the neutrino flux νe, we must have

sin2(2θ) sin2(

π�

L

)> 0.1.

(a) For themaximummixing θ = π/4, i.e. sin2(2θ) = 1, this impliesπ�/L > 0.32or � > L/10. For E = 4MeV and Δm2c4 = 10−4 eV2, one finds � > 10km. Thetypical distances necessary to observe this phenomenon are of the order of a fractionof the oscillation length.

(b) If the mixing is not maximum, one must operate at distances � greater thanL/10. Note that is the mixing angle is too small, (sin2(2θ) < 0.1 i.e. θ < π/10),the oscillation amplitude is too weak to be detected, whatever the distance �. In thatcase, one must improve the detection efficiency to obtain a positive conclusion.

7.3.7 (a) In all experiments except KamLAND, the distance is smaller than 1km.Therefore, in all of these experiments |1 − Pe| ≤ 10−3. The oscillation effect is notdetectable if the estimate |Δm2| c4 ∼ 10−4 eV2 is correct.

(b) For |Δm2| c4 = 7.1 × 10−5 eV2, tan2 θ = 0.45 and � = 180km, we obtainPe = 0.50 which agrees with the measurement. The theoretical prediction takinginto account the effects due to the dispersion in energy is drawn on Fig. 7.14. We seeincidentally how important it is to control error bars in such an experiment.

Oscillations of Three Species: Atmospheric Neutrinos

7.3.1 At time t = 0, we have:

|ν(0)〉 = |να〉 =∑

j

Uα j |ν j 〉,

and therefore at time t :

|ν(t)〉 = e−ipct/�∑

j

Uα j e−im2

j c3t/(2�p) |ν j 〉 .

7.9 Problem: Neutrino Oscillations 179

Fig. 7.14 Experimental points of Fig. 7.12 and the theoretical prediction of (7.50) (sinusoidalfunction damped by energy dispersion affects). This curve is a best fit of solar neutrino data. Wenotice that the KamLAND data point corresponds to the second oscillation of the curve

We conclude that the probability Pα→β to observe a neutrino of flavor β at time t is

Pα→β(t) = ∣∣〈νβ |ν(t)〉∣∣2 =∣∣∣∣∣∣

∑

j

U ∗β j Uα j e

−im2j c

3t/(2�p)

∣∣∣∣∣∣

2

.

7.3.2 We have Li j = 4π�E/(|Δm2i j | c3). The oscillation lengths are proportional

to the energy. We can use the result of question 1.3, with a conversion factor of 1000to go from 4MeV to 4GeV.

• For |Δm212| c4 = 7.1 × 10−5eV2, we find L12 = 140,000km.

• For |Δm223| c4 = 2.5 × 10−3eV2, we find L23 = 4000km.

7.3.3 We want to know the minimal distance necessary in order to observe oscil-lations. We assume that both mixing angles θ12 and θ23 are equal to π/4, whichcorresponds to maximum mixing. We saw in the first part that if this mixing is notmaximum, the visibility of the oscillations is reduced and that the distance which isnecessary to observe the oscillation phenomenon is increased.

By resuming the argument of the first part, we find that the modification ofthe neutrino flux of a given species is detectable beyond a distance �i j such thatsin2(π�i j/Li j ) ≥ 0.1 i.e. �i j ≥ Li j/10. This corresponds to �12 ≥ 14,000km forthe oscillation resulting from the superposition 1 ↔ 2, and �23 ≥ 400km for theoscillation resulting from the superposition 2 ↔ 3.

7.3.4 The factor of 2 between the expectedmuon and electron neutrino fluxes comesfrom a simple counting. Each particle π− (resp. π+) gives rise to a νμ, a νμ and aνe (resp. a νμ, a νμ and a νe). In practice, part of the muons reach the ground before

180 7 Two-State Systems

decaying, which modifies this ratio. Naturally, this effect is taken into account in anaccurate treatment of the data.

7.3.5 For an energy of 4GeV, we have found that the minimum distance to observethe oscillation resulting from the 1 ↔ 2 superposition is 14,000km. We thereforeremark that the oscillations νe � νμ, corresponding to the mixing 1 ↔ 2 which westudied in the first part cannot be observed at terrestrial distances. At such energies(4GeV) and for evolution times corresponding at most to the diameter of the Earth(0.04 s), the energy difference E1 − E2 and the oscillations that it induces can beneglected.

However, if the estimate |Δm223| c4 > 10−3eV2 is correct, the terrestrial distance

scales allow in principle to observe oscillations resulting from 2 ↔ 3 and 1 ↔ 3superpositions, which correspond to νμ � ντ or νe � ντ .

7.3.6 The angular distribution (therefore the distribution in �) observed for the νe’sdoes not show any deviation from the prediction made without any oscillation. How-ever, there is a clear indication for νμ oscillations: there is a deficit of muon neutrinoscoming from below, i.e. those which have had a long time to evolve.

The deficit in muon neutrinos is not due to the oscillation νe � νμ of the first part.Indeed, we have seen in the previous question that this oscillation is negligible at timescales of interest. The experimental data of Fig. 7.13 confirm this observation. Thedeficit in muon neutrinos coming from below is not accompanied with an increaseof electron neutrinos. The effect can only be due to a νμ � ντ oscillation.7

No oscillation νe � ντ appears in the data. In the framework of the present model,this is interpreted as the signature of a very small (if not zero) θ13 mixing angle.

7.3.7 Going back to the probability (7.50) written in question 1.4, the probabilityfor an atmospheric muon neutrino νμ to be detected as a νμ is:

P(�) = 1 − sin2(2θ23) sin2(

π�

L23

), (7.51)

where the averaging is performed on the energy distribution of the neutrino. If wemeasure the neutrino flux coming from the top, we have � L23, which givesPtop = 1. If the neutrino comes from the bottom, the term sin2(π�/L23) averages to1/2 and we find:

Pbottom = 1 − 1

2sin2(2θ23).

The experimental data indicate that for −1 ≤ cosα ≤ −0.5, Pbottom = 1/2. Thedistribution is very flat at a value of 100 events, i.e. half of the top value (200 events).

We deduce that sin2(2θ23) = 1, i.e. θ23 = π/4 and a maximum mixing angle forνμ � ντ . The results published by Super-Kamiokande fully agree with this analysis.

7For completeness, physicists have also examined the possibility of a “sterile” neutrino oscillation,i.e. an oscillation with a neutrino which would have no detectable interaction with matter.

7.9 Problem: Neutrino Oscillations 181

7.9.4 Comments

The difficulty of such experiments comes from the smallness of the neutrino interac-tion cross sections with matter. The detectors are enormous water tanks, where aboutten events per day are observed (for instance νe + p → e+ + n). The “accuracy” ofa detector comes mainly from the statistics, i.e. the total number of events observed.

An amazing feature of neutrino oscillations is that they are genuine quantumeffects whose wavelengths are macroscopic, owing to the smallness of the neutrinomasses or, equivalently, energies, and therefore take place over distances of the sizeof the earth.

The unexpected phenomenon in this physics is that quantum oscillations are notrestricted to “geometrical” structures such as the NH3 molecule. Oscillations appearbetween elementary pointlike particles of different species!

We notice that in the case of neutrinos, quantum oscillation phenomena can onlybe observed at macroscopic distances because of the smallness of mass differences.It is amusing that phenomena that are so far from our usual intuition can only beobserved at very large distance scales.

In 1998, the first undoubted observation of the oscillation ντ � νμ was announcedin Japan by the Super-Kamiokande experiment (Y. Fukuda et al., Phys. Rev. Lett.81, 1562 (1998)). This experiment uses a detector containing 50,000 tons of water,inside which 11,500 photomutipliers detect the Cherenkov light of the electrons ormuons produced. About 60 ντ ’s were also detected, but this figure is too small togive further information. An accelerator experiment confirmed the results afterwards(K2K collaboration, Phys. Rev. Lett. 90, 041801 (2003)).The KamLAND experiment is a collaboration between Japanese, American and

Chinese physicists. The detector is a 1000m3 volume filled with liquid scintil-lator (an organic liquid with global formula C-H). The name means KAMiokaLiquid scintillator Anti-Neutrino Detector. Reference: KamLAND Collaboration,Phys. Rev. Lett. 90, 021802 (2003); see also http://kamland.lbl.gov/.

Very many experimental results come from solar neutrinos, which we have notdealt with here. This problem is extremely important, but somewhat too complex forour purpose. The pioneering work is due to Davis in his celebrated paper of 1964(R. Davis Jr., Phys. Rev Lett. 13, 303 (1964)).

Davis operated on a 37Cl perchlorethylene detector and counted the number of37Ar atoms produced. In 25years, his overall statistics has been 2200 events, i.e.one atom every 3days! This was much smaller than what was expected from solar-model calculations. At first, it seemed that some neutrinos were being lost duringtheir travel between the Sun and the Earth. In 1991, the SAGE experiment donewith Gallium confirmed the deficit (A. I. Abasov et al., Phys. Rev Lett. 67, 3332(1991) and J. N. Abdurashitov et al., Phys. Rev Lett. 83, 4686 (1999)). In 1992, theGALLEX experiment, using a Gallium target in the Gran Sasso, also confirmed thesolar neutrino deficit (P. Anselmann et al., Phys. Lett. B285, 376 (1992)).

In 2001 the Sudbury Neutrino Observatory (SNO), whose detector uses 1000tonnes of heavy water, and which is built 2Km deep in a mine near Sudbury, Ontario

182 7 Two-State Systems

in Canada, gave decisive experimental results on solar neutrinos (Q.R. Ahmad et al.,Phys. Rev. Lett. 87, 071307 (2001) and 89, 011301 (2002); see also M.B. Smy, Mod.Phys. Lett. A 17, 2163 (2002)).

The 2002 Nobel prize for physics was awarded to Raymond Davis Jr., the pioneerof this chapter of neutrino physics, and toMasatoshi Koshiba the leader of the reactorneutrino KamLAND collaboration.

The 2015 Nobel prize for physics was awarded to Takaaki Kajita (Super-Kamiokande) and Arthur B. McDonald (SNO).

Chapter 8Algebra of Observables

Our axioms may be very elegant and profound, but they seem to be deprived ofsubstance. Within this framework, how does one treat a particular case?What are theHamiltonian and the observables for a given system? In wave mechanics, we haveoperators px = −i�(∂/∂x), H = p2/2m + V and we solve differential equations,but what should we do here? It is true that for the ammonia molecule in the previouschapter, we have guessed quite easily how to proceed, butwhat is the generalmethod?

The answer lies in the first major discovery of Paul Adrien Maurice Dirac, whowas an unusual person and who arrived on the scene of quantum mechanics duringthe summer of 1925.

Dirac’s answer seemed surprising: in the general formalism of Hilbert space, thestructures that play a key role and allow us to perform calculations are not the specialform of such or such an observable, but rather the algebraic relations between theobservables, in particular their commutation relations. These algebraic relations havethe same form in all representations, and, as we shall see, they enable us to performactual calculations.

8.1 Commutation of Observables

Weknow that in general two observables A and B do not commute. Their commutator[ A, B] is defined by

[ A, B] = A B − B A. (8.1)

© Springer International Publishing Switzerland 2016J.-L. Basdevant, Lectures on Quantum Mechanics,Graduate Texts in Physics, DOI 10.1007/978-3-319-43479-7_8

183

184 8 Algebra of Observables

8.1.1 Fundamental Commutation Relation

We have already seen that between position and momentum observables, we have,in wave mechanics

[x, px ] = i� I , (8.2)

where I is the identity operator.This relation is independent of the representation we use. In Heisenberg’s matrix

mechanics it was found by Born and Jordan in July 1925, Eq. (6.60).This fundamental commutation relation is actually the definition of position and

momentum observables along the same axis. In three-dimensional space, if we notex j ( j = 1, 2, 3) the components of a position vector and similarly for momentump j , the set

[x j , pk] = i�δ j,k I , [x j , xk] = [ p j , pk] = 0 (8.3)

defines a couple of vector observables of position x and momentum p. And that’swhat we use in order to do calculations.

8.1.2 Other Commutation Relations

For observables that have a classical analogue, we still use the correspondence prin-ciple. For instance, for the angular momentum

L = r × p

one can easily obtain, using the fundamental relation (8.2), the commutation relations

[L x , L y] = i�L z, (8.4)

and two others related by cyclic permutations. The three relations can be put togetherin the form:

L × L = i�L. (8.5)

We show in the next chapter that what defines in general an angular momentumobservable J is not the particular form L = r × p (valid for a particle in space), butthe algebraic relations

J × J = i� J . (8.6)

We show how this algebra allows us to calculate numbers, and furthermore that it ismore general than L = r × p and leads to the existence of purely quantum angularmomenta, which have no classical analogues, such as the spin 1/2 of the electron.

8.1 Commutation of Observables 185

When there is no classical analogue, there is no substitute to guessing or postu-lating the corresponding algebra of observables, by keeping in mind the symmetriesof the problem.

Here, we illustrate this on some simple and important results. First, we give thegeneral proof of uncertainty relations for any couple of variables. Then we considerthe time evolution of the expectation value of a physical quantity and the Ehrenfesttheorem, which enables us to see how the classical limit appears in quantummechan-ics. Next, we give an algebraic solution of the harmonic oscillator problem, due toDirac, which plays a crucial role in quantum field theory. Next, we say a few wordsabout observables that commute, in order to have the concept of a complete set ofcommuting observables, which is essential in order to treat problems in several vari-ables, in particular, three-dimensional problems. Finally we say a few words aboutDirac’s first major discovery in quantum mechanics.

8.1.3 Dirac in the Summer of 1925

Before we start, let us say a few words about how Dirac entered quantum mechanicsin 1925. Dirac was a student in Cambridge. He was born on August 8, 1902 so hewas barely 23. One can talk for hours about him. He was completely unusual. Verydiscreet, very polite, and careful, he thought a lot. He had a great culture, in particularin mathematics. He knew about noncommutative algebras, but let’s not anticipate.

At that time, Cambridge had an impressive intellectual richness and was similar toGöttingen in that respect. The professors were Keynes in economics, J.J. Thomson,E. Rutherford, A.S. Eddington, and Fowler in physics, as well as others.

And there, by accident, everything got started. On July 28, 1925, Heisenberg wasinvited to give a talk in Cambridge at the very fashionable Kapitza club.

That talk had no influence on Dirac (who kept thinking instead of listening);neither of them remembered seeing the other on that day. But Heisenberg gave hisarticle to Fowler who gave it to Dirac. He read the paper with some difficulty: it wasfull of philosophical ideas, but the mathematical formalism was clumsy. Dirac hadnothing against philosophy, but he didn’t particularly care for it in that context; whathe wanted was good mathematics. Very rapidly he said, “This will lead us nowhere.”

Two weeks later, he walked into Fowler’s office and said, “It’s remarkable; itcontains the key of quantum mechanics!”

What had happened is that Heisenberg had cracked up!Since 1924, he had elaborated a system of calculational rules and a theory that

worked well. Then Born had gotten involved in the whole business, and he had foundthat there were matrices. They had seen Hilbert. And Heisenberg, a positivist, hadlearned something tragic about his theory: some physical quantities did not commute!The product of a and b was not the same as the product of b and a! And that wascontrary to any physical sense. The product of two quantities had never dependedon the order. So Heisenberg was terrified, “My theory is not beautiful!” He was so

186 8 Algebra of Observables

terrified that he had put all the dust under the carpet and he wrote his papers in sucha way to hide this noncommutativity as well as he could.

Dirac, very meticulously, had redone the calculations step by step, and he realizedthat everything boiled down to this noncommutativity. In particular, he proved inde-pendently from Born and Jordan the fundamental relation [x, px ] = i� I in August1925 without mentioning matrices.

He knew the existence of noncommutative algebras. So he made an attempt tomodify the classical equations in order to take into account this noncommutativity.After all nothing dictates that physical quantities should commute.

Some time after, Dirac said, “You know, Heisenberg was scared. He was scaredby these foreign mathematics. I can understand that he was scared: all his theory wasat stake. I had an enormous advantage over him: I had no risk.”

So, Dirac constructed his own version of quantum mechanics, based on noncom-mutative algebras, that is, on commutators.

8.2 Uncertainty Relations

In 1927 Heisenberg stated his uncertainty principle. When Dirac saw it, he simplysaid, “Oh yes, indeed, I proved that in 1925.”

The general proof of uncertainty relations for any couple of physical quantities isthe following.

Consider two quantities A and B, and the corresponding observables A and B. Let|ψ〉 be the state of the system. The measurement of A and of B gives the expectationvalues 〈a〉 and 〈b〉, and the dispersions Δa and Δb. We want to relate Δa and Δbfor two given observables A and B.

A simple calculation based on the Schwarz inequality and the manipulation ofDirac’s formalism shows that, if |ψ〉 is the state of the system, then

Δa Δb ≥ 1

2|〈ψ|[ A, B]|ψ〉|. (8.7)

Therefore

1. If A and B do not commute, [ A, B] �= 0, then, in general the two dispersions onA and B cannot be made simultaneously as small as possible except in specialcases where the state |ψ〉 is such that 〈ψ|[ A, B]|ψ〉 = 0.

2. For x and px , using (8.2), we obtain Heisenberg’sΔ x Δ px ≥ �/2.We no longerrefer to the Fourier transformation, we can forget about it.

3. The uncertainty relations are therefore generalized to any couple of observables.They come from the noncommutativity of the corresponding observables.

Proof We first center the variables, that is, we set A′ = A − 〈a〉 and B ′ = B − 〈b〉,so that 〈 A′〉 = 〈B ′〉 = 0. We then have

8.2 Uncertainty Relations 187

(Δa)2 = 〈ψ| A′2|ψ〉, and (Δb)2 = 〈ψ|B ′2|ψ〉.

Consider any state |ψ〉 and the vector ( A′ + iλB ′)|ψ〉, with λ real. The square ofthe norm of this vector is:

‖( A′ + iλB ′)|ψ〉‖2 = 〈ψ|( A′ − iλB ′)( A′ + iλB ′)|ψ〉= 〈ψ| A′2|ψ〉 + λ2〈ψ|B ′2|ψ〉 + iλ〈ψ|[ A′, B ′]|ψ〉= Δa2 + λ2Δb2 + iλ〈ψ|[ A′, B ′]|ψ〉.

Because A′ and B ′ are Hermitian, the operator i[ A′, B ′] is also Hermitian and thelast term is real. The above expression is the square of the norm of a vector, thus itmust be positive whatever the value of λ. Because it is positive for λ → ∞ it mustnot change sign. Therefore the discriminant of the trinomial in λ must be negative,and, since [ A′, B ′] = [ A, B],

Δa Δb ≥ 1

2|〈ψ|[ A, B]|ψ〉| Q.E.D. (8.8)

8.3 Evolution of Physical Quantities

We now address the following problem: what is the time variation of the expectationvalue of a physical quantity?

8.3.1 Evolution of an Expectation Value

Consider the expectation value 〈a〉 of a physical quantity

〈a〉 = 〈ψ| A|ψ〉.

We take the time derivative:

d

dt〈a〉 =

(d

dt〈ψ|

)A|ψ〉 + 〈ψ|

(∂

∂tA

)|ψ〉 + 〈ψ| A

(d

dt|ψ〉

).

Using the Schrödinger equation and its Hermitian conjugate expression:

i�d|ψ〉dt

= H |ψ〉, and − i�d〈ψ|dt

= 〈ψ|H ;

188 8 Algebra of Observables

we obtain:d

dt〈a〉 = 1

i�〈ψ|[ A, H ]|ψ〉 + 〈ψ|∂ A

∂t|ψ〉. (8.9)

This formula is due to Ehrenfest (1927) (but Dirac had found it in 1925).We remark that, in a similar way to the Schrödinger equation, the Hamiltonian

governs the time evolution of a physical expectation value. Here it does so throughits commutator with the observable.

If the operator A does not explicitly depend on time, we have

d

dt〈a〉 = 1

i�〈ψ|[ A, H ]|ψ〉. (8.10)

8.3.2 Particle in a Potential, Classical Limit

We call qi (i = 1, 2, 3) the three position variables x, y, z and pi (i = 1, 2, 3) thecoordinates of the momentum px , py, pz . The operators qi and pi obey the canonicalcommutation relations:

[qi , q j ] = 0, [ pi , p j ] = 0, [q j , pk] = i�δ j,k . (8.11)

From these relations one can prove the commutation relations

[q j , pmj ] = m(i�) pm−1

j , [ p j , qnj ] = −n(i�)qn−1

j , (8.12)

which we can generalize to any function F = F(qi , pi ) of the operators qi and piwhich can be expanded in a power series

[q j , F] = i�∂ F

∂ p j, and [ p j , F] = −i�

∂ F

∂q j. (8.13)

If the Hamiltonian does not depend explicitly on time, choosing F = H , we obtainthe evolution equations:

d

dt〈qi 〉 =

⟨∂ H

∂ p j

⟩,

d

dt〈p j 〉 = −

⟨∂ H

∂q j

⟩. (8.14)

These forms have a great similarity to the equations ofHamilton’s analyticalmechan-ics, as we shall see in Chap. 15.

8.3 Evolution of Physical Quantities 189

The Hamiltonian of a particle in a potential is

H = p2

2m+ V (r). (8.15)

Substituting in (8.14), we obtain

d〈r〉dt

= 〈 p〉m

(8.16)

d〈 p〉dt

= −〈∇V (r)〉. (8.17)

Equation (8.16) is the correct definition of the group velocity of a wave packet. Itis the same for expectation values as the classical relation between the position andthe velocity. Actually, Eq. (8.17) differs slightly from the classical equation, whichwould be for expectation values

d〈 p〉dt

= −∇V (r)∣∣∣r=〈r〉

,

because, in general, f (〈r〉) �= 〈 f (r)〉. However, if the distribution in r is peakedaround somevalue r0, then 〈r〉 ∼ r0 and f (〈r〉) ∼ 〈 f (r)〉, and in this caseEqs. (8.16)and (8.17) are, for the expectation values, essentially the same as the classical equa-tions of the motion. This observation is the Ehrenfest theorem (1927).

One-Dimensional Classical Limit

In one dimension we have

d

dt〈p〉 =

⟨−dV

dx

⟩�= −dV

dx

∣∣∣x=〈x〉

.

We expand the function f (x) = −∂V/∂x in the vicinity of x = 〈x〉 and we obtain

f (x) = f (〈x〉) + (x − 〈x〉) f ′(〈x〉) + 1

2(x − 〈x〉)2 f ′′(〈x〉) + · · ·

that is, taking the expectation value,

〈 f 〉 = f (〈x〉) + Δx2

2f ′′(〈x〉) + · · ·

where Δx2 = 〈(x − 〈x〉)2〉.The nonclassical term in the evolution of the expectation valuewill be negligible if

|Δx2 f ′′(〈x〉)/ f (〈x〉)| � 1,

190 8 Algebra of Observables

or, if we come back to the potential V ,

∣∣∣∣Δx2∂3V

∂x3

∣∣∣∣ �∣∣∣∣∂V

∂x

∣∣∣∣ ,

that is, if the potential varies slowly on the extension of the wave packet (or on aninterval of the order of the de Broglie wavelength). That is the actual content of theEhrenfest theorem.

We recall that a macroscopic system (i.e., a systemwhose orders of magnitude aresuch that � seems negligible) is not necessarily a classical system. We can perfectlywell consider a wave packet having two peaks located at a large macroscopic dis-tance from each other. As we said in Chap. 4 such a state, macroscopic but quantummechanical, is very unstable.

One can check that for a harmonic oscillator, V = mω20x

2/2, or, more generally ifthe potential is a second degree polynomial, the Ehrenfest theorem gives identicallythe classical equation of motion d2〈x〉/dt2 = −ω2

0〈x〉.

8.3.3 Conservation Laws

There are two cases where d〈a〉/dt = 0.

• Either the observable commutes with the Hamiltonian, [ A, H ] = 0. The quantityA is always conserved, it is a constant of the motion.

• Or the state |ψ〉 is such that 〈ψ|[ A, H ]|ψ〉 = 0. This is in particular the case forstationary states, that is eigenstates of H for which no quantity evolves in time.

Here are some examples of applications of these results.

Conservation of the Norm

Consider the identity operator A = I ; this gives the conservation of the norm

d

dt〈ψ|ψ〉 = 0,

that is to say, conservation of probability.

Conservation of Energy for an Isolated System

Consider a time-independent problem; the choice A = H gives

d

dt〈E〉 = 0.

8.3 Evolution of Physical Quantities 191

Conservation of Momentum

Consider the motion of a free particle of Hamiltonian H = p2/2m. The observablespx , py, pz commute with H and we obtain conservation of the momentum:

d

dt〈pi 〉 = 0, i = x, y, z.

This can be generalized to an N particle system whose Hamiltonian is translationinvariant (pairwise interactions V (xi − x j )). The total momentum 〈P〉 = 〈∑ pi 〉 isconserved.

8.4 Algebraic Resolution of the Harmonic Oscillator

In order to see how the principles work, and to see the importance of the commutationrelations, we show how the one-dimensional harmonic oscillator problem can besolved using the algebra of observables. This calculation is also due to Dirac.

In order to simplify the proof, we make use of some qualitative results we alreadyknow. Consider the Hamiltonian

H = p2

2m+ 1

2mω2 x2.

With the change of observables

X = x

√mω

�, P = p√

m�ω, (8.18)

we obtainH = �ωH, (8.19)

with

H = 1

2

(X2 + P2

). (8.20)

We must solve the eigenvalue problem

H|ν〉 = εν |ν〉,

where we have assumed that the eigenvalues εν are not degenerate. We know thisresult from Chap.5, but this can be proven in the present context.

The commutation relation of X and P can be deduced from (8.2),

[X , P] = i. (8.21)

192 8 Algebra of Observables

Operators a, a†, and N

In order to solve the eigenvalue problem we introduce the following operators

a = 1√2(X + i P), a† = 1√

2(X − i P) (8.22)

whose commutator is[a, a†] = 1. (8.23)

Consider now the operator

N = a†a = 1

2(X2 + P2 − 1) (8.24)

that satisfies the commutation relations

[N , a] = −a, [N , a†] = a†. (8.25)

We have

H = N + 1

2I ,

therefore H and N have the same eigenvectors. Let ν be the eigenvalues of N and|ν〉 its eigenvectors; we have, coming back to the initial Hamiltonian

H |ν〉 =(

ν + 1

2

)�ω|ν〉. (8.26)

Determination of the Eigenvalues

The determination of the eigenvalues comes from the following lemmas.

Lemma 1 The eigenvalues ν of the operator N are positive or zero.

In fact, consider a vector |ν〉 and the norm of the vector a|ν〉:

‖a|ν〉‖2 = 〈ν|a†a|ν〉 = 〈ν|N |ν〉 = ν〈ν|ν〉 = ν‖|ν〉‖2. (8.27)

Therefore ν ≥ 0 anda|ν〉 = 0, if and only if ν = 0. (8.28)

Lemma 2 The vector a|ν〉 is either an eigenvector of N , corresponding to the eigen-value ν − 1, or the null vector.

In fact, consider the vector N a|ν〉. If we use the commutation relation of N and a,we obtain

8.4 Algebraic Resolution of the Harmonic Oscillator 193

N (a|ν〉) = a N |ν〉 − a|ν〉 = νa|ν〉 − a|ν〉 = (ν − 1)(a|ν〉).

Therefore

• Either ν − 1 is an eigenvalue of N and a|ν〉 is an associated eigenvector;• Or ν − 1 is not an eigenvalue of N and a|ν〉 is the null vector.

By an analogous argument one can show that:

Lemma 3 If ν + 1 is an eigenvalue of N , a†|ν〉 is an eigenvector associated withthe eigenvalue ν + 1 and

‖a†|ν〉‖2 = (ν + 1)‖|ν〉‖2. (8.29)

It is now simple to prove that:

Theorem The eigenvalues of N are the nonnegative integers.

In fact, because the eigenvalues are nonnegative, one of themmust be smaller thanthe others. Let us call it νmin. Since νmin is the smallest eigenvalue, νmin − 1 is notan eigenvalue. Therefore, a|νmin〉 is the null vector and its norm is zero. Inasmuch aswe have shown that ‖a|ν〉‖2 = ν‖|ν〉‖2 for all values of ν, this means that νmin = 0.

Starting with this eigenvalue and the corresponding eigenvector |νmin = 0〉, wecan generate all other eigenvalues and corresponding eigenvectors by repeatedlyapplying the operator a†, at each step we add 1: (. . . ν → ν + 1 → ν + 2 . . .).We recover the energy levels of the harmonic oscillator.

Eigenstates

Ground State

The ground state |0〉 satisfies (8.28)

a|0〉 = 0, or(X + i P

)|0〉 = 0. (8.30)

In the language of wave functions, this amounts to

(mω

�x + d

dx

)ϕ0(x) = 0, (8.31)

where ϕ0(x) is the ground state wave function. The solution is

ϕ0(x) = C0 e−(mω/�)x2/2, (8.32)

where C0 is a normalization constant. We recover the result of Chap. 5.

194 8 Algebra of Observables

Excited States

We assume the eigenstates are normalized: 〈n|n〉 = 1. Owing to the above Lemmas1 and 2 and to Eqs. (8.27) and (8.29),

a|n〉 = √n|n − 1〉, a†|n〉 = √

n + 1|n + 1〉. (8.33)

Hence the name of annihilation operator (for a) and of creation operator (for a†)because they allow us to get from an energy state (n + 1/2)�ω to energy states(n + 1/2 ∓ 1)�ω, and they respectively annihilate or create an energy quantum �ω.Similarly, the operator N corresponds to the number of quanta in the state |n〉. Thesequence of states |n〉 is generated from the ground state |0〉 by applying repeatedlythe operator a†,

|n〉 = 1√n! (a

†)n|0〉. (8.34)

This allows us to find the wave function ϕn(x) of the energy state (n + 1/2)�ω interms of the ground state wave function:

ϕn(x) = 1√n!

1√2n

[x

√mω

�−

√�

mω

d

dx

]n

ϕ0(x). (8.35)

This is an explicit and compact formula for the Hermite functions.Similarly, one can show with (8.33) and the definition of a and a†, that we have

x |n〉 =√

�

2mω

(√n + 1|n + 1〉 + √

n|n − 1〉)

(8.36)

p|n〉 = i

√m�ω

2

(√n + 1|n + 1〉 − √

n|n − 1〉)

. (8.37)

We see, in this example, the elegance and the power of Dirac’s algebraic method.This treatment of the harmonic oscillator, and the operators a, a†, and N , are fun-damental tools in many branches of physics such as quantum field theory, statisticalmechanics, and the many-body problem.

8.5 Commuting Observables

When two observables commute, there is no constraint such as the uncertainty rela-tions. This case is, however, important in practice.

8.5 Commuting Observables 195

8.5.1 Theorem

We know that if two matrices commute, one can diagonalize them simultaneously.This remains true in the infinite-dimensional case. If two observables A and B

commute, there exists a common eigenbasis of these two observables.This theorem is generalized immediately to the case of several observables A, B,

C , which all commute.

Proof Let {|α, rα〉} be the eigenvectors of A, where the index rα means that an eigen-vector associated with an eigenvalue aα belongs to an eigensubspace of dimensiondα ≥ 1,

A|α, rα〉 = aα|α, rα〉, r = 1, . . . , dα. (8.38)

By assumption, we have [ A, B] = 0, i.e.

A B|α, rα〉 = B A|α, rα〉 = aα B|α, rα〉, r = 1, . . . , dα. (8.39)

Therefore, the vector B|α, rα〉 is an eigenvector of A with the eigenvalues aα. Ittherefore belongs to the corresponding eigensubspace.We call this vector |α,β, kαβ〉;the index kαβ means that again this vector may be nonunique. Therefore, this vectoris a linear combination of the vectors {|α, rα〉}, that is,

B|α, rα〉 =∑

rα

brα |α, rα〉,

which can be diagonalized with no difficulty. In other words, if A and B commute,they possess a common eigenbasis.

The reciprocal is simple. The Riesz theorem says that the orthonormal eigenvec-tors of an observable form a Hilbert basis. Suppose A and B have in common thebasis {|ψn〉} with eigenvalues an and bn:

A|ψn〉 = an|ψn〉 and B|ψn〉 = bn|ψn〉. (8.40)

If we apply B to the first expression and A to the second, and subtract, we obtain

( A B − B A)|ψn〉 = (anbn − bnan)|ψn〉 = 0.

Because {|ψn〉} is a Hilbert basis, we therefore have

[ A, B]|ψ〉 = 0, whatever |ψ〉;

which means that [ A, B] = 0.

196 8 Algebra of Observables

8.5.2 Example

Actually, we have not yet seen examples of this because we have considered onlyone-dimensional problems. But the statement is nothing very complicated.

Consider, for instance, an isotropic two-dimensional harmonic oscillator. Theeigenvalue problem of the Hamiltonian is a priori a difficult problem because itseems to be a partial differential equation in two variables. But the Hamiltonian canbe written as the sum of two independent Hamiltonians acting on different variables:

H = − �2

2m

∂2

∂x2+ 1

2mω2x2 − �

2

2m

∂2

∂y2+ 1

2mω2y2 = Hx + Hy . (8.41)

The two operators Hx and Hy , which are both operators in one variable and which acton different variables commute obviously. One can solve the eigenvalue problemsof Hx and Hy separately:

Hxϕn(x) = Enϕn(x); Hyϕn(y) = Enϕn(y).

The eigenvalues of H are the sums of eigenvalues of Hx and Hy with eigenfunctionsthat are the products of corresponding eigenfunctions:

En = En1 + En2 = (n1 + n2 + 1)�ω, ψn(x, y) = ϕn1(x)ϕn2(y).

In other words, a sum of Hamiltonians that commute has for eigenvalues the sumof eigenvalues of each of them, and for eigenfunctions the product of correspondingeigenfunctions.

8.5.3 Tensor Structure of Quantum Mechanics

This example has another interest. It shows that in the case of systems with severaldegrees of freedom, the Hilbert space is the tensor product of the Hilbert spaces inwhich each individual degree of freedom is described.

In the above example, it is the fashionable way of saying that the products ofeigenfunctions of the one-dimensional harmonic oscillator

ψn,m(x, y) = ϕn(x)ϕm(y) (8.42)

form a basis of square integrable functions in two variables (x, y), or that the spaceof square integrable functions in two variables L2(R2) is the tensor product of twospaces L2(R) of square integrable functions in one variable.

8.5 Commuting Observables 197

The tensor structure of quantum mechanics is important and useful in complexsystems. One can find it at various degrees of sophistication in the literature.1 InDirac’s notations the elements of the basis ϕn1(x)ϕm2(y) are written as

|ψn,m〉 = |1 : ϕn〉 ⊗ |2 : ϕm〉, (8.43)

where (1) and (2) stand for the two degrees of freedom (or subsystems) and n andmare the corresponding eigenstates. The symbol ⊗ stands for “tensor product,” whichis just an ordinary product in Eq. (8.42). Any state of the global system |ψ〉 can bewritten as

|ψ〉 =∑

n,m

cn,m |1 : ϕn〉 ⊗ |2 : ϕm〉. (8.44)

An important property is that the Hermitian scalar product of two factorizablevectors |u〉 ⊗ |v〉 and |u′〉 ⊗ |v′〉 factorizes as the product

(〈u′| ⊗ 〈v′|)(|u〉 ⊗ |v〉 = 〈u′|u〉〈v′|v〉. (8.45)

In Chap.12 we show, with spin 1/2, an example just as simple but not as trivial ofa tensor product of Hilbert spaces.

8.5.4 Complete Set of Commuting Observables (CSCO)

This brings us to a notion that is useful both conceptually and technically.A set of operators A, B, C, . . ., is said to form a complete set of commuting

observables (CSCO) if the common eigenbasis of this set of operators is unique.In other words, to each set of eigenvalues aα, bβ, cγ, . . . there corresponds a singleeigenvector |αβγ . . .〉 (up to a phase factor). If an operator O commutes with allof the operators of the CSCO, then it is a function of these operators. For a givensystem, there exists an infinite number of CSCOs.We show in the following chaptersthat one chooses a CSCO according to criteria of convenience. Neither the naturenor the number of observables that form a CSCO are fixed a priori.

For a one-dimensional harmonic oscillator, the Hamiltonian

Hx = p2x2m

+ 1

2mω2 x2

is a CSCO by itself. There is only one eigenbasis of Hx formed by the Hermitefunctions ψn(x).

For the two-dimensional isotropic oscillator above (8.41), this is not the case. Apossible basis is formed by the set of functions {ψn1(x)ψn2(y)}, where ψn1(x) and

1See J.-L. Basdevant and J. Dalibard, Quantum Mechanics, Chap.5, Sect. 6.

198 8 Algebra of Observables

ψn2(y) are, respectively, the eigenfunctions of Hx and Hy . The eigenvalue corre-sponding to ψn1(x)ψn2(y) is

En1,n2 = �ω(n1 + n2 + 1),

which is degenerate, except for n1 = n2 = 0. This implies that there are actuallyseveral eigenbases of H (actually an infinite number). For instance, in the subspaceassociated with 2�ω, the possible base elements are:

{cos θψ1(x)ψ2(y) + sin θψ2(x)ψ1(y),− sin θψ1(x)ψ2(y) + cos θψ2(x)ψ1(y)}.

Therefore, H is not a CSCO by itself. A possible CSCO is the set of twoHamiltonians {Hx , Hy}. In fact, the two eigenvalues {Enx = (nx + 1/2)�ω, Eny =(ny + 1/2)�ω} uniquely specify an eigenvector.

This does not imply that in this problem a CSCO is necessarily composed oftwo observables. A rigorous theoretician may object that the operator Hπ = Hx +π Hy forms a CSCO by itself. Indeed, its eigenvalues are nπ = (nx + 1/2) + π(ny +1/2), and since π is transcendental there is a unique couple of integers (nx , ny)

corresponding to a given eigenvalue. An experimentalist will reply that it is simplerto measure two numbers (nx , ny) rather than investing in a measuring apparatus thatdirectly gives the answer nπ .

8.5.5 Completely Prepared Quantum State

Why is the notion of a completely prepared quantum state important physically? Ifwe want to specify as accurately as possible the initial conditions of an experiment,wemust knowwhetherwe start from a specific quantum state orwith some ill-definedsituation. In the above case of the isotropic harmonic oscillator, if we know the initialenergy n�ω, we only know that the initial state belongs to a subspace of dimension n,generated by the n functionsψn1(x)ψn2(y)with n1 + n2 + 1 = n. A measurement ofthe energy is not sufficient to specify unambiguously the initial state. If we measurethe vibrational energies along both the x and y axes, we know the state. One saysthat one deals with a completely prepared quantum state.

More generally, consider two observables A and B that commute. There exists acommon eigenbasis {|α,β, γ〉} of A and B, where the index γ indicates that A andB are not necessarily a CSCO by themselves.

Consider a state |ψ〉, and suppose that by measuring A on |ψ〉, we find the valuea0. After the measurement, the state of the system has changed. It is

|ψ〉 → |ψ0〉 = λ

⎛

⎝∑

β,γ′|α0;β, γ′〉〈α0;β, γ′|

⎞

⎠ |ψ〉,

where λ is a normalization factor such that 〈ψ0|ψ0〉 = 1.

8.5 Commuting Observables 199

On this new state |ψ0〉, we measure B. Suppose the result is b1. Similarly, weobtain another state |ψ1〉 after the measurement:

|ψ0〉 → |ψ1〉 = λ′(

∑

αγ

|α,β1; γ〉〈α,β1; γ|)

|ψ0〉.

If we insert the expression of |ψ0〉 in this formula, we obtain

|ψ1〉 = λλ′ ∑

αβγ′|α,β1; γ〉〈α,β1; γ|α0,β; γ′〉〈α0,β; γ′|ψ〉.

But by assumption, 〈α,β; γ|α′,β′; γ′〉 = δαα′δββ′δγγ′ .Therefore, the expansion of |ψ1〉 reduces to

|ψ1〉 = λλ′ ∑

γ

|α0,β1; γ〉〈α0,β1, γ|ψ〉.

In other words, |ψ1〉 is still an eigenvector of A with the same eigenvalue a0.This is an important result. If two observables A and B commute, if we succes-

sively measure A, with a result a0, and then B with a result b1, this second measure-ment does not affect the value found previously for A. If we redo a measurement ofA, we find the same result a0 with probability one.

This result can be extended to any number of commuting observables. If onemeasures all the physical quantities of a CSCO, any (“immediate”) newmeasurementof one of these quantities will always give the same result. The state vector definedby this series of measurements is defined uniquely. One says that with this series ofmeasurements one obtains a state that is completely prepared or completely known.

8.6 Sunday September 20, 1925

Let us get back to Dirac, in the summer of 1925. His problem was to find how toincorporate noncommutativity in classical mechanics.

Dirac knew that in analytical mechanics, which had been developed one centurybefore, Hamilton had found a quite fruitful formulation.2

In this version of mechanics, the state of a particle is described at any time by itsposition x and its momentum p.

The system is characterized by a Hamilton function or Hamiltonian H which, fora particle of mass m in a potential V (x) is

2See, for instance, J.-L. Basdevant, Variational Principles in Physics, Chap.4. New York: Springer(2006).

200 8 Algebra of Observables

H = p2

2m+ V (x). (8.46)

The evolution equations of the state variables x and p are then given by the canonicalequations

dx

dt= ∂H

∂ p, and

dp

dt= −∂H

∂x. (8.47)

Dirac wanted to construct his quantummechanics with noncommutative algebras,that is, with commutators.

On Sunday, September 20, 1925, Dirac committed a crime! He was very wellorganized and he had the habit of working hard all week and of relaxing on Sundaysby talking a walk. But on that Sunday he committed a crime. In fact, during his walkhe started thinking. He had given himself the right to walk when he was thinking, butnever to think on his Sunday walk! He thought about commutators, and, suddenly,he thought about something he had been told of, and about which he had read. Hethought about theworks of a former student of Polytechnique, in Paris, Siméon-DenisPoisson! And Dirac rushed to the library, but there was some suspense because “OnSundays it’s closed.” He had to wait. On Monday morning, he rushed in and readabout what Carl Gustav Jacob Jacobi had called the greatest discovery of Poisson:the Poisson brackets.

Consider two physical quantities f and g, which are functions of the state variables(x, p); the Poisson bracket of f and g is the quantity

{ f, g} =(

∂ f

∂x

∂g

∂ p− ∂ f

∂ p

∂g

∂x

). (8.48)

For the state variables (x, p) we find the relation

{x, p} = 1, (8.49)

and

{x, f } = ∂ f

∂ p, {p, f } = −∂ f

∂x. (8.50)

In three dimensions, (xi , pi ) (i = 1, 2, 3), (8.49) generalizes as

{xi , x j } = 0, {pi , p j } = 0, {xi , p j } = δi j . (8.51)

The time evolution of a physical quantity f (x, p) is

f = d f

dt= (

∂ f

∂xx + ∂ f

∂ pp). (8.52)

8.6 Sunday September 20, 1925 201

But, using Hamilton’s equations (8.47), one obtains

f = { f, H}. (8.53)

In particular, the equations (8.47) can be written in the symmetric way

x = {x, H}, p = {p, H}. (8.54)

Diracwas fascinated. The quantum commutators, divided by i�, play a completelysimilar role as the Poisson brackets do in analytical mechanics. Just compare thefundamental commutation relation (8.2) and the relation (8.49), and, similarly, theEhrenfest theorem (8.10) and Eq. (15.20).

That is howDirac understood the actual formof the correspondence principle.Onemust, in the classical equations, replace the Poisson brackets by the quantum com-mutators divided by i�. Dirac called the quantum physical quantities “q-numbers”which are noncommutative contrary to the classical commutative “c-numbers”.

He finished his work on November 7, 1925, and he published his article “TheFundamental Equations of Quantum Mechanics” in December 1925. In November,he wrote to Heisenberg who replied that his work was “most beautiful and remark-able.” He was apparently insensitive to the publication of the works of Born andJordan in November 1925 and of Born, Heisenberg, and Jordan in January 1926,where one can find a number of his results (proven independently).

8.7 Exercices

1. Commutator algebra

Show the following equalities:

[ A, BC] = [ A, B]C + B[ A, C] , [ A, Bn] =n−1∑

s=0

Bs[ A, B]Bn−s−1

[ A, [B, C]] + [B, [C, A]] + [C, [ A, B]] = 0 (Jacobi identity) .

2. Classical equations of motion for the harmonic oscillator

Show that for a harmonic oscillator V (x) = mω2x2/2, the Ehrenfest theorem givesidentically the classical equation of motion:

d2〈x〉dt2

= −ω2 〈x〉.

202 8 Algebra of Observables

3. Conservation law

Consider a system of two particles interacting through a potential V (r1 − r2). Checkthat the total momentum P = p1 + p2 is conserved. Show that this property can beextended to a system of n interacting particles.

4. Hermite functions

Prove from (8.33) and the definition of a and a†, the recursion relations for theHermite functions:

x |n〉 =√

�

2mω

(√n + 1|n + 1〉 + √

n|n − 1〉)

, (8.55)

p|n〉 = i

√m�ω

2

(√n + 1|n + 1〉 − √

n|n − 1〉)

. (8.56)

5. Generalized uncertainty relations

a. Consider, in 3 dimensions, the radial variable r = √x2 + y2 + z2 and a real func-

tion f (r) of this variable. Show that the commutator of px with f (r) is:

[ px , f ] = −i�x

rf ′(r),

where f ′(r) is the derivative of f .b. Consider the operator Ax = px − iλx f (r) where λ is a real number.

• Calculate the square of the norm of Ax |ψ〉 for an arbitrary vector |ψ〉.• Add the analogous relations for Ay and Az , and derive an inequality relating

〈p2〉, 〈r2 f 2〉, 〈 f 〉 and 〈r f ′〉, which holds for any function f and any state |ψ〉.c. Considering the cases f = 1, f = 1/r , and f = 1/r2 show that one has the

following relations in three dimensions:

〈p2〉〈r2〉 ≥ 9

4�2, 〈p2〉 ≥ �

2〈1r〉2, 〈p2〉 ≥ �

2

4〈 1r2

〉.

d. Harmonic oscillator. The Hamiltonian of a three-dimensional harmonic oscillatoris H = p2/2m + mω2 r2/2.

• Using the first inequality, find a lower bound for the ground state energy ofthis oscillator, and explain why this bound is equal to the ground state energy.

• Write the differential equation satisfied by the corresponding ground statewave function and calculate this wave function.

e. Hydrogen atom. TheHamiltonian of the hydrogen atom is, considering the protonmass as very large compared to the electron mass,

8.7 Exercices 203

H = p2

2me− e2

r

where, for simplicity, we set e2 = q2/4πε0.

• Using the second inequality, find a lower bound for the ground state energyof the hydrogen atom, and explain why this bound is equal to the ground stateenergy.

• Write the differential equation satisfied by the corresponding ground statewave function φ(r) and calculate this wave function.

7. Time-energy uncertainty relation

Consider a state |ψ〉 of a system whose energy dispersion is ΔE , and an observableA whose expectation value and dispersion are 〈a〉 and Δa. Using the commutationrelations, show that following inequality holds:

Δa ΔE ≥ �

2

∣∣∣∣d〈a〉dt

∣∣∣∣ .

Deduce from this that if the typical evolution time-scale τ of the system is definedby τ = |Δa/(d〈a〉/dt)|, one has the inequality τ ΔE ≥ �/2.

8. Virial Theorem

Consider a one-dimensional system of Hamiltonian H = p2/2m + V (x) whereV (x) = λxn .

a. Calculate the commutator [H , x p].b. By taking the expectation value of this commutator, show that in any eigenstate

of H , one has the relation:2〈T 〉 = n〈V 〉 ,

where T = p2/2m is the kinetic energy operator. Check this relation on the har-monic oscillator.

c. Generalize this result to three dimensions by calculating [H , r · p] and consid-ering a potential V (r) which is a homogeneous function of the variables x, y, z,of degree n. A homogeneous function of degree n satisfies V (αx,αy,αz) =αnV (x, y, z) and r · ∇V = nV .

d. Show that for an arbitrary potential V (r) one has the general relation:

2〈T 〉 = 〈r ∂V

∂r〉.

204 8 Algebra of Observables

8.8 Problem. Quasi-Classical States of the HarmonicOscillator

We consider a one-dimensional harmonic oscillator of frequency ω and we study theeigenstates |α〉 of the annihilation operator:

a|α〉 = α|α〉

where α is a complex number. We expand |α〉 on the basis {|n〉}:

|α〉 =∑

n

Cn|n〉.

1. Determination of α:

(a) Write the recursion relation between the coefficients Cn .(b) Express the Cn’s in terms of the first coefficient C0.(c) Calculate the coefficients Cn by normalizing |α〉, i.e. 〈α|α〉 = 1.(d) What are the allowed values for the number α ?(e) In an energy measurement on the state |α〉, what is the probability to find the

value En = (n + 1/2)�ω?

2. Consider a state |α〉. Starting from the expression of the Hamiltonian and thedefinition of this state:

(a) Calculate the expectation value 〈E〉,(b) Calculate the expectation value of the square of the energy 〈E2〉 (use the

commutator of a and a†),(c) Deduce the value of the dispersion ΔE in that state.(d) In what sense can one say that the energy is definedmore andmore accurately

if |α| � 1?

3. Calculate 〈x〉,Δx, 〈p〉,Δp in a state |α〉. In that state, what is the value of theproduct Δx Δp?

4. We assume that at t = 0, the oscillator is in the state |α〉.(a) Write the state |ψ(t)〉 of the system at time t .(b) Show that the state |ψ(t)〉 is also an eigenstate of the operator a and give the

corresponding eigenvalue.(c) We set α = α0 eiφ with α0 real positive. What are, at time t , the values of

〈x〉, 〈p〉 and Δx Δp?

5. We now determine the wave functions corresponding to |α〉.(a) Check that the change of variables from x and p to X and P leads to the

expression of the operator P acting on wave functions: ψ(X, t)

8.8 Problem. Quasi-Classical States of the Harmonic Oscillator 205

P = −i∂

∂X.

Give the corresponding expression for the operator X acting on functionsϕ(P, t).

(b) Calculate the wave function ψα(X) of the state |α〉.(c) Calculate the Fourier transform ϕα(P) of this wave function.(d) Starting from the time dependence of |ψα(X, t)|2 and |ϕα(P, t)|2, explain

the results obtained previously.

8.8.1 Solution

1. By definition, we have:

a|α〉 =∞∑

n=1

Cn√n |n − 1〉 = α

∞∑

n=0

Cn |n〉.

The ensuing recursion relationCn√n = αCn−1 allows to calculate the coefficients

Cn in terms of C0 and α, whatever the value of the complex number α:

Cn = αn

√n!C0.

For any α, we therefore obtain:

|α〉 = C0

∞∑

n=0

αn

√n !

|n〉

and, by normalizing the result:

〈α|α〉 = |C0|2∞∑

n=0

|α|2nn! = e|α|2 |C0|2 ⇒ C0 = e−|α|2/2

up to an arbitrary phase factor.The probability p(En) to find En is:

p(En) = |〈n|α〉|2 = e−|α|2 |α|2n/n!

which is a Poisson distribution.

206 8 Algebra of Observables

2. The expectation value of the energy is obtained by using:

〈E〉 = 〈α|H |α〉 = �ω〈α|(a†a + 1/2)|α〉 = (|α|2 + 1/2)�ω

〈E2〉 = �2ω2〈α|[(a†a)2 + a†a + 1/4]|α〉.

We have [a, a†] = 1 and therefore:

(a†a)2 = a†aa†a = a†a†aa + a†a,

hence:〈E2〉 = �

2ω2(|α|4 + 2|α|2 + 1/4)

and the variance:

ΔE2 = 〈E2〉 − 〈E〉2 = �2ω2|α|2, ΔE = �ω|α| ,

or, equivalently ΔE/〈E〉 = |α|/(|α|2 + 1/2) ∼ 1/|α| for |α| � 1. The relativedispersion of the energy ΔE/〈E〉 goes to zero as |α| increases.

3. The calculation of these expectation values yields:

〈X〉 = 〈α|(a + a†)/√2|α〉 = (α + α∗)/

√2

〈P〉 = 〈α|(a − a†)/ i√2|α〉 = i(α∗ − α)/

√2

〈X2〉 = 〈α|(a2 + a+2 + aa† + a†a)|α〉/2= (

α2 + α∗2 + 2|α|2 + 1)

ΔX2 = 〈X2〉 − 〈X〉2 = 1/2

therefore:Δx = √

�/2mω and 〈x〉 = (α + α∗)√

�/2mω.

Similarly, we obtain:

Δp = √m�ω/2 and 〈p〉 = i(α∗ − α)

√m�ω/2 ,

and for all α:Δx Δp = �/2 .

Since the lower bound of the Heisenberg inequality is attained whatever the valueof α, the X or P representation of |α〉 is a Gaussian function of X or P . We willcheck it explicitly in the following. We remark that 〈x〉 and 〈p〉 can be as large asone wants if we increase |α| whereas Δx and Δp remain constant (and of coursecompatible with the uncertainty relations). As for the energy, the position andmomentum become well defined in relative value as |α| becomes large.

8.8 Problem. Quasi-Classical States of the Harmonic Oscillator 207

4. Time evolution: we start with |ψ(0)〉 = |α〉, hence:

|ψ(t)〉 = e−|α|2/2∞∑

n=0

αn

n !e−i(n+1/2)ωt |α〉 = e−iωt/2|αe−iωt 〉.

Therefore |ψ(t)〉 is an eigenstate of a with the eigenvalue β = αe−iωt . From theresult of question (3), we obtain:

〈X〉t = (αe−iωt + α∗eiωt )/√2 〈P〉t = i(α∗eiωt − αe−iωt )/

√2 .

Setting α = α0eiϕ with α0 > 0, we obtain:

〈x〉t = α0

√2�

mωcos(ωt − ϕ) = x0 cos(ωt − ϕ)

〈p〉t = −α0

√2m�ω sin(ωt − ϕ) = −p0 sin(ωt − ϕ)

with, naturally, Δxt Δpt = �/2. The time evolution of 〈x〉t and 〈p〉t is the sameas for a classical oscillator (cf. Exercise 3).

5. We have

P = p√m�ω

= −i

√�

mω

∂

∂x= −i

∂

∂X

and similarly

X =√mω

�x = i

√m�ω

∂

∂ p= i

∂

∂P.

In terms of the variable X , we find:

1√2

(X + ∂

∂X

)ψα(X) = αψα(X)

whose solution is ψα(X) = C exp(−(X − α

√2)2/2

). In terms of the variable

P , we have:i√2

(P + ∂

∂P

)ϕα(P) = αϕα(P)

and the solution is ϕα(P) = C ′ exp(−(P + iα

√2)2/2

).

Thewave function is a real Gaussian centered at 〈X〉multiplied by a planewave ofwave vector 〈P〉. This wave function is called a minimal wave packet because theprobability distribution is the same as for the ground state of the oscillator, exceptthat it is shifted by X0 = �(α

√2). In particular the Heisenberg inequality is

saturated at any time. The time evolution consists in replacing α by α exp(−iωt).

208 8 Algebra of Observables

The oscillation of the center of the wave function is the same as that of a classicaloscillator.

The statesα are genuine quantum states in the sense that they satisfy all conditionsof quantum mechanics. However, the physical properties of an oscillator preparedin the state |α〉, with |α| � 1, are very similar to those of a classical oscillator.Traditionally, these states are called “quasi-classical” or “coherent” states of theharmonic oscillator and they play a key role in the quantum theory of radiation.

8.9 Problem. Benzene and C8 Molecules

Consider the states of an electron in a hexagonal molecule C6 formed of 6 equal-spaced atoms. The distance between two neighboring atoms is denoted d. We note|ξn〉, n = 1, . . . , 6 the states localized respectively in the vicinity of the atomsn = 1, . . . , 6. We assume that 〈ξn|ξm〉 = δn,m . The Hamiltonian H of this systemis defined in the basis {|ξn〉}, by H = E0 I + W with:

W |ξn〉 = −A(|ξn+1〉 + |ξn−1〉)

and A > 0. We use here the cyclic conditions |ξ7〉 ≡ |ξ1〉 and |ξ0〉 ≡ |ξ6〉. We note|ψn〉 and En , n = 1, . . . , 6 the eigenstates of W and the corresponding eigenvalues.For simplicity, we chose the origin of energies such that E0 = 0.

We define the rotation operator R by R|ξn〉 = |ξn+1〉.1. What are the eigenvalues λk k = 1, . . . , 6 of R?2. The eigenvector corresponding to λk is noted |φk〉 = ∑6

p=1 ck,p|ξp〉. Write therecursion relation between the coefficients ck,p and determine these coefficientsby normalizing |φk〉.

3. Check that the vectors |φk〉 form an orthonormal basis of the six-dimensionalspace under consideration.

4. Check that the same vectors |φk〉 are eigenvectors of the operator R−1 = R†

defined by R|ξn〉 = |ξn−1〉 and calculate the corresponding eigenvalues.5. Show that W and R commute. What conclusions can we draw from that?6. Express W in terms of R and R−1. Deduce the eigenstates of W and the corre-

sponding eigenvalues. Discuss the degeneracies of the energy levels.7. Consider now a regular 8-center closed chain of atoms (called the cyclo-

octatetraene molecule).

(a) Using a method similar to the preceding one, deduce the energy levels for anelectron moving on this chain. Discuss the degeneracies of these levels.

(b) At time t = 0 the electron is assumed to be localized on the site n = 1,|ψ(t = 0)〉 = |ξ1〉. Calculate the probability p1(t) to find the electron againon the site n = 1 at a later time t ; we set ω = A/�.

8.9 Problem. Benzene and C8 Molecules 209

(c) Does there exist a time t �= 0 for which p1(t) = 1? Explain why. Is thepropagation of an electron on the chain periodic?

8. Consider now an electron on a closed chain of N sites, located regularly on acircle with a distance d between two adjacent sites. The states localized in thevicinity of each center n = 1, . . . , N are denoted |ξn〉. The Hamiltonian is definedas above by H = E0 I + W with W |ξn〉 = −A(|ξn+1〉 + |ξn−1〉) and A > 0. Byextending the argument above, calculate the energy levels and the correspondingeigenstates. What happens in the limit of a chain of infinite length?

8.9.1 Solution

1. One has obviously R6 = I . Therefore, λ6k = 1 and λk = e2ikπ/6 k = 1, . . . , 6.

2. The definition R|φk〉 = e2ikπ/6|φk〉 gives the recursion relation

e2ikπ/6ck,p = ck,p−1 .

Therefore, we have up to an arbitrary phase factor ck,p = e−2ikpπ/6/√6.

3. One has 〈φk |φk ′ 〉 = (1/6)∑

p e−2i(k ′−k)pπ/6 = δk,k ′ .

4. A direct calculation gives R−1|φk〉 = e−2ikπ/6|φk〉. The eigenvalues of R−1 = R†

are λ−1k = λ∗

k .5. A direct calculation gives W R|ξk〉 = |ξk〉 + |ξk+2〉 = RW |ξk〉. Therefore W and

R, which are respectively Hermitian and unitary operators, commute and possessa common eigenbasis.

6. We have W = −A(R + R−1). The eigenvectors of W are therefore:

|φk〉 = 1√6

6∑

p=1

e−2ikpπ/6|ξp〉 k = 1, . . . , 6 ,

with eigenvalues Ek = −2A cos(2kπ/6). The ground state E6 = −2A is non-degenerate, the levels E5 = E1 = −A and E4 = E2 = A are twice degenerate,the level E3 = 2A is non-degenerate.

7. (a) Using a method similar to the preceding one, we obtain the eight followingenergy levels:• E8 = −2A is the ground state (non degenerate).• E7 = E1 = −A

√2 (two-fold degenerate),

• E6 = E2 = 0 (two-fold degenerate),• E5 = E3 = A

√2 (two-fold degenerate),

• E4 = 2A (non degenerate).

210 8 Algebra of Observables

(b) Using 〈φk |ξ1〉 = eikπ/4/√8, we write:

|ψ(t = 0)〉 = |ξ1〉 = 1√8

8∑

k=1

eikπ/4|φk〉

and therefore:

|ψ(t)〉 = 1√8

8∑

k=1

eikπ/4 e−i Ek t/� |φk〉 .

The probability to find the electron again on the site n = 1 is p1(t) =|〈ξ1|ψ(t)〉|2, i.e. p1(t) = |(1/8)∑

k e−i Ek t/�|2. This yields, puttingω = A/�:

p1(t) =∣∣∣∣∣1

8

8∑

k=1

e−i Ek t/�

∣∣∣∣∣

2

=∣∣∣∣1

4

(1 + 2 cos(ωt

√2) + cos(2ωt)

)∣∣∣∣2

.

(c) We find of course p1(0) = 1. To get p1(t) = 1 at a later time, one shouldfind t �= 0 such that cos(ωt

√2) = 1 and cos(2ωt) = 1. This would mean

that ωt√2 = 2Nπ and 2ωt = 2N ′π with N and N ′ integers. Taking the ratio

of these two quantities, we see that one should have√2 = N ′/N , i.e.

√2

rational! Consequently, the particle never reaches again its initial state on thesite 1, and the evolution of the state of the system is not periodic. However onecan show that the system comes back as close to the initial state as onewishes,if one waits for a long enough time, which is called ergodicity. This type oftime evolution is said to be quasi-periodic, it has similarities with quasi-cristalline spatial structures. Notice that with 2 centers (NH3 molecule), 4centers, or with Benzene, with 6 centers, this does not happen because theenergy levels have rational ratios and the evolution is periodic. Above n = 6centers, one always observes the same ergodic phenomenon; the values ofcos(2Kπ/n) have always irrational ratios for integer values of k.

8. These results can readily be extended to N centers, as we shall examine inthe next problem. The levels are En = −2A cos(2nπ/N ) which are all twicedegenerate except the ground state n = N and the highest level n = N/2 ifN is even. These levels populate an energy band of fixed width 4A, whichbecomes a continuum in the limit N → ∞. The eigenstates are of the form|φn〉 = (1/

√N )

∑Np=1 e

−2inpπ/N |ξp〉.

8.10 Problem. Conductibility of Crystals; Band Theory

The previous results above on cyclic molecules give an explanation for the electricconductibility of metals. At first sight, this phenomenon is paradoxical. Metals arecompact cristalline structures, with atomic distances of the order of an angström,

8.10 Problem. Conductibility of Crystals; Band Theory 211

and, from a mechanical point of view, one would imagine that the mean free path ofan electron in a medium of such a large density is extremely small (a few angströms).

The quantum result that we want to elicit here is that, contrary to first intuition,an electron in such a periodic structure can propagate freely provided its energy islocated inside an energy band that generalizes, in the case of a periodic potential, thenotion of energy levels in simple potentials. That free propagation entails the electricconductibility of materials.

Electrons in a Periodic Potential

Instead of a circular structure as for benzene and octene, we consider the behaviorof an electron in a long linear periodic chain of potential wells. We consider a one-dimensional problem.Weassume there are on the x axis an infinite set of equal-spacedattractive atomic centers. A given center is at a distance xn = n d from the origin(Fig. 8.1).

In that set of potentials, we place an electron of mass m. The detailed structure ofthe potential is of no importance here. Let V (x) be the potential of the center n = 0.In the absence of other centers the hamiltonian of the electron in this potential is

h = p2/2m + V (x).

We assume V (x) has only a single bound state of energy E0 and wave function ϕ(x)

hϕ(x) = E0ϕ(x).

This wave function has an extension of the order of λ0, i.e. it decreases as e−|x |/λ0

(or faster) when |x | → ∞.We want to find the eigenfunctions and eigenvalues of the hamiltonian

H = p2/2m ++∞∑

n=−∞V (x − xn).

The method used for the ammonia molecule in Chap. 7 is going to simplify greatlythe problem. We will make a matrix model for the hamiltonian similar to what we

Fig. 8.1 Periodic potential seen by an electron in a crystal

212 8 Algebra of Observables

Fig. 8.2 One-dimensional square wells. Wave function φ(x) localized on one of the wells

have seen for the ammonia molecule NH3.

In fact we can approximate the row of atoms to the limit as N → ∞ of a set of Nsquare well potentials of width a and depth V0 (Fig. 8.2).We assume that the distanced between the centers is much larger than the extension λ0 of the wave function ϕ(x)in a single center, d � λ0.

In other words, we place ourselves in conditions such that the tunneling betweentwo neighboring centers is small but non-zero, but that the tunneling between moredistant centers is negligible.

Basis of Localized States

(i) The eigenfunctions of the hamiltonian H are equally distributed on the differentpotential wells. In the energy eigenstates, the electron has the same probability to beon all sites.(ii) Appropriate linear combinations of the eigenfunctions localize the electron in agiven potential well; we note by ψn the orthonormal combination that localizes theelectron in the vicinity of the potential well n.

The state |n〉 of an electron localized on the potential well n corresponds to thewave function ψn(x), with 〈n|n′〉 = δn,n′ .

The periodicity of the problem (invariance of the hamiltonian in the transformationx → x + m d, m integer) implies that

ψn(x) = ψ0(x − xn)

where xn = n d.(iii) Since, by assumption, the tunneling is weak, the functions ψn(x) are close to theatomic wave functions ϕ(x − xn) that the electron would have on the site n in theabsence of the other sites: ψn(x) ∼ ϕ(x − xn).

Of course, the set {ϕ(x − xn)} is not orthogonal, but, within our assumptions,∫ϕ∗(x − xn)ϕ(x − xm)dx is small for m �= n.

(iv) The set of states {|n〉} is an orthonormal basis of the space of the electron’s states,which we call: the basis of localized states (this can be considered as a definition).

8.10 Problem. Conductibility of Crystals; Band Theory 213

It is in that basis that we make a model for the hamiltonian in a very similar manneras what we did in Eq. (7.16).

Model Hamiltonian

Themodel hamiltonian in the basis {|n〉} of localized states is as follows The diagonalelements are simply equal to E0, the energy of an electron in a center in the absenceof coupling of the centers by quantum tunneling:

〈n|H |n〉 = E0

We set equal to −A, where A is a (small) real constant, the elements of the two firstparallels to the main diagonal

〈n|H |n + 1〉 = 〈n − 1|H |n〉 = −A.

This energy A characterizes the possibility for the electron to jump from site n to theimmediate neighboring sites: n + 1 and n − 1.

Finally, we assume that all other matrix elements of H vanish, since the tunneleffect decreases exponentially with the distance.

〈n|H |n + m〉 = 0, |m| ≥ 2

In the basis {|n〉} of localized states, the hamiltonian is represented by infinitematrix:

H =

∣∣∣∣∣∣∣∣∣∣∣∣∣∣

Column n − 2 n − 1 n n + 1 n + 2.. . . . . . . . . . . . . . . . . . . . . . .

. . . . . . 0 −A E0 −A 0 0 0 . . . . . .

. . . . . . 0 0 −A E0 −A 0 0 . . . . . .

. . . . . . 0 0 0 −A E0 −A 0 . . . . . .

. . . . . . 0 0 0 0 −A E0 −A . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . .

∣∣∣∣∣∣∣∣∣∣∣∣∣∣

Row:

n-1n

n+1n+2:

(8.57)

Stationary States, Energy Bands

1. We write the stationary states as

|Ψk(t)〉 = e−iEk t

� |Ψk〉, where |Ψk〉 =∑

n

Ckn |n〉,

and Ek is the corresponding eigenvalue.

214 8 Algebra of Observables

Write the eigenvalue problem which determines the eigenvalues Ek and thecorresponding complex coefficients {Ck

n }.2. Show that these equations are all satisfied if one sets

Ckn = eikxn (xn = n d)

where k is an arbitrary wave number (or a wave vector).3. What is the eigenvalue Ek of the energy corresponding to a given value of k?4. Why can one restrict the value of k to the interval

−π/d ≤ k ≤ π/d ?

5. What is the degeneracy of a given energy level Ek?6. Show that in a stationary state, the probability |Ck

n |2 to find the electron on agiven site is the same on all sites. (Since the linear chain is of infinite length, thewave functions Ψk(x) are not normalizable.)

7. Show that the energy E of the electron can take any value in an interval betweenE0 − 2A and E0 + 2A, called an energy band of width 4A.

8. Write the wave functionΨk(x) of an eigenstate, called a Bloch function, in termsof the localized wave functions ψn(x). Make use of the translational invarianceto express that in terms of ψ0(x − xn).

9. Replacing the variable k by a more physical quantity

q = �k

and writing E(q) instead of E(k), rewrite the form of a stationary wave func-tion and its approximation in terms of the atomic wave functions ϕn(x). Showthat the coefficients of the atomic wave functions are the values at x = xn of amonochromatic plane wave Φ(x, t).

10. Interpret the result as the free propagation of a particle of momentum q andenergy E(q).

11. Build a wave packet of such free waves.12. Assuming the energy E(q) is close to the minimum of the energy band,

kd = qd/� � 1

show that this corresponds to the propagation of a particle of effective mass meff

which is different from the electron mass:

meff = �2/2A d2.

8.10 Problem. Conductibility of Crystals; Band Theory 215

8.10.1 Solution

1. The stationary states|ψ(t)〉 = e

−iEt�

∑

n

Cn|n〉

are obtained by solving the eigenvalue problem

H |ψ〉 = E |ψ〉

or ∑

m

HnmCm = E Cn

which determines the eigenvalues E and the corresponding complex coefficients{Cn}.This is an infinite set of linear equations, part of which is

⎧⎪⎨

⎪⎩

−A Cn−2 + E0 Cn−1 − A Cn = E Cn−1

−A Cn−1 + E0 Cn − A Cn+1 = E Cn

−A Cn + E0 Cn+1 − A Cn+2 = E Cn+1

These equations are all satisfied if one sets

Cn = eikxn (xn = n d)

where k is an arbitrary wave number (or a wave vector)2. This corresponds to the energy eigenvalue

E(k) = E0 − 2A cos(k d)

3. The solutions are invariant under the transformation kd → kd + 2mπ (m inte-ger), so one can restrict kd to the interval [−π,+π] or

−π/d ≤ k ≤ π/d.

4. E(k) is invariant under k → −k; there is a twofold degeneracy.5. For a givent value of k, we have

|Ψ (t)〉 = e−iE(k)t/�|Ψk〉 with |Ψk〉 =∑

n

eikxn |n〉.

In a stationary state, the probability |Cn|2 to find the electron in the vicinity ofeach center is the same.

216 8 Algebra of Observables

Fig. 8.3 Energy band in an infinite one-dimensional periodic structure

The state |Ψk〉 is therefore not normalizable, because we have assumed the crys-tal is infinite.

6. The energy E of the particle can take any value between E0 − 2A and E0 + 2A.This is called an allowed energy band, of width 4A (Fig. 8.3).

7. The stationary state |Ψk〉 is

Ψk(x) =∑

n

eikxnψn(x)

and since ψn(x) = ψ0(x − xn), this can be rewritten as

Ψk(x) = eikx∑

n

e−ik(x−xn)ψ0(x − xn),

called a Bloch function. In this form, the stationary states appear as the productof eikx by a periodic function of x , of period d , invariant under the transformationx → x + md, m integer. This reflects the invariance property of the problem.

8. Setting q = �k and writing E(q) instead of E(k), we obtain

ψq(x, t) =∑

n

e−i(E(q)t−qxn)/�ψn(x) ∼∑

n

e−i(E(q)t−qxn)/�ϕn(x)

where ϕ(x) is the atomic wave function (in the absence of coupling) and wehave set ϕn(x) = ϕ(x − xn).

9. In this form the stationary wave function appears as a superposition of atomicwave functions of each site, whose coefficients are the values at x = xn , i.e. atthe site, of the function

8.10 Problem. Conductibility of Crystals; Band Theory 217

Fig. 8.4 Propagation of a plane wave inside an energy band

Φ(x, t) = e−i(E(q)t−qx)/�.

This function Φ is a monochromatic plane wave propagating alongs +Ox (q >

0) or −Ox (q < 0), which, by itself would represent the freemotion in space ofa particle of momentum q and energy E(q).The function ψ is a sum of the local, atomic, functions ψn modulated by theenveloping function Φ(x, t) that propagates freely.The real parts of ψq(x, t) and of Φ(x, t) at t = 0 are represented (Fig. 8.4).

10. In order to observe a propagation of the electron we must build a wave packetof the type ψ(x, t) = ∑

n γn(t)ψn(x) where

γn(t) =∫

g(q)e−i(E(q)t−qxn)/�dq/(2π�)1/2

where γn(t) is the value at x = xn of the function

Γ (x, t) =∫

g(q)e−i(E(q)t−qx)/�dq/(2π�)1/2.

which is a wave packet where the relation between E(q) and q is more compli-cated than for a free particle in space.

11. Assuming that the wave packet consists in a superposition of states near theminimum of the energy band, i.e. kd = qd/� � 1, only the small values of qwill contribute and one can expand E(q) as

E(q) � E0 − 2A + Ad2q2

�2.

where the relation between the energy and the momentum is the same as for afree particle of effective mass different from the electron mass

218 8 Algebra of Observables

meff = �2/2A d2.

In general, for electrons in a good metal conductor, the effective mass is of thesame order as the electron mass; meff/me is of the order of 1, 6 for Be, 1, 2for Na, 1, 0 for Cu. That is not at all the case in a semiconductor, for instancemeff/me = 0, 067 for Gallium Ga As.Note that if we had enriched our model by adding terms−B on the next parallelsto the diagonal, the result would be similar

E(k) = E0 − 2A cos kb − 2B cos 2kb.

Chapter 9Approximation Methods

In quantum mechanics the number of problems for which there exist analyticalsolutions is rather restricted. In general one must resort to approximation meth-ods. In this chapter, we present two of these methods: perturbation theory and thevariational method. We are mainly interested in applications.

9.1 Perturbation Theory

9.1.1 Definition of the Problem

Consider the eigenvalue problem:

H |ψ〉 = W |ψ〉 (9.1)

where the Hamiltonian H is the sum of a dominant term H0, whose eigenvalues andeigenstates are known, and a perturbation which we write as λH1, where λ is a realparameter:

H = H0 + λH1. (9.2)

The solution of the eigenvalue problem of H0 is:

H0|n, r〉 = En|n, r〉, r = 1, 2, . . . , pn (9.3)

where the degeneracy of the eigenvalue En is pn , and where the pn orthonormaleigenstates |n, r〉 with r = 1, 2, . . . , pn span the eigensubspace En . We assume thatthe term λH1 is sufficiently weak to bring only small perturbations to the spectrumof H0.

© Springer International Publishing Switzerland 2016J.-L. Basdevant, Lectures on Quantum Mechanics,Graduate Texts in Physics, DOI 10.1007/978-3-319-43479-7_9

219

220 9 Approximation Methods

Power Expansion of Energies and Eigenstates

We assume that the energy levels W of H vary analytically in λ. Therefore, if λ issmall, these levels and the corresponding states will be close to those of the non-perturbed Hamiltonian H0.

Perturbation theory consists in expanding |ψ〉 and W in powers of λ:

|ψ〉 = |ψ0〉 + λ |ψ1〉 + λ2 |ψ2〉 + . . . (9.4)

W = W (0) + λ W (1) + λ2 W (2) + . . . (9.5)

and in calculating the coefficients of the expansion.The method consists of inserting these expansions in the eigenvalue equa-

tion (9.1):

(H0 + λ H1

) (|ψ0〉 + λ |ψ1〉 + . . .) =

(W (0) + λ W (1) + . . .

) (|ψ0〉 + λ |ψ1〉 + . . .)

(9.6)

and in identifying each order in powers of λ:

H0|ψ0〉 = W (0)|ψ0〉 (9.7)

H0|ψ1〉 + H1|ψ0〉 = W (0)|ψ1〉 + W (1)|ψ0〉 (9.8)

H0|ψ2〉 + H1|ψ1〉 = W (0)|ψ2〉 + W (1)|ψ1〉 + W (2)|ψ0〉 (9.9)

. . . = . . .

We have to take into account the normalization condition:

1 = 〈ψ|ψ〉 = 〈ψ0|ψ0〉 + λ(〈ψ0|ψ1〉 + 〈ψ1|ψ0〉) + . . . (9.10)

which yields:

〈ψ0|ψ0〉 = 1 (9.11)

Re〈ψ0|ψ1〉 = 0 (9.12)

. . . = 0

Since the description of the perturbed state is done in the same Hilbert space asfor the unperturbed state, each term |ψi 〉 can be expanded in the original eigenbasisof H0:

|ψi 〉 =∑

n

pn∑

r=1

γin,r |n, r〉 (9.13)

9.1 Perturbation Theory 221

The series of Eqs. (9.7)–(9.11) provides recursion relations for calculating all terms|ψi 〉 andW (i). At a given order, we get the corresponding approximation to the exactsolution.

We note that Eq. (9.7) implies that |ψ0〉 is an eigenvector of H0, and that W (0) isan eigenvalue of H0. Therefore, to lowest order:

W (0) = En (9.14)

and |ψ0〉 is a vector of the corresponding eigensubspace En .

9.1.2 First Order Perturbation Theory

First Order Perturbation in the Non Degenerate Case

If the level En is not degenerate, we simply note |n〉 the corresponding eigenvector.The solution to first order is particularly simple. Let |ψn〉 = |ψ0

n〉 + λ|ψ1n〉 + . . . be

the perturbed state and Wn = W (0)n + λ W (1)

n + . . . the corresponding energy level.Equation (9.7) then implies:

|ψ0n〉 = |n〉 W (0)

n = En, (9.15)

i.e. the perturbed state and energy level are close to the unperturbed ones. We takethe scalar product of Eq. (9.8) with the vector 〈n|. Taking into account (9.15) and thefact that 〈n|H0 = En〈n|, we obtain, by setting ΔE (1)

n = λ W (1)n ,

ΔE (1)n = 〈n|λH1|n〉. (9.16)

To first order, the energy shift ΔEn of the level En is equal to the expectation valueof the perturbing Hamiltonian in the unperturbed state |n〉.First Order Perturbation in the Degenerate Case

Suppose that the level En of H0 has a pn-fold degeneracy. We note |n, r〉, r =1, . . . , pn an orthonormal basis of the corresponding eigensubspace. In general theperturbation λH1 will lift the degeneracy and the level En will be split in pn sub-levels En + λW (1)

n,q , q = 1, . . . , pn . We denote |ψn,q〉 the corresponding eigenstatesand |ψ0

n,q〉 the zeroth order in λ of each of these eigenstates.Each |ψ0

n,q〉 belongs to the eigensubspace En . There is no reasonwhy |ψ0n,q〉 should

coincide with one of the basis vectors |n, r〉, since these can been chosen arbitrarily.It is rather a linear combination of them, and we have in general:

|ψ0n,q〉 =

pn∑

r=1

Cq,r |n, r〉 (9.17)

where we want to determine the coefficients Cq,r .

222 9 Approximation Methods

We multiply (9.8) on the left by 〈n, r ′| and we obtain:

pn∑

r=1

〈n, r ′|λH1|n, r〉Cq,r = λ W (1)n,q Cq,r ′ . (9.18)

For each value of q , this is nothing but the eigenvalue problem for the pn × pnmatrix 〈n, r ′|λH1|n, r〉. The pn shifts ΔE (1)

n,q = λW (1)n,q of the level En are given by

the solutions of the so-called secular equation1:

∣∣∣∣∣∣∣

〈n, 1|λH1|n, 1〉 − ΔE . . . 〈n, 1|λH1|n, pn〉... 〈n, r |λH1|n, r〉 − ΔE

...

〈n, pn|λH1|n, 1〉 . . . 〈n, pn|λH1|n, pn〉 − ΔE

∣∣∣∣∣∣∣= 0.

We also obtain theCq,r and therefore the eigenstates to zeroth order in λ correspond-ing to these eigenvalues.

Summary

In all cases, degenerate or not, the first order energy shift to a level En is obtained bydiagonalizing the restriction of the perturbing Hamiltonian to the correspondingsubspace.

First Order Perturbation to the Eigenstates

Consider the non degenerate case. Using (9.8) and taking the scalar product with theeigenstate |k〉 for k �= n, we obtain:

(En − Ek) 〈k|ψ1n〉 = 〈k|H1|n〉.

Therefore we can write |ψ1n〉 as:

|ψ1n〉 = |n〉 〈n|ψ1

n〉 +∑

k �=n

〈k|H1|n〉En − Ek

|k〉. (9.19)

Equation (9.11) implies Re(〈n|ψ1

n〉) = 0. By making the change of phase |ψn〉 →

eiλα|ψn〉 in (9.4) we can choose α such that Im(〈n|ψ1

n〉) = 0 without loss of

1Perturbation theory was first used in celestial mechanics by Laplace and Lagrange. The initialpurpose was to calculate the long term (secular) perturbations of the motions of planets around thesun (dominant term), due to the gravitational field of the other planets (perturbation). Poisson andCauchy showed that the problem was basically an eigenvalue problem (6 × 6 matrices for Saturn,up to 8 × 8 for Neptune).

9.1 Perturbation Theory 223

generality. The first order perturbation |ψ1n〉 to the state vector is then completely

determined as:

|ψ1n〉 =

∑

k �=n

〈k|H1|n〉En − Ek

|k〉. (9.20)

9.1.3 Second Order Perturbation to the Energy Levels

We consider the non-degenerate case for simplicity. Using the above result for thefirst order perturbation to the state vector, and taking the scalar product of Eq. (9.9)with the eigenstate |n〉, one obtains the second order correction to the energy of theeigenstates:

ΔE (2)n = λ2W (2)

n = λ2∑

k �=n

|〈k|H1|n〉|2En − Ek

. (9.21)

Example: Anharmonic Potential

Consider a harmonic potential which is perturbed by a quartic potential:

H0 = p2

2m+ 1

2mω2 x2, λH1 = λ

m2ω3

�x4. (9.22)

where λ is a dimensionless real parameter. Using the expression of x in terms ofannihilation and creation operators, we find the shift of the energy level En = (n +1/2) �ω at first order in λ:

ΔE (1)n = λ

m2ω3

�〈n|x4|n〉 = 3λ

4�ω (2n2 + 2n + 1). (9.23)

Remarks on the Convergence of Perturbation Theory

In using the expansions (9.4) and (9.5), we implicitly assumed that the solution couldbe expanded in a power series in λ, therefore that it is analytic in the vicinity of λ = 0and that the series converges for λ sufficiently small.

The case of the anharmonic potential is somewhat pathological in the sense thatone can prove that the power series expansion in λ never converges: the series has avanishing radius of convergence! Nevertheless, the result (9.23) is a good approxi-mation as long as the correction to the unperturbed term (n + 1/2)�ω is small. For afixed value of λ (small compared to unity) this will only occur for values of n smallerthan some value nmax(λ) since the correction increases as n2. This can be understoodphysically since:

224 9 Approximation Methods

• The term proportional to x4 is only small if the extension of the wave function isnot too large; it becomes dominant as soon as 〈x2〉 is large.

• For λ ≥ 0 the potential mω2 x2/2 + λH1 has bound states, while for λ < 0 (evenarbitrarily small) the force becomes repulsive for x sufficiently large. The Hamil-tonian is no longer bounded from below and there are no bound states. Thereforewhen one crosses the value λ = 0, the physical nature of the problem changesdramatically. This is reflected in the mathematical properties of the solution; thereis a singularity at λ = 0 and the power series expansion around the origin has avanishing radius of convergence.

A well-known example of a series which does not converge, but whose first termsgive an excellent approximation of the exact answer is Stirling’s formula, used toapproximate Euler’s Gamma function:

Γ (x) =√2π

x

( xe

)x(1 + 1

12 x+ 1

288 x2+ . . .

).

This is called an asymptotic series, which is safely used in computers, although it isnot convergent.

9.2 The Variational Method

The variational method, which is very convenient for estimating the approximatevalue of energy levels (mostly the ground state) and which is frequently used inquantum chemistry.

The Ground State

The first use of the variational method is to derive an upper bound on the groundstate energy of a quantum system. It is based on the following theorem:Let |ψ〉 be any normalized state; the expectation value of a Hamiltonian H in thisstate is always larger than or equal to the ground state energy E0 of this Hamiltonian:

〈ψ|H |ψ〉 ≥ E0 f or any |ψ〉. (9.24)

To prove this result, we expand |ψ〉 on an eigenbasis of H :

|ψ〉 =∑

n

Cn|n〉 ,∑

n

CnC∗n = 1

with H |n〉 = En|n〉 and by definition E0 ≤ En . Calculating 〈ψ|H |ψ〉 − E0, weobtain:

9.2 The Variational Method 225

〈ψ|H |ψ〉 − E0 =∑

n

EnCnC∗n − E0

∑

n

CnC∗n =

∑

n

(En − E0)|Cn|2 ≥ 0

which proves (9.24). Alternatively, onemay simply observe that if the spectrum of anoperator is bounded from below, the expectation value of this operator is necessarilygreater than or equal to the lower bound of the spectrum.

In practice this result is used in the following way. We choose a state |ψ〉 whichdepends on some parameters and we calculate 〈E〉 in this state. The minimum valuewe find by varying the parameters gives an approximation for the ground state energy,which is furthermore an upper bound for this energy level.

Example

Consider the harmonic oscillator H = p2/2m + mω2 x2/2 and the normalized testfunction:

ψa(x) =√2a3

π

1

x2 + a2.

In this case there is a single variational parameter a and we obtain:

E(a) = 〈ψa|H |ψa〉 =∫

ψa(x)

(− �

2

2m

d2

dx2+ 1

2m ω2x2

)ψa(x) dx .

We can compute E(a) by using:

∫ +∞

−∞dx

x2 + a2= π

a

and its derivatives with respect to a. We obtain:

E(a) = �2

4ma2+ 1

2mω2a2

which is minimum for a2 = �/(mω√2), hence:

Emin = �ω√2.

This gives an upper bound to the exact result �ω/2. The difference between the exactresult and the value derived from the variational method could be further reduced bychoosing more elaborate test functions with several variational parameters. Had wechosen Gaussian functions as the set of test functions, we would have obtained theexact result of course, since the true ground state of H would have been an elementof this set.

226 9 Approximation Methods

Relation with Perturbation Theory

The following result is useful.

The first order of perturbation theory is an upper bound for the ground state energy.Indeed, to first order perturbation theory, the ground state energy is:

W0 = 〈ψ0| (H0 + λH1) |ψ0〉

where |ψ0〉 is the ground state wave function of H0. Because of theorem (9.24), W0

is an upper bound to the ground state energy of H0 + λH1.

Other Levels

One can generalize the variational method to other states by using the followingtheorem:

The function

|ψ〉 −→ Eψ = 〈ψ|H |ψ〉〈ψ|ψ〉

is stationary in |ψ〉 if and only if |ψ〉 is an eigenstate of H .To prove this result we consider a variation |δψ〉 of |ψ〉, i.e. |ψ〉 → |ψ〉 + |δψ〉.

Expanding the above formula to first order, we find:

〈ψ|ψ〉 δEψ = 〈δψ|(H − Eψ)|ψ〉 + 〈ψ|(H − Eψ)|δψ〉.

If |ψ〉 is an eigenstate of H with eigenvalue E , then Eψ = E and (H − Eψ)|ψ〉 = 0.Consequently δEψ = 0 whatever the infinitesimal variation |δψ〉.

Conversely, if δEψ = 0 whatever the variation |δψ〉, we must have:

〈δψ|(H − Eψ)| ψ〉 + 〈ψ|(H − Eψ)|δψ〉 = 0.

This must happen in particular if we make the choice:

|δψ〉 = η (H − Eψ)|ψ〉,

where η is an infinitesimal number. Inserting this in the above formula, we obtain:

〈ψ|(H − Eψ)2|ψ〉 = 0.

The norm of the vector (H − Eψ)|ψ〉 vanishes, therefore:

(H − Eψ)|ψ〉 = 0.

This means that |ψ〉 is an eigenvector of H with the eigenvalue Eψ .

9.2 The Variational Method 227

In practice, we can use this result in the following way. We choose a set ofwave functions (or state vectors) which depend on a set of parameters which wecall α collectively. We calculate the expectation value of the energy E(α) for thesewave functions. All the extrema of E(α) with respect to the variations of α will beapproximations to the energy levels. Of course, these extrema will not be in generalexact solutions, since the choice of test wave functions does not cover the entireHilbert space.

9.3 Exercises

1. Calculations of energy levels

Consider a particle of mass m placed in the isotropic 3D potential V (r) ∝ rβ . Wechoose the normalized Gaussian test function:

ψa(r) = (a/π)3/4 exp(−ar2/2). (9.25)

We find in this state:

〈p2〉 = 3

2a�

2 〈rβ〉 = a−β/2 Γ (3/2 + β/2)

Γ (3/2)

This gives an upper bound for the ground state of:

• The harmonic potential (β = 2) for which we recover the exact result.• The Coulomb potential V (r) = −e2/r ; one finds:

E0 = − 4

3π

me4

�2to be compared with the exact result: − 1

2

me4

�2.

• The linear potential V (r) = gr ; one finds:

E0 =(81

2π

)1/3 (�2g2

2m

)1/3

2.345

(�2g2

2m

)1/3

to be compared with the coefficient 2.338 of the exact result.

2. Uncertainty relations

Using the inequality (9.24) for systems whose ground state is known, we can deriveuncertainty relations between 〈p2〉 and 〈rα〉, where α is a given exponent.

a. The 〈r2〉 〈p2〉 uncertainty relation. Consider a one-dimensional harmonic oscil-lator, whose ground state is �ω/2. Show that the inequality 〈p2〉〈x2〉 ≥ �

2/4 holdswhatever the state |ψ〉. Extend the result to three dimensions.

228 9 Approximation Methods

b. The 〈1/r〉 〈p2〉 uncertainty relation. Consider the hydrogen atom HamiltonianH = p2/2m − e2/r and its ground state energy which is E0 = −me4/(2�

2) (seeChap.11). Show that 〈p2〉 ≥ �

2〈 1r 〉2 for all |ψ〉.3. Comparison of the ground states of two potentials

Consider two potentials V1(r) and V2(r) such that V1(r) < V2(r) in any point r .Show that the energy of the ground state of a particle moving in the potential V1 isalways lower than the energy of the ground state for the particle moving in V2.

4. Existence of a bound state in a potential well

Consider a particle moving at one dimension in a potential V (x) which tends to zeroin ±∞ and which is such that V (x) ≤ 0 for all x . Show that there is always at leastone bound state for this motion. Is this result still valid in three dimensions?

5. Generalized Heisenberg inequalities

Consider the Hamiltonian H = p2/2m + grα where g and α have the same signand where α > −2. The energy levels En of H can be derived from the eigenvaluesεn of the operator (−Δρ + ηρα) (where ρ is a dimensionless variable and whereη = |α|/α) by the scaling law:

En = εn |g|2/(α+2)

(�2

2m

)α/(α+2)

,

as one can check directly by making the scaling r = ρ(�2/(2m|g|))1/(α+2)

.Show, using the variational method, that the following general relation holds:

〈p2〉 〈rα〉2/α ≥ κ �2 with κ = |α| 22/α

( |ε0|α + 2

)(α+2)/α

,

where ε0 is the smallest eigenvalue of the operator −Δρ + ηρα.

Chapter 10Angular Momentum

The rotations of systems play a fundamental role in physics, be it in atomic spectra,in magnetic resonance imaging as well as in the stabilization of space crafts. Theconservation laws of angular momentum play a central role, as important as energyconservation. We make use of this fact when we study the hydrogen atom.

The quantization of angular momentum displays a special feature in that it is uni-versal. Contrary to energy levels, which depend on the system under consideration,the possible discrete values of the angular momentummust be chosen in a given uni-versal cataloguewhichwe establish. They depend on the system under consideration,but they are “ready to wear” and not “made-to-measure.” The reason is that Planck’sconstant has the dimension of an angular momentum, and that � is the natural unitof angular momenta. Hence the quantization involves dimensionless factors.

The field of applications of our results is very broad.

• We use these results in order to study the hydrogen atom or the spin of the electronand other particles.

• The rotation spectra of molecules play an important role, both in chemistry and inastrophysics.

• The angular distributions of final particles in collisions or decays in nuclear andelementary particle physics allow us to determine the structure of particles and thenature of fundamental interactions.

• Last, and not least, all ofmagnetism, onwhichwe base the experimental analysis ofour results, comes from rotating charges. We will discover that there exist angularmomenta that do not have any classical analogues, in particular spin 1/2 to whichwe devote Chap.12 and which is one of the most revolutionary discoveries of the1920s.

© Springer International Publishing Switzerland 2016J.-L. Basdevant, Lectures on Quantum Mechanics,Graduate Texts in Physics, DOI 10.1007/978-3-319-43479-7_10

229

230 10 Angular Momentum

10.1 Fundamental Commutation Relation

10.1.1 Classical Angular Momentum

Our starting point is the correspondence principle. The angular momentum L of aparticle of momentum p and position r with respect to the origin is classically thevector product

L = r × p.

Quantum mechanically we therefore postulate that the vector observable L (i.e.,the set of three observables) corresponding to this angular momentum is L = r× p,or, in the wave function formalism,

L = �

ir × ∇.

However, as said previously, more than this particular representation, it is thecommutation relations of these three observables Lx, Ly, and Lz, that are of interestbecause they are independent of the representation.

These algebraic relations are, as pointed out in (8.2):

[Lx, Ly] = i�Lz, (10.1)

and two other relations obtained by cyclic permutations. These can be put togetherin the compact form

L × L = i�L. (10.2)

10.1.2 Definition of an Angular Momentum Observable

However we shall discover that there exist in nature angular momenta which do nothave a classical analog. Therefore, we take as a general definition of any angularmomentum observable J the algebraic relation between its coordinates

J × J = i�J, or [Jx, Jy] = i�Jz. (10.3)

By definition, any triplet of observables corresponding to the components of a vectorquantity that satisfy this relation is an angular momentum observable. We keep theletter L in the specific case of orbital angular momenta.

We first show how one can obtain numbers (i.e., eigenvalues) by manipulatingthis algebra. This was done by Heisenberg and Jordan in 1925 (but they missed themost interesting part of their results). This method is quite similar to Dirac’s methodfor the harmonic oscillator (Chap.8 Sect. 8.4). The results are obtained in a muchsimpler way than with wave functions.

10.1 Fundamental Commutation Relation 231

10.1.3 Results of the Quantization

The results we obtain are as follows.Consider a vector observable J that has the algebraic properties (10.3) and the

corresponding physical quantity. It is an experimental result that

1. Whatever the system under consideration, a measurement of the square of theangular momentum

J2 = J2x + J2y + J2z (10.4)

gives one of the values

j(j + 1)�2, where 2j is an integer.

2. After finding this value for the square of the angularmomentum, themeasurementof one of the components of J along any axis gives one of the (2j + 1) values

m�, where m ∈ {j, j − 1, . . . ,−j}.

3. In the case of orbital angular momenta (which have a classical analogue) thenumbers j and m are integers.

Actually, the algebra (10.3) is very important in mathematics. It is called the Liealgebra of the group of rotations in three dimensions. And these results were provenas soon as 1913 by Elie Cartan in his classification of Lie groups. Why the physicistsof the 1920s didn’t know it is a mystery! Elie Cartan did not forget to point out theirignorance in his book Lectures on the Theory of Spinors in 1935.

How did Cartan proceed, inasmuch as he did not know the Planck constant �?Because � plays no role in this game. � is the natural unit of angular momenta.In (10.3), if we divide both sides by �

2 we obtain an algebraic relation betweendimensionless operators. Let K = J/�, we obtain

[Kx, Ky] = iKz or K × K = iK.

10.2 Proof of the Quantization

10.2.1 Statement of the Problem

Elie Cartan’s proof is crystal clear. The problem is to find all matrices that satisfythe relation (10.3) and to calculate their eigenvalues and eigenvectors.

1. Jx, Jy, and Jz do not commute. We can only diagonalize one of them. However,they obviously have the same eigenvalues.

232 10 Angular Momentum

2. Looking at the physics of the problem enables us to understand a methodologythat the mathematicians Sophus Lie and Elie Cartan knew at the end of the 19thcentury.

The square of the angular momentum J2 commutes with the three components

[J2, Jx] = [J2, Jy] = [J2, Jz] = 0.

This is a simple and direct calculation. It is easy to understand because J2 does notdepend on the system of axes. It is rotation invariant and it makes no discriminationamong the components.

Therefore, we can diagonalize simultaneously J2 and any one of the componentsof J, for instance, Jz.

Note that J2 and one of the components is the maximal set of operators that arefunctions of only Jx, Jy, and Jz, which can be diagonalized simultaneously. J2 and Jzform a CSCO if one considers only angular momentum.

Our problem is to find the eigenvectors common to J2 and Jz and the correspondingeigenvalues.

Analytic solution

At this point, we could use wave functions. Because L is invariant under dilatation,it acts only on angular variables. The calculation of eigenfunctions and eigenvalueshad been done in the 19th century by Legendre and Fourier.

Ifwework in spherical variableswithOz as the polar axis, r is the radial coordinate,θ the colatitude, andφ the azimuth. The expression of L2 and Lz is not too complicated

Lz = �

i

∂

∂ϕ, (10.5)

L2 = −�2

(1

sin θ

∂

∂θsin θ

∂

∂θ+ 1

sin2 θ

∂2

∂ϕ2

). (10.6)

This is what Legendre and Fourier did.We look for eigenfunctionsYm� (θ,ϕ) common

to L2 and Lz. These are called the spherical harmonics:

L2Ym� = λ2

�2Ym

� , and LzYm� = μ�Ym

� ,

where λ2 and μ are the eigenvalues.However, this resolution of differential equations is tedious and, more important,

we would miss half of the results. The Ym� will be given below.

Cartan’s algebraic relation is much more elegant.

10.2 Proof of the Quantization 233

10.2.2 Vectors |j,m > and Eigenvalues j and m

Welook for the eigenvectors common to J2 and Jz, and the corresponding eigenvalues.We define the vectors |j,m > and the numbers j and m as

J2|j,m〉 = j(j + 1)�2|j,m〉 (10.7)

Jz|j,m〉 = m�|j,m〉 (10.8)

with j ≥ 0 (the eigenvalues of J2 are positive because 〈ψ|J2|ψ〉 ≥ 0). For themoment, no other constraint exists on the possible values of j and m. We assumethese vectors are orthonormal

〈j,m|j′,m′〉 = δj,j′ δm,m′ .

This eigenbasis is unique, because J2 and Jz form a CSCO.There are two indices because two operators are diagonalized simultaneously.The method consists of first finding the eigenvalues of J2, then for a given value

of J2, we search the eigenvalues of Jz in the corresponding eigensubspace.This is geometrically intuitive. We fix the norm |j| of a vector, and we seek

the values of its components. Classically, all values between j and −j are allowed;quantum mechanically, there is only a finite discrete set of values for the projection.

10.2.3 Operators J± = Jx ± iJy

We now follow Elie Cartan. The technique is similar to Dirac’s creation and annihi-lation operators.

i. Consider the two operators J+ and J−:

J+ = Jx + iJy, and J− = Jx − iJy. (10.9)

J+ and J− are Hermitian conjugates of each other J†+ = J−, J†− = J+.We show that these operators enable us to move from one vector to another in the

eigensubspace of J2 by increasing by one unit of � the eigenvalues of Jz.

ii. Commutation relation of J± with J2 and JzBecause J± are linear combinations of Jx and Jy, which commute with J2, they

commute with J2

[J2, J±] = 0. (10.10)

However, J± do not commute with Jz. In fact, using the relations (10.3):

[Jz, J±] = [Jz, Jx] ± i[Jz, Jy] = i�Jy ± i(−i�Jx)

234 10 Angular Momentum

therefore[Jz, J±] = ±�J±. (10.11)

iii. The states J±|j,m〉 and their normalizationWe apply these commutation relations to a vector |j,m >. We see that

J2J±|j,m〉 = J±J2|j,m〉 = j(j + 1)�2J±|j,m〉 (10.12)

JzJ±|j,m〉 = (J±Jz ± �J±)|j,m〉 = (m ± 1)�J±|j,m〉. (10.13)

Therefore the vectors J±|j,m〉 are either eigenvectors of J2 and Jz with eigenvaluesj(j + 1)�2 and (m ± 1)�, or equal to the null vector (J±|j,m〉 = 0) if j(j + 1)�2 and(m ± 1)� are not eigenvalues of J2 and Jz.

Thus, the operators J± can be repeatedly applied to any eigenvector and thisincreases or decreases the value ofm by any integer in the eigensubspace of J2. Theyallow us to move around in this subspace.

However, as our physical intuition tells us, to a given value of j, the projection mis bounded on both sides: −j ≤ m ≤ j.

In fact, the square of the norm of the vector J±|j,m〉 is

‖J±|j,m〉‖2 = 〈j,m|J†±J±|j,m〉 = 〈j,m|J∓J±|j,m〉.

However

J∓J± = (Jx ∓ iJy)(Jx ± iJy) = J2x + J2y ± i[Jx, Jy] = J2 − J2z ∓ �Jz.

Therefore, owing to (10.7), we have:

‖J±|j,m〉‖2 = j(j + 1)�2 −m2�2 ∓m�

2 = (j(j + 1) −m(m ± 1))�2 ≥ 0. (10.14)

We therefore conclude that:

1. Because the square of the norm of a vector is positive, we necessarily have theinequalities:

−j ≤ m ≤ j. (10.15)

2. Ifm+1 ≤ j (orm−1 ≥ −j), the vector J±|j,m〉 is nonzero and it is proportionalto the vector |j,m± 1〉. The proportionality coefficient is deduced from the normcalculated above:

J±|j,m〉 = √j(j + 1) − m(m ± 1) � |j,m ± 1〉 (10.16)

(we make an implicit choice for the phase).

10.2 Proof of the Quantization 235

3. If the eigenvalues m = j and m = −j exist, we have

J+|j, j〉 = 0, J−|j,−j〉 = 0.

10.2.4 Quantization

Consider the maximum value of m, mmax. This means that mmax + 1 is not an eigen-value. Therefore the vector J+|j,mmax〉 is the null vector, and its norm is zero. There-fore, according to (10.14),

mmax = j

and the vector |j, j〉 exists.If we apply repeatedly the operator J− to this vector |j, j〉, we generate a whole

series of eigenvectors of Jz corresponding to the eigenvalues �(j − 1), �(j − 2),and so on. However, there exists a minimum value mmin such that mmin − 1 is notan eigenvalue and the vector J−|j,mmin〉 is the null vector. Therefore, because of(10.14),

mmin = −j.

Consequently, in the repeated application of J− to the vector |j, j〉, there exists aninteger N such that

j − N = −j.

In other words, the eigenvalues of the square of the angular momentum (10.7) aresuch that 2j is an integer

j = N/2. (10.17)

For a given value of j = N/2, the corresponding eigensubspace is of dimension2j + 1 = N + 1. The eigenvalues of Jz corresponding to the set of the N + 1 values

m ∈ {−j,−j + 1, . . . , j − 1, j} (10.18)

are either integers or half integers according to the value of j.There we are! We have found the eigenvalues, that is, the catalogue for which we

were looking. If J is an observable such that J × J = i�J, the eigenvalues of theobservable J2 = J2x + J2y + J2z are of the form j(j + 1)�2, where j is either an integer

or a half integer, positive or zero. The eigenvalues of the observable Jz are of theform m�, where m is an integer or a half-integer.

For a system in an eigenstate of J2 corresponding to the value j, the only possiblevalues of m are the 2j + 1 numbers {−j,−j + 1, . . . , j − 1, j}.

236 10 Angular Momentum

Construction of the States |j,m〉This algebraic analysis allows us to construct the eigenstates |j,m〉 for a given j,starting from (10.16). At this stage, we are on a formal level. Concrete applicationscome later.

The state |j, j〉 satisfiesJ+|j, j〉 = 0,

which defines it, as we show. The states |j,m〉,m = j − n, are obtained by using(10.16) and by repeatedly applying the operator J−:

|j, j − n〉 = γn(J−)n|j, j〉,

where the γn are calculated with (10.16) and the normalization of the |j,m〉.

10.3 Orbital Angular Momenta

Consider now the orbital angular momentum of a particle with respect to the originL = r × p, which we mentioned in the beginning. After the results of the previoussection, if �(�+ 1)�2 are the eigenvalues of L2(� ≥ 0) and m� the eigenvalues of Lz,then 2� and 2m are integers.

However, in this case � and m are integers.

10.3.1 Formulae in Spherical Coordinates

In addition to the formulae (10.5) and (10.6) above, a useful formula for our purposein spherical coordinates is the following. The operators L± have the form:

L± = Lx ± iLy = �e±iϕ

(± ∂

∂θ+ i cot θ

∂

∂ϕ

). (10.19)

10.3.2 Integer Values of m and �

The state of a particle in space can be described by a wave function ψ(r) =ψ(x, y, z). In spherical coordinates, this wave function becomes a function�(r, θ,ϕ). The operator Lz has the very simple form

Lz = �

i

∂

∂ϕ. (10.20)

10.3 Orbital Angular Momenta 237

Consider an eigenstate of the projection on the z-axis of the angularmomentum of theparticle with the eigenvalues m�. The corresponding wave function ψm(r) satisfies

Lzψm(r) = m�ψm(r).

The form (10.20) of Lz gives us the simple ϕ dependence of the wave function

ψm(r) = Φm(r, θ)eimϕ,

where Φm(r, θ) is arbitrary at this point. The function ψm(r) is a particular case ofa wave function ψ(x, y, z). In the change ϕ → ϕ + 2π, x, y and z do not changeand the function ψm is unchanged. It must therefore be a periodic function of ϕ withperiod 2π. Therefore

eimϕ = eim(ϕ+2π) ⇒ ei2πm = 1.

Therefore, in the case of an orbital angular momentum, m must be an integer.In the above analysis, we have seen that m and j differ by an integer. Therefore,

for an orbital angular momentum the value of � is a nonnegative integer.

10.3.3 Spherical Harmonics

The eigenfunctions common to L2 and Lz, denoted Ym� (θ,ϕ), are called the spherical

harmonics. The spherical harmonics associated with the eigenvalues �(� + 1)�2 andm� satisfy:

L2Ym� (θ,ϕ) = �(� + 1)�2 Ym

� (θ,ϕ) (10.21)

LzYm� (θ,ϕ) = m� Ym

� (θ,ϕ). (10.22)

We have just seen that their ϕ dependence is simply (eimϕ), so that they factorizeas

Ym� (θ,ϕ) = F�,m(θ) eimϕ.

The spherical harmonics form a Hilbert basis of square integrable functions onthe sphere of radius one. They are completely defined in the following way.

1. They are normalized:

∫ ∫ (Ym

� (θ,ϕ))∗

Ym′�′ (θ,ϕ) sin θ dθ dϕ = δ�,�′ δm,m′ .

2. Their phases are such that the recursion relation (10.16) which we repeat below,is satisfied and that Y 0

� (0, 0) is real and positive (this is a convention):

238 10 Angular Momentum

L±Ym� (θ,ϕ) = √

�(� + 1) − m(m ± 1) � Ym±1� (θ,ϕ). (10.23)

3. Starting from the relationL+Y �

� (θ,ϕ) = 0 (10.24)

we obtain with (10.19)Y �

� (θ,ϕ) = C(sin θ)�ei�ϕ ,

where the normalization constant C is determined by the above constraints.

Examples of Spherical Harmonics

The spherical harmonics play an important role in atomic and molecular physics.They formwith their linear combinations the atomic orbitals of one-external electronatoms, in particular atomic hydrogen which we describe in the next Chapter. Thefirst are:

� = 0 Y 00 (θ,ϕ) = 1√

4π(10.25)

� = 1 Y 11 (θ,ϕ) = −

√3

8πsin θ eiϕ (10.26)

Y 01 (θ,ϕ) =

√3

4πcos θ (10.27)

Y−11 (θ,ϕ) =

√3

8πsin θ e−iϕ. (10.28)

Fig. 10.1 Graph as afunction of the polar angle θof |Ym

� (θ,ϕ)|2 = |F�,m(θ)|2for � = 0, 1, 2 and |m| ≤ l

10.3 Orbital Angular Momenta 239

In Fig. 10.1, the squares |Ym� (θ,ϕ)|2 = |F�,m(θ)|2 of spherical harmonics cor-

responding to the first values of � and m are represented in polar coordinates interms of θ.

Wave Function of a Particle in an Eigenstate of the Orbital AngularMomentum

Thewave functionsψ�,m(r) of particles in an eigenstate of the orbital angularmomen-tum are therefore of the form

ψ�,m(r) = R�,m(r)Ym� (θ,ϕ).

The radial dependence of these functions is contained in the radial wave functionR�,m(r), which can have any form a priori.

10.4 Rotation Energy of a Diatomic Molecule

It seems that the first one who understood empirically the quantization of angularmomentum was Ehrenfest (in an article published on June 15, 1913) just beforeBohr’s article on the hydrogen atom (published in July 1913). Ehrenfest noticed that� had the dimension of an angular momentum, and he postulated the quantization,without giving any proof, in order to explain the variation with temperature of thespecific heats of diatomic molecular gases (he found an improvement on the theoryof Einstein and Stern).

Diatomic Molecule

A simple illustration of the quantization of the values of L2 is obtained through therotational energy spectrumof amolecule. Such a spectrum is presented inFig. 10.2 forthe diatomic cesiummolecule Cs2. It has been obtained1 by measuring the frequencyof thephotons needed to ionize theCs2 molecules that are formed in avery cold atomicvapor of cesiumatoms (temperature∼100µK).Thedata of Fig. 10.2,which representonly a small fraction of the total spectrum, exhibit a series of peaks characteristic ofa quantized rotational energy.

One can visualize a diatomic molecule formed by two atoms of massM separatedby a distance R as a two-body system bound by a harmonic potential. Classically, ifthe interatomic distance R is in its equilibrium position, the molecule has a rotationalenergy:

Erot = L2

2I, (10.29)

where I = MR2/2 is the moment of inertia of the system and L is its angularmomentum with respect to its center of gravity. In quantum mechanics, this resulttransposes into:

1These data, corresponding to the seventeenth excited vibrational state, are extracted fromA. Fiorettiet al., Eur. Phys. J. D 5, 389 (1999).

240 10 Angular Momentum

Erot(�) = �2�(� + 1)

2I, (10.30)

where the rotational energy is quantized. Formula 10.30 gives a very good accountfor the series of peaks of Fig. 10.2. The distance between two consecutive peaksincreases linearly with the peak index, as expected from:

Erot(�) − Erot(� − 1) = �2

I�.

The moment of inertia deduced from this spectrum corresponds to a distance R =1.3nm between the two cesium atoms. This distance, much larger than the usualinteratomic spacing in diatomic molecules, indicates that the Cs2 dimer is actuallyprepared in a long-range molecular state.

If one investigates the absorption spectrum of the cold molecular gas on a muchwider range, one finds several series of lines such as the one of Fig. 10.2. Each seriescorresponds to a given vibrational state of the molecule. The moments of inertiaassociated with these series differ slightly from one another: this is a consequence ofthe variation of the average distance between the two atoms in the various vibrationalstates of the molecule.

The study of rotational excitations of molecules is an important field of researchin physics, chemistry, and astrophysics.

Fig. 10.2 Rotational spectrum of cold Cs2 molecules, showing the quantization of L2. This spec-trum is obtained by measuring the number of molecular ions produced by a laser beam crossingthe assembly of cold molecules, as a function of the laser frequency ν. The height of each peak isproportional to the population of the corresponding rotational level �. (Courtesy of Pierre Pillet)

10.5 Interstellar Molecules, the Origin of Life 241

10.5 Interstellar Molecules, the Origin of Life

The observation of molecular clouds is of considerable interest because it has led inthe last decades to the discovery of more and more interstellar molecules. At first,these were simply unusual astrophysical observations, but quite rapidly they raisedthe problem of the origin of life.

Rotation Spectra of Molecules

We have mentioned above the rotation spectra of molecules. Considering a moleculeas a rigid rotator with principal axes x, y and z, and corresponding moments of inertiaIx, Iy, and Iz, the energy spectrum comes from the Hamiltonian:

HR = L2x

2Ix+ L2

y

2Iy+ L2

z

2Iz.

If two moments of inertia are equal, for instance, Ix = Iy ≡ I , and the third one isverymuch smaller, as is the case for a diatomicmolecule, the spectrum is particularlysimple. The energy levels are

El,m = �2

(l(l + 1) − m2

2I+ m2

2Iz

).

The energy difference of consecutive levels increases linearly with the angularmomentum �,

Erot(�) − Erot(� − 1) = �2

I�. (10.31)

The CO Molecule

The CO molecule plays an important role in astrophysics. It emits in the millimeterrange.

The symmetric molecules such as H2 or O2 have the defect that they do not havean electric dipole moment (such as the molecule NH3) and therefore they do not havean electric dipole emission.

The nonsymmetricmoleculeCOdoes possess a permanent electric dipolemomentand it emits intensively in these transitions. This molecule has a size R = 0.1128nmand the masses MC = 12amu, MO = 15.99492amu, which leads to a frequency ofthe transition � = 2 → � = 1 of 230.54GHz, or equivalently a series of transitionsof frequencies ν� = 115.27 � with � = 1, 2, . . ..

Carbon and oxygen are comparatively abundant in the interstellarmediumbecausethese elements are synthesized in most stars. The relative abundance of the COmole-cule is 10−5 compared the hydrogen atom, themost abundant element in the universe.In general, its distribution is more dense near the center of galaxies than on the edge.This molecule radiates strongly because it has a permanent electric dipole moment,and it is easy to observe. It is a very useful radioastronomical indicator.

242 10 Angular Momentum

Carbon monoxide CO has a length R = 0.1128nm and masses MC = 12amu,MO = 15.99492amu, which corresponds to a transition frequency � = 2 → � = 1of 230.54GHz.

A famous example, which is very rich and was one of the first to be analyzedis the group of three galaxies M81, M82, NGC 3077. This group is located in theconstellation Ursa Major at 2.5 Mpc. These three galaxies (and other smaller ones)seem to orbit around each other quietly. M82 has an anomalous shape with a centralprominence, perpendicular to its plane.

For a long time the galaxy M82 was considered as irregular, and that its nucleuswas undergoing an eruption or an explosion. In fact, the visible picture, on the left-hand side of Fig. 10.3, shows the unusual shape of a jellyfish. The radioastronomicalobservation of carbon monoxide CO, shown on the right, reveals that actually every-thing is perfectly normal. The central deformation of M82 results from importanttidal effects produced on M82 by the presence of the nearby much more massiveM81 galaxy.

Interstellar Molecules

Many interstellar molecules have been identified by now in molecular clouds, which,in turn, generate stars. Figure10.4 shows one of these natural sources of millimetricemission, a molecular interstellar cloud in the Orion nebula, the closest region of starformation, 1500 light-years from the sun, in the same spiral arm.

At present, more than 200 molecules have been detected. Some of them are notobserved in laboratories because they are too unstable at room temperature, such asacetylenic nitriles HC3N, HC5N, and so on, up to HC11N. These are linear moleculeswhose moments of inertia can be calculated quite simply, therefore their spectra canbe predicted with (10.31). (These molecules are identified by the quantum theory ofangular momentum!). A well-known example is the case of fullerenes, spherical C60

molecules that were identified in 1985 thanks to their spectrum, which is calculable.They were also found in meteorites, and they were synthesized in laboratories. Suchmolecules generated a major breakthrough in nanotechnologies. The 1996 Nobelprize in chemistry was awarded to Harold Kroto, Robert Curl, and Richard Smalleyfor the discovery of this new chapter of the chemistry of carbon.

Fig. 10.3 M82 galaxy. Left “Anomalous” aspect in the optical range with the shape of a jellyfish.Right Radio emission at 230.54GHz of carbon monoxide which shows a perfectly normal distrib-ution in a ring around the galactic nucleus. (Photo credit: John Stawn Ward, Thesis www.jsward.com/publications/thesis.pdf.)

10.5 Interstellar Molecules, the Origin of Life 243

Fig. 10.4 Left Orion nebula. (Photo credit: NASA, C. R. O’Dell and S. K. Wong, http://hubblesite.org/newscenter/newsdesk/archive/releases/1995/45/.) Right Level curves of formaldehyde HCHOin the Trifida nebula, where stars also form. (Photo credit: James E. Brau, University of Oregonhttp://physics.uoregon.edu/~jimbrau/BrauImNew/Chap18/FG18_21.jpg.)

Fig. 10.5 Molecular spectrum from theOrion nebula in the frequency range 215–235GHz. (Credit:Craig Kulesa, http://loke.as.arizona.edu/~ckulesa/research/overview.html.)

Coming back to the Orion nebula, the great discovery lies in a large family oforganic molecules. Figure10.5 shows this diversity in the (small) frequency range213–233GHz. One observes:

• The strong intensity of the 12CO peak and its isotope 13CO• Numerous organic molecules such as HCHO, CH3OH, C2H5OH, and glycine, thesimplest aminoacid (outside the spectral range of the Figure)

• Molecules that are unknown in laboratories such as acetylenic nitriles mentionedabove

244 10 Angular Momentum

The Origin of Life

The observation of such molecules is interesting in many respects. It raises thequestion of the origin of life. In the spectrum of Orion, in Fig. 10.5, two types ofmolecules are particularly interesting: HCN and HCHO, because these are veryreactive molecules with which one can construct quite easily aminoacids, thereforebiological molecules.

These organic molecules and the mechanism of their formation obviously raisesthe question of the place from where life originates. The presence of an aminoacid,glycine, whose formula is NH2CH2COOH is quite significant in this respect. It isquite possible that not only amino acids, but DNA can form in extrastellar regions.

However, this idea must be compared with the amazing observation made inSan Diego, in 1953, by Stanley Miller, who was at that time a young 19-year-oldchemistry student. He set up the initial conditions of the terrestrial atmosphere. Ina mixture of NH3, CH4, H2O, and H2, he provoked artificial thunderstorms withelectric discharges of 60,000V. Within a week, he obtained 10 of the 20 amino acidsthat form living matter! This is a strong argument in favor of the terrestrial beginningof prebiotic chemistry.

This latter result has to be put together with measurements made on objects thatbelong to the solar system, in particular, comets and meteorites. Comets are difficultto observe. They were formed 4.5 billion years ago, at the same time as the sun, inthe external regions of the original nebula. Since then, they have spent most of theirtime in the outer and cold regions of the solar system. Comet nuclei have evolvedvery little since they were formed. Their chemical composition gives an access to thechemical composition of the solar nebula when it was formed, 4.5 billion years ago.In 1997, the Hale–Bopp comet, which came close to the sun, allowed a considerableimprovement of the catalogue of comet molecules. Observations at the MillimetricRadioastronomy Institute (IRAM), and at the Caltech Submillimeter Observatory(CSO) displayed the existence of seven new molecules: sulphur monoxide SO anddioxide SO2, formic acid HCOOH, formamide NH2CHO, cyanoacetylene HC3N,methyl formiate HCOOCH3, and ethanal CH3CHO. These observations confirmedthe presence of HNCO and OCS, which had been identified one year before in thecometHyakutake. Two dozen interesting organicmolecules have now been identifiedin comets.

The chemical composition of meteorites is easier to analyze. In 1969, 10 aminoacids, out of the 20 which form DNA and RNA, were discovered in a meteorite thatlanded inMurchison,Australia.Most of themwere the same as those found byMiller.They were all present in the racemic form (equal amounts of enantiomers) whichexcludes any terrestrial contamination. It is difficult to avoid relating this observationwith the richness of interstellar molecules.

Therefore, the hypothesis that life can have an extraterrestrial or an interstellarorigin cannot be dismissed. Itmaybe that a conjunctionof both processes is evenmorefavorable. For instance, life could originate in the encounter of extraterrestrial aminoacids and terrestrial nucleic acids. This field of research, quite close to fundamentalresearch, opens fascinating perspectives.

10.5 Interstellar Molecules, the Origin of Life 245

Observations of the Rosetta Probe on Churyumov Gerasimenko

In October 2015, ESAs Rosetta spacecraft has made the first in situ detection ofoxygen molecules outgassing from a comet, a surprising observation that suggeststhey were incorporated into the comet during its formation, more than 4 billionyears ago.

Rosetta has been studying Comet 67P for over a year and has detected an abun-dance of different gases pouring from its nucleus. Water vapour, carbon monoxideand carbon dioxide are the most prolific, with a rich array of other nitrogen-, sulphur-and carbon-bearing species, and even noble gases also recorded.

Oxygen is the third most abundant element in the Universe, but the simplestmolecular version of the gas, O2, has proven surprisingly hard to track down, evenin star-forming clouds, because it is highly reactive and readily breaks apart to bindwith other atoms and molecules.

Despite its detection on the icy moons of Jupiter and Saturn, O2 had been missingin the inventory of volatile species associated with comets until now.

It was a great surprise to find an unexpectedly high abundance of O2 from 1–10%relative to H2O, at least an order of magnitude higher than expected from molecularclouds.

This observation means in particular that one must revise seriously our models ofSolar System evolution and the formation of the earth.

10.6 Angular Momentum and Magnetic Moment

How can we directly compare our results with experiment? Molecular spectra areinteresting, but they do not give us access to nonclassical angular momenta.

In fact, we have discovered possible half-integer values of the angular momentum,and we know that angular momenta with classical analogues correspond to integervalues of (j,m). Are such half-integer values mathematical artifacts or do they existin nature?

It is quite possible to measure angular momenta directly, but at our level thedescription of the corresponding experiments would be complicated.

We rely on a phenomenon that is intuitively related to angular momentum, thatis, magnetism, which has numerous applications from nuclear magnetic resonanceto superconductivity.

The experimental evidence for the quantization of angular momenta relies to alarge extent on the fact that when charged particles rotate, they possess magneticmoments.

246 10 Angular Momentum

10.6.1 Classical Model

Classically, a rotating charge distribution has a magnetic moment proportional to itsangular momentum.

We can make a classical model of an atom (let’s say hydrogen for simplicity) byconsidering a particle with mass me and charge −q (the electron), rotating with auniform velocity v along a circle of radius r centered on a charge +q. This positivefixed charge represents the nucleus and it is supposed to be much heavier than theelectron. The angular momentum of this system is:

L = r × p = merv u, (10.32)

where u is the unit vector orthogonal to the orbital plane of the electron. Themagneticmoment of this elementary current loop is:

µ = IS u, (10.33)

where I = −qv/(2πr) is the intensity in the loop, and S = πr2 is the loop area.We then find a remarkably simple relation between the angular momentum and themagnetic moment of this classical system:

µ = γ0L, with γ0 = −q

2me. (10.34)

Note that the proportionality coefficient, called the gyromagnetic ratio, does notdepend on the radius r of the trajectory of the electron, nor on its velocity v. Strictlyspeaking, the presence of an external magnetic field perturbs the electronic motionand modifies this very simple relation, but one can show that this perturbation is veryweak for realistic fields, and we neglect it here.

If we place this magnetic moment in a magnetic fieldB the system has a magneticenergy

WM = −µ · B, (10.35)

and a torqueΓ = µ × B (10.36)

is exerted on the magnetic moment.From (10.35) one could naively expect that the magnetic moment of the atom

would get aligned with the local magnetic field, as does the needle of a compass.However, the proportionality between themagnetic moment and the angular momen-tumgives rise to a radically different phenomenon, analogous to the gyroscopic effect.The evolution equation of the angular momentum is dL/dt = Γ . The proportionalitybetween L and µ then implies:

10.6 Angular Momentum and Magnetic Moment 247

Fig. 10.6 Time evolution ofthe components of amagnetic moment placed in afield B(r) along the z-axis.

dµ

dt= −γ0 B × µ. (10.37)

Consequently, for an atom at r, the magnetic moment does not align with the axisof the local magnetic field B(r), but it precesses around this axis with the angularfrequency:

ω0 = −γ0B(r), (10.38)

as shown in Fig. 10.6. The quantity ω0 is called the Larmor frequency.This precession phenomenon is very important in practice. It is a particular case

of a general theorem of electrodynamics2 proven by Larmor in 1897. This problemwas considered independently the same year by H. A. Lorentz.

10.6.2 Quantum Transposition

The quantum transposition of this result consists of assuming that the same propor-tionality relation remains true in quantum mechanics. Any system in an eigenstateof the square of the angular momentum J2 possesses a magnetic moment µ that isproportional to J

µ = γJ. (10.39)

This is an hypothesis. It is verified experimentally.

2See, for example, J. D. Jackson, Classical Electrodynamics. New York: Wiley, (1975).

248 10 Angular Momentum

For an orbital electron, the gyromagnetic factor is the same as in classical physics

γ0 = −q

2me. (10.40)

In general the gyromagnetic factor γ of (10.39) depends on the value of j in a complexsystem, such as a nucleus.

10.6.3 Experimental Consequences

In a magnetic field B, the system has a magnetic potential energy HM which wededuce from (10.35),

HM = −µ · B. (10.41)

The quantization of angular momenta therefore leads to a perfectly analogousquantization of magnetic moments, up to a coefficient. But measuring magneticmoments is easier and more intuitive than measuring angular momenta.

Orbital Angular Momenta

From the properties of L, we deduce the following results, anticipating the conclu-sions of the next Chapter.

1. Consider an electronmoving in a central potential.We suppose that the electronis in a given energy level En and in an eigenstate of the orbital angular momentumwith eigenvalue �(� + 1)�2. As a consequence of rotation invariance, in the absenceof an external magnetic field, the 2� + 1 states corresponding to m = −�, . . . ,+�

have the same energy En. Let us call these states |n, �,m〉; they are eigenstates ofLz with eigenvalues m�. Using the assumption (10.41), the state |n, �,m〉 is also aneigenstate of μz. The corresponding eigenvalue is μz = γ0m�. The negative quantity

μB = γ0� = −q�

2me∼ −9.27 10−24 J T−1 (10.42)

is called the Bohr magneton. From the properties of orbital angular momenta L, wecan deduce the following.

2. If we place the system in a magnetic field B parallel to z, the degeneracy islifted. The state |n, �,m〉 is an eigenstate of the observable HM , with eigenvalue:

Wm = −mμBB.

10.6 Angular Momentum and Magnetic Moment 249

We therefore expect to observe a splitting of the atomic energy level En into 2� + 1sublevels, equally spaced by the interval �E = −μBB. This is called the Zeemaneffect. It can be observed in a transition En → En′ . In the absence of a magneticfield, this transition occurs at a single frequency (En − En′)/2π�. If we apply a fieldB, several lines appear. The number of such lines is directly related to the angularmomenta � and �′ of the initial and final levels.

Notice that if all angular momenta were orbital, then 2j+1 would always be odd.This simple qualitative prediction existed in classical physics. It wasmade byLorentzand byLarmor in 1897. (They both had inmind a three-dimensional harmonicmotionof electrons.)

10.6.4 Larmor Precession

Another consequence of this proportionality relation between J and µ is the Larmorprecession phenomenon, which takes place at the quantum level for the expectationvalues 〈µ〉. The Ehrenfest theorem yields:

d

dt〈µ〉 = 1

i�〈[µ, HM]〉,

where we assume that only the term HM in the Hamiltonian does not commute withµ. Indeed the other terms of the Hamiltonian are supposed to be rotation invariant.Hence they commute with J and with µ. Owing to (10.3), the commutation relationsof µ are:

µ × µ = i�γµ.

Therefore, a simple calculation yields:

d

dt〈µ〉 = −γB × 〈µ〉.

The expectation value 〈µ〉 satisfies the same equations of motion as those foundabove for the classical quantity (10.37). This comes from the fact that the Hamil-tonian is linear in µ. The measurement of the Larmor precession frequency providesa direct determination of the gyromagnetic ratio γ and a consistency test of theresults. We therefore have an experimental means to measure angular momenta viathe measurements of magnetic moments.

250 10 Angular Momentum

10.6.5 What About Half-Integer Values of j and m?

To conclude this Chapter, we come back to the half-integer values of j andm that wefound in the general derivation of the eigenvalues of angular momenta. Concerningan orbital angular momentum, we did not accept such values. Nevertheless, onemay wonder whether these values appear in Nature, or whether they are simply amathematical artifact.

If all angularmomenta are orbital angularmomenta, j is an integer and 2j+1 is odd.Initially, Zeeman performed his experiments with cadmium and zinc (atomswith twoexternal electrons), and he saw, as expected, an odd number of lines. But then therewas a drama. In fact, Zeeman continued his experiments on alkali atoms, sodium,potassium, and he found an even number of lines. This was called the anomalous Zee-man effect. It is astonishing that, for 25 years, nobody was able to explain those evennumbers. The anomalous Zeeman effect seemed to be the greatest challenge givento the physics community. Of course we come back to this when we study spin 1/2.

10.7 Exercises

1. Rotation invariant operator

Show that if an operator A commutes with two components of the angular momentum(e.g. Jx and Jy), it also commutes with the third component (e.g. Jz).

2. Commutation relations for r and p

Prove the following commutation relations:

[Lj, xk] = i�εjk�x� [Lj, pk] = i�εjk�p�,

where εijk = 1 (resp. −1) if (i, j, k) is an even (resp. odd) permutation of (x, y, z),and εijk = 0 otherwise. Deduce the following identity:

[L, p2] = [L, r2] = 0.

3. Rotation invariant potential

Consider a particle in a potential V (r). What is the condition on V (r) in order for Lto be a constant of the motion?

4. Unit angular momentum

Consider a system in an eigenstate of L2 with eigenvalue 2�2, i.e. � = 1.

a. Starting from the action of the operators L+ and L− on the basis states {|�,m〉}common to L2 and Lz, find the matrices which represent Lx, Ly and Lz.

b. Give in terms of the angles θ and ϕ the probability density for a system in theeigenstate of L2 and Lx corresponding to the eigenvalues � = 1 and mx = 1.

10.7 Exercises 251

5. Commutation relations for J2x , J2y and J2z

a. Show that [J2x , J2y ] = [J2y , J2z ] = [J2z , J2x ].b. Show that these three commutators vanish in j = 0 or j = 1/2 states; for example:

〈j,m1|[J2z , J2x ]|j,m2〉 = 0

for any relevant pair m1,m2 in the case j = 0 or j = 1/2.c. Show that they also vanish in j = 1 states. Find the common eigenbasis to J2x , J

2y

and J2z in this case.

Chapter 11The Hydrogen Atom

The explanation of spectroscopic data was one of the first great victories of quantumtheory. In modern science and technology, the mastery of atomic physics is respon-sible for decisive progress ranging from laser technology to the exploration of thecosmos.

The particular case of the hydrogen atom is perhaps the most striking. Its particu-larly simple spectrum delivered the first clues of quantum laws. It has been used as atestbed for the development of quantum theory. Its hyperfine structure is responsibleboth for the hydrogen maser and for a revolution in astrophysics, because the hydro-gen atom is the most abundant element in the universe and its 21-cm line has beenextensively studied in radioastronomy to probe the structure of the interstellar andintergalactic media. Furthermore, the hydrogen atom is probably the physical systemthat is known with the greatest accuracy. It can be calculated “completely” in thesense that the accuracy of present experimental results is the same as the accuracyof theoretical computer calculations, of the order of 10−12 to 10−13 relative accuracy(the only competitor is celestial mechanics).

Here, we first consider technical points, that is, how a two-body problem, wherethe potential depends only on the distance of the particles, reduces to a one-particleproblem. In the case of a central potential, we use the invariance properties of theproblem in order to choose the CSCO made up with the Hamiltonian and the angu-lar momentum, H , L2 and L z , and we show how the traditional quantum numbers{n, l,m} used in atomic physics appear. We study the Coulomb potential and wecalculate the bound state energies of hydrogen in the nonrelativistic approximationand recover the En = −EI /n2 formula obtained by Bohr in 1913.We end with someconsiderations on similar atoms, muonic atoms, where the electron is replaced by aheavier sibling, the muon whose mass is 207 times larger.

Taking into account relativistic kinematics and spin effects requires a formalismnot covered in this book: the Dirac equation. These corrections are small comparedto the leading terms. Up to that point, the problem can be solved analytically. Otherfine structure effects, such as the Lamb shift, require the more elaborate formalismof quantum field theory.

The theoretical treatment of complex atoms (i.e., atoms with more than one elec-tron) involves serious computational problems, even at the nonrelativistic stage. Thehelium atom with its two electrons can only be calculated numerically. Actually

© Springer International Publishing Switzerland 2016J.-L. Basdevant, Lectures on Quantum Mechanics,Graduate Texts in Physics, DOI 10.1007/978-3-319-43479-7_11

253

254 11 The Hydrogen Atom

this calculation was considered as the first true test of quantum mechanics, becausethe much simpler case of hydrogen could be treated successfully by several otherapproaches issued from the “old” Bohr–Sommerfeld quantum theory. Owing to theaccuracy of present numerical calculations, the helium atom is considered to beknown exactly.1

11.1 Two-Body Problem; Relative Motion

Consider a system of two particles, of masses M1 and M2, and of positions r1 and r2,whose mutual interaction is given by a potential V (r1 − r2). The potential dependsonly on the relative position of the particles. The Hamiltonian is:

H = p12

2M1+ p2

2

2M2+ V (r1 − r2), (11.1)

and the system is described by wave functions Ψ (r1, r2).We can separate the global motion of the center of mass of the system and the rela-

tive motion of the two particles. We introduce the position and momentum operatorsof the center of mass

R = M1 r1 + M2 r2M1 + M2

, P = p1 + p2, (11.2)

and the relative position and momentum operators:

r = r1 − r2, p = M2 p1 − M1 p2M1 + M2

. (11.3)

We can rewrite the Hamiltonian as

H = Hc.m. + Hrel, (11.4)

with:

Hc.m. = P2

2M, Hrel = p2

2μ+ V (r). (11.5)

We have introduced the total mass M and the reduced mass μ:

M = M1 + M2, μ = M1M2

M1 + M2. (11.6)

1T. Kinoshita, “Ground state of the helium atom”, Phys. Rev., 105, 1490 (1957).

11.1 Two-Body Problem; Relative Motion 255

Just as in classical mechanics, the Hamiltonian H separates in the sum of (i) theHamiltonian Hc.m. describing the free motion of the center of mass (momentum P ,total mass M) and (ii) the Hamiltonian Hrel which describes the relative motion ofthe two particles in the potential V (r) (momentum p, reduced mass μ).

Let {Xi } and {Pi } be the coordinates of R and P , and {xi } and { pi } those of r andp. The commutation relations are:

[X j , Pk] = i�δ jk, [x j , pk] = i�δ jk (11.7)

and[X j , pk] = 0, [x j , Pk] = 0. (11.8)

In other words, the position and momentum operators of the center-of-mass andof the relative variables obey the canonical commutation relations (11.7), and anyvariable associatedwith the center-of-massmotion commuteswith any other variableassociated with the relative motion (11.8).

These commutation relations imply:

[ P, Hrel] = 0, [ P, H ] = 0, [H , Hrel] = 0. (11.9)

Consequently there exists a basis of eigenfunctions of H that are simultaneouslyeigenfunctions of P and Hrel. The eigenfunctions of P are the plane waves eiK ·R,where K is an arbitrarywavevector. Consequently the desired basis of eigenfunctionsof H has the form:

Ψ (R, r) = eiK ·R ψ(r),

where ψ(r) is an eigenfunction of Hrel:

Hrel ψ(r) = E ψ(r). (11.10)

The eigenvalues Etot of H are

Etot = �2K 2

2M+ E, (11.11)

the sum of the kinetic energy of the global system (Hc.m.) and the internal energy(Hrel).

One consequence of this is the conservation of the total momentum d〈P〉/dt = 0.This is due to the fact that the potential depends only on the relative variable r =r1 − r2; in other words, the Hamiltonian of the system is translation invariant.

Because r and p have canonical commutation relations, the problem reduces tothe quantum motion of a particle of mass μ in the potential V (r). For an atomicsystem made of an electron (M1 = me) and the rest of the atom (M2), we have

256 11 The Hydrogen Atom

M2 � me. Therefore we can neglect the small difference between the reduced massμ and the electron mass me, remembering that it is easy to correct for reduced masseffects if necessary. We are interested in the eigenvalue problem of Hrel.

11.2 Motion in a Central Potential

The Coulomb potential is central, that is, it only depends on the distance r = |r| ofthe two particles. The problem is therefore rotation invariant.

Spherical Coordinates, CSCO

Owing to the symmetry of the problem, it is appropriate to work in spherical coor-dinates.

The LaplacianΔ has the following expression (which is easily obtained bywritingthe square of the angular momentum L = −i�r × ∇ and expanding)

Δ = 1

r

∂2

∂r2r − 1

r2�2L2. (11.12)

Equation (11.10) is then written as

(− �

2

2me

1

r

∂2

∂r2r + L2

2mer2+ V (r)

)ψ(r) = E ψ(r). (11.13)

The Hamiltonian Hrel commutes with the three angular momentum operators L i ,i = x, y, z. Each L i commutes with L2. In addition L i only acts on the variablesθ and ϕ, and it commutes with r, ∂/∂r, V (r). In other words, the HamiltonianHrel, which from now on is denoted H for simplicity, commutes with the angularmomentum:

[H , L] = 0.

Consequently H , L2 and a given component of L (e.g., L z) form a set of commutingobservables. We verify a posteriori that this set is complete by checking that the basiscorresponding to the common eigenfunction is unique.

The relation [H , L] = 0 implies the conservation of the angular momentum:d〈L〉/dt = 0. This is due to the fact that the Hamiltonian of the system is rotationinvariant.

Eigenfunctions Common to H , L2, and L z

Separation of the Angular Variables

Part of the eigenvalue problem (11.13) is already solved because we know the formof the eigenfunctions common to L2 and L z . These are the spherical harmonics. Weseparate the variables in the following way

11.2 Motion in a Central Potential 257

ψ�,m(r) = R�(r) Y�,m(θ,ϕ), (11.14)

L2ψ�,m(r) = �(� + 1)�2 ψ�,m(r), (11.15)

L zψ�,m(r) = m� ψ�,m(r), (11.16)

where � and m are integers, with |m| ≤ �. Substituting in (11.13), the eigenvalueequation becomes:

(− �

2

2me

1

r

d2

dr2r + �(� + 1)�2

2mer2+ V (r)

)R�(r) = E R�(r). (11.17)

This equation is independent of the quantum numberm. This is why we have not putan indexm for the unknown function R�(r) in Eq. (11.14). This differential equationis the radial equation and R�(r) is called the radial wave function.

The normalization of the wave function, which we must impose on finding boundstates, is

∫ |ψ(r)|2 d3r = 1; in spherical coordinates,

∫d2Ω

∫ ∞

0dr r2 |ψ(r, θ,ϕ)|2 = 1.

Here Ω is the solid angle with d2Ω = sin θ dθ dϕ. Because the spherical harmonicsare normalized, we obtain, for the radial wave function R�(r), the condition

∫ ∞

0dr r2 |R�(r)|2 = 1. (11.18)

Introducing the reduced wave function u�(r) = r R�(r), the Schrödinger equationbecomes:

(− �

2

2me

d2

dr2+ �(� + 1)�2

2mer2+ V (r)

)u�(r) = E u�(r), (11.19)

with∫ ∞0 |u�(r)|2 dr = 1. One can prove that any normalizable solution R�(r) is

bounded at the origin, and therefore u�(0) = 0.This equation has the structure of the Schrödinger equation describing the one

dimensional motion of a particle of mass me in the potential:

Veff(r) = V (r) + �(� + 1)�2

2mer2. (11.20)

This effective potential is the superposition of the interaction potential between thetwo particles 1 and 2, and a centrifugal barrier term which is repulsive and increasesas the angular momentum increases, Fig. 11.1.

258 11 The Hydrogen Atom

Fig. 11.1 Effective potential that enters the one-dimensional Schrödinger equation for the reducedradial wave function u�(r). For � = 0 (left) the motion occurs in the “bare” potential V (r); for� �= 0 (right) the effective potential is the superposition of V (r) and of the centrifugal barrier�(� + 1)�2/(2mer2). The Figure is drawn for a Coulomb potential V (r) ∝ 1/r

Quantum Numbers

The energy levels depend on the parameter �, but they do not depend on the projectionm. For each value of �, corresponding to a given value of the square of the angularmomentum, each level has a degeneracy of degree (2� + 1). For a given �, we dealwith a one-dimensional problem of the same type as studied in Chap. 5. The boundstate energy levels (E < 0) correspond to solutions R�(r) that satisfy (11.18).

The Radial Quantum Number n′

For a given �, we can arrange the possible values of the bound state energies inan increasing sequence, which we label by an integer n′ (n′ = 0, 1, 2 . . .), the staten′ = 0 being themost strongly bound. Depending on the potential, this sequencemaybe finite (as for a square well potential) or infinite (as for the Coulomb potential).

The general mathematical properties of the differential equation (11.17), togetherwith the conditions that R�(0) is finite and that R�(r) can be normalized (11.18), showthat this number n′ corresponds to the number of nodes of the radial wave function;the number of times it vanishes between r = 0 and r = ∞. This is independent ofthe form of the potential V (r) (provided it is not too pathological).

The quantum number n′ is called the radial quantum number. A radial wavefunction, defined by the two quantum numbers � and n′ and normalized to unity,is unique (up to a phase factor). The eigenvalues of the Hamiltonian are thereforelabeled in general by the two quantum numbers n′ and �. They do not depend onthe quantum number m as a consequence of the rotation invariance of the system.This means that the 2� + 1 states corresponding to given values of n′ and � and todifferent values of m, have the same energy and are degenerate.

These general considerations apply to any two-body system with a central poten-tial: the hydrogen atom and also to a certain extent alkali atoms, diatomic molecules,the deuteron, and quark systems.

11.2 Motion in a Central Potential 259

The Principal Quantum Number n

In the following section, we solve Eq. (11.19) in the case of a Coulomb potentialV (r) = −q2/4πε0r . In this particular case, the energy levels only depend on thequantity n′ + � + 1. It is therefore customary to label atomic levels with the threequantum numbers �,m, and the positive integer n, called principal quantum number,defined by the relation:

n = n′ + � + 1.

The energy eigenstates are then classified by increasing values of n, (n = 1, 2, 3, . . .).The classification of atomic states by the three integers (n, �,m) is just a redefinitionof a catalogue with respect to the classification in terms of (n′, �,m). For a givenvalue of n, there are only n possible values of �: � = 0, 1, . . . , n − 1. For eachvalue of �, there are 2� + 1 possible values of m. The wave function of an energyeigenstate is labeled with the three corresponding quantum numbers (ψn,�,m(r)) andthe corresponding energy is denoted En,�.

Spectroscopic notation (states s, p, d, f , . . .)

The measurement of the energy levels of an atom often comes from the observationof the wavelengths of its spectral lines. We show in Fig. 11.2 the energies En,� of thevalence electron of sodium and some of the observed transitions. Each horizontalline represents a state; the number on the left is the value of the principal quantumnumber n. Each column corresponds to a given value of �. The energy of the state isgiven on the vertical axis (for instance, E3,0 = −5.13 eV). On the right, we give theenergy levels En of hydrogen, which, as we show, only depend on n.

The quantum theory of the emission of a photon by an excited atom imposesselection rules. In the transition from a state (n, �) to a state (n0, �0) by emissionof a photon of energy �ω = En,� − En0,�0 , all transitions are not allowed. Only thetransitions for which � = �0 ± 1 are intense (see Chap.16, Sect. 16.2.4).

Experimental observations in the 19th century showed that one can group the linesin series which were given names according to their aspect. In the case of sodium,after the theory had been understood, it turned out that these series correspond to thefollowing transitions,

the sharp series �ω = En,�=0 − E3,1

the principal series �ω = En,�=1 − E3,0

the diffuse series �ω = En,�=2 − E3,1

the fundamental series �ω = En,�=3 − E3,2.

Each of these four series corresponds to transitions from a state of given � (andvarious values of n) to a well-defined state. Consequently, the tradition consists ofattributing to a given value of � the initial of the corresponding series (spectroscopicnotation):

Symbolic letter: s p d f g hCorresponding value of� : 0 1 2 3 4 5.

260 11 The Hydrogen Atom

Fig. 11.2 Energy levels of the external electron of sodium (left) and energy levels of hydrogen(right)

A state of well-defined energy is then denoted by a number (the value of n) followedby a letter (corresponding to the value of �):

n = 1, � = 0 : state 1s ; n = 3, � = 2 : state 3d.

11.3 The Hydrogen Atom

We now consider the specific case of the hydrogen atom. Here, we consider theproblem in its first approximation, where we neglect spin effects. We consider theproblem of a particle of mass me in the Coulomb field of the proton, which is con-sidered infinitely massive (the reduced mass correction is straightforward):

V (r) = − q2

4πε0r= −e2

r.

11.3 The Hydrogen Atom 261

q is the elementary charge and we set e2 = q2/4πε0. The radial equation is:

(− �

2

2me

1

r

d2

dr2r + �(� + 1)�2

2mer2− e2

r

)R�(r) = E R�(r). (11.21)

11.3.1 Atomic Units; Fine Structure Constant

The above equation involves three constants: � (action), me (mass), and e2 (productof an energy by a length). Using these three constants, it is useful to form a lengthunit and an energy unit relevant for our problem.

Fine Structure Constant

We notice that e2/� has the dimension of a velocity. Unless the differential equationhad pathologies (which is not the case), it must represent the typical velocity v ofthe electron in the lowest energy levels of the hydrogen atom. This velocity mustbe compared with the velocity of light c, which is the absolute velocity standard inphysics. The ratio between these two velocities is a dimensionless constant α, whichis a combination of the fundamental constants q, �, and c:

α = e2

�c= q2

4πε0�c∼ 1

137.

The smallness of this constantα guarantees that the nonrelativistic approximationis acceptable up to effects of the order of v2/c2 ∼ 10−4. The constant α is called,for (unfortunate) historical reasons, the fine structure constant. A more appropriateterminologywould have been: fundamental constant of electromagnetic interactions.

Any charge Q is an integer multiple of the elementary charge Q = Zq (or an inte-ger multiple of q/3 if one incorporates quarks). Therefore the fundamental form ofCoulomb’s lawbetween two charges Q = Zq and Q′ = Z ′q isV (r) = αZ Z ′(�c/r),with Z and Z ′ integers, which only involves mechanical quantities. The introductionof electric units and of ε0 is only a convenient manner to describe macroscopic caseswhere Z and Z ′ are very large. The experimental determination of the fundamentalconstant α is a key point in physics: 1/α = 137.035 999 139 (031).

The fact that this is a dimensionless number stirred minds at the beginning. Onecannot change the value of α by changing units.

For a long time, after this discovery (i.e., the discovery of the universality ofPlanck’s constant) people have tried to obtain it starting from transcendental num-bers e, π, the Euler constant γ, and so on. For instance e(−π2/2) � 1/139, or better(1/2)π(−e2/2) � 137.3.

The truth came with the great revolution of quantum field theory introduced byKen Wilson in the 1970s. What is called the renormalization group enabled us tounderstand that the value of the fine structure constant, as well as other dimensionless

262 11 The Hydrogen Atom

constants in elementary interactions, depends on the energy, or on the distance. Infact, experiments in the LEP collider at CERN have shown that around 100 GeVcenter of mass energy, the value of the fine structure constant is larger, α � 1/129.There is some hope to calculate it in unified field or superstring theories.

Atomic Units

The length unit of the problem is the Bohr radius:

a1 = �2

mee2= 1

α

�

mec∼ 0.53Å,

where �/mec is the Compton wavelength of the electron. The Bohr radius is thetypical size of an atom.

The energy unit relevant for the hydrogen atom is:

EI = mee4

2�2= 1

2mec

2α2 ∼ 13.6 eV

which, as we show, is the ionization energy of the atom. The electron-volt is a typicalenergy for external atomic electrons.

The atomic time scale is 2π�3/(mee4) ∼ 1.5 10−16 s. It represents the period of

the classical circular motion of the electron around the proton for the energy −EI .

11.3.2 The Dimensionless Radial Equation

Having identified the relevant length and energy scales of the problem, we introducethe dimensionless quantities ρ = r/a1 and ε = −E/EI . We define ε with a minussign so that this quantity is positive if we are dealing with a bound state, whoseenergy is negative. We obtain the following dimensionless equation,

(1

ρ

d2

dρ2ρ − �(� + 1)

ρ2+ 2

ρ− ε

)R�(ρ) = 0. (11.22)

This equation is well known to mathematicians. Laguerre gave the solutions in1860. The following properties can be proven.

1. For each value of �, we obtain a infinite set of normalizable solutions labeled byan integer n′ = 0, 1, . . .:

R(ρ) = e−√ερ ρ� Qn′,�(ρ), (11.23)

where Qn′,�(ρ) = C0 + C1ρ + · · · + Cn′ρn′is called a Laguerre polynomial of

degree n′. It has n′ real zeros between ρ = 0 and ρ = +∞.

11.3 The Hydrogen Atom 263

Table 11.1 Radial wavefunctions Rn,�(ρ) for theCoulomb problem, forn = 1, 2, 3

n = 1 � = 0 2 e−ρ

n = 2 � = 01√2

(1 − ρ

2

)e−ρ/2

� = 11

2√6

ρ e−ρ/2

n = 3 � = 02

33/2

(1 − 2

3ρ + 2

27ρ2

)e−ρ/3

� = 125/2

37/2ρ

(1 − ρ

6

)e−ρ/3

� = 223/2

39/2√5

ρ2 e−ρ/3

2. These normalizable solutions correspond to the eigenvalues

ε = 1

(n′ + � + 1)2. (11.24)

As alreadymentioned above, the integer n′ corresponds to the number of nodes of theradial wave function and is called the radial quantum number. The principal quantumnumber is the integern = n′ + � + 1.Thefirst radialwave functions Rn,�(ρ) are givenin Table11.1. We remark that, owing to (11.24), εn = 1/n2 is an eigenvalue of allradial equations corresponding to values of � smaller than n : � = 0, 1, . . . , n − 1.

We do not give a rigorous proof here that the normalizable solutions of (11.22)can indeed be cast in the form (11.23), but we can check that these solutions makesense both around the origin and at infinity.

Around ρ = 0

The Coulomb term 1/ρ and the constant term ε are negligible compared withthe centrifugal term �(� + 1)/ρ2 (for � �= 0). Assuming a power law dependenceRn,�(ρ) ∝ ρs around ρ = 0, one finds that the only possible exponent s compati-ble with a normalizable solution is s = � (s = −� − 1 is not square-integrable for� ≥ 1). This corresponds to the expansion of (11.23) around ρ = 0. Notice that inan s-wave (� = 0), a function behaving as 1/r is square-integrable; however, it doesnot satisfy the Schrödinger equation because Δ(1/r) = −4πδ(r) where δ(r) is theDirac distribution which is not a square integrable function.

At Infinity

Keeping only the leading terms in the expansion in Rn,�, we have:

Rn,�(ρ) ∼ e−√ερ

(Cn′ρn−1 + Cn′−1ρ

n−2 + · · · )

If one injects this expansion into the differential equation (11.22), it is immediateto check that the term in e−√

ερ ρn−1 always cancels out of the equation, and the

264 11 The Hydrogen Atom

coefficient of the next term e−√ερ ρn−2 is proportional to Cn′(1 − n

√ε). Therefore

this term also cancels out for the particular choice ε = 1/n2. The subsequent termsof the expansion, which depend on the centrifugal barrier, allow the determinationof the coefficients Cn′ ,Cn′−1, . . . ,C0.

Coming back to the initial variables for length and energy, we can summarize theabove results.Each solution of the Schrödinger equation (11.13) corresponding to a bound statefor the Coulomb problem is labeled by three integers (or quantum numbers):

n = 0, 1, 2, . . . , � = 0, 1, . . . , n − 1, m = −�, . . . , �.

The energy of a solution depends only on the principal quantum number n:

En = − EI

n2, with EI = mee4

2�2∼ 13.6 eV.

To each energy level, there correspond several possible values of the angular momen-tum. The total degeneracy (in � and m) of a level with given n is

n−1∑

�=0

(2� + 1) = n2.

The wave function corresponding to a given set n, �,m is unique (up to a phasefactor) and it reads:

ψn,�,m(r) = Y�,m(θ,ϕ) e−r/(n a1)

(r

a1

)�

×(C0 + C1

r

a1+ · · · + Cn−�−1

(r

a1

)n−�−1)

. (11.25)

where the Ck’s (k = 0, . . . , n − � − 1) are the coefficients of the Laguerre polyno-mials and where a1 = �

2/(mee2) ∼ 0.53 Å.

11.3.3 Spectrum of Hydrogen

In Fig. 11.3 we represent the energies En of the hydrogen atom. Each line representsan energy level, the number on the right is the value of n, the column corresponds toa given value of �, and we give the value of the energy on the vertical axis.

The selection rule given for the observable spectral lines of the sodium atom� = �0 ± 1 still holds. The most famous series is the Balmer series. It correspondsto transitions from states ns to the state 2p:

11.3 The Hydrogen Atom 265

Fig. 11.3 Energy levels of hydrogen

�ω = En − E2 = 13.6n2 − 4

4n2eV.

The first lines of the Balmer series are in the visible part of the spectrum (�ω ∼ 2to 3 eV; λ ∼ 0.5 µm). The Lyman series, corresponding to transitions to the groundstate, lies in the ultraviolet (λ ≤ 121.5 nm).

11.3.4 Stationary States of the Hydrogen Atom

The Ground State (1s)

The ground state corresponds to n = 1, therefore � = 0 and m = 0 (1s state in thespectroscopic language). Because the spherical harmonic Y0,0(θ,ϕ) is a constantequal to 1/

√4π, the normalized wave function of this state is:

ψ1,0,0(r) = e−r/a1√

πa31

.

The probability of finding the electron in a spherical shell of thickness dr , representedin Fig. 11.4, is:

266 11 The Hydrogen Atom

Fig. 11.4 Radial probability density P(r), giving the probability of finding the electron between rand r + dr in a hydrogen atom prepared in its ground state

P(r)dr = |ψ1,0,0(r)|2 4πr2 dr.

The probability density per unit volume is proportional to the exponential functione−2r/a1 , and it ismaximumfor r = 0.Themost probable distance between the electronand the proton is the Bohr radius a1 = 0.53 Å.

Other States

Figure11.5 represents the radial probability density Pn,�(r) = r2 |Rn,�(r)|2 for vari-ous states n, �. We note the reduction of the number of nodes of the radial wavefunction as � increases for a given n. For a level n, l, the function Pn,�(r) hasn′ = n − � − 1 zeroes, where n′ is the degree of the corresponding Laguerre poly-nomial. In particular, for � = n − 1, we remark that P(r) has a single maximum,located at a distance r = n2a1 (Eq. 11.25).

Figure11.6 represents some spatial probability densities |ψn,�,m(r)|2 in a planey = 0 (these are axial symmetric functions around the z-axis). For large quantumnumbersn � 1,wenotice that one gets closer to “classical” situations, correspondingto a localized particle.

Fig. 11.5 Radial probability density r2 |Rn,�(r)|2 of the states n = 2, 3, 4 of hydrogen

11.3 The Hydrogen Atom 267

Fig. 11.6 Probability density |ψn,�,m(r)|2 in the y = 0 plane for n = 6, � = 5 (mesh size: 60 a1 ×60 a1). For m = 0, the particle is localized in the vicinity of the z axis. For large |m| (in particularm = ±5), the particle is localized in the plane z = 0, in the vicinity of a circle centered at the origin,of radius r = 30 a1 (circular state). The vertical scale of the surface m = 0 has been reduced by afactor of 2 with respect to the five other surfaces in order to improve the visibility

11.3.5 Dimensions and Orders of Magnitude

Consider a hydrogen atom prepared in a stationary state |n, �,m〉. Using the virialtheorem, one can show that the classical relation between the kinetic energy and thepotential energy still holds for the expectation values of these quantities:

E (kin)n = 〈 p2

2me〉 = −En = EI

n2, (11.26)

E (pot)n = 〈−e2

r〉 = 2 En = −2EI

n2. (11.27)

Using the properties of Laguerre polynomials, one finds that the mean radius hasthe following variation with n and �,

268 11 The Hydrogen Atom

〈r〉 = a12

(3n2 − �(� + 1)

), (11.28)

as well as

〈1r〉 = 1

n2a1, 〈 1

r2〉 = 2

n3 (2� + 1) a21, 〈r2〉 = n2 a21

2

(5n2 + 1 − 3�(� + 1)

).

Also, setting ρ = r/a1, one gets for p > −2� − 1:

p + 1

n2〈ρp〉 − (2p + 1)〈ρp−1〉 + p

4

((2� + 1)2 − p2

) 〈ρp−2〉 = 0.

Remark Because of the n2 variation of the mean atomic radius, the maximal prob-ability density for a given radial wave function decreases like 1/n4. This is why areadjustment of scales is necessary in order to visualize Fig. 11.5 properly.

11.3.6 Historical Landmarks

The importance of this calculation done by Schrödinger in 1926, is not so much inthe result, which was known, but in the fact that it was obtained by a systematicmethod, which could be generalized to other cases.

Spectroscopy

The law involving integers was a crowning achievement for spectroscopy.After Newton’s discovery of the decomposition of white light in 1680, Wollaston,

in 1802, and mainly Fraunhofer, after 1814, had discovered dark and shining linesin the solar spectrum and in stars.

People had realized that spectral lines observed in the light emitted by a bodywerecharacteristic of the elements in this body. The observation of regularities allowedthe attribution of a series of lines to a given element (there was a catalogue of 5000lines in 1890). Very soon, people had the idea that spectral lines were analogous toharmonics in acoustics and that there should exist some simple relations betweenthem.

But that didn’t lead very far. Fundamental physicists considered spectroscopy tobe too complicated and outside the scope of physics, such as the harmonics of apiano, which depend on the shape of the instrument as a whole. Why? Because thespectra were too complicated.

The Spectrum of Hydrogen

One had to start with the spectrum of the simplest atom, hydrogen. But people didn’tknow that, because the spectrum of atomic hydrogen was found late. It was difficultto obtain the spectrum of atomic hydrogen. One had to operate with a discharge tube,

11.3 The Hydrogen Atom 269

and disentangle the spectra of atomic and molecular hydrogen, and of other speciessuch as nitrogen and water.

The first line of hydrogenwas discovered in 1853 by the Swedish physicist AndersJonas Ångström. Then in 20 years, he found three others that were called, includingthe first, α, β, γ, δ.

In 1881 Huggins discovered in the star α of the Lyre, ten other lines “whosesequence seemed to be following the visible lines of hydrogen.” Hydrogen is abun-dant in stars, but one had to guess it at that time!

Balmer

Integers played an important role in science in the 19th century. Examples can befound in chemical reactions, atomic theory, classification and evolution of speciesin zoology and in botany, and so on. It is by chance that, in 1885, Balmer, who wasa high school teacher in Basel, and who was fascinated by numerology, learned thepositions of the first four lines of hydrogen. He realized that the wavelengths of thelines could be represented with an accuracy of 10−3 by a formula involving integers

λ = An2

n2 − 4, with A = 0.3646 µm.

And this law was good to one part in a thousand when applied to the ten lines ofHuggins! Balmer predicted the next ones and the limit.

Although he was not a physicist, Balmer found the simplicity of the formulaquite striking. He called that constant A the “fundamental number” of hydrogen. Inhis 1885 paper, he wrote, “It appears to me that hydrogen . . . more than any othersubstance is destined to open new paths to the knowledge of the structure of matterand its properties.”

In 1912, Niels Bohr, who was 27, was working with Rutherford on an atomicmodel. He was not aware of Balmer’s formula and of analogous results obtained byRydberg for alkali atoms. One day, by chance, he learned the existence of Balmer’sformula; it only took him a few weeks to construct his celebrated model of thehydrogen atom, which is one of the turning points of quantum physics.

11.4 Muonic Atoms

To end this chapter, we describe an application of what we have done in nuclear andelementary particle physics.

The μ lepton, or muon, discovered in 1937, has physical properties of a heavyelectron. As is the electron, it is elementary or pointlike; it has the same electriccharge, the same spin, but it is 200 timesmoremassive:mμ = 206.8me. It is unstableand it decays into an electron and two neutrinos: μ → e + νe + νμ with a lifetime of2 × 10−6 s. Therefore, physically it is a true heavy electron except that it is unstable.

270 11 The Hydrogen Atom

In particle accelerators, one can produce muons, slow them down, and have themcaptured by atoms where they form hydrogenlike systems. In a complex atom, themuon is not constrained with respect to electrons by the Pauli principle. The muonexpels electrons, cascades from one level to another, and eventually falls in thevicinity of the nucleus at a distance aμ = �

2/Zmμe2, which is 200 times smallerthan the distance of the internal electrons. Therefore, it forms a hydrogenlike atom,because it does not feel the screened electron electric field. The lifetime of themuon ismuch longer than the total time of the cascades (∼10−14 s). It is alsomuch longer thanthe typical atomic time �

3/mμe4 ∼ 10−19 s. The muon can therefore be consideredas stable on these time scales.

It then forms a muonic atom. The Bohr radius of a muonic atom is of the sameorder as a nuclear radius. The effect of external electrons is screened. Consider lead(Z = 82), whose nuclear radius is R ≈ 8.5 fm. One finds aμ ≈ 3.1 fm, which meansthat the μ penetrates the nucleus noticeably. In fact, in the ground state, it has a 90%probability of being inside the nucleus. The description of the nucleus as a pointparticle is inadequate. Consequently, spectra of muonic atoms provide informationon the structure of nuclei, in particular concerning their charge distributions.

For a spherical nucleus, the potential is harmonic inside the nucleus (assuminga constant charge density), and Coulomblike outside the nucleus. If the nucleus isdeformed, flattened, or cigar-shaped, spherical symmetry is broken, and the levelswill no longer be degenerate in the magnetic quantum number m. This results in asplitting of the spectral lines, which enables us to determine the charge distribution,that is the proton distribution inside the nucleus.

Figure11.7, obtained at CERN, shows the spectra of muonic atoms in the casesof gold (Z = 79), which is a spherical nucleus, and of uranium (Z = 92), which isa deformed nucleus. We notice the more complicated structure of the higher energy

Fig. 11.7 Transition line from the 2p level (actually split into two sublevels 2p1/2 and 2p3/2, tothe 2s level in muonic atoms of gold (Z = 79, A = 197) and uranium (Z = 92, A = 238) (scalein keV). Gold is spherical, and its spectrum has a simple shape; uranium is deformed, and the upperpeak is split in four lines (CERN document)

11.4 Muonic Atoms 271

line for uranium. This is a very accurate method to determine the deformations ofnuclei.

The existence of the muon has been a mystery for more than 40 years. When itwas discovered, Rabi said, “Who ordered that?” Why a heavy electron? All matterwe know about around us can be built starting with protons, neutrons, electrons,and neutrinos, or, in terms of fundamental constituents, with the family of quarksand leptons {u, d, e, ν}, using the Schrödinger equation. (Of course, to construct azebra or a humming bird, there are simpler methods, and we are interested in simplefeatures such as 2 ears, 4 feet, 1 tail for a rabbit, 2 humps for a camel. But with apowerful enough computer, it’s in principle feasible). So, why a heavy electron, withwhich one can imagine a Gulliver universe: atoms, molecules, a chemistry, a biology,200 times smaller, but 200 times more energetic than the matter we know?

Muons are useful. They have many applications, such as probing nuclei, prob-ing crystals, and probing pyramids, but why do they exist? What role is the muonsupposed to play?

In 1974, with the discovery of a new quark, the charm quark c, it was realizedthat the muon forms, with its neutrino, the charm quark, and the strange quark s,discovered in the 1940s, a new family of quarks and leptons (c, s,μ, νμ). This familygenerates at higher scales, a new atomic and nuclear physics, but its members areunstable. In 1975–1976, with the discovery of a new lepton τ , another new quark wasdiscovered, the b quark (beautiful or bottom), and, in 1995, the top quark t , hencea third family (t, b, τ , ντ ).2 By July 2000, the direct evidence for the τ neutrino bythe DONUT collaboration in Fermilab, filled completely the family of quarks andleptons of the standard model.

In 1989 at the LEP electron collider, it was proven that the (light) fundamentalconstituents of matter consist of these three families only. Present ideas are thata more fundamental theory could not exist, and that the big bang could not havehappened so nicely, if these two extra (useless) families of quarks and leptons didnot exist. They are of importance in order to create the world. But, after that, theyare merely toys for physicists.

However, we still do not understand themasses of these quarks and leptons (there-fore their stability). The origin ofmass is one of the great problems ofmodern physics.The discovery of the Higgs boson at CERN in June 2012 is a celebrated first step inthat direction.

To end this chapter, we note that in January 2014, the ASACUSA experiment3 atCERN succeeded in producing a beam of antihydrogen atoms, i.e., positrons boundto antiprotons. The comparisons of hydrogen and antihydrogen atoms constitute oneof the best ways to perform highly precise tests of matter/antimatter symmetry.

2M. Perl, “The leptons after 100 years,” Phys. Today, October (1997), p. 34.3Kuroda, N. et al. A source of antihydrogen for in-flight hyperfine spectroscopy. Nat. Commun.5:3089 doi:10.1038/ncomms4089 (2014); and further references therein.

272 11 The Hydrogen Atom

11.5 Exercises

1. Expectation value of r for the Coulomb problem

Consider the dimensionless radial equation for the hydrogen atom:

(d2

dρ2− �(� + 1)

ρ2+ 2

ρ

)un,�(ρ) = ε un,�(ρ), (11.29)

where un,�(ρ) = ρRn,�(ρ) is the reduced wave function and satisfies the conditions∫ ∞0 |un,�(ρ)|2dρ = 1 and un,�(0) = 0.

a. By multiplying this equation by ρ un,�(ρ) and by integrating over ρ, show that:

〈ρ〉n2

+ �(� + 1)

n2− 2 =

∫ +∞

0ρ un,�(ρ) u′′

n,�(ρ) dr.

One can use the result 〈1/ρ〉 = 1/n2 deduced from the virial theorem (11.27).b. By multiplying the Schrödinger equation by ρ2 u′

n,�(ρ) and by integrating over ρ,show that: 〈ρ〉

n2− 1 = −

∫ +∞

0ρ un,�(ρ) u′′

n,�(ρ) dr.

c. Deduce from the above results that 〈ρ〉 = (3n2 − �(� + 1))/2.

2. Three-dimensional harmonic oscillator in spherical coordinates

We treat the three-dimensional harmonic oscillator problem as a central potentialproblem. Consider the Hamiltonian:

H = p2

2m+ 1

2mω2r2

with r2 = x2 + y2 + z2.

a. Introduce the dimensionless quantities ρ = r√mω/� and ε = E/�ω. Show that

the radial equation (11.17) becomes:

(−1

ρ

d2

dρ2+ �(� + 1)

ρ2+ ρ2 − 2ε

)R�(ρ) = 0. (11.30)

b. One can prove that the normalisable solutions of (11.30) are labelled by an integern′:

Rn′,�(ρ) = ρ� Pn′,�(ρ) e−ρ2/2,

where Pn′,�(ρ) is a polynomial of degree n′. These solutions correspond to par-ticular values of ε:

11.5 Exercises 273

ε = 2n′ + � + 3/2.

In the following we set n = 2n′ + �.Show that one recovers the levels E = �ω(n1 + n2 + n3 + 3/2) (ni integer ≥ 0)obtained in Cartesian coordinates in Chap. 4 (Exercise 3), and associated to theeigenstates |n1; n2; n3〉. To what values of the angular momentum do the energylevels En correspond?

c. Give the explicit correspondence between the states |n1; n2; n3〉 and |n, �,m〉 forn = n1 + n2 + n3 = 1.

3. Relation between the Coulomb problem and the harmonic oscillator

Consider the three-dimensional harmonic oscillator problem treated in the previousexercise, but writing the potential as V (r) = K 2mω2r2/2 where K is dimensionless,in order to keep track of the parameters. Using the dimensionless variable ρ of theprevious exercise, the radial equation for the reduced wave function u = ρR(ρ) is:

(d2

dρ2− �(� + 1)

ρ2− K 2ρ2 + 2

E

�ω

)u(ρ) = 0. (11.31)

Similarly, consider the Coulomb problem with a potential V (r) = −Ze2/r , whereZ is dimensionless. The radial equation for the variable ρ = r/a1 is:

(d2

dρ2− �(� + 1)

ρ2+ 2Z

ρ+ E

EI

)u(ρ) = 0. (11.32)

a. Show that under the transformation u(ρ) = xα f (x) where x = √ρ, and with an

appropriate choice of α, one can cast the hydrogen problem (11.32) in the sameform as the harmonic oscillator problem (11.31).

b. Discuss the correspondence between the parameters of the two problems.c. Recalling the results of the previous exercise, find the energy levels of the hydro-

gen atom.

4. Confirm or invalidate the following assertions

a. If [H , L] = 0, the energy levels do not depend on m (i.e., on the eigenvalues ofthe projection of one of the components of the angular momentum L).

b. If [H , L2] = 0, the energy levels do not depend on �.

5. Centrifugal barrier effects

Consider a central potential. We note E� the lowest energy-level for a given �. Showthat E� increases with �.

6. Algebraic method for the hydrogen atom

We consider the radial dimension less equation for the Coulomb problem (11.29)and we introduce the operators:

274 11 The Hydrogen Atom

A−� = d

dρ+ � + 1

ρ− 1

� + 1A+

� = d

dρ− � + 1

ρ+ 1

� + 1.

a. Calculate A−� A+

� . Show that (11.29) can be written:

(A−

� A+�

)u� =

(ε − 1

(� + 1)2

)u�. (11.33)

b. Show that:

A+� A−

� = A−�+1A

+�+1 − 1

(� + 2)2+ 1

(� + 1)2.

By multiplying (11.33) by A+� , show that A+

� u�(ρ) satisfies the radial equationwith the same eigenvalue ε but for an angular momentum �′ = � + 1.

c. Similarly, show that A−�−1u�(ρ) satisfies the radial equation with the same eigen-

value ε but for an angular momentum �′ = � − 1.d. Calculate the expectation value of A−

� A+� with the radial function u�(ρ), and show

that ε ≤ 1/(� + 1)2.e. Show that, for a given value of ε, there exists amaximum value �max of the angular

momentum such that ε = 1/n2, where we have set n = �max + 1. Show that thecorresponding radial wave function u�max(ρ) satisfies the differential equation:

(d

dρ− n

ρ+ 1

n

)u�max(ρ) = 0.

f. Deduce from these results the energy levels and the correspondingwave functionsof the hydrogen atom.

7. Molecular potential

Consider a central potential of the form:

V (r) = A/r2 − B/r (A, B > 0).

We want to calculate the energy levels of a particle of mass me in this potential.

a. Write the radial equation.b. By a change of notations, reduce this equation to an eigenvalue problem which is

formally identical to the Kepler problem. Check that one can solve this equationusing the same arguments as for the hydrogen atom.

c. Give the explicit values of the energy levels in terms of A and B.

11.6 Problem. Decay of a Tritium Atom 275

11.6 Problem. Decay of a Tritium Atom

In all this problem, we consider nuclei as infinitely massive compared to the electron,of mass m. We note a1 the Bohr radius a1 = �

2/me2 and EI = mc2α2/2 ∼ 13.6eVthe ionisation energy of the hydrogen atom, where α is the fine structure constant.We note e the unit charge, and we work with units such that 4πε0 = 1.The nucleus of the tritium atom is the isotope 3H , of charge Z = 1. In the groundstate |ψ0〉 of this atom, the wave function of the electron (n = 1, l = 0,m = 0) isthe same as in the usual hydrogen atom:

ψ0(r) = 1√(πa13)

e−r/a1 . (11.34)

The tritium nucleus is radioactive and transforms into helium 3 through β decay:

3H → 3He + e− + ν (11.35)

(ν is an antineutrino), where the emitted electron has an energy of the order of 15 keVand the the helium nucleus 3He has charge Z = 2. The decay is an instantaneousprocess, the β electron is emitted with a large velocity and leaves the atomic systemvery rapidly. Consequently, an ionized 3He+ atom is formed, for which, at the timet0 of the decay, the wave function of the electron is practically the same as in tritium,and we shall assume it is still given by Eq. (11.35). We note |n, l,m〉 the states ofthe ionized helium atom which is a hydrogen-like system, i.e., one electron placedin the Coulomb field of a nucleus of charge 2.

The Energy Crisis in Tritium Decay

(1.1) Write the hamiltonian H1 of the atomic electron before the decay and thehamiltonian H2 of this electron after the decay (when the potential term has suddenlychanged.)

(1.2) What are, in terms of EI , the energy levels of the 3He+ atom? Give its Bohrradius and its ground state wave function ϕ110(r).

(1.3) Calculate the expectation value 〈E〉 of the energy of the electron after thedecay. One can, for instance make use of the fact that 〈ψ0|(1/r)|ψ0〉 = 1/a1 and thatH2 = H1 − e2/r . Give the value of 〈E〉 in eV.(1.4) Express in terms of |ψ0〉 and |n, l,m〉 the probability amplitude c(n, l,m) andthe probability p(n, l,m) to find the electron in the state |n, l,m〉 of 3He+ after thedecay. Show that only the probabilities pn = p (n, 0, 0) do not vanish.

(1.5) Calculate the probability p1 to find the electron in the ground state of 3He+.What is the corresponding contribution to 〈E〉?(1.6) A numerical calculation gives the following values:

276 11 The Hydrogen Atom

p2 = 1

4,

∞∑

n=3

pn = 0.02137,∞∑

n=3

pnn2

= 0.00177.

Calculate the probability∑∞

n=1 pn to find the atomic electron in a bound state of3He+ and the corresponding contribution to 〈E〉. What do you think of the result?

(1.7) Experimentally, in the β decay of the tritium atom one observes that, in about3% of the events, there are two outgoing electrons, one with a mean kinetic energy〈Ec〉 ∼ 15 keV, the other with 〈Ec〉 ∼ 34.3 eV, thus leaving a completely ionized3He2+ nucleus, as if the β decay electron “ejected” the atomic electron. Can youexplain this phenomenon?

11.6.1 Solution

(2.1) The two hamiltonians are H1 = p2/2m − e2/r , and H2 = p2/2m − 2e2/r.

(2.2) The levels of a hydrogen-like atom of nuclear charge Z are En = −Z2EI /n2

and, in this case, En = −4EI /n2. The new Bohr radius is a2 = a1/2, and the wavefunction ϕ100(r) = e−r/a2/

√(πa23).

(2.3) The expectation value of the electron energy in the new nuclear configurationis: 〈E〉 = 〈ψ0|H2|ψ0〉 = 〈ψ0|H1|ψ0〉 − e2〈ψ0|(1/r)|ψ0〉, which amounts to 〈E〉 =−EI − e2/a1 = −3EI ∼ −40.8 eV.

(2.4) By definition, the probability amplitude is c(n, l,m) = 〈n, l,m|ψ0〉, and theprobability p(n, l,m) = |〈n, l,m|ψ0〉|2.

The analytic form is c(n, l,m) = ∫Rnl(r) Ym∗

l (θ,φ)ψ0(r)d3r, where Rnl(r)are the radial wave functions of the 3He+ hydrogen-like atom. Since ψ0 is ofthe form ψ0(r) = χ(r)Y 0

0 (θ,φ), the orthogonality of spherical harmonics impliesp(n, l,m) = 0 if l,m �= 0.

(2.5) The probability amplitude in the lowest energy state is

(p1)1/2 = 4π

∫e−r/a2√

πa32

e−r/a1√

πa31

r2dr = 16√2/27.

Hence the probability p1 = 0.70233 and the contribution to the energy p1E1 =−38.2 eV.

(2.6)With the numerical values given in the text, one has p2E2 = −EI /4 = −3.4 eV,and p = ∑∞

1 pn = 0.9737. The contribution to 〈E〉 is 〈EB〉 = ∑∞1 pnEn =

−3.0664 EI = −41.7 eV.The total probability is smaller than 1; there exists a non-zero probability

(1 − p) = 0.026 that the atomic electron is not bound in the final state.

11.6 Problem. Decay of a Tritium Atom 277

The contribution of bound states 〈EB〉 = −41.7 is smaller than the total expec-tation value of the energy 〈E〉. The probability (1 − p) corresponds therefore toa positive electron energy, that is to say an ionization of 3He+ into 3He2+ withemission of the atomic electron.

(2.7) There is necessarily a probability 1 − p = 0.026 for the atomic electron notto be bound in helium, therefore that the atom be ionized in the decay. If the meankinetic energy of the expelled electron is Ec ∼ 34.3 eV, this represents a contributionof the order of (1 − p)Ec ∼ +0.89 eV to the mean energy which compensates theapparent energy deficit noted above.

Comment: This type of reaction has been intensively studied in order to determinethe neutrino mass. If M1 and M2 are the masses of the two nuclei, Eβ the energyof the β electron, E the energy of the atomic electron, and Eν the neutrino energy,energy conservation is, for each event: M1c2 − EI = M2c2 + Eβ + Eν + E . For agiven value of E , the determination of the maximum energy of the β electron (whichcovers all the spectrum up to 19 keV in the tritium atom case) provides a methodfor determining the minimum value m νc2 of Eν through this energy balance. Animportant theoretical problem is that current experiments are performed onmoleculartritium (HT or TT molecules) and that molecular wave functions are not knownexplicitly, contrary to the atomic case considered here.

Chapter 12Spin 1/2

Spin 1/2 is the first truly revolutionary discovery of quantum mechanics. Theproperties of this physical quantity in itself, the importance of its existence, andthe universality of its physical effects were totally unexpected.

The physical phenomenon is the following. In order to describe completely thephysics of an electron, one cannot use only its degrees of freedom corresponding totranslations in space. One must take into account the existence of an internal degreeof freedom that corresponds to an intrinsic angular momentum. In other words, theelectron, which is a pointlike particle, “spins” on itself. We use quotation marks forthe word “spins”. One must be cautious with words, because this intrinsic angularmomentum is purely a quantum phenomenon. It has no classical analogue, exceptthat it is an angular momentum.

One can use analogies and imagine that the electron is a sort of quantum top.But we must keep in mind the word “quantum”. The electron is a pointlike objectdown to distances of 10−18 m. One must admit that a pointlike object can possess anintrinsic angular momentum. (As we have already pointed out, in this respect, thephoton, which is pointlike, has an intrinsic angular momentum, and is a zero massparticle, is from this point of view even more strange.)

12.1 Experimental Results

Experimentally, this intrinsic angular momentum, called spin, has the followingmanifestations (we do not enter in any technical detail).

© Springer International Publishing Switzerland 2016J.-L. Basdevant, Lectures on Quantum Mechanics,Graduate Texts in Physics, DOI 10.1007/978-3-319-43479-7_12

279

280 12 Spin 1/2

1. If we measure the projection of the spin along any axis, whatever the state of theelectron, we find either of two possibilities

+�/2, or − �/2.

There are two and only two possible results for this measurement.2. Consequently, if one measures the square of any component of the spin, the result

is �2/4 with probability one.

3. Therefore, a measurement of the square of the spin S2 = S2x + S2y + S2z gives theresult

S2 = 3�2

4.

4. A system that has a classical analogue, such as a rotatingmolecule, can rotatemoreor less rapidly on itself. Its intrinsic angular momentum can take various values.However, for the electron, as well as for many other particles, it is an amazingfact that the square of its spin S2 is always the same. It is fixed: all electrons inthe universe have the same values of the square of their spins S2 = 3�2/4. Theelectron “spins” on itself, but it is not possible to make it spin faster.

One can imagine that people did not come to that conclusion immediately. Thediscovery of the spin 1/2 of the electron is perhaps the most breathtaking story ofquantum mechanics.

The elaboration of the concept of spin was certainly the most difficult step of allquantum theory during the first quarter of the 20th century. It is a real suspense thatcould be called the various appearances of the number 2 in physics. There are manynumbers in physics; it is difficult to find a simpler one than that.

And that number 2 appeared in a variety of phenomena and enigmas that seemedto have nothing to do a priori with one another, or to have a common explanation.The explanation was simple, but it was revolutionary. For the first time people werefacing a purely quantum effect, with no classical analogue. Nearly all the physicalworld depends on this quantity, the spin 1/2.

The challenge existed for a quarter of a century (since 1897). Perhaps, there wasnever such a long collective effort to understand a physical structure. It is almostimpossible to say who discovered spin 1/2, even though one personality dominates,Pauli, who put all his energy into finding the solution.We show that in order to manipulate spin 1/2, and understand technicalities we

essentially know everything already. We have done it more or less on two-statesystems.

But for anybody, it is a complicated matter to have a really intuitive representationof spin 1/2. It is really a personal affair, as can be seen in Fig. 12.1.These gentlemenare simply discussing spin effects.

12.2 Spin 1/2 Formalism 281

Fig. 12.1 Three physicistsdiscussing the optimal wayto measure spin effects inproton collisions at theArgonne ZGS accelerator(CERN document)

12.2 Spin 1/2 Formalism

The measurement results fit perfectly with the general framework of the theoryof angular momenta. The electron spin is a half-integer angular momentum thatcorresponds to the quantum numbers j = 1/2, m = 1/2. The corresponding Hilbertspace is two-dimensional.

Any spin state is a linear superposition of two basis states and the degree offreedom corresponding to spin is described in a two-dimensional Hilbert space:Espin.

Representation in a Particular Basis

We choose a basis of states in which both S2 and Sz are diagonal, which we denote{|+〉, |−〉}:

Sz|+〉 = �

2|+〉, Sz|−〉 = −�

2|−〉, S2|±〉 = 3�2

4|±〉. (12.1)

Using the notation of Chap. 10, the states |±〉 would be |j = 1/2,m = ±1/2〉. Theaction of Sx and Sy on the elements of the basis is written as (see Eq. (10.16))

Sx|+〉 = �/2|−〉, Sx|−〉 = �/2|+〉 (12.2)

Sy|+〉 = i�/2|−〉, Sy|−〉 = −i�/2|+〉. (12.3)

An arbitrary spin state |Σ〉 can be written as:

|Σ〉 = α |+〉 + β |−〉, |α|2 + |β|2 = 1. (12.4)

282 12 Spin 1/2

The probabilities of finding +�/2 and −�/2 in a measurement of Sz on this state areP(+�/2) = |α|2, P(−�/2) = |β|2.

We will often call |+〉 and |−〉 respectively the “spin up” and “spin down” stateswith respect to the z axis, but we keep in mind that it has very little geometricalmeaning.

Matrix Representation

It is convenient to use matrix representations for the states and the operators:

|+〉 =(10

), |−〉 =

(01

), |Σ〉 =

(αβ

). (12.5)

We can use the Pauli matrices σ ≡ {σx, σy, σz}

σx =(0 11 0

), σy =

(0 −ii 0

), σz =

(1 00 −1

)(12.6)

which satisfy the commutation relations

σ × σ = 2i σ. (12.7)

The spin observables are represented as

S = �

2σ. (12.8)

In this basis, the eigenstates |±〉x of Sx and |±〉y of Sy are:

|±〉x = 1√2

(1

±1

), |±〉y = 1√

2

(1±i

). (12.9)

The major complication, compared to the ammonia molecule of Chap.7 is to dealwith a vector degree of freedom, and a vector observable. But by studying angularmomenta, we understood how to deal with that situation.

We remark that, as we already know in different systems, the state |+〉 for whichwe are sure that Sz = �/2 can also be considered as a linear superposition with equalweights of Sx = ±�/2 or Sy = ±�/2.

If the spin “points” upwards, it points both on the left and on the right with equalprobabilities. This is again the superposition principle.

Notice that we just said that the spin points upwards, and we have spoken classi-cally of a purely quantum property. That is not too severe, because we have a math-ematical dictionary to translate it properly. But remember that Galileo just escapedbeing burned to death, and Giordano Bruno was burned to death, because they saidthat it is the earth that rotates and that the sun is fixed. Nevertheless, this does not

12.2 Spin 1/2 Formalism 283

prevent people nowadays from saying that the sun rises and sets everyday. Oncethings have been understood, it’s much simpler, in everyday life, to come back totraditional ways of visualizing things.

12.3 Complete Description of a Spin 1/2 Particle

Wenow turn to the complete description of the state of an electron taking into accountboth the space degrees of freedom and the spin.

Here, we follow amore direct path than the traditional way of showing a nontrivialtensor product of Hilbert spaces.

The result is quite intuitive. We deal with two random variables, position and spinalong the z-axis. We need the probability law of this couple of variables.

The state of a particle in space is described by a square-integrable function thatbelongs to what we call the “external” Hilbert space Eexternal. At any point in space,the spin state is described by a two-dimensional vector of Espin, the “internal” degreeof freedom.

We therefore need twowave functions to describe the state of the electron in spacewith its spin. We must double the number of wave functions (or the “dimension” ofthe Hilbert space). A tensor product of spaces is as simple as that in this particularcase.

There are several possible representations of the states, together with correspond-ing representations of observables. Choosing a particular representation is a matterof convenience.

Mixed Representation

The state is represented by a vector of Espin whose components are square-integrablefunctions:

ψ+(r, t) |+〉 + ψ−(r, t) |−〉. (12.10)

The physical interpretation of this representation is as follows. |ψ+(r, t)|2 d3r (resp.,|ψ−(r, t)|2 d3r) is the probability of finding the particle in a volume d3r around thepoint r, with a spin component +�/2 (resp., −�/2) along the z-axis.

Two-Component Wave Function

The state vector is represented in the form:

(ψ+(r, t)ψ−(r, t)

), (12.11)

The physical interpretation of ψ+ and ψ− as probability amplitudes for the coupleof random variables (r, Sz) is the same as above.

284 12 Spin 1/2

12.3.1 Observables

The space observables (x, id/dx) only act on wave functions ψ±(r, t). They do notmake any difference between spin states. In the matrix representation the spaceobservables are diagonal 2 × 2 matrices whose elements are operators and that donot act on spin variables.

Spin observables are 2 × 2 matrices with numerical coefficients seen above. Theydo not act on space variables.

In the general case, there exist observables that can act on both sets of variables.These are simply the products of the 2 × 2 matrices by the operators; for instance,

r · S = xSx + ySy + zSz.

More generally, the products of operators AextBsp follow the rule

(AextBsp

)(ψσ(r, t)|σ〉) =

(Aextψσ(r, t)

) (Bsp|σ〉

), σ = ±. (12.12)

Functions of a Two-Valued Variable.

In some problems, it is useful to use a single wave function depending on fourvariables (plus time) ψ(r,σ; t), where the fourth variable σ can take the values ±1.One has obviously

ψ(r,σ; t) = ψσ(r, t), σ = ±1.

Atomic States

In many problems of atomic physics, it is useful to use the quantum numbers n, �,mto classify the states |n, �,m〉 that form a basis of Eexternal. The introduction of spinis done in the space spanned by the family {|n, �,m〉 |σ〉} where the spin quantumnumber can take the two values ±1. It is convenient to use the compact notation

|n, �,m,σ〉, (12.13)

where the states of an electron are described by four quantum numbers. The actionof space operators on the states |n, �,m〉 is known (see Chap. 11), therefore theaction of general operators on the states |n, �,m,σ〉 can readily be inferred from theconsiderations developed above.

12.4 Physical Spin Effects 285

12.4 Physical Spin Effects

Physical spin effects belong to the following main categories.

1. There are angular momentum effects, in particular in nuclear and particlephysics. It is compulsory to take into account spin in order to observe conserva-tion of the total angular momentum. The structure of fundamental interactions relieson the spin 1/2 of the electron, the quarks, and the neutrinos.

This is too difficult to examine here.

2. There are surprising and revolutionary effects compared to classical physicsthat are due to the Pauli principle. Such effects are fundamental in order to understandthe structure of matter, atoms, molecules, solids, and liquids. We show some of themin Chap.14. The Pauli principle plays a central role in statistical physics and inchemistry.

3. There are magnetic effects that actually are at the origin of the discovery of spin1/2. We mainly use those effects in order to develop our experimental discussion.

12.5 Spin Magnetic Moment

Indeed, all what has been said above is for the moment simple matrix calculus. But itwill acquire a much greater physical flavor because to the spin of the electron (or anyspin 1/2 particle) there corresponds an intrinsic magnetic moment, a spin magneticmoment, that is proportional to the spin.

μ = γ S = μ0 σ. (12.14)

The gyromagnetic ratio γ, the value of which we come back to, is a characteristic ofthe particle under consideration.

The action of μ, which is proportional to S, on the state vectors is obvious.It is through measurements of this magnetic moment that we can find the easiest

access to spin measurements.For instance, if we place this spin magnetic moment in a field B, there is an

interaction HamiltonianW = −μ · B, (12.15)

which leads to a number of experimental observations.

Hamiltonian of a One-Electron Atom

In what follows we consider an atom with one external electron in its ground state.Globally, the atom is electrically neutral. However, it carries the spin 1/2 of theexternal electron, and the corresponding magnetic moment.

286 12 Spin 1/2

Suppose the atom is moving in space and that it is placed in a potential V (r), andthat furthermore it is placed in a magnetic field B. The magnetic potential energy isgiven by (12.15).

The Hamiltonian is the sum of two terms

H = Hext + W , (12.16)

where

Hext = p2

2m+ V (r)

is of the type seen in previous Chapters. In particular, Hext does not act on the spinvariables. The operator W is given by (12.15). It acts in Espin through the threeoperators μx, μy, μz. If the field is inhomogeneous, it also acts on space variablesthrough the three functions Bx(r), By(r), Bz(r).

The Schrödinger equation is

i�d

dt|ψ〉 = H|ψ〉. (12.17)

If we choose the representation of states (12.11) and if we decompose on the ortho-normal basis {|+〉, |−〉}, we obtain the coupled differential system

i�∂

∂tψ+(r, t) =

(− �

2

2mΔ + V (r )

)ψ+(r, t)

+〈+|W |+〉 ψ+(r, t) + 〈+|W |−〉 ψ−(r, t),

i�∂

∂tψ−(r, t) =

(− �

2

2mΔ + V (r )

)ψ−(r, t)

+〈−|W |+〉 ψ+(r, t) + 〈−|W |−〉 ψ−(r, t).

The matrix elements of W in the basis {|+〉, |−〉} are functions of the external spacevariables. They add to the usual potential energy terms, which are diagonal, and, ingeneral, they couple the evolution equations of the components ψ+ and ψ−.

We now have all the tools to be able to do physics and to discover the suspenseof the discovery of spin 1/2.

12.6 The Stern–Gerlach Experiment

The first measurement of the intrinsic magnetic moment of the electron, and the firstundeniable appearance of the number 2 came in 1921–1922 with the experiment ofStern and Gerlach.

12.7 Principle of the Experiment 287

12.7 Principle of the Experiment

A collimated beam of silver atoms is sent in a region where an inhomogeneousmagnetic field is applied along the z-direction, perpendicular to the initial velocity ofthe atoms (Fig. 12.2a). The possible deflection of the beamby the field gradient is thenmeasured by observing the impacts of the atoms on a detection plate perpendicularto the initial direction of the beam. The silver atom has one external electron in anorbital angular momentum state � = 0. Therefore, the atom’s magnetic moment isequal to the intrinsic magnetic moment of the valence electron.

12.7.1 Semi-classical Analysis

We first analyze this experiment within classical mechanics. The atoms are neutraland they are not subject to a magnetic Lorentz force. However, because they have anonvanishing magnetic moment μ, the force

Fz = μz∂Bz

∂z, (12.18)

parallel to the z-direction, acts on them and deflects their trajectory.

Fig. 12.2 a The Stern–Gerlach experiment: atoms from a collimated beam are deflected as theycross a regionwhere an inhomogeneousmagnetic field is applied. This experiment can be interpretedas a measurement of the component of the atomic magnetic moment along the direction of the field(z in the Figure). bMagnetic gradient between the polar pieces of the magnet

288 12 Spin 1/2

We recall that when a magnetic moment μ is placed in a magnetic field B themagnetic interaction energy is

W = −μ · B, (12.19)

and the torqueΓ = μ × B (12.20)

is exerted on the magnetic moment. In addition, if the magnetic field is inhomoge-neous, the force (12.18), or more generally

F = ∇(μ · B) =∑

i=x,y,z

μi(t)∇Bi, (12.21)

acts on the dipole.As we have seen in (10.37), the magnetic moment of an atom at r does not align

with the axis of the local magnetic field B(r), but it precesses along this axis withthe Larmor frequency

ω0 = −γ0B(r). (12.22)

We assume that the classical trajectory of the atoms lies in the plane of symmetryx = 0 of themagnet (see Fig. 12.2b).Along this trajectory themagnetic field is alwaysparallel to the z-axis, so that the Larmor precession takes place around z. Also, owingto the symmetry of the device, the quantities ∂Bz/∂x and ∂Bz/∂y vanish along theatomic beam trajectory (we neglect possible edge effects). If the displacement of themagnetic moment during a single precession period 2π/ω0 is small compared withthe typical variation scale of the magnetic field, we can average the force (12.21)over the Larmor period. The contributions of μx and μy to (12.21) then vanish, andone is left only with the z-component of the force Fz = μz(t) ∂Bz/∂z. In addition wededuce from (10.37) that μz stays constant as the atom moves in the magnetic fieldgradient, which is helpful to justify the result (12.18).

12.7.2 Experimental Results

In the absence of a magnetic field gradient one observes a single spot on the detectingplate, in the vicinity of x = z = 0 (Fig. 12.3a). Themagnetic field gradient provides away to measure the z-component of the magnetic moment of the atoms as they enterthe field zone. Let us assume that all the atoms carry the same magnetic moment ofnorm μ0, and that this moment is oriented at random when an atom enters the fieldzone. Classically, this should produce some continuous distribution of μz betweenthe two extreme values−μ0 and+μ0, so that onewould expect that the impacts of theatoms on the screen form an extended line parallel to z (Fig. 12.3b). The endpointsof the line correspond to atoms whose magnetic moments are oriented respectivelysuch as μz = +μ0 and μz = −μ0.

12.7 Principle of the Experiment 289

Fig. 12.3 Possible results of a Stern–Gerlach experiment. a In the absence of a magnetic gradientno deflection of the atomic trajectories occurs, and the atoms form a single spot around the pointx = z = 0; each dot represents the impact of an atom on the detection screen. b Simulation of theresult expected from classical mechanics, assuming that all atoms carry the same magnetic momentμ0 with a random orientation; the distribution of the z-component of the magnetic moment isuniform between −μ0 and +μ0. c Simulation of the result found experimentally with silver atoms:the experiment, which can be considered as a measurement of the z-component of the magneticmoment, yields only the two results +μ0 and −μ0

The experiment is difficult. But the observation differs radically from this classicalprediction. The set of impacts never forms a continuous line on the screen. For atomssuch as silver, the impacts are grouped in two spots corresponding to μz = +μ0 andμz = −μ0, with μ0 = 9.27 10−24 JT−1 (Fig. 12.3c).

The result μ0 of the Stern–Gerlach experiment is consistent with:

μ0 = � |γ0| = �q

2me, (12.23)

which amounts to taking L = � in (10.34). The quantity (12.23) is the absolute valueof the Bohr magneton.

12.7.3 Explanation of the Stern–Gerlach Experiment

Our theory explains the observed spatial separation of the states |±〉z. We consider anincident atomic beam propagating along y; each atom possesses a magnetic moment.In a region of length L, a magnetic field B parallel to z is applied with a gradientalong z:

B(r) = Bz(r)uz, with Bz(r) = B0 + b′z. (12.24)

In full rigor, the Eq. (12.24) is incorrect because the field B(r) does not satisfy∇.B = 0. A more realistic calculation can be done with a field B = B0uz + b′(zuz −xux) which satisfies Maxwell equations. If the dominant part of the field B0uz ismuch larger than the transverse field −b′xux on the transverse extension Δx of the

290 12 Spin 1/2

atomic wave packet (i.e., B0 � b′Δx), the eigenstates of −μ · B remain practicallyequal to |±〉z and the present approach is valid.

Under these conditions, the Schrödinger equation (12.17) can be decoupled intotwo equations:

i�∂

∂tψ+(r, t) =

(p2

2m− μ0B

)ψ+(r, t), (12.25)

i�∂

∂tψ−(r, t) =

(p2

2m+ μ0B

)ψ−(r, t). (12.26)

These two equations are both of the same type as the Schrödinger equation seen inChap.5, but the potential is not the same for ψ+ and ψ−. In order to proceed furtherwe set:

π± =∫

|ψ±(r, t)|2 d3r, π+ + π− = 1, (12.27)

whereπ+ andπ− are the probabilities of findingμz = +μ0 andμz = −μ0.Wededucefrom (12.25) and (12.26) that

dπ+dt

= dπ−dt

= 0. (12.28)

We define the functions:

φ±(r, t) = ψ±(r, t)/√

π±, (12.29)

which are the conditional probability amplitudes of particles for which μz = ±μ0.These normalized functions also satisfy the Schrödinger-type equations (12.25) and(12.26).

We now define:

〈r±〉 =∫

r |φ±(r, t)|2 d3r, (12.30)

〈p±〉 =∫

φ∗±(r, t)

�

i∇φ±(r, t) d3r, (12.31)

where 〈r+〉 (resp., 〈r−〉) is the average position of particles for which μz = +μ0

(resp., μz = −μ0), and 〈p±〉 are their average momenta. A simple application of theEhrenfest theorem gives:

(d/dt)〈r±〉 = 〈p±〉/m, (12.32)

(d/dt)〈px±〉 = (d/dt)〈py±〉 = 0, (12.33)

(d/dt)〈pz±〉 = ±μ0b′. (12.34)

12.7 Principle of the Experiment 291

At t = 0 we assume that

〈r±〉 = 0, 〈px±〉 = 〈pz±〉 = 0, 〈py±〉 = mv.

We obtain at time t:

〈x±〉 = 0, 〈y±〉 = vt, 〈z±〉 = ±μ0b′t2/2m. (12.35)

Therefore there is a spatial separation along z of the initial beam into two beams.One beam corresponds to μz = +μ0, and the other to μz = −μ0. When the beamsleave the magnet of length L, their separation is

δz = 〈z+〉 − 〈z−〉 = μ0b′

m

L2

v2. (12.36)

If the field gradient is sufficiently strong so that δz > Δz (separation larger than thespatial extension of each wave packet), we obtain two beams: one in the internalstate |+〉, the other in the state |−〉. Therefore, the formalism we have developedin this section explains completely the Stern–Gerlach experiment and its results. Aswas announced in the case of the population inversion for the ammonia molecule inChap.7, this experiment elicits two fundamental aspects of a measurement processin quantum mechanics:

• A measurement requires a finite spatial extension (δz = 0 if L = 0).• A measurement is never instantaneous (δz = 0 if T = L/v = 0).

These two aspects were absent in the formulation of the principles of quantummechanics presented in Chapter 6.

Finally a simple inspection of the evolution of the expectation value of the sepa-ration of the two spots, and of their dispersion at the exit of the magnet, leads to thefollowing result. Let T = L/v be the time the atoms spend in the inhomogeneousmagnetic field, and let us note E⊥ = 〈p2z 〉/2m the transverse energy communicatedto the atom by the field gradient. In order to observe the splitting, the followingcondition must be satisfied,

T E⊥ ≥ �/2.

This condition, where the value of the field gradient has disappeared, is one importantaspect of the so-called time-energy uncertainty relation that appears in any quantummeasurement.

12.7.4 Successive Stern–Gerlach Setups

Consider now the experimental situation shown in Fig. 12.4. We place two consecu-tive magnets. The first has a field gradient directed along z and it splits the incident

292 12 Spin 1/2

Fig. 12.4 A beam of silver atoms crosses twomagnetic field zones. The first creates a field gradientalong z, the second a field gradient along x. After the first magnet, a shutter only lets the atoms inthe internal state |+〉z pursue. The second magnet allows us to perform a measurement of the x-component of themagneticmoment. One finds the two results+μ0 and−μ0 with equal probabilities

beam into two beams corresponding to the two internal states |+〉z and |−〉z. Whenthe beams leave the field zone, we stop the beam corresponding to the state |−〉z andwe keep only the beam in the state |+〉z. This latter beam is then sent into anotherStern–Gerlach device whose axis is along the x-axis, orthogonal to z. We thereforeperform a measurement of the x-component of the atomic magnetic moment, whosecorresponding observable is μx. The result observed experimentally is that the beamis again split in twobeams of equal intensities corresponding to values of themagneticmoment along x equal to +μ0 and −μ0, respectively.

This, naturally, is themanifestation of the superposition principle. The eigenstatesof μx and of μy are

|±〉x = 1√2

(|+〉z ± |−〉z) , |±〉y = 1√2

(|+〉z ± i|−〉z) . (12.37)

There is nothing surprising.

12.7 Principle of the Experiment 293

12.7.5 Measurement Along an Arbitrary Axis

Supposewe are interested inmeasuring the component of themagneticmoment alongan arbitrary axis. This is shown in Fig. 12.5. We place a Stern–Gerlach apparatusalong an arbitrary direction defined by the unit vector uθ such that:

uθ = ux sin θ + uz cos θ. (12.38)

This corresponds to a measurement of the component μθ of the magnetic momentalong uθ (i.e., μθ = μx sin θ + μz cos θ). By the correspondence principle, the corre-sponding observable is:

μθ = μx sin θ + μz cos θ = μ0

(cos θ sin θsin θ − cos θ

). (12.39)

This choice guarantees that the expectation values 〈μx〉, 〈μy〉, and 〈μz〉 of the compo-nents of the magnetic moment transform as the components of a usual three-vectorunder rotations, which is essential.

Just as μx, μy, μz, the operator μθ has the eigenvalues +μ0 and −μ0. Its eigen-vectors are:

Fig. 12.5 A beam of silver atoms is prepared in the state |+〉z . It then crosses a field gradientdirected along uθ . In this measurement of the component of the magnetic moment along uθ , the twopossible results are +μ0 and −μ0 with respective probabilities cos2 θ/2 and sin2 θ/2. The graph inthe lower right corner shows a typical result for θ = π/4

294 12 Spin 1/2

|+〉θ = |+〉z cos (θ/2) + |−〉z sin (θ/2) =(cos(θ/2)sin(θ/2)

), (12.40)

|−〉θ = −|+〉z sin (θ/2) + |−〉z cos (θ/2) =(− sin(θ/2)

cos(θ/2)

). (12.41)

The experimental observations are the following. If a beam prepared in the state |+〉zis sent in the field gradient directed along uθ, one finds that this beam is split intotwo beams corresponding to a magnetic moment along uθ equal to +μ0 and −μ0,with relative intensities I+(θ) = I+(0) cos2(θ/2) and I−(θ) = I+(0) sin2(θ/2).

In order to account for this result, we apply the principles of Chap. 6. A measure-ment of μθ can give two possible values, the eigenvalues +μ0 and −μ0; if the initialsystem is in the state |+〉z, the respective probabilities for these two issues are:

p+ = |θ〈+|+〉z|2 = cos2(θ/2) (12.42)

p− = |θ〈−|+〉z|2 = sin2(θ/2). (12.43)

Therefore, Eq. (12.39) explains why the experimental measurement, which involvesa large number of atoms, gives two spots with relative intensities cos2(θ/2) andsin2(θ/2). The measurement only gives a result with probability 1 when θ is 0 orπ, that is, when the preparation axis uz and the measurement axis uθ are parallel orantiparallel.

12.8 The Discovery of Spin

12.8.1 The Hidden Sides of the Stern–Gerlach Experiment

Coming back to Stern andGerlach, one naturally thinks that they have discovered spin1/2 ! They have found the quantization of the magnetic moment, the superpositionprinciple, and the explanation of the electron’s magnetic moment if not its spin.

It is with that type of reasoning and observations that Fresnel had founded thewave theory of light, in particular, the laws of polarization, which was a similar anddifficult problem in the 19th century.

Absolutely not! The result was considered

1. As perfectly natural2. And as a brilliant confirmation of the old quantum theory of Bohr and Som-

merfeld (remember this happened in 1922)

If people had thought about it, they might have found the answer. But at that time,physicists were concerned with quite a different problem: they wanted to prove theold quantum theory and the quantization of trajectories that Bohr had used in hismodel of the hydrogen atom.

12.8 The Discovery of Spin 295

Actually the Stern–Gerlach experiment had been induced by theorists, Born, Bohr,Sommerfeld, and Pauli, for months if not for years, and they had predicted the result.

In physics, there is some sort of credibility principle for an experimental result.One believes in a result if and only if a theory has predicted it. An example isthe discovery of the 3K cosmic background radiation by Penzias and Wilson. It wasconsidered as a background noise until they were told about the prediction of Gamowof the existence of background radiation in the big bang theory.

The quantization of Lz had been guessed very early.Sommerfeld, in the old quantum theory, had predicted the spatial quantization

of trajectories and the directional quantization in a magnetic field B. He knew thatμ = γL, and that the orbital gyromagnetic ratio is q/2m. Sommerfeld understood theprinciple of the experiment as soon as 1918. And he expected a lot from it becauseit would have been the first proof of quantization in a nonradiative process.

But one could argue that there should be three spots and not two. Imagine anelectron in a circular uniform motion around a proton. The quantization of angularmomentum is an integer multiple of �. In a magnetic field, the plane of the trajec-tory could have three directions corresponding, respectively, to an angular momen-tum parallel, antiparallel, or perpendicular to the field B with Lz = �, Lz = −�, orLz = 0.

Not at all! As soon as 1918, Bohr proved that the trajectory Lz = 0 was unstable.One should therefore observe only two spots Lz = ±1.1

Max Born insisted in 1920 (he was 30), “This experiment must absolutely bedone.” At that time, Born was a professor in Frankfurt, where there was an artist ofatomic andmolecular beams, Otto Stern (32 at that time), but Stern wasn’t interested.

So, Born, who was a mathematician, decided to do experiments. And he managedto do so thanks to a talented assistant Fraulein Elizabeth Bormann. This new activityof Born was a surprise to all physicists. (One day, Rutherford asked him if he had arelative doing experiments. Born answered, “No, but I have a good assistant.”)

But Born had to face the facts; he suffered from the Pauli effect: the better you areas a theorist, the more you are a disastrous experimentalist. Whenever Pauli entereda laboratory, everything went wrong. One day, in Göttingen, an experimental setupof Franck exploded. Everyone looked for Pauli, but there was no trace of him. Sometime later, someone learned that at the precise time of the explosion, Pauli was ona train, which had stopped in Göttingen, on the way from Munich to Hanover. ThePauli effect acted at a distance!

Born eventually convinced Stern. Actually, Stern did not know what to think. Atfirst he proposed the experiment, but some time later he was skeptical, “Quantumrestrictions on trajectories are simply calculational rules. I’m going to show oncefor all that what theorists say is nonsense.” However, Stern suffered somewhat fromthe Pauli effect. All his experimental setups were constructed by his technician. Heknew remarkably how to conceive them, but he wasn’t very skillful.

1It is one of the few times that Bohr predicted correctly the result of an experiment. He did it witha wrong argument.

296 12 Spin 1/2

And that time, it was too difficult. Neither the technician nor Fraulein Bormannsucceeded. Fortunately, Gerlach, who was a very talented 21-year-old experimen-talist, had just arrived in Frankfurt, after graduating in Tübingen. Born said “ThankGod, now we have at last someone who knows how to do experiments!” Gerlachtook care of everything–the technician, Fraulein Bormann, and Stern’s ideas–and hedid the experiment. He was successful, and found the two spots. It seemed to be atriumph for Sommerfeld.

Pauli (22 at that time) congratulated Gerlach and said to: “Let us hope now thatthe old unbeliever Stern will now be convinced of directional quantization!”

The triumph was even greater because, by measuring μ0 they found to a fewpercent

μ0 = |γorb0 |� = q�

2me, (12.44)

exactly the prediction of Bohr and Sommerfeld!At that time, nobody could suspect that Nature had played a bad trick. Equa-

tion (12.44) must be read as

μspin = (q

me)(

�

2), and not (

q

2me)� !

In other words, the spin gyromagnetic ratio is twice the orbital gyromagnetic ratio,and the angular momentum is �/2.

Dirac proved that in 1927, for any charged pointlike spin 1/2 particle, in his theoryof a relativistic electron.

Einstein used to say that the Lord is not mean, but he is subtle. On that point, theLord had really been nasty!

12.8.2 Einstein and Ehrenfest’s Objections

However, there were some skeptical people who thought about the physics. Becausethe experiment is amazing, it is completely opposed to classical conceptions. Einsteinand Ehrenfest performed a critical analysis of the experiment in semi-classical termsas in Sommerfeld’s theory.

They calculated the time it would take a loop of current to orient itself in themagnetic field, and they found a value of t � 109 s, that is, 30 years! Yet the atomsstay in the inhomogeneous field for 10−4 s.

Einstein and Ehrenfest concluded: “We must make a complete revisal of ourclassical ideas …this experimental shows a conceptual dead end …we must find anew quantum idea.”

Indeed, for us, in quantum mechanics, this is not a problem. In the initial beamthere is a probability amplitude for the spin to point upwards or downwards, and theorientation of the magnetic moment is “already there,” to speak in classical terms.

12.8 The Discovery of Spin 297

Now, one can think that with Einstein getting involved in that business, and withthat conclusion of his, people started wondering and found spin 1/2. The discovery ofthe number 2 must have occurred at that time. Not at all; nobody paid any attention.

12.8.3 Anomalous Zeeman Effect

At that time, physicists were all concerned with a true intellectual challenge thatseems to have nothing to do with all that, the anomalous Zeeman effect.

We place an atom, prepared in a state of energy E and angular momentum j, in amagnetic field B parallel to z. The magnetic energy is:

W = −μ · B. (12.45)

The corresponding level is split in 2j + 1 sublevels of respective energies:

E − γ�B0m, m = −j, . . . , j.

A corresponding splitting of each line is observed in the spectrum. If all angularmomenta are orbital angular momenta (i.e., they have a classical interpretation), jmust be an integer. In that case 2j + 1 is odd and we expect a splitting in an oddnumber of levels. The splitting of spectral lines in a magnetic field, first observed byZeeman in the period 1896–1903, showed that in many cases, in particular for alkaliatoms, this is not true. There is a splitting into an even number of levels.

Faraday had been convinced as early as 1845 that there was a deep connectionbetween optical and magnetic phenomena. In one of the last experiments of his life,in 1862, he attempted to find the influence of magnetic fields on radiation. Manytechnical problems prevented him from obtaining a positive answer. It was only in1896 that these experiments were redone successfully by Zeeman. Already at thattime, theorists, in particular H.A. Lorentz, predicted one should observe a splittingin an odd number of lines (1—no splitting—or 3). Zeeman first confirmed this resulton the spectra of cadmium and zinc. The discovery, in the particular case of sodium,of what was to be called the “anomalous Zeeman effect”, namely an even number oflines, remained for more than 25 years a real challenge for the scientific community,which was totally confused by this phenomenon. It was only after many struggles,in the years 1925–1926, with the ideas of Pauli, and of Uhlenbeck and Goudsmit,that the introduction of the notion of spin completely clarified the problem. The“anomalous Zeeman effect” then appeared as a perfectly normal phenomenon.

It was a real antique tragedy. One must realize that if we have the theory in frontof us (i.e. that there exist half-integer angular momenta for which 2j + 1 is even)physicists did not have it at that time, despite the works of Elie Cartan in 1913.

298 12 Spin 1/2

12.8.4 Bohr’s Challenge to Pauli

The anomalous Zeeman effect seemed to be an impenetrable wall. Bohr thought therewas only one person capable of solving that problem: Pauli, whom Bohr invited toCopenhagen in 1922. Pauli was 22, he had been Sommerfeld’s student in Munich,and he was known for his remarkable book on relativity. He had a difficult temper,as we said; for him most things were either obvious or stupid. One day, Ehrenfesthad said to him, “You know, Herr Pauli, I find your paper on relativity much nicerthan you are.” Pauli replied, “That’s funny! For me it’s the opposite, I find you muchnicer than all the rubbish you write.”

Of course, Pauli’s colleagueswere terrified. Franck told him; “Please behavewhenyou get to Copenhagen.” “I’ll have to learn Danish”, said Pauli.

Niels Bohr asked Pauli two questions.

• Why are there shells in complex atoms, containing 2, 8, 18, and so on; that is 2n2

electrons, n = 1, 2, 3 . . .? This remark had been made by Rydberg in 1903. Whyaren’t the electrons all in the ground state?

• What is the cause of the Zeeman effect?

And Pauli started thinking. He thought very hard. So hard that he said, in hisNobel prize Chapter in 1945, that one day, as he was strolling in the beautiful streetsof Copenhagen, a colleague had said to him in a friendly manner, “You look veryunhappy;” whereupon Pauli replied fiercely, “How can one look happy when heis thinking about the anomalous Zeeman effect.” Pauli understood that both theanomalous Zeeman effect and the effect of the inner core of electrons are a conceptualdead end in classical terms.

And, at the beginning of 1924, W. Pauli, Jr. wrote “The electron has a new,specifically quantum property, which corresponds to a two-valued physical quantity,which cannot be described classically. In an atom, an electron is characterized byfour quantum numbers, (n, l, m, σ) where σ = ±1.” And Pauli stated the principleof exclusion of equivalent electrons, called by Heisenberg the “Pauli verbot,” whichis very deep and also revolutionary.

At this point, one is tempted to say, “Fine, Pauli found the answer.” Not at all.

12.8.5 The Spin Hypothesis

The hypothesis of the electron spin, such as presented above, that the electron hasan intrinsic angular momentum S whose measurement along an axis gives one ofthe two values ±�/2, and whose associated magnetic moment μS = γSS is suchthat γS = q/m = 2γorbital, was proposed by two young Dutch physicists in 1925,Uhlenbeck and Goudsmit. Uhlenbeck hesitated between a career in physics and inhistory, and Goudsmit did not have his degree yet. As soon as they understood thattheir hypothesis explained many experimental facts which were obscure up to then,

12.8 The Discovery of Spin 299

they discussed it with their professor, P. Ehrenfest, who encouraged them to publishtheir work. Their idea was received with various reactions in the physics community.Bohr was very enthusiastic, whereas Pauli and Lorentz found serious objections.The objection of Lorentz comes from relativity. If one imagines that the electron is asphere whose electrostatic energy is equal to its mass mec2, one finds a radius of theorder of e2/(mec2) and the equatorial velocity of the sphere is much larger than thevelocity of light, if the angular momentum is �/2 (one obtains veq ∼ c/α = 137c).We know that this objection is not valid, because spin is a purely quantum concept.

12.8.6 The Fine Structure of Atomic Lines

In 1925, Pauli did not believe in spin. In January 1926 he called it a heresy, “Irrlehre.”Pauli’s criticism concerned what is known as fine structure effects. It is one of

these effects of fine splitting of atomic lines that constitutes a world record in theagreement between experiment and theory. This splitting comes from the interactionof the electron’s spinmagnetic moment with its orbital angular momentum. Considerthe hydrogen atom and let us imaginewe sit on the electron. It sees the proton circlingaround it with an angular momentum L and producing a magnetic field that interactswith the spin magnetic moment. The resulting interaction is W = 2α2EI(a0/r)3

(L · S)/�2. And, once all calculations are performed, this gives a splitting that is

twice too large!Fortunately everything was saved by an English physicist. In March 1926, L. H.

Thomas remarked that the rest frame of the electron is not an inertial frame, and thata correct relativistic calculation introduces a factor of 1/2 in the formula, called theThomas precession.2 (Remember Pauli was a great specialist of relativity!)

On March 12, 1926, Pauli wrote to Bohr: “I must surrender!”So there is the incredible story of the number two in physics. It appeared every-

where, but with traps. For a long time, nobody thought that all these numbers two hada common origin. There are lots of numbers in physics, but that one is quite simple.Spin was a two-valued quantum number. Its gyromagnetic ratio was twice the orbitalgyromagnetic ratio. The “normal” Zeeman effect appeared in two-electron atoms.The anomalous Zeeman effect was a splitting in an even number of spectral lines.Complete electron shells in atoms contained 2n2 electrons. The exclusion principlesaid that two electrons could not be in the same state. The fine structure of atomsconsisted of a splitting in two lines. The Thomas precession introduced a factorof 1/2.

Maybe theCreator thought those physicistswere going too far and hewas annoyedto see them discover the structure of the world He had created. So He introduced thisnumber two with all possible traps, thinking that they would stop teasing him. ButHe had forgotten that He had created Pauli. One cannot think of every detail, evenwhen one is a Creator.

2See, for example, J. D. Jackson, Classical Electrodynamics, Sect. 11.8. New York: Wiley (1975).

300 12 Spin 1/2

12.9 Magnetism, Magnetic Resonance

In his Nobel Lecture, on December 12, 1946, Otto Stern spoke about his experiment,but he insisted much less on its revolutionary aspects than on the measurement of theparameter μ0 which he called the elementary magneton, the elementary quantum ofmagnetism.

The discovery of magnetism goes very far back in time. It is probable that thefirst physicists of mankind knew this extraordinary phenomenon, which appeared asa force at a distance at a human scale. One can find examples of compasses in Chinain 2600 B.C., and with the Vikings in the 12th century.

The treatise “De Magnete” of William Gilbert, in 1600, classifies magnetic phe-nomena into three categories.

• The strong and permanentmagnetism of substances such as iron, cobalt, and nickel• The weaker paramagnetism, induced by strong fields in crystals and in fluids• The diamagnetism, weak and repulsive, that appears in all substance and thatappears as the square of the applied field

All these forms of magnetism come from electrons.Paramagnetism is the response of a magnetic moment, be it orbital or spin, to an

external field.Ferromagnetism does not come from the orbital motion of electrons, but from a

subtle collective consequence of Pauli’s principle. In the transition metals, such asFe, Co, Ni, the wave functions of external electrons are such that the electron spinspoint spontaneously in the same direction. This results in a permanent macroscopicmagnetization.

Stern knew he had discovered the elementary quantum of magnetism proposed byPierreWeiss in 1911. For a long time, physicists had understood that ferromagnetismcannot be explained by currents. Weiss introduced the idea of an “elementary mag-neton,” analogous to the elementary charge, as being the “Greatest common divisorof molecular magnetic moments.”3

But, in the same type of experiment, Sternwentmuch further. In 1933, hemanagedtomeasure themagneticmoment of the proton. Thiswasmuchmore difficult, becauseit is 1000 times smaller because of the mass factor (q/m) and one must suppresselectronic effects. Stern usedH2 orHDmolecules,where the effect of paired electronscancel. Stern found a gyromagnetic ratio roughly 2.5 times what one expects from apointlike particle (i.e., the nuclear magneton μN = q�/2mp = 3.1525 10−8 eVT−1).

This observation was the starting point of the discovery of a fourth type of mag-netism, nuclear magnetism, which is a major discovery of the 20th century. Thereexists a nuclear ferromagnetism, a nuclear paramagnetism. They are very weak, buttheir consequences are huge.

This is when Rabi appears. He was born in 1899, he had worked on a thesis onmagnetism at Cornell; then he had gone to Columbiawhere he learned some quantummechanics. In the course of hiswork, hemade a little calculationwhich had enormous

3The Weiss magneton was actually one fifth of the value found by Stern.

12.9 Magnetism, Magnetic Resonance 301

consequences. In particular, between 1933 and 1939, he gained a factor of 1000 inaccuracy compared to Stern’s measurements.

12.9.1 Spin Effects, Larmor Precession

Uncorrelated Space and Spin Variables

In most physical situations, such as the Stern–Gerlach experiment, the space andthe spin variables are correlated. For instance, we only studied in Chap. 11 a firstapproximation to the hydrogen atom where we neglected spin effects. If we includethe spin degree of freedom in this approximation, we find that the states |n, �,m,+〉and |n, �,m,−〉 are degenerate. Actually, there exist corrections to this approxima-tion, such as the fine structure of the hydrogen atom, that wementioned above, whichis due to the interaction between the spin magnetic moment and the electromagneticfield created by the proton. The degeneracy is then partially lifted and the new statesare combinations of the initial states |n, �,m,σ〉. In other words, in a given eigenstateof the total Hamiltonian, the spatial wave function of an electron depends on its spinstate, and the two random variables r and Sz are correlated.

There are many cases where this correlation is extremely weak. In such cases, thetwo variables r and Sz can be considered as independent and their probability law isfactorized. Such a physical situation is represented by a factorized state vector:

Φ(r, t)(

α+(t)α−(t)

). (12.46)

If one performs spin measurements in this case, the results are independent of theposition of the particle. The only relevant observables are 2 × 2 Hermitian matriceswith numerical coefficient (which can depend on time).

Such cases happen in practice, in particular in magnetic resonance experiments,where one can deal with electrons but also with other spin 1/2 particles, in particularprotons. We then use terms such as spin state of the proton instead of state of theproton because the position of the proton in space does not play any role in theexperiment under consideration.

12.9.2 Larmor Precession in a Fixed Magnetic Field

We choose the z-axis parallel to a magnetic field B0. Ignoring space variables, theHamiltonian is:

H = −μ · B0 = −μ0B0 σz. (12.47)

302 12 Spin 1/2

We set− μ0B0/� = ω0/2, that is, ω0 = −γB0. (12.48)

The eigenstates of H are the eigenstates |+〉 and |−〉 of σz.Consider an arbitrary state |ψ(t)〉 such that |ψ(0)〉 = α|+〉 + β|−〉 with |α|2 +

|β|2 = 1. Its time evolution is:

|ψ(t)〉 = α e−iω0t/2 |+〉 + β eiω0t/2 |−〉. (12.49)

The expectation value 〈μ〉 reads:

〈μx〉 = 2μ0 Re(α∗β eiω0t

) = C cos(ω0t + ϕ), (12.50)

〈μy〉 = 2μ0 Im(α∗β eiω0t

) = C sin(ω0t + ϕ), (12.51)

〈μz〉 = μ0(|α|2 − |β|2) , (12.52)

where C and ϕ are, respectively, the modulus and phase of the complex numberα∗β. We recover the Larmor precession that we derived for an arbitrary angularmomentum in Chap.10, Sect. 10.6.4. The projection 〈μz〉 of the magnetic momentalong the field is time-independent, and the component of 〈μ〉 perpendicular to Brotates with the angular velocity ω0. The fact that μz is a constant of motion is aconsequence of the commutation relation [H, μz] = 0 and of the Ehrenfest theorem.

This provides a simple method to measure the angular frequency ω0. We placea coil in a plane parallel to B0 and we prepare a macroscopic quantity of spins allin the same spin state |ψ(0)〉. The precession of 〈μ〉 at the frequency ω0 causes aperiodic variation of the magnetic flux in the coil, and this induces an electric currentat the same frequency. This method is, however, not as accurate as the resonanceexperiment we present below.

12.9.3 Rabi’s Calculation and Experiment

Superposition of a Fixed Field and a Rotating Field

A technique invented by Rabi in the 1930s, allows us to perform a very accuratemeasurement of ω0 by a resonance phenomenon. With this technique, he was ableto gain a factor of 1000 in the accuracy of nuclear magnetic moments.

We place the magnetic moment in a known field B0, on which we superimposea weak field B1 which rotates at a variable angular velocity ω in the xy plane. Sucha field can be obtained with two coils along the x- and y-axes, with a current at afrequency ω and phase-shifted by π/2 (one works with radio frequencies). At theresonance, for ω = ω0, the spin flips between the two possible states |±〉. Noticethat this calculation, characteristic of a driven two-state system, is similar to thatperformed for the ammonia maser in Chap. 7.

12.9 Magnetism, Magnetic Resonance 303

The form of the Hamiltonian is:

H = −μ · B = −μ0B0σz − μ0B1 cosωt σx − μ0B1 sinωt σy. (12.53)

We set:|ψ(t)〉 = a+(t)|+〉 + a−(t)|−〉. (12.54)

The Schrödinger equation yields the differential system:

i a+ = ω0

2a+ + ω1

2e−iωt a−, (12.55)

i a− = ω1

2eiωt a+ − ω0

2a−, (12.56)

where we have defined μ0B0/� = −ω0/2, μ0B1/� = −ω1/2.The change of functions b±(t) = exp(±iωt/2)a±(t) leads to:

i b+ = −ω − ω0

2b+ + ω1

2b−. (12.57)

i b− = ω1

2b+ + ω − ω0

2b−. (12.58)

The above transformation is the quantum form of a change of reference frame. Ittransforms from the laboratory frame to the frame rotating with the magnetic field atan angular velocityω around the z-axis.With this change of reference frame, the basisof Hilbert space is time-dependent, whereas the Hamiltonian is time-independent:

ˆH = �

2

(ω0 − ω ω1

ω1 ω − ω0

)= −�

2(ω − ω0)σz + �

2ω1σx.

In this rotating reference system, the problem becomes time-independent!Equations (12.57), and (12.58) imply b± + (Ω/2)2 b± = 0 with:

Ω2 = (ω − ω0)2 + ω2

1 . (12.59)

Suppose the spin is initially in the state |+〉; that is, b−(0) = 0. One finds:

b−(t) = − iω1

Ωsin

(Ωt

2

)(12.60)

b+(t) = cos

(Ωt

2

)+ i

ω − ω0

Ωsin

(Ωt

2

). (12.61)

304 12 Spin 1/2

Fig. 12.6 Rabi oscillations a slightly off resonance ω − ω0 = 3ω1; b at resonance ω = ω0

The probability that a measurement of Sz at time t gives the result −�/2 is:

P+→−(t) = |〈−|ψ(t)〉|2 = |a−(t)|2 = |b−(t)|2

=(ω1

Ω

)2sin2

(Ωt

2

). (12.62)

This formula, which is due to Rabi, exhibits the resonance phenomenon:

• If the frequency ω of the rotating field is noticeably different from the frequencyω0 we want to measure, more precisely, if |ω − ω0| � ω1, the probability that thespin flips, that is, that we measure Sz = −�/2, is very small for all t.

• If we choose ω = ω0, then the probability for a spin flip is equal to one at timestn = (2n + 1)π/ω1 (n integer) even if the amplitude of the rotating field B1 is verysmall.

• For |ω − ω0| ∼ ω1, the probability amplitude oscillates with an appreciable ampli-tude, smaller than one.

In Fig. 12.6, we have drawn the time oscillation of the probability P+→− offresonance and at resonance. For a typical magnetic field of 1 Tesla, the resonancefrequency is ωe/2π ∼ 28Ghz for an electron, and 2.79 ωN/2π ∼ 43MHz for a pro-ton. These frequencies correspond to decametric waves in the nuclear case, andcentimetric waves in the electronic case.

At the resonance, the system periodically absorbs and emits the energy 2µ0B byabsorption and stimulated emission.

Notice that, here,we have closed the loop, concerning the starting point inChap. 2.We have proven in a particular case that absorption and stimulated emission ofradiation do occur at the Bohr frequency of the system ν = ΔE/h.

Rabi’s Experiment

The resonance effect described above was understood by Rabi in 1939. It provides avery accurate measurement of a magnetic moment. The device used by Rabi consists

12.9 Magnetism, Magnetic Resonance 305

Fig. 12.7 Apparatus developed by Rabi for observing themagnetic resonance effect. In the absenceof magnetic resonance, all particles emitted in the state |+〉 reach the detector. If resonance occurs,the spins of the particles flip between the two magnets, and the signal drops

of a source, two Stern–Gerlach deflectors with magnetic fields in opposite directions,and a detector (Fig. 12.7). Between the two Stern–Gerlachmagnets, one places a zonewith a superposition of a uniform field B0 and a rotating fieldB1, as described above.

Consider first the effect of the two Stern–Gerlach magnets in the absence of thefields B0 and B1. A particle emitted by the source in the spin state |+〉 undergoes twosuccessive deflections in two opposite directions and it reaches the detector. Whenthe fields B0 and B1 are present, this is not true anymore. If the frequency ω of therotating field is close to the Larmor frequency ω0, the resonance phenomenon willchange the component μz of the particle. When such a spin flip occurs between thetwo Stern–Gerlach magnets, the two deflections have the same direction (upward inthe case of Fig. 12.7), and the particle misses the detector. The signal registered onthe detector as a function of the frequency of the rotating field B1 undergoes a sharpdrop for ω = ω0 (Fig. 12.8). This leads to a measurement of the ratio |μ|/j = �ω0/B0

for a particle of angular momentum j. Actually, this measurement is so precise thatthe main source of error comes from the determination of B0. In practice, as shownin Fig. 12.8, the frequency ω stays fixed and one varies the magnitude of the field B0,or equivalently the frequency ω0.

Fig. 12.8 Signal (obtainedby Rabi) recorded on thedetector of Fig. 12.7 with abeam of HD molecules, as afunction of the field B0(B1 = 10−4 T,ω/2π = 4MHz)

306 12 Spin 1/2

In 1933 Stern, with his apparatus, had measured the proton magnetic momentwith 10% accuracy. That was a very difficult experiment, because nuclear magneticmoments are 1000 times smaller than electronic ones. In 1939, with his resonanceapparatus, Rabi gained a factor of 1000 in accuracy. The resonance is very selec-tive in frequency and the presence of other magnetic moments causes no problem.Rabi’s result struck the minds of people. It was greeted as a great achievement. Sternremarked that Rabi had attained the theoretical accuracy of the measurement, whichis fixed by the uncertainty relations.WhenHulthén announced that Rabiwas awardedthe Nobel prize on December 10, 1944, on the Stockholm radio, he said that “By thismethod Rabi has literally established radio relations with the most subtle particlesof matter, with the world of the electrons and of the atomic nucleus.”

The Nobel prize for 1943 was awarded to Stern, that of 1944 to Rabi, and that of1945 to Pauli. Rabi did not go to the reception. The world had undergone a terriblyhard period.

12.9.4 Nuclear Magnetic Resonance

Stern and Rabi agreed in 1943 on the fact that there was an intrinsic limitation withatomic andmolecular beams, both in time and in intensity. They knew that eventuallyone would have to operate on condensed matter.

The great breakthrough of the applications of nuclear magnetic resonance(N.M.R.) came with the works of Felix Bloch at Stanford and of Edward Purcellat MIT, in 1945. Owing to the development of radiowave technologies, Bloch andPurcell were able to operate on condensed matter, and not on molecular beams. Oneuses macroscopic numbers of spins, thereby obtaining much more intense signalsand more manageable experiments. The resonance is observed, for instance, by mea-suring the absorption of the wave generating the rotating field B1. The imbalancebetween the populations of the two states |+〉 and |−〉, which is necessary in order toget a signal, results from the conditions of thermal equilibrium. In a field B0 = 1T,the magnetic energy for a proton is 2.79µNB ∼ 10−7 eV and the relative populationdifference between the two spin states due to the Boltzmann factor at room temper-ature is π+ − π− ∼ 4 10−6. This relative difference is small, but quite sufficient toobserve a significant signal because one dealswith samples containing amacroscopicnumber of spins (typically 1023).

Bloch and Purcell discovered nuclear paramagnetism, a fourth type ofmagnetism.The applications of magnetic resonance are numerous in domains ranging from

solid-state physics and low temperatures, chemistry, biology, or medicine. By itsmagnetic effects, spin can play the role of a local probe inside matter. NMR hastransformed chemical analysis and the determination of the structure of molecules(see, for instance, Fig. 12.9). It has become an invaluable tool in molecular biology.Since 1980, NMR has also caused a revolution in medical diagnosis and physiology.Under the less frightening name of M.R.I. (Magnetic Resonance Imaging, sincethere is no dangerous nuclear reaction taking place) it allows us to measure and

12.9 Magnetism, Magnetic Resonance 307

Fig. 12.9 One of the first examples of nuclear magnetic resonance applied to chemistry: the reso-nance signal obtained with the protons of the ethanol molecule CH3CH2OH consists of a three-peakstructure. These peaks are associated respectively with the three protons of the CH3 group, with thetwo protons of the CH2 group, and with the unique proton of the OH group. The magnetic field B0is ∼0.8T, and the total trace is 7.5µT wide

visualize in three dimensions and with a spatial precision better than a millimeter theconcentration in water of “soft matter” (muscles, brain, and so on), which, in contrastwith bones, is difficult to observe with X-rays. One studies in this way the structureand the metabolism of living tissues; one can detect internal injuries, tumors, andso on. The nuclear spin, which was a curiosity for some visionary physicists of the1940 and 1950s, has become one of the great hopes of modern medicine.

One can, by now, visualize the activity of the brain in real-time. It is possibleto localize and register the response of the visual cortex to some stimulation. Thefollowing step, after submitting a volunteer to a sequence of such excitations, is toask her to think about the signal. The NMR response of the brain is the same as thatobtained by an external stimulation. This may be considered a direct experimentalproof that we think, which is somewhat comforting for the mind.

12.9.5 Magnetic Moments of Elementary Particles

The electron, the proton, and the neutron are spin 1/2 particles. The correspondingspin magnetic moment is related to the spin S by the relation μ = γS. Experimentsgive the following values of the gyromagnetic ratios

electron γ � 2γ0 = −q/me,

proton γ � +2.79 q/mp,

neutron γ � −1.91 q/mp.

308 12 Spin 1/2

Thepossible results of themeasurement of the component of thesemagneticmomentsalong a given axis are therefore:

electron μz = ±μB = ∓q�/2me,

proton μz = ±2.79 q�/2mp,

neutron μz = ∓1.91 q�/2mp.

The quantity μB = −9.274 10−24 JT−1 is called the Bohr magneton. The quantityμN = q�/2mp = 5.051 10−27 JT−1 is called the nuclear magneton.

Dirac’s relativistic theory of the electron explains the value of the electron mag-netic moment:

μ = ge

(q

2me

)S, with ge = 2.

The value measured experimentally for the gyromagnetic factor ge nearly coincideswith this prediction. One can account for the slight difference between the exper-imental result and Dirac’s prediction by taking into account the coupling of theelectron with the quantized electromagnetic field (quantum electrodynamics). Thisconstitutes one of the most spectacular successes of fundamental physics. The exper-imental and theoretical values of the quantity ge coincide within the accuracy limitsof experiments and of computer calculations. At present, one has for the electron,setting ge = 2(1 + a),

atheo. = 0.001 159 652 18178 (77), (12.63)

aexp. = 0.001 159 652 18073 (28) ; (12.64)

the errors in parentheses bear on the two last digits. To lowest order in the finestructure constant, quantum electrodynamics gives the result a = α/2π = 0.00116.

The coefficients +2.79 and −1.91 for the proton and for the neutron are due tothe internal structure of these particles. They can be measured with great accuracyby magnetic resonance experiments: μp/μN = 2.792 847 386 (63) and μn/μN =−1.913 042 75 (45); they can be calculated with 10% accuracy in the quark model.

12.10 Entertainment: Rotation by 2π of a Spin 1/2

It seems obvious and of common sense geometrically that the rotation of 2π of asystem around a fixed axis is equivalent to the identity. However, strictly speakingthis is not true for a spin 1/2 particle.

12.10 Entertainment: Rotation by 2π of a Spin 1/2 309

Let us return to the calculation of Sect. 12.9.2 and suppose that, at time t = 0 thestate of the system is | + x〉:

|ψ(t = 0)〉 = 1√2

(|+〉 + |−〉) .

Themean value of the magnetic moment, given by (12.50), through (12.52), is 〈μ〉 =μ0 ux. Equation (12.49) gives the evolution of this state. After a time t = 2π/ω0,classically, the system has precessed by an angle 2π aroundB. In quantummechanicsone verifies that the expectation value is back to its initial value 〈μ〉 = μ0 ux. Whatcan we say, however, about the state vector? We can check that |ψ(t)〉 is still aneigenvector of Sx (or of μx) with an eigenvalue +�/2. However, we notice that, quitesurprisingly, the state vector has changed sign:

|ψ(t = 2π/ω0)〉 = − 1√2

(|+〉 + |−〉) = −|ψ(0)〉.

A2π rotation is therefore not equivalent to the identity for a spin 1/2.Only rotations of4nπ give back the initial state identically. This property can also be guessed from thedependence eimϕ for orbital angular momenta: using the same formula for m = 1/2and ϕ = 2π would give eiπ = −1.

This peculiarity was understood as soon as spin 1/2 was discovered in 1926. Itremained a controversial point for more than 50 years. Does the phase of the statevector after a rotation of 2π have a physical meaning? The positive experimentalanswer was only given in the 1980s in a series of remarkable experiments.4 The spin1/2 particles are sent in a two-channel interferometer. In one of them a magneticfield rotates the spins by multiples of 2π. The change in sign of the wave function isobserved by a displacement of the interference fringes and the experimental signalconfirms that a rotation of 4π is needed to recover the fringe pattern that is measuredin the absence of rotation.

This property reflects an important mathematical structure that relates the twoLie groups SO(3) and SU(2). Rotations in Euclidian space R3 form the well-knowngroup SO(3). Mathematically, one says that there is a local isomorphism betweenthe Lie algebras of the two groups SO(3) and SU(2), but that these two groupsare not globally isomorphic. This formalism, which was called spinor theory, wasdeveloped in the early 20th century by the mathematician Elie Cartan. The minussign is completely equivalent to the fact that on a Moebius strip one must make aneven number of rotations in order to get back to the starting point.

4A.W. Overhauser, A.R. Collela and S.A.Werner,Phys. Rev. Lett. 33, 1237 (1974); 34, 1472 (1975);35, 1053 (1975).

310 12 Spin 1/2

12.11 Exercises

1. Determination of the magnetic state of a silver atom

Consider a silver atom in an arbitrary state of its magnetic moment:

α|+〉z + β|−〉z with |α|2 + |β|2 = 1. (12.65)

a. Show that this is an eigenstate of u · μ with the eigenvalue +μ0, where u is a unitvector whose direction will be determined.

b. Alice sends to Bob one silver atom in the unknown state (12.65). Can Bob deter-mine this state using Stern et Gerlach measurements?

c. Alice sends to Bob N (� 1) silver atoms, all prepared in the same unknown state(12.65). Give a possible strategy for Bob to determine this state (within statisticalerrors).

2. Repeated measurement; quantum Zeno paradox

The magnetic moment μ of a neutron can be described in the same way as themagnetic moment of a silver atom in the Stern-Gerlach experiment. If a neutron isplaced in a uniform magnetic field B parallel to the z axis, it can be considered atwo-state system formagneticmomentmeasurements (disregarding space variables).

We note |+〉 and |−〉 the eigenstates of the observable μz. These eigenstatescorrespond to the two eigenvalues +μ0 and −μ0. The Hamiltonian of the system inthe field B is H = −Bμz. We set ω = −2μ0B/�.

a. Give the energy levels of the system.b. At time t = 0 the neutron is prepared in the state: |ψ(0)〉 = (|+〉 + |−〉)/√2.

What results can be obtained by measuring μx on this state, with which probabil-ities?

c. Write the state |ψ(T)〉 of the magnetic moment at a later time T .d. We measure μx at time T . What is the probability to find +μ0?e. We now perform on the same system a sequence of N successive measurements

at times tp = pT/N p = 1, 2, . . . ,N . What is the probability that all these mea-surements give the result μx = +μ0?

f. What does this probability become if N → ∞? Interpret the result; do you thinkit makes sense physically?

3. Products of Pauli matrices

Show that

σj σk = δj,k + iεj,k,� σl (12.66)

where εj,k,� = 1 (resp. −1) if (j, k, �) is an even (resp. odd) permutation of (x, y, z),and εj,k,� = 0 otherwise.

12.11 Exercises 311

4. Algebra with Pauli matrices

Consider the Pauli matrices σ, and two vectors A and B. Show that:

(σ · A)(σ · B) = A · B + iσ · (A ∧ B).

5. Spin and orbital angular momentum

Consider a spin 1/2 particle whose state is |ψ〉 = ψ+(r, t)|+〉 + ψ(r, t)|−〉. Let S bethe spin observable and L the orbital angular momentum. We assume that:

ψ+(r) = R(r)

(Y0,0(θ,ϕ) + 1√

3Y1,0(θ,ϕ)

)

ψ−(r) = R(r)√3

(Y1,1(θ,ϕ) − Y1,0(θ,ϕ)

).

a. What is the normalization condition on R(r)?b. What are the probabilities to find ±�/2 in measurements of Sz or Sx?c. What are the possible results of a measurement of Lz? Give the corresponding

probabilities?

Chapter 13Addition of Angular Momenta

When atomic spectral lines are observed with sufficient resolution, they appear tohave in general a complex structure, each line being in fact a group of nearby compo-nents. The (numerous) fine and hyperfine splittings of atomic levels are of particularimportance because of what they revealed on atomic structure and in terms of appli-cations. The origin of such structures lies in the magnetic interactions of the electroninside the atom, due to interactions of orbital and/or spin magnetic moments.

Technically, this boils down to the notion of the addition of angular momentain quantum mechanics, and the total angular momentum of a system. This notion isuseful in many physical problems and we will give its basic elements in Sect. 13.1. InSect. 13.2 we give a qualitative description of the spin—orbit interaction of the elec-tron spinmagneticmomentwith themagnetic field originating from its orbitalmotionaround the nucleus. A known example of the corresponding splitting is the yellowline (D-line) of sodium. In Sect. 13.3 we will describe more thoroughly the hyperfineinteraction between the spin magnetic moments of the electron and of the proton inthe ground state of hydrogen. This interaction produces a splitting which is respon-sible for the 21cm line of hydrogen, of considerable interest in astrophysics. Itsmany analogs in alcali atoms have important practical applications, for instance inatomic clocks.

13.1 Addition of Angular Momenta

The Total Angular Momentum Operator

Consider two angular momentum observables J1 and J2 which, by definition, act intwo different Hilbert spaces E1 and E2. This may concern, for instance, a system oftwo particles: E1 (resp. E2) is then the space L2(R3) of square-integrable functions

© Springer International Publishing Switzerland 2016J.-L. Basdevant, Lectures on Quantum Mechanics,Graduate Texts in Physics, DOI 10.1007/978-3-319-43479-7_13

313

314 13 Addition of Angular Momenta

in r1 (resp. r2). It can also concern a particle moving in space (E1 = L2(R3)), whichpossesses an intrinsic angular momentum (E2 = Espin).

The Hilbert space of the total system is the (tensor) product:

E = E1 ⊗ E2.

By definition, the total angular momentum observable of the system is:

J = J1 ⊗ I2 + I1 ⊗ J2 ≡ J1 + J2, (13.1)

where I1 (resp. I2) is the identity operator in E1 (resp. E2). This observable acts inE and it is an angular momentum observable. Indeed it satisfies the commutationrelations:

J × J = i� J (13.2)

since J1 and J2 commute. Therefore we know that we can diagonalize simultane-ously J 2 and Jz . We also know the set of their possible eigenvalues: �2 j ( j + 1) with2 j integer for J 2, and �m with m = − j, . . . , j − 1, j for Jz for a given j .

The four angular momentum observables:

J 21 , J 2

2 , J 2, Jz

commute. Moreover this set forms a CSCO in the sense of Chap.10, Sect. 10.2. Theircommon eigenbasis is therefore unique. We write their common eigenvectors as

| j1, j2; j,m〉.

One has by definition:

J 21 | j1, j2; j,m〉 = j1( j1 + 1)�2 | j1, j2; j,m〉 (13.3)

J 22 | j1, j2; j,m〉 = j2( j2 + 1)�2 | j1, j2; j,m〉 (13.4)

J 2| j1, j2; j,m〉 = j ( j + 1)�2 | j1, j2; j,m〉 (13.5)

Jz| j1, j2; j,m〉 = m� | j1, j2; j,m〉. (13.6)

As in Chap.10, we omit the presence of other possible quantum numbers whichwould be cumbersome to write and are irrelevant in the present discussion.

Uncoupled and Coupled Bases

The Hilbert space E corresponding to the degrees of freedom associated with theangular momenta J1 and J2 is generated by the family of factorized states:

{| j1,m1〉 ⊗ | j2,m2〉} ≡ {| j1,m1; j2,m2〉}.

13.1 Addition of Angular Momenta 315

In this basis, the observables J 21 , J1z, J 2

2 , J2z are diagonal. Consider the eigen-subspace of the two observables J 2

1 and J 22 corresponding to given values j1 and

j2. The dimension of this subspace is (2 j1 + 1)(2 j2 + 1) and we ask the followingquestion:

What are, in this subspace, the eigenvectors of J 2 and Jz , and the correspondingeigenvalues j ( j + 1)�2 and m�?

In other words, we want to perform in each eigen-subspace of J 21 and J 2

2 a changeof basis to go from the uncoupled factorized eigenbasis common to { J 2

1 , J1z, J 22 , J2z}

to the coupled eigenbasis common to { J 21 , J 2

2 , J 2, Jz}. The eigenvalues of J 2 and Jzwill be expressed as functions of j1, j2,m1 and m2. Once the determination of thevalues of j is performed, we shall express the eigenstates | j1, j2; j,m〉 in terms ofthe states | j1,m1; j2,m2〉:

| j1, j2; j,m〉 =∑

m1m2

C j,mj1,m1; j2,m2

| j1,m1; j2,m2〉 (13.7)

C j,mj1,m1; j2,m2

= 〈 j1,m1; j2,m2| j1, j2; j,m〉. (13.8)

The coefficients C j,mj1,m1; j2,m2

of the change of basis (13.7) are called Clebsch–Gordancoefficients.

13.1.1 A Simple Case: The Addition of Two Spins 1/2

The (simplest) case of two spin 1/2 particles will be of particular interest in thefollowing (21cm line of hydrogen, Pauli principle, etc.). We will first treat it inan elementary way, before considering the general problem of the coupling of twoarbitrary angular momenta.

The Hilbert Space of the Problem

Consider a system of two spin 1/2 particles, for instance the electron and the protonin a hydrogen atom, or the two electrons of a helium atom. We note the particles 1and 2. The Hilbert space of the two spins Es is the (tensor) product of the two spinspaces (we omit here the external position variables)

Es = E1spin ⊗ E2

spin. (13.9)

Es is a four-dimensional spacegeneratedby the family {|σ1〉⊗|σ2〉},σ1 = ±,σ2 = ±,which we note in the simpler form:

{|+ ; +〉, |+ ; −〉, |− ; +〉, |− ; −〉}. (13.10)

316 13 Addition of Angular Momenta

The total spin operator isS = S1 + S2.

If we need to include external variables, he most general state (space + spin) |ψ〉 ofthis two spin 1/2 particle system can be written as:

|ψ〉 = ψ++(r1, r2)|+ ; +〉 + ψ+−(r1, r2)|+ ; −〉+ ψ−+(r1, r2)|− ; +〉 + ψ−−(r1, r2)|− ; −〉. (13.11)

Matrix Representation

We can use a matrix representation of the spin states and spin operators for thissystem. In the basis (13.10) a state is represented by a four-component vector. Theobservables S1 and S2 (extended to the tensor product space) can easily be written,using the Pauli matrices, and a block 2 × 2 notation for the 4 × 4 matrices:

S1x = �

2

⎛

⎜⎜⎝0

... I. . . . . . . . .

I... 0

⎞

⎟⎟⎠ S2x = �

2

⎛

⎜⎜⎝σx

... 0. . . . . . . . .

0... σx

⎞

⎟⎟⎠

S1y = �

2

⎛

⎜⎜⎝0

... −i I. . . . . . . . .

i I... 0

⎞

⎟⎟⎠ S2y = �

2

⎛

⎜⎜⎝σy

... 0. . . . . . . . .

0... σy

⎞

⎟⎟⎠

S1z = �

2

⎛

⎜⎜⎝I

... 0. . . . . . . . .

0... − I

⎞

⎟⎟⎠ S2z = �

2

⎛

⎜⎜⎝σz

... 0. . . . . . . . .

0... σz

⎞

⎟⎟⎠

where I stands for the 2 × 2 identity matrix.

Total Spin States

We consider Es and we note |S, M〉 the eigenstates of S2 and Sz with respectiveeigenvalues S(S + 1)�2 and M�. Since Sz = S1z + S2z , the largest possible value ofM is 1

2 + 12 = 1. The corresponding state is unique; it is the state |+ ; +〉. Similarly,

the smallest possible value of M is − 12 − 1

2 = −1, and the corresponding eigenstateis |− ; −〉.

13.1 Addition of Angular Momenta 317

Let us calculate the action of the square of the total spin on these two vectors:

S2|+ ; +〉 =(S21 + S22 + 2S1 · S2

)|+ ; +〉

=(3

4�2 + 3

4�2 + �

2

2

(σ1x σ2x + σ1y σ2y + σ1zσ2z

)) |+ ; +〉= 2�

2|+ ; +〉.

Similarly:S2|− ; −〉 = 2�

2|− ; −〉.

The two states |+ ; +〉 and |− ; −〉 are therefore eigenstates of S2 with the eigenvalue2�

2, which corresponds to an angular momentum equal to 1. With the notations ofthe first paragraph, we have therefore:

|s1 = 12 , m1 = 1

2 ; s2 = 12 , m2 = 1

2 〉 = |s1 = 12 , s2 = 1

2 ; S = 1, M = 1〉

and:

|s1 = 12 , m1 = − 1

2 ; s2 = 12 , m2 = − 1

2 〉 = |s1 = 12 , s2 = 1

2 ; S = 1, M = −1〉.

Since we have recognized two states |S = 1, M = ±1〉 of angular momentum 1, wenow look for the third one |S = 1, M = 0〉. In order to do this, we use the generalrelation found in Chap. 10: S−| j,m〉 ∝ | j,m − 1〉 and we obtain:

S−|S = 1, M = 1〉 =(S1− + S2−

)|+ ; +〉 ∝ |− ; +〉 + |+ ; −〉.

After normalization, we obtain the state:

|S = 1, M = 0〉 = 1√2

(|+ ; −〉 + |− ; +〉) .

One can check that this state is indeed an eigenstate of S2 and Sz with respectiveeigenvalues 2�

2 and 0.We have identified a 3-dimensional subspace in Es corresponding to a total angular

momentum equal to 1. The orthogonal subspace, of dimension 1, is generated by thevector:

1√2

(|+ ; −〉 − |− ; +〉) .

One can readily verify that this vector is an eigenvector of S2 and of Sz with botheigenvalues equal to zero.

To summarize, the total spin in the particular case j1 = j2 = 1/2 corresponds to:

S = 1 or S = 0

318 13 Addition of Angular Momenta

and the four corresponding eigenstates, which form a basis of EH , are:

|1, M〉 :⎧⎨

⎩

|1, 1〉 = |+ ; +〉|1, 0〉 = (|+ ; −〉 + |− ; +〉)/√2|1, − 1〉 = |− ; −〉

(13.12)

|0, 0〉 : |0, 0〉 = (|+ ; −〉 − |− ; +〉)/√2. (13.13)

In this particular case of two spin 1/2 particles, we have solved the problem ofSect. 1.2 by decomposing the 4 = 2× 2 dimensional space Es (tensor product of thetwo 2-dimensional spaces) into a direct sum of a one-dimensional space (S = 0) anda 3-dimensional space (S = 1), i.e. 2 × 2 = 1 + 3.

One can check that any rotation of the (x, y, z) axes which transforms the m1 =±1 into m ′

1 = ±1 and, similarly, m2 = ±1 into m ′2 = ±1 will end up with a

similar |1, M ′ = m ′1 + m ′

2〉, |0, 0〉 quadruplet. Notice that the state |0, 0〉 is rotationinvariant.

Symmetry Properties

The following symmetry properties will be important when we consider identicalparticles and the Pauli principle.

The three states |1, M〉 are called collectively the triplet state of the two-spinsystem. They are symmetric with respect to the interchange of the z projections ofthe spins of the two particles, σ1 and σ2. The state |0, 0〉 is called the singlet stateand it is antisymmetric in the same exchange. In mathematical terms, if we define apermutation operator Ps

12 in Es by the relation:

Ps12|σ1 ; σ2〉 = |σ2 ; σ1〉, (13.14)

the triplet and singlet states are eigenvectors of this operator:

Ps12|1, M〉 = |1, M〉 Ps

12|0, 0〉 = −|0, 0〉. (13.15)

13.1.2 Addition of Two Arbitrary Angular Momenta

We now want to establish the following results:

1. Consider two angular momentum observables J1 and J2. In the subspace corre-sponding to given values j1 and j2, the possibles values for the quantum numberj associated with the total angular momentum J are, assuming for definitenessthat j1 ≥ j2:

j ∈ { j1 + j2, j1 + j2 − 1, j1 + j2 − 2, . . . , j1 − j2 + 1 , j1 − j2}.

13.1 Addition of Angular Momenta 319

2. For a given value of j , the quantum number m takes the values

m ∈ { j, j − 1, j − 2, . . . ,− j}.

3. The ensuing states | j,m〉 ≡ | j1, j2; j,m〉 are linear combinations of the factorizedstates | j1,m1; j2,m2〉. Assuming again, for definiteness, that j1 ≥ j2, their totalnumber

S =k=− j2∑

k= j2

(2( j1 + k) + 1) = (2 j1 + 1)(2 j2 + 1) (13.16)

is equal to the initial number of factorized states, as expected.

Construction of the States Such that j = j1 + j2

Any vector | j1,m1; j2,m2〉 is an eigenvector of Jz = J1z + J2z with the eigenvaluem� and m = m1 + m2. One therefore deduces the following result:

The vector | j1, j2; j,m〉 corresponding to j = j1 + j2 and m = j1 + j2 existsand it is unique.

Indeed the maximum values of m1 and m2 are j1 and j2, therefore the maximumvalue of m is mmax = j1 + j2. We deduce that the maximal value of j is alsojmax = j1 + j2 since the index m can take all the values m = − j, − j + 1 . . . , j ina given eigen-subspace of J 2.

The only normalized vector in the Hilbert space which fulfills the condition m =mmax (up to a phase factor) is:

| j1,m1 = j1; j2,m2 = j2〉.

This vector is also an eigenstate of J 2 with the eigenvalue j ( j+1)�2 and j = j1+ j2,as can be checked directly using the following expression:

J 2 = J 21 + J 2

2 + J1+ J2− + J1− J2+ + 2 J1z J2z .

Consequently we can write:

| j = j1 + j2,m = j1 + j2〉 = |m1 = j1;m2 = j2〉. (13.17)

Remark Here and in what follows, we omit the indices j1 and j2 in the left andright-hand sides of (13.17); these are implicit in the form: | j,m〉 ≡ | j1, j2; j,m〉 and|m1;m2〉 ≡ | j1,m1; j2,m2〉.

We now define as in Chap.10 the raising and lowering operators:

J+ = J1+ + J2+ J− = J1− + J2−.

320 13 Addition of Angular Momenta

We have:J±| j,m〉 ∝ | j,m ± 1〉,

J1±|m1;m2〉 ∝ |m1 ± 1;m2〉 J2±|m1;m2〉 ∝ |m1;m2 ± 1〉.

where the proportionality coefficients are given in (10.16). Starting with | j = j1 +j2,m = j1 + j2〉 we can generate a series of 2( j1 + j2) + 1 states | j = j1 + j2,m ′〉for m ′ ∈ j1 + j2 − 1, . . . ,−( j1 + j2) by applying repeatedly the operator J−. Forinstance, using the normalization coefficients given in (10.16) we find:

|ψa〉 = | j = j1 + j2,m = j1 + j2 − 1〉 (13.18)

∝ √j1 |m1 = j1 − 1;m2 = j2〉 + √

j2 |m1 = j1;m2 = j2 − 1〉.

Eigensubspaces of J z

Agraphical representation of the uncoupled basis states |m1;m2〉 is given in Fig. 13.1in the particular case j1 = 3/2 and j2 = 1. In the m1,m2 plane, each dot representsa basis state. A fixed m = m1 + m2, corresponding to an eigensubspace E(m)

of Jz , is represented by a straight dashed line. The point in the upper right cornercorresponds to the state (13.17). As we already noted the dimension of this particulareigensubspace E( j1 + j2) is 1. The next dashed line corresponds tom = j1 + j2 − 1,and the corresponding eigensubspace E( j1 + j2 − 1) has dimension 2, with thepossible basis:

|m1 = j1 − 1;m2 = j2〉 |m1 = j1;m2 = j2 − 1〉 (13.19)

In general the eigenvaluem� of Jz has some degeneracy, except form = ±( j1 + j2).By construction, each eigensubspace E(m) of Jz is invariant under the action of

the Hermitian operators J+ J− and J− J+. Indeed J+ (resp. J−) globally increases(resp. decreases) the value of m1 + m2 by 1. Using the expression:

J 2 = 1

2

(J+ J− + J− J+

)+

(J1z + J2z

)2,

it follows that E(m) is also globally invariant under the action of J 2.

Construction of All States of the Coupled Basis

The total dimension of the Hilbert space is (2 j1 + 1)(2 j2 + 1). Inside this space wehave already identified the 2 j + 1 vectors of the coupled basis with j = j1 + j2. Wenow give the principle of the determination of all remaining states of the coupledbasis.

Consider the subspaceE( j1+ j2−1) of Jz , whose possible basis is given in (13.19).Inside this subspacewe have already identified the vector |ψa〉 given in (13.18), whichis by construction an eigenvector of Jz and J 2 with eigenvalues ( j1 + j2 − 1)� and( j1 + j2)( j1 + j2 + 1)�2. Consider the vector of E( j1 + j2 − 1) orthogonal to |ψa〉:

13.1 Addition of Angular Momenta 321

Fig. 13.1 Representation of the uncoupled basis states |m1;m2〉. The dashed lines represent theeigensubspaces E(m) of Jz . These subspaces are globally invariant under the action of J 2. Thepresent figure corresponds to j1 = 3/2 and j2 = 1

|ψb〉 = √j2 |m1 = j1 − 1;m2 = j2〉 − √

j1 |m1 = j1;m2 = j2 − 1〉.

Since E( j1 + j2 − 1) is globally invariant under the action of J 2, we can diagonalizethis operator inside E( j1 + j2 − 1), and the corresponding eigenbasis is orthogonal.We know that |ψa〉 is an eigenvector of J 2. Therefore |ψb〉, which is orthogonal to|ψa〉, is also an eigenvector of J 2:

J 2 |ψb〉 = j ( j + 1)�2|ψb〉, (13.20)

and we want to determine the value of j . On one hand we have Jz|ψb〉 = m�|ψb〉with m = j1 + j2 − 1; therefore j ≥ j1 + j2 − 1 since one always has j ≥ m.On the other hand, we cannot have j = j1 + j2, since this would mean there existstwo independent vectors (|ψa〉 and |ψb〉) corresponding to the same values of j andm (i.e. j1 + j2 and j1 + j2 − 1 respectively). This cannot be true since (i) there isonly one vector corresponding to j = m = j1 + j2 and (ii) there is a one-to-onecorrespondence, through the action of J±, between states associated with ( j,m) andstates associated with ( j,m ± 1). Consequently we must have j = j1 + j2 − 1 in(13.20):

|ψb〉 ∝ | j = j1 + j2 − 1,m = j1 + j2 − 1〉.

By applying repeatedly the operator J− to |ψb〉, we generate a new series of states,which are labelled | j = j1 + j2 − 1,m ′〉.

We have now identified all vectors in the two subspaces E( j1 + j2) and E( j1 +j2 − 1). We can repeat the same operation for the subspace E( j1 + j2 − 2) (whose

322 13 Addition of Angular Momenta

dimension is 2( j1 + j2) − 3, with two vectors already identified), etc., until all theeigenstates |m1;m2〉 are used. This occurs when we reach a quantum numberm suchthat the dimension of E(m) is smaller than or equal to the dimension of E(m + 1)(m = −1/2 for the values chosen in Fig. 13.1). Altogether, we find 2 jmin + 1 seriesof states, where jmin = min( j1, j2). The possible values for j are therefore:

j = j1 + j2, j = j1 + j2 − 1, . . . , j = j1 + j2 − 2 jmin = | j1 − j2|

The total number of states of the coupled and uncoupled bases coincide as seen in(13.16).

(2 j1 + 1)(2 j2 + 1) = 2( j1 + j2) + 1 + 2( j1 + j2) − 1 + . . . + 2| j1 − j2| + 1.

Mathematically we have decomposed the (2 j1 + 1) × (2 j2 + 1) dimensional space(tensor product of a 2 j1 + 1 and a 2 j2 + 1 dimensional space) into the direct sum ofa 2( j1 + j2) + 1, a 2( j1 + j2) − 1, . . ., etc. dimensional space.

Using this general procedure, one can determine the coefficients which relate thevectors of the uncoupled basis and those of the coupled one (the Clebsch–Gordancoefficients, defined in (13.7)). The general expression of a Clebsch–Gordan coeffi-cient is quite involved and can be found elsewhere.1

13.2 One-Electron Atoms, Spectroscopic Notations

In Chap.11, we neglected spin effects in the hydrogen atom. If one takes spininto account, the classification of atomic states requires four quantum numbers:|n, �,m,σ〉,σ = ±. The states σ = ± are degenerate in energy in the Coulombapproximation. The spin-orbit interaction which we will discuss later on liftsthis degeneracy, and we will see that the energy eigenstates are the eigenstates|n, �, j,m j 〉 of the total angularmomentum J = L+S. Their energies do not dependon the quantum number m j giving the projection on z of J , since the Hamiltonianis rotation invariant. In the case of an electron of orbital angular momentum � andspin 1/2, the values of j are therefore:

j = � ± 1/2,

except for � = 0 in which case j = 1/2.One classifies the states according to the above quantum numbers. The spectro-

scopic notation consists in adding, on the right of the symbol (n�) of Chap.11, thevalue of j , for instance:

1See e.g. exercise 4 and A.R. Edmonds, Angular Momentum in Quantum Mechanics (PrincetonUniversity Press, 1950).

13.2 One-Electron Atoms, Spectroscopic Notations 323

2p3/2 ⇔ n = 2, � = 1, j = 3

2= � + 1

2,

3d3/2 ⇔ n = 3, � = 2, j = 3

2= � − 1

2.

13.2.1 Fine Structure of Monovalent Atoms

The resonance line of monovalent atoms appears to be split into two components.One example is the yellow line of sodium, corresponding to the transition 3p → 3s,which is used in highly efficient sodium vapor street lights. It is split into two linescalled respectively D1 and D2, of wavelengths λ1 589.6 nm and λ2 589.0 nm.The same effect is observed in the hydrogen atom: the Lyman α line, correspondingto the transition 2p → 1s, is also split into two components.

This splitting is due to the so-called spin-orbit interaction: the first excited level,which has an orbital angularmomentum � = 1 (p state), is cleaved into two sub-levelsbecause of this interaction. One level corresponds to j = 3/2, the other to j = 1/2.The splitting is weak compared to the main effect, i.e. the energy difference betweenthe initial levels (1s and 2p for hydrogen). The 2p3/2 − 2p1/2 energy differencefor hydrogen is roughly 4.5 × 10−5 eV; the 3p3/2 − 3p1/2 splitting in sodium is∼2 × 10−3 eV.

The physical origin of the spin-orbit coupling can be understood with a classicalargument. Suppose we model the hydrogen atom as an electron orbiting with avelocity v around a proton. The proton is much heavier than the electron and it isassumed to be at rest in the laboratory frame. It creates an electrostatic field actingon the electron:

E = q

4πε0r3r. (13.21)

In the rest frame of the electron, the proton moves at velocity −v and this gives rise,in addition to the electric field (13.21), to a magnetic field:

B = −v × E/c2 = q

4πε0mec2r3L. (13.22)

Here L = mer×v stands for the angular momentum of the electron in the laboratoryframe. In order to derive (13.22), we assume that |v| � c and we consider only thedominant terms in v/c. The spin magnetic moment of the electron μs = −(q/me)Sinteracts with this magnetic field and this gives rise to a magnetic hamiltonian. Thequantum Hamiltonian can be rewritten in the laboratory frame using the naturalatomic units, i.e. the Bohr radius a1 and the ionization energy EI , together with thefine structure constant α2:

2 For all details, see e.g. J.D. Jackson, Classical Electrodynamics, sect. 11.8 (Wiley, 1975).

324 13 Addition of Angular Momenta

Ws.o. = α2 EI

(a1r

)3 L · S�2

. (13.23)

Remarks

1. The spin-orbit coupling is a relativistic effect. We notice on (13.23) that, sincea1/r and L · S/�

2 are of the order of 1, the spin-orbit coupling is of orderα2 (1/137)2 compared to the main effect. This is indeed of order v2/c2, sincev/c ∼ α.

2. For the s states (� = 0), the term (13.23) vanishes. However one can show thatthere exists a relativistic shift of the levels, called theDarwin term,whose value is:

WD = π2e2�2

2m2ec

2|ψ(0)|2.

This term vanishes for � �= 0, since ψ(0) = 0 in that case. It only affects s waves.

All the above terms can be obtained directly and exactly in the framework ofthe relativistic Dirac equation. When one solves this equation for the Coulombpotential, one finds that the states 2s1/2 and 2p1/2 of hydrogen are degenerate,and that the state 2p3/2 lies 10GHz above these two states. Experimentally, oneobserves a splitting between the two states 2s1/2 and 2p1/2, called the Lamb shiftin the name of its discoverer. This splitting, of the order of 1 GHz, is due to thecoupling of the electron with the quantized electromagnetic field. The calculationof the Lamb shift by Bethe in 1947 was the first spectacular success of quantumelectrodynamics.

The spin-orbit splitting is small and we can calculate it by perturbation theory(Chap. 9). For instance, for the level n = 2 of hydrogen, we have to diagonalizethe restriction of WS.O. (13.23) to the subspace generated by the 6 states |n = 2,� = 1,m,σ〉. This coupling WS.O. involves the scalar product L · Swhich is diagonalin the basis |n, �, j,m j 〉 of the eigenstates of the total angular momentum. We havethe equality:

L · S = 1

2

((L + S)2 − L2 − S2

)= 1

2

(J 2 − L2 − S2

),

with eigenvalues ( j ( j + 1) − �(� + 1) − 3/4)�2/2. Using (13.23) one thereforeobtains the splitting of the j = � + 1/2 and j = � − 1/2 states (for instance 2p3/2and 2p1/2):

ΔE(n, �) ≡ E( j = � + 1/2) − E( j = � − 1/2) = (� + 1/2)An,�,

with:

An,� = α2EI

∫|ψn,�,m(r)|2

(a1r

)3d3r.

Onecan check easily that this quantity is independent ofm and that its numerical valuecoincides with the experimental result for the 2p1/2–2p3/2 splitting of hydrogen. For

13.2 One-Electron Atoms, Spectroscopic Notations 325

other atoms, one may observe more complicated effects. For instance, in sodium,there is inversion of the spin-orbit effect: E(3d3/2) > E(3d5/2). This comes from aneffect of the core of internal electrons.

Finally, one can understand the origin of the name fine structure constant forα which governs the order of magnitude of fine structure effects. The name hadbeen introduced in 1920 by Sommerfeld who had calculated the fine structure ofhydrogen in the framework of the old quantum theory, by considering the relativisticeffect due to the eccentricity of the orbits, and not to the (yet to be discovered) spineffects. Sommerfeld’s calculation gave the good result, but this was simply anotherawkward coincidence due to the particular symmetries of the hydrogen problem andto the ensuing degeneracies in �.

13.3 Hyperfine Structure; The 21 cm Line of Hydrogen

An even smaller effect (splitting of the order of 6 × 10−6 eV) has very importantpractical applications. This effect comes from the magnetic interaction between thetwo spin magnetic moments of the electron and the proton:

μe = γe Se γe = −q/me, (13.24)

μp = γp Sp γp 2.79 q/mp. (13.25)

This interaction is called the spin-spin, or hyperfine interaction.Wewill only considerits effect in the ground state of hydrogen n = 1, � = 0.

Interaction Energy

We neglect here effects due to the internal structure of the proton and we treat it asa point-like particle. The calculation of the magnetic field created at a point r bya magnetic dipole μp located at the origin, is a well known problem in magneto-statics.3 The result can be written as:

B(r) = − μ0

4πr3

(μp − 3(μp · r) r

r2

)+ 2μ0

3μp δ(r). (13.26)

The interaction Hamiltonian between the magnetic moment μe of the electronand this magnetic field reads:

W = −μe · B.

For r �= 0, W reduces to the usual dipole-dipole interaction (be it electric ormagnetic):

r �= 0 Wdip = μ0

4πr3

(μe · μp − 3(μe · r)(μp · r)

r2

).

3See e.g. J.D. Jackson, Classical Electrodynamics, sect. 5.6 (Wiley, 1975).

326 13 Addition of Angular Momenta

This interaction will not contribute to our calculation because of the following math-ematical property. For any function g(r), r = |r| regular at r = 0, the angularintegration yields: ∫

g(r) Wdip(r) d3r = 0. (13.27)

The field (13.26) is singular at r = 0, owing to the term proportional to δ(r). Thisleads to a contact interaction:

Wcont = −2μ0

3μe · μp δ(r).

The origin of this singularity at r = 0 lies in the assumption that the proton ispoint-like in our analysis. This entails that all magnetic field lines converge to thesame point. A calculation taking into account the internal structure of the proton,its finite size, and the corresponding modification of the field is leads to a similarresult, because the size of the proton is very small compared with the size of theprobability distribution of the electron in the 1s state. This further calculation shedslight on the internal structure of the proton, and other nuclei such as deuterium ortritium. The point-like model is strictly valid for positronium, which is an atommade of an electron and a positron, both being point-like, and for muonium madeof an electron and a positive muon both being point-like. The theoretical interestof Quantum Electrodynamics calculations of such transition frequencies, and thecomparisonwith high accuracy experimental data can be found in S. G. Karshenboimand V.D. Ivanov.4

Perturbation Theory

The observable W acts on space and on spin variables.We consider the orbital groundstate of the hydrogen atom which, owing to spin variables, is a four state system. Astate of this 4-dimensional subspace can be written as:

|ψ〉 = ψ100(r) |Σ〉, (13.28)

where ψ100(r) is the ground state wave function found in Chap. 11: ψ100(r) =e−r/a1/

√πa31 .

We first take the expectation value over space variables, which results in an oper-ator acting only on spin variables,

H1 =∫

ψ∗100(r) W ψ100(r) d3r.

then we diagonalize this latter operator. The contact term is readily evaluated as:

4S.G. Karshenboim and V.D. Ivanov, Hyperfine structure in hydrogen and helium ion, Phys. Lett.B 524, p. 259–264, 2002.

13.3 Hyperfine Structure; The 21 cm Line of Hydrogen 327

H1 = −2μ0

3μe · μp |ψ100(0)|2. (13.29)

H1 is an operator which acts only on spin states. It can be cast in the form:

H1 = A

�2Se · Sp, (13.30)

where the constant A can be inferred from the values of γe, γp, and ψ100(0):

A = −2

3

μ0

4π

4

a31γeγp�

2 = 16

3× 2.79

me

mpα2EI .

One obtains:

A 5.87 × 10−6 eV ν = A

h 1417 MHz λ = c

ν∼ 21 cm. (13.31)

Diagonalization of H1

The diagonalization of H1 in the Hilbert space of spin states is straightforward.Considering the total spin S = Se + Sp, one has:

Se · Sp = 1

2

(S2 − S2e − S2p

),

which is diagonal in the basis of the eigenstates |S, M〉 of the total spin, with eigen-values:

�2

2(S(S + 1) − 3/2) with S = 0 or S = 1.

The ground state E0 = −EI of the hydrogen atom is therefore split by the hyperfineinteraction in two sublevels corresponding to the triplet |1, M〉 and singlet |0, 0〉states:

E+ = E0 + A/4E− = E0 − 3A/4

triplet state |1, M〉,singlet state |0, 0〉. (13.32)

The difference of these two energies is equal to A, i.e. 5.87×10−6 eV; it correspondsto the characteristic line of hydrogen at a wavelength λ ∼ 21 cm.

Remarks

(i) In its ground state, the hydrogen atom constitutes a four level system with twoenergy levels. By a method whose principle is similar to what we have discussedin Chap.6, it is possible (but technically more complicated) to devise a hydrogen

328 13 Addition of Angular Momenta

maser.5 Among other things, this allows to measure the constant A, or equivalently,the frequency ν = A/h with an impressive accuracy:

ν = 1 4︸︷︷︸A

2 0︸︷︷︸B

4 0︸︷︷︸C

5 7 5 1︸ ︷︷ ︸D

. 7 6 8 4︸ ︷︷ ︸E

±0. 0 0 1 7 Hz.

• In this result, we have underlined several groups of digits. The first two (A) wereobtained by Fermi in 1930; they correspond to the contact term considered above.The two following ones (B) are calculated using the Dirac equation, and the exper-imental value for the anomalous magnetic moment of the electron (deviation ofthe order of 10−3). Other corrections account for the two following decimals (C):relativistic vacuum polarization corrections, finite size of the nucleus, polarizationof the nucleus, etc. The set (D,E) is out of range for theorists at present.

Such an accuracy has, in particular, provided a means to test the predictions ofgeneral relativity.6 A hydrogenmaser was sent in a rocket at an altitude of 10,000km,and the variation of its frequency as the gravitational field and the velocity vary wasmeasured. Despite numerous difficulties, it was possible to check the predictionsof relativity with an accuracy of 7 × 10−5, which is still one of the most accurateverifications of the theory (actually of the equivalence principle).

(ii) The hyperfine splitting of alkali atoms has the same origin as that of hydrogen,although it is more difficult to calculate theoretically. It is measured with impressiveaccuracy. One observes the following frequencies:

7Li 0.83 GHz 2s state23Na 1.77 GHz 3s state39K 0.46 GHz 4s state85Rb 3.04 GHz 5s state87Rb 6.83 GHz ”133Cs 9.19 GHz 6s state

This leads to the achievement of masers and atomic clocks.7 One of the many appli-cations is the definition of the time standard based on the hyperfine effect of theisotope 133 of cesium in its ground state (ΔE ∼ 3.8 × 10−5 eV). One second isdefined as being equal to 9 192 631 770 periods of the corresponding line. The rel-ative accuracy of the practical realization of this definition is 10−15, which we havementioned in Sect. 7.7.3. Such an impressive precision has been made possible withthe use of laser cooled atoms.

However, this field is undergoing an impressive breakthrough with the use ofoptical wavelengths which at present have already achieved a 10−18 level of stability

5H.M. Goldenberg, D. Kleppner, and N.F. Ramsey, Phys. Rev. Lett. 8, 361 (1960).6R. Vessot et al., Phys. Rev. Let. 45, 2081 (1980).7For the recent developments of atomic clocks, see e.g. W. Itano and N. Ramsey, Accurate mea-surement of time, Scientific American, p. 46, July 1993.

13.3 Hyperfine Structure; The 21 cm Line of Hydrogen 329

and precision.8 It is quite probable that a new international time standard will bedefined within a few years from now.

The effect of an external magnetic field

Ifweplace the hydrogen atom in an externalmagnetic field, themagneticHamiltonianbecomes:

HM = A

�2Se · Sp − μe · B0 − μp · B0. (13.33)

Here we do not take into account space variables, assuming that the Zeeman splittingis small enough so that first order perturbation theory is valid.

The nuclear magnetonμN is much smaller than the BohrmagnetonμB . Therefore,we can neglect the last term in the expression (13.33). In this approximation, thediagonalization of HM is simple. We set η = q�B0/(2me) and tan 2θ = 2η/A, andwe obtain the following splitting:

(A/4) + η → |1, 1〉(A/4) − η → |1,−1〉

−(A/4) +√A2/4 + η2 → cos θ |1, 0〉 + sin θ |0, 0〉

−(A/4) −√A2/4 + η2 → − sin θ |1, 0〉 + cos θ |0, 0〉

The levels are represented on Fig. 13.2.

Fig. 13.2 Zeeman splitting of the 21cm line

8B.J. Bloom, T.L. Nicholson, J.R. Williams, S.L. Campbell, M. Bishof, X. Zhang, W. Zhang, S.L.Bromley, J. Ye, An optical lattice clock with accuracy and stability at the 10−18 level, Nature 506,71-75, 2014, doi:1038/nature12041.

330 13 Addition of Angular Momenta

One observes, as for NH3, a competition between the hyperfine coupling and thepresence of the field. For weak fields, the states |1, 0〉 and |0, 0〉 are unaffected,whereas the energies of the states |1, 1〉 and |1,−1〉 vary linearly with B. There isa splitting of the 21 cm line into three components. For strong fields, the eigenstatesare the factorized states |σe ; σp〉. The transition region (η ∼ A) is around B ∼ 0.1 T.

13.4 Radioastronomy

The main stages of the progress of astronomy are primarily due to technologicalprogress in the means of observation: Galileo’s and Newton’s telescopes, photog-raphy, space telescopes, X-ray and γ-ray astronomy, space probes, cosmic neutrinodetectors, and so on. This happened in parallel with progress in fundamental researchthat enables a more and more refined analysis and interpretation of the data.

Fig. 13.3 The Milky Way. Top separate views of the northern and southern hemispheres. Bottomthe pictures have been put together in order to see the angular position of stars with respect to theplane of the galaxy, which is at a large angle from the plane of the solar system (this is what provokesthe apparent asymmetries in the upper figure). One can see on the bottom right the two Magellanicclouds which are only visible in the southern hemisphere. On both pictures, the contrast has beenstrongly amplified (Photo credit AxelMellinger; http://home.arcor-online.de/axel.mellinger/allsky.html.)

13.4 Radioastronomy 331

It seems easy to observe the cosmos. Our own galaxy, the Milky Way which is sobeautiful to look at in the summer, is paradoxically one of the most difficult galaxiesto explore. There are so many objects that one only sees a few of them: 6000 withthe naked eye, 100 million with telescopes, among the 200 billion stars which itcontains. The nearby stars are very bright but they screen everything that lies behindthem.

Since the 1950s, the development of radioastronomy has brought a fascinatingdevelopment of our knowledge of the Universe. Radiowaves are more penetratingthan light waves and they carry a lot of complementary information.

In galaxies, matter exists in two main forms (let aside dark matter which amountsto 20% of the mass of the universe). The first, which is directly visible, is condensedmatter: stars at various stages of their evolution and planets which are now beingdiscovered in other solar systems than ours. However there also exists a diffuseinterstellar medium, composed mainly of atomic hydrogen, whose total mass is quiteimportant (from 10 to 50% of the total visible galactic mass).

The temperature of these interstellar clouds is typically 100 K. Since the corre-sponding thermal energy kT ∼ 10−2 eV is much smaller than EI , hydrogen atomscannot be appreciably excited by thermal collisions from the 1s ground state to theother states of the Lyman series. However the transitions between the two hyperfinestates S = 1 and S = 0 occur easily. The emission of 21 cm radiation correspondsto the spontaneous transition from the S = 1 state to the S = 0 state. This emis-sion is very weak, because the lifetime of the triplet S = 1 state is extremely long:τ ∼ 3.5 × 1014 s ∼ 107 years.9 Nevertheless the amount of atomic hydrogen in theinterstellar medium is so large that an appreciable signal is observed.

The observation of this line of hydrogen has deeply modified our understandingof the interstellar medium. The intensity of the line in a given direction gives themass distribution of the amount of hydrogen. TheDoppler shift allows tomeasure thevelocities of the hydrogen clouds. The splitting of the line and its polarization providea measurement of the magnetic field inside the interstellar medium. By analyzing thestructure of our galaxy, the Milky Way (which is difficult to observe because we arein its plane), it has been possible to show that it is a spiral galaxy, of radius 50,000light-years, and that we are at 30,000 light-years from the center (see Fig. 13.4).One can also measure the density of the interstellar medium (0.3 atoms.cm−3 onthe average), its temperature (20–100K), its structure (roughly one interstellar cloudevery 1000 light-years along a line of sight), and its extension outside the plane ofthe galaxy (roughly 1000 light-years).

The interstellar medium is a diffuse medium in the galaxies where the physicaland chemical life of galaxies takes place. This medium has an overall large mass(10–50% of the total mass of a galaxy). It is very dilute (1–20 atoms per cm3) andcold (50–100 K).

9This very long lifetime is due to the combination of two facts; the energy difference is very small,and the emission proceeds through a magnetic dipole transition (while the atomic resonance linescorrespond to electric dipole transitions).

332 13 Addition of Angular Momenta

Fig. 13.4 Left Spiral structure of the Milky Way as deduced from radioastronomical observationsat a 21cm wavelength. (Courtesy Frédéric Zantonio.) Right Barred spiral structure of the MilkyWay

This medium consists primarily of atoms: 90% atomic hydrogen and 10% atomichelium (i.e., respectively, 75 and 25% in mass) that were formed just after the bigbang, during the first three minutes of the Universe. There are other atoms, suchas carbon, calcium, potassium, and others, but in smaller amounts. They have beenformed in stellar nucleosynthesis and have been ejected in the interstellar medium,for instance in supernovae explosions. The rest of the interstellar medium (from 1 to50%) consists of molecules such as H2O, CO2, C2H5OH which we have mentionedin Chap. 10, and interstellar dust.

The interstellar medium causes spectacular phenomena in astronomical observa-tions. Figure13.5 shows veils ofmatter in the Cygnus nebula and between the Pleiadestars, which are young stars of our galaxy.

Fig. 13.5 Left veil of interstellar matter in the Cygnus (or Veil) nebula (Photo credit: HSTWFPC2 http://casa.colorado.edu/~maloney/CygnusLoop.gif.) Right veil of matter between twoof the Pleiade stars. The parallel wisps extending from lower left to upper right, discovered by G.Herbig and T. Simon in 1999, are due to radiation pressure of the strong starlight shining fromMerope—outside frame—on the dust particles (Photo credit: NASA and Hubble Heritage Teamhttp://heritage.stsci.edu/2000/36/big.html.)

13.4 Radioastronomy 333

One can understand from these examples that this type of observation is quitelimited. In order to see such objects or mass distributions, they must either emit lightor be illuminated by light sources, and furthermore they must not be hidden by otherobjects (after all, we see a horsehead, but there may also be woodlice or lobsters).

The interstellar medium, which is cold, does not emit in the visible part of thespectrum, but it abundantly emits radiofrequency waves. The radiowaves emitted bytheMilkyWay, as well as by other galaxies, penetrate matter much more deeply thanlight waves, and they bring new information compared to what we can see directly.

However, this radiation comes from sources that are of extremely low densitiesand the interstellar medium has a small brilliance compared to stars. In order toobserve such emissions, one needs high-performance selective noiseless amplifiers.This is where a major technical contribution of quantum mechanics to astrophysicswas made. Masers, which we described in Chap.7, have precisely these features.The introduction of masers in astrophysics is due to Townes himself, who turnedto astrophysics in the mid 1960s and caused a true revolution. In fact, with masersone can observe the radiation coming from weak sources, that would be drownedin the background noise if one were to use traditional amplifiers. The masers usedin astrophysics are solid-state masers, in particular, ruby masers that can deliverimportant amplification factors.

The observation of the interstellar medium requires powerful radiotelescopes, andinterferometric setups in order to reach an acceptable resolution (radio wavelengthsare 100,000 times larger than optical ones). At present, a maximum resolution isobtained with the technique of very-large-baseline interferometry (VLBI). The datarecorded with radiotelescopes located on different continents can be put togetherand synchronized with atomic clocks. The baseline is of the order of 10,000km.Astrophysicists are contemplating putting space-radiotelescopes on large orbits inorder to improve the resolution.

Radioastronomical observations represent a formidable task. The total energyreceived up to now in radioastronomy is of the order of the kinetic energy of theashes of a cigar falling from 1m high.

One aims at various directions in the sky. By a sophisticated analysis of theobserved signal, one can reconstruct the position, the density, the velocity (byDopplereffect), and the composition of interstellar clouds.

The pioneers of centimetric radioastronomy, which is fundamental in order toobserve hydrogen, were Purcell and Van de Hulst in 1947.

13.5 The 21-cm Line of Hydrogen

Coming “back home”, that is, to our galaxy and its present neighborhood, Onefrequency dominates the radiowave emission and plays a key role: 1420 MHz. It ison that frequency and only on that one, that hydrogen, the most abundant element inthe Universe, emits radiation.

334 13 Addition of Angular Momenta

This effect is one of the best-known phenomena of atomic physics. As we haveseen,when the spins of the proton and the electron flip fromone relative configurationto the other, there is emission or absorption on 1420 MHz, or at a wavelength of 21-cm. (This is in a domain close to television frequencies, and it is protected by aninternational agreement.)

Importance of the 21-cm Line

The importance of the hydrogen 21-cm line in astrophysics comes from the followingfacts.

• Hydrogen is very abundant.• It absorbs any emission in the visible part of the spectrum (the sky is opaque forhydrogen lines in the visible and ultraviolet regions).

• On the other hand, cold interstellar atomic hydrogen emits this radiation abun-dantly. Furthermore, the interstellar medium and the entire galaxy are transparentfor this radiowave. The temperature of interstellar clouds is typically 50–100K.The corresponding thermal energy kT ∼ 10−2 eV is much smaller than EI , andthe atoms cannot be excited appreciably from the 1s state to other levels of theLyman series. But thermal transitions between the two hyperfine levels S = 1and S = 0 are quite easy. The 21-cm emission corresponds to the spontaneoustransition from the S = 1 state to the S = 0 state. This emission is very weakbecause the lifetime of the triplet state S = 1 is very long: τ ∼ 3, 5×1014 s ∼ 107

years!10 Nevertheless, there are enormous amounts of hydrogen in the interstellarmedium and the signals received on earth are appreciable.

• It is the only emission of hydrogen that can be collected, except for a few nearbystars. Its detection was predicted by Van de Hulst in 1947; the first observation isdue to Purcell in 1951.

Its observation gives information on galactic and extragalactic structure and dynam-ics, and on the formation of galaxies.

The intensity distribution of the emission of the 21-cm line is shown in Fig. 13.6.We notice immediately that the emission extends in directions far beyond those ofvisible light (Fig. 13.3). The hydrogen clouds extend far beyond stars (i.e., condensedmatter) in the sky. Their study is bound to bring new and abundant information.

The observation of this atomic hydrogen line (and that due to carbon monoxide,less abundant but very luminous) profoundly changed our understanding of the inter-stellar medium.

10This long lifetime is due to two effects: the energy difference between the two levels is small; andit is a magnetic dipole transition much weaker than the electric dipole transitions. The advantage isthat, symmetrically, these radiowaves are absorbed weakly.

13.5 The 21-cm Line of Hydrogen 335

Fig. 13.6 Emission density of the 21-cm line in the sky. The emission is naturally brighter in theplane of the Milky Way where matter is closer and more dense on the average (Photo credit:J. Dickey (UMn), F. Lockman (NRAO), SkyView, http://antwrp.gsfc.nasa.gov/apod/ap010113.html.)

The Milky Way

There are countless results. By probing our own galaxy, the Milky Way, (which isdifficult because the sun lies in its plane), one has been able to reconstruct it structureand to show that it has a spiral structure, as does its sister galaxy, Andromeda, M31.It has been possible to show that it has a radius of 50,000 light-years, and that the sunlies at some 35,000 light-years from the center (see Fig. 13.4). Today, astronomerstend to think it has the structure of a barred spiral galaxy (see the right-hand side ofFig. 13.4). The measured average density of the interstellar medium is 0.3 atoms percm3, the temperature varies from 20 to 100K, and there is on average one interstellarcloud every 1000 light-years along a line of sight. The thickness perpendicular to thegalactic plane is roughly 1000 light-years.

A strange discovery occurred in 2003. The Milky Way might be performing anact of cannibalism on its satellite galaxy, Canis Major. The detailed analysis ofexternal hydrogen clouds, and infrared matter, reveals a closed winding filament-shaped structure that passes through a dwarf satellite galaxy of theMilkyWay, CanisMajor, as one can see in Fig. 13.7. Canis Major is at a distance of 25,000 light-yearsfrom the sun, it is the satellite galaxy closest to the center of the Milky Way.

Therefore the Milky Way would be surrounded by a tidal stream of stars andmatter. Such a feature is observed in other galaxies.

336 13 Addition of Angular Momenta

Fig. 13.7 Left Global structure of atomic hydrogen and stars in and around the Milky Way. RightResidual filament going through Canis Major, the brighter region on the left, once the spiral armsnearest to the center of theMilkyWay have been removed (Photo credit: R. Ibata, Strasbourg Obser-vatory, Two-Micron All Sky Survey or “2MASS” experiment, http://astro.u-strasbg.fr/images_ri/canm-e.html.)

13.6 The Intergalactic Medium; Star Wars

A number of results have been obtained by observing outside our galaxy.First, we can observe our sister galaxy, Andromeda, at two million light-years

from us, twice as massive as the Milky Way. The pictures in the visible part ofthe spectrum are shown in Fig. 13.8. One can see the spiral structure. In radiowave

Fig. 13.8 Andromeda nebula, M31. Left In visible light (Photo credit: NASA and RobertGendler, http://antwrp.gsfc.nasa.gov/apod/ap021021.html.) Right In 21cm radioastronomy (Photocredit: NRAO, Group of Astronomers at the Robert C. Byrd Green Bank Telescope (GBT) ofthe National Science Foundation, http://www.universetoday.com/am/publish/clouds_hydrogen_swarm_andromeda. html 422004.)

13.6 The Intergalactic Medium; Star Wars 337

emission, one can see extended and powerful sources that do not correspond to anyvisible matter. Astrophysicists tend to think that these clouds are materials whichhave not been used up to now in the construction of galaxies, but that will play (orare playing) a major role in galactic dynamics.

This reveals an interesting feature. In fact, atomic hydrogen extends much beyondstars in a galaxy. And that feature allows us to learn about the relationship of galaxieswith respect to each other. This could not be guessed with usual telescopes on humantime scales.

In the sameway as stars live in an interstellar medium, there exists an intergalacticmedium.Galaxies possess links. Contrary to the interstellarmedium, the intergalacticmedium contains practically only primordial hydrogen and helium from the big bang(there is no intergalactic nucleosynthesis).

The presence of intergalactic hydrogen clouds is obvious in the Cartwheel galaxy,which is 200Mpc from us (1pc = 3.26 light-years), Fig. 13.9, where the contour linesof hydrogen clouds are superimposed on the optical photograph. This galaxy surviveda head-on collision with a smaller galaxy 300 million years ago. This provoked theannular structure of the Cartwheel, with a rim of the size of the Milky Way and anucleus. Numerical simulations confirm this idea. The contour lines of the importanthydrogen cloud surrounding all these galaxies aremarked. It leaves some doubt aboutwhich galaxy is responsible for this hit-and-run offense.

Fig. 13.9 The Cartwheel galaxy, A 0035. On the left, the ring shape, with a wheel of the size ofthe Milky Way and a nucleus. A hydrogen cloud joins this structure to the right hand side galaxy,and also to the middle galaxy, which is not a spiral galaxy. It is thought that this latter object, whichis now in the axis of the Cartwheel at 100 kpc, had a head-on collision with it 300 million yearsago, which gave rise to the structure. The extension of the hydrogen cloud leaves some doubt on theobject responsible for the collision which could be the upper-right galaxy (Photo credit: J. Higdon(NRAO), C. Struck, P. Appleton (ISU), K. Borne (Hughes STX), and R. Lucas (Stsci), NASA;http://antwrp.gsfc.nasa.gov/apod/ap970224.html.)

338 13 Addition of Angular Momenta

Fig. 13.10 Group of galaxies around M81. In the center, one can see M81, which is a large spiralgalaxy, on the right, M82 which was thought to be very irregular, and on the left NGC 3077 (Photocredit Robert Gendler, http://antwrp.gsfc.nasa.gov/apod/ap000209.html.)

A famous example, which is very rich and was one of the first to be analyzed isthe group of three galaxies M81, M82, NGC 3077 shown in Fig. 13.10. This group islocated in the constellation Ursa Major at 2.5 Mpc. These three galaxies (and othersmaller ones) seem to orbit around each other quietly. M82 has an anomalous shapewith a central prominence, perpendicular to its plane.

The 21-cm radioastronomical observation transforms these impressions, as onecan see in Fig. 13.11. One sees that the group bathes in hydrogen clouds. We noticethat the spiral structure ofM81 appears clearly, and that the emission at 21-cm, whichis abundant in cold regions, is weak in the hot regions of galactic nuclei.

These galaxies appear as floating islands in a common ocean of intergalactichydrogen gas of a considerable extension. One observes the umbilical cords ofhydrogen, in particular betweenM81 and NGC 3077, which show that these galaxieshave a common history. One can see that the large galaxy M81 is in the process ofperforming cannibalism on a smaller galaxy, on its left, and of absorbing it. Thisphenomenon is very hard to see in visible light. Such observations are confirmed bynumerical simulations. (We have mentioned interstellar organic molecules on thisexample in 10.)

Finally, the Doppler picture of the 21-cm emission of M81 (Fig. 13.11, Left.)exhibits an unexpected feature. The plane of this galaxy rotates globally around anaxis at roughly—30 ◦. The upper part of the galaxy is redshifted and the lower partis blueshifted. At this level of accuracy, no individual motion appears in the spiralarms themselves, whereas the overall rotation of the plane of the galaxy is manifest.

13.6 The Intergalactic Medium; Star Wars 339

Fig. 13.11 Left The group of galaxies M81, M82, NGC3077 as seen on the 21-cm wavelength.The extension of the hydrogen cloud inside which the group is evolving is very large. One can seethat M81 is eating up, on the left, a smaller galaxy, which is more difficult to see in visible light. Wenotice the absence of radio emissions in the vicinity of the hotter cores of galaxies. The orientationof this figure is not the same as the optical picture of Fig. 13.10; it is rotated by roughly 45◦. RightDoppler shift of the top and bottom of M81. One can see that it corresponds to a global motion ofthe entire system around an axis and shows no motion of spiral arms (Photo credit: Greydon Moorehttp://www.cosmicastronomy.com/bodes4.htm.)

Spiral Arms, Birthplaces of Stars

In other words, stars and gas clouds turn around the galactic nucleus, but the spiralarms themselves do not move! The spiral structure is a stationary situation of thegravitational field (the calculation of the field is a complicated nonlinear problem).Stars and interstellar matter turn around the center, and they slow down when theycross the spiral arms. This is why the spiral arms can be seen both in the visible partof the spectrum and in radiowaves. The spiral arms are brighter precisely because ofthe jamming phenomenon that happens in them, and matter is more abundant there.

Spiral galaxies, an example of which is shown in Fig. 13.12, resemble huge rotat-ing fireworks. But, as we have just seen, the spiral arms do not move. They do notturn around!

Although it is not known at present how spiral arms are formed, one can show thatif they exist, they are stable. It is a very difficult mathematical problem to explainhow they arise.

Nevertheless, this observation shows a fundamental type ofmechanism of star for-mation. During their slowing down, stars and the interstellar medium are compressed

340 13 Addition of Angular Momenta

Fig. 13.12 Left Spiral galaxy NGC 1232 (Photo credit: FORS1, 8.2m VLT Antu, ESO, http://www.star.ucl.ac.uk/~apod/apod/ap040125.html.) Right Schematic star formation mechanism bycompression of molecular clouds in spiral arms

Fig. 13.13 The “Pillars of Creation” in the Eagle nebula, in the constellation Serpens. The twopictures of this star formation region are taken with different filters (and the prints do not have thesame orientation). On the left, one can see the general shape of the molecular cloud in the nebula.The region of interest is a bit below the center (Photo credit: ESA, ISO, ISOGAL Team, http://antwrp.gsfc.nasa.gov/apod/ap010914.html.) On the right, one can see a detailed picture of thatregion, where young bright stars are formed and ejected in the medium (Photo credit: Jean-CharlesCuillandre (CFHT), Hawaiian Starlight, CFHT http://antwrp.gsfc.nasa.gov/apod/ap030213.html.)

as sketched in Fig. 13.12. The compression results in an increase in temperature andin density. Locally, this leads to a gravitational collapse followed by an ignition ofthermonuclear reactions and the birth of stars in the spiral arms. This is how manystars form, in particular, massive stars.

Regions of star formation are among the most spectacular objects that one can seein the sky. In Fig. 13.13 one can see one of these “star cradles,” called the “Pillarsof Creation” in the Eagle nebula, in the constellation Serpens. The compressedmolecular cloud literally spits out bursts of young stars in the interstellar medium.

13.6 The Intergalactic Medium; Star Wars 341

Fig. 13.14 Zeemansplitting of the ground stateof deuterium.

13.7 Exercises

1. Permutation operator

Show that the permutation operator defined in (13.14) can be written as Ps12 =

(1 + σ1 · σ2)/2.

2. The singlet state

Consider two spins 1/2, and the eigenbasis {|±〉u ⊗|±〉u} of the two operators S1 · uand S2 · u, u being any unit vector of R3. Show that the singlet state is written in thatbasis as:

1√2

(|+〉u ⊗ |−〉u − |−〉u ⊗ |+〉u) .

3. Spin and magnetic moment of the deuteron

We note J the total angular momentum observable of the electron cloud of anatom, and I the angular momentum of the nucleus. The respective magnetic momentobservables are μJ = gJμB J/� and μI = gIμN I/� where gJ and gI are dimen-sionless factors. The magnetic interaction Hamiltonian of the electron cloud with thenucleus is of the form W = aμJ · μI where a is a constant which depends on theelectron distribution around the nucleus.

a. Suppose that the state of the nucleus (energy EI , square of the angular momentumI (I +1)�2) and the state of the electronic cloud (energy EJ , square of the angularmomentum J (J +1)�2) are both fixed. What are the possible values K (K +1)�2

of the total angular momentum K of the atom?

b. Express W in terms of I2, J

2and K

2. Express the hyperfine energy levels of the

atom in terms of I, J and K .c. Calculate the splitting between two consecutive hyperfine levels.

342 13 Addition of Angular Momenta

d. When one applies a uniform weak magnetic field B on a deuterium atom, oneobserves that the two hyperfine levels (EK and EK ′) of the ground state are splitas a function of B as shown on the Fig. 13.14. Knowing that the single electronof the atom is in its orbital ground state � = 0, what is the value of the deuteronspin?

e. Assuming that the proton and neutron inside the deuteron have a zero orbitalangular momentum, what is their spin state?

f. We have a = −8μ0/12πa31 where a1 is the Bohr radius and ε0μ0c2 = 1. Giventhat ge = 2 and gI = 0.86, to what frequency must a radiotelescope be tuned inorder to detect deuterium in the interstellar medium?

Chapter 14Identical Particles, the Pauli Principle

The origin of geometry, from Pythagoras to Euclid, lies in our environment, and inthe observation that one can model the world in which we live by a space where eachobject is described by a point or a set of points. The concept of space itself cameafter the simpler concept of “place” of an object. The idea of space arose with thequestion as to whether the place exists independently of the fact that some objectoccupies it. In this context, by definition, two objects cannot have the same positionat the same time.

In this chapter, we address the quantum transposition of this problem. In the prob-abilistic quantum description, there is no reason a priori why the density probabilityfor two particles to be at the same point in space should vanish, contrary to theclassical observation. It is therefore legitimate to elevate the above question to statevectors (or wave functions) rather than positions. Can two particles be in the samestate at the same time?

Naturally, two particles of different kinds, such as an electron or a proton, willnever be in the same state: even if their wave functions coincide, their mass differenceimplies differences in the values of various physical quantities and one can always tellthem from each other. However, there exist in Nature identical particles: all electronsin the universe have the same mass, the same charge, and so on. Can such particles,whose intrinsic properties are all the same, be in the same state? The answer lies inone of the simplest, but most profound, principles of physics, whose consequenceson the structure of matter are numerous and fundamental: the Pauli principle.

The depth, the intellectual upheaval, and the philosophical implications of thePauli principle are considerable. If this principle did not provoke the same interestas relativity among philosophers, and even among physicists, it is probably becauseit explained so many experimental facts (many more than relativity) that Fermi andDirac had incorporated it immediately in the general theory of quantum mechanics.

The result is the following. The state of a state of N identical particles is eithertotally symmetric or totally antisymmetric if one exchanges any twoof these particles.It is symmetric if the spin of these particles is integer; it is antisymmetric if the spin is

© Springer International Publishing Switzerland 2016J.-L. Basdevant, Lectures on Quantum Mechanics,Graduate Texts in Physics, DOI 10.1007/978-3-319-43479-7_14

343

344 14 Identical Particles, the Pauli Principle

half-integer. This relation between the symmetry of states and the spin of particles isan experimental fact. This property can be proven theoretically but the proof cannotbe explained in a simple way.

We show inSect. 14.1.1 that there is a genuine physical problemwith the principlesof Chap.6. Some predictions are ambiguous, so that these principles do not sufficewhen one deals with systems containing identical particles. A new fundamentalprinciple must be added in order to get rid of this ambiguity.

The essential point is that, bydefinition, two identical particles canbe interchangedin a physical system without modifying any property of this system. The mathemati-cal tool that corresponds to the interchange of two particles is the exchange operatorwhich we introduce in Sect. 14.2. In Sect. 14.3 we express the Pauli principle as anadditional axiom. Finally, in Sect. 14.4, we discuss some consequences of the Pauliprinciple.

14.1 Indistinguishability of Two Identical Particles

14.1.1 Identical Particles in Classical Physics

By definition, two particles are identical if all their intrinsic properties are the same.In classical mechanics, for a two-particle system, it is always possible to measureat a given time the position of each particle. At that instant, we can define whichparticle we call 1 and which one we call 2. It is also possible to follow the trajectoryof particle 1 and that of particle 2. We can keep on distinguishing unambiguouslyeach particle at any later time. For instance, in the collision of two billiard balls ofthe same color, we can unambiguously tell the difference between the two processesof Fig. 14.1. Therefore, for any system that is described by classical physics, twoparticles are always distinguishable, whether or not they are identical (the notion ofidentity of classical macroscopic objects is anyway an idealization).

Fig. 14.1 Collision between two identical particles

14.1 Indistinguishability of Two Identical Particles 345

14.1.2 The Quantum Problem

The situation is different in quantummechanics. At a given time we can still measurethe positions of the particles and label them with the indices 1 and 2. However,because the notion of a trajectory is not defined, it may be impossible to follow thetwo particles individually as time goes on. For instance, one cannot tell the differencebetween the two processes sketched in Fig. 14.1 if the twowave functions of particles1 and 2 overlap. It is impossible to know whether particle 1 has become particle 1′or particle 2′. In quantum mechanics two identical particles are indistinguishable.

Here, physics falsifies the famous “principle of the identity of indistinguishables,” which isa basic principle in Leibniz’s philosophy, where two real objects are never exactly similar.We show that there exist cases where N identical particles can be in the same state (Bose–Einstein condensate) although they are not a single entity. The number N of these particlesis a measurable quantity, although they are indistinguishable from each other.

14.1.3 Example of Ambiguities

In the framework of the principles of Chap. 6, this indistinguishability leads to ambi-guities in the predictions of physical measurements. Consider, for instance, twoidentical particles moving in a one-dimensional harmonic potential. We label theparticles 1 and 2 and we assume that the Hamiltonian is:

H = p212m

+ 1

2mω2 x21 + p22

2m+ 1

2mω2 x22 = h(1) + h(2).

For simplicity we suppose that the particles have no mutual interaction. Let (n +1/2)�ω and φn(x) (n = 0, 1, . . .) be the eigenvalues and eigenfunctions of theone-particle Hamiltonian h = p2/2m + mω2 x2/2.

There is no problem in describing the physical situation where both particles arein the ground state of h. The corresponding state is:

Φ0(x1, x2) = φ0(x1)φ0(x2),

and its energy is E0 = �ω.On the contrary, the description of the first excited state of the system is ambigu-

ous. This corresponds to one of the particles being in the first excited state of h and theother in the ground state. The total energy is 2�ω. One possible state isφ1(x1)φ0(x2);another possible state is φ0(x1)φ1(x2). Because these two states are possible candi-dates, then, according to the superposition principle, any linear combination:

Φ(x1, x2) = λφ1(x1)φ0(x2) + μφ0(x1)φ1(x2)

also corresponds to an energy 2�ω.

346 14 Identical Particles, the Pauli Principle

Therefore there are several different states that appear to describe the same phys-ical situation. This might not be a problem, provided no measurement could makethe difference. Alas this is not true! These various states lead to different predictionsconcerning physically measurable quantities. Consider, for instance the product ofthe two positions, the observable x1 x2, for which the labeling of the two particles isirrelevant. Its expectation value is:

〈x1 x2〉 = �

mωRe(λ∗μ).

This prediction depends on λ and μ. However, nothing in the theory that we havepresented tells us the values of these parameters. Therefore there is a basic ambiguityin the predictions of our principles, and we need a prescription in order to fix thevalues of λ and μ.

It is a remarkable fact of Nature that the only allowed values are λ = ±μ, andthat the sign only depends on the nature of the particles under consideration. Theallowed states for a system of identical particles are therefore restrictions of the mostgeneral states that one could imagine if the particles were distinguishable.

14.2 Two-Particle System; The Exchange Operator

14.2.1 The Hilbert Space for the Two-Particle System

Within the framework that we have used up to now, we describe a two-particle system(distinguishable or not) by labeling these particles. A state of the system is therefore

|ψ〉 =∑

k,n

Ck,n |1 : k ; 2 : n〉. (14.1)

Implicitly the vectors {|1 : k〉} form a basis of one-particle states. The states {|1 :k ; 2 : n〉} are factorized states where the first particle has the quantum numbers kand the second n. They form a basis of the two-particle Hilbert space.

14.2.2 The Exchange Operator Between Two IdenticalParticles

The labeling of the particles that we used above has no absolute meaning if they areidentical. Consequently, the predictions of experimental results must be independentof this labeling. In order to describe this property due to the exchange symmetry, weintroduce the exchange operator P12 such that for any couple (k, n):

14.2 Two-Particle System; The Exchange Operator 347

P12 |1 : k ; 2 : n〉 = |1 : n ; 2 : k〉. (14.2)

One can verify that this operator is Hermitian and that it satisfies:

P212 = I . (14.3)

Example

a. For two spinless particles, we have

P12 ≡ P (external)12 , that is, P12 Ψ (r1, r2) = Ψ (r2, r1).

b. For two spin 1/2 particles, then P12 exchanges both the orbital and the spin vari-ables of the two particles:

P12 = P (external)12 P (spin)

12 .

c. Permutation of two spin 1/2 particles. In this case, one canwrite down explicitlythe action of P12 by using the representation (13.11),

P12∑

σ1,σ2

Ψσ1,σ2(r1, r2) |1 : σ1; 2 : σ2〉 =∑

σ1,σ2

Ψσ1,σ2(r2, r1) |1 : σ2; 2 : σ1〉

where σi = ± with i = 1, 2.In order to discuss the properties of this permutation, it is convenient to work withthe eigenbasis of the square of the total spin S = S1 + S2 and Sz

|S = 1, m = 1〉 = |1 : + ; 2 : +〉,|S = 1, m = 0〉 = 1√

2(|1 : + ; 2 : −〉 + |1 : − ; 2 : +〉) ,

|S = 1, m = −1〉 = |1 : − ; 2 : −〉,|S = 0, m = 0〉 = 1√

2(|1 : + ; 2 : −〉 − |1 : − ; 2 : +〉) .

We already noticed that:

• The triplet states (S = 1) are symmetric under the interchange of σ1 and σ2:

P (spin)12 |S = 1,m〉 = |S = 1,m〉,

• The singlet state (S = 0) is antisymmetric in this interchange:

P (spin)12 |S = 0,m = 0〉 = − |S = 0,m = 0〉.

348 14 Identical Particles, the Pauli Principle

14.2.3 Symmetry of the States

How can one fulfill the requirement that the experimental results must be unchangedas one goes from |Ψ 〉 to P12|Ψ 〉? These two vectors must represent the same physicalstate, therefore they can only differ by a phase factor, that is, P12|Ψ 〉 = eiδ |Ψ 〉.Because P2

12 = I , we have e2iδ = 1 and eiδ = ±1. Therefore:

P12|Ψ 〉 = ±|Ψ 〉. (14.4)

We reach the following conclusion.

The only physically acceptable state vectors for a system of two identical particlesare either symmetric or antisymmetric under the permutation of the two particles.

Referring to (14.1), this implies Ck,n = ±Cn,k . The only allowed states are eithersymmetric under the exchange of 1 and 2:

|ΨS〉 ∝∑

k,n

Ck,n (|1 : k ; 2 : n〉 + |1 : n ; 2 : k〉) ; P12|ΨS〉 = |ΨS〉 (14.5)

or antisymmetric:

|ΨA〉 ∝∑

k,n

Ck,n (|1 : k ; 2 : n〉 − |1 : n ; 2 : k〉) ; P12|ΨA〉 = −|ΨA〉, (14.6)

where the coefficients Ck,n are arbitrary.This restriction to symmetric or antisymmetric state vectors is a considerable

step forward in order to solve the ambiguity pointed out in the previous section.For instance, the expectation value 〈x1x2〉 considered in Sect. 14.1.2 can now takeonly two values ±�/(2mω), corresponding to the two choices λ = ±μ = 1/

√2.

However, it is not yet sufficient and some essential questions are still open.

a. Can a given species, such as electrons, behave in some experimental situationswith the plus sign in (14.4) and in other situations with the minus sign?

b. Assuming that the answer to the first question is negative, what decides whichsign should be attributed to a given species?

The answers to these two questions lead us to one of the simplest and most fun-damental laws of physics: the Pauli principle. (The general formulation was derivedfrom Pauli’s ideas by Fermi and Dirac.)

14.3 The Pauli Principle 349

14.3 The Pauli Principle

14.3.1 The Case of Two Particles

All particles in Nature belong to one of the two following categories:

• Bosons for which the state vector of two identical particles is always sym-metric under the operation P12.

• Fermions for which the state vector of two identical particles is always anti-symmetric under the operation P12.

All particles of integer spin (including 0) are bosons (photon, π meson, αparticle, gluons, etc.).All particles of half-integer spin are fermions (electron, proton, neutron, neu-trino, quarks, He3 nucleus, etc.).

The state vectors of two bosons are of the form |ΨS〉 (Eq. (14.5)), those of fermionsare of the form |ΨA〉 (Eq. (14.6)). The Pauli principle therefore consists of restrictingthe set of accessible states for systems of two identical particles. The space of physicalstates is no longer the (tensor) product of the basis states, but only the subspace formedby the symmetric or antisymmetric combinations.

The Pauli principle also applies to composite particles such as nuclei or atoms,provided experimental conditions are such that they are in the same internal state(be it the ground state or a metastable excited state). For instance, hydrogen atomsin their electronic ground state have a total spin S = 0 or S = 1 and they behaveas bosons. Deuterium, the isotope of hydrogen is a fermion (nuclear spin 1, electronspin 1/2, total spin half-integer).

As we announced, this connection between the symmetry of states and the spinof the particles is an experimental fact. However, it is both a triumph and a mysteryof contemporary physics that this property, called the “spin–statistics connection,”can be proven starting from general axioms of relativistic quantum field theory.1 Itis a mystery because it is probably the only example of a very simple physical lawfor which the proof exists but cannot be explained in an elementary way.

Example

a. The wave function of two identical spin zero particles must be symmetric:Ψ (r1, r2) = Ψ (r2, r1).

b. The state of two spin 1/2 particles must be of the form:

1Markus Fierz (1939). “Über die relativistische Theorie Kräftefreier Teilchenmit beliebigem Spin”.Helvetica Physica Acta 12 (1): 337. doi:10.5169/seals-110930. Wolfgang Pauli (15 October 1940).“The Connection Between Spin and Statistics”. Physical Review 58 (8): 716722. doi:10.1103/PhysRev.58.716. See also: I. Duck and E.C.G. Sudarshan, Pauli and the Spin-Statistics Theorem,World Scientific, Singapore, (1997).

350 14 Identical Particles, the Pauli Principle

|Ψ 〉 = Ψ0,0(r1, r2) |S = 0,m = 0〉 +∑

m

Ψ1,m(r1, r2) |S = 1,m〉,

where Ψ0,0 and Ψ1,m are, respectively, symmetric and antisymmetric:

Ψ0,0(r1, r2) = Ψ0,0(r2, r1) , Ψ1,m(r1, r2) = −Ψ1,m(r2, r1).

Therefore the orbital state and the spin state of two identical fermions are corre-lated.

14.3.2 Independent Fermions and Exclusion Principle

Consider a situationwhere two fermions, for instance, two electrons, are independent,that is, they do not interact with each other. The (effective) total Hamiltonian thenreads H = h(1) + h(2). In such conditions, the eigenstates of H are products ofeigenstates |n〉 of h: |1 : n ; 2 : n′〉. We remark that, if n = n′, that is, if thetwo particles are in the same quantum state, the state |1 : n ; 2 : n〉 is necessarilysymmetric. This is forbidden by the Pauli principle, which results in the (weaker)formulation:

Two independent fermions in the same system cannot be in the same state.

If |n〉 and |n′〉 are orthogonal, the only acceptable state is the antisymmetriccombination:

|ΨA〉 = 1√2

(|1 : n ; 2 : n′〉 − |1 : n′ ; 2 : n〉) .

In this simplified form, the Pauli principle appears as an exclusion principle.

14.3.3 The Case of N Identical Particles

For a system of N identical particles, we proceed in a similar manner. We introducethe exchange operator Pi j of the two particles i and j . The indistinguishabilityimposes that Pi j |Ψ 〉 leads to the same physical results as |Ψ 〉. The general form ofthe Pauli principle is as follows.

The state vector of a system of N identical bosons is completely symmetricunder the interchange of any two of these particles.

The state vector of a systemof N identical fermions is completely antisymmetricunder the interchange of any two of these particles.

14.3 The Pauli Principle 351

For instance, for N = 3, one has:

Ψ±(u1, u2, u3) ∝ ( f (u1, u2, u3) + f (u2, u3, u1) + f (u3, u1, u2))

± ( f (u1, u3, u2) + f (u2, u1, u3) + f (u3, u2, u1)) ,

where f is any function of the three sets of variables u1, u2, u3. The plus sign(resp., minus sign) corresponds to a function Ψ that is completely symmetric (resp.,completely antisymmetric).

More generally, let us consider an orthonormal basis {|n〉} of the one-particlestates, and the N ! permutations P of a set of N elements. We want to describe thefollowing physical situation, “One particle in the state |n1〉, one particle in the state|n2〉, . . ., one particle in the state |nN 〉.” In order to do this, we number in an arbitraryway the N particles from 1 to N .

The Case of Bosons

The state is:

|Ψ 〉 = C√N !

∑

P

|1 : nP(1) ; 2 : nP(2) ; . . . ; N : nP(N )〉, (14.7)

where∑

P denotes the sum over all permutations. Notice that two (or more) indicesni , n j , . . . labeling the occupied states may coincide. The normalization factor C isexpressed in terms of the occupation numbers Ni of the states |ni 〉:

C = (N1! N2! . . .)−1/2 .

The Case of Fermions

In the case of fermions, the result is physically acceptable if and only if the N states|ni 〉 are pairwise orthogonal. The state |Ψ 〉 is then:

|Ψ 〉 = 1√N !

∑

P

εP |1 : nP(1) ; 2 : nP(2) ; . . . ; N : nP(N )〉, (14.8)

where εP is the signature of the permutation P: εP = 1 if P is an even permutationand εP = −1 if P is odd.This state vector is oftenwritten in the formof a determinant,called the Slater determinant:

|Ψ 〉 = 1√N !

∣∣∣∣∣∣∣∣∣

|1 : n1〉 |1 : n2〉 . . . |1 : nN 〉|2 : n1〉 |2 : n2〉 . . . |2 : nN 〉

......

...

|N : n1〉 |N : n2〉 . . . |N : nN 〉

∣∣∣∣∣∣∣∣∣

. (14.9)

352 14 Identical Particles, the Pauli Principle

If two particles are in the same state, two columns are identical and this determinantvanishes.

The set of states that can be constructed using (14.7) (resp., (14.8)) forms a basisof the Hilbert space of an N boson (resp., N fermion) system.

14.4 Physical Consequences of the Pauli Principle

We give below a few of the many physical consequences of the Pauli principle thatconcern both few-body systems and the macroscopic properties of a large numberof bosons or fermions.

14.4.1 Exchange Force Between Two Fermions

Consider the helium atom and neglect magnetic effects, as we did for hydrogen inChap.11. We label the electrons 1 and 2, and the Hamiltonian is:

H = p212me

+ p222me

− 2e2

r1− 2e2

r2+ e2

r12, with r12 = r1 − r2.

The eigenvalue problem is technically complicated and can only be solved numer-ically, but the results of interest here are simple (Fig. 14.2). The ground state(E0 = −78.9 eV) corresponds to a symmetric spatial wave function, whereas thefirst two excited states E1A = −58.6 eV and E1S = −57.8 eV have, respectively,antisymmetric and symmetric spatial wave functions. The symmetry of the wavefunction implies a specific symmetry of the spin state: E0 and E1S are singlet spinstates, and E1A is a triplet spin state. In the ground state, the two spins are antiparallel.In order to flip one of them to make them parallel, a considerable amount of energy(∼20 eV) is necessary.

This corresponds to a “force” that maintains the spins in the antiparallel state.It is not a magnetic coupling between the spins: this magnetic interaction can becalculated, and it corresponds to an energy of the order of 10−2 eV. The “force”that we are facing here has an electrostatic origin, the Coulomb interaction, and it

Fig. 14.2 The first threelevels of the helium atom

14.4 Physical Consequences of the Pauli Principle 353

is transferred into a constraint on the spins via the Pauli principle. Such an effect iscalled an exchange interaction. The same effect is the basic cause of ferromagnetism(unpaired valence electrons in, e.g., iron, cobalt, nickel).

14.4.2 The Ground State of N Identical IndependentParticles

Consider N identical independent particles. The Hamiltonian is therefore the sum ofN one-particle Hamiltonians:

H =N∑

i=1

h(i). (14.10)

Let {φn, εn} be the eigenfunctions and corresponding eigenvalues of h: hφn = εnφn ,where we assume that the εn are ordered: ε1 ≤ ε2 · · · ≤ εn · · · .

From the previous considerations, we see that the ground state energy of a systemof N bosons is

E0 = Nε1,

whereas for a system of fermions, we have

E0 =N∑

i=1

εi .

In this latter case, the highest occupied energy level is called the Fermi energy ofthe system and it is denoted εF . The occupation of the states φn is represented onFig. 14.3 both for a bosonic and a fermionic assembly.

Consider, for instance, N independent fermions of spin s confined in a cubic boxof size L . We choose here a basis of states corresponding to periodic boundary con-ditions. Each eigenstate of the Hamiltonian h is a plane wave φ p(r) = ei p·r/�/

√L3,

associated with one of the 2s + 1 spin states corresponding to a well-defined com-ponent ms� of the spin on a given axis (ms = −s,−s + 1, . . . , s). The momentump can be written p = (2π�/L)n, where the vector n = (n1, n2, n3) stands for atriplet of positive or negative integers. The ground state of the N fermion system isobtained by placing 2s + 1 fermions in each plane wave φ p, as long as | p| is lowerthan the Fermi momentum pF . This Fermi momentum pF is determined using:

N =∑

p (p<pF )

(2s + 1).

354 14 Identical Particles, the Pauli Principle

Fig. 14.3 Ground state of a system of N independent identical particles. Left bosonic case, withall the particles in the ground state of the one-body Hamiltonian. Right fermionic case, where thefirst N/(2s + 1) states of the one-body Hamiltonian are occupied (here s = 1/2)

For a large number of particles, we can replace this discrete sum by an integral thatyields

N � (2s + 1)L3

(2π�)3

∫

p<pF

d3 p = 2s + 1

6π2

L3 p3F�3

. (14.11)

This equation relates the density ρ = N/L3 of the gas and the Fermi momentum,independently of the size of the box:

ρ = 2s + 1

6π2(pF/�)3 . (14.12)

The average kinetic energy per particle can also be easily calculated:

〈p2〉2m

= 1

N

∑

p (p<pF )

(2s + 1)p2

2m� 2s + 1

N

L3

(2π�)3

∫

p<pF

p2

2md3 p,

which leads to〈p2〉2m

= 3

5

p2F2m

. (14.13)

14.4.3 Behavior of Fermion and Boson Systems at LowTemperatures

The difference between the ground states of N -fermion or N -boson systems inducesradically different behaviors of such systems at low temperature.

14.4 Physical Consequences of the Pauli Principle 355

Fermions

In a system of fermions at zero temperature and in the absence of interactions, wehave just seen that all the energy levels of the one-body Hamiltonian are filled up tothe Fermi energy εF . This simplemodel describes the conduction electrons in ametalremarkably well, and it accounts for many macroscopic properties of a solid, such asits thermal conductibility. Using the result (14.12), the Fermi energy εF = p2F/2me

can be written in terms of the number density ρe of conduction electrons,

εF = �2(3ρeπ2

)2/3

2me,

where we used 2s + 1 = 2 for electrons. This energy can reach large values (εF =3 eV for sodium). This is much larger than the thermal energy at room temperature(kBT � 0.025 eV). This explains the success of the zero-temperature fermion gasmodel for conduction electrons. At room temperature, very few electrons participatein thermal exchanges.

The application of the Pauli principle to fermionic systems has many conse-quences, ranging from solid-state physics to the stability of stars such as white dwarfsor neutron stars. In nuclear physics, the Pauli principle explains why neutrons are sta-ble under β decay inside nuclei. An isolated neutron is unstable and decays throughthe process n → p + e + ν with a lifetime of the order of 15 min. Inside a nucleus,a neutron can be stabilized if all the final states allowed by energy conservation forthe final proton are already occupied. Therefore, the decay cannot occur because ofenergy conservation.

Bosons

Concerning bosonic systems, a spectacular consequence of quantum statistics is theBose–Einstein condensation. In the absence of interactions between the particles, ifthe number density ρ = N/V is such that:

ρΛ3T > 2.612 with ΛT = h√

2πmkBT, (14.14)

a macroscopic accumulation of particles occurs in a single quantum state, i.e. theground state of the confining potential of the particles. Such congregations of bosonscontrast with the “individualistic” character of fermions.

Until June1995, the usual example of aBose–Einstein condensation thatwasgivenin textbooks was the transition normal liquid → superfluid liquid of helium whichhappens at a temperature of T = 2.17 K. However the complicated interactionsinside the liquid make the quantitative treatment of the superfluid transition quiteinvolved, and different from the simple theory of the Bose–Einstein condensation ofan ideal gas.

356 14 Identical Particles, the Pauli Principle

Fig. 14.4 Bose–Einstein condensation of a gas of 87Rb atoms observed via the evolution of themomentum distribution of the particles in the xy plane. The atoms are confined in an anisotropicharmonic trap (ωx < ωy) and cooled by evaporation. The trap is then turned down suddenly andthe momentum distribution is measured using a time-of-flight technique. Left the temperature Tis noticeably larger than the condensation temperature Tc. The momentum distribution is isotropicand close to a Maxwell–Boltzmann distribution (m〈v2i 〉 = kBT for i = x, y). Middle images(T ≤ Tc): a noticeable fraction of the atoms accumulate in the ground state of the magnetic trap.Right image (T Tc): a very large fraction of the atoms is in the ground state of the trap. Themomentum distribution of the trap reflects the trap anisotropy. (m〈v2i 〉 = �ωi/2). Strictly speakingone should take into account the interactions between atoms to explain quantitatively the details ofthis distribution (photographs by F. Chevy et K. Madison, ENS Paris)

We now have at our disposal experiments2 performed on gases of alkali atoms(lithium, sodium, rubidium) which are initially cooled by laser inside a vacuumchamber with a very low residual pressure (below 10−11 mBar). The atoms are thenconfined by an inhomogeneous magnetic field and they are cooled further down byevaporation until they reach Bose–Einstein condensation, at a temperature below1µK. The evaporative cooling technique consists in eliminating the more energeticatoms, in order to keeponly the slower ones. The collisions between the trapped atomsmaintain permanently the thermal equilibrium. Starting from 109 atoms, one canobtain, after evaporation, a situation where the 106 remaining atoms are practicallyall in the ground state of the system. Figure14.4 shows the time evolution of themomentumdistributionof a gas of bosons (rubidium87atoms) confined in amagnetictrap and cooled down to the condensation point. These Bose–Einstein condensatespossess remarkable coherence and superfluid properties, and this has been a veryactive field of research in recent years.3

The 2001Nobel prize in Physicswas awarded jointly to EricA. Cornell,WolfgangKetterle and Carl E. Wieman “For the achievement of Bose–Einstein condensation

2See M.H. Anderson et al., Science 269, 198 (1995), where the first Bose–Einstein condensate isshown in an experiment carried out with rubidium atoms.3E. Cornell and C. Wieman, Bose-Einstein condensation, Scientific American, March 1998, p. 26;W. Ketterle, Experimental studies of Bose-Einstein condensation, Physics Today, December 1999,p. 30; K. Helmerson et al., Atom Lasers, Physics World, August 1999, p. 31; Y. Castin et al., Bose-Einstein condensates make quantum leaps and bounds, Physics World, August 1999, p. 37; M. W.Zwierlein, C. A. Stan, C. H. Schunck, S. M. F. Raupach, S. Gupta, Z. Hadzibabic, and W. Ketterle(2003). Observation of BoseEinstein Condensation of Molecules, Physical Review Letters 91 (25):250401.

14.4 Physical Consequences of the Pauli Principle 357

in dilute gases of alkali atoms, and for early fundamental studies of the properties ofthe condensates”.

14.4.4 Stimulated Emission and the Laser Effect

Consider a system of independent bosons (cf. Sect. 14.4.2) on which we apply fora finite length of time the one-body potential V = ∑

i v(i). Each potential v(i) acts

only on the particle i and it can induce transitions between the various eigenstatesof h(i). We want to show that the probability for a particle to reach a given final state|φl〉 is increased if this state is already occupied.

Consider first the case where we are dealing with a single particle initially in thestate |φk〉. If we assume that the effect of v is weak, the probability that the particlereaches the state |φl〉 under the action of v is proportional to |vkl |2 = |〈φk |v|φl〉|2.This result of time dependent perturbation theory will be proven in Chap. 16 (see e.g.Eq. (16.15)).

Suppose now that the state |φl〉 is already occupied by N particles, but there isonly one particle in the state |φk〉. The properly symmetrized initial state of the N +1boson system is:

|Ψi 〉 = 1√N + 1

(|1 : φk ; 2 : φl ; . . . ; N : φl ; N + 1 : φl〉+ |1 : φl ; 2 : φk ; . . . ; N : φl ; N + 1 : φl〉 + · · ·+ |1 : φl ; 2 : φl ; . . . ; N : φl ; N + 1 : φk〉

)

and we are interested in the probability to reach the final state:

|Ψ f 〉 = |1 : φl ; 2 : φl ; . . . ; N : φl ; N + 1 : φl〉.

The transition probability is now proportional to:

|Vi f |2 = |〈Ψi |V |Ψ f 〉|2 = (N + 1) |vkl |2.

The presence of N particles in the state |φl〉 increases by a factor N+1 the probabilitythat the particle initially in |φk〉 reaches this state. The transition probability is thesum of the rate for a spontaneous transition, proportional to |vkl |2 and independentof N , and of the rate stimulated by the presence of the N bosons in the state |φl〉and proportional to N |vkl |2.

This gregarious behavior also manifests itself for photons, which are masslessbosons. This explains the phenomenon of stimulated emission of light which is atthe basis of the principle of lasers. An excited atom decays preferentially by emittinga photon in the quantum state occupied by the photons already present in the laser

358 14 Identical Particles, the Pauli Principle

cavity. This leads to a chain reaction in the production of photons which is the keypoint in the mechanism of lasers.4

14.4.5 Uncertainty Relations for a System of N Fermions

Consider N independent fermions of spin s, each placed in a potential V (r) centeredat the origin. The Hamiltonian is therefore:

H =N∑

i=1

h(i) with h = p2

2m+ V (r).

We note εn the energy levels of h and gn their degeneracies. The ground state E0 ofH is obtained by filling the lowest lying levels εn up to the Fermi energy εF , eachwith (2s + 1) particles per state:

E0 = (2s + 1)k∑

n=0

gnεn,

where the number k is determined by the relation N = (2s + 1)∑k

n=0 gn . For aharmonic potential V (r) = mω2r2/2, we have:

εn = (n + 3/2)�ω, gn = (n + 2)(n + 1)/2.

Therefore we find for N � 1:

k �(

6N

2s + 1

)1/3

, E0 � ξ N 4/3�ω,

with ξ = (3/4)61/3(2s + 1)−1/3.Consider now an arbitrary state |Ψ 〉 of these N fermions. We define 〈r2〉 = 〈r2i 〉

and 〈p2〉 = 〈p2i 〉, with i = 1, . . . , N . In this state, 〈H〉 ≥ E0 and consequently:

〈H〉 = N〈p2〉2m

+ N

2mω2〈r2〉 ≥ ξN 4/3

�ω,

4C.H. Townes,How the LaserHappened: Adventures of a Scientist, OxfordUniversity Press (1999);A. Siegman, Lasers (University Science Books, Mill Valley, 1986); O. Svelto, Principles of LasersPlenum Press, New-York (1998).

14.4 Physical Consequences of the Pauli Principle 359

which gives:

〈p2〉 + m2ω2〈r2〉 − 2ξN 1/3�mω ≥ 0 for N � 1.

This second degree trinomial in mω is positive for all values of mω. Therefore weobtain that in any state of these N fermions:

〈r2〉 〈p2〉 ≥ ξ2 N 2/3�2 for N � 1. (14.15)

This relation is valid in particular in the center-of-mass frame, where 〈 p〉 = 0. If wechoose the origin of space at the center of the cloud, we obtain:

Δx Δpx ≥ ξ

3N 1/3

� for N � 1. (14.16)

For spin 1/2 particles, ξ/3 ∼ 0.36.A similar calculation for fermions placed in a 1/r potential leads to:

〈p2〉 ≥ γ �2

⟨1

r

⟩2N 2/3 for N � 1 (14.17)

with γ = 3−1/3(2s + 1)−2/3. The relation (14.17) plays an important role in thestability of self-gravitating systems, in particular neutron stars.

The Pauli principle therefore modifies the uncertainty relations. If we place Nidentical fermions in a volume V ∼ (Δx)3, each of these fermions must occupya different quantum state. We can thus consider that the space accessible to eachparticle is a region of linear extension ∼(V/N )1/3, so that the de Broglie wavelengthof each particle is reduced by a factor N 1/3.

This brings a different, but equivalent, point of view on the physics of an Nfermion system that we investigated in Sect. 14.4.3. In particular, consider again anideal Fermi gas confined in a cubic box of size L at zero temperature. The positiondistribution in the box is uniform and the average momentum per particle can bededuced from (14.13) so that:

Δx2 = 1

L

∫ L/2

−L/2x2 dx = L2

12,

Δp2x = Δp2

3= p2F

5= �

2N 2/3

5L2

(6π2

2s + 1

)2/3

,

fromwhich one can check that (14.16) is well satisfied. Actually the productΔx Δpxcalculated for the ideal Fermi gas confined in a square box exceeds by ∼10% therigorous lower bound (14.16).

360 14 Identical Particles, the Pauli Principle

Fig. 14.5 An incident wavepacket φ1(r) or φ2(r)crosses a 50–50% beamsplitter. After the crossing,this provides a coherentsuperposition of twooutgoing wave packets φ3(r)and φ4(r)

Exercises

1. Identical Particles on a Beam Splitter

Consider a particle prepared at initial time ti in a wave packet ψ(r, ti ) = φ1(r)incoming on a 50–50% beam splitter (Fig. 14.5). At a later time t f , the wavepacket has crossed the beam splitter and the state of the particle can be writtenψ(r, t f ) = (φ3(r) + φ4(r)) /

√2, where φ3 and φ4 correspond to normalized wave

packets propagating in each of the output ports. We assume 〈φ3|φ4〉 � 0.

a. We prepare the particle in the stateψ(r, ti ) = φ2(r), which is deduced fromφ1(r)by symmetry with respect to the beam splitter plane. The state of the particle attime t f can then be written:

ψ(r, t f ) = αφ3(r) + βφ4(r).

Determine (within a global phase factor) the coefficients α and β. Make use ofthe fact that the interaction of the particle with the beam splitter is described bya Hamiltonian (which needs not to be explicitly written). We take 〈φ2|φ1〉 = 0.

b. We prepare at time ti two fermions in the same spin state, one in the external stateφ1(r), the other one in the state φ2(r). What is the final state of the system? Canone detect both fermions in the same output port?

c. Consider the same problem with two bosons, also prepared in the same spin state,one in the external state φ1(r), the other one in the state φ2(r). Show that the twobosons always exit in the same port. This experiment has been performed withphotons by C.K. Hong et al., Phys. Rev. Lett. 59, 2044 (1987).

2. Fermions in a Square Well

Consider two spin 1/2 particles of massm confined in a one dimension infinite squarewell of size L (Chap. 5, Sect. 5.3).

a. We neglect the interactions between the particles. Determine the four lowestenergy levels.

14.4 Physical Consequences of the Pauli Principle 361

b. We suppose that the particles interact together via a contact potential V (x1−x2) =g δ(x1 − x2), where the coupling constant g does not depend on spin. Determineto first order in g the four lowest energy levels of this two particle system.

14.5 Problem: Discovery of the Pauli Principle

Pauli understood what is known as his Principle in 1925. He proceeded by first gath-ering a whole set of independent and sometimes paradoxical pieces of experimentalspectroscopic data which had accumulated in the beginning of the 1920s. Putting thattogether with his personal intuition the symmetry properties of the wave functions,he discovered a common feature in all those experimental results and he understoodwhat was general in that discovery.

As Pauli did, we shall focus mainly on the case of the helium atom. However, weshall assume that we know what has been developed in Chap. 12 about the existenceof the electron spin 1/2 and its properties (which Pauli did not know completely...)

Experimental Data

In the beginning of the 1920s the following aspects of helium spectroscopy wereknown.

a. The lines observed in the optical spectroscopy of this atom were measured. In thepresence of a magnetic field B, some of them were split by the Zeeman effectwith a given multiplicity. A non-split was calle a singlet, split levels were calledtriplets if they were split into three sub-levels.Now, spectroscopy had clearly observed a very intriguing fact,—early in the20th century—. There were actually two sorts of helium, called orthoheliumand parahelium which coexisted in all experiments and were characterized bydifferent energy levels. All helium spectra exhibited both categories of lines, invariable proportions according to what kind of helium was studied. For instance,the Solar helium contained abundant parahelium.Figure14.6 represents schematically the arrangement of the lowest lying levelsin the two cases together with their multiplicities. All the parahelium states aresinglets, all the orthohelium ones are triplets.

b. One notices that the ortho and parahelium are stablewith respect to one another:they do not transform into each other in experimentally accessible time scales(say a few months). However, in its ground state, the orthohelium has a muchlarger energy than that of the parahelium (about 20 eV: see Fig. 14.6). Why can’tit transform into parahelium? The reaction energy, 20 eV, is quite considerable.Pauli is deeply intrigued by that observation.

c. In its ground state the parahelium has a practically vanishing magnetic suscepti-bility. It is not paramagnetic, it does not posses any internal magnetic moment,as opposed to orthohelium.

362 14 Identical Particles, the Pauli Principle

Fig. 14.6 Lowest lying energy levels of heliumas seen experimentally.Only the three first ones (twosinglets of parahelium and the triplet ground state of orthohelium) are represented. The transitionsactually observed are represented by vertical arrows

Theory of the Helium Atom.

Non-perturbed Description.

Let r1 and r2 be the positions (x1, y1, z1) and (x2, y2, z2) of the two electrons withrespect to the nucleus, which we assume fixed.

In first approximation, we neglect the electrostatic repulsion e2/r12 (r12 = |r1 −r2|) between the two electrons, and the electron spins. The Hamiltonian of the twoelectron system of in the field of the nucleus of charge 2 is

H0 = H1 + H2 , with Hi = p2i2m

− 2e2

ri(i = 1, or 2).

If ψa and ψb are two normalized eigenfunctions of the hydrogen-like hamiltonianHi (with eigenenergies Ea and Eb) the function ψ(r1, r2) = ψa(r1)ψb(r2) is aneigenfunction of H0.

Since electrons (1) and (2) are not distinguishable,we introduce thewave functions

ψ± = 1√2[ψa(r1)ψb(r2) ± ψa(r2)ψb(r1)] (14.18)

14.5 Problem: Discovery of the Pauli Principle 363

which are respectively symmetric and antisymmetric in the permutation of the twoelectrons.

What are the eigenenergies E0a,b corresponding to ψ+ and ψ−?

Introduction of the Perturbation e2/r12.

We are now interested in the two hydrogen-like states |1s〉 (n = 1, � = 0,m = 0)and |2s〉 (n = 2, � = 0,m = 0) and therefore in the two following cases:

ψa = ψb = ψ1s on one side and, (14.19)

ψa = ψ1s ψb = ψ2s on the other side. (14.20)

(a) How can one evaluate, in first order perturbation theory, the influence of theCoulomb repulsion e2/r12 on the energy E0

a,b?Note One will not take into account the states ψ2p (n = 2, � = 1) in questions(a), (b), (c) and (d).

(b) Show that once this perturbation is taken into account, the energies of the statesψ+ and ψ− can be written

E±a,b = E0

a,b + Ca,b ± Aa,b

where Ca,b and Aa,b are integrals that one will write but not calculate explicitly.

(c) Show the the integral Ca,b is positive and give its simple physical meaning.

(d) Show qualitatively the the integral Aa,b is most likely positive.Note One can remark that the term e2/r12 takes large values if the electrons areclose to one another.

(e) The values of the following integrals are

I1 =∫

|ψ1s(r1)|2|ψ1s(r2)|2 e2

r12d3r1 d

3r2 = 34 eV

I2 =∫

|ψ1s(r1)|2|ψ2s(r2)|2 e2

r12d3r1 d

3r2 = 15 eV

J =∫

ψ∗1s(r1)ψ

∗2s(r2)ψ1s(r2)ψ2s(r1)

e2

r12d3r1 d

3r2 = 0.4 eV

Represent on the same diagram, starting from the levels E0a,b which will be

calculated, the levels E±a,b of the two states ψ+ and ψ− in first order perturbation

theory in the cases (14.19) and (14.20) above, i.e., a = 1s, b = 1s and a =1s, b = 2s. Note that in the case (14.19) only the state ψ+ remains.

364 14 Identical Particles, the Pauli Principle

Introduction of the Electron Spins

For each of the electrons, (1) and (2), wemust now introduce two possible spin states.We use the notation of Chap. 13, Sect. 13.1.1 |σ1,σ2〉 with σ1,2 = ±1.

We consider the states of total spin S defined by

|Σ+〉 = | + +〉|Σ0〉 = (| + −〉 + | − +〉)/√2|Σ−〉 = | − −〉

(14.21)

| σ 〉 = (| + −〉 − | − +〉)/√2 (14.22)

a. To what values of the total spin S and of its projection Sz along Oz (in � units)do the states |Σ+〉, |Σ0〉, |Σ−〉 et |σ〉 correspond to?

b. Show that the states |Σ+〉, |Σ0〉, |Σ−〉 and |σ〉 are eigenstates of the spin exchangeoperators Psp

12 defined by Psp12 |σ1,σ2〉 = |σ2,σ1〉, and give the corresponding

eigenvalues.

Zeeman Effect. Singlet and Triplet States

When the two-electron system interacts with amagnetic field B (along Oz) its energylevels are modified proportionally to the projection on Oz of the magnetic moment,therefore of Sz .

A system of total spin S = 1 with components Sz = +�, 0, −� has, in thepresence of the field B, the three Zeeman interaction energies:

+2μe B, 0, −2μe B μe: Bohr magneton.

Draw the variation of the energy levels of a system of two electrons in a magneticfield B in the spin states |Σ+〉, |Σ0〉, |Σ−〉 and |σ〉. Justify the names triplet andsinglet.

Total Wave Function of the Helium Atom

We now incorporate both the space variables and the spin variables. For instance, wenote |ψ+,Σ0〉 ≡ ψ+(r1, r2)|Σ0〉 the state of the atom where the electrons are in theorbital state |ψ+〉 and in the spin state Σ0〉.

Make a list of the a priori possible states of the atom in the two orbital states (14.19)and (14.20), without imposing any exclusion criterion (the problem is considered in1924–1925).

Note, for each of these states whether it is globally symmetric or antisymmetricunder a permutation of the two electrons.

Transition Probabilities

We now consider the possible transitions between the states of this list of a prioripossible states of the helium atom.

14.5 Problem: Discovery of the Pauli Principle 365

It can be shown that the transition probability amplitude (emission or absorption)of an atom between a state |i〉 and a state | f 〉 is proportional to the matrix elementof a transition hamiltonian Hint between these two states: 〈 f |Hint |i〉.

In an electromagnetic transition of an atom, the transition hamiltonian is propor-tional to the electric dipole moment D = er . In our two-electron problem (1) and(2), the electric dipole moment is D = e(r1 + r2).

Note that this hamiltonian is invariant under permutations of the two electrons.

(a) Consider the matrix elements 〈ψ′,Σ |Hint |ψ,σ〉 or 〈ψ′,σ|Hint |ψ,Σ〉 where ψand ψ′ represent any orbital state and Σ one of the triplet states. Show that thesematrix elements vanish.What conclusion can one draw on the triplet-singlet or singlet-triplet transitions?

(b) Consider the non-zero matrix elements found in the previous question.Let ψ(r1, r2) and ψ′(r1, r2) be the orbital wave function of electrons in statesψ and ψ′. Express the matrix elements as integrals over r1 and r2.Since the orbital wave functions are either symmetric or antisymmetric underthe exchange of r1 and r2, don’t some matrix elements also vanish?What can we deduce about the possible transitions in the helium atom?

Confrontation Between Theory and Experiment

We now go back to the set of experimental observations of Sect. 1, that we noterespectively a, b and c. Show that a certain elimination among the states which area priori possible in Sects. 2 and 5 allows to account for these experimental results ina consistent way. Give an explanation to the experimental result of figure (14.6) andindicate to which quantum states these levels correspond.

Generalize that elimination in the form of a Principle.

14.5.1 Solution

Theory of the Helium atom.

2.1 The eigenenergy of ψ+ and ψ− is E0a,b = Ea + Eb

2.2

(a) and (b) We have a perturbation of degenerate states. However in the basis{ψ+,ψ−} the perturbation is diagonal. In first order, the perturbation ΔE is thesolution of

C − A − ΔE 00 C + A − ΔE

= 0

with

C =∫

ψ2a(r1)ψ

2b(r2)

e2

r12d3r1d

3r2

366 14 Identical Particles, the Pauli Principle

Fig. 14.7 Theoretical lowest energy levels of helium: left, case (14.19), right, case (14.20)

A =∫

ψ∗a(r1)ψ

∗b(r2)ψa(r2)ψb(r1)

e2

r12d3r1d

3r2

Therefore ΔE± = C ± A, i.e. E± = E0 + C ± A.(c) The integral C is positive since −eψ2

a(r1)d3r1 and −eψ2

b(r2)d3r2 represent

respectively the charge densities of electron (1) and of electron (2). The integralC is the electrostatic repulsion of the two electrons.

(d) If r12 is small, ψ∗a(r1)ψa(r2) is positive, since r1 is close to r2. Similarly for

ψ∗b(r2)ψb(r1). Now, small values of r12 correspond to large values of the function

to integrate.The integral A (called exchange integral) is most probably positive.

(e) The energy diagram therefore has the following structure shown onFig. (14.7), with:

• Case (14.19), 1s2.

Ea + Ea = 2 × 4 × (−13.5) eV � −108 eV, Caa = I1 = 34 eV

=⇒ E+aa = −74 eV

• Case (14.20), 1s 2s.

Ea + Eb = 5 × (−13.5) eV � −67.5 eV, Cab = I2 = 15 eV A = J = 0.4 eV

=⇒ E+ab − E−

ab = 0.8 eV , (E+ab + E−

ab)/2 = 52.5 eV

2.3

a. The total spin S is equal to 1 (states |Σ〉) with projections (+�, 0, −�), or zero(state |σ〉).

b. One has obviouslyPsp12 |Σ〉 = |Σ〉 , Psp

12 |σ〉 = −|σ〉 .

14.5 Problem: Discovery of the Pauli Principle 367

The eigenvalues are ±1, the states |Σ〉 are symmetric and the state |σ〉 is anti-symmetric.

2.4From the previous results, the states |Σ〉 and |σ〉 give rise to the following Zeemanlevels

The levels |Σ〉 are associated in three levels and give a triplet, the state |σ〉 is asinglet, insensitive to the magnetic field.

2.5 There are eight possible states in case (14.20), ψa �= ψb:

|ψ+ Σ+〉 , |ψ+ Σ0〉 , |ψ+ Σ−〉 are symmetric (triplet),

|ψ+ σ〉 is antisymmetric (singlet),

|ψ− Σ+〉 , |ψ− Σ0〉 , |ψ− Σ−〉 are antisymmetric (triplet),

|ψ− σ〉 is symmetric (singlet).

In case (14.19) the function ψ− vanishes identically. Only four states are left;

|ψ+ Σ+〉 , |ψ+ Σ0〉 , |ψ+ Σ−〉 symmetric (triplet),

|ψ+ σ〉 antisymmetric (singlet).

2.6 (a) The operator Hint does not involve spin variables. The orthogonality of |Σ〉and |σ〉 : 〈Σ |σ〉 = 0, is such that the transitions singlet↔triplet are forbidden.

(b)

〈ψ′Σ |Hint |ψΣ〉 = 〈ψ′σ|Hint |ψσ〉

=∫

ψ′∗(r1, r2)Hint (r1, r2)ψ(r1, r2)d3r1d3r2.

The integral of an antisymmetric function of r1 and r2 vanishes, we therefore have

〈ψ+|Hint |ψ−〉 = 〈ψ−|Hint |ψ+〉 = 0.

The only allowed transitions take place between orbital states of same symmetry.

368 14 Identical Particles, the Pauli Principle

3. Confrontation Between Theory and Experiment

If we look back at the states (2,5) and if we note their global symmetry as follows:

STATE Space Spin GLOBAL|1s2s+,Σ〉 sym sym S|1s2s+,σ〉 sym antisym A|1s2s−,Σ〉 antisym sym A|1s2s−,σ〉 antisym antisym S|1s2,Σ〉 sym sym S|1s2,σ〉 sym antisym S

• the result 2.6 (a) says the transitions between singlet and triplet states are forbidden.If helium is a singlet, it stays so, it is parahelium. If it is a triplet it stays so, it isorthohelium. We recover the experimental result [1-2].

• The only allowed transitions between the above states are

|1s2s+,Σ〉 ↔ |1s2,Σ〉 and |1s2s+,σ〉 ↔ |1s2,σ〉.

• The ground state of helium is a singlet (result [1-1]). We know (result [1-3] redun-dant with [1-1]) that this ground state does not have a magnetic moment. It istherefore the state |1s2,σ〉 of parahelium.

• The first para excited state must be |1s2s+,σ〉 since one observes the transitionfrom this state to the ground state.

• The levels ortho |1s2,Σ〉 and |1s2s+,Σ〉 do not exist! Otherwise, one would seethe transition |1s2s+,Σ〉 → |1s2,Σ〉.The absence of that line in the spectrum of helium is what gave Pauli the idea ofhis exclusion principle!

• The interpretation of Fig. 14.6 is therefore that the only states that appear are(increasing values of energy)

|1s2,σ〉 (ground state—para)|1s2s−,Σ〉 (lowest energy ortho)|1s2s+,σ〉 (first excited state para)

which are all totally antisymmetric.

The states that account for these experimental results are all totally antisymmetric.One generalizes that, after further investigations, by postulating that the only possiblestates of an electron system are totally antisymmetric.

14.6 Problem. Heisenberg Relations … 369

14.6 Problem. Heisenberg Relations for Fermions.The Way to Macroscopic Systems

14.6.1 Uncertainty Relations for N Fermions

Consider N independent fermions of spin s, (N � 1), each placed in a potentialV (r) centered at the origin. The Hamiltonian is therefore:

H =N∑

i=1

h(i) with h = p2

2m+ V (r).

We note εn the energy levels of h and gn their degeneracies. The ground state E0 ofH is obtained by filling the lowest lying levels εn up to the Fermi energy εF , eachwith (2s + 1) particles per state:

E0 = (2s + 1)k∑

n=0

gnεn,

where the number k is determined by

N = (2s + 1)k∑

n=0

gn.

In all this problem, we are interested in systems of a very large number of particles:

N � 1.

Harmonic Interactions

We first consider N independent fermions of spin s interacting with a harmonicpotential centered at the origin, so that the total hamiltonian is

H N =N∑

i=1

p2i2m

+ mω2

2

N∑

i=1

r2i . (14.23)

We have:εn = (n + 3/2)�ω, gn = (n + 2)(n + 1)/2.

a. Calculate k and the ground state energy EN0 of this system (for N � 1).

b. Consider an arbitrary state |Ψ 〉 of these N fermions. In this state, we define theexpectation values 〈r2〉 = 〈r2i 〉 and 〈p2〉 = 〈p2i 〉, with i = 1, . . . , N . Using the

370 14 Identical Particles, the Pauli Principle

variational theorem, 〈Ψ |H |Ψ 〉 ≥ E0 ,∀ω, show that

〈r2〉 〈p2〉 ≥ ξ2 N 2/3�2 for N � 1, (14.24)

where ξ is a parameter. Since the potential is centered at the origin, show that thisleads to the uncertainty relation

Δx Δpx ≥ ξ

3N 1/3

� for N � 1. (14.25)

Calculate ξ/3 for spin 1/2 particles.c. We now turn to pairwise interactions. Justify briefly the Lagrange identity, valid

for N euclidean vectors {pi }, i = 1, N

NN∑

i=1

p2i =N∑

i≤ j=1

( pi − p j )2 +

(N∑

i=1

pi

)2

.

d. Using this identity, show that the previous hamiltonian of N independent spin 1/2fermions can be rewritten as the sum

H N = H Ncm + H N

rel , (14.26)

where

H Ncm = P2

2mN+ mω2

2N(

N∑

i=1

r i )2 with P =N∑

i=1

pi

is the center of mass hamiltonian, and

H Nrel = 1

2mN

N∑

i≤ j=1

( pi − p j )2 + mω2

2N

N∑

i≤ j=1

(r i − r j )2

is the translation invariant hamiltonian of the N pairwise interacting fermions.– Justify that H N

cm and H Nrel commute.

– Calculate the ground state energy ENrel of H

Nrel in terms of EN

0 .e. Setting the relative momentum of particle i and j as qi j = ( pi − p j )/2, 〈q2〉 =

〈q2i j 〉, and 〈r2〉 = 〈(ri − r j )2〉, and proceeding as before in Eqs. (14.24) and

(14.25) prove the uncertainty relation

Δx Δqx ≥ ξ′

3N 1/3

� for N � 1. (14.27)

Calculate ξ′/3 for spin 1/2 particles.– Compare this inequality with (14.25) (for all values of N ).

14.6 Problem. Heisenberg Relations … 371

Gravitational Interactions

We now turn to a gravitating system of N fermions of massm and spin s with interac-tions centered at the origin: V (ri ) = −Gm2/(ri ) where G isNewton’s constant. The hamiltonian is

H (N ) =N∑

i=1

pi2/2m +

N∑

i=1

V (ri ).

The energy levels εn of the one-particle hamiltonian and their degeneracies gn are

εn = −m(Gm2)2

2�2n2, gn = n2.

a. Calculate the ground state energy E0 of HN , (N � 1).b. Setting, as previously, 〈p2〉 = 〈 p2i 〉, and 〈(1/r)〉 = 〈(1/ri )〉, use the above result

to obtain a Heisenberg inequality of the form

〈p2〉 ≥ γ �2

⟨1

r

⟩2N 2/3 for N � 1. (14.28)

and calculate the value of γ.

14.6.2 White Dwarfs and the Chandrasekhar Mass

The sun burns quietly its hydrogen and transforms it into helium. In 5 billion years,the fuel will be exhausted and there will no longer exist a thermal pressure to balancethe gravitational pressure. The system will implode until it reaches temperature anddensity conditions such that the fusion reactions of helium into carbon and oxygenoccur: 3 4He→ 12C and 4 4He→ 16O. After this second phase, which is much shorter(∼108 years) than the previous one, a new effect appears which prevents furtherthermonuclear reactions to start, such as burning carbon and oxygen into 23Na, 28Siand 31P, and these latter nuclei into 56Fe, the most tightly bound nucleus. Actually,the density is so high that the quantum pressure of the degenerate electron gas canstop the gravitational collapse. The system then becomes a white dwarf star, whoseonly fate is to lose its heat by radiating.

All stars do not end up as white dwarfs. For initial masses noticeably largerthan M�, the star can reach the nuclear stages 28Si and 56Fe where fusion can nolonger take place. Beyond some critical value of the final mass of the star, called theChandrasekhar mass, the pressure of the degenerate electron gas cannot compete anylonger with gravitation, and this leads to a gravitational catastrophe, the explosionof a supernova and the formation of a neutron star in its center.

372 14 Identical Particles, the Pauli Principle

White dwarfs have masses of the order of the solar mass M�, sizes comparableto the Earth radius (i.e. 0.01 R� ∼ 104 km), and densities 106 g cm−3 (i.e. ∼ 106

larger than usual densities). Neutron stars are much more compact objects, withmasses ∼ M� and radii ∼ 10 km. The densities then reach 1015 g cm−3. We wish tounderstand these orders of magnitude.

White Dwarfs

Consider N nuclei of mass Amp and charge Z , surrounded by N Z electrons. Thecomplete calculation is somewhat involved5 but a simple and quite accurate estimatecan be obtained because, to good approximation, the mass in the star has a constantdensity distribution inside a sphere of radius R. This holds for the N nuclei of massAmp and charge Z (Z ∼ A/2 in our case), as well as for the distribution of the ZNelectrons. The dominant terms in the total energy of the system are the following:

• The potential energy Ep is dominated by the gravitational attraction between thesenuclei. For a spherical object of radius R and with a uniform spatial density, onehas

Ep = −3

5

GM2

R,

where M = N Amp is the mass of the star.Note that the electrostatic interaction of the electrons is negligible since the localcharge density inside the star is zero. Since the electron density is also uniform,one has, for the electrons,

〈1r〉 = 3

2Rso that Ep = −2

5GM2

⟨1

r

⟩. (14.29)

• The kinetic energy Ek is that of the electrons, Ek = p2/2me. However, nuclearenergies are far above the electron mass mec2 ∼ 0.5MeV. Therefore, the properapproach is to treat Ek in its relativistic form6:

Ek =N Z∑

i=1

√p2i c

2 + m2ec

4 ⇒ 〈Ek〉 ∼ N Zc√

〈p2〉 + m2ec

2. (14.30)

• Putting together (14.29) and (14.30) we obtain

E = Ep + Ek ∼ −2

5GM2

⟨1

r

⟩+ N Zc

√〈p2〉 + m2

ec2 , (14.31)

5S. Chandrasekhar, An introduction to the study of Stellar Structure, Dover, New York, 1967.6Concerning the use and properties of the operator

√p2c2 + m2c4 in the Schrödinger equation, see

J.L. Basdevant and S. Boukraa, Z. Phys. C28, 413 (1985); C30, 103 (1986), and further referencestherein.

14.6 Problem. Heisenberg Relations … 373

where we further make use of the Heisenberg inequality (14.28) in the form

〈p2〉 ∼ �2

⟨1

r

⟩2(N Z)2/3 for (N Z) � 1. (14.32)

(We set the constant γ of (14.28) equal to 1. This simplifies calculations, andalso has the advantage of slightly moving apart from the strict minimum of theinequality.) Of course, both terms are approximate. Inequality (14.28) has beenobtained in a non-relativistic approach, and 〈√p2 + m2

ec2〉 �= √〈p2〉 + m2

ec2.

However, it can be shown that for an ideal Fermi gas, the total energy derivedwith this approximation coincides to within 10% with the exact result,7 in the twolimits of non-relativistic and ultra-relativistic electron systems.As such this yields acceptable orders of magnitude (more complete procedures areavailable of course).

a. Express the energy (14.31) as a function of the parameter 〈p2〉.b. Express the condition for the position of the minimum of the energy in the form

〈p2〉〈p2〉 + m2

ec2

=(

M

MCh

)4/3

where M is the mass of the star, and MCh , called the Chandrasekhar mass,is expressed in terms of the dimensionless gravitation “structure” constantαG = Gm2

p/(� c) � 5.9× 10−39 (where G = 6.7× 10−11 m3/kg/s2 is Newton’sconstant), the proton mass mp = 1.67 × 10−27 kg, and Z and A.

c. Calculate the numerical value of MCh . Compare with the solar massm� = 2.0×1030 kg.

d. Deduce that the Chandrasekhar mass MCh is an upper bound for the mass M ofthe star, beyond which the star collapses gravitationally.

e. Give the expression of the equilibrium radius R. For M = M�, compare with thesolar radius R� = 7 × 105 km and the earth’s radius R⊕ = 6.4 × 104 km.

14.6.3 Neutron stars

At higher densities, it become energetically favorable for protons to capture electronsaccording to an inverse β process : p + e− → n + ν. The neutrinos escape from thestar which forms a neutron star.

Neutron stars were discovered in the middle 1960s as pulsars. They are giganticnuclei, in the sense of nuclear physics. They are made of neutrons (electrically neu-tral) bound by the gravitational force, and packed up on one another not far fromnuclear distances (∼10−15 m). The size of such objects is of the order of 10 km,

7See e.g. K. Huang, Statistical Mechanics, Chap. 11 (John Wiley, New-York, 1963).

374 14 Identical Particles, the Pauli Principle

their mass is of the order of a few solar masses M�, and their density is of the orderof 1014–1015 g cm−3. We only consider their main structure, and do not take intoaccount the external proton and electron shells that stabilize them.

Following a procedure very similar to the previous one, we shall assume that thedominant terms in the energy of a neutron star containing N neutrons are:

• The potential energy Ep dominated by the gravitational attraction (nuclear reac-tions are absent). Assuming for simplicity a uniform distribution of the N neutrons,we take as before:

Ep = −3

5

GN 2m2n

R∼ −2

5GN 2m2

n

⟨1

r

⟩,

• and the relativistic kinetic energy of the neutrons:

Ek =N∑

i=1

√p2i c

2 + m2nc

4 ⇒ 〈Ek〉 ∼ Nc√

〈p2〉 + m2nc

2.

Since neutrons are fermions, Eq. (14.32) which relates 〈p2〉 and 〈(1/r)〉 still holds.a. Express the total energy Ep + Ek in terms of 〈p2〉.b. Express the condition for the position of the minimum of the energy in the form

〈p2〉〈p2〉 + m2

nc2

=(

N

N1

)4/3

and give the expression and the numerical value of N1. Comment on the role ofN1.

c. If the number of neutrons is below the critical number N1, show that the radiusR = 〈(1/r)〉 of the star is given by:

R = R1

(N1

N

)1/3(1 −

(N

N1

)4/3)1/2

d. Show that because of relativistic effects, the mass Ms of the star differs fromthe mass Nmn of its constituents. Calculate the mass Ms using the total energy(Ms = (Ep + Ek)/c2).

e. Calculate the maximum value of the mass Ms , the corresponding value of N andof the radius R.

14.6 Problem. Heisenberg Relations … 375

14.6.4 Mini-boson Stars

To appreciate the power of the Pauli principle in generating “macroscopic” physicalobjects, it is interesting to consider the following (purely academic) problem.

Consider a set of N bosons of equal mass m with pairwise gravitational interac-tions V (r i − r j ) and a hamiltonian

H (N ) =N∑

i=1

pi2/2m +

N∑

i≤ j=1

V (r i − r j ).

a. Using the Lagrange identity, show that the N-body hamiltonian H (N ) can bewritten as

H (N ) = P2/2Nm + H (N )rel ,

where P = ∑Ni=1 pi is the total N-body momentum, and where the relative hamil-

tonian H (N )rel is a sum of two-body hamiltonians of the type

H 2rel(i, j) = 4

((pi − p j

)/2)2

2Nm+ V (r i − r j )

where we write the kinetic energy in a form such that (pi − p j )/2 and (r i − r j )

have a canonical commutation relation.b. We place ourselves in the total c.m. system so that 〈P2/2Nm〉 = 0.

Denoting by |Ω〉 the normalized ground state of H (N )rel , and EN

0 the correspondingground state energy. Show that

EN0 ≥

(N (N − 1)

2

)〈Ω|H 2

rel(i, j)|Ω〉 ≥(N (N − 1)

2

)E (2)(μ′), (14.33)

where (i, j) are any two different indices, and E (2)(μ′) is the two-body ground-state energy corresponding to a mass mu′ = mN/4.

c. A boson star is, to first approximation, a systemof N bosons ofmassmwhich havepairwise gravitational interactions, i.e. V (r) = −Gm2/r where G is Newton’sconstant.Write an expression for the lower bound of EN

0 = 〈Ω|H (N )|Ω〉 in terms of thetwo-body energy E (2)(μ′) in the c.m.s.

d. Deduce from that an inequality relating EN0 = 〈Ω|H (N )|Ω〉 to the two-body

energy E (2)(μ′) in the c.m.s (i.e. 〈P2/2Nm〉 = 0).

376 14 Identical Particles, the Pauli Principle

e. Show that in the non-relativistic approximation this leads the Heisenberginequality

〈p2〉 ≥ 2 �2

⟨1

r

⟩2, (14.34)

where 〈p2〉 = 〈p2i 〉 and 〈1/r〉 = 〈1/(|(r i − r j )|〉.f. We assume that in first approximation, the total energy of the mini-boson star is

the sum of a potential term

Ep = 〈N∑

i≤ j=1

V (r i − r j )〉 = −N (N − 1)

2Gm2〈1

r〉

and a relativistic kinetic term

Ek = Nc√

〈p2〉 + m2c2 .

Using the inequality (14.34) express the total energy of the star in terms of 〈p2〉.g. Express the condition for the position of the minimum of the energy in the form

〈p2〉〈p2〉 + m2c2

=(

N

N1

)2

and give the expression of N1.h. What is the limiting upper value of N for which the star becomes gravitationally

unstable and collapses in amini-boson black hole?Calculate its value form = mn .i. If the number of bosons is below the critical number N1, show that the radius

R = 〈(1/r)〉 of the star is given by:

R = R1

(N1

N

) (1 −

(N

N1

)2)1/2

j. Show that because of relativistic effects, the mass Ms of the star differs fromthe mass Nmn of its constituents. Calculate the mass Ms using the total energy(Ms = (Ep + Ek)/c2).

k. Calculate the maximum value of the mass Ms , the corresponding value of N andof the radius R.

l. Calculate Ms and N1 in the academic case m = mn , and in the hypothetical caseof a sneutrino of larger mass.

14.6 Problem. Heisenberg Relations … 377

14.6.5 Solution

Uncertainty Relations for N Fermions

Harmonic Interactions

a. We find for N � 1:

k �(

6N

2s + 1

)1/3

, E0 � ξ N 4/3�ω,

with ξ = (3/4)61/3(2s + 1)−1/3.

b. In the state |Ψ 〉, 〈H〉 ≥ E0, therefore:

〈H〉 = N〈p2〉2m

+ N

2mω2〈r2〉 ≥ ξN 4/3

�ω,

which gives:

〈p2〉 + m2ω2〈r2〉 − 2ξN 1/3�mω ≥ 0 for N � 1.

This second degree trinomial in mω is positive for all values of mω. Thereforewe obtain that in any state |Ψ 〉 of these N fermions:

〈r2〉 〈p2〉 ≥ ξ2 N 2/3�2 for EN

0 (14.35)

This relation is valid in particular in the center-of-mass frame, where 〈 p〉 = 0. Ifwe choose the origin of space at the center of the cloud, we obtain:

Δx Δpx ≥ ξ

3N 1/3

� for N � 1. (14.36)

For spin 1/2 particles, ξ/3 ∼ 0.36.

c. The identity is simple: the diagonal terms p2i have the same coefficient N , andthe crossed terms pi · p j cancel.

d. The decomposition follows from the Lagrange identity.–H N

cm and H Nrel commute since [P, (r i − r j )] = 0 and vice versa.

– Since P/√N and

∑r i/

√N have canonical commutation relations, the grounds

state energy of H Ncm is EN

cm = (3/2)�ω.Therefore, since the ground state energy of H N is EN

0 = ENrel + EN

cm we obtain

ENrel = EN

0 − ENcm � ξ N 4/3

�ω − (3/2)�ω ∼ EN0 for N � 1.

378 14 Identical Particles, the Pauli Principle

e. The definitions are such that q i j = ( pi − p j )/2 and (ri − r j ) have canonicalcommutation relations. Therefore, we can proceed as previously and obtain:

Δx Δqx ≥ ξ′

3N 1/3

� for N � 1. (14.37)

Where ξ and ξ′ coincide for large N , since the term (3/2)�ω is negligible. Thetwo inequalities are equivalent for large N .The Pauli principle modifies profoundly the uncertainty relations.

Gravitational Interactions

The energy levels εn of the one-particle hamiltonian and their degeneracies gn are

εn = −m(Gm2)2

2�2n2, gn = n2.

a. We therefore obtain that there exists an integer k such that:

N � (2s + 1)k3

3E0 � −(2s + 1)

m(Gm2)2

2�2n2k.

Therefore E0 and N are related by:

E0 � −m(Gm2)2

2�2(2s + 1)2/3 (3N )1/3

b. Consider an arbitrary state |Ψ 〉 of this N fermion system, we therefore obtain

〈p2〉 ≥ γ �2

⟨1

r

⟩2N 2/3 for N � 1 (14.38)

with γ = 3−1/3(2s + 1)−2/3.

Again, the Pauli principle modifies profoundly the uncertainty relations whenfermions are involved.

White Dwarfs and the Chandrasekhar Mass

The potential energy Ep is dominated by the gravitational attraction between thenuclei, we assume a spherical object of radius R with a uniform density, and

Ep = −3

5

GM2

R,

where M = N Amp is the mass of the star.

14.6 Problem. Heisenberg Relations … 379

We have

〈1r〉 = 3

2Rso that Ep = −2

5GM2

⟨1

r

⟩. (14.39)

We consider the kinetic energy Ek in its relativistic form

Ek =N Z∑

i=1

√p2i c

2 + m2ec

4 ⇒ 〈Ek〉 ∼ N Zc√

〈p2〉 + m2ec

2. (14.40)

Putting together (14.29) and (14.40) we obtain

E = Ep + Ek ∼ −2

5GM2

⟨1

r

⟩+ N Zc

√〈p2〉 + m2

ec2, (14.41)

We use a simplified Heisenberg–Pauli inequality

〈p2〉 ∼ �2

⟨1

r

⟩2(N Z)2/3 for (N Z) � 1. (14.42)

a. The energy of the star, as a function of 〈p2〉 is therefore

E = −2

5

G(N A)2m2p

�cc(N Z)−1/3

√〈p2〉 + N Zc

√〈p2〉 + m2

ec2

b. The radius R of the star and the average squaredmomentum 〈p2〉 are nowobtainedby minimizing the total energy Ep + Ek with respect to 〈p2〉. This leads to:

〈p2〉〈p2〉 + m2

ec2

=(

M

MCh

)4/3

, (14.43)

where we have introduced the Chandrasekhar mass MCh:

MCh ∼ 4

α3/2G

Z2

A2mp. (14.44)

Where αG = (Gm2p)/(�c) = 5.9 × 10−39 is dimensionless.

c. The numerical value of the Chandrasekhar mass is, in these approximations,

MCh ∼ 3.7 × 1030 kg ∼ 1.8M�.

The exact value, which requires solving a non-linear differential equation, is∼1.4M�.

d. We notice that the Eq. (14.43) has a solution only if the mass M is smaller than theChandrasekhar mass. For masses larger than the Chandrasekhar mass, the Fermi

380 14 Identical Particles, the Pauli Principle

pressure of the electron gas cannot compete with the gravitational pressure: thesystem is unstable and it undergoes a gravitational collapse.

e. For a mass lower than the Chandrasekhar mass, the equilibrium radius is givenin our model by:

R = RCh

(MCh

M

)1/3(1 −

(M

MCh

)4/3)1/2

, (14.45)

where we have put:

RCh ∼ 2.5Z

A

1√αG

�

mec(∼6300 km for A = 2Z).

This result is in good agreement with predictions obtained from more elaboratetreatments. The predicted value for MCh is then 1.4M�. For a relatively lowmass (M MCh), we obtain the scaling law R ∝ N−1/3 of the non relativistictreatment:

R ∼ 7800 km ×(M�M

)−1/3

for A = 2Z . (14.46)

The star of van Maanen, was one of the first white dwarfs to be discovered. Itsradius is∼8900 km (78 times smaller than the sun radius) and itsmass is 0.68 M�,in good agreement with (14.46).The equilibrium radius (14.45) is a decreasing function of the mass of thewhite dwarf, and, owing to relativistic effects, it shrinks to zero when the massapproaches the Chandrasekhar mass. Therefore, the most massive white dwarfscorrespond to ultra relativistic electrons and they all have the same mass MCh.For M = M�, compare with the solar radius R� = 7 × 105 km and the earth’sradius R⊕ = 6.4 × 104 km.

Neutron Stars

At higher densities, it become energetically favorable for protons to capture electronsaccording to an inverse β process : p + e− → n + ν. The neutrinos escape from thestar which forms a neutron star.

The neutron stars were discovered in the middle 1960s as pulsars. They are gigan-tic nuclei, in the sense of nuclear physics. They are made of neutrons (electricallyneutral) bound by the gravitational force, and packed up on one another at nucleardistances ∼10−15 m. The size of such objects is of the order of 10 km, their mass isof the order of a few solar masses M�, and their density is of the order of 1014–1015g cm−3.

The dominant terms in the energy of a neutron star containing N neutrons are:

• The potential energy Ep dominated by the gravitational attraction. Assuming forsimplicity a uniform distribution of the N neutrons, we take as before:

14.6 Problem. Heisenberg Relations … 381

Ep = 3

5

GN 2m2n

R.

• The relativistic kinetic energy of the neutrons:

Ek =N∑

i=1

√p2i c

2 + m2nc

4 ⇒ 〈Ek〉 ∼ Nc√

〈p2〉 + m2nc

2.

a. Express the total energy Ep + Ek in terms of 〈p2〉. The total energy is

Ep + Ek = −2

5

GN 2m2n

�cc(N )−1/3

√〈p2〉 + Nc

√〈p2〉 + m2

nc2

b. Taking 〈p2〉 as the variable, the minimum energy is obtained for:

〈p2〉〈p2〉 + m2

nc2

=(

N

N1

)4/3

with N1 ∼ 4

α3/2G

∼ 9 × 1057.

As for the case of a white dwarf star, this equation has a solution only if thenumber of neutron is below a critical number N1.

c. In this case, the radius of the star is given by:

R = R1

(N1

N

)1/3(1 −

(N

N1

)4/3)1/2

with:

R1 ∼ 2.5�

mnc

1√αG

∼ 7 km .

d. Because the neutrons have an average velocity close to c, the mass Ms of the stardiffers from the mass Nmn of its constituents. The mass Ms is obtained using thetotal energy (Ep + Ek = Msc2), which yields:

Ms = Nmn

(1 −

(N

N1

)4/3)1/2

.

e. The mass is maximal when N ∼ 0.7 N1 and it takes the value:

Mmaxs ∼ 6.5 × 1030 kg ∼ 3 M�, (14.47)

corresponding to a radius ∼5 km.Beyond the critical number N1, a gravitational catastrophe happens: the systemcan lose energy by radiating neutrinos, it can fall into this “catastrophic” ground

382 14 Identical Particles, the Pauli Principle

state where a new kind of physics applies, and it becomes a black hole.One can refine this extremely simple model, in particular by taking into accountthe inhomogeneous density profile in the neutron star. The theory, which is due toLandau, Oppenheimer and Volkov, leads to a critical mass very similar to (14.47).

Again, the collapse comes from a simple property of degenerate Fermi gases. Inthe non-relativistic regime, the pressure P of a Fermi gas is related to the densityρ by P ∝ ρ5/3. This can always balance the gravitational inward pressure which isPgrav ∝ ρ4/3, provided the density is large enough. If the conditions are such thatthe Fermi gas is relativistic, then its pressure is P ∝ ρ4/3, and there is a value of themass such that the gravitational pressure prevails and the system collapses.

Mini-boson Stars

To appreciate the power of the Pauli principle in generating “macroscopic” physi-cal objects, it is interesting to consider the following problem which is yet purelyacademic.

Consider a set of N bosons of equal mass m with pairwise gravitational interac-tions V (r i − r j ) and a hamiltonian

H (N ) =N∑

i=1

pi2/2m +

N∑

i≤ j=1

V (r i − r j ).

a. The expression follows immediately from the Lagrange identity:

H (N ) = P2/2Nm + H (N )rel ,

where P = ∑Ni=1 pi is the total N-body momentum, and where the relative

hamiltonian H (N )rel is a sum of two-body hamiltonians of the type

H (N )rel =

N∑

i≤ j=1

H 2rel(i, j) =

N∑

i≤ j=1

4((pi − p j

)/2)2

2Nm+ V (r i − r j )

where we write the kinetic energy in a form such that (pi − p j )/2 and (r i − r j )

have a canonical commutation relation.b. We place ourselves in the total c.m. system so that 〈P2/2Nm〉 = 0.

By the variational principle, we know that if E (2)(μ′ = nM/4) is the two-bodyground state energy, then the expectation value

〈4((pi − p j

)/2)2

2Nm+ V (r i − r j )〉 ≥ E (2)(μ′ = Nm/4)

holds for any state.Therefore, since all two-body sub-systems are the same, it follows immediatelythat

14.6 Problem. Heisenberg Relations … 383

EN0 =

(N (N − 1)

2

)〈Ω|H 2

rel(i, j)|Ω〉 ≥(N (N − 1)

2

)E (2)(μ′ = nM/4),

(14.48)c. From the above expression, we have

E (2)(μ′) = −1

2

Nmc

4

(Gm2)2

�2c2

and thus for N � 1,

EN0 ≥ −N 3

16mc2

(Gm2)2

(�c)2.

d. Therefore, since

⟨N∑

i=1

pi2/2m +

N∑

i≤ j=1

V (r i − r j )

⟩≥ EN

0

which amounts to

N 〈p2〉2m

− N 2

2Gm2

⟨1

r

⟩≥ −N 3mc2

16

(Gm2)2

(�c)2.

e. Since this holds whatever G, we obtain the inequality:

〈p2〉 ≥ 2�2

⟨1

r

⟩2(14.49)

where 〈p2〉 = 〈p2i 〉 and 〈1/r〉 = 〈1/(|(r i − r j )|〉 .f. Assuming that in first approximation, the total energy of the mini-boson star is

the sum of a potential term

Ep = 〈N∑

i≤ j=1

V (r i − r j )〉 = −N (N − 1)

2Gm2〈1

r〉

and a relativistic kinetic term

Ek = Nc√

〈p2〉 + m2c2 .

Using the inequality (14.49) the total energy of the star can be expressed as afunction of 〈p2〉 as

E = Nc

(√〈p2〉 + m2c2 − N

23/2(Gm2)

(�c)

√〈p2〉

)/c2

384 14 Identical Particles, the Pauli Principle

g. The minimum of the energy occurs for

〈p2〉〈p2〉 + m2c2

=(

N

N1

)2

with N1 ∼ �c

23/2Gm2.

h. N1 is the limiting upper value of N for which the star becomes gravitationallyunstable and collapses in a mini-boson black hole? For m = mn , we obtainN1 = 6 × 1037 which is very small compared to the fermionic neutron case (forinstance we have N1mn ∼ 1011 kg). This is due to the effect of the term α

3/2G

instead of αG . The power 3/2 is characteristic of fermion systems. It enhancesenormously the number of constituents, the sizes and the masses of fermionsystems as compared to what they would be in “similar” boson systems.

i. If the number of bosons is below N1, the radius R = of the star is given by:

R = R1

(N1

N

) (1 −

(N

N1

)2)1/2

j. The mass Ms = (Ep + Ek)/c2 is Ms = Nm√1 − (N/N1)2 and the radius

R = �√2

mc

N1

N

√1 − (N/N1)2.

k. The maximum value of Ms occurs for (N/N1)2 = 1/2 : Mmax = Nm/

√2 and

R = 2�/(mc).l. In the academic casem = mn , N1 = 6×1037, Ms ∼ 7×1010 kg, and R ∼ 0.6 fm

(a nuclear size!). The hypothetical case of a sneutrino of larger mass can becalculated easily.Such a mini boson-star or black hole (for N ≥ N1) would indeed be a terriblysmall (and comparatively massive) stellar object. Notice that the formation ofsuch an object would imply some mechanism of energy loss by radiation (ofphotons, neutrinos, gravitons etc.).

Chapter 15Lagrangian and Hamiltonian, LorentzForce in Quantum Mechanics

The Lorentz force q v × B acting on a particle of charge q moving with a velocityv in a magnetic field B, does not derive from a potential. Therefore, the formulationof quantum mechanics we have used up to now does not apply. The aim of thisChapter is to generalize the correspondence principle, in order to obtain the form ofthe Hamiltonian in such a problem.

Naturally, it would suffice to simply introduce this Hamiltonian with no otherjustification than the fact that it accounts for observed phenomena. However, it isinstructive to first develop some considerations on classical mechanics. In fact, thedevelopment of analytic mechanics in the 18th and 19th centuries, which is due tothe works of d’Alembert, Bernoulli, Euler, Lagrange, Hamilton, etc., had brought outan amazing geometric structure of the theory, based on a variational principle, theprinciple of least action. Itwas one of the first remarkable discoveries ofDirac in 1925and 1926, to understand that the same basic structure underlies quantum mechanics.Starting from this observation, the correspondence principle can be stated in a muchmore profound manner, which enables one to treat complex problems that would bedifficult to attack without this analysis.

In Sect. 15.1 we shall recall the basic elements of the Lagrangian formulationof mechanics, based on the principle of least action. In Sect. 15.2, we shall presentthe “canonical” formulation of Hamilton and Jacobi, which will allow us to exhibit,in Sect. 15.3, the parallelism between classical and quantum mechanics. The wordHamiltonian which we have used so often will then take its full significance. InSect. 15.4, we shall give the Lagrangian and Hamiltonian formulations of the prob-lem of interest, i.e. the motion of a charged particle in a magnetic field. Finally, inSect. 15.5, we shall transpose the result to quantum mechanics and we will also takeinto account the possibility that the particle has spin.

© Springer International Publishing Switzerland 2016J.-L. Basdevant, Lectures on Quantum Mechanics,Graduate Texts in Physics, DOI 10.1007/978-3-319-43479-7_15

385

386 15 Lagrangian and Hamiltonian, Lorentz Force in Quantum Mechanics

Fig. 15.1 Examples oftrajectories starting from x1at time t1 and arriving at x2at time t2. Among all thesepossible trajectories, thetrajectory actually followedby the particle is the one forwhich the action S isextremal

15.1 Lagrangian Formalism and the Least Action Principle

In his Méchanique analytique published in 1787, one century after Newton’s Prin-cipia, Lagrange proposed a new way to consider mechanical problems. Instead ofdetermining the position r(t) and the velocity v(t) of a particle at time t knowing itsinitial state {r(0), v(0)}, Lagrange asks the following equivalent but different ques-tion. What is the trajectory actually followed by the particle if, leaving r1 at time t1,it reaches r2 at t2?

Least Action Principle

In order to simplify the discussion, consider first a one-dimensional problem. Amongthe infinite number of possible trajectories (see Fig. 15.1) such that:

x(t1) = x1 x(t2) = x2, (15.1)

what is the law that determines the good one? Lagrange made use of the “princi-ple of natural economy”,1 which is an expression due to Fermat, that was adoptedby Maupertuis and Leibniz (who called it the principle of “the best”). Lagrange’sprescription is the following:

a. Any mechanical system is characterized by a Lagrange function, or LagrangianL(x, x, t),which depends on the coordinate x,on its derivative with respect to timex = dx/dt,and possibly on time. The quantities x and x are called state variables.For instance, considering a particle in a one-dimensional potential, one has:

L = 1

2mx2 − V (x, t). (15.2)

b. For any trajectory x(t) satisfying (15.1), one defines the action S:

1In full rigor, the variational principle as we present it here was formulated by Hamilton in 1828.In order to simplify the presentation, we have shrunk parts of History.

15.1 Lagrangian Formalism and the Least Action Principle 387

S =∫ t2

t1

L(x, x, t) dt. (15.3)

The principle of least action states that the physical trajectory X(t) is such that S isminimal, or, more generally, extremal.

Lagrange Equations

Let X(t) be the physical trajectory. Consider a trajectory x(t) infinitely close toX(t), and also starting from x1 at time t1, and arriving at x2 at time t2:

x(t) = X(t) + δx(t), x(t) = X(t) + δx(t), δx(t) = d

dtδx(t), (15.4)

with, by assumption:δx(t1) = δx(t2) = 0. (15.5)

To first order in δx, the variation of S is:

δS =∫ t2

t1

(∂L∂x

δx(t) + ∂L∂x

δx(t)

)dt.

Integrating the second term by parts and taking into account (15.5), we obtain:

δS =∫ t2

t1

(∂L∂x

− d

dt

(∂L∂x

))δx(t) dt. (15.6)

It follows from the principle of least action that δS must vanishwhatever the infinites-imal function δx(t). Therefore, the equation which determines the physical trajectoryis the Lagrange equation:

∂L∂x

= d

dt

(∂L∂x

). (15.7)

We can check readily that we recover the usual equation of motion mx = −dV / dxfor a point particle placed in the potential V (x, t) if we consider the Lagrangian(15.2).

The generalization to s degrees of freedom: {xi, xi}, i = 1, . . . , s (with, forinstance, s = 3N for N particles in a three-dimensional space) is straightforward.One uses a Lagrangian L({xi}, {xi}, t) and obtains the set of Lagrange equations:

∂L∂xi

= d

dt

(∂L∂xi

)i = 1, . . . , s. (15.8)

388 15 Lagrangian and Hamiltonian, Lorentz Force in Quantum Mechanics

Remarks

(1) The Lagrange equations keep the same form in any coordinate system. This isparticularly useful in order to make changes of variables, e.g. to get from cartesiancoordinates (x, y, z) to spherical coordinates (r, θ,ϕ). The xi are called generalizedcoordinates.

(2) It is remarkable that the laws of mechanics can be derived from a variational prin-ciple, which states that the physical trajectory minimizes a certain quantity, here theaction. Almost all physical laws can be formulated in terms of variational principleslike the Fermat principle in geometrical optics. The situations which are physicallyencountered appear to result as “optimizing” the effects of various “conflicting”contributions.

(3) In the absence of forces (i.e. a uniform potential in (15.2)), S is minimum forx = constant, corresponding to a linear uniformmotion. The presence of the potentialcan be visualized as a property of space which bends the trajectories. Forces andinertia appear to be in conflict. The particle follows a trajectory of minimal “length”,this length being measured by the action S. Therefore one can understand that amechanical problem is reduced to a geometric problem. The motion of a particle in afield of force deriving from a potential in a flat Euclidean space, can be transformedinto the free motion of a particle in a curved space, where it follows geodesics.Einstein had this idea inmind in 1908when he started to construct General Relativity.It took him seven years to elaborate the mathematical structure of the theory.

The elaboration of the fundamental concepts and principles of mechanics was performed duringthe 17th century. Copernicus had given the notion of reference frames. Galileo had understoodthe principle of inertia: uniform linear motion is a state relative to the observer, and not a process.It is the modification of the velocity which constitutes a process. The final lines were written byNewton.After theNewtonian synthesis and the publication in 1687of thePhilosophiae Naturalis PrincipiaMatematica, the 18th and the 19th centuries were marked by a fascinating endeavor. Through theimpetus of d’Alembert, Maupertuis, the Bernoulli brothers (in particular Daniel), of Euler andof Lagrange, the basic structure of mechanics, i.e. a geometric structure, was discovered. A largeclass of problems could be reduced to problems of pure geometry. D’Alembert, who was thefirst to understand the importance of the abstract concepts of mass and of momentum, attackedthe concept of force introduced by Newton. For d’Alembert, motion is the only observablephenomenon, whereas the “causality of motion” remains an abstraction. Hence the idea to studynot a particular trajectory of the theory, but the set of all motions that it predicts (to characterizea force by the set of all its effects is actually a very modern point of view.) In 1787, one centuryafter the Principia, Lagrange published his Méchanique analytique, and gave a new formulationof mechanics where the geometric and global structure of the theory is emphasized.The first formulation of a physical law in terms of the least action principle originated froma dispute between Fermat and Descartes, around 1640, about the notion of proof (Descartes’s“proof” of Snell’s laws for refraction was actually wrong). Fermat, who was a mathematician,and who knew little, if any, physics, got interested in the laws of geometrical optics, in particularthe equality between the angles of incidence and of reflection. He proved the results as being ageometrical property of the optical length of the light rays (in the case of reflection, this had beenunderstood by Heron of Alexandria in 100 a.d.). The Snell–Descartes laws predicted which pathwould be followed by a light ray with given initial properties. In the more general point of viewof Fermat, one determines the path effectively followed by a light ray which goes from A to B.

15.1 Lagrangian Formalism and the Least Action Principle 389

At the end of his life, in 1661, Fermat stated his “principle of natural economy” which led tonumerous fields of research, still very active nowadays, on variational principles.

Energy

Weassume the system is isolated, i.e. ∂L/∂t = 0, andwe calculate the time evolutionof the quantity L(x, x) along the physical trajectory x(t):

dLdt

(x, x) = x(t)∂L∂x

+ x(t)∂L∂x

= d

dt

(x(t)

∂L∂x

),

where we have transformed the first term using the Lagrange equation (15.7). There-fore:

d

dt

(x(t)

∂L∂x

− L)

= 0.

For an isolated system, the quantity:

E = x(t)∂L∂x

− L ,

(resp. E =

s∑

i=1

xi(t)∂L∂xi

− L)

(15.9)

is conserved. It is a constant of the motion, called the energy of the system. In thesimple case (15.2) we recover E = mx2/2 + V (x).

15.2 Canonical Formalism of Hamilton and Jacobi

Conjugate Momenta

The quantity:

p = ∂L∂x

(resp. pi = ∂L

∂xi

), (15.10)

which appears in the definition of the energy (15.9), is called the conjugate momen-tum, or generalized momentum, of the variable x. In the simple case (15.2), it reducesto the linear momentum p = mx, but this is no longer true in non-Cartesian coor-dinates or, as we shall see, when the forces are velocity dependent. We notice thatEq. (15.7) implies:

p = ∂L∂x

,

(resp. pi = ∂L

∂xi

). (15.11)

390 15 Lagrangian and Hamiltonian, Lorentz Force in Quantum Mechanics

Canonical Equations

The description of the state of the particle (or the system) by x and the conjugatemomentum p, instead of x and the velocity x, has some advantages. We assume thatwe can invert Eq. (15.10) and calculate x in terms of the new state variables x andp. The equations of motion are obtained by performing what is called a Legendretransformation. Let us introduce the Hamilton function, or Hamiltonian:

H(x, p, t) = px − L(resp. H(xi, pi, t) =

∑

i

pixi − L)

. (15.12)

We write the total differential of H:

dH = p dx + x dp − ∂L∂x

dx − ∂L∂x

dx − ∂L∂t

dt.

If we take into account (15.10) and (15.11), the first and fourth term cancel, and thethird term is nothing but −p dx, therefore:

dH = x dp − p dx − ∂L∂t

dt. (15.13)

This gives the equations of motion:

x = ∂H

∂p, p = −∂H

∂x

(resp. xi = ∂H

∂pi, pi = −∂H

∂xi

)(15.14)

which are called the canonical equations of Hamilton and Jacobi. They are firstorder differential equations in time, and they are symmetric in x and p (up to aminus sign). They have the big technical advantage to present the time evolution ofthe state variables directly in terms of theses state variables. More generally, if wenote X = (r, p) the coordinates of the system in phase space, these equations havethe form X = F(X). Such a problem, called a dynamical system, is of considerableinterest in many fields, including mathematics.

Notice that in quantummechanics, the Ehrenfest theorem derived in Chap. 7 givesthe time evolution of expectation values as:

d

dt〈xi〉 =

⟨∂H

∂pi

⟩d

dt〈pi〉 =

⟨−∂H

∂xi

⟩,

where we can see some similarity with the canonical Hamilton–Jacobi equations.

15.2 Canonical Formalism of Hamilton and Jacobi 391

Poisson Brackets

Consider two functions f and g of the state variables x, p and possibly of time, forinstance two physical quantities. One defines the Poisson bracket of f and g as thequantity:

{f , g} = ∂f

∂x

∂g

∂p− ∂f

∂p

∂g

∂x

({f , g} =

s∑

i=1

∂f

∂xi

∂g

∂pi− ∂f

∂pi

∂g

∂xi

). (15.15)

We find immediately:

{f , g} = −{g, f } {x, p} = 1, (15.16)

and more generally:

{xi, xj} = 0 {pi, pj} = 0 {xi, pj} = δij, (15.17)

and:

{x, f } = ∂f

∂p{p, f } = −∂f

∂x. (15.18)

Let us now calculate the time evolution of a quantity f (x, x, t):

f = df

dt= ∂f

∂xx + ∂f

∂pp + ∂f

∂t. (15.19)

Using Hamilton’s equations (15.14), we obtain:

f = {f , H} + ∂f

∂t. (15.20)

In particular, the canonical equations (15.14) are written in the completely symmet-ric way:

x = {x, H} p = {p, H}. (15.21)

In the canonical formalism, theHamiltonian governs the time evolution of the system.If a physical quantity f does not depend explicitly on time, i.e. ∂f /∂t = 0, then itstime evolution is obtained through the Poisson bracket of f and the Hamiltonian:f = {f , H}. If this Poisson bracket vanishes, f is a constant of the motion.

15.3 Analytical Mechanics and Quantum Mechanics

The results derived in the previous section reveal an amazing property. There is astrong analogy in the structures of analytical mechanics and quantummechanics. Let

us associate to any quantum observable A, the observable ˆA such that by definition:

392 15 Lagrangian and Hamiltonian, Lorentz Force in Quantum Mechanics

〈 ˆA〉 ≡ d

dt〈a〉

for any state of the system. The Ehrenfest theorem then implies:

ˆA = 1

i�[A, H] + ∂A

∂t(15.22)

to be compared with (15.20). Similarly, the canonical commutation relations:

[xj, pk] = i�δjk (15.23)

are highly reminiscent of Eq. (15.17).This identity in the structures of the two kinds of mechanics was one of the

first great discoveries of Dirac. Of course, the mathematical nature and the physicalinterpretation of the objects under consideration are different. But the equationswhich relate them are the same provided we make the following correspondence,which was understood by Dirac during the summer of 1925:

Quantization rule.One replaces the Poisson brackets of analytical mechanics by thecommutators of the corresponding observables, divided by i�:

Analytical mechanics {f , g} −→ 1

i�[f , g] Quantum mechanics (15.24)

This is the genuine form of the correspondence principle. In general, for com-plex systems (large number of degrees of freedom, constraints, etc.), the systematicmethod to obtain the form and the commutation relations of observables consists inreferring to the Poisson brackets of the corresponding classical systems. We will seean example below when we treat the Lorentz force in quantum mechanics.

One can now understand why the name of Hamilton (1805–1865) appears so often in quan-tum mechanics, although Hamilton lived one century before its invention. Hamilton is one ofthe great geniuses of science. He made decisive contributions to analytical mechanics, and heinvented vector analysis; he also invented the same year as Cayley and Grassmann (1843), non-commutative algebras and matrix calculus (the elements of Hamilton’s quaternions are called... Pauli matrices in quantum mechanics). He is the author of the synthesis of the geometricaland wave theories of light. He proved in what limit the first is an approximation of the sec-ond. Hamilton was fascinated by variational principles, in particular by the similarity betweenMaupertuis’s principle in mechanics and Fermat’s principle in optics. In 1830 he made theremarkable statement that the two formalisms of optics and of mechanics were basically thesame and that Newtonian mechanics corresponds to the same limit as geometrical optics, whichis only an approximation. This remarkwas ignored by his contemporaries, and themathematicianFelix Klein said in 1891 that it was a pity. It is true that, in 1830, no experimental fact couldreveal the existence of Planck’s constant. However, in many ways, Hamilton can be consideredas a precursor of quantum mechanics. Louis de Broglie refers to Hamilton’s work in his thesis.

15.4 Classical Charged Particle in an Electromagnetic Field 393

15.4 Classical Charged Particle in an Electromagnetic Field

Classically a particle of charge q, placed in an electromagnetic field, is subject to theLorentz force:

f = q (E + v × B) .

This force is velocity dependent and it does not derive from a potential. Furthermore,the magnetic force qv × B does no produce any work and the energy of the particleis E = mv2/2 + qΦ, where Φ is the scalar potential associated with the electricfield E.

TheHamiltonian is certainly different from p2/2m + qΦ. Otherwise the equationsof motion would be strictly the same as in the absence of a magnetic field.We can usethe previous developments in order to determine the correct form of the HamiltonianH .

Maxwell’s equations, specifically the couple of equations:

∇ · B = 0 , ∇ × E = −∂B∂t

, (15.25)

allow to express the fields E and B in terms of the scalar and vector potentials Φ andA:

B = ∇ × A , E = −∇Φ − ∂A∂t

. (15.26)

Consider a particle of mass m and charge q placed in this electromagnetic field. Wenote r and r = v the position and the velocity of this particle. A possible Lagrangianfor this particle can be written in terms of the potentials A and Φ:

L = 1

2mr2 + q r · A(r, t) − q Φ(r, t). (15.27)

Indeed, starting from the Lagrange equations and using:

d

dtA(r, t) = ∂A

∂t+ x

∂A∂x

+ y∂A∂y

+ z∂A∂z

,

we can check that we obtain the desired equation of motion:

mdv

dt= q(E + v × B).

Consider now the conjugate momentum p. From the definition (15.10) we have:

p = mr + qA(r, t). (15.28)

394 15 Lagrangian and Hamiltonian, Lorentz Force in Quantum Mechanics

In otherwords the conjugatemomentumpno longer coincideswith the linearmomen-tum, i.e. the product of the mass times the velocity mr!

Equation (15.28) is easily inverted: r = (p − qA(r, t))/m, which gives the Hamil-tonian we are looking for:

H = 1

2m(p − qA(r, t))2 + qΦ(r, t). (15.29)

As the Lagrangian, it is expressed in terms of the potentials A and Φ, and not thefields E and B.

15.5 Lorentz Force in Quantum Mechanics

15.5.1 Hamiltonian

Wenow followDirac’s quantization rules of Sect. 15.3. TheHamiltonian of a chargedparticle in an electromagnetic field is:

H = 1

2m(p − qA(r, t))2 + qΦ(r, t) (15.30)

where the position and the conjugate momentum operators r and p satisfy the canon-ical commutation relations:

[xj, xk] = 0 [pj, pk] = 0 [xj, pk] = i� δjk .

In the wave function formalism, we can still choose p = −i�∇. The velocity observ-able is no longer p/m, but:

v = 1

m(p − qA(r, t)). (15.31)

Note that two components of the velocity (e.g. vx and vy) do not, in general, com-mute in presence of a magnetic field. Using the Ehrenfest theorem, one can verifythat (15.30) provides the appropriate structure of the equations of motion for theexpectation values.

15.5.2 Gauge Invariance

One thing, though, seems surprising. The potentials Φ and A are not unique. Twosets (Φ,A) and (Φ ′,A′) related to each other by a gauge transformation:

15.5 Lorentz Force in Quantum Mechanics 395

A′ = A + ∇χ(r, t) Φ ′ = Φ − ∂χ

∂t, (15.32)

where χ(r, t) is an arbitrary function, correspond to the same electric and magneticfields E and B. Since the energy observable H is expressed in terms of A and Φ,the energy seems to depend on the gauge. However, we know that physical resultsshould not depend on the gauge!

The answer to this problem is simple and remarkable. In a gauge transformation,the wave function also changes:

ψ(r, t) → ψ′(r, t) = eiqχ(r,t)/� ψ(r, t). (15.33)

One can check that if ψ is a solution of the Schrödinger equation for the choice ofpotentials (A, Φ), then ψ′ is a solution for the choice (A′, Φ ′). For time independentproblems, this guarantees that the energy spectrum of the Hamiltonian for the choice(A, Φ) coincides with the one obtained with (A′, Φ ′).

The transformation (15.33) does not modify the probability density:

|ψ(r, t)|2 = |ψ′(r, t)|2,

which is of course crucial. It simply affects the phase of the wave function by anamount which depends on the point in space.

One can verify more generally that the expectation values of all measurablequantities are gauge invariant. Consider, for instance, the velocity operator v =(p − qA

)/m. We find:

(p − qA

′)ψ′ =

(−i�∇ − qA − q∇χ

)eiqχ/� ψ

= eiqχ/�

(−i�∇ − qA

)ψ = eiqχ/�

(p − qA

)ψ,

from which we deduce:

ψ′∗(p − qA

′)ψ′ = ψ∗

(p − qA

)ψ.

This proves that the probability current is the same in both gauges. If we integrate thisrelation over space, we find that the expectation value of the velocity is also gaugeindependent. On the contrary, the momentum p is not a gauge invariant physicalquantity.

If one postulates that the laws of physics are invariant under all gauge transformations (15.33)whereχ(r, t) is arbitrary, one canderive that theHamiltonianhas the structure (15.30). In quantumfield theory, gauge invariance plays a crucial role in the physics of fundamental interactions andelementary constituents of matter.

The fact that the Hamiltonian (15.30) depends on the potentials and not on thefields can be verified experimentally following a suggestion of Aharonov and Bohm

396 15 Lagrangian and Hamiltonian, Lorentz Force in Quantum Mechanics

Fig. 15.2 TheBohm–Aharonov effect: in aYoung slit experiment performedwith charged particles,the interference pattern is shifted when a current runs in the solenoid. However, the magnetic fieldcreated by this solenoid is zero everywhere except inside the solenoid itself. The correspondingphase shift has no classical counterpart (in terms of a force acting on the particles for instance) andone refers to it as a topological phase shift

in 1956. In a Young slit interference device, one places between the two slits asolenoid of small diameter, parallel to the slits (Fig. 15.2). When a current flows inthe solenoid, one can observe a modification of the system of fringes. However, themagnetic field is zero everywhere outside the solenoid, in particular near the slits.Conversely, the vector potential is non zero outside the solenoid. This experimenthas been performed and it has confirmed the quantum mechanical predictions.2

15.5.3 The Hydrogen Atom Without Spin in a UniformMagnetic Field

Weplace a hydrogen atom in a constant uniformfieldB, which derives from the vectorpotentialA = B × r/2, and we neglect spin effects for the moment. The Hamiltonian

H = 1

2me(p + qA)2 + V (r),

where V (r) = −q2/4πε0r and −q is the electron charge, can be expanded as:

H = H0 + q

2me

(p · A + A · p

)+ q2

2meA2

with H0 = p2

2me+ V (r).

2A. Tonomura et al., Phys. Rev. Lett 56, 792 (1986).

15.5 Lorentz Force in Quantum Mechanics 397

The first term H0 is simply the Hamiltonian studied in Chap.11. The second term iscalled the paramagnetic term. We remark that, since p · A = A · p in this gauge, wecan rewrite this term as:

q

2me

(B × r

) · p = q

2me

(r × p

) · B = −γ0 L · B = −µL · B (15.34)

with γ0 = −q/2me. We recover the magnetic dipole interaction term introduced inChap. 10 (Eq.10.41).

The third term q2A2/2me is called the diamagnetic term. One can check that

for the lowest lying levels En of the hydrogen atom, and for magnetic fields below1 Tesla, the diamagnetic term is negligible: it is much smaller (by a factor of ∼10−4)than the paramagnetic term, which is itself small (by a factor of ∼10−4) comparedto |En|.

15.5.4 Spin 1/2 Particle in an Electromagnetic Field

Consider a spin 1/2 particle, charged or not, with an intrinsic magnetic momentµS = γSS where S is the spin observable. If we place this particle in an electromag-netic field, and possibly in another potential V (r), its Hamiltonian is:

H = 1

2m

(p − qA(r, t)

)2 + qΦ(r, t) + V (r) − µS · B(r, t) (15.35)

where q is the charge of the particle, and A and Φ are the electromagnetic potentials.This Hamiltonian is called the Pauli Hamiltonian. It acts on the Hilbert space

Eexternal ⊗ Espin described in Chap.12. For an electron, the form (15.35) is directlyobtained as the nonrelativistic limit of theDirac equation, which predicts γS = 2γ0 =−q/me.

15.6 Exercises

1. The Lorentz force in quantum mechanics

In this exercise, we want to check that with the prescription:

H = 1

2m(p − qA(r))2

for the Hamiltonian of a charged particle in a magnetic field, one recovers, using theEhrenfest theorem, the classical equations of motion.

398 15 Lagrangian and Hamiltonian, Lorentz Force in Quantum Mechanics

Fig. 15.3 Study of the Bohm–Aharonov effect: classical paths in a two-hole Young interferenceexperiment

a. We assume in all what follows that the field B is constant, uniform and directedalong the z axis. We set B = |B|. We introduce the vector potential A = B × r/2.Check that this choice gives the appropriate value of B.

b. Write the classical equation of motion of a particle of charge q and mass min this field. Give the expression of the energy E of the particle. Describe thecharacteristics of the motion of the particle.

c. Consider the observable u = p − qA with A = A(r). Here p is the usual momen-tum operator, i.e. p = −i�∇ (which yields [x, px] = i�). Show that p.A = A.p.Write the commutation relations [ux, uy], [uy, uz] and [uz, ux] for the three com-ponents of the observable u.

d. We assume that the quantum Hamiltonian has the form H = u2

2m . Calculated〈r〉 / dt and d〈u〉 / dt. Compare the result with the classical equations of motion.

e. Deduce from this result the form of the velocity observable v.f. Can the three components of the velocity be defined simultaneously in a magnetic

field? Write the corresponding uncertainty relations.

2. The Aharonov–Bohm effect

Consider the two-hole Young interference experiment shown in Fig. 15.3. A solenoidwhose axis is perpendicular to the plane of the figure is placed between the two holesB and B′. One sends a particle from the source point O at time t1 and one detects theimpact of this particle on the detection screen at a later time t2. We shall assume thatthe probability amplitudeA(C) to detect the particle inC is approximately given by3 :

A(C) = AOBC + AOB′C ∝ eiS/� + eiS′�,

where S et S′ are the classical actions calculated along the paths OBC et OB′Crespectively.

a. In absence of current in the solenoid, recover the fringe spacing xs found inChap.1(Sect. 2.2).

3This prescription can be deduced from the formulation of quantum mechanics based on pathintegrals; see R.P. Feynman and A.R. Hibbs, Quantum Mechanics and Path Integrals (McGraw-Hill, New-York, 1965).

15.6 Exercises 399

b. Determine the change in the interference signal when a current flows in thesolenoid. Express the result in terms of the total magnetic flux πr2B (where ris the radius of the solenoid and B the magnetic field inside the solenoid) and themagnetic flux quantum h/q.

15.7 Problem. Landau Levels

Energy Levels in a Magnetic Field

In this problemwewant to determine the energy levels of a spinless particle of chargeq and mass m, which is placed in a constant and uniform magnetic field B = B uz.We use here the Landau gauge A(r) = Bx uy.

a. Write the eigenvalue equation for the Hamiltonian H. The eigenfunction isdenoted Ψ (r) and the corresponding eigenvalue Etot .

b. We look for particular solutions which are factorized:

Ψ (x, y, z) = eikzz ψ(x, y).

Show that ψ(x, y) is solution of the eigenvalue equation:

−�2

2m

(∂2

∂x2+

(∂

∂y− i

qB

�x

)2)

ψ(x, y) = E ψ(x, y) (15.36)

where we set E = Etot − �2k2z /2m.

c. Equation (15.36) describes the motion of the charge in the xy plane. We look forparticular solutions of this equation which are also factorized with respect to xand y:

ψ(x, y) = eikyy χ(x)

(i) Write the equationwhich determinesχ(x). Towhich physical problem does itcorrespond? One will introduce the cyclotron angular frequency ωc = qB/m.

(ii) Show that the possible eigenvalues for the energy E are:

E =(

n + 1

2

)�ωc. (15.37)

Do the eigenvalues depend on the wavevector ky? The corresponding energylevels are called the Landau levels.

d. We now determine the degeneracy of a given Landau level, assuming that thexy motion of the particle is confined in a rectangle [0, X] × [0, Y ]. We shallneglect any edge effect, assuming that a0 = (2�/qB)1/2 � X, Y and we restrictto relatively low values of the quantum number n.

400 15 Lagrangian and Hamiltonian, Lorentz Force in Quantum Mechanics

(i) We choose periodic boundary conditions for the motion along the y axis.Show that the wave vector ky is quantized: ky = 2πj/Y , where j is an integer.

(ii) What are the relevant values of j, such that the wave function ψ(x, y) islocalized in the rectangle X × Y (and is thus physically acceptable)?

(iii) Express the degeneracy of a Landau level as a function of the flux BXY andof the magnetic flux quantum h/q.

The Lowest Landau Level (LLL)

We now consider the quantum motion of a particle of charge q and mass m in auniformmagnetic fieldB = B uz. Herewe choose the symmetric gaugeA = B × r/2.We restrict ourselves to the motion of the particle in the xy plane (kz = 0) and we setas above ωc = qB/m.

a. Write the eigenvalue equation for the energy. One will introduce Lz = xpy − ypx.b. Consider the Landau level with the lowest energyELLL = �ωc/2 (see Eq. (15.37)).

Show that the functions:

ψ�(x, y) = (x + iy)� e−(x2+y2)/(2 a20),

where � is an arbitrary integer and a0 = (2�/qB)1/2 are all energy eigenstates forthe eigenvalue ELLL.

c. Recover for the LLL the degeneracy calculated in the previous exercise, assumingthat the particle is confined in a disk centered in x = y = 0, with a radius R a0.

This eigenstate basis plays an important role for the study of the fractional quantum Hall effect,which was discovered for a two-dimensional electron gas placed in a magnetic field.

15.7.1 Solution

a. The eigenvalue equation for the energy reads:

−�2

2m

(∂2

∂x2+

(∂

∂y− i

qB

�x

)2

+ ∂2

∂z2

)Ψ (x, y, z) = Etot Ψ (x, y, z)

b. This eigenvalue equation is separable and one can check immediately that thefunctions Ψ (x, y, z) = eikzz ψ(x, y) satisfying (15.36) with E = Etot − �

2k2z /2mare eigenfunctions of the energy. As for the classical case, the motion along z islinear and uniform.

c. (i) The substitutionψ(x, y) = eikyy χ(x) leads to the following equation forχ(x):

− �2

2m

d2χ

dx2+ 1

2mω2

c (x − xc)2 χ = Eχ

15.7 Problem. Landau Levels 401

where we set xc = �ky/(qB). This is the Schrödinger equation for a onedimension harmonic oscillator of frequency ωc/2π, centered in xc.

(ii) The energy eigenvalues are E = (n + 1/2) �ωc, where n is a non negativeinteger. These eigenvalues do not depend on ky.

d. (i) The periodic boundary conditions for the y axis entail eikyY = 1, i.e. ky =2πj/Y , where j is an integer which can be a priori positive or negative.

(ii) In order to have a wave function localized in the desired rectangle, the centerof the oscillator corresponding to themotion along the x axis has to be between0 and X:

0 < xc < X ⇒ 0 < j < jmax = qBXY

2π�

Since the extension of the wave function along x is of the order of a few a0for the first Landau levels, the hypothesis a0 � X implies that the particle isindeed localized with a probability close to 1 in the rectangle X × Y .

(iii) The number of independent states corresponding to a given Landau level isjmax = Φ/Φ0, where we set Φ = BXY and Φ0 = 2π�/q. This is the degen-eracy of the level. The functions ψn(x, y) = eikyy χn(x) constitute a basis forthis level. Another possible basis is obtained by exchanging the roles of xand y, using the gauge A(r) = −By ux. A third possible gauge choice is thesymmetric gauge, studied in the next exercise, and it leads to a third possiblebasis for each of the Landau level (for simplicity, the next exercise is actuallyrestricted to the lowest Landau level).

The Lowest Landau Level (LLL)

a. For the gauge chosen in the text, the eigenvalue equation for the motion in the xyplane is:

(−�2

2m

(∂2

∂x2+ ∂2

∂y2

)− ωc

2Lz + 1

8mω2

c

(x2 + y2

))Ψ (x, y) = E Ψ (x, y)

b. We introduce the polar coordinates ρ, θ in the xy plane. The functions ψ�(x, y) =(x + iy)� e−(x2+y2)/(2 a20) = ρ� ei�θ e−ρ2/(2a20) are eigenstates of Lz = −i� ∂ /∂θwiththe eigenvalue ��. Inserting this result into the above eigenvalue equation, onereaches the desired result, after a relative long, but basic, calculation.

c. The stateψ�(x, y) is relevant if thiswave function is essentially localized inside thedisk of radiusR. The probability density |ψ�(x, y)|2 ∝ ρ2� e−ρ2/a20 has amaximumlocated a distance �1/2 a0 from the origin, with a width �1/4 a0 (for � 1). There-fore the quantum numbers � must be located between 0 and �max = R2/a2

0, whichcan also be written �max = Φ/Φ0, whereΦ0 = 2π�/q andΦ = πR2 B representsthe field flux across the accessible surface. We recover indeed the degeneracyfound in the previous exercise.

Chapter 16The Evolution of Systems

Any experimental observation or any practical use of quantum phenomena relieson processes where a system evolves from a know initial state, and one performsmeasurements on it at some later time. It is therefore important to understand thevarious types of evolution a system can have. Up to now, we have been mainlyinterested in isolated systems whose nature did not change, i.e. there was no creationor annihilation of particles.

In this Chapter, we present two characteristic processes: the oscillatory behaviorof a two-state system under the influence of an external field, constant or oscillating,and the irreversible decay process of a system coupled to a continuum. In Sect. 16.1,we introduce the notion of a transition probability and a basic tool: time-dependentperturbation theory. In Sect. 16.2, we consider the atomic transitions induced by anexternal electromagnetic field, i.e. absorption and induced emission. In Sect. 16.3,we consider the problem of the decay of a system, such as an excited atom or anexcited nucleus. We show how the exponential decay law emerges, and how one cancalculate the lifetime of a system.We also introduce the notion ofwidth of an unstablesystem. Finally, in Sect. 16.4, we discuss a few aspects of the time-energy uncertaintyrelation, ΔE Δt ≥ �/2, which differs quite radically from the uncertainty relationswe established in Chap.8, and which illustrates the special role played by time innon-relativistic quantum theory.

16.1 Time-Dependent Perturbation Theory

Transition Probabilities

Consider a system whose evolution can be derived from the Hamiltonian:

H = H0 + H1(t), (16.1)

© Springer International Publishing Switzerland 2016J.-L. Basdevant, Lectures on Quantum Mechanics,Graduate Texts in Physics, DOI 10.1007/978-3-319-43479-7_16

403

404 16 The Evolution of Systems

where H0 is time-independent. The eigenvectors |n〉 and eigenvalues En of H0 areassumed to be known:

H0|n〉 = En|n〉. (16.2)

The operator H1(t) is an interaction term which may depend explicitly on time, andwhich a priori does not commute with H0. This term can induce transitions betweentwo eigenstates |n〉 and |m〉 of H0.

Our aim is the following. Assuming the system is prepared at time t0 in a givenstate |ψ(t0)〉 = |n〉, we want to calculate the probability

Pn→m(t) = |〈m|ψ(t)〉|2 (16.3)

to find the system in the eigenstate |m〉 of H0 at a later time t .

Example: A Collision Process

In addition to the examples that we shall meet in this Chapter, a collision is a typicalsituationwhere this problem appears. Consider two particles a and b, each ofwhich isprepared in a wave packet sharply defined in momentum p. At initial time the centersof these twowave packets propagate towards each other. In the absence of interactionthe particles propagate freely: 〈pa〉 and 〈pb〉 are constants of the motion. If we takeinto account the interaction potential H1 between the particles, a scattering processtakes place. A measurement of the distribution of final momenta of the particles willgive useful information on the forces themselves. The problem is to calculate theprobability distribution of final momenta, knowing the interaction potential.

Evolution Equations

At any time, the state of the system |ψ(t)〉 can be expanded in the basis {|n〉} ofeigenstates of H0:

|ψ(t)〉 =∑

γn(t)e−i En t/�|n〉. (16.4)

In this expression, we explicitly write the time evolution factors e−i En t/� that wouldbe present for H1 = 0. This simplifies the evolution equations of the coefficientsγn(t). Using the Schrödinger equation we obtain:

i�∑

n

(γn(t) − i

�Enγn(t))e

−i En t/�|n〉 =∑

n

γn(t)e−i En t/�(H0 + H1)|n〉,

and therefore:

i�∑

n

γn(t)e−i En t/�|n〉 =

∑

n

γn(t)e−i En t/� H1|n〉. (16.5)

Multiplying by 〈k|, we get:

i�γk(t) =∑

n

γn(t) e−i(En−Ek )t/� 〈k|H1|n〉. (16.6)

16.1 Time-Dependent Perturbation Theory 405

The problem is completely determined by this set of coupled differential equationsand by the initial condition which specifies |ψ(t0)〉.Perturbative Solution

In general, this set of equations does not have an analytic solution. In order to makeprogress, we assume as in Chap.9 that H1 is “small” compared to H0. In other words,we consider the Hamiltonian Hλ = H0 + λH1 andwe assume that the correspondingcoefficients γk(t) are analytic functions of λ around the origin including at λ = 1:

γk(t) = γ(0)k (t) + λ γ(1)

k (t) + · · · + λp γ(p)k (t) + · · · (16.7)

Inserting this expansion in (16.6), we identify the coefficients of each power of λand we obtain:

to order 0 : i�γ(0)k (t) = 0, (16.8)

to order 1 : i�γ(1)k (t) =

∑

n

γ(0)n (t) e−i(En−Ek )t/� 〈k|H1|n〉 , (16.9)

to order r : i�γ(r)k (t) =

∑

n

γ(r−1)n (t) e−i(En−Ek )t/� 〈k|H1|n〉. (16.10)

This system can be solved by iterations. The terms γ(0)k (t) are determined by

the knowledge of the initial state of the system. Inserting these into (16.9) we cancalculate the terms of order 1, γ(1)

k (t), which in turn give the terms of order 2 throughEq. (16.10), and so on. One therefore determines successively all the terms in theexpansion.

First Order Solution: The Born Approximation

The zeroth order equation is solved immediately. We find that γ(0)k (t) is a constant.

If we choose the initial condition |ψ(t0)〉 = |i〉, we obtain:

γ(0)k (t) = δk,i . (16.11)

Inserting this in (16.9), we have for f �= i :

i�γ(1)f (t) = ei(E f −Ei )t/� 〈 f |H1|i〉. (16.12)

Taking into account the assumption γ(1)f (t0) = 0, this gives:

γ(1)f (t) = 1

i�

∫ t

t0

ei(E f −Ei )t/� 〈 f |H1|i〉 dt. (16.13)

In this approximation, called the Born approximation, the transition probability froman initial state |i〉 to a final state | f 〉 (with f �= i) is given by:

406 16 The Evolution of Systems

Pi→ f (t) = |γ(1)f (t)|2.

This approximation is acceptable if Pi→ f � 1 (necessary condition).

Particular Cases

Constant Perturbation

We suppose the perturbation H1 is “switched on” at t0 = 0 and “switched off” at alater time T . We suppose also that it is constant between 0 and T . Setting �ω0 =E f − Ei , we obtain:

γ(1)f (t ≥ T ) = 1

i�〈 f |H1|i〉 e

iω0T − 1

iω0, (16.14)

and consequently:

Pi→ f (t ≥ T ) = 1

�2|〈 f |H1|i〉|2 y(ω0, T ). (16.15)

We shall often make use of the above function y(ω, T ), defined as:

y(ω, T ) = sin2(ωT/2)

(ω/2)2with

∫ +∞

−∞y(ω, T ) dω = 2πT . (16.16)

Its graph is given on Fig. 16.1.

Sinusoidal Perturbation

Consider a coupling H1(t) such that H1(t) = H1 e−iωt for 0 < t < T and H1(t) = 0otherwise. A simple calculation gives:

Pi→ f (t ≥ T ) = 1

�2|〈 f |H1|i〉|2 y(ω − ω0, T ). (16.17)

Fig. 16.1 Graph of thefunction y(ω, T )

16.1 Time-Dependent Perturbation Theory 407

There is a resonance if the angular frequency ω of the perturbation is equal to theBohr frequency ω0 = (E f − Ei )/� of the system. The resonance curve giving thevariation of Pi→ f as a function of ω has a full width at half maximum Δω ∼ 2π/T .It gets sharper as the interaction time T increases.

Perturbative and Exact Solutions

We already met this in the case of a two-state system when we studied in Chap.12the magnetic resonance of a spin 1/2 placed in a rotating magnetic field. For thisparticular problem we know the exact solution of the evolution equations and it isinstructive to compare it with the approximate result we have derived above.

Consider the specific example of theRabi experiment.We note T the time spent bythe molecular beam inside the cavity where the magnetic field rotating at frequencyω/2π is applied. The time-dependent coupling H1 is (cf. Eq. (12.53)):

〈+|H1|−〉 = �ω1

2e−iωt (ω1 = −γB1).

The exact formula obtained by Rabi is (cf. (12.62)):

P+→−(T ) = ω21

Ω2sin2 (ΩT/2) with Ω = (

(ω − ω0)2 + ω2

1

)1/2

and the approximation (16.17) gives:

P+→−(T ) = ω21

(ω − ω0)2sin2((ω − ω0)T/2).

We notice that both formulas nearly coincide in two cases.

• If the excitation frequency is sufficiently far from resonance: |ω − ω0| ω1. Inthis case Ω � |ω − ω0| and the two results coincide for all times.

• If the excitation is close to resonance (|ω − ω0| � ω1) and if the interaction timeis short enough: ω1T/2 � 1.

16.2 Interaction of an Atom with an Electromagnetic Wave

Electromagnetic transitions play a central role in atomic and molecular physics.Three basic processes are involved. Under the influence of an electromagnetic wave,an atom or a molecule can absorb energy. If it is in an excited state, it can also decayinto a lower energy state either by spontaneous emission of radiation, or throughthe emission of radiation induced by an external electromagnetic wave. These threeprocesses were introduced in 1917 by Einstein who understood, in a remarkableintuition, how a collection of atoms and photons can reach thermal equilibrium.

408 16 The Evolution of Systems

Here we want to study the behavior of an atom in a monochromatic wave whoseelectric field is:

E(r, t) = E0 ε cos(ωt − k · r). (16.18)

This plane travelling wave has an amplitude E0, a wave vector k, and a polarizationε orthogonal to k. We want to calculate the probabilities for the two processes ofabsorption and induced emission of light by the atom. Spontaneous emission cannotbe treated quantitatively in this book, because the proper approach requires the quan-tization of the electromagnetic field. We will simply make a few qualitative remarksconcerning this latter process (see also Sect. 16.3 and Exercise 2 at the end of thischapter).

16.2.1 The Electric Dipole Approximation

Weassume that we know the energy levels of the atomic system.We note respectively|1〉 and |2〉 the ground state of energy E1 and an excited state of energy E2.Wewant tostudy here the absorption of light which results in the transition of the atom from theinitial state |1〉 to the final state |2〉. The induced emission of light can be calculatedin the same way, assuming the initial state is |2〉 and the final state is |1〉.

In order to describe this phenomenon, we consider the simple case of a one-electron atom. We note D = q r the electric dipole moment operator, which is pro-portional to the position of the external electron with respect to the core of the atom.We treat the atom as infinitely heavy, and we note R0 the position of the core. Thecoupling between the atom and the electric field (16.18) is given by:

H1(t) = − D.E(R0, t). (16.19)

This coupling is called the electric dipole interaction Hamiltonian.

16.2.2 Justification of the Electric Dipole Interaction

The complete interaction of an atom with an external electromagnetic field (E, B)deriving from the potentials (A, Φ) is obtained using the considerations developedin Chap.15. Let r i and pi be the position and momentum operators of the electrons(i = 1, . . . , Z). Assuming the nucleus of charge Z is fixed, and omitting the spinmagnetic interactions, theHamiltonian of the system in presence of external fields is:

16.2 Interaction of an Atom with an Electromagnetic Wave 409

H =N∑

i=1

1

2m

(pi − qeA(r i , t)

)2 + qeΦ(r i , t) − Zq2e

4πε0ri

+ 1

2

∑

i

∑

j �=i

q2e

4πε0|r i − r j | . (16.20)

As such, t his expression is much too complicated. In practice, one must expand(16.20) and make approximations.

In a systematic expansion of the Hamiltonian (16.20), there exist terms due to theelectric field of the incident wave, and others due to the magnetic field. We neglectthis second type of interactions. In fact, one has |B| = |E|/c for a plane wave invacuum. Since the typical velocity of an external electron in an atom is of the orderof αc ∼ c/137, i.e. much smaller than the velocity of light, the Lorentz force, andthe ensuing magnetic effects, are very small compared with the electric part. If wewere considering X rays and internal electrons, these magnetic effects would becomparable to the electric ones.

Even if we limit ourselves to the electric dipole interaction of a one-electron atom,we should keep in full rigor the dependence of the incident field on r . However,the typical extension of the electron orbit is the atomic scale (〈r〉 ∼ 1Å). This ismuch smaller than the wavelengths of a radiation corresponding to the infrared, thevisible or the ultraviolet part of the spectrum, (λ = 2π/k ≥ 103 Å). Consequentlythe variation of E with r is negligible and it is legitimate to replace E(R0 + r, t) byE(R0, t).

To summarize, the simple expression we choose for H1 in the case of a one-electron atom, is the dominant term of the interaction between the electromagneticfield (E, B), and the charge and current density inside the atom. It is the first term ofamultipole expansion which contains smaller effects, of magnetic and/or relativisticorigin.

16.2.3 Absorption of Energy by an Atom

In order to simplify the notations, we assume that the center of mass of the atom isat R0 = 0. At time t , the atomic state is:

|ψ(t)〉 = γ1(t) e−i E1t/� |1〉 + γ2(t) e

−i E2t/� |2〉 +∑

n �=1,2

γn(t) e−i En t/� |n〉

with the initial conditions: γ1(0) = 1 and γ2(0) = · · · = γn(0) = 0. Injecting theexpression (16.19) into the general result (16.13), we find:

γ2(t) = qeE02�

〈2|r · ε|1〉(ei(ω0+ω)t − 1

ω0 + ω+ ei(ω0−ω)t − 1

ω0 − ω

)(16.21)

410 16 The Evolution of Systems

with �ω0 = E2 − E1.A resonance phenomenon appears for ω ∼ ω0. In the above expression, the first

term is of the order of 1/ω = T0/2π, where T0 is the period of the exciting field(T0 ∼ 10−15 s in the optical domain). For ω = ω0 the second term increases linearlywith the interaction time t . If t T0, we can neglect the first term compared to thesecond and we obtain:

P1→2(t) = q2e E2

0

4�2|〈2|r · ε|1〉|2 y(ω − ω0, t). (16.22)

In this expression, the presence of the square of the matrix element |〈2|r|1〉|2 is ofgreat importance in order to determine which transitions are allowed, as we shall see.We also remark the presence of the function y(ω − ω0, t) (16.16). This transitionprobability has a resonant behavior in the vicinity of ω = ω0, and the width of theresonance is of the order of 1/t .

Contribution of Spontaneous Emission

At resonance, the time t must be sufficiently small so that |γ2(t)| � 1, which is anecessary condition for the perturbative approach to be valid. Also the time t has tobe much smaller than the lifetime τ of the level |2〉 due to spontaneous emission.Otherwise this process has to be taken into account in the above calculation and itgives a finite width to the resonance line (see Sect. 16.3 and Exercise 2 at the endof this chapter). We shall see in Sect. 16.2.5 that τ T0, so that it is possible tofulfill simultaneously the conditions t T0, so that (16.22) holds, and t � τ so thatspontaneous emission can be neglected.

The Concept of Photon

We remark on the result (16.22) that the transitions are important only when thefrequency of the light wave is close to a Bohr frequency of the atom: �ω = E2 − E1.This phenomenon is analogous to the photoelectric effect: an electron jumps froma state to another provided the incoming frequency is tuned to a Bohr frequency. Inthe case of the photoelectric effect, an electron is emitted and the final state belongsto the continuum of {ionized atom + electron} states.

Contrary to a common prejudice due to the chronology of the discoveries, thisprovides an explanation of the photoelectric effect although we have not quantizedthe electromagnetic field and we have not introduced the concept of a photon. Thisconcept becomes necessary when one studies the properties of radiation itself andthe spontaneous emission of radiation.

Validity of the Perturbative Treatment

The electrostatic Coulomb field seen by an electron in an atom is of the order of∼1011V/m, which is enormous compared to the electric field of a “standard” lightwave. In order to compete with the Coulomb field, one must use laser beams withan intensity of ∼1015W/cm2, which is considerable. In most usual situations, theuse of perturbation theory is justified, i.e. the external field appears as a very smallfluctuation compared with the Coulomb field.

16.2 Interaction of an Atom with an Electromagnetic Wave 411

16.2.4 Selection Rules

We now derive from (16.22) the selection rules for electric dipole absorption andinduced emission. Consider the matrix element:

〈2|r|1〉 ≡ 〈n2, �2,m2|r|n1, �1,m1〉.

In spherical coordinates, we have z = r cos θ, x ± iy = r sin θe±iϕ, i.e. the coordi-nates of r are expressed linearly in terms of r Y1,m(θ,ϕ). In the abovematrix element,the contribution of interest is the angular integral:

∫ (Y�2,m2(Ω)

)∗Y1,m(Ω) Y�1,m1(Ω) d2Ω.

Owing to the properties of spherical harmonics seen in Chap. 10, this integral isnon-zero if and only if:

�2 = �1 ± 1 and m2 − m1 = 1, 0,−1. (16.23)

This is the case for instance for the Lyman α line of hydrogen: 2p → 1s, or theresonance line of sodium: 3p → 3s; in both cases, �1 = 1 and �2 = 0. For a pair oflevels which does not fulfill (16.23), the transition is forbidden. An example is thetransition corresponding to the 21cm line of hydrogen, for which both levels havezero orbital angularmomentum (�1 = �2 = 0). The dominant coupling between thesetwo levels is a magnetic dipole interaction, whose matrix element is much smallerthan for an electric dipole coupling.

16.2.5 Spontaneous Emission

The complete calculation of spontaneous emission requires the quantization of theelectromagnetic field and we shall not treat it here. However it is interesting to givethe main results and to discuss them.

Consider an excited atomic state |i〉 which is coupled by an electric dipole transi-tion to a state | f 〉 with a lower energy. The atom prepared in the state |i〉 may decayto the state | f 〉 by emitting spontaneously a photon with an energy �ωi f = Ei − E f .One can show that the probability dPi→ f that the decay takes place during an arbitrar-ily short time interval dt is proportional to dt . Therefore one defines a probability perunit time dPi→ f /dt , which is independent of dt and which is given by the formula:

dPi→ f

dt= ω3

i f

3πε0�c3|〈i | D| f 〉|2, (16.24)

412 16 The Evolution of Systems

where D is the electric dipole moment introduced above. Since each photon carriesan energy �ωi f , the energy radiated per unit time d I/dt is therefore:

d I

dt= ω4

i f

3πε0c3|〈i | D| f 〉|2.

Wenotice that these transitions follow the same selection rules as found inSect. 16.2.4,since the same matrix element is concerned.

We can compare this result with the classical formula giving the total intensityradiated per unit time by an electric dipole of moment p(t) = P cosωt :

d I

dt= 1

6πε0c3, | p(t)|2.

After a time average over a period 2π/ω, we obtain:

d I

dt= ω4

12πε0c3P2.

We notice the analogy between the classical and quantum formulas, with thecorrespondence:

Classical QuantumFrequency ω → ωi f = (Ei − E f )/�

Amplitude P → 2 |〈i | D| f 〉|This substitution was made by Heisenberg in 1925. It was a basic ingredient of hismatrix mechanics.

Lifetime of an Atomic Level; Orders of Magnitude

Consider an assembly of N0 atoms all in state |i〉 at time 0. Since the probabilitythat a given atom decays in a time step dt is proportional to dt , the number of atomsN (t) still in the state |i〉 at time t follows an exponential decay law:

N (t) = N0 e−t/τ with

1

τ= dPi→ f

dt.

The quantity τ is called the lifetime of the level |i〉.For a monovalent atom, we know that the size a of an outer electron orbit is of

the order of �2/me2 � e2/�ω, which gives:

1

τ∼ ω3

i f

3πε0�c3q2e a

2 ∼ ωi f α3 with α = e2

�c� 1

137. (16.25)

16.2 Interaction of an Atom with an Electromagnetic Wave 413

For an optical radiation, the order of magnitude of the lifetime τ of atomic levels is10−7 to 10−9s. This is much longer than a typical Bohr period 2π/ωi f , owing to thesmallness of the coefficient α3 entering into (16.25).

16.3 Decay of a System

After studying the resonant or quasi-resonant coupling between two levels, we turnto another class of problems where an initial state is coupled to a continuum of finalstates, i.e. a collection of states whose energies are very close and can be consideredas a continuum. This is a central problem in collision physics and in the descriptionof the decay of a system.

16.3.1 The Radioactivity of 57Fe

In order to present in a concrete way the issues addressed in this section, we considerthe specific case of a radioactive nucleus. We start with cobalt 57. This has thepeculiarity that the isolated nucleus 57Co is stable, but the atom is not. A proton ofthe nucleus can absorb an electron of the internal K shell. This gives rise to what iscalled an electronic capture, or, equivalently, an inverse β reaction:

57Co + e− → 57Fe∗∗ + ν .

The atom 57Co has a lifetime of 270 days.The excited Fe nucleus 57Fe∗∗ produced in this reaction emits a first photon γ1,

of energy 123 keV, with a very short lifetime τ1 � 10−10 s. This leaves the nucleusin another excited state 57Fe∗. Then the 57Fe∗ emits a second photon γ2, of energy14 keV, with a lifetime τ2 � 1.4 10−7 s, leaving the 57Fe nucleus in its ground state:

57Fe∗∗ → 57Fe∗ + γ1 �ω1 = 123 keV57Fe∗ → 57Fe + γ2 �ω2 = 14 keV

It is possible to measure for each decay the time interval between the emissions ofthe two photons γ1 and γ2.

By convention, we denote t0 = 0 the time when the photon γ1 is detected. Wewant to calculate the probability P(t) that the nucleus decays and falls back in itsground state between the time 0 and t . Experimentally the answer to this question iswell known; the decay 57Fe∗ → 57Fe + γ obeys the exponential law:

P(t) = 1 − e−t/τ ,

414 16 The Evolution of Systems

with in the present case τ � 1.4 10−7 s. For t � τ , the probability that the systemdecays between 0 and t is proportional to t :

t � τ → P(t) � t

τ. (16.26)

The Hilbert Space of the Problem

For this example, we must consider the Hilbert space which describes the state of anironnucleus, accompanied by a certain number of photons. This is a situation differentfrom those we have met up to now. Strictly speaking it requires the formalism ofquantum field theory. Here we simply assume that some matrix elements betweenthe relevant states exist, but we will not attempt to calculate them explicitly.

There are two types of states to be considered:

• The initial state |i〉 ≡ |Fe∗〉 prepared at time t = 0 in the absence of photons (thephoton γ1, which is the signature that the Fe∗ is prepared, is absorbed by thedetector at t = 0).

• The possible final states | f 〉 ≡ |Fe + γ, E f 〉, representing the Fe nucleus in itsground state accompanied by one photon. E f represents the sum of the energiesof the γ photon and of the nucleus in its ground state. In full rigor, we must specifyalso the direction of propagation of the outgoing photon and its polarization, inorder to define | f 〉 completely.

The states |i〉 and | f 〉 are eigenstates of the Hamiltonian H0 which describesnuclear forces on one hand, and freely propagating photons on the other hand. Thesestates are not eigenstates of the coupling H1 between the nucleus and the quantizedelectromagnetic field. In particular, a nucleus prepared in the state |i〉will not remainin this state indefinitely. We want to calculate the evolution of the system assumingwe know the matrix elements 〈 f |H1|i〉.Density of Final States

For simplicity, we neglect the recoil of the nucleus. The energy of the 57Fe is thereforefixed. The emitted photon can be in a whole series of energy states, which form adiscrete set if we assume that the system is contained in a finite volume. Considersome energy band dE . Inside this band, there is a number dN of photon states. Wedefine the density of states ρ(E):

ρ(E) = dN

dE. (16.27)

This allows us to replace a discrete sumover the possible final states, by an integralover the final state energy E f , which is much easier to handle:

∑

f

−→∫

ρ(E f ) dE f .

16.3 Decay of a System 415

Although we have not defined precisely the relation between photons and theelectromagnetic field, all we need to know here is that photons are massless particles.The corresponding density of states can then be easily calculated. It is sufficient tomake use of the fact that photons of momentum p have an energy E = c | p|. As forthe non relativistic particles that we studied in Chap.4, the momenta of the photonare quantized if we suppose that the experiment takes place in a (arbitrarily large)cubic box of size L , with periodic boundary conditions:

p = 2π�

Ln, n = (n1, n2, n3) n1, n2, n3 integers. (16.28)

Combining E = cp and (16.28) we obtain:

ρ(E) = L3

2π2�3

E2

c3,

to be compared with (16.27), for non relativistic massive particles. It remains to bechecked in the end of the calculation that the predictions for any measurable quantitydoes not depend on the volume L3 of the fictitious box that we have introduced. Inthe present case this is ensured by the expression for the matrix element 〈 f |H1|i〉,which scales as L−3/2.

16.3.2 The Fermi Golden Rule

We now come back to our decay problem. The nucleus 57Fe∗ can decay in a con-tinuous set of Fe + γ states. We are not interested in the probability that it decaysin a specific state, but in the probability that it decays to some domain D f of finalstates, characterized by their direction Ω (within a small solid angle d2Ω). We musttherefore sum the formula which gives Pi→ f (16.15) on all possible final states ofthe domain D f :

d2Pi→D f (t) = 1

�2

∑

f ∈D f

|〈 f |H1|i〉|2 y(ωi f , t)

= 1

�2

∫

D f

|〈 f |H1|i〉|2 y(ω f i , t) ρ(E f ) dE fd2Ω

4π, (16.29)

with ω f i = (E f − Ei )/�. We now use the fact that, as the time t increases, thequantity y considered as a function of E f becomes more and more peaked in thevicinity of E f = Ei . Using (16.16) we obtain:

1

2πty(ω f i , t) ≈ δ(ω f i ) = � δ(E f − Ei ). (16.30)

416 16 The Evolution of Systems

We can therefore neglect the variations of ρ(E) and of thematrix element 〈 f |H1|i〉 inthe integral over E f . In other words, we extract the matrix element 〈 f |H1|i〉 and thedensity of states ρ from the integral, and evaluate them at the central point E f = Ei .Using (16.16), this leads to:

d2Pi→D f (t) = 2π

�|〈 f, E f = Ei |H1|i〉|2 ρ(Ei )

d2Ω

4πt. (16.31)

We recover the linear dependence in time observed experimentally for short times(cf. (16.26)). Let us assume for simplicity that the matrix element 〈 f, E f = Ei |H1|i〉does not depend on the direction Ω of the emitted photon. Summing over all solidangles, we deduce the lifetime of state |i〉:

1

τ= 2π

�|〈 f, E f = Ei |H1|i〉|2 ρ(Ei ). (16.32)

The fundamental relation (16.31) is called the Fermi golden rule. The range oftimes t for which it can be applied is limited by two constraints:

• The time t should be short enough so that Pi→all f (t) � 1:

t � τ . (16.33)

This is a necessary condition for the validity of first order perturbation theory.• The time t should be long enough so that the frequency width∼1/t of the function

y(ωi f , t) in (16.29) is much smaller than the typical scale of variation of the twoother terms, 〈 f |H1|i〉 and ρ. Denoting κ this scale of variation in the frequencydomain, the second constraint reads:

t−1 � κ. (16.34)

In any problem where the Fermi golden rule is used, one must check that there existsa time interval during which these two constraints are simultaneously satisfied.

16.3.3 Orders of Magnitude

Wehave already given in (16.25) the scaling laws for the lifetime of an atomic excitedlevel, which can decay by spontaneous emission with an electric dipole transition.Aside from geometric factors, this decay rate reads:

1

τ∼ α

a21ω3i f

c2= �

3ω3i f

m2ee

2c3, (16.35)

16.3 Decay of a System 417

where me is the electron mass, a1 = �2/mee2 is the Bohr radius and �ωi f is the

energy of the emitted photon.We can discuss the consistency of the Fermi Golden Rule on this example. The

frequency scale κ for the variations of 〈 f |H1|i〉 and ρ is typically κ ∼ ωi f . Therefore(16.33) and (16.34) can be simultaneously verified if:

ω−1i f � τ ⇒ �ω2

i f

m2ec

3� 1.

A typical Bohr frequency is EI /�, where EI = mee4/2�2 is the ionization energy

of the hydrogen atom. The consistency of our approach then reads:

α3 � 1.

Since α � 1/137 � 1, this inequality is well satisfied. The smallness of the finestructure constant guarantees that the perturbative treatment of the effect of electro-magnetic interactions onto the atomic levels is a good approximation.

In going from atomic systems to nuclear systems, considering the expression1/τ ∼ α a21ω

3i f /c

2,we expect that the electric dipole decay rates should be (i) reducedby a factor of order 10−10 owing to the change of size (10−15 m instead of 10−10 m),(ii) enhanced by a factor of order 1018 owing to the change of energy scale (1 MeVinstead of 1 eV).

We can therefore transpose (16.35) to the nuclear scale using R ∼ r0A1/3 for theradius of a nucleus, where r0 ∼ 1.2 fm and where A is the number of nucleons. Weobtain:

1

τ∼ α

r20 A2/3 ω3

c2. (16.36)

One can check that the energies and lifetimes of the 57Fe excited states agree accept-ably with this estimate. In particular we verify immediately that τ2/τ1 ∼ (ω1/ω2)

3 ∼103. In the case of nitrogen 13, there exists an excited state with �ω = 2.38MeV anda lifetime τ ∼ 10−15s. Using these parameters and (16.36), we obtain τ ∼ 2 10−15s,a good order of magnitude.

16.3.4 Behavior for Long Times

We have just found how the notion of lifetime for an excited atomic or nuclear levelemerges using the short-time approximation of the decay law. For longer times, firstorder perturbation theory no longer applies since we no longer have Pi→all f � 1.In this case, one can recover the measured exponential decay law using anotherapproximation due to Wigner and Weisskopf.

418 16 The Evolution of Systems

In order to illustrate this, we consider the following simple model. We assumethat the only non-vanishing matrix elements are 〈i |H1| f 〉 and 〈 f |H1|i〉:

〈i |H1|i〉 = 〈 f |H1| f 〉 = 0.

The initial state is |ψ(0)〉 = |i〉. Using the above form of the coupling, we can writethe state of the system at a later time t as:

|ψ(t)〉 = γi (t) e−i Ei t/� |i〉 +

∫γ( f, t) e−i E f t/� | f 〉 ρ(E f ) dE f . (16.37)

Here we assume for simplicity that the energy is the only quantum number whichcharacterizes the final states. We set H( f ) ≡ 〈i |H1| f 〉 (H( f ) is simply a functionof E f ), and the Schrödinger equation gives:

i�γi (t) =∫

ei(Ei−E f )t/� H( f ) γ( f, t) ρ(E f ) dE f , (16.38)

i�γ( f, t) = ei(E f −Ei )t/� H∗( f ) γi (t), (16.39)

with the initial conditions γi (0) = 1 , γ( f, 0) = 0.We integrate formally (16.39):

γ( f, t) = H∗( f )i�

∫ t

0ei(E f −Ei )t ′/� γi (t

′) dt ′, (16.40)

andwe insert this result into (16.38).We then obtain the integro-differential equation:

γi (t) = − 1

�2

∫dE f ρ(E f )

∫ t

0ei(Ei−E f )(t−t ′)/� |H( f )|2 γi (t

′) dt ′, (16.41)

which can be rewritten as:

γi (t) = −∫ t

0N (t ′′) γi (t − t ′′) dt ′′,

with

N (t ′′) = 1

�2

∫ei(Ei−E f )t ′′/� |H( f )|2 ρ(E f ) dE f .

The function N (t ′′) is proportional to the Fourier transform of the function of thefinal energyG(E f ) = |H( f )|2 ρ(E f ). By definition of a continuum, the width of thefunction G(E f ) is large. ThereforeN (t ′′) has a narrow width and it is non-vanishingonly if t ′′ is close enough to 0.We note t ′′ = τc the characteristic time abovewhich theintegrand oscillates so rapidly thatN (t ′′) is negligible. We make the approximation(to be justified a posteriori in each case) that γi (t − t ′′) varies slowly in the time

16.3 Decay of a System 419

interval 0 < t ′′ < τc. We can then replace γi (t − t ′′) by γi (t) in the right hand sideof the integro-differential equation and we obtain:

γi (t) = −γi (t)∫ t

0N (t ′′) dt ′′.

For times t large compared to τc, the upper bound of the integral can be extended toinfinity. Finally we use the relation:

∫ +∞

0dt ′′ ei(ω−ω0)t ′′ = πδ(ω − ω0) + iPP

(1

ω − ω0

),

where PP is the principal value integral, and we obtain:

γi (t) = −(

1

2τ+ iδωi

)γi (t)

with

1

τ= 2π

�|H( f )|2 ρ(Ei ) and δωi = PP

(∫ |H( f )|2Ei − E f

ρ(E f ) dE f

). (16.42)

The differential equationwhich gives the evolution of γi (t) is integrated immediately.This gives the probability that the system has decayed at time t :

P(t) = 1 − |γi (t)|2 = 1 − e−t/τ , (16.43)

i.e. the exponential law.One can check that the value for τ derived in (16.42) coincideswith the value (16.32) calculated previously using perturbation theory.

The quantity �δωi corresponds to an energy shift of the excited state due tothe coupling of the nucleus and the electromagnetic field. This shift is exactly thesame as what one obtains in second order time-independent perturbation theory (cf.(9.21)). Note that to first order the energy-shift vanishes because of our assumptionsconcerning the diagonal elements of H1. In the case of atomic levels, this secondorder shift is called the Lamb shift (see Chap. 13, Sect. 13.2.1).

We now insert the result for γi (t) in the Eq. (16.40) giving γ( f, t). We obtain theenergy distribution of the final states:

p(E f ) = |γ( f, t = ∞)|2 = |H( f )|2 1

(E f − Ei )2 + Γ 2/4, (16.44)

where we have set Ei = Ei + �δωi and Γ = �/τ . If we assume that |H( f )|2 variesslowly, this probability law is a Lorentz function, centered at Ei , with a full width athalf maximum Γ = �/τ (see Fig. 16.2). In other words the energy of final states ison the average Ei with a dispersion ΔE :

420 16 The Evolution of Systems

Fig. 16.2 Energydistribution of the photon γ2emitted in the decay of 57Fe∗

ΔE = Γ/2 = �/2τ . (16.45)

This dispersion in energy of the final state is characteristic of any unstable system:beta decay of nuclei, radiative decay of atomic states, etc. It originates from thefact that the initial state |i〉 is an eigenstate of the Hamiltonian in the absence ofinteraction, but it is not an eigenstate of the full Hamiltonian. Therefore, this initialstate does not have a well-defined energy.

16.4 The Time-Energy Uncertainty Relation

One of the great controversial questions in the 1930’s concerned the time-energyuncertainty relation:

ΔE Δt ≥ �/2. (16.46)

Although this relation is commonly accepted, its interpretation varies considerablyfrom one author to the other. It is indeed quite different from the uncertainty relationsthat we derived in Chap. 8. The relation Δx Δp ≥ �/2 for instance, follows directlyfrom the principles and the commutation relation of the operators x and p. It istherefore an intrinsic property of any system. On the opposite, in the Schrödingerequation, time is not an operator, but a parameter which has a well defined value inthe equations. Although we can measure it physically, time is not an observable.

We will not give an exhaustive review of all points of view, neither will we adoptone attitude rather than another.1 We simply wish to make a few observations whichcan be used as a starting point for further reflection.

1See for instance the article of Aharonov and Bohm, Physical Review, vol. 122, p. 1649, (1961).

16.4 The Time-Energy Uncertainty Relation 421

16.4.1 Isolated Systems and Intrinsic Interpretations

Werecall some results presented in the first Chapters for systemswhoseHamiltoniansdo not depend on time.

Stationary States

These are eigenstates of the energy, whose time evolution reduces to a multiplicativeglobal phase factor: |ψ(t)〉 = e−i Et/�|ψ(0)〉. If the system is prepared in such a state,the expectation value 〈a〉 of any observable A does not change with time. This agreeswith the relation (16.46):

An isolated system whose energy is well-defined (ΔE = 0) does not evolve fromt = −∞ to t = +∞.

Evolution of a System

The state of a system |ψ(t)〉 can be a superposition of two ormore energy eigenstates.For instance in Chap. 2, we constructed a wave packet as an infinite sum of stationarystates. Such a system does not have a well defined value of the energy, and theexpectation values of observables evolve with time, except if they correspond to aconserved quantity.

Consider a physical quantity A associated with the system, such as the positionof the needle on a wrist watch. We denote 〈a〉t and Δa the mean position and themean-square deviation of this quantity at time t . Let v = d〈a〉t/dt be the velocityassociated to 〈a〉t . The characteristic time τ it takes the “wave packet” to cross acertain point, for instance a = 0, is τ = Δa/|v|. In Chap. 8 we proved the followingproperties:

• Δa is related to the energy dispersion ΔE by:

Δa ΔE ≥ 1

2|〈ψ|[ A, H ]|ψ〉|,

where |ψ〉 is the state of the system at time t .• v is given by:

v = d〈a〉tdt

= 1

i�〈ψ|[ A, H ]|ψ〉.

Combining these two relations, we obtain:

τ ΔE = Δa

|v| ΔE ≥ �

2.

We recover a relation similar to (16.46), and it appears here as an intrinsic property ofthe quantum system: the larger the dispersionΔE on the energy, the shorter the char-acteristic time of evolution of any quantity. This formulation is due to Mandelstammand Tamm.

422 16 The Evolution of Systems

Decay of an Unstable System

We have seen in Sect. 16.3.4 above that when a system is unstable and decays, itsenergy is not well defined. The energy distribution of the final products is peakedaround some value with a dispersion related to the lifetime τ by ΔE = �/(2τ ). Thisis also of the form (16.46), and it is again an intrinsic property of the system.

16.4.2 Interpretation of Landau and Peierls

This interpretation2 comes from the analysis of the measurement of the energy Eof a system. In order to perform such a measurement, we must couple the system ofHamiltonian Hs , to a detector of Hamiltonian Hd . The detector is initially in a stateof known energy εd , and the coupling takes place for a lapse of time T . When thecoupling is switched off, the state of the set system + detector is a superposition ofeigenstates of Hs + Hd , with an average energy E ′ + ε′

d close to E + εd , up to anuncertainty �/T :

|E ′ + ε′d − E − εd | ∼ �

T.

This results from the shape of the function y(ω = E/�, t) introduced in Sect. 16.1.Suppose that we know precisely the initial and final energies of the detector εd andε′d . We can therefore deduce the uncertainty on E ′ − E :Δ(E ′ − E) � �/T . In otherwords, even if the system is in a well-defined energy state before the measurement,an observer has access to this value only up to an uncertainty �/T .

16.4.3 The Einstein–Bohr Controversy

In 1930, Einstein presented the following argument. A clock is placed in a box,hanging on a spring. It is set to open a shutter at time t1 and to close it at timet2 = t1 + T , the interval T being determinedwith great accuracy. A radiation escapesfrom the box when the shutter is open and we measure the corresponding energy Eby weighing the box before and after the experiment (E = δm c2). Since we have allour time to do this weighing, it can be very precise. Therefore this procedure seemsto be a counter-example to the relation ΔE T ≥ �/2. Bohr disproved the argumentin the following way:

(1) The position of the box which contains the clock is defined up to some quantumuncertainty Δz. Since the clock is placed in a gravitational field, its rate depends onthe gravitational potential and, owing to general relativity, there is an uncertainty:

2See, for instance, L. Landau and E. Lifshitz,QuantumMechanics (Pergamon Press, Oxford, 1965).

16.4 The Time-Energy Uncertainty Relation 423

ΔT

T= g Δz

c2(16.47)

on how long the shutter stays open.

(2) At time t2, the determination of the decrease of the weight δm g = Eg/c2 of thebox is made by measuring the momentum the box acquires during this time intervalT :

pz = δm g T = Eg

c2T .

Here we assume that T is shorter than the oscillation period of the spring; a similarargument could be elaborated in the reverse case. Owing to the quantum uncertaintyΔpz on the initial momentum of the box, the accuracy on the measurement of theenergy is:

ΔE = c2

gTΔpz . (16.48)

Combining the two Eqs. (16.47) and (16.48) and using Heisenberg’s inequalityΔz Δpz ≥ �/2 for the position and the initial momentum of the box, we recover thedesired inequality. The story says that Bohr, who had spent an entire night to find thiscounter argument, was quite proud of using Einstein’s General Relativity to solvethe problem.

16.5 Exercises

1. Excitation of an Atom with Broad Band Light

Consider n two-level atoms driven by the electric field E(t) = E0 ez f (t) cosωt ,where f (t) is a function which is zero outside the interval [−τ , τ ]. We consider anelectric dipole coupling between the atoms and the field. The ground and excitedatomic states are denoted a and b respectively, and we set by convention Ea = 0and Eb = �ω0. We suppose that (i) the typical scale of variation of f (t) is very largecompared with the period 2π/ω and (ii) the excitation frequency ω/2π is close tothe Bohr frequency ω0/2π. One will neglect the contribution of non resonant terms.

a. We define �Ω1 = −dE0 with d = 〈b|D · ez|a〉 (Ω1 is supposed to be real) andwedenote g(Ω) the Fourier transform of f (t). Using perturbation theory, calculatethe average number of excited atoms at time τ . This number is denoted nb(τ ).

b. The electric field now consists in a succession of wave packets:

E(t) = E0 ez∞∑

p=1

f (t − tp) cos(ω(t − tp)) with t1 < t2 < · · ·

Consider T such that t� + τ < T < t�+1 − τ . Calculate nb(T ).

424 16 The Evolution of Systems

c. We suppose that the successive wave packets arrive in a randomway, with γ wavepackets per unit time in average. We note nb(T ) the statistical mean of nb(T ).Calculate nb(T ) for γT 1. Show that one can define a transition probabilityper unit time from level a to level b. This quantity will be denoted Γa→b.

d. We put w(ω + Ω) = (ε0c/2) E20 γ |g(Ω)|2 and we denote Φ the incident flux of

energy. Relate w and Φ. Express Γa→b in terms of w(ω0).e. We suppose now that all atoms are initially in the state b. How can on transpose

the previous reasoning?f. Write the evolution equations for the mean populations na(t) and nb(t). What is

the steady state of the system?

2. Atoms in Equilibrium with Black-Body Radiation

We consider again themodel of the previous exercise and we suppose that the atomicassembly is irradiated by the radiation of a black body of temperature T . We recallthat one has in this case:

w(ω) = μω3

e�ω/kBT − 1,

where μ depends only on fundamental constants. What must one add to the previousmodel in order to ensure the consistency of Statistical Physics (Einstein, 1917)?

3. Ramsey Fringes

A neutron (which is a spin 1/2 particle) propagates along the z axis. We denote|±〉 the eigenstates of the operator Sz , projection of the neutron spin on the z axis.The neutron is initially prepared in the state |+〉 and it crosses two radio-frequencycavities of length L , separated by a distance D L (Fig. 16.3). One applies in eachcavity a rotating magnetic field:

B1 = B1(cosωt ux + sinωt uy).

The whole experimental setup is placed in a constant and uniform magnetic fieldB0 parallel with the z axis. The motion of the neutron is treated classically as a uni-form linear motion with velocity v. We are interested only in the quantum evolution

Fig. 16.3 Experimental setup for the observation of Ramsey fringes

16.5 Exercises 425

of the spin state of the neutron. The magnetic moment operator of the neutron isdenoted μ = γ S and we set ω0 = −γB0 et ω1 = −γB1.

Calculate at first order in B1 the probability amplitude to find the neutron in thespin state |−〉 at the output of the device. Show that the spin flip probability variesrapidlywith the detuningω − ω0, and that one can determine the resonance frequencyω0 with a much better precision than if only one of the two cavities is used.

16.6 Problem. Molecular Lasers

Preliminaries

Laser operation is based on population inversion, i.e. a situation where an excitedenergy level e of a quantum system is more populated than a lower level g. This canindeed lead to the amplification of radiation resonant with the e − g transition. Westudy here a possible way to achieve population inversion in amolecule. The physicalprinciple of the inversion is based on the large difference between the time scale forthe relaxation of the vibrations of the molecule, and the lifetime of its electronicallyexcited state.

a. Consider a one-dimensional problem and a wave function ψ(x) which can beexpanded in a Taylor series. Show that the operator T (d) = e−id p/�, where d isa length and p is the momentum operator, is such that:

T (d) ψ(x) = ψ(x − d)

Note: the expansion eiu = ∑∞n=0 (i u)

n/n! is mathematically legitimate.

b. Consider a one dimensional harmonic oscillator ofmassM and frequencyΩ/(2π)

centered at x = 0. Its energy eigenstates are denoted |n〉. Show that T (d) =eα(a−a†) where a† and a are the usual creation and annihilation operators. Calculatethe real constant α.

c. Let |n〉d be the eigensates of the same oscillator, but centered at x = d. Show thatthe ground state |0〉d can be expanded on the states |n〉 as

|0〉d = e−λ2/2∞∑

n=0

λn

√n! |n〉 with λ = d

√MΩ

2�. (16.49)

The following identity is useful:

eα(a−a†) = e−α2/2 e−αa† eαa .

426 16 The Evolution of Systems

Molecular Lasers

Weconsider amolecule formed by a core of two nuclei, each accompanied by internalelectrons, surrounded by a single external electron. The position r = {x, y, z} of theelectron is taken with respect to the centre of gravity of the core.

In good approximation, the external electron is placed in an effective potentialV (r) and has a series of energy levels εn and corresponding wave functions ψn(r).

The core, consisting in the two nuclei and the internal electrons, has both vibra-tional and rotational motions. For simplicity we consider here only the vibrationaldynamics, that can be approximated by a harmonic one-dimensional motion. Wedenote u the distance between the nuclei, M the reduced mass andΩ/(2π) the oscil-lation frequency. The key point in the following is that the mean distance of the twonuclei depends on the state of the outer electron.

In the electronic ground state of energy ε1, the eigenfunctions of the molecularHamiltonian are ψ1(r) φn(u), where φn(u) is a solution of the Schrödinger equation

− �2

2M

d2

du2φn(u) + 1

2MΩ2(u − b)2φn(u) = En φn(u).

Similarly, in the first excited state of the outer electron, of energy ε2, the eigenfunc-tions are ψ2(r)χm(u) with

− �2

2M

d2

du2χm(u) + 1

2MΩ2(u − c)2χm(u) = Emχm(u).

As mentioned above, the constants b and c, which can be determined experimentallyor in a more sophisticated theoretical approach, are different.

a. What are the total energy levels (i.e. electronic plus vibrational energy) of themolecule, corresponding to the above states?What is the relation between the functions φn(u) and χn(u)?

b. We now study the electromagnetic transitions between the abovemolecular states.Themolecule is placed in an oscillating electric field F, polarized along the z axis,of angular frequency ω, and of amplitude F . The dominant part of the interactionHamiltonian is H1 = −qzF cosωt , i.e. the dipole electric interaction with theexternal electron. This interaction does not depend on the variable u.Assume the molecule is initially (t = 0) in its ground state (both electronic andvibrational). Using first order perturbation theory, write the transition probabilityto an arbitrary final state at time t .Show that a discrete number of transitions are allowed. Give the value ωn of theangular frequency of the external field corresponding to these allowed transitions,and give the wave functions of the corresponding excited states.

c. Show that the probability P(ωn) for a field of frequency ωn/(2π) to excite themolecule factorizes into the product of an electronic excitation probability, and aprobability pn to excite the vibrational states of the nuclei. Show that

16.6 Problem. Molecular Lasers 427

pn = |An,0|2 where An,0 =∫

χ∗n(u) φ0(u) du.

Making use of the results obtained in the first section, calculate An,0. What typeof probability law is pn?What is the value n0 = 〈n〉 for which the excitation of the molecular vibrationoccurs on the average? What is the root mean square deviation Δn?

d. Calculate n0 and Δn for a molecule where M = 4 10−27 kg, Ω = 2 1014 s−1,b = 2Å, c = 3Å.

e. In molecular physics, one frequently uses a principle based on semi-classicalarguments, called the Franck–Condon principle, which states that when a tran-sition occurs between two different electronic levels, “the distance between thenuclei does not change”. To be more specific, the electromagnetic transition issufficiently fast so that, if the nuclei were classically bound objects, neither theirmomenta nor their positions would change during the time interval of the transi-tion.Consequently, only the potential energy of the nuclei changes. This can be takeninto account in a quantum mechanical calculation by assuming that, when thetransition occurs, the parameter b suddenly changes into c, the motion remainingharmonic.Within this model, calculate in terms of b and c the classical variation of thevibration energy of the nuclei, assuming they are at rest initially.Compare the result with the calculation of question 3, and show that this com-parison provides a justification of the Franck–Condon principle.

f. Suppose there exists a very rapid relaxation mechanism in the excited electroniclevel so that the electronically excited molecules fall in the state of smallestvibrational energy ψ2(r)χ0(u). Towards which sub-levels ψ1(r)φn(u) will elec-tric dipole transitions occur preferentially?What are the corresponding emission angular frequencies ω′

n?Let n′

0 = 〈n′〉; calculate with the data of question 4 the Boltzmann factorN (En′

0)/N (E0) = exp(−(En′

0− E0)/kT ) at room temperature (kT ≈ 0.025eV).

Assuming that ε2 > ε1 + En′0, and that roughly one or two percent of the mole-

cules are excited by the oscillating field from the ground state of ε1 to the excitedstates of ε2, can you explain why such a system is well suited for achieving a lasersource?

16.6.1 Solution

Section 1

a. The proof is straightforward. We have:

p = �

i

∂

∂x⇒ T (d) = e−d ∂

∂x .

428 16 The Evolution of Systems

Therefore

T (d) ψ(x) =∞∑

n=0

(−d)n

n!

(∂

∂x

)n

ψ(x),

which is simply the Taylor expansion of ψ(x − d).b. The creation and annihilation operators are defined as

a = 1√2(X + i P), a† = 1√

2(X − i P)

with P = p/√

�MΩ . Therefore:

p = −i

√�MΩ

2

(a − a†

)and T (d) = exp

(−d

√MΩ

2�(a − a†)

)

which is the desired result, with

α = −d

√MΩ

2�.

c. We have|0〉d = T (d)|0〉 = eα(a−a†)|0〉 = e−α2/2e−αa†eαa|0〉.

with α = −λ. In addition:

a|0〉 = 0 ⇒ eαa|0〉 = |0〉 (a†)n |0〉 = √n! |n〉.

Therefore:

|0〉d = e−λ2/2∞∑

n=0

λn

(a†

)n

n!|0〉 = e−λ2/2

∞∑

n=0

λn

√n!

|n〉

Section 2

a. The energy levels of the total system (core+external electron) are

E (1)n = ε1 + (n + 1/2)�Ω

E (2)n = ε2 + (n + 1/2)�Ω.

Let { fn(u)} be the wave functions of a harmonic oscillator centered at the originu = 0. We have

φn(u) = fn(u − b) χn(u) = fn(u − c) φn(u) = χn(u − (b − c))

therefore

16.6 Problem. Molecular Lasers 429

φn(u) = T (b − c) χn(u).

b. The transition probability from the initial state i , of wave function ψ1(r) φ0(u),to a final state f is at lowest order in F :

Pi→ f (t) = q2F2

4�2|〈 f |z|i〉|2 sin2

(ω−ω02 t

)(

ω−ω02

)2

where we have setω0 = (E f − Ei )/�.

Consider the matrix element 〈 f |z|i〉. It can be written as

〈 f |z|i〉 =∫

ψ∗f (r) z ψ1(r) d3r

∫ζ∗f (u) φ0(u) du

where ζ f is some vibrational state of the nuclei. Therefore, the transitions insidethe same electronic level are doubly forbidden:

• First,∫ |ψ1(r)|2 z d3r = 0 since the electronic wave function has generally a

well defined symmetry (odd or even) with respect to the z axis.• Secondly one has:

∫φ∗n(u) φ0(u) du = δn,0.

The only transitions allowed a priori are those between the electronic levels ε1and ε2. The corresponding values of ωn are given by

�ωn = ε2 − ε1 + n�Ω n = 0, 1, . . . .

The wave function correponding to the excited level ωn is ψ2(r) χn(u).c. We have at resonance:

Pi→ f (t) = Pe pn t2

with

Pe = q2F2

4�2

∣∣∣∣∫

ψ2(r)∗ z ψ1(r) d3r

∣∣∣∣2

and pn = |An,0|2 with An,0 =∫

χ∗n(u) φ0(u) du.

From the results of the first section, we obtain

φ0(u) = χ0(u − d) = e−λ2/2∞∑

n=0

λn

√n!

χn(u)

with d = b − c and:

430 16 The Evolution of Systems

λ = (b − c)

√MΩ

2�.

Since the χn are orthogonal, we therefore obtain

An,0 = e−λ2/2 λn

√n!

, pn = |An,0|2 = e−λ2 λ2n

n!

which is a Poisson law.The mean value and root mean square deviation are n0 = 〈n〉 = λ2 =(b − c)2MΩ/(2�) and Δn = √

n0 = |λ|.d. For c − b = 1Å= 10−10 m, M = 4 10−27 kg, Ω = 2 1014 s−1, we obtain

n0 � 38 Δn � 6.

e. A classical oscillator centered on u0 has an energy:

E = p2

2M+ 1

2MΩ2(u − u0)

2 (u0 = b or c).

The initial state is: p = 0, u = u0 = b, Ei = ε1, and the final state, if p and uhave not changed, has an energy:

E f = ε2 + 1

2MΩ2(b − c)2.

From the Franck–Condon principle, the energy of the final state should be E f . Inthe quantum calculation, we have 〈E f 〉 = ε2 + (n0 + 1/2)�Ω and therefore:

〈E f 〉 = ε2 + 1

2(b − c)2MΩ2 + �Ω

2

which is precisely, up to the zero-point energy term 1/2, the classical result. Notethat since n0 and Δn are large, the 1/2 does not play any crucial role.

f. The calculation of Sects. 16.2 and 16.3 can be transposed symmetrically for emis-sion. The probability to emit on the transition

ψ2(r)χ0(u) −→ ψ1(r) φn(u)

at the frequency ω′n = ε2 − ε1 − n�Ω , is proportional to |An,0|2 and it is maxi-

mum for the sub-energy levels n′0 ∼ 38 of the electronic ground state.

The Boltzmann factor N (E38)/N (E0) ∼ 10−87 is extremely small at room tem-perature. The population of the sublevels n ∼ 38 of the electronic ground state atroom temperature is negligible.

16.6 Problem. Molecular Lasers 431

Fig. 16.4 Principle of a molecular laser: the excitation by the oscillating field followed by a rapidrelaxation mechanism to ψ2(r) χ0(u) generates the required population inversion

The process under consideration therefore allows to achieve a population inver-sion between the vibrational ground state of the electronically excited manifoldE (2)0 and the excited states of the electronic ground state manifold E (1)

n (withn ∼ 38) since an appreciable fraction of the molecules will have been excited toE (2)0 by the incident radiation (see Fig. 16.4). This population inversion can be

used to generate stimulated emission, and, therefore, to create a laser oscillation.

Chapter 17Entangled States. The Way of Paradoxes

The way of paradoxes is the way of truth.To test Reality we must see it on the tight-rope.

When the Verities become acrobats, we can judge them.Oscar Wilde, The portrait of Dorian Gray

At the end of Chap. 3 we mentioned Einstein’s revolt against the probabilistic aspectof quantum mechanics and the uncertainty relations. As we said, Einstein was wor-ried about two aspects. One is the notion of a complete description of reality. Hethought that a complete description is possible in principle, and that the probabilisticdescription is simply easier to handle. The other aspect is the notion of determinism:same causes produce same effects. Einstein wrote the worldwide famous:

The theory produces a good deal but hardly brings us closer to the secret of the Old One.I am at all events convinced that He does not play dice.1

which is usually contracted to “God does not play dice.” Because this theoryworks, itmust be an intermediate step toward an underlying, more sophisticated theory (whichcould, for instance, involve “hidden variables”2) that is not available at present, andwhere we have to perform averages that lead to the present version of quantummechanics.

1“Die Theorie liefert viel, aber dem Geheimnis des Alten bringt sie uns kaum näder. Jedenfalls binich überzeugt, dass Der nicht würfelt.”2Again, Einstein never used that word.

© Springer International Publishing Switzerland 2016J.-L. Basdevant, Lectures on Quantum Mechanics,Graduate Texts in Physics, DOI 10.1007/978-3-319-43479-7_17

433

434 17 Entangled States. The Way of Paradoxes

17.1 The EPR Paradox

In 1935, Einstein, Podolsky, and Rosen pointed out, in a celebrated paper,3 a paradoxthat has stirred the world of physics since then. Its starting point is a “gedankenexperiment.” The spirit of the original version can be presented in a relatively simpleform. Consider a particle that decays into two particles. One of them comes in mydirection, the other in yours. One can manage to make the momentum P = p1 + p2of the initial particle, that is, the total momentum, as well defined and as small asone wishes. For simplicity, we assume it vanishes. We therefore do not know exactlywhere the particle is, but this is not a problem; we assume we use large enoughdetectors. Similarly, we can measure with all wanted accuracy the relative positionof the final particles r = r1 − r2, inasmuch as P and r commute.

On the other hand, one can prove quite generally that in the decay, there is conser-vation of the total momentum. If P = 0, the two final particles have exactly oppositemomenta.

Let’s check this experimentally. On a large number N of decays (all of which aresuch that both counters are activated), we place detectors sufficiently far from eachother so that no information can be transmitted between both detectors when theymeasure their respective signals. This amounts to saying that the relative position islarge enough. We proceed and perform, say, 1000 measurements. We notice after-wards, by comparing the set of time-ordered data, that systematically if you find avalue p of the momentum of your particle, I always find − p for my particle. Wehave checked momentum conservation.We become convinced that we can repeat theoperation as many times as we wish; we will always come to the same conclusion.

After a while, say event number 1001, I’m fed up and I let you measure themomentum of your particle while I measure the position of mine with as great anaccuracy as I wish. After that measurement, I call you on the phone; you tell methe value p1 of the momentum that you have found. Therefore, I know exactly boththe position r1, measured by me and the momentum − p1, measured by you, ofmy particle with as great an accuracy as I wish, and this is in contradiction withHeisenberg’s inequalities.

Notice, and it is important, that this scheme works because I have used a systemsuch that I have information on the quantum state of the set of the two particles inorder to obtain physical information from you on my particle.

One says that the state of the particles is entangled or correlated. This means thatinformation on one of the particles is directly connected to information on the other,wherever they are located in space. However, EPR did not coin the word entangle-ment, nor did they generalize the special properties of the state they considered.Following the EPR paper, Erwin Schrödinger wrote a letter (in German) to Einsteinin which he used the word Verschränkung (translated by himself as entanglement)to describe the correlations between two particles that interact and eventually getseparated in space as in the EPR experiment.

3A. Einstein, B. Podolsky, and N. Rosen, “Can quantum-mechanical description of physical realitybe considered complete?” Phys. Rev. 47, 777 (1935).

17.1 The EPR Paradox 435

The following assertion is a key statement of the EPR article:

If, without in any way disturbing a system, we can predict with certainty (i.e.with a probability equal to unity) the value of a physical quantity, then thereexists an element of physical reality corresponding to this physical quantity.

Notice the implicit and fundamental role of locality4 in the argument. Your mea-surement, which is done at a time and distance such that one cannot transmit anyinformation to my particle, cannot by any means perturb the results of my measure-ments. Einstein, Podolsky, and Rosen claim that no reasonable definition of realitycan allow the contrary.

Their conclusion is that the description of reality by quantum mechanics is notcomplete.

17.2 The Version of David Bohm

There are, however, some loopholes in the gedanken experiment of EPR, and theexperimental verification seemed problematic.

What leads to measurable criteria is the remarkable version of the EPR argumentgiven by David Bohm5 in 1952. This presentation is more convenient to work withand to treat mathematically than the initial version, although it is basically equivalentfrom the conceptual point of view.

Suppose we prepare two spin 1/2 particles a and b in the singlet spin state:

|Ψs〉 = 1√2

(|a : +z ; b : −z〉 − |a : −z ; b : +z〉) . (17.1)

Particle a is detected by Alice, who measures the component of its spin along an axisof unit vector ua (Fig. 17.1); similarly, particle b is detected by Bob who measuresits spin component along an axis of unit vector ub.

Alice and Bob’s measurements are strongly correlated.If they both agree to make only one measurement of the projection of the spin of

their particles along a common axis, each of them has a probability 1/2 to find+�/2or −�/2.

Here, the law of interest is angular momentum conservation. The total angularmomentum is zero in the singlet state. All its components are zero.

Let us assume first that Alice and Bob both choose the same axis z to do theirmeasurements: ua = ub = ez . The argument presented in the introduction applies:

4In the present context, locality means that some action at a point in space can have a detectableeffect only at some other point in space within the light cone of that action. At a distance r , onemust wait for a time at least equal to r/c in order to observe such an effect, it cannot be immediate.5Bohm, D. (1951).Quantum Theory, Prentice-Hall, Englewood Cliffs, page 29, and Chap.5 Sect. 3,and Chap.22 Sect. 19.

436 17 Entangled States. The Way of Paradoxes

Fig. 17.1 Gedanken experiment corresponding to the EPR argument. Two spin 1/2 particles a andb are prepared in the singlet state. Alice measures the component of the spin of particle a along anaxis ua . Bob measures the component of the spin of particle b along an axis ub

with a probability 1/2, Alice will find +�/2 and Bob will find −�/2, and with thesameprobability 1/2,Alicewill find−�/2 andBobwill find+�/2.Alice andBobcannever obtain the same result. There is a perfect correlation, or rather anticorrelation,of the two results.

This will occur if they decide to make their measurements along any axis (samefor both) since |Ψs〉 is a state of total spin zero, invariant under rotations. For instancein the basis of spin eigenstates along the x-axis. Using

| ± z〉 = 1√2

(| + x〉 ± | − x〉) , (17.2)

the singlet state (17.1) is written as

|Ψs〉 = 1√2

(|a : +x ; b : −x〉 − |a : −x ; b : +x〉) . (17.3)

Such correlations appear frequently in daily life. Suppose we have two cards,one is red and the other one is yellow. We place each of them in a sealed envelope,we mix the envelopes at random in a closed box, and we give one of them to Aliceand the other one to Bob. When Alice opens her envelope, she sees the color of hercard (red with a probability 1/2, yellow with a probability 1/2). There is obviouslya perfect anticorrelation with Bob’s subsequent result. If Alice’s card is red, Bob’scard is yellow and vice versa. There is no paradox in these anticorrelations: the factthat Alice looks at the color of her card does not affect the color of Bob’s card.

According to the EPR claim above, there is an element of physical reality asso-ciated with the color of Bob’s card, because, without perturbing it in any manner,one can determine the color of this card by simply asking Alice what her result is.Similarly, there is an element of physical reality associated with the component Sbz ,because without perturbing in any manner particle b, one can determine the value ofSbz that one would measure in an experiment: it is sufficient to ask Alice to measurethe component Saz and to tell Bob the result. If Alice finds +�/2, Bob is sure to find−�/2 by measuring Sbz , and vice versa.

Actually, the EPR argument goes one step further. We can transpose theargument about the z-axis to the x-axis, therefore there must also exist an element

17.2 The Version of David Bohm 437

of physical reality associated with the component Sbx of particle b. If one preparesa two-particle system in the singlet state, Bob can determine the component Sbxwithout “touching” particle b. It is sufficient for him to ask Alice to measure thecomponent Sax and to tell him her result. Although the term “element of physicalreality” is somewhat vague up to this point, we feel that we are reaching interest-ing grounds. In fact, the observables Sbx and Sbz do not commute. How can theysimultaneously possess this element of physical reality? (Notice that in this debate,no attention is paid to what one could do with that information experimentally insubsequent operations.)

Obviously, the above argument is contrary to the basic principles of quantummechanics. When particles a and b are in an entangled state, such as the singletstate, it is risky to claim that one doesn’t “act” on particle b when performing ameasurement on a. Taken separately, particles a and b are not in well-defined states;only the global system a + b is in a well defined quantummechanical state. It is onlyfor factorized states, of the type

|Φ〉 = |a : +α ; b : −α〉 ≡ |a : +α 〉|b : −α〉 (17.4)

that the EPR argument can be applied safely. However, in that case there is noparadox: a measurement on a gives no information on a measurement that would beperformed on b.

At this stage, we can have either of the following attitudes. We can stick to thequantum description that bears this paradoxical nonlocal character: the two particlesa and b, as far as they may be from each other (a on Earth, b on the moon), do nothave individual realities when their spin state is an entangled state. It is only afterAlice (on the Earth) has measured Saz that the quantity Sbz (for the particle on themoon) acquires a well-defined value.6 However, if a measurement on Earth affectsinstantaneously a measurement on the moon, there is something we do not reallyunderstand in the theory.

On the contrary, we can adopt the point of view of Einstein, and hope that someday one will find a more “complete” theory than quantum mechanics. In that theory,the notion of locality will have the same meaning as it has in classical physics, andso will the notion of reality.

17.2.1 Bell’s Inequality

In 1964, John Bell, an Irish physicist working at CERN, made a decisive theoreticalbreakthrough.7 This allowed to carry the debate between two radically antagonisticconceptions of the physical world onto experimental grounds.

6One can check that this formulation does not allow the instantaneous transmission of information.In order to see the correlations with Alice’s result, Bob must ask Alice what her result is, and thecorresponding information travels (at most) at the velocity of light.7J.S. Bell, Physics 1, 195 (1964).

438 17 Entangled States. The Way of Paradoxes

Bell’s formulation is the following. Suppose the super-theory that Einstein washoping for exists, and suppose it involves “hidden variables” in the following sense.For any pair (a, b) of the EPR problem described above, that theory involves a para-meterλ that determines completely and ahead of time the results of themeasurementsof Alice and Bob. For the moment, we know nothing about the parameter λ, whichis absent in an orthodox quantum description.

We denote Λ the manifold in which the parameter λ evolves. In the super-theoryframework, there must exist a function A(λ, ua) = ±�/2 for Alice and a functionB(λ, ub) = ±�/2 for Bob, which give the results of their measurements. Theseresults therefore depend on the value of λ: for instance, if λ pertains to some subsetΛ+(ua), then A(λ, ua) = �/2; if λ is in the complementary subset Λ − Λ+(ua),then A(λ, ua) = −�/2. Locality plays a crucial role in the previous assumptions.In fact, we have assumed that the function A depends on the value of λ and on thedirection of analysis ua chosen by Alice, but not on the direction of analysis ub

chosen by Bob.The parameterλ of the super-theory varies fromone pair (a, b) to another,whereas

in quantum mechanics, all pairs are prepared in the same state |Ψs〉 and nothing canmake any difference between them. This parameter is therefore not accessible toa physicist who uses quantum mechanics: it is a hidden variable. All the beauty ofBell’s argument is to prove that there exist strong constraints on the theorieswith localhidden variables, and that these constraints can be established without any furtherassumptions than the ones given above. Notice that all correlations encountered indaily life can be described in terms of hidden variable theories. On the previousexample of cards with different colors, the hidden variable comes from the shufflingof the cards. If a careful observer memorizes the motion of the cards in this shuffling,they can predict with probability 1 the result of Alice (red or yellow) and that of Bob(yellow or red).

In order to get to Bell’s result, we introduce the correlation function E(ua, ub).Quite generally, considering two random variables x and y, corresponding to a

probability density p(x, y), one defines the linear correlation coefficient r(x, y) by

r = 〈xy〉 − 〈x〉〈y〉Δx Δy

.

If the two variables are correlated, that is x = ay + b, then r = ±1 (according to thesign of a) and, if they are independent, that is p(x, y) = p1(x)p2(y), then r = 0.

The function E(ua, ub) is equal to the expectation value of the product of theresults of Alice and Bob, for given directions of analysis ua and ub, divided by �

2/4in order to have a dimensionless quantity. Whatever the underlying theory, one hasthe following property:

|E(ua, ub)| ≤ 1. (17.5)

Indeed, for each pair the product of Alice’s and Bob’s results is ±�2/4.

For a hidden variable theory, the function E(ua, ub) can be written as

17.2 The Version of David Bohm 439

E(ua, ub) = 4

�2

∫P(λ) A(λ, ua) B(λ, ub) dλ, (17.6)

where the function P(λ) describes the (unknown) distribution law of the variable λ.The only constraints on P are:

for any λ , P(λ) ≥ 0 , and∫

P(λ) dλ = 1. (17.7)

Here we assume that the functionP(λ) does not depend on the directions of analysisua and ub. Indeed these directions can be chosen by Alice and by Bob after the pairwith hidden parameter λ has been prepared. (The derivation given here is actuallydue to Clauser, Horne, Shimony, and Holt.8)

In the framework of quantum mechanics, one can check that the value of thefunction E(ua, ub) is:

E(ua, ub) = 4

�2〈Ψs |(Sa .ua) (Sb.ub)|Ψs〉 = −ua .ub. (17.8)

Bell’s theorem can be stated in the following way.

Theorem 8 1. For a local hidden variable theory, the quantity:

S = E(ua, ub) + E(ua, u′b) + E(u′

a, u′b) − E(u′

a, ub) (17.9)

always satisfies the inequality:|S| ≤ 2. (17.10)

2. This inequality can be violated by the predictions of quantum mechanics.

We first prove the inequality satisfied by hidden variable theories. We introducethe quantity:

S(λ) = A(λ, ua) B(λ, ub) + A(λ, ua) B(λ, u′b)

+ A(λ, u′a) B(λ, u′

b) − A(λ, u′a)B(λ, ub),

which enters into the definition of S:

S = 4

�2

∫P(λ) S(λ) dλ.

8J.F. Clauser, M.A. Horne, A. Shimony, and R.A. Holt, Phys. Rev. Lett. 23, 880 (1969).

440 17 Entangled States. The Way of Paradoxes

This quantity S(λ) can be rewritten as

S(λ) = A(λ, ua)(B(λ, ub) + B(λ, u′

b))

+ A(λ, u′a)

(B(λ, u′

b) − B(λ, ub)), (17.11)

which is always equal to ±�2/2. Indeed the quantities B(λ, ub) and B(λ, u′

b) canonly take the two values ±�/2. Therefore they are either equal or opposite. In thefirst case, the second line of (17.11) vanishes, and the first one is equal to ±�

2/2. Inthe second case, the first line of (17.11) vanishes, and the second term is ±�

2/2. Wethen multiply S(λ) by P(λ) and we integrate over λ in order to obtain the inequalitywe were looking for.

Concerning the second point of Bell’s theorem, it suffices to find an example forwhich the inequality (17.10) is explicitly violated. Consider the vectors ua , u′

a , ub,and u′

b represented on Fig. 17.2:

ub.ua = ua .u′b = u′

b.u′a = −ub.u′

a = 1√2. (17.12)

Using (17.8) we find:S = −2

√2, (17.13)

which obviously violates the inequality (17.10).After this remarkable step forward due to Bell, which transformed a philosophical

discussion into an experimental problem, experimentalists had to find the answer. Isquantum mechanics always right, even for a choice of angles such as in Fig. 17.2,which would eliminate any realistic and local hidden variable super-theory, or, onthe contrary, are there experimental situations where quantum mechanics can befalsified, which would allow for a more complete theory, as Einstein advocated?

Fig. 17.2 Choice ofdirections of measurementsof Alice and Bob that leadsto a violation of Bell’sinequality

17.2 The Version of David Bohm 441

17.2.2 Experimental Tests

The first experimental attempts to find a violation of Bell’s inequality started at thebeginning of the 1970s. These experiments were performed on photon pairs ratherthan on spin 1/2 particles, because it is experimentally simpler to produce a two-photon entangled state of the type (17.1).

The previous argument can be transposed with no difficulty to photon pairs. Thespin states | + z〉 and | − z〉 are replaced by the polarization states of the photon| ↑〉 and | →〉, corresponding to vertical and horizontal polarizations. The states| + x〉 and | − x〉, which are symmetric and antisymmetric combinations of |± : z〉,are replaced by photon states linearly polarized at ±45 degrees from the verticaldirection:

| ↗〉 = 1√2

(| ↑〉 + | →〉) , | ↖〉 = 1√2

(−| ↑〉 + | →〉) . (17.14)

The first experimental tests, in the early 1970s in the United States and in Italy, ledto contradictory results concerning the violation ofBell’s inequality. The experimentsof Freedman and Clauser in Berkeley in 1972,9 of Fry and Thompson in Texas in1976,10 and particularly the “two channel” experiments of Aspect and his group inOrsay between 1980 and 1982, led to the undeniable violation of Bell’s inequality ina situation close to the gedanken experiment presented above.11 The experiments ofAspect use pairs of photons emitted in an atomic cascade of calciumatoms (Fig. 17.3).These calcium atoms are prepared by lasers in an excited state e1. This excited statehas a lifetime of 15 ns, and decays toward an excited state e2 by emitting a photon a, ofwavelength λa = 551 nm. This latter level e2 has a lifetime of 5 ns, and decays itselfto the ground state f by emitting a second photon b, of wavelength λb = 422 nm.The initial level e1 and the final level f have zero angular momentum, whereas theintermediate level e2 has angularmomentum 1.Under these conditions, one can showthat the polarization state of the emitted photon pair is:

|Ψp〉 = 1√2

(|a :↑ ; b :↑〉 + |a :→ ; b :→〉) . (17.15)

This entangled state leads to the same type of correlations as the singlet spin stateconsidered above. The transposition of Bell’s argument shows that some quantity S′,involving correlation functions between the polarizations of detected photons, mustverify |S′| ≤ 2 for any local hidden variable theory. The Orsay result, S′ = 2.697 ±0.015 violates this inequality, but it is in agreement with the quantum mechanical

9S.J. Freedman, J.F. Clauser (1972), “Experimental test of local hidden-variable theories”, Phys.Rev. Lett. 28 (938): 938–941.10E.S. Fry and R.C. Thompson (1976) Phys. Rev. Lett. 37, 465.11A. Aspect, P. Grangier, and G. Roger, Phys. Rev. Lett. 49, 91 (1982); A. Aspect, J. Dalibard, andG. Roger, Phys. Rev. Lett. 49, 1804 (1982).

442 17 Entangled States. The Way of Paradoxes

Fig. 17.3 Left levels of atomic calcium used in order to produce photon pairs with correlatedpolarizations. Right the photons a and b are first filtered in frequency: Fa transmits photons a andstops photons b, and vice versa for Fb. They are then detected on photomultipliers PMa+, PMa−,PMb+, and PMb−. The polarizing cubes are analogues of Stern–Gerlach devices. They transmitphotons with the polarization | ↑〉 towards the detectors PMa+, PMb+, and they deflect photonswith polarization | →〉 toward the detectors PMa−, PMb−. Because the photons are emitted nearlyisotropically, only a small fraction (∼10−5) of the emitted pairs is actually used

prediction S′ = 2.70. Therefore the universal realistic local hidden variable super-theory which was supposed to replace quantum mechanics, as Einstein believed,cannot exist, at least for this system. Physicists must learn to live with the genuineindeterminism of quantum mechanics.

The truth is that many physicists were disappointed with these results. They wereof course hoping that the proof of the existence of a super-theory would give excitingresearch themes. But what is more disconcerting is that the result confirms the wavepacket reduction. Something does happen instantaneously at a distance. We werediscussing this with my great friend Albert Messiah on February 28, 2001 and hetold me

Entangled states and Bell’s inequality are what make quantummechanics unbearable!Whenwe observe long distance entanglement, we are not sure that we really understand what’sgoing on. When Newton faced the problem of instantaneous action at a distance, he calledupon God’s finger. Quantum entanglement causes a similar reaction. Bell, with his inequal-ities, did hope to falsify quantum mechanics. The problem with that theory, is that it alwaysmanages to accommodate any situation, even with consequences that seem unbearable forus!12

Contrary to what many people say, Einstein was quite right when he thought andsaid that the interpretation of quantum mechanics causes problems. Here, the localhidden variables assumption is contradicted by experiment. This is a confirmationof Feynman’s premonitory claim of 1965: “I think I can safely say that nobodyunderstands quantum mechanics”.

12Albert Messiah actually used the term “undrinkable”, which is much stronger than unbearablefor a Frenchman.

17.3 The GHZ Experiment 443

17.3 The GHZ Experiment

Needless to say that Bell’s initial discovery, which can be considered as one ofthe great intellectual achievements in the history of science, generated a numberof subsequent works, some of which have already been mentioned. But in 1998,Greenberger,Horne andZeilinger (GHZ), found an evenmore spectacular example ofquantumentanglementwhich reveals unambiguously the non-locality of the quantumworld. They first analysed systems of four particles but, together with Shimony, andupon a suggestion of David Mermin, they subsequently applied their arguments tomeasurements involving three observers.13

The key point which makes that approach starkling is that it does not involve theevaluation of any correlation, but, as we shall see, a simple inspection of the setsof individual results recorded by the observers, after the experiment is finished. Theexperimental technique was presented in 1998.14

The final conclusive experimental results were published in 2000 by Jian-WeiPan, D. Bouwmeester, M. Daniell, H. Weinfurter and A. Zeilinger15.

Zeilinger and his collaborators worked with entangled photon states, measuringpolarizations. The original GHZ three photon state is

|ΨGHZ 〉 = | ↑,↑,↑〉 + | →,→,→〉√2

(17.16)

There are several excellent papers which explain in a simple way the issue of theGHZ experiment, among which that of Wilczek16 and that of Herbert Bernstein.17

Here, we choose, for simplicity, to discuss the case of three spin 1/2 particles18

which has been considered byHerbert Bernstein.We dealwith three spin 1/2 particles

13D.M. Greenberger, M. Horne, and A. Zeilinger, in Bells Theorem, Quantum Theory, and Concep-tions of the Universe, edited by M. Kafatos, (Kluwer, Dordrecht, 1989); D.M. Greenberger, M.A.Horne, A. Shimony, and A. Zeilinger, Bell’s theorem without inequalities, Am. J. Phys. 58, 1131(1990); N.D. Mermin, Phys. Today 43 (6),9 (1990).14Jian-wei Pan and Anton Zeilinger, Greenberger-Horne-Zeilinger-state analyzer, Phys. Rev. 57(3), 1998.15Jian-Wei Pan,D. Bouwmeester,M.Daniell, H.Weinfurter andA. Zeilinger (2000). “Experimentaltest of quantum nonlocality in three photon GHZ entanglement”. Nature 403 (6769): 515–519;Jian-Wei Pan and Anton Zeilinger, (2002)Multi-Photon Entanglement and Quantum Non-Localityhttps://vcq.quantum.at/fileadmin/Publications/2002-12.pdf.16Frank Wilczek, Entanglement made simple, Quanta Magazine, April 28, 2016.17Herbert J.Bernstein,SimpleVersionof theGreenberger-Horne-Zeilinger (GHZ)ArgumentAgainstLocal Realism, Foundations of Physics 29 (4):521–525 (1999).18I am deeply gratefull to James Rich for explaining to me the whole issue, and I reproduce hispresentation.

444 17 Entangled States. The Way of Paradoxes

named respectively a, b, c and we are interested in their spin components along anorthonormal reference frame (x, y, z).

Quantum Situation

Consider the GHZ state:

|GHZ〉 ≡ 1√2|saz = +, sbz = +, scz = +〉 − 1√

2|saz = −, sbz = −, scz = −〉

(17.17)where we omit the �/2 in the values of these spin components.

We know (see Chap.12) that we can decompose the sz components into sx and sycomponents of a spin 1/2 in the following ways

|sz = +〉 = (|sx = +〉 + |sx = −〉)/√2 = (|sy = +〉 + |sy = −〉)/√2

and

|sz = −〉 = (|sx = +〉 − |sx = −〉)/√2 = (|sy = +〉 − |sy = −〉)/(i√2)

There are several ways to decompose the |GHZ〉 state.a. We can decompose the sz components into one sx and two sy components and

obtain, after some simple arithmetics, the following decomposition of the |GHZ〉state, Eq. (17.17), such as

|GHZ〉 = 1

2(|sax = +, sby = +, scy = +〉 + |sax = +, sby = −, scy = −〉

+ |sax = −, sby = +, scy = −〉 + |sax = −, sby = −, scy = +〉 )

We remark that if one measures, on this state, two sy’s and one sx , one alwaysfinds and odd number of +�/2. (All four terms have that property.)

b. Alternatively, we can also decompose the sz components into three sx componentsand obtain:

|GHZ〉 = 1

2(|sax = +, sbx = +, scx = −〉 + |sax = +, sbx = −, scx = +〉

+ |sax = −, sbx = +, scx = +〉 + |sax = −, sbx = −, scx = −〉 )

We remark that if one measures 3 sx ’s, one always finds an even number of+�/2.(All four terms have that property.)

17.3 The GHZ Experiment 445

Notice that in both cases, if b and c find the same result then sax = −, and if theyhave opposite results then sax = +whichmeans that sax has an EPR element of reality.A similar property holds for b and for c.

Local Realistic Situations

Suppose that pairs of particles (a, b) possess local predetermined values of measure-ments of (sx , sy): ⎡

⎣sax say

sbx sby

⎤

⎦

There are, a priori, 16 types of pairs:

[+ ++ +

] [+ ++ −

] [+ −+ +

] [+ −+ −

] [+ +− +

] [+ +− −

] [+ −− +

] [+ −− −

]

[− ++ +

] [− ++ −

] [− −+ +

] [− −+ −

] [− +− +

] [− +− −

] [− −− +

] [− −− −

]

For example, for a pair of type [− −− +

]

a measurement of (sax , sby ) yields (−+).

Turning now to particle triplets, there are 64 kinds of them, represented below:

⎛

⎝+ ++ ++ +

⎞

⎠

⎡

⎣+ ++ ++ −

⎤

⎦

⎡

⎣+ ++ −+ +

⎤

⎦

⎡

⎣+ −+ ++ +

⎤

⎦

⎡

⎣+ ++ −+ −

⎤

⎦

⎡

⎣+ −+ ++ −

⎤

⎦

⎡

⎣+ −+ −+ +

⎤

⎦

⎛

⎝+ −+ −+ −

⎞

⎠

⎡

⎣+ ++ +− +

⎤

⎦

⎡

⎣+ ++ +− −

⎤

⎦

⎡

⎣+ ++ −− +

⎤

⎦

⎡

⎣+ −+ +− +

⎤

⎦

⎡

⎣+ ++ −− −

⎤

⎦

⎡

⎣+ −+ +− −

⎤

⎦

⎡

⎣+ −+ −− +

⎤

⎦

⎡

⎣+ −+ −− −

⎤

⎦

⎡

⎣+ +− ++ +

⎤

⎦

⎡

⎣+ +− ++ −

⎤

⎦

⎡

⎣+ +− −+ +

⎤

⎦

⎡

⎣+ −− ++ +

⎤

⎦

⎡

⎣+ +− −+ −

⎤

⎦

⎡

⎣+ −− ++ −

⎤

⎦

⎡

⎣+ −− −+ +

⎤

⎦

⎡

⎣+ −− −+ −

⎤

⎦

⎡

⎣− ++ ++ +

⎤

⎦

⎡

⎣− ++ ++ −

⎤

⎦

⎡

⎣− ++ −+ +

⎤

⎦

⎡

⎣− −+ ++ +

⎤

⎦

⎡

⎣− ++ −+ −

⎤

⎦

⎡

⎣− −+ ++ −

⎤

⎦

⎡

⎣− −+ −+ +

⎤

⎦

⎡

⎣− −+ −+ −

⎤

⎦

446 17 Entangled States. The Way of Paradoxes

⎡

⎣+ +− +− +

⎤

⎦

⎡

⎣+ +− +− −

⎤

⎦

⎡

⎣+ +− −− +

⎤

⎦

⎛

⎝+ −− +− +

⎞

⎠

⎛

⎝+ +− −− −

⎞

⎠

⎡

⎣+ −− +− −

⎤

⎦

⎡

⎣+ −− −− +

⎤

⎦

⎡

⎣+ −− −− −

⎤

⎦

⎡

⎣− ++ +− +

⎤

⎦

⎡

⎣− ++ +− −

⎤

⎦

⎛

⎝− ++ −− +

⎞

⎠

⎡

⎣− −+ +− +

⎤

⎦

⎡

⎣− ++ −− −

⎤

⎦

⎛

⎝− −+ +− −

⎞

⎠

⎡

⎣− −+ −− +

⎤

⎦

⎡

⎣− −+ −− −

⎤

⎦

⎡

⎣− +− ++ +

⎤

⎦

⎛

⎝− +− ++ −

⎞

⎠

⎡

⎣− +− −+ +

⎤

⎦

⎡

⎣− −− ++ +

⎤

⎦

⎡

⎣− +− −+ −

⎤

⎦

⎡

⎣− −− ++ −

⎤

⎦

⎛

⎝− −− −+ +

⎞

⎠

⎡

⎣− −− −+ −

⎤

⎦

⎡

⎣− +− +− +

⎤

⎦

⎡

⎣− +− +− −

⎤

⎦

⎡

⎣− +− −− +

⎤

⎦

⎡

⎣− −− +− +

⎤

⎦

⎡

⎣− +− −− −

⎤

⎦

⎡

⎣− −− +− −

⎤

⎦

⎡

⎣− −− −− +

⎤

⎦

⎡

⎣− −− −− −

⎤

⎦

The types of triplets in curved brackets ( ) are the ones for which a measurementof 2 sy’s and 1 sx always yields an odd number of +�/2. But for these eight types, ifone measures 3 sx ’s, one always finds an odd number of +�/2. Therefore there is noparticle triplet that imitates the GHZ state. In other words, there exists no predeter-mined values of the three spins that agree with quantum mechanical measurements.Notice that no reference to hidden variables is made here.

The superposition of states is compulsory to explain the experimental results. Insome sense, we are facing a situation which has similarities with Schrödinger’s catas was pointed out by several authors.

17.4 Quantum Cryptography; How to Take Advantageof an Embarrassment

The ambition of cryptography is to transmit a message from an issuer (Alice) toa receiver (Bob) and to minimize the risk that a spy may intercept and decipherthe message. In order to do that, classical cryptography uses sophisticated methodsthat cannot be “broken” in a reasonable amount of time, with the present capacitiesof computers. Quantum cryptography is based on a somewhat different principle.It allows Alice and Bob to make sure no spy has intercepted the message beforeactually sending it!

The principle of this technique is to profit from the fact that in quantummechanicsa measurement perturbs the state of the system, in particular with entangled states.One therefore devises a procedure thatwill prove the existence of a spy before sendingthe actual message!

17.4 Quantum Cryptography; How to Take Advantage of an Embarrassment 447

The Communication Between Alice and Bob

A message can always be coded in binary language, that is, by a succession of 0and 1. Each number, 0 or 1, represents a piece of elementary information, or bit.In order to transmit her message, we assume that Alice sends Bob a beam of spin1/2 particles in a well-controlled sequence, and that Bob detects these particles oneafter the other in a Stern–Gerlach type of apparatus. Each particle carries a bit codedthrough its spin state.

Suppose first that Alice sends each particle in the state | + z〉 or | − z〉. By con-vention, | + z〉 represents the value 1 and | − z〉 the value 0. Bob orients his Stern–Gerlach apparatus also along the z-axis, he measures the spin states of the particlesthat arrive, and he reconstructs Alice’s message. Such a procedure has no quantumfeature and it is simple to spy upon. The spy just sits between Alice and Bob andhe places his own Stern–Gerlach apparatus in the z-direction. He measures the spinstate of each particle, and re-emits it toward Bob in the same spin state. He thereforereads the message and neither Alice nor Bob can detect his presence.

The situation changes radically if Alice chooses at random, for each particle sheis sending, one of the four states | + z〉, | − z〉, | + x〉, or | − x〉, without tellinganyone which axis she has chosen (x or z) for a given particle. Suppose Alice sendsBob a series of particles without trying, for the moment, to give it any intelligibleform. There are 16 particles in the examples shown in Figs. 17.4 and 17.5, but inpractice one works with much larger numbers. It is only at the end of the procedure,as we show, that Alice will decide which particles should be taken into account inorder to construct the message she wants to transmit.

What can Bob do in this situation? He can orient the axis of his Stern–Gerlachapparatus in an arbitrary way, x or z. On average, for half of the particles, his choiceis the same as Alice’s, in which case the bit he detects is significant. Indeed, if Alicesends a particle in the state | + x〉 and if Bob chooses the x-axis, he does measure +with probability 1. For the other half of the particles, Alice and Bob choose differentaxes and Bob’s results are useless: if Alice sends | + x〉 and if Bob chooses thez-axis, he will detect + with probability 1/2 and − with probability 1/2.

In order to make sure that no spy has intercepted the transmission, Bob announcesopenly the set of axes he has chosen, x or z, for each event. He also says what resultshe has obtained (i.e., + or −) for a fraction of the particles. For instance, in thecase of 16 particles shown in Figs. 17.4 and 17.5, Bob announces publicly his 16choices of axes, and his first 8 results. Alice examines the results, and she can detectwhether a spy has operated. Her argument is the following. The spy does not knowthe directions x or z she has chosen for each particle. Suppose that the spy orients hisStern–Gerlach apparatus in a randomway along x or z, and that he re-emits a particlewhose spin state is the same as what he has measured. If he chooses the x-axis andhe gets the result +, he sends Bob a particle in the | + x〉 state. This operation isdetectable, because it induces errors in Bob’s observations.

Consider, for instance, the case where Alice has sent a particle in the state | + z〉,and Bob has also oriented his detector along the z-axis, but where the spy has orientedhis own Stern–Gerlach apparatus along the x-axis. The spy can measure + with aprobability 1/2 and − with a probability 1/2. According to his result, he re-emits to

448 17 Entangled States. The Way of Paradoxes

Number of the particle 1 2 3 4 5 6 7 8Axis chosen by Alice z z x z z x x z

(kept secret)State chosen by Alice + − + − − − + −

(kept secret)Axis chosen by Bob z x x z x z x x(broadcast openly)

State measured by Bob + − + − − + + +(broadcast openly)

Useful measurement ? yes no yes yes no no yes no

Fig. 17.4 Detection of a possible spy among Bob’s results done along the same axis (particles 1,3, 4, and 7), Alice looks for a possible difference that would mean a spy has operated. No anomalyappears here. In practice, in order to have a sufficient confidence level, one must use a number ofevents much larger than 8

Fig. 17.5 After making surethat there is no spy, Alicechooses among the usefulmeasurements those thatallows communication of themessage. For instance, tocommunicate the message“1,1”, that is, “+,+”, sheopenly asks Bob to look atthe results of hismeasurements 11 and 15

Number of the particle 9 10 11 12 13 14 15 16Axis chosen by Alice x z x z z x z z

(kept secret)State chosen by Alice + − + + − − + −

(kept secret)Axis chosen by Bob z z x x z z z x(broadcast openly)

State measured by Bob − − + + − + + +(kept secret)

Useful measurement ? no yes yes no yes no yes no

Bob a particle in the state | + x〉 or | − x〉. In both cases, because Bob’s detector isoriented along z, Bob can measure+with a probability 1/2 and−with a probability1/2. If the spy had not been present, Bob would have found + with probability 1.

Therefore, among all the results announced by Bob, Alice looks at those whereher own choice of axes is the same as Bob’s (Fig. 17.4). If no spy is acting, Bob’sresults must be identical to hers. If a spy is present, there must be differences in 25%of the cases. Therefore, if Bob announces publicly 1000 of his results, on average 500will be useful for Alice (same axes), and the spy will have induced an error in 125of them (on average). The probability that a spy is effectively present, but remainsundetected by such a procedure is (3/4)500 ∼ 3 × 10−63, which is negligible.

Once Alice hasmade sure that no spy has intercepted the communication, she tellsBob openly which measurements he must read in order to reconstruct the messageshe wants to send him. She simply chooses them among the sequence of bits forwhich Bob and her have made the same choice of axes, and for which Bob did notannounce his result openly (Fig. 17.5).

The Quantum Non-cloning Theorem

In the previous paragraph, we have assumed that the spy chooses at random theaxis of his detector for each particle, and that he sends to Bob a particle in the state

17.4 Quantum Cryptography; How to Take Advantage of an Embarrassment 449

corresponding to his measurement. One may wonder whether this is his best strategyin order to remain unseen. In particular if the spy could clone each incident particlesent by Alice into two particles in the same state, it would be possible to send one ofthem to Bob and to measure the other one. The spy would then become undetectable!

Fortunately for Alice and Bob, this cloning of an unknown state is impossible inquantummechanics.19 One cannot generate in a reliable way one or several copies ofa quantum state unless some features of this state are known in advance. In order toprove this result, let us note |α1〉 an initial quantum state which we want to copy. Thesystemonwhich the copywill be “printed” is initially in a known state, whichwe note|φ〉 (the equivalent of a blank sheet of paper in a copying machine). The evolutionof the total system original + copy during the cloning operation must therefore be:

cloning : |original : α1〉 ⊗ |copy : φ〉 −→ |original : α1〉 ⊗ |copy : α1〉 .

(17.18)

The evolution is governed by some Hamiltonian which we need not specify, butwhich cannot depend on |α1〉 since this state is unknown by assumption. For another state |α2〉 of the original, orthogonal to |α1〉, we must also have:

cloning : |original : α2〉 ⊗ |copy : φ〉 −→ |original : α2〉 ⊗ |copy : α2〉 .

(17.19)

The impossibility of cloning is then obvious if we consider the initial state:

|α3〉 = 1√2

(|α1〉 + |α2〉) . (17.20)

If the cloning were successful for this state, we would find:

cloning : |original : α3〉 ⊗ |copy : φ〉 −→ |original : α3〉 ⊗ |copy : α3〉 . (17.21)

However, the linearity of the Schrödinger equation imposes, by linear superpositionof (17.18) and (17.19):

|original : α3〉 ⊗ |copy : φ〉 −→1√2

(|original : α1〉 ⊗ |copy : α1〉 + |original : α2〉 ⊗ |copy : α2〉) .

This final state is an entangled state. It is therefore different from the desired state(17.21).

19W.K. Wooters and W.H. Zurek, Nature 299, 802 (1982).

450 17 Entangled States. The Way of Paradoxes

The inspection of this proof allows to understand the contribution of quantummechanics to cryptography. If we limit ourselves to a two-state transmission, |α1〉 =| + z〉 and |α2〉 = | − z〉, then the spy can remain invisible as we have explained inthe previous section. The two operations (17.18) and (17.19) are possible; we simplyneed to measure the spin state of the incident particle along the z axis, and to re-emitone or more particles in the same state. It is the fact that we can use simultaneouslythe states |α1〉, |α2〉 and linear combinations of these states |α3,4〉 = |± : x〉 whichmakes the originality of quantum cryptography, and forbids any reliable duplicationof a message intercepted by a spy.

Present Experimental Setups

As for experimental tests of Bell’s inequality, current physical setups use photonsrather than spin 1/2 particles. Various methods can be used to code information onphotons. We only consider the coding of polarization, which is effectively used inpractice. Alice uses four states that define two non-orthogonal bases, each of whichcan code the bits 0 and 1, for instance in the form:

| ↑〉 : 1 ; | →〉 : 0 ; | ↗〉 : 1 ; | ↖〉 : 0. (17.22)

In present quantum cryptography devices, the challenge is to obtain sufficientlylarge distances of transmission. One currently reaches distances of the order of tenkilometers, by using optical telecommunication techniques, in particular photons inoptical fibers.

An important point is the light source. The non-cloning theorem, which is crucialin order to provide the safety of the procedure, only applies to individual photons.On the contrary, usual light pulses used in telecommunications contain very largenumbers of photons, typically more than 106. If one uses such pulses for coding thepolarization, the noncloning theorem no longer applies. Indeed it is sufficient for thespy to remove a small part of the light in each pulse and to let the remaining partpropagate to Bob. The spy can measure in this way the polarization of the photonsof the pulse without modifying the signal noticeably.

In order to guarantee the safety of the procedure, each pulse must contain asingle photon. This is a difficult condition to satisfy in practice and one uses thefollowing alternative as a compromise. Alice strongly attenuates the pulses so thatthe probability p of finding one photon in each pulse is much smaller than one. Theprobability of having two photons will be p2 � p, which means that there will bevery few two (or more) photon pulses. Obviously, most of the pulses will containno photons, which is a serious drawback of the method because Alice must code theinformation redundantly. In practice, a value of p between 0.01 and 0.1 is consideredto be an acceptable compromise.

Once this basic question is solved, the essential part of the system uses opticaltelecommunication technologies. The source is a strongly attenuated pulsed laser,and the coding polarization takes place directly in the optical fiber using integratedmodulators. The attenuated pulses are detected with avalanche photodiodes, whichtransform a single photon into amacroscopic electrical signal by an electronmultipli-cation process. In order to identify unambiguously the photons emitted by Alice and

17.4 Quantum Cryptography; How to Take Advantage of an Embarrassment 451

detected by Bob, electric pulses synchronized with the laser pulses are sent to Bobby conventional techniques, and they play the role of a clock. Finally, a computerizedtreatment of the data, involving a large number of pulses, fulfills the various stagesof the procedure described above, in particular to test the absence of a possible spyon the line.

At present the systems that have been built are more demonstration prototypesrather than operational systems. Several relevant parameters have been tested, such asthe distance and the transmission rate, the error rate, and so on. Actually, developingthese systems has for the moment a prospective character, because conventional(nonquantum) cryptographic systems are considered to be very reliable by civilianor military users. This confidence was a little bit shaken in 1994, as we shall see inthe next section.

17.5 The Quantum Computer

The Quantum Bits, or “q-bits”

In the previous section, we have seen that one can code a bit of information(0 or 1) with two orthogonal states of a spin 1/2 particle or with a polarized photon.At this stage of a quantum mechanics course, a question rises naturally: in terms ofinformation theory, what is the significance of a linear superposition of these twostates? In order to account for this possibility, one introduces the notion of “q-bit”which, contrary to a classical bit, allows the existence of such intermediate states.The notion of a q-bit in itself is not very rich; however it has interesting implicationsif one considers a quantum computer, based on the manipulation of a large numberof q-bits.

We take the very simplified definition of a computer as a system which is capableof performing operations on sets of N bits called “registers”. The content of a registeris a binary word, which represents a number memorized by the computer. For N = 3,we therefore have 8 possible words:

(+,+,+) (+,+,−) (+,−,+) (+,−,−) (−,+,+) (−,+,−) (−,−,+) (−,−,−)

Consider now a q-register, made of a set of N q-bits. The 2N possible states of theclassical register will define a basis of the space of states of the q-register, which canitself be in a linear superposition of all the basis states:

|Ψ 〉 =∑

σ1=±

∑

σ2=±

∑

σ3=±Cσ1,σ2,σ3 |σ1,σ2,σ3〉 for N = 3.

Suppose that the computer calculates, i.e. it performs an operation on the state of theq-register. Since this operation is performed on a linear superposition of states, wecan consider that it is done “in parallel” on the 2N classical numbers. This notion of

452 17 Entangled States. The Way of Paradoxes

quantum parallelism is the basis of a gain in efficiency of the computer. The gain maybe exponential if the 2N calculations corresponding to N q-bits are indeed performedsimultaneously.

Naturally, many questions arise. From the fundamental point of view, what kindof calculations can one perform and what kind of algorithms can one use with sucha device? In practice, how can one construct it?

The Algorithm of Peter Shor

In the previous section,we referred to non-quantumcryptography. The correspondingsystems are often called algorithmic protocols. One of these protocols is based onthe fact that some arithmetic operations are very easy to perform in one way, andvery difficult in the reverse way. For instance it is simple to calculate the productof two numbers, but it takes much more time to factorize a number in its primedivisors. If one considers the product P of two large primenumbers, onemust performapproximately

√P divisions in order to identify the factors. The computing time

increases exponentially with the number of digits (or of bits) of P: the factorizationoperation becomes impossible in practice for numbers with more than 300 digits,while the product operation leading to P can still be performed easily with a smallcomputer. This “non reversibility” is at the origin of a cryptographic method dueto Rivest, Shamir and Adelman (RSA), which is commonly used at present (creditcards, electronic transactions, etc.), and which is considered to be extremely reliable.

This is why the 1994 paper of Peter Shor20 created a shock in the community. Shorshowed that a quantum computer could factorize the product of two prime numberswith a number of operations reduced exponentially as compared to known algorithmsrunning on classical computers! The turmoil has now calmed down and the presentsituation is the following. The algorithm proposed by Shor is correct in its principle,and it does have the expected gain in efficiency. However the practical elaborationof a quantum computer seems outside the range of present technology, although nophysical law forbids it.

Principle of a Quantum Computer

We will not attempt to explain Shor’s algorithm here, and we will only give someintuitive ideas on how a quantum computer can perform a calculation. The basicprinciple is that the calculation must reduce to the evolution of a system with aHamiltonian. It starts from some initial state and it ends by a “measurement” whichdetermines the state of the q-register, and which interrupts the evolution. Accordingto the principles of quantum mechanics, the value found in the measurement is oneof the eigenvalues corresponding to the eigenstates of the measured observable. Inthis context, it corresponds to the state of a classical register, i.e. a binary word. Inorder to perform the successive operations, the Hamiltonian of the system is time-dependent and it evolves under the action of a clock which determines the rhythmof the calculation. At first sight, the determination of this Hamiltonian for a given

20Peter Shor Polynomial-time algorithms for prime factorization and discrete logarithms on aquantum computer, SIAM Journal on Computing 26 (1997), 1484–1509.

17.5 The Quantum Computer 453

practical calculation seems to be a formidable task. Actually, one can show that theconstruction of the Hamiltonian can be done relatively easily. An actual calculationcan be decomposed into a succession of simple operations which affect only one ortwo bits. These simple operations are performed by logical gates, such as the wellknown classical NOT, AND, OR.

The quantum gates required for the Shor algorithm must have some particularfeatures:

• They must be reversible, since they follow from a Hamiltonian evolution of theinitial bits.

• They must handle q-bits, on which one can perform certain logical operationswhich are classically inconceivable.

A simple example of a quantum gate is the gate√NOT . It transforms the q-bits

0 and 1 into the symmetric and antisymmetric linear superpositions of 0 and 1 (it isa rotation of π/2 for a spin 1/2). If one applies this gate twice, one inverts 0 and 1,which corresponds to the gate NOT, hence the name of gate

√NOT .

One could think that itwould be sufficient to let the computer evolve towards a one-component state, which would be the desired value. Actually, very few algorithmsgive rise to such a simple manipulation. In general, the final state of the computer isstill a linear superposition, and the result of the calculation is therefore probabilistic.For instance in Shor’s algorithm, the result is rather an indication of a possible result.It is easy to check by a conventional method whether the answer is correct, and if not,to pursue the calculation. Peter Shor has proven that this trial and error proceduregives the correct answer with a probability arbitrarily close to 1, using a number oftrials which increases linearly (and not exponentially) with the number of digits ofthe number we want to factorize.

Decoherence

The principle of a quantum computer is compatible with the laws of physics, and itseems possible to construct such a computer, at least if one only considers simplecalculations with a small number of gates. If one wants to attain large computingsizes, the global state of the computer has to be a quantum superposition of a largenumber of states, whose evolution must be controlled in such a way that all theproperties of a linear superposition are preserved. It is not clear at present whethersuch a system can be devised. Studies are underway in two main directions:

• First the register which evolves must be extremely well protected from the outsideenvironment. Indeed the couplingwith this environment will induce a decoherenceeffect, which may destroy the interferences between the various terms of the linearsuperposition.

• Second, in case perturbations occur, one must prepare error correction codes, inorder to place the computer in the same state where it was before the externalperturbation occurred.

These two directions—the choice of the system and the correction codes—areunder intensive investigation and the questions raised have been stimulating both foralgorithmic and for experimental quantum mechanics. At present it is very difficult

454 17 Entangled States. The Way of Paradoxes

to predict the outcome of theses investigations. However one spin-off is that it is quitepossible that simple logical operations will, in the mean run, lead to applications insystems of quantum cryptography.

17.6 Quantum Teleportation

Quantum entanglement allows an amusing operation, “quantum teleportation.” Sup-pose Alice has a spin 1/2 particle A in the spin state:

α|+〉 + β|−〉 , with |α|2 + |β|2 = 1,

that she wants to teleport, or fax, to Bob in a simple manner without measuring αand β.

Bell States

Consider a system of two particles of spin 1/2, whose spin state is written:

α|+;+〉 + β|+;−〉 + γ|−;+〉 + δ|−;−〉 (17.23)

with |α|2 + |β|2 + |γ|2 + |δ|2 = 1.The probability of finding in ameasurement (+�/2,+�/2) is |α|2, the probability

of finding (+�/2,−�/2) is |β|2, and so on. The four states of theBell basis are definedas follows,

|Ψ+〉 = 1√2

(|+;+〉 + |−;−〉) |Φ+〉 = 1√2

(|+;−〉 + |−;+〉)

|Ψ−〉 = 1√2

(|+;+〉 − |−;−〉) |Φ−〉 = 1√2

(|+;−〉 − |−;+〉) .

The projectors PΨ = |Ψ 〉〈Ψ | on one of these states |Ψ 〉 can be called occupationoperators of one of these states. In fact, the eigenvalues of these operators are 1(occupied state) and 0 (vacant state).

A direct calculation gives the probabilities of finding (17.23) in each Bell state.

State |Ψ+〉 occupied: probability |α + δ|2/2 ,

State |Ψ−〉 occupied: probability |α − δ|2/2 ,

State |Φ+〉 occupied: probability |β + γ|2/2 ,

State |Φ−〉 occupied: probability |β − γ|2/2 .

Because the basis of Bell states is orthonormal, the sum of these four probabilitiesis indeed 1.

17.6 Quantum Teleportation 455

Fig. 17.6 Principle of the teleportation of the quantum state of a particle

Teleportation

In order for Alice to “teleport” the state α|+〉 + β|−〉 of particle A to Bob, theprocedure is as follows.

Alice and Bob also have in common a pair of spin 1/2 particles B andC , preparedin the singlet state (Fig. 17.6):

1√2

(|+;−〉 − |−;+〉) .

a. The state of the system of three spins (A, B,C) is:

|Ψ 〉 = α√2|+;+;−; 〉 + β√

2|−;+;−; 〉 − α√

2|+;−;+; 〉 − β√

2|−;−;+; 〉.

This state can be decomposed on the Bell basis of the spins A and B; the basisfor spin C remains |±〉:

|Ψ 〉 = 1

2|Ψ+〉 (α|−〉 − β|+〉) + 1

2|Ψ−〉 (α|−〉 + β|+〉)

−1

2|Φ+〉 (α|+〉 − β|−〉) − 1

2|Φ−〉 (α|+〉 + β|−〉) .

If Alice measures the spin state of the couple AB that projects this state on oneof the four vectors of the Bell basis of AB, the probability of finding the pair ABin each of the Bell states is the same (|α|2 + |β|2)/4 = 1/4.

b. If Alice finds the pair AB in the state |Φ−〉, the state of the spin C is the originalstate α|+〉 + β|−〉.

c. In order to teleport the a priori unknown state α|+〉 + β|−〉 of particle A toparticle C , Alice must not try to measure this state. She must simply make a Bellmeasurement on the pair AB and tell the result to Bob. When she finds that thestate |Φ−〉 is occupied (in 25% of the cases), Bob need not do anything. The spinstate of C after the measurement is the state of A before the measurement. Inall other cases, Bob can reconstruct the initial state by a simple transformation.For instance, if Alice finds the pair AB in the Bell state |Φ+〉, the spin state of C

456 17 Entangled States. The Way of Paradoxes

is α|+〉 − β|−〉, which can be transformed into the initial state α|+〉 + β|−〉 byperforming a rotation of π around the z-axis.

d. One cannot use this method to transmit information more rapidly than with usualmeans. As long as Alice doesn’t tell Bob the result of her Bell measurement, Bobhas no available information. It is only after he knows Alice’s results, and he hasrejected or reconstructed the fraction of experiments (75%) which do not give|Φ−〉 that he can make profit of the “teleportation” of the spin state of particle A.

Further Reading

• On the EPR problem: Quantum theory and measurement, edited by J. A. Wheelerand W. H. Zurek (Princeton University Press, 1983). The conceptual implicationsof Quantum Mechanics, Workshop of the Hugot fundation of Collège de France,J. Physique 42, colloque C2 (1981).

• J. S. Bell, Physics 1, 195 (1964); see also J. Bell, Speakable and unspeakable inquantummechanics, CambridgeUniversity Press (1993). Franck Laloë (2012),DoWe Really Understand Quantum Mechanics, Cambridge University Press, ISBN978-1-107-02501-1

• C. Bennett, G. Brassard et A. Ekert,Quantum Cryptography, Scientific American,October 1992; R. Hughes and J. Nordholt, Quantum cryptography takes the air,Physics World, May 1999, p. 31.

• S. Lyod, Quantum mechanical computers, Scientific American, November 1995,p. 44; S. Haroche and J.-M. Raimond, Quantum computing: dream or nightmare,Physics Today, August 1996, p. 51; Special issue of Physics World on QuantumInformation, March 1998; N. Gershenfeld and I.L. Chuang, Quantum computingwith molecules, Scientific American, June 1998, p. 50; J. Preskill, Battling deco-herence: the fault-tolerant quantum computer, Physics Today, June 1999, p. 24.

17.7 Concluding Remarks

In writing the previous edition of this book, my last chapter was devoted to thedevelopments of astrophysics and cosmology. Needless to say that these remain atthe forefront of physics. The recent observation of gravitational waves, detectedon September 14, 2015 by both of the twin Laser Interferometer Gravitational-wave Observatory (LIGO) detectors, located in Livingston, Louisiana, and Hanford,Washington, USA, and operated by Caltech and MIT is undoubtedly a major scien-tific event of our time. There is no doubt that enormous progress and impressive newresults are to be expected in the field.

I have decided to end this book with an introductory chapter of what is becomingan impressive byproduct of quantum mechanics: Quantum information. A simpleinspection of the literature on the subject shows that it is becoming a major scientificand technological field of research. Quantum information science includes impres-sive theoretical as well as experimental topics in quantum physics. It is fascinating to

17.7 Concluding Remarks 457

see the developments of Quantum computing, of Quantum cryptography, Quantumcommunication Quantum teleportation etc. One of the very immediate proofs of thatlies in the Springer collection “Quantum Information Processing”, whose Volume 1appeared in 2002, and whose Volume 15, January 2016 - June 2016, was just issued(12 issues per year). The list of titles of the contributions is impressive, as is theirnumber and diversity. I hope everyone will forgive me for ducking in front of anexhaustive review of that part of quantum physics.

All those developments are impressive by themselves, their topics and their goals.But, what is perhaps more impressive is that their roots lie on purely intellectualquestions raised by Einstein in 1935, and pursued by a few jewelers such as JohnBell. Huge technological progress is made because of intellectual paradoxes! Thingsexploded after the 1970s with experiments which were devised to get an answer tothe question: do we understand QuantumMechanics? Answer which, in my opinion,is still lacking. Is that really crucial? Above all, everyone is convinced that quantummechanics is a good theory, even a very impressive theory, given all the efforts andall the means that mankind has made to find its limits (aside from saying that beforethe Planck time of, say,

√(�G/c5) ∼ 5.4 10−44 s, “something else” replaced it). So

one should learn and use quantum physics! But, as so many people say or have said:we are not at all sure that we understand it.

In some sense, I have a tendency to feel that we should consider that quantummechanics is a “complete” theory, with a beginning and an end, just as electromag-netism was after 1864, Maxwell’s equations and the discovery of electromagneticradiation by Hertz. The invariance of the velocity of light under Galilean transfor-mations was inscribed inMaxwell’s equation. Some nasty jerk could (and may) haveclaimed that all that was bullshit because of that stupid remark, which actually tookso many years forMichelson to verify experimentally. Of course, nowadays, it seemsquite natural for us to imagine that Einstein, Lorentz, Poincaré and others just thoughtthat one should question the unshakeable notion of the nature of time, that mankindhad always taken for granted. And then, was relativity the end of the story? Not atall, since the same Einstein imagined the photon etc. etc. and people went on withquantum mechanics and QED.

This time, so much serious and hard work has been done by so many competentpeople on “thinkable” approaches or alternatives, that the answer, once it comes, willbe unexpected, be it simple or terribly subtle. In any case fascinating.

Chapter 18Solutions to the Exercises

Chapter 1

Probabilities

1. Distribution of impacts

Going to polar coordinates gives:

f (ρ) = 1

σ2ρ e−ρ2/2σ2

(ρ ≥ 0).

One can check that∫ +∞

0 f (ρ) dρ = 1.

2. Is this a fair game?

The probability that 6 does not appear is(

56

)3; that it appears once, 3 × (

56

)2 ×(

16

); twice, 3 × (

56

) × (16

)2; thrice

(16

)3. The respective gains are −1, 1, 2, 5, the

expectation value of the gain is therefore:

1

63

(−53 + 3 × 52 + 2 × 3 × 5 + 5) = − 15

216.

Don’t play ! In order for the game to be fair, the third gain should be 20 euros!

3. Spatial distribution of the molecules in a gas

We start with the simple alternative: either a molecule is inside v, (p = v/V ) or it isnot (q = 1 − v/V ). The probability for finding k molecules in v is therefore:

PN (k) =(Nk

) ( v

V

)k (1 − v

V

)N−k.

© Springer International Publishing Switzerland 2016J.-L. Basdevant, Lectures on Quantum Mechanics,Graduate Texts in Physics, DOI 10.1007/978-3-319-43479-7_18

459

460 18 Solutions to the Exercises

The expectation value is 〈k〉 = Np = Nv/V , and the dispersion is σ = √Npq ∼√

Nv/V = √〈k〉. In our numerical example 〈k〉 ∼ 3 × 1016, therefore σ ∼ 1.7 ×108. The relative dispersion σ/〈k〉 is very small (∼10−8). The probability for findinga number of molecules outside the interval 〈k〉 ± 2σ is 5 % (Gaussian law).

Chapter 2

1. Spreading of a wave packet of a free particle

a. Using a procedure similar to the previous one, we obtain from the Schrödingerequation:

d〈x2〉tdt

= i�

m

∫x

(ψ

∂ψ∗

∂x− ψ∗ ∂ψ

∂x

)dx.

b. The Schrödinger equation gives:

dA

dt= �

2

2m2

∫x

(ψ

∂3ψ∗

∂x3− ∂2ψ

∂x2

∂ψ∗

∂x

)dx + c.c.

Using

ψ∂3ψ∗

∂x3− ∂2ψ

∂x2

∂ψ∗

∂x+ c.c. = ∂

∂x

(ψ

∂2ψ∗

∂x2+ ψ∗ ∂2ψ

∂x2− 2

∂ψ

∂x

∂ψ∗

∂x

)

and an integration by part, we obtain:

dA

dt= − �

2

2m2

∫ (ψ

∂2ψ∗

∂x2+ ψ∗ ∂2ψ

∂x2

)dx + �

2

m2

∫∂ψ

∂x

∂ψ∗

∂xdx,

and a second integration by parts of the first member of the right hand side yieldsthe result for B(t).

c. Using the Schrödinger equation again, one gets:

dB

dt= i�3

m3

∫ (∂3ψ

∂x3

∂ψ∗

∂x− ∂ψ

∂x

∂3ψ∗

∂x3

)dx,

and an integration by part shows that this is zero. The coefficient B is a constantand we shall put in the following B = 2v2

1 , where v1 has the dimension of avelocity.

d. The integration of the equation of evolution for A(t) yields A(t) = 2v21 t + ξ0 and

〈x2〉t = 〈x2〉0 + ξ0t + v21 t

2.e. Using 〈x〉t = 〈x0〉0 + v0t, we obtain for Δx2

t = 〈x2〉t − 〈x〉2t the wanted result.

Note that these results can be recovered in a simple way using Ehrenfest’s theorem(Chap. 7).

18 Solutions to the Exercises 461

2. The Gaussian wave packet

a. For t = 0, we obtain:

ψ(x, 0) = (σ2/π

)1/4eip0x/� e−x2σ2/2

and:

〈x〉0 = 0 〈p〉0 = p0 Δx0 = 1

σ√

2Δp0 = σ�√

2.

Hence the result Δx0 Δp0 = �/2. The Heisenberg inequality is saturated in thecase of a Gaussian wave packet.

b. At time t, the wave function ψ(x, t) is the Fourier transform ofe−ip2t/(2m�) ϕ(p), which is still the exponential function of a second order polyno-mial in the variable p (with complex coefficients). The general results concerningthe Fourier transform of Gaussian functions apply and one obtains after a rathertedious calculation:

|ψ(x, t)|2 = 1

Δx(t)√

2πexp

(−

(x − p0t

m

)2 1

2Δx2(t)

),

with Δx2(t) given by (3.24). One recovers on this particular case the generalresults of Chap. 2 and of the previous exercise: propagation of the center of thewave packet at the velocity 〈p〉0/m, quadratic variation of the variance of the wavepacket.

3. Characteristic size and energy in a linear or quadratic potential

For a quadratic potential, the relation (3.21) with γ = 1 yields as a typical energy Efor the ground state:

E = �2

2m Δx2+ 1

2mω2 Δx2

which is minimum for Δx = √�/mω, with E = �ω in this case. This is indeed the

correct order of magnitude for the extension and the energy of the ground state: theexact result is �ω/2 (Chap. 4).

For the linear potential α|x|, the order of magnitude of the energy of the groundstate is obtained as the minimum of

E = �2

2mΔx2+ αΔx

when the extension Δx varies. It corresponds to Δx = (�

2/mα)1/3

, and the corre-

sponding energy is (3/2)(�

2α2/m)1/3

.

462 18 Solutions to the Exercises

4. Laplacian operator in three dimensions

We notice that for r = 0, then Δ(1/r) = 0. Using an integration by parts, we findfor any ϕ in S:

∫Δ

(1

r

)ϕ(r) d3r = −

∫∇

(1

r

)· ∇ϕ(r) r2 dr d2Ω

= +∫

d2Ω

∫ ∞

0

1

r2

(∂

∂rϕ(r)

)r2 dr

= −∫

d2Ω ϕ(0) = −4π ϕ(0),

which demonstrates the very useful identity Δ(1/r) = −4πδ(r).

5. Fourier transform and complex conjugation

The Fourier transform of f ∗(k) is g∗(−x). If f (k) is real, then g∗(−x) = g(x). If f (k)is even, then g(x) is also even. We can then conclude:

f (k) real and even ←→ g(x) real and even.

Chapter 3

1. Expectation values and variances

〈x〉 = a/2, Δx = a√

1/12 − 1/2π2, 〈p〉 = 0, Δp = π�/a, therefore:

Δx Δp = �π√

1/12 − 1/2π2 ∼ 0.57�.

2. The mean kinetic energy is positive

Using an integration by parts, we find:

〈p2x〉 = −�

2∫

ψ∗(x)∂2ψ

∂x2dx = �

2∫ ∣∣∣∣

∂ψ

∂x

∣∣∣∣2

dx ≥ 0.

3. Real wave functions

If ψ is real, we have:

〈p〉 = �

i

∫ +∞

−∞ψ(x)

∂ψ

∂xdx = �

2i

[ψ2

]+∞−∞

= 0

since ψ2 vanishes at infinity for a physical state.

18 Solutions to the Exercises 463

4. Translation in momentum space

Using the general properties of the Fourier transform, we obtain 〈p〉 = p0 + q andΔp = σ.

5. The first Hermite function

We use:

d

dx

(e−x2/2

)= −x e−x2/2 and

d2

dx2

(e−x2/2

)= (x2 − 1) e−x2/2,

hence the result.

Chapter 4

1. Uncertainty relation for the harmonic oscillator

The functions ψn(x) are either odd or even, so that the density probability |ψn(x)|2is even, hence 〈x〉 = 0. From this well defined parity of the ψn’s, we also deduce〈p〉 = 0, since the operator ∂/∂x changes the parity of the function. We have

〈x2〉 = Δx2 =∫ +∞

−∞ψ∗n(x) x

2 ψn(x) dx.

Therefore, by applying (5.16), 〈x2〉 = Δx2 = (n + 1/2) �/(mω). In order to calcu-late 〈p2〉, we can just refer to the initial eigenvalue equation (5.9) and notice that〈p2〉/2m + m ω2〈x2〉/2 = En, therefore 〈p2〉 = Δp2 = (n + 1/2) m�ω. Altogether,one finds that in the eigenstate n: Δx Δp = (n + 1/2)�. For n = 0, Δx Δp = �/2:the eigenfunction is a Gaussian and the Heisenberg inequality is saturated.

2. Time evolution of a one-dimensional harmonic oscillator

The initial wavefunction of the system is given as a linear combination of the eigen-functions of the Hamiltonian, which makes the calculations quite simple.

a. The wave function at time t is:

ψ(x, t) = cos θ φ0(x) e−iωt/2 + sin θ φ1(x) e

−3iωt/2.

b. One deduces the expectation values:

〈E〉 = (cos2 θ + 3 sin2 θ) �ω/2 〈E2〉 = (cos2 θ + 9 sin2 θ) �2ω2/4,

and the variance ΔE2 = sin2(2θ) �2ω2/4. The expectation values of functions

of the energy are all time independent as a consequence of energy conservation(Ehrenfest’s theorem, see Chap. 7).

464 18 Solutions to the Exercises

c. For the position distribution, we get:

〈x〉 =√

�

2mωcos ωt sin 2θ 〈x2〉 = �

2mω(1 + 2 sin2 θ).

3. Three-dimensional harmonic oscillator

a. The reasoning is similar to the one for the three dimensional square well, and weobtain:

En1,n2,n3 = (n1 + n2 + n3 + 3/2) �ω,

where the ni’s are non negative integers. The eigen-energies can therefore bewritten En = (n + 3/2) �ω, where n is a non negative integer. The degeneracy gnof the level En is the number of triplets (n1, n2, n3) whose sum n1 + n2 + n3 isequal to n; one finds that gn = (n + 1)(n + 2)/2.

b. If the φn’s are the eigenfunctions of the one-dimensional harmonic oscillator withangular frequency ω, we have:

Φn1,n2,n3(r) = φn1(x) φn2(y) φn3(z).

c. If the oscillator is not isotropic, the energy levels read:

En1,n2,n3 = (n1 + 1/2) �ω1 + (n2 + 1/2) �ω2 + (n3 + 1/2)�ω3,

and the corresponding wavefunctions are:

Φn1,n2,n3(r) = φ(1)n1

(x) φ(2)n2

(y) φ(3)n3

(z),

where φ(i)n represents the nth eigenstate in the potential of frequency ωi. These

energy levels are generally not degenerate, except if the ratio of two frequenciesωi and ωj is a rational number.

4. One-dimensional infinite potential well

a. For symmetry reasons, we get 〈x〉 = a/2. For 〈x2〉, we find:

〈x2〉 = 2

a

∫ a

0x2 sin

nπx

adx = a2

(1

3− 1

2n2π2

),

therefore:

Δx2 = a2

12

(1 − 6

n2π2

).

18 Solutions to the Exercises 465

b. We first normalize the wavefunction ψ(x) = Ax(a − x):

∫ a

0|ψ|2 dx = |A|2

∫ a

0x2(a − x)2 dx = |A|2 a

5

30.

hence |A|2 = 30/a5. The probabilities are pn = |αn|2 with:

αn = A

√2

a

∫ a

0x(a − x) sin

nπx

adx = 4

√15

n3π3(1 − (−1)n) .

For symmetry reasons, αn = 0 if n is even. We find:

∞∑

n=1

pn = 960

π6

∞∑

k=0

1

(2k + 1)6= 1

〈E〉 =∑

Enpn = �2π2

2ma2

960

π6

∑ 1

(2k + 1)4= 5�

2

ma2

〈E2〉 =∑

E2npn = �

4π4

4m2a4

960

π6

∑ 1

(2k + 1)2= 30�

4

m2a4

ΔE =√

5�2

ma2.

c. We can calculate directly the average value and the variance of the kinetic energyoperator. For the average value we recover the result:

〈E〉 = − �2

2m

∫ a

0ψ(x)

d2ψ

dx2dx = �

2a

m|A|2

∫ a

0x(a − x) dx = 5�

2

ma2.

In order to calculate 〈E2〉 one meets a difficulty since, in the sense of functions (butnot of distributions), d4ψ/dx4 = 0. This would lead to the absurd result 〈E2〉 = 0and to a negative variance σ2 = 〈E2〉 − 〈E〉2.In fact, in this problem, the Hilbert space is defined as the C∞ periodic functions,of period a, which vanish at x = 0 and x = a. For such functions, one has:

∫ a

0φ1(x)

d4φ2

dx4dx =

∫ a

0

d2φ1

dx2

d2φ2

dx2dx

for all pairs of functions φ1(x),φ2(x). This relation can serve as a definition ofthe operator H2 in this problem. With this prescription, one obtains:

〈E2〉 = �4

4m2

∫ a

0

(∂2ψ

∂x2

)2

dx = A2�

4a

m2= 30�

4

m2a4.

466 18 Solutions to the Exercises

Actually, the correct general definition of the operator E2K consists in using

the spectral representation E2Kψ(x) ≡ ∫

K(x, y) ψ(y) dy where the kernel K isdefined as:

K(x, y) = 2

a

∞∑

n=1

E2n sin(nπx/a) sin(nπy/a).

This amounts to using the probabilities pn as above in the calculation of theexpectation values of powers of the energy 〈Eq〉.

5. Isotropic states of the hydrogen atom

a. The Schrödinger equation is:

− �2

2mψ′′ − A

xψ = Eψ.

Therefore:

− �2

2m

(−2

a+ x

a2

)e−x/a − Ae−x/a = E x e−x/a.

Equating the terms in e−x/a, we obtain:

A = �2/ma i.e. a = �

2/mA = �/mcα.

The terms in x e−x/a then give: E = −�2/2ma2, i.e. E = −mc2α2/2.

b. E = −13.6 eV, a = 5.3 × 10−2 nm.c. The condition

∫ ∞0 ψ2dx = 1 entails C2a3/4 = 1, i.e. C = 2/a3/2.

d. We find 〈1/x〉 = C2∫ ∞

0 x e−2x/adx = 1/a and therefore:

〈V 〉 = −A〈1

x〉 = −mc2α2 = 2E.

The expectation value 〈p2/2m〉 can be calculated directly by noticing that in thestate under consideration:

E = 〈 p2

2m〉 − 〈A

x〉 therefore 〈 p

2

2m〉 = 1

2mc2α2 = −E.

We obtain the relation 2〈p2/2m〉 = +〈A/x〉, which is also true in classicalmechanics if the averaging is performed on a closed orbit (virial theorem).

6. δ-function potentials

a. (i) One obtains limε→0

(ψ′(+ε) − ψ′(−ε)) = (2mα/�2)ψ(0).

18 Solutions to the Exercises 467

(ii) There is only one bound state, with ψ(x) = λ−1/20 e−|x|/λ0 . One has Kλ0 = 1

and E = −�2/(2mλ2

0).b. The wave function can be written:

ψ(x) = B eK(x+d/2) for x < −d/2

ψ(x) = C e−K(x+d/2) + C′ eK(x−d/2) for − d/2 ≤ x ≤ +d/2

ψ(x) = B′ e−K(x−d/2) for x > d/2

The solutions can be classified in terms of their symmetry with respect to x = 0:

• Symmetric solution ψS : B = B′, C = C′,• Antisymmetric solution ψA : B = −B′, C = −C′.

The quantization condition is (Kλ0 − 1)2 = (e−Kd)2.The energy levels E± are therefore obtained by solving the equation Kλ0 = 1 ±e−Kd . E− corresponds to ψS and E+ to ψA. If λ0 > d, there is only one bound stateψS; if λ0 < d, there are two bound states ψS and ψA. One can compare this withthe model of the ammonia molecule, in particular if Kd � 1.

Chapter 5

1. Translation and rotation operators

a. The proof is straightforward. We have:

p = �

i

d

dx⇒ T(x0) = e−x0

ddx .

Therefore:

T(x0) ψ(x) =∞∑

n=0

(−x0)n

n!

(d

dx

)n

ψ(x),

which is simply the Taylor expansion of ψ(x − x0) around point x.b. Similarly, one finds

R(ϕ) ψ(r, θ) =∞∑

n=0

(−ϕ)n

n!(

∂

∂θ

)n

ψ(r, θ),

where we recognize the Taylor expansion of ψ(r, θ − ϕ).

2. The evolution operator

The formula giving the derivative of the exponential of a function remains validfor the operator U since the Hamiltonian is time independent. Hence U(τ ) and Hcommute. We have therefore:

468 18 Solutions to the Exercises

d

dtU(t − t0) = − i

�H U(t − t0).

This entails that |ψ(t)〉 = U(t − t0) |ψ(t0)〉 is a solution of the Schrödinger with theproper initial condition since U(0) = 1. The unitarity of U is a direct consequenceof the fact that H is Hermitian:

U†(τ ) = eiHτ/� = U(−τ ) = U−1(τ ).

3. Heisenberg representation

We have by definition:

a(t) = 〈ψ(t)|A|ψ(t)〉 = 〈ψ(0)|U†(t)AU(t)|ψ(0)〉.

We can define A(t) = U†(t)AU(t), i.e.

A(t) = eiHt/� A e−iHt/�,

which is indeed a solution of the differential equation given in the text of the exerciseand which is such that a(t) = 〈ψ(0)|A(t)|ψ(0)〉.

4. Dirac formalism with a two-state problem

a. 〈ψ2|H|ψ1〉 = E1〈ψ2|ψ1〉 = E2〈ψ2|ψ1〉, therefore (E1 − E2)〈ψ2|ψ1〉 = 0 and〈ψ2|ψ1〉 = 0.

b. 〈E〉 = (E1 + E2)/2, ΔE2 = ((E1 − E2)/2)2, ΔE = �ω/2.

c. |ψ(t)〉 = (e−iE1t/�|ψ1〉 − e−iE2t/�|ψ2〉

)/√

2.

d. a = ±1.e. |ψ±〉 = (|ψ1〉 ± |ψ2〉) /

√2.

f. p = |〈ψ−|ψ(t)〉|2 = cos2(ωt/2).

Chapter 6

1. Linear three-atom molecule

a. The three energy levels and the corresponding eigenstates are:

E1 = E0, |ψ0〉 = 1√2

⎛

⎝10

−1

⎞

⎠

and

E± = E0 ± a√

2, |ψ±〉 = 1

2

⎛

⎝1

∓√2

1

⎞

⎠ .

18 Solutions to the Exercises 469

b. The probabilities are PL = PR = 1/4 and PC = 1/2.c. One has:

|ψL〉 = 1√2|ψ0〉 + 1

2|ψ+〉 + 1

2|ψ−〉

hence 〈E〉 = E0 and ΔE = a.

2. Crystallized Violet and Malachite Green

a. The Hamiltonian is:

H =⎛

⎝0 −A −A

−A 0 −A−A −A 0

⎞

⎠ .

As in the case of NH3, the off-diagonal elements of the matrix H written in thebasis of classical configurations represent the quantum effects, i.e., the passageby tunnelling from one configuration to the another one.

b. 〈E〉1 = 〈φ1|H|φ1〉 = −2A, 〈E2〉1 = 4A2, therefore ΔE2 = 0 in the state |φ1〉.Similarly, 〈E〉2 = 〈φ2|H|φ2〉 = +A, 〈E2〉2 = A2, therefore ΔE2 = 0 in the state|φ2〉.Consequently, these two states are energy eigenstates with respective eigenvalues−2A and +A. Naturally, this can be seen directly by letting H act on these twostates.

c. Knowing two eigenvectors of H, it suffices to look for a vector orthogonal to bothof them. One finds |φ3〉 = (2|1〉 − |2〉 − |3〉)/√6. Altogether, we obtain:eigenvalue λ = −2A eigenvector |φ1〉 = (|1〉 + |2〉 + |3〉)/√3;eigenvalue λ = +A, eigenvector |φ2〉 = (|2〉 − |3〉)/√2;eigenvalue λ = +A, eigenvector |φ3〉 = (2|1〉 − |2〉 − |3〉)/√6.The eigenvalue λ = A has a degeneracy equal to 2. This eigenbasis is not unique.

d. Absorption of light occurs with ΔE = 3A = 2.25 eV, in the yellow part of thespectrum. The ion appears with the complementary color, i.e. violet.

e. The Hamiltonian is now:

H =⎛

⎝Δ −A −A−A 0 −A−A −A 0

⎞

⎠

and the eigenvalue equation can be written:

det(H − λI) = −(λ − A)(λ2 + (A − Δ)λ − A(Δ + 2A)

),

hence the eigenvalues:

E0 = A, E± =((Δ − A) ±

√(Δ + A)2 + 8A2

)/2.

• If Δ � A : E0 = A , E+ ∼ A + 2Δ/3 , E− ∼ −2A + Δ/3.• If Δ � A : E0 = A , E+ ∼ Δ, E− ∼ −A.

470 18 Solutions to the Exercises

f. There are two possible transitions from the ground state E−:

hν1 = hc/λ1 = hc/(450 nm) ∼ 2.75 eV =√

(Δ + A)2 + 8A2

which corresponds to a value Δ = 1 eV, and an absorption in the violet part. Withthis value of Δ one obtains

hν2 = E0 − E− = (3A − Δ)/2 + 1/2√

(Δ + A)2 + 8A2 = 2 eV.

This corresponds to an absorption in the red-orange part of the spectrum andto a wavelength λ2 = hc/hν2 = 620 nm in good agreement with experimentalobservation.

Chapter 7

1. Commutator algebra

The first relation is immediate. To show the second relation, one can start with:

ABn − BnA = ABn − BABn−1

+ BABn−1 − B2ABn−2

+ B2ABn−2 − . . .

+ Bn−1AB − BnA,

and each line of the right hand side can be written as Bs[A, B]Bn−s−1 with s =0, . . . , n − 1, hence the result. Finally the Jacobi identity is derived by expanding allcommutators, and checking that the twelve terms which appear cancel each other.

2. Classical equations of motion for the harmonic oscillator

For the harmonic oscillator, the Ehrenfest theorem gives:

d〈x〉dt

= 〈p〉m

d〈p〉dt

= −mω2〈x〉

hence the result: d2〈x〉 / dt2 + ω2 〈x〉 = 0.

3. Conservation law

For a system of n interacting particles, the total Hamiltonian reads:

H =n∑

i=1

p2i

2mi+ 1

2

∑

i

∑

j =i

V (ri − rj).

18 Solutions to the Exercises 471

The total momentum of the system is P = ∑i pi. This operator commutes both with

the kinetic energy term of the Hamiltonian and with the interaction term:

[Px, V (ri − rj)] = [px,i, V (ri − rj)] + [px,j, V (ri − rj)]= �

i

∂V

∂x(ri − rj) − �

i

∂V

∂x(ri − rj) = 0.

Therefore 〈P〉 is a constant of motion.

4. Hermite functions

The expression of the position and momentum operators in terms of creation andannihilation operators are:

x =√

�

2mω

(a + a†

)p = i

√m�ω/2

(a† − a

)

hence the result.

5. Generalized uncertainty relations

a. Applying the commutator to any function Ψ (r), one has for all Ψ :

[px, f ] Ψ (r) = �

i

(∂

∂x(fΨ ) − f

∂Ψ

∂x

)= −i�

x

rf ′(r) Ψ (r)

hence the relation [px, f ] = −i�(x/r) f ′(r).b. • The square of the norm of Ax|ψ〉 is:

‖ Ax|ψ〉 ‖2 = 〈ψ|(px + iλxf )(px − iλxf |ψ〉= 〈ψ|p2

x + λ2x2 f 2 − iλ[px, xf ]|ψ〉.

By a direct calculation, one has: [px, xf ] = −i�(f + x2

r f′), therefore

‖ Ax|ψ〉 ‖2= 〈p2x〉 + λ2〈x2f 2〉 − �λ〈f + x2

rf ′〉.

• Adding the analogous relations for Ay and Az, we obtain for any state |ψ〉

〈p2〉 + λ2〈r2f 2〉 − �λ〈3f + rf ′〉 ≥ 0.

This second degree trinomial in λ must be non-negative for all λ. Therefore,the discriminant is negative or zero, i.e.

4〈p2〉〈r2f 2〉 ≥ �2〈3f + rf ′〉2.

472 18 Solutions to the Exercises

c. For f = 1, we obtain 〈p2〉 〈r2〉 ≥ (9/4)�2;for f = 1/r, 〈p2〉 ≥ �

2〈r−1〉2;for f = 1/r2, 〈p2〉 ≥ (�2/4) 〈r−2〉.

d. Harmonic oscillator For any state |ψ〉, one has 〈E〉 ≥ �2/(2m〈r2〉) + mω2 〈r2〉/2.

Minimizing with respect to 〈r2〉, we find the lower bound 〈E〉 ≥ (3/2)�ω for theenergy of the oscillator. Since there is a value of λ and a corresponding valueof 〈r2〉, for which the trinomial has a double root and vanishes, this means thereexists a state for which this lower bound is attained. It is therefore the groundstate since no state can have a lower energy than the ground state. We haveAx|φ〉 = 0 → (px + iλx)|φ〉 = 0. In terms of wave functions, this corresponds to:

(�

∂

∂x+ λx

)φ(r) = 0,

and a similar equation for y and z. The solution of this set of three equations is:

φ(r) = N exp(−λr2) = N exp(−3r2/4r20),

with 〈r2〉 = r20 , N−1 = (2πr2

0/3)3/4 and λ = 3�/4r20 .

e. Hydrogen atom

• Similarly, the lower bound for the energy of the hydrogen atom is Emin =−mee4/(2�

2), and this bound is attained, therefore it is the ground state energy.• The equation Ax|ψ〉 = 0 leads to the differential equation:

�∂ψ

∂x+ λ

x

rψ(r) = 0

and a similar equation for y and z. The solution of this equation is ψ(r) =N exp (−r/r0) with N = 1/

√πr3

0 , 〈r−1〉 = 1/r0, and λ = 1/r0. This is indeedthe ground state of the hydrogen atom.

7. Time-energy uncertainty relation

Uncertainty relations give ΔaΔE ≥ |〈ψ|[A, H]|ψ〉|/2. The Ehrenfest theorem givesd〈a〉/dt = (1/i�)〈ψ|[A, H]|ψ〉 hence the inequality and the result.

8. Virial Theorem

a. We have:

[H, xp] = [H, x]p + x[H, p] = i�

(− p2

m+ x

∂V

∂x

)

and, for the potential under consideration:

18 Solutions to the Exercises 473

[H, xp] = i�

(− p2

m+ nV (x)

).

b. For an eigenstate |ψα〉 of H one has 〈ψα|[H, xp]|ψα〉 = 0. Hence:

〈ψα| − p2

m+ nV (x)|ψα〉 = 0 ⇒ 2〈T〉 = n〈V 〉.

The harmonic oscillator corresponds to the case n = 2: the average kinetic andthe potential energies are equal when the system is prepared in an eigenstate |n〉of the Hamiltonian.

c. We obtain in three dimensions:

[H, r · p] = i�

(−p2

m+ r · ∇V (r)

)

therefore 2〈T〉 = n〈V 〉 as obtained previously. This applies to the Coulomb prob-lem (V (r) = −e2/r). In this case, n = −1 and we have 2〈T〉 = −〈V 〉. For theharmonic oscillator, one has 〈T〉 = 〈V 〉.

d. From the equation obtained above, we have for a central potential V (r):

r · ∇V (r) = r∂V

∂rhence 2〈T〉 = 〈r ∂V

∂r〉.

Chapter 8

1. Calculations of energy levels

Consider a particle of mass m placed in the isotropic 3D potential V (r) ∝ rβ . Wechoose the normalized Gaussian test function:

ψa(r) = (a/π)3/4 exp(−ar2/2). (E.1)

We find in this state:

〈p2〉 = 3

2a�

2 〈rβ〉 = a−β/2 Γ (3/2 + β/2)

Γ (3/2)

This gives an upper bound for the ground state of:

• The harmonic potential (β = 2) for which we recover the exact result.• The Coulomb potential V (r) = −e2/r; one finds:

E0 = − 4

3π

me4

�2to be compared with the exact result: − 1

2

me4

�2.

• The linear potential V (r) = gr; one finds:

474 18 Solutions to the Exercises

E0 =(

81

2π

)1/3 (�

2g2

2m

)1/3

� 2.345

(�

2g2

2m

)1/3

to be compared with the coefficient 2.338 of the exact result.

2. Uncertainty relations

Using the inequality (9.24) for systems whose ground state is known, we can deriveuncertainty relations between 〈p2〉 and 〈rα〉, where α is a given exponent.

a. The 〈r2〉 〈p2〉 uncertainty relation. Consider a one-dimensional harmonic oscil-lator, whose ground state is �ω/2. Whatever the state |ψ〉 we have:

〈p2〉2m

+ 1

2mω2〈x2〉 ≥ �ω

2⇒ 〈p2〉 + m2ω2〈x2〉 − �mω ≥ 0.

We recognize a second degree polynomial inequality in the variable mω. Thenecessary and sufficient condition for this to hold for all values of mω is:

〈p2〉〈x2〉 ≥ �2

4. (E.2)

In three dimensions, noting r2 = x2 + y2 + z2, we obtain in the same manner:

〈p2〉〈r2〉 ≥ 9�2

4. (E.3)

b. The 〈1/r〉 〈p2〉 uncertainty relation. The hydrogen atom Hamiltonian is H =p2/2m − e2/r and its ground state energy is E0 = −me4/(2�

2) (see Chap. 11).Consequently we have for all |ψ〉:

〈p2〉2m

− e2〈1

r〉 ≥ −me4

2�2.

We have again a second degree polynomial in the variable me2 which is alwayspositive, from which we deduce:

〈p2〉 ≥ �2〈1

r〉2. (E.4)

3. Comparison of the ground states of two potentials

We denote |ψi〉 (i = 1, 2) the ground state of the Hamiltonian H = p2/2m + Vi(r)with energy Ei. Since V2(r) > V1(r) for any r we have:

18 Solutions to the Exercises 475

〈ψ2|V2(r)|ψ2〉 > 〈ψ2|V1(r)|ψ2〉⇒ E2 = 〈ψ2| p

2

2m+ V2(r)|ψ2〉 > 〈ψ2| p

2

2m+ V1(r)|ψ2〉.

The second step of the reasoning consists in noticing that, because of the theorem atthe basis of the variational method, one has:

〈ψ2| p2

2m+ V1(r)|ψ2〉 ≥ E1 = 〈ψ1| p

2

2m+ V1(r)|ψ1〉,

hence the result.

4. Existence of a bound state in a potential well

To show the existence of a bound state in a one-dimensional potential V (x) which isnegative everywhere and tends to zero at ±∞, we use the variational method withGaussian trial functions:

ψσ(x) = (σ2/(2π))1/4 exp(−σ2x2/4).

The mean kinetic energy is Tσ = �2σ2/(8m). This positive quantity tends to zero

quadratically when σ tends to zero. The mean potential energy is:

〈V 〉σ = σ√2π

∫V (x) e−σ2x2/2 dx.

By assumption this quantity is negative. When σ tends to zero, it tends to zero linearlywith σ if the integral

∫ ∞−∞ V (x) dx converges, or it may even diverge if the integral

is itself divergent. In any case, there exist a range of value for the variable σ suchthat |〈V 〉σ| > Tσ . For these values of σ, the total mean energy Eσ = Tσ + 〈V 〉σ isnegative. The ground state energy is necessarily lower than Eσ and it is also negative:the corresponding state is a bound state.

This result cannot be extended to three dimensions. If one still takes Gaussianfunctions e−σ2r2/4 as a trial set, the kinetic energy varies as σ2 as before, while thenegative potential energy term now scales as σ3. When σ tends to zero, the potentialenergy tends to zero faster than the kinetic energy, and the total energy may be alwayspositive so that one cannot infer the existence of a bound state.

Actually one can find simple three-dimensional potentials negative or zero every-where which have no bound state. Consider for instance an isotropic square potentialwell: V (r) = −V0 for r < r0 (with V0 > 0) and V (r) = 0 otherwise. One finds thatthere is no bound state if V0 < �

2π2/(8mr20). (To show this result, consider first states

with zero angular momentum, and then generalize to arbitrary angular momenta usingthe previous exercise).

476 18 Solutions to the Exercises

5. Generalized Heisenberg inequalities

For any state |ψ〉, we have:

1

2m〈p2〉 + g〈rα〉 − ε0 g2/(α+2)

(�

2

2m

)α/(α+2)

≥ 0.

Minimizing with respect to g, one obtains the result mentioned in the text (it is safeto treat separately the case α > 0, g > 0 in which case ε0 > 0, and the case α < 0,g < 0 in which case ε0 < 0). One recovers the usual results for α = 2 and α = −1.For the linear potential, α = 1, we have ε0 = 2.33811 hence the uncertainty relation:

〈p2〉〈r〉2 ≥ 4

27ε3

0 �2 ∼ 1.894 �

2.

Chapter 9

1. Rotation invariant operator

[A, Lx] = 0 and [A, Ly] = 0 ⇒ [A, [Lx, Ly]] = i�[A, Lz] = 0.

2. Commutation relations for r and p

A straightforward calculation gives:

[Lz, x] = −y[px, x] = i�y [Lz, y] = x[py, y] = −i�x [Lz, z] = 0,

[Lz, px] = py[x, px] = i�py [Lz, py] = −px[y, py] = −i�px [Lz, pz] = 0.

Therefore, since [A, B2] = [A, B]B + B[A, B], we obtain [L, p2] = [L, r2] = 0.

3. Rotation invariant potential

IfV (r) depends only on r = |r|, then H = p2/2m + V (r) (which is rotation invariant)commutes with the angular momentum Lwhich is a constant of the motion (Ehrenfesttheorem).

[H, L] = 0 ⇒ d〈L〉dt

= 0.

This is not true if V (r) depends not only on r, but also on θ and ϕ.

18 Solutions to the Exercises 477

4. Unit angular momentum

a. The action of Lz on the basis set |� = 1,m〉 is Lz|1,m〉 = m� |1,m〉:

Lz = �

⎛

⎝1 0 00 0 00 0 −1

⎞

⎠ .

The action of Lx and Ly is obtained using the operators L+ and L−, whose matrixelements are deduced from the recursion relation (10.16):

L+|1, 1〉 = 0 L+|1, 0〉 = �

√2 |1, 1〉 L+|1,−1〉 = �

√2 |1, 0〉

L−|1, 1〉 = √2 |1, 0〉 L−|1, 0〉 = �

√2 |1,−1〉 L−|1,−1〉 = 0

Hence the matrix of Lx = (L+ + L−)/2 and Ly = i(L− − L+)/2:

Lx = �√2

⎛

⎝0 1 01 0 10 1 0

⎞

⎠ Ly = �√2

⎛

⎝0 −i 0i 0 −i0 i 0

⎞

⎠ .

b. The eigenvectors of Lx are:

|1,±1〉x = 1

2

(|1, 1〉 ± √

2|1, 0〉 + |1,−1〉)

eigenvalues ± �

|1, 0〉x = 1√2

(|1, 1〉 − |1,−1〉) eigenvalue 0.

The eigenfunction corresponding to mx = +1 is:

ψ(θ,ϕ) = 1

2(Y 1

1 (θ,ϕ) + Y−11 (θ,ϕ)) + 1√

2Y 0

1 (θ,ϕ)

=√

3

8π(cos θ − i sin θ sin ϕ).

Therefore:

I(θ,ϕ) = |ψ(θ,ϕ)|2 = 3

8π(1 − sin2 θ cos2 ϕ).

5. Commutation relations for J2x , J2

y and J2z

a. Since [J2, Jz] = 0, we have also [J2, J2z ] = 0. Hence [J2

x + J2y , J

2z ] = 0, and there-

fore [J2z , J

2x ] = [J2

y , J2z ]. The third equality is obtained using a circular permutation

of x, y and z.

478 18 Solutions to the Exercises

One can also calculate explicitly these commutators:

[J2x , J

2y ] = i�{Jx, {Jy, Jz}}

= i�

3({Jx, {Jy, Jz}} + {Jy, {Jz, Jx}} + {Jz, {Jx, Jy}}).

where we set {A, B} = AB + BA.b. For j = 0 the result is obvious since |0, 0〉 is an eigenstate of all components with

eigenvalue zero.For j = 1/2, J2

x , J2y and J2

z are proportional to the unit 2 × 2 matrix, with eigen-value +�

2/4. They obviously commute.c. For j = 1, we consider the matrix elements:

〈1,m2|[J2x , J

2z ]|1,m1〉 = (m2

1 − m22)〈1,m2|J2

x |1,m1〉.

For m21 = m2

2, this is obviously zero. We only have to consider the cases m1 =0,m2 = ±1. Since Jx|1, 0〉 ∝ (|1, 1〉 + |1,−1〉) and Jx|1,±1〉 ∝ |1, 0〉 the cor-responding scalar products, i.e. the matrix elements under consideration, vanish.

Owing to the (x, y, z) symmetry, the common eigenbasis is {|j = 1,mx = 0〉,|j = 1,my = 0〉, |j = 1,mz = 0〉} where |j = 1,mi = 0〉 is the eigenvector of Ji(i = x, y, z) associated with the eigenvalue 0:

|j = 1,mx = 0〉 = 1√2

(|1, 1〉 − |1,−1〉) ,

|j = 1,my = 0〉 = 1√2

(|1, 1〉 + |1,−1〉) ,

|j = 1,mz = 0〉 = |1, 0〉.

In spherical coordinates, the corresponding (angular) wave functions are x/r, y/rand z/r, with a normalization coefficient (4π/3)−1/2.

Chapter 10

1. Expectation value ofr for the Coulomb problem

a. Using∫ ∞

0 |un,�(ρ)|2dρ = 1 and ε = 1/n2, we obtain the first identity.b. To show the second identity we use:

∫u′n,� un,� dρ = 0

∫ρ2 u′

n,� u′′n,� dρ =

∫ρ un,� u

′′n,� dρ

∫ρ u′

n,� un,� dρ = −1/2∫

ρ2 u′n,� un,� dρ = −〈ρ〉,

which can be shown by integrating by parts.

18 Solutions to the Exercises 479

c. Summing the two equations derived before, we obtain the desired expressionfor 〈ρ〉.

2. Three-dimensional harmonic oscillator in spherical coordinates

a. The radial equation is

(− �

2

2m

1

r

d2

dr2r + 1

2mω2r2 + �(� + 1)

2mr2�

2 − En,�

)Rn,�(r) = 0.

We set ρ = r√mω/� and ε = E/�ω and we obtain (11.30).

b. The energy levels are indeed of the form

En = (n + 3/2)�ω with n = 2n′ + �,

and the corresponding eigenstates can be labelled as |n, �,m〉. There exists adegeneracy in �, but it is different from the case of the hydrogen atom. For agiven value of the energy, i.e. of n, � has the same parity as n. Therefore thesuccessive levels correspond alternatively to even and odd angular momenta:

n = 0 E = 3�ω/2 � = 0

n = 1 E = 5�ω/2 � = 1

n = 2 E = 7�ω/2 � = 0, 2

n = 3 E = 9�ω/2 � = 1, 3 etc.

For a given value of n, the (n + 1)(n + 2)/2 states |n, �, m〉 with (� = 0, 2 . . . n),or (� = 1, 3 . . . n), are linear combinations of the (n + 1)(n + 2)/2 states|n1; n2; n3〉 with n1 + n2 + n3 = n.

c. For n = 1, three orthogonal wave functions are

ϕ100(r) = Cxe−αr2/2 , ϕ010(r) = Cye−αr2/2 , ϕ001(r) = Cze−αr2/2

where C is a normalization constant. We therefore obtain, by expressing x, y, zin terms of r and Y1,m(θ,ϕ):

|n = 1, � = 1, m = 0〉 = |n1 = 0; n2 = 0; n3 = 1〉,

|n = 1, � = 1, m = ±1〉 = ∓ 1√2|n1 = 1; n2 = 0; n3 = 0〉

− i√2|n1 = 0; n2 = 1; n3 = 0〉.

480 18 Solutions to the Exercises

3. Relation between the Coulomb problem and the harmonic oscillator

a. The change of variable ρ → x and of unknown function u(ρ) → f (x) leads to thefollowing equation for the Coulomb problem:

d2f

dx2+ 2α − 1

x

df

dx+

(α(α − 2) − 4�(� + 1)

x2+ 8Z + 4E

EIx2

)f (x) = 0.

The choice α = 1/2 eliminates the term df /dx and leads to an equation with thesame structure as the radial equation for the harmonic oscillator:

(d2

dx2− (2� + 1/2)(2� + 3/2)

x2+ 8Z + 4E

EIx2

)f (x) = 0.

b. The correspondence between the parameters of, respectively, the harmonic oscil-lator and the Coulomb problem, is:

�harm. ↔ (2�coul. + 1/2) K2 ↔ −4Ecoul./EI Eharm./�ω ↔ 4Z.

In other words, the roles of the coupling constant and of the energy eigenvalueare interchanged! The shift in � ensures the proper �-degeneracy of the hydrogenlevels.

c. From the result of exercise 2, we know that the eigenvalues for the energy are:

Eharm = K(2n′ + �harm + 3/2

)�ω.

Using the correspondence that we just found, this yields:

4Z =√

−4Ecoul.

EI

(2n′ + 2�coul. + 2

)

which can also be writtenEcoul. = Z2 EI/(n′ + �coul. + 1)2. We recover indeed theenergy levels of the Coulomb problem. Notice that this provides an expression ofthe Laguerre polynomials in terms of the Hermite polynomials.

4. Confirm or invalidate the following assertions

a. True. In fact, if [H, L] = 0 one has [H, Lz] = 0 and [H, L2] = 0. We consider aneigenbasis common to H , L2 and Lz, |E�,m, �,m〉. Since [H, L] = 0 implies that[H, L±] = 0, we obtain:

L±H|E�,m, �,m〉 = E�,mL±|E�,m, �,m〉 = HL±|E�,m, �,m〉.

18 Solutions to the Exercises 481

We know that L±|E�,m, �,m〉 is an eigenstate of Lz with the eigenvalue (m + 1)�.Therefore the set of states {|E�,m, �,m〉,m = −� . . . �} are eigenstates of H withthe same eigenvalue E�,m ≡ E�.

b. Wrong. The square of the angular momentum commutes with the Hamiltonianand the energy levels can be labelled with �. It is only the Coulomb and harmonicpotentials which have special symmetry properties which produce degeneraciesin �.

5. Centrifugal barrier effects

Consider the Hamiltonians:

H� = p2r

2m+ �(� + 1)�2

2mr2+ V (r) where p2

r = −�2

(1

r

∂

∂rr

)2

which act only on the variable r. The state |n′ = 0, �,m〉 ≡ |ψ�〉, which is an eigen-state of:

H = p2r

2m+ L2

2mr2+ V (r),

is an eigenstate of H� with the eigenvalue E�, which is the smallest eigenvalue of H�.We obviously have:

H�+1 = H� + (� + 1)�2

mr2.

Taking the expectation value of this expression in the state |ψ�+1〉 = |n′ = 0, � +1,m〉, we obtain:

E�+1 = 〈ψ�+1|H�|ψ�+1〉 + 〈 (� + 1)�2

mr2〉.

We have 〈ψ|H�|ψ〉 ≥ E� for all ψ, and (� + 1)�2/mr2 is a positive operator: its expec-tation value in any state is positive. Therefore, E�+1 ≥ E�, and more quantitatively:

E�+1 − E� ≥ (� + 1)�2

m〈ψ�+1| 1

r2|ψ�+1〉.

6. Algebraic method for the hydrogen atom

a. The expression for A−� A

+� is:

A−� A

+� = d2

dρ2− �(� + 1)

ρ2+ 2

ρ− 1

(� + 1)2.

482 18 Solutions to the Exercises

Therefore the radial equation can be written as

(A−

� A+�

)u� =

(ε − 1

(� + 1)2

)u�.

b. One finds for A+� A

−� :

A+� A

−� = d2

dρ2− (� + 1)(� + 2)

ρ2+ 2

ρ− 1

(� + 1)2.

By multiplying (11.33) by A+� , we obtain

(A+

� A−�

)A+

� u� =(

ε − 1

(� + 1)2

)A+

� u�,

which can also be written:

(d2

dρ2− (� + 1)(� + 2)

ρ2+ 2

ρ

)A+

� u�(ρ) = ε A+� u�(ρ).

Therefore, A+� u�(ρ) satisfies the radial equation with the same eigenvalue ε but

for an angular momentum �′ = � + 1.c. Similarly one can write the equation for the angular momentum � as:

(A+

�−1A−�−1

)u� =

(ε − 1

�2

)u�.

One then obtains that A−�−1u�(ρ) satisfies the radial equation with the same eigen-

value ε but for an angular momentum �′ = � − 1.d. By multiplying (11.33) by u∗

� and integrating over ρ we find:

∫ ∞

0u∗

�(ρ)(A−

� A+� u�(ρ)

)dρ =

(ε� − 1

(� + 1)2

) ∫ ∞

0|u�(ρ)|2dρ.

We integrate by parts the left hand side of this equation:

∫ ∞

0u∗

�(ρ)(A−

� A+� u�(ρ)

)dρ = −

∫ ∞

0|A+

� u�(ρ)|2dρ

We deduce that the quantity ε − 1/(� + 1)2 is necessarily negative:

ε ≤ 1

(� + 1)2.

18 Solutions to the Exercises 483

e. The argument is then analogous to the case of the harmonic oscillator or to thequantization of angular momentum. By repeatedly applying A+

� one can increasethe value of � by an integer. This is limited from above since ε� ≤ 1/(� + 1)2 andthere is a maximum value �max of � such that:

ε = 1

(�max + 1)2≡ 1

n2.

The function A+�max

u�max(ρ) is identically zero. Therefore, u�max satisfies:

(d

dρ− n

ρ+ 1

n

)u�max(ρ) = 0.

f. The energy levels are En = −EI/n2. The solution of the above equation isu�max(ρ) ∝ ρn exp−ρ/n, up to a normalization factor. Coming back to the func-tion R�, we recover (11.23) for � = n − 1 (i.e. n′ = 0): R�max(ρ) ∝ ρn−1e−ρ

√ε.

By repeatedly applying A−�−1 one obtains the other solutions of the same energy

ε = 1/n2 for � = n − 2, n − 3, . . . , 0.

7. Molecular potential

a. The radial equation is:

(− �

2

2me

1

r

d2

dr2r + �(� + 1)�2 + 2meA

2mer2− B

r

)R(r) = E R(r).

We defineS as the positive real number such thatS(S + 1) = �(� + 1) + 2meA/�2,

i.e.:

S = −1

2+ 1

2

√(2� + 1)2 + 8meA/�2.

Note that S is generally not an integer. We set as usual a1 = �2/Bme, E =

−εmB2/2�2 and r = ρa1. The radial equation becomes:

(1

ρ

d2

dρ2ρ − S(S + 1)

ρ2+ 2

ρ− ε

)R(ρ) = 0.

As in the case of hydrogen, the normalisable solutions are labelled by an integern′ ≥ 0 and they are of the form: R(ρ) = e−ρ

√ε ρS Pn′,�(ρ), where Pn′,�(ρ) is a

polynomial of degree n′. and one must have ε = (n′ + S + 1)−2. The energies arethen:

En′,� = −B2me

2�2

1

(n′ + S + 1)2.

484 18 Solutions to the Exercises

Note that this potential is quantitatively very different from a molecular potentialeven though it has the same global features (attractive at long distances and repulsiveat short distances). The long range attractive force in a molecule is not a Coulombforce and the repulsion at short distances is much stronger than a r−2 potential.

Chapter 11

1. Determination of the magnetic state of a silver atom

a. It is always possible to write the coefficients α et β as α = cos(θ/2),β = eiϕ sin(θ/2) (if necessary, one can multiply the state (12.65) by a globalphase factor in order to have a real value for α). Consider now a Stern–Gerlachmagnet, which is oriented along the unit vector u, defined by the polar anglesθ,ϕ:

u = sin θ cos ϕ ex + sin θ sin ϕ ey + cos θ ez.

This gives:

u · μ = μ0

(cos θ e−iϕ sin θeiϕ sin θ cos θ

).

One can check easily that cos(θ/2)|+〉 + eiϕ sin(θ/2)|−〉 is an eigenstate of u · μwith the eigenvalue +μ0.

b. Bob has to choose an axis u′ for his Stern–Gerlach measurement, but he doesnot know the value of u. His measurement yields a binary answer ±μ0, andthe state of the system is then |±〉u′ . It is different from the initial state andsubsequent measurements will not bring any new information on the state sent byAlice. Consequently, Bob cannot determine the initial state from a measurementperformed on a single magnetic moment. The only certainty that Bob can havewhen he obtains the result +μ0 (resp. −μ0) in his measurement with the u′ axis isthat the initial state was not |−〉u′ (resp. |+〉u′), which is a very poor information.

c. If Alice sends a large number N of magnetic moments, all prepared in the sameunknown state (8.67), Bob can split this ensemble into three sets. For the firstN/3 magnetic moments, he measures μz. From the relative intensity of the twospots μz = ±μ0, he deduces |α|2 and |β|2. For the second set of N/3 magneticmoments, he measures μx. The relative intensities of the two spots correspondingto μx = ±μ0 yield |α ± β|2. Finally Bob measures μy for the last set of N/3magnetic moments, which yields |α ± iβ|2. From these three sets of results, Bobdeduces α and β, within a global phase factor which is unimportant. Of course,this determination of α and β is only approximate and the relative statistical erroris N−1/2.

2. Results of repeated measurements; quantum Zeno paradox

a. The energy levels of H are E± = ±�ω/2.b. The state |ψ(0)〉 is an eigenstate of μx with eigenvalues +μ0. Therefore, one

obtains +μ0 with probability 1 in a measurement of μx.

18 Solutions to the Exercises 485

c. The evolution of the state under consideration is:

|ψ(T)〉 = (|+〉e−iωT/2 + |−〉eiωT/2)/√

2.

d. The corresponding probability is:

P(T) = |x〈+|ψ(T)〉|2 = |〈ψ(0)|ψ(T)〉|2 = cos2 ωT

2.

e. After a measurement giving the result μx = +μ0, the system is again in thesame state as initially |ψ(0)〉 = (|+〉 + |−〉)/√2. The probability for all the Nsuccessive measurements to give the same result +μ0 is therefore PN (T) =cos2N (ωT/2N).

f. In the (mathematical) limit N → ∞, we obtain

PN = exp

(N ln

(cos2

(ωT

2N

)))

∼ exp

(2N ln(1 − ω2T 2

8N2)

)∼ exp

(−ω2T 2

4N

)→ 1.

This result may seem paradoxical: observing the system prevents it from evolving!Some people claim that watching water prevents it from boiling. However, thesolution of the “Quantum Zeno Paradox” lies in the fact that any measurementhas a finite extension both in space and in time. In practice, one cannot divide Tin infinitely small parts except by interacting permanently with the system, whichis another problem.

3. Products of Pauli matrices

One first checks that σ2j = 1. For j = k, a direct calculation yields the result (12.66).

4. Algebra with Pauli matrices

We first develop the products: (σ · A)(σ · B) = ∑jk σjσkAjBk . The result (12.66) of

the previous exercise then yields the desired formula.

5. Spin and orbital angular momentum

a. One must have∫ ∞

0 |R(r)|2r2dr = 1/2 since the Ym� ’s are orthonormal and∫

(|ψ+|2 + |ψ−|2) d3r = 1.b. We find for Sz: p(+�/2) = 2/3, p(−�/2) = 1/3, and for Sx: p(+�/2) = 1/3,

p(−�/2) = 2/3.c. Lz = � and Lz = 0 with p(+�) = 1/6, p(0) = 5/6.

486 18 Solutions to the Exercises

Chapter 12

1. Permutation operator

We use the eigenbasis of S1z and S2z. In this basis we find:

(σ1xσ2x + σ1yσ2y

) |σ;σ〉 = 0(σ1xσ2x + σ1yσ2y

) |σ;−σ〉 = 2| − σ;σ〉(1 + σ1zσ2z

) |σ;σ〉 = 2|σ;σ〉(1 + σ1zσ2z

) |σ;−σ〉 = 0,

where σ = ±1. Hence the result:

1

2

(1 + σ1σ2

) |σ;σ〉 = |σ;σ〉, 1

2

(1 + σ1σ2

) |σ;−σ〉 = | − σ;σ〉.

2. The singlet state

This invariance of the decomposition of the singlet state when u varies comes fromthe fact that the state is of angular momentum zero, and therefore that it is rotationinvariant.

3. Spin and magnetic moment of the deuteron

a. The eigenvalues of K2 are K(K + 1)�2 with:

K = J + I, J + I − 1, J + I − 2, . . . , |J − I|.

b. We have W = AJ · I/�2 where we set A = agIgJμBμN . One has J · I = (K

2 −J

2 − I2)/2, hence the expression of W .

c. EI,J,K = EI + EJ + A (K(K + 1) − J(J + 1) − I(I + 1)) /2.d. EI,J,K − EI,J,(K−1) = AK .e. The electron has a total angular momentum 1/2, its spin. The possible values ofK

are therefore I ± 1/2. One of the levels is split into 4 sub-levels (angular momen-tum 3/2), while the other is split in two (angular momentum 1/2). ThereforeI = 1.

f. Triplet state (S = 1).g. One obtains for the hyperfine splitting in deuterium

ΔE = 3A

2= 1.72

1836α4mec

2 ∼ 1.36 × 10−6 eV.

Hence a wavelength and a frequency of the emitted radiation

λ ∼ 91 cm , ν ∼ 328 MHz.

18 Solutions to the Exercises 487

The experimental value is λ = 91.5720 cm. In a more accurate calculation, onemust incorporate a more accurate value of gI (0.8574), and take into accountreduced mass effects and relativistic effects.

Chapter 13

1. Identical particles on a beam splitter

a. The final wave packet must be normalized: |α|2 + |β|2 = 1. In addition the twofinal states (φ3(r) + φ4(r)) /

√2 and αφ3(r) + βφ4(r) must be orthogonal since

the two initial states φ1(r) and φ2(r). This entails α + β = 0. We shall take in thefollowing α = −β = 1/

√2.

b. The initial state for two fermions reads:

|Ψ (ti)〉 = 1√2

(|1 : φ1 ; 2 : φ2〉 − |1 : φ2 ; 2 : φ1〉) .

We neglect the interaction between the fermions when they cross the beam splitter.The final state is then obtained by linearity:

|Ψ (tf )〉 = 1

2√

2(|1 : φ3〉 + |1 : φ4〉) ⊗ (|2 : φ3〉 − |2 : φ4〉)

− 1

2√

2(|1 : φ3〉 − |1 : φ4〉) ⊗ (|2 : φ3〉 + |2 : φ4〉) .

It can also be written:

|Ψ (tf )〉 = 1√2

(|1 : φ4 ; 2 : φ3〉 − |1 : φ3 ; 2 : φ4〉) .

The two fermions never come out on the same side of the beam splitter, which isdirect consequence of the exclusion principle.

c. The initial state for two bosons is:

|Ψ (ti)〉 = 1√2

(|1 : φ1 ; 2 : φ2〉 + |1 : φ2 ; 2 : φ1〉) ,

hence the final state:

|Ψ (tf )〉 = 1

2√

2(|1 : φ3〉 + |1 : φ4〉) ⊗ (|2 : φ3〉 − |2 : φ4〉)

+ 1

2√

2(|1 : φ3〉 − |1 : φ4〉) ⊗ (|2 : φ3〉 + |2 : φ4〉)

= 1√2

(|1 : φ3 ; 2 : φ3〉 + |1 : φ4 ; 2 : φ4〉) .

488 18 Solutions to the Exercises

The two bosons are always detected on the same side of the beam splitter. Thissurprising conclusion results from the destructive interference between the twoquantum paths: {

φ1 → φ3

φ2 → φ4and

{φ1 → φ4

φ2 → φ3,

which would both lead to the final state (|1 : φ3 ; 2 : φ4〉 + |1 : φ4 ; 2 : φ3〉)/√

2,corresponding to one boson in each output port.

2. Fermions in a square well

a. The energy levels for the one body Hamiltonian are En = n2E1 with E1 =π2

�2/(2mL2). The four lowest levels correspond to:

• the state |1+, 1−〉, of energy 2E1,• the four states |1±, 2±〉, of energy 5E1,• the state |2+, 2−〉, of energy 8E1,• the four states |1±, 3±〉, of energy 10E1.

b. One must diagonalize the restriction of the potential to each of the eigensubspacesfound above. For the non degenerate subspaces, one just needs to calculate thematrix element V1 = 〈α+,α − |V |α+,α−〉,, which gives after a simple algebraV1 = 3g/(2L). Each level |α+,α−〉 is displaced by this amount.The case of the eigensubspaces Eα,β , spanned by the four vectors |α±,β±〉 withα = β, is more subtle. The problem can be simplified by noticing that V doesnot affect the spin variables. The diagonalization of V inside Eα,β then amountsto three distinct eigenvalue problems:

• The dimension 1 subspace corresponding to |α+,β+〉 is not coupled to theother vectors of Eα,β . A simple calculation yields:

〈α+,β + |V |α+,β+〉 = 0.

This energy level is not displaced by V .• The same conclusion is reached for |α−,β−〉.• The restriction of V to the two dimensional subspace spanned by |α+,β−〉

and |α−,β+〉 reads:g

L

(1 −1−1 1

),

whose eigenvalues are 2g/L and 0.

To summarize, an energy level corresponding to a 4 dimension eigensubspace issplit into two sublevels: one sublevel is 3 fold degenerate and it is not shifted byV , the other level is not degenerate and its shift is 2g/L.

18 Solutions to the Exercises 489

Chapter 14

1. The Lorentz force in quantum mechanics

a. One checks easily that ∇ × A = B.b. The classical equations of motion mr = f and r = v give: r = (q/m)r × B. The

Lorentz force does not produce any work. The energy boils down to the kineticenergy E = mv2/2, which is a constant of the motion.The longitudinal velocity, parallel to B, is constant. The transverse velocity, per-pendicular to B, rotates around B with the angular velocity ω = −qB/m. Thetrajectory is a helix of axis z: the motion along z is uniform and the motion in thexy plane is circular.

c. We have Ax = −By/2, Ay = Bx/2, Az = 0. Therefore:

[px, Ax] = [py, Ay] = [pz, Az] = 0 ⇒ p.A = A.p.

The commutation relations between the components of u are:

[ux, uy] = [px − qAx, py − qAy] = −q[px, Ay] − q[Ax, py] = i�qB

and [ux, uz] = [uy, uz] = 0.d. One has [B, C2] = [B, C]C + C[B, C]. Therefore:

[r, H] = i�

m(p − qA) = i�

mu,

and, by applying the Ehrenfest theorem,

d

dt〈r〉 = 1

i�〈ψ|[r, H]|ψ〉 = 〈u〉

m.

On the other hand, one calculates the following commutators:

[ux, H] = 1

2m[ux, u2

y] = i�qB

muy [uy, H] = 1

2m[uy, u2

x] = −i�qB

mux,

and [uz, H] = 0. In other words:

[u, H] = i�q

mu × B ⇒ d

dt〈u〉 = q

m〈u〉 × B.

Altogether, we obtain:d2

dt2〈r〉 = q

m

d

dt〈r〉 × B,

which is identical to the classical equations of motion. Note that one recoversthe classical equations of motion identically, and not approximately, because the

490 18 Solutions to the Exercises

Hamiltonian is a second degree polynomial in the dynamical variables.Setting v = u/m, the observable v corresponds to the velocity operator. In termsof that observable, the Hamiltonian can be written as H = mv2

/2 which is theindeed the kinetic energy. Therefore, in a magnetic field, the linear momentummv does not coincide with the conjugate momentum p. These are related bymv = p − qA, which corresponds to the classical relation (15.28).

e. The commutation relations between the components of u show that in the presenceof a magnetic field, the various components of the velocity cannot be definedsimultaneously. One can define simultaneously the longitudinal component andone of the transverse components. The two transverse components satisfy theuncertainty relation:

Δvx Δvy ≥ �

2

|qB|m2

.

2. The Aharonov–Bohm effect

a. We denote respectively D and D′ the distances OBC and OB′C. If the vectorpotential is zero, the classical action corresponds to a uniform motion at velocityD/Δt (or D′/Δt) between O and B (or B′) and then between B (or B′) and C:

S0 = mD2

2ΔtS′

0 = mD′2

2Δt,

withΔt = t2 − t1. For a pointC located at a distance x from the center of the screenO′, we find D2 − D′2 � 2D0ax/L, where D0 represents the distance OBO′ =OB′O′, a is the distance between the holes, and L the distance between the planepierced by the two holes and the detection screen (we assume x � L). The quantityD0/Δt represents the average velocity v of the particles and we set λ = h/(mv).We find:

|A(C)|2 ∝∣∣∣eiS0/� + eiS

′0/�

∣∣∣2 ∝ 1 + cos

((S0 − S′

0)/�) = 1 + cos(2πx/xs),

which corresponds to the usual signal found in a Young double slit experiment,with the fringe spacing xs = λL/a.

b. When a current flows in the solenoid, the vector potential is not zero anymore, andthe classical action is changed because of the term qr · A(r) in the Lagrangian(15.27). The classical trajectories are not modified since no force acts on theparticle. We have thus:

S = S0 +∫

OBCq r · A(r) dt S′ = S′

0 +∫

OB′Cq r · A(r) dt

and the intensity in C is:

|A(C)|2 ∝ 1 + cos((S0 − S′

0)/� + Φ)

18 Solutions to the Exercises 491

where the phase Φ reads:

Φ = q

�

(∫

OBCr · A(r) dt −

∫

OB′Cr · A(r) dt

)

= q

�

(∫

OBCA(r) · dr −

∫

OB′CA(r) · dr

)= q

�

∮A(r) · dr.

The last integral is calculated along the closed contour OBCB′O. Its value doesnot depend on the position of C. It is equal to the flux of the magnetic fieldinside this contour, i. e. πr2B0, where r is the radius of the solenoid and B0 themagnetic field inside the solenoid. Therefore the current induces a global shift ofthe interference pattern, corresponding to the phase change Φ = πr2B0q/�.

Chapter 15

1. Excitation of an atom with broad band light

a. The state vector of an atom reads at time t: α(t) |a〉 + β(t) e−iω0t |b〉. The evolu-tions of α and β are given by:

α = −iΩ1 f (t) cos(ωt) e−iω0t β(t) � −iΩ1

2f (t) eiδt β(t)

β = −iΩ1 f (t) cos(ωt) eiω0t α(t) � −iΩ1

2f (t) e−iδt α(t),

where we neglected the non resonant terms and define δ = ω − ω0. The initialconditions are α(0) = 1, β(0) = 0. The solution of the second equation is, atlowest order in Ω1:

β(τ ) = −iΩ1

2

∫ τ

−τ

f (t) e−iδt dt.

The bounds of the integral can be extended to ±∞ since f (t) is zero outside theinterval [−τ , τ ]. This gives:

nb = n |β(τ )|2 = nπ

2Ω2

1 |g(−δ)|2 .

b. We generalize the preceding calculation and we obtain:

nb = nπ

2Ω2

1 |g(−δ)|2∣∣∣∣∣∣

�∑

p=1

eiω0tp

∣∣∣∣∣∣

2

.

c. We calculate the statistical average of∣∣∣∑�

p=1 eiω0tp

∣∣∣2:

492 18 Solutions to the Exercises

∣∣∣∣∣∣

�∑

p=1

eiω0tp

∣∣∣∣∣∣

2

=�∑

p=1

�∑

p′=1

eiω0(tp−t′p).

Since the various times tp are uncorrelated, the statistical average of eiω0(tp−t′p)

is zero unless p = p′, in which case this term is equal to 1. Therefore there are� ∼ γT non zero terms in this sum and we get:

nb(T) = nπ

2Ω2

1 |g(−δ)|2 γT .

The average number of atoms in state b increases linearly with time and we candefine a transition rate from a to b:

Γa→b = π

2Ω2

1 |g(−δ)|2 γ.

d. The energy contained in a wave packet is (ε0c/2)E20

∫f 2(t) dt, and the mean

energy flux reads:

Φ = ε0c

2E2

0 γ

∫f 2(t) dt = ε0c

2E2

0 γ

∫|g(Ω)|2 dΩ =

∫w(ω + Ω) dΩ,

where we used the Parseval–Plancherel equality. The function w(ω) is the energyspectral density. It is related to Γa→b by:

Γa→b = πd2

�2ε0c3w(ω0).

e. The preceding reasoning can be transposed to the case where the atoms are initiallyin the state b. One defines in this way a transition rate from b to a: Γb→a = Γa→b.

f. For this incoherent excitation, the evolution of the atom numbers na and nb isobtained by adding simply the two transition rates that we just found (this can beproven rigorously using the density operator formalism.

na = −Γa→bna + Γb→anb nb = Γa→bna − Γb→anb.

The steady state is simply na = nb = n/2 since Γb→a = Γa→b.

2. Atoms in equilibrium with black-body radiation

Within the framework of the previous exercise, we predict that the atom numbersna and nb are equal. On the opposite, Statistical Physics imposes the result nb/na =exp(−�ω0/kBT). Indeed one knows that a system (here the atomic assembly) incontact with a thermostat at temperature T (the black body radiation) must reach athermodynamical equilibrium characterized by the same temperature T .

18 Solutions to the Exercises 493

Einstein’s hypothesis consists in adding a second decay process from level b tolevel a, which creates a asymmetry between the populations of these levels. Supposethat this second process is characterized by the rate Γ ′

b→a. The evolutions of the atomnumbers na and nb are now:

na = −Γa→bna + (Γb→a + Γ ′b→a)nb nb = Γa→bna − (Γb→a + Γ ′

b→a)nb,

and the equilibrium state is:

nbna

= Γa→b

Γb→a + Γ ′b→a

avec Γa→b = Γb→a.

Let us impose the value exp(−�ω0/kBT) to this ratio. This leads to the value of therate Γ ′

b→a:Γ ′b→a = Γb→a

(e�ω0/kBT − 1

).

We have obtained in the previous exercise the relation between Γb→a and the energyspectral density w(ω). For the case of black body radiation, this relation entails:

Γ ′b→a = πd2

�2ε0c3w(ω0)

(e�ω0/kBT − 1

) = μπd2

�2ε0c3.

We find that the rate Γ ′b→a is independent of temperature, hence of the state of the

electromagnetic field. It corresponds to the spontaneous emission process describedqualitatively in Chap. 16, Sect. 16.3. Einstein’s reasoning gives an account for twoimportant characteristics of the spontaneous emission rate. It is proportional to thesquare of the average dipole d of the transition, and it varies as the cube of the Bohrfrequency ω0 of this transition (cf. Eq. (16.24)).

3. Ramsey fringes

The state vector of the neutron is γ+(t) e−iω0t/2 |+〉 + γ−(t) eiω0t/2 |−〉. Inside thecavities, the evolutions of the coefficients γ± are given by:

iγ+ = ω1

2ei(ω0−ω)t γ−(t) iγ− = ω1

2ei(ω−ω0)t γ+(t).

At the entrance of the first cavity, one has γ+(0) = 1 and γ−(0) = 0. At the exit ofthe first cavity (t1 = L/v), γ−(t1) is given by:

γ−(t1) = ω1

2(ω − ω0)

(1 − ei(ω−ω0)t1

),

where we restrict to first order term inB1. The coefficient γ− does not evolve anymoreuntil the neutron enters the second cavity at time T = D/v. The evolution equation

494 18 Solutions to the Exercises

of γ− during the crossing of the second cavity can be integrated similarly and weobtain finally:

γ−(T + t1) = ω1

2(ω − ω0)

(1 − ei(ω−ω0)t1

) (1 + ei(ω−ω0)T

),

hence the spin flip probability:

P+→− = ω21

(ω − ω0)2sin2((ω − ω0)t1/2) cos2((ω − ω0)T/2).

When one varies ω around the resonance frequency ω0, one gets a resonance witha width ∼ π/T . This is much narrower than the width that one would obtain with asingle cavity (∼ π/t1). This setup, which allows to accurately measure the resonancefrequency ω0, is currently used in metrology and high resolution spectroscopy.

Index

AAbasov, A.I. et al., 181Abdurashitov, J.N. et al., 181Absorption, 170Action, 386, 398Addition of angular momenta, 313Adjoint operators, 126Aharonov, Y., 395, 398, 490Ahmad, Q.R. et al., 182Alley, 167Ammonia molecule, 92Andromeda, 335Ångström, A.J., 269Angular distributions, 229Angular momentum

spin, 397Anharmonic oscillator, 223, 481Annihilation operator, 194, 273Anomalous Zeeman effect, 250, 297Anselmann, P. et al., 181Antineutrino, 171Antiparticle, 171Antisymmetric state, 351Approximation

Born, 405electric dipole, 408Wigner–Weisskopf, 417

Aspect, A., 52, 441Atmospheric neutrinos, 174Atom

cold, 356helium, 254, 352hydrogen, 202, 260

Atomic clock, 165Avogadro’s number, 12

BBalmer, J., 265, 269Balmer series, 265Beam splitter, 360, 487Bell basis, 454Bell, John, 51Bell state, 454Benzene molecule, 208Bernoulli, D., 385, 388Bernoulli sequence, 31β decay, 171Bienaymé-Tchebycheff (inequality), 32Binomial law (probability), 30Black body, 424, 492Black hole, 382Bloch, F., 306B mesons, 157Bohm, David, 51, 398, 435, 490Bohr

frequency, 410magneton, 12, 248, 289, 308radius, 12, 262

Bohr–Einstein controversy, 422Bohr, N., 17, 254, 269, 422Bohr–Sommerfeld theory, 254Boltzmann constant, 12Bormann, E., 295Born approximation, 405Born, M., 44, 137, 201Bose–Einstein condensate, 356Bose–Einstein condensation, 355Boson, 349Bottom quark, 271Bound state, 84, 228Broad band field, 423, 491Broglie (de), Louis, 19, 44, 392Broglie (de), waves, 26

© Springer International Publishing Switzerland 2016J.-L. Basdevant, Lectures on Quantum Mechanics,Graduate Texts in Physics, DOI 10.1007/978-3-319-43479-7

495

496 Index

Bruno, Giordano, 282

CCambridge, 185Canonical commutation relation, 392Canonical equation, 390Cartan, Elie, 231, 309Cartwheel galaxy, 337Cascade (atomic), 441Castin, Y., 356Cauchy, A., 222Cayley, 392Central potential, 256Centrifugal barrier, 257Chandrasekhar (mass), 371Classical configurations, 153Clauser, J.F., 52, 441Clebsch–Gordan coefficient, 315Closure relation, 128Coherent state, 204, 205Cold atoms, 7, 356Collision, 404Commutation relation, 183, 392CO molecule, 241Complete set of commuting observables

(CSCO), 197Completely prepared quantum state, 198Composite materials, 11Compton wavelength, 12, 262Conjugate momentum, 389, 394Conservation

of angular momentum, 256of energy, 190, 422of momentum, 255of probability, 190

Conservation laws, 190Constant of motion, 389Constant perturbation, 406Contact interaction, 326Contact potential, 361Continuous spectrum, 135Continuum of final states, 413Copernicus, 388Cornell, E., 356Correspondence principle, 68, 293, 392Coulomb interaction, 260Creation operator, 194, 273Cryptography, 446

Dd’Alembert, 385, 388Dalibard, J., 52

Darwin term, 324Davis, R., 182Davisson, C., 20Decay, 413Decoherence, 81, 453Degeneracy, 129Density of states, 414

photon, 415Descartes, R., 388Destruction operator, 273Deuterium, 341Diamagnetic term, 397Diatomic molecule, 239Dirac

equation, 324function, 53

Dirac formalism, 124Dirac, P.A.M., 119, 183, 348, 385, 392Directional quantization, 295Dispersion, 38Distribution theory, 55DONUT, 271Double well, 92Dye molecule, 169, 469

EEhrenfest, P., 188, 189, 239, 296Ehrenfest theorem, 190, 249, 290, 390, 392Eigenfunction, 72Eigenstate, 72Eigenvalue, 72, 73, 129Eigenvector, 129Eigler, D., 102Einstein, Albert, 6, 16, 296, 407, 422, 424Einstein–Podolsky–Rosen, 51, 434Electric dipole approximation, 408Electric dipole transition, 259, 331

selection rule, 411Electromagnetic transition, 407Electron, 12Electron gas, 355Electron magnetic moment, 308Elementary magneton, 300Emission

spontaneous, 407, 411stimulated (or Induced), 357, 407

Energy, 74, 389conservation, 422Fermi, 353width of a level, 419

Energy levels, 83Energy quantization, 84

Index 497

Entangled state, 434Entanglement, 434EPR paradox, 51Equation

canonical, 390Lagrange, 387

Equilibriumthermodynamic, 424, 492

Error function, 32Euler, 385, 388Evolution operator, 144, 467Exchange force, 352Exchange operator, 346Exclusion principle, 298, 350Expectation value, 38

FFermat, 386, 388Fermi

energy, 353golden rule, 415hyperfine structure, 328

Fermi, E., 348Fermion, 349Ferromagnetism, 353Feynman path integrals, 398Feynman, R.P., 16, 45, 52Fine structure

constant, 12monovalent atom, 323

Fine structure constant, 261, 325, 417First order perturbation theory, 221Flavor, 174Flux quantum, 399Force

exchange, 352Lorentz, 393

Fourier transformation, 42, 58, 65Fraunhofer, 268Freedman, S.J., 52, 441Fry, E.S., 441Fukuda, Y. et al., 181Fullerene, 242Function

Dirac, 53Heaviside, 57

GGaAs, 88Galileo, 282, 388GALLEX Collaboration, 181

Garcia, C.P., 5Gauge invariance, 394Gaussian law (probability), 30Gaussian wave packet, 53Gedanken experiment, 70Geometric law (probability), 32Germer, L., 20Glycine, 244Göttingen, 137GPS, 165Grangier, Ph., 52Grassmann, 392Gravitational catastrophe, 371Greeberger, D.M., 443Ground state, 91, 202, 224Gyromagnetic ratio, 246, 307

HHale–Bopp comet, 244Hall effect (quantum), 400Hamilton, 385, 392Hamiltonian, 75, 390Hamilton–Jacobi, 390Harmonic oscillator, 85

quasi-classical state, 204, 205three dimensional, 105, 272, 464, 479

Haroche, S., 82Heaviside function, 57Heisenberg

inequality, 227, 228, 474, 476pressure, 50representation, 144uncertainty relations, 46

Heisenberg, W., 46, 136, 201, 358, 369Helium atom, 254, 352, 361Herbig, G., 333Hermite functions, 122, 202Hermitian operators, 127Hidden variable, 438Hilbert, D., 119Hilbert space, 120Horne, M., 443Huggins, W., 269Huygens probe, 166Hydrogen

21-cm line, 333Hydrogen atom, 105, 202, 260

spectrum, 264Hydrogen maser, 328Hyperfine structure, 325

498 Index

IIdentical particles, 343Incoherent field, 423, 491Indistinguishability, 344Induced emission, 407Inequality (Heisenberg), 227, 474Interaction

contact, 326hyperfine, 325spin-orbit, 323

Interstellar medium, 331Interstellar molecules, 241Invariance

Gauge, 394rotation, 250, 476

Inversion of the ammonia molecule, 98

JJacobi, C.G.J., 200Jacobi identity, 201Jian-Wei Pan, 443Jordan, P., 201

KK2K collaboration, 181KamLAND collaboration, 173, 181Ketterle, W., 356Klein, Felix, 75, 392Koshiba, Masatoshi, 172, 182

LLagrange, 222, 385, 386, 388

equation, 387Lagrangian, 386Laguerre, 262Lamb shift, 324, 419Landau, L., 382, 422Landau level, 399, 400Langevin, P., 44Laplace, P., 222Laplacian, 57Larmor frequency, 288Larmor precession, 247, 249, 288, 301Laser, 358Least action principle, 386Leibniz, 345, 386LEP, 171, 271Leptons, 8, 171Lie

algebra, 231, 309group, 231, 309

Lie, Sophus, 231Lifetime, 413

atomic level, 412Line (21cm), 325Linear momentum, 394Linear operators, 126Logical gate (quantum), 453Lorentz force, 393Lorentzian line shape, 419Lyman series, 265

MM81, 242, 338M82, 242, 338Magnetic dipole transition, 331Magnetic moment, 245, 287

spin, 285Magnetic resonance, 300, 302

nuclear, 306Magnetism, 245, 300Magneton

Bohr, 248Maser, 333

hydrogen, 328Mass

Boson star, 376, 384Chandrasekhar, 371neutron star, 374, 381

Matrices, 138Maupertuis, 386, 388McDonald, Arthur B., 172, 182Mean square deviation, 32Measurement, 291

repeated, 310, 484Meson, 171Messiah, Albert, 442Metal, 355Microlasers, 5Milky Way, 331, 335Miller, Stanley, 244Moebius, 309Molecular binding, 101Molecule, 169, 468, 484

Benzene, 208Momentum, 45

conjugate, 389, 394linear, 394

Momentum conservation, 255Momentum observable, 67Muon, 171, 269Muonic atom, 269Muonium, 326Murchison, 244

Index 499

NNanotechnologies, 6, 102Neutrino, 170, 269, 381, 413Neutrino masses, 172Neutrino oscillations, 147, 170Neutron, 12Neutron magnetic moment, 308Neutron star, 373, 380Newton, I., 386, 388NGC3077, 242, 338Nuclear

magnetic resonance, 306magnetism, 300magneton, 12, 308

Nucleusunstable, 413, 417

OObservable, 66, 151Operator

annihilation (or destruction), 273creation, 273Hermitian, 127Laplacian, 57rotation, 143, 208, 467self-ajoint, 127translation, 143, 467

Oppenheimer, 382Optical lattice clocks, 329Origin of life, 244Orion nebula, 242, 243Oscillation lengths, 175Oscillator, 164

anharmonic, 223

PParamagnetic term, 397Parseval-Plancherel (theorem), 59Particle in a box, 91Pauli

effect, 295Hamiltonian, 397matrices, 282verbot, 298

Pauli principle, 298, 343, 361Pauli, W., 139, 280, 348, 361Peierls, R., 422Pellegrini, V., 5Penzias, A., 164Perl, M., 271Permutation, 351

Perturbationconstant, 406sinusoidal, 406time dependent, 403time independent, 219

Photoelectric effect, 410Photon, 140, 357, 410

density of states, 415Physical quantities, 35, 66Physical system, 35Pillars of Creation, 340Planck constant, 12Planck, M., 7, 16Poisson, 222Poisson bracket, 391Poisson distribution, 205Poisson law (probability), 32Polarization of light, 140Polarization of the photon, 157Polaroid, 141Population inversion, 160Position observable, 67Positronium, 326Potential

anharmonic, 223, 481contact, 361delta function, 106, 466harmonic, 105linear, 53, 461scalar, 393, 408vector, 393, 398, 408

Potential barriers, 101Prebiotic chemistry, 244Principal quantum number, 259Principle

correspondence, 392of quantization, 134of spectral decomposition, 134of wave packet reduction, 134symmetrization, 350

Probabilities (Notions on), 27Probability, 130

density, 28Probability amplitudes, 26, 64Proton, 12Proton magnetic moment, 308Purcell, E., 306Pythagorean theorem, 123

QQuantization of energy, 83Quantization rule, 392

500 Index

Quantum bit (or q-bit), 446, 451Quantum computer, 451Quantum Hall effect, 400Quantum logic, 142Quantum superposition, 81Quark, 8, 271Quasi-classical state, 204, 205Quaternions, 392

RRabi

experiment, 304formula, 304oscillations, 162, 304

Rabi, I., 271, 300Radial quantum number, 258Radial wave function, 257Radiation, 407Radioactivity, 413Radius

Bohr, 12Ramsauer effect, 106, 107Ramsey fringes, 424, 493Random variable, 29Reduced mass, 254Reduced wave function, 257Reduction of the wave packet, 73Relative motion, 254Relativistic effect, 324, 328Restriction of an operator, 222Result, 64Riesz, F., 125, 131Roger, G., 52Rotating field, 302Rotation invariance, 256Rotation operator, 143, 208, 467Rotation spectrum, 229Rutherford, 269Rydberg, 269Rydberg constant, 12

SSAGE collaboration, 181Saint Augustine, 76Scalar potential, 393, 408Scanning tunneling microscope, 102Scattering, 84Scattering state, 84Schrödinger equation, 27, 38, 134Schrödinger’s cat, 78Schrödinger, E., 44, 119

Second order perturbation theory, 223Secular equations, 222Selection rule, 259, 411Self-ajoint operators, 127Shimony, A., 443Shor, Peter, 452Simon, T., 333Singlet state, 318, 347, 435Sinusoidal perturbation, 406Slater determinant, 351Smart materials, 11Smy, M.B., 182Snell’s laws, 388SO(3) group, 309Sodium

yellow line, 323Sodium spectrum, 259Solvay congress, 50, 71Sommerfeld, A., 254Space

curved, 388Spectral

decomposition, 132theorem, 131

Spectroscopic notation, 259Spherical coordinates, 256Spherical harmonics, 237Spin, 279

total, 316Spin 1/2, 157, 229, 397Spin 1/2 particle, 283Spin magnetic moment, 285Spin-orbit interaction, 323Spin-statistics connection, 349Spinor, 309Spiral arms, 339Spontaneous emission, 160, 407, 411

atom, 417nucleus, 417

Square integrable functions, 121Stability of matter, 48Standard model, 8Star

neutron, 373, 380white dwarf, 372, 378

State, 35coherent (or quasi-classical), 204, 205singlet, 318, 347, 435stationary, 421triplet, 318, 347unstable, 413

State variable, 386Stationary states, 77, 190, 421

Index 501

Sterile neutrino, 180Stern–Gerlach experiment, 287Stern, O., 306Stimulated emission, 161, 357, 407Strange neutral mesons, 157Strange quark, 271Structure

fine, 323hyperfine, 325

SU(2) group, 309Super-Kamiokande, 175Supernovae, 172Superposition principle, 40, 133Suzuki, A., 174Symmetric state, 351Symmetrization principle, 350Symmetry, 89System, 35

Tτ lepton, 171Teleportation, 454Tensor product, 196, 315Theorem

non cloning, 448Parseval-Plancherel, 59virial, 203

Thermodynamic equilibrium, 424, 492Thomas precession, 299Thompson, R.C., 441Three-dimensional harmonic oscillator, 105,

464Time dependent perturbation, 403Time-energy uncertainty relation, 160, 203,

420Top quark, 271Total spin, 316Transition

electric dipole, 331electromagnetic, 407magnetic dipole, 331

Transition probability, 357, 403Translation invariance, 255Translation operator, 143, 467Triplet state, 318, 347Tunnel effect, 83, 94Two-state system, 148

UUncertainty relation, 152, 186, 227, 474

fermions, 358, 369

time-energy, 203, 420Unitary operator, 144Unstable state, 413

VVacuum polarization, 328Variable

random, 29Variance, 32Variational method, 224Vector potential, 393, 398, 408Verifications of general relativity, 167Vessot, 167Virial theorem, 203, 267VLBI, 333Volkov, 382

WWave

function, 37mechanics, 35packet, 41packet reduction, 73

WavelengthCompton, 12

Wave packetGaussian, 53spreading, 52

Wavesde Broglie, 26

Weak interactions, 171Weiss, P., 300White dwarf, 372, 378Wiemann, Carl E., 356Wigner–Weisskopf approximation, 417Wilson, K., 261Wilson, R.W., 164Wineland, D.J., 82

YYoung (Interferences), 396, 398

ZZeeman effect (anomalous), 249, 297, 329,

342Zeilinger, A., 20, 443Zeno paradox, 310, 484Zero point energy, 86, 104