mathschoolinternational.commathschoolinternational.com/.../Schaums...by-Hecht-and-Peleg-and-Pnini.pdf · v Preface The main purpose of this second edition of Quantum Mechanics is

Quantum Mechanics

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Quantum Mechanics

Second Edition

Yoav Peleg, Ph.D.Reuven Pnini, Ph.D.

Elyahu Zaarur, M.Sc.Eugene Hecht, Ph.D.

Schaum’s Outline Series

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Singapore Sydney Toronto

YOAV PELEG, Ph.D., received his doctorate in physics from the Technion Institute of Technology in Haifa, Israel. He has published to date dozens of articles, mostly in the area of general relativity and quantum cosmology. Currently he is working as a researcher with Motorola Israel.

REUVEN PNINI, Ph.D., received his doctorate in physics from the Technion Institute of Technology in Haifa, Israel. He has published to date several articles, mostly in the area of condensed matter physics. He is currently the Chief Scientifi c Editor of Rakefet Publishing Ltd.

ELYAHU ZAARUR, M.Sc., received his master of science in physics from the Technion Institute of Technology in Haifa, Israel. He has published more than a dozen books on physics. He is currently the Managing Director of Rakefet Publishing Ltd.

EUGENE HECHT, Ph.D., is a full-time member of the Physics Department of Adelphi University in New York. He has authored nine books including Optics, 4th edition, published by Addison Wesley, which has been the leading text in the fi eld, worldwide, for over three decades. Dr. Hecht has also written Schaum’s Outline of Optics and Schaum’s Outline of College Physics.

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v

Preface

The main purpose of this second edition of Quantum Mechanics is to make an already fine book more usable for the student reader. Accordingly, a great deal of effort has been given to simplifying and standardizing the notation. For example, a number of modern QM textbooks now distinguish operators from other quantities by placing a cap (^) over the corresponding symbol for the operator. This simple emendation can nonethe-less be very helpful to the student and that practice has been adopted throughout this edition. Similarly I have avoided using the same symbol to represent different quantities, inasmuch as this can be unduly confusing. Wherever necessary, discussions have been extended and the prose has been clarified. The all-but-unavoidable typographical and other minor first-edition errors have been corrected. Additionally, all of the art has been redrawn to improve visual readability, content, clarity, and accuracy. A substantial number of new introductory-level solved problems have been added to ensure that the student can gain a good grasp of the basics before approaching a more challenging range of questions. Indeed, it is my intention to add more such problems in future editions.

If you have any comments or suggestions, or favorite problems you’d like to share, send them along to Prof. E. Hecht, Physics Department, Adelphi University, Garden City, NY 11530 or if you prefer e-mail, [email protected].

EUGENE HECHT

Freeport, NY


vii

Contents

CHAPTER 1 Introduction 1

1.1 The Particle Nature of Electromagnetic Radiation 1.2 Quantum Particles 1.3 Wave Packets and the Uncertainty Relation

CHAPTER 2 Mathematical Background 13

2.1 The Complex Field C 2.2 Vector Spaces over C 2.3 Linear Operators and Matrices 2.4 Eigenvectors and Eigenvalues 2.5 Fourier Series and the Fourier Transform 2.6 The Dirac Delta Function

CHAPTER 3 The Schrödinger Equation and Its Applications 25

3.1 Wavefunctions of a Single Particle 3.2 The Schrödinger Equation 3.3 Particle in a Time-Independent Potential 3.4 Scalar Product of Wave-functions: Operators 3.5 Probability Density and Probability Current

CHAPTER 4 The Foundations of Quantum Mechanics 61

4.1 Introduction 4.2 Postulates in Quantum Mechanics 4.3 Mean Value and Root-Mean-Square Deviation 4.4 Commuting Observables 4.5 Function of an Operator 4.6 Hermitian Conjugation 4.7 Discrete and Continuous State Spaces 4.8 Representations 4.9 The Time Evolution 4.10 Uncertainty Relations 4.11 The Schrödinger and Heisenberg Pictures

CHAPTER 5 Harmonic Oscillator 98

5.1 Introduction 5.2 The Hermite Polynomials 5.3 Two- and Three-Dimensional Harmonic Oscillators 5.4 Operator Methods for a Harmonic Oscillator

CHAPTER 6 Angular Momentum 117

6.1 Introduction 6.2 Commutation Relations 6.3 Lowering and Raising Operators 6.4 Algebra of Angular Momentum 6.5 Differential Represen-tations 6.6 Matrix Representation of an Angular Momentum 6.7 Spherical Symmetry Potentials 6.8 Angular Momentum and Rotations

CHAPTER 7 Spin 145

7.1 Definitions 7.2 Spin 1/2 7.3 Pauli Matrices 7.4 Lowering and Raising Operators 7.5 Rotations in the Spin Space 7.6 Interaction with a Magnetic Field

Contents

CHAPTER 8 Hydrogen-like Atoms 164

8.1 A Particle in a Central Potential 8.2 Two Interacting Particles 8.3 The Hydrogen Atom 8.4 Energy Levels of the Hydrogen Atom 8.5 Mean Value Expressions 8.6 Hydrogen-like Atoms

CHAPTER 9 Particle Motion in an Electromagnetic Field 179

9.1 The Electromagnetic Field and Its Associated Potentials 9.2 The Hamil-tonian of a Particle in the Electromagnetic Field 9.3 Probability Density and Probability Current 9.4 The Magnetic Moment 9.5 Units

CHAPTER 10 Solution Methods in Quantum Mechanics—Part A 204

10.1 Time-Independent Perturbation Theory 10.2 Perturbation of a Non-degenerate Level 10.3 Perturbation of a Degenerate State 10.4 Time-Dependent Perturbation Theory

CHAPTER 11 Solution Methods in Quantum Mechanics—Part B 232

11.1 The Variational Method 11.2 Semiclassical Approximation (The WKB Approximation)

CHAPTER 12 Numerical Methods in Quantum Mechanics 249

12.1 Numerical Quadrature 12.2 Roots 12.3 Integration of Ordinary Differential Equations

CHAPTER 13 Identical Particles 264

13.1 Introduction 13.2 Permutations and Symmetries of Wavefunctions 13.3 Bosons and Fermions

CHAPTER 14 Addition of Angular Momenta 273

14.1 Introduction 14.2 {ˆ , ˆ , ˆ , ˆ }J J J J12

22 2

z Basis 14.3 Clebsch–Gordan Coefficients

CHAPTER 15 Scattering Theory 296

15.1 Cross Section 15.2 Stationary Scattering States 15.3 Born Approximation 15.4 Partial Wave Expansions 15.5 Scattering of Identical Particles

CHAPTER 16 Semiclassical Treatment of Radiation 330

16.1 The Interaction of Radiation with Atomic Systems 16.2 Time-Dependent Perturbation Theory 16.3 Transition Rate 16.4 Multipole Transitions 16.5 Spontaneous Emission

viii

Contents ix

APPENDIX Mathematical Appendix 347

A.1 Fourier Series and Fourier Transform A.2 The Dirac d-Function A.3 Hermite Polynomials A.4 Legendre Polynomials A.5 Associated Legendre Functions A.6 Spherical Harmonics A.7 Associated Laguerre Polynomials A.8 Spherical Bessel Functions

Index 355


Quantum Mechanics


1

CHAPTER 1

Introduction

1.1 The Particle Nature of Electromagnetic RadiationIsaac Newton considered light to be a beam of particles that set the pervading aether vibrating. The resulting aether waves guided the light particles along their way. During the nineteenth century, numerous experiments involving interference and diffraction demonstrated that light was some sort of wave. Optics was integrated into electromagnetic theory by James Maxwell who showed that light is electromagnetic. Nonetheless, the phenomenon of blackbody radiation, which was studied toward the end of the nineteenth century, could not be explained within the classical framework of electromagnetic theory. In 1900, Max Planck arrived at a formula that matched the blackbody radiation curves. He subsequently derived that formula by assuming that the oscillators in the walls of the emitting chamber could have only certain “quantized” energies. Planck’s analysis was basically wrong but it introduced the powerful idea of energy quantization; energy can only appear in whole-number multiples of some basic amount.

In 1905, Einstein proposed a return to the particle theory of light. He asserted that a beam of light of frequency n consists of energy quanta, or photons, each possessing energy hn, where h = 6.62 × 10−34 joules second (J⋅s) (i.e., Planck’s constant). Einstein showed how the introduction of the photon could account for the unexplained characteristics of both blackbody radiation and the photoelectric effect. About 20 years later, the photon was shown to exist as a distinct entity (the Compton effect; see Problem 1.3).

The photoelectric effect was discovered by Heinrich Hertz in 1887. It is one of several processes by which electrons can be removed from a metal surface. A schematic drawing of the apparatus for studying the pho-toelectric effect is given in Fig. 1.1.

Fig. 1.1

V

A Photocurrent

Radiant energy

Collector

Liberated electrons

Metalplate

+–

A beam of radiant energy (usually ultraviolet) impinges on the metal plate and knocks electrons free. The electrons fly off with a range of kinetic energies (KE ). By putting a negative potential (V ) on the collector some of the electrons are turned back. The critical potential VS such that eVS = KEmax (the maximum kinetic

CHAPTER 1 Introduction2

energy of the emitted electrons) is called the stopping potential. The experimental results of the photoelectric effect are summarized in Fig. 1.2 and below.

(i) When radiant energy shines on a metal surface, a current flows almost instantaneously, even for a very weak incident beam.

(ii) For fixed frequency and retarding potential, the photocurrent is directly proportional to the incident intensity, or more accurately the incident irradiance (W/m2) as shown in Fig. 1.2(c).

(iii) For constant frequency and irradiance, the photocurrent decreases with the increase of the retarding potential V, and finally reaches zero when V = −VS .

(iv) For any given surface, the stopping potential VS depends on the frequency of the light but is independent of irradiance. For each metal there is a threshold frequency n0 that must be exceeded for photoemis-sion to occur; no electrons are emitted from the metal unless n > n0, no matter how large the incident irradiance.

The experimental correlation between the stopping potential VS and the frequency of radiant energy can be represented by

eVS = hn − hn0 = hn − f (1.1)

where h is the same for all metals (Planck’s constant) and f is the work function. The work function is the minimum energy needed to liberate an electron from the surface of the metal target. It is different for each metal (see Table 1.1). In Fig. 1.2(d), each line intersects the n axis at a value of n0, and the eVS axis at a value of −f, both of which are characteristic of the particular target metal.

eVS

Max

imum

kin

etic

ene

rgy

(eV

)

0

–f

0.5 1.0 1.5n

n0

1

2

3

Frequency (× 1015 Hz)

(d)

Pota

ssiu

mSo

dium

Zinc

Tung

sten

Plat

inum

Phot

ocur

rent

0Incident irradiance (W/m2)

(c)

Low intensity

0Potential (V )

(b)

–Vs

High intensity

Metal 2

Phot

ocur

rent

Low intensity

0Potential (V )

(a)

–Vs

High intensity

Metal 1

Phot

ocur

rent

Fig. 1.2


1.2 Quantum ParticlesQuantum particles (photons, electrons, etc.) are not “particles” in our usual sense of the word; they do not behave like mini-cannonballs. Quantum particles are more like oscillating puffs of matter possessed of both wavelike and particlelike properties that defy conceptualization. The dynamic parameters of quantum par-ticles (energy E and momentum p) are linked to their wave parameters (frequency n and wave or propagation vector k) by the relations

E h= =

=

⎧⎨⎪

⎩⎪

n �

�

ω

p k (1.2)

where � = h/2π . In the second half of the 1800s, it was discovered that atoms emit or absorb electromagnetic radiation at

only well-determined frequencies. This fact can be explained by assuming that the energy of an atom can take on only certain discrete values. In other words, the energy of an atom is quantized. This was one of the central assertions made by Niels Bohr in 1913 when he proposed his theory of the hydrogen atom. The existence of discrete energy levels was demonstrated in 1914 by the Frank–Hertz experiment. Bohr supposed that the electron moves in orbits restricted by the requirement that its angular momentum be an integral multiple of h/2p. For a circular orbit of radius r, the electron velocity v is given by

m rnh

ne n nv = =2

1 2π , , . . . (1.3)

The relation between the Coulomb force and the centrifugal force can be written in MKS units as

ke

r

mr

n

e n

n0

2

2

2

=v

(1.4)

where e is the charge of the electron and k0 is the Coulomb constant. We assume that the nuclear mass is infinite. Combining Eq. (1.3) and Eq. (1.4) we obtain

vn

e knh

=2 2

0π (1.5)

and

rn h

m e kne

= 1

4 2

2 2

20π

(1.6)

The orbital energy is then

E mer

m e k

n hn e nn

e= − = −12

222 2 4

02

2 2vπ

(1.7)

Table 1.1 Representative Work Function Values

Work Function Metal (f in eV )

Na 2.28Co 3.90Al 4.08Pb 4.14Zn 4.31Fe 4.50Cu 4.70Ag 4.73Pt 6.35


Bohr postulated that electrons in these orbits do not radiate, despite their acceleration; they are in stationary states. Electrons can make discontinuous transitions from one allowed orbit to another. The change in energy will appear as radiation of frequency

n = − ′E Eh (1.8)

The physical basis of the Bohr model remained unclear until 1923, when de Broglie put forth the hypothesis that material particles have wavelike characteristics; a particle of energy E and momentum p is associated with a wave of angular frequency ω = E /� and a wave vector k = p/h. The corresponding wavelength is therefore

λ π= =2k

hp

(1.9)

This is the de Broglie relation.

1.3 Wave Packets and the Uncertainty RelationThe wave and particle aspects of electromagnetic radiation and matter can be united through the concept of a wave packet. A wave packet is a superposition of waves resulting in a sinusoidal pulse. We can construct a wave packet in which the component harmonic waves interfere with each other almost completely outside a given spatial region (Fig. 1.3). We thus obtain a localized wave packet that is a useful representation of a classical particle. A three-dimensional wave packet consisting of a superposition of plane waves may be written as

f g e di( )( )

( )r k k= ⋅∫1

2 3 2π /k r (1.10)

or in one dimension,

f x g k e dkikx( ) ( )=−∞

∞

∫1

2π (1.11)

The evolution of wave packets is determined by the Schrödinger equation (see Chap. 3). When a wave packet evolves according to the postulates of quantum mechanics (see Chap. 4), the widths of the curves f(x) and g(k) are related by

Δx Δk > 1 (1.12)

Using the de Broglie relation p k= � , we have

Δ Δp x > � (1.13)

This is the Heisenberg uncertainty relation: if we try to construct a highly localized wave packet in space, then it is impossible to associate a well-defined momentum with it. In contrast, a wave packet with a defined momentum within narrow limits must be spatially very broad. Note that since � is very small, the notions of

Fig. 1.3

Δx

x

f (x)


classical physics will fail only for a microscopic system (see Problem 1.17). The uncertainty relation acts to reconcile the wave–particle duality of matter and radiation (see Problem 1.4).

Any wave pulse can be imagined to be composed of an infinite number of superimposed sinusoidal com-ponent waves. Each of these travels with a velocity known as the phase velocity whereas the pulse or wave packet travels with the group velocity. For a wave of angular temporal frequency w = 2p n and angular spatial frequency (or propagation number) k = 2p /l, the phase velocity is

vp k= =ω λn (1.14)

This is the rate at which a point of constant phase on any one of the constituent harmonic waves travels through space. By contrast, the pulse travels with the speed vg which is related to the angular frequency w and propagation number k of the component waves by the relation

vg

ddk

= ω (1.15)

SOLVED PROBLEMS

1.1. Consider the experimental results of the photoelectric effect described in Sec. 1.1, i–iv. For each result discuss whether it would be expected on the basis of the classical properties of electromagnetic waves.

SOLUTION

(i) An electron in a metal will be free to leave the surface only after the incident beam provides its bind-ing energy. Because of the continuous nature of the electromagnetic radiation, we expect the energy absorbed on the metal’s surface to be proportional to the irradiance of the beam (energy per unit time per unit area), the area illuminated, and the time of illumination. A simple calculation (see Problem 1.14) shows that in the case of an irradiance of 10−10 W/m2, photoemission can be expected only after 100 h. Experimentally, the delay times that were observed for the same irradiance were not longer than 10−9 s. Classical theory is thus unable to explain the instantaneous emission of electrons from the metal.

(ii) With the increase of radiant energy, the energy absorbed by the electrons in the metal increases. There-fore, classical theory predicts that the number of electrons emitted (and thus the current) will increase proportionally to the irradiance. Here classical theory is able to account for the experimental result.

(iii) The result described in Sec. 1.1, iii shows that there is a distribution in the energies of the emitted elec-trons. The distribution in itself can, within the framework of the classical theory, be attributed to the varying degrees of binding of electrons to metal, or to the varying amount of energy transferred from the beam to the electrons. But the fact that there exists a well-defined stopping potential independent of irradiance indicates that the maximum energy of released electrons does not depend on the amount of energy reaching the surface per unit time. Classical theory is unable to account for this.

(iv) According to the classical point of view, emission of electrons from the metal depends on the irradiance of the incident beam but not on its frequency. The existence of a frequency below which no emission occurs, however large the irradiance, cannot be predicted within the framework of classical theory.

In conclusion, the classical theory of electromagnetic radiation is unable to fully explain the photoelectronic effect.

1.2. Interpret the experimental results of the photoelectric effect in view of Einstein’s hypothesis of the quantization of radiant energy.

SOLUTION

(i) According to the hypothesis that light consists of photons, we expect that a photon will be able to transfer its energy to an electron in a metal, and therefore it is feasible that photoemission occurs instantaneously even at a very small irradiance. This is contrary to the classical view, which proposes that the emission of electrons depends on continuous accumulation of energy absorbed from the incident beam.


(ii) According to quantum theory, irradiance is equal to the energy of each photon multiplied by the number of photons crossing a unit area per unit time. It is reasonable that the number of emitted electrons per unit time (which is equivalent to the current) will be proportional to the incident irradiance.

(iii) The frequency of the electromagnetic radiation determines the energy of the photons hn. Therefore, the energy transferred to electrons in a metal due to light absorption is well defined, and thus for any given frequency there exists a maximum kinetic energy of the photoelectrons. This explains the effect described in Fig. 1.2.

(iv) Equation (1.1) can be given a simple interpretation if we assume that the binding energy of the elec-trons that are least tightly bound to the metal is f = hn0. The maximum kinetic energy of emitted electrons is hn − f. Using the definition of stopping potential, eVS is the maximum kinetic energy; therefore, eVS = hn − hn0 .

1.3. Consider the Compton effect (see Fig. 1.4). According to quantum theory, a monochromatic electro-magnetic beam of frequency n is regarded as a collection of particlelike photons, each possessing an energy E = hn and a momentum p = hn/c = h/l, where l is the wavelength. The scattering of electromagnetic radiation becomes a problem of collision of a photon with a charged particle. Suppose that a photon moving along the x axis collides with a particle of mass m. As a result of the collision, the photon is scattered at an angle q, and its frequency is changed. Find the increase in the photon’s wavelength as a function of the scattering angle.

Fig. 1.4

E0

h/l

h /l′

hn′

hn

Before collision After collision

x

y

q

x

p, E

y

j

SOLUTION

First, since the particle may gain significant kinetic energy, we must use it by relativistic dynamics. Applying energy conservation we obtain

(before collision) h E h En n��

photon particle photon+ = ′ +0

pparticle (after collision) (1.3.1)

where E0 is the rest energy of the particle (E0 = mc2). The magnitudes of the momenta of the incident and scattered photons are, respectively,

phc

hp

hc

hλ λλ λ= = = ′ = ′′

n nand (1.3.2)

The scattering angle q is the angle between the directions of pl and pl′. Applying the law of cosines to the triangle in Fig. 1.5, we obtain

p p p p p2 2 2 2= + −′ ′λ λ λ λ θcos (1.3.3)

Recall that for a photon pc = hn; therefore, multiplying both sides of Eq. (1.3.3) by c2, we obtain

h h h p c2 2 2 2 2 2 22n n nn+ ′ − ′ =cosθ (1.3.4)


Using Eq. (1.3.1) we have

h h E E h h h E E EEn n n n nn− ′ = − ⇒ + ′ − ′ = + −02 2 2 2 2 2

02

02 2 (1.3.5)

Relying on relativity theory, we replace E2 with E p c02 2 2+ . Subtracting Eq. (1.3.4) from Eq. (1.3.5), we obtain

− ′ − = −2 1 2 2202

0h E EEnn ( cos )θ (1.3.6)

Therefore, using Eq. (1.3.1),

h E E E m c h he2

0 021nn n n′ − = − = − ′( cos ) ( ) ( )θ (1.3.7)

We see that h

m cc

c c

e( cos ) .1 − = − ′

′ = ′ − = ′ −θ λ λn n

nn n n Therefore, the increase in the wavelength Δl is

Δλ λ λ θ= ′ − = −hm ce

( cos )1 (1.3.8)

This is the basic equation of the Compton effect.

1.4. Consider a beam of light passing through two narrow, vertical, parallel slits that are very close together. When either one of the slits is closed, the irradiance distribution observed on a screen placed far beyond the barrier is a broad diffraction peak (see Fig. 1.6). When both slits are open, the peaks almost completely overlap and the pattern is as shown in Fig. 1.7: an interference pattern within the diffraction envelope. Note that this pattern is not the two single-slit diffraction patterns superposed. Can this phenomenon be explained in terms of classical particlelike photons? Is it possible to demonstrate particle aspects of light with this experimental setup?

Fig. 1.6

Double slits

Incident light

Distant screen

Irradiance

Distant screen

Double slits

Incident light

Fig. 1.5

j

q

p

pl

pl′


30

400

800

1200

1600

2000

2400

2800

4 5 6

Detector position

Phot

on c

ount

rat

e (c

ount

s/se

cond

)

7 8

Diffraction, left

Interference modelBoth slits openOne slit openOther slit open

Diffraction, right

Fig. 1.7b Courtesy TECHSPIN Inc.

Fig. 1.7a

max

max

max

max

max

min

min

min

min


SOLUTION

Suppose that the beam of light consisted of a stream of pointlike classical particles. If we consider each of these particles separately, we note that each one must pass through one of the slits. Therefore, the pattern obtained when the two slits are open must be the superposition of the patterns obtained when each of the slits is open separately. This is not what is observed in the experiment. The pattern actually obtained can be explained only in terms of interference of the light passing simultaneously through both of the slits (see Fig. 1.7). However, it is possible to observe particle aspects of light in this system: if the light intensity is very weak, the photons will reach the screen at a low rate. If a photographic plate is placed at the screen, the pat-tern will be formed slowly, one point at a time. This indicates the arrival of separate photons on the screen. Note that it is impossible to determine which slit each of these photons passes through; such a measurement would destroy the interference pattern.

1.5. Figure 1.8 describes schematically an experimental apparatus known as Heisenberg’s microscope whose purpose is to measure the position of an electron. A beam of electrons of well-defined momentum px moving in the positive x direction scatters light shining along the negative x axis. An electron will scatter a photon that will be detected through the microscope.

According to optics theory, the precision with which the electron can be localized is

Δx ∼ λθsin

(1.5.1)

where l is the wavelength of the light. Show that if we minimize Δ x by reducing l, this will result in a loss of information about the x-component of the electron momentum.

ElectronsPhotons

Lens

Screen

q

Fig. 1.8 Heisenberg microscope.

SOLUTION

According to quantum theory, recording light consists of photons, each with a momentum hn/c. The direction of the photon after scattering is undetermined within the angle subtended by the aperture, i.e., 2q. Hence the magnitude of the x-component of the photon is uncertain by

Δphcx ∼ 2n

sinθ (1.5.2)

Therefore,

Δ Δx phcx ∼ ∼2 4n

sinsin

θ λθ π� (1.5.3)


We can attempt to overcome this difficulty by measuring the recoil of the screen in order to determine more precisely the x-component of the photon momentum. But we must remember that once we include the micro-scope as part of the observed system, we must also consider its location. The microscope itself must obey the uncertainty relations, and if its momentum is to be specified, its position will be less precisely determined. Thus, this apparatus gives us no opportunity for violating the uncertainty relation.

1.6. Prove that the Bohr hydrogen atom approaches classical conditions when n becomes very large and small quantum jumps are involved.

SOLUTION

Let us compute the frequency of a photon emitted in the transition between the adjacent states

nk = n and nl = n − 1 when n >> 1. We define the Rydberg constant Rm e k

h ce= = × −2

1 093 102 4

02

37 1π

. .m So,

Ech

nRk

k

= 2 and Ech

nRl

l

= 2 . Therefore, the frequency of the emitted photon is

n =−

=+ −n n

n ncR

n n n n

n ncRk l

k l

k l k l

k l

2 2

2 2 2 2

( )( ) (1.6.1)

nk − nl = 1, so for n >> 1 we have

n n n n n nk l k l+ ≅ ≅2 2 2 4 (1.6.2)

Therefore, n ≅ 2 3cR n/ . According to classical electromagnetic theory a rotating charge with a frequency f will emit radiation of frequency f. On the other hand, using the Bohr hydrogen model, the orbital frequency of the electron around the nucleus is

fr

m e k

n hnn

n

e= =v

24 2 4

02

3 3ππ

(1.6.3)

or fn = 2cR/n3, which is identical to n.

1.7. Show that the uncertainty relation Δ Δx p > � forces us to reject the semiclassical Bohr model for the hydrogen atom.

SOLUTION

In the Bohr model we deal with the electron as a classical particle. The allowed orbits are defined by the quantization rules: the radius r of a circular orbit and the momentum p = mv of the rotating electron must satisfy the quantization of angular momentum pr n n= =� ( , . . .1 2, ). To consider an electron’s motion in classical terms, the uncertainties in its position and momentum must be negligible when compared to r and p; in other words, Δ x r<< and Δp p<< . This implies

Δ Δxr

pp

<< 1 (1.7.1)

On the other hand, the uncertainty relation imposes

Δ Δ Δ Δxr

pp rp

x prp n

≥ ⇒ ≥� 1 (1.7.2)

So, Eq. (1.7.1) is incompatible with Eq. (1.7.2), unless n >> 1.

1.8. (a) Consider a thermal neutron, that is, a neutron with speed v corresponding to the average thermal energy at the temperature T = 300 K. Is it possible to observe a diffraction pattern when a beam of these neutrons falls on a crystal? (b) In a large accelerator, an electron can be provided with an energy exceeding 1 GeV = 109 eV. What is the de Broglie wavelength corresponding to such electrons?

SOLUTION

(a) The average thermal energy of an absolute temperature T is E k TBav = 32

where kB is the Bolzmann con-stant (kB = 1.38 × 10−23 J/K). Therefore, we have

12 2

32

22

mpm

k Tnn

Bv = = (1.8.1)


According to the de Broglie relation the corresponding wavelength is

λ = =hp

h

m k Tn B3 (1.8.2)

For T = 300 K, we have

λ = ×

× × × × ×≅

−

− −

6 63 10

3 1 67 10 1 38 10 3001 4

34

27 23

.

. .. Å (1.8.3)

This is the order of magnitude of the spaces between atoms in a crystal, and therefore a diffraction phe-nomenon analogous to that of x-rays.

(b) We note that the electron’s rest energy is m ce2 60 5 10≅ ×. eV. Therefore, if an energy of 109 eV is

imparted to the electron, it will move with a velocity close to the speed of light, and it must be treated

using relativistic dynamics. The relation l = h/p remains valid, but we have E p c m ce= +2 2 2 4 . In this example, mec

2 is negligible when compared with E, and we obtain

λ ≅ = × × ××

= ×−

−−hc

E6 6 10 3 10

1 6 101 2 10

34 8

1015.

.. m == 1.2 fm (1.8.4)

With electrons accelerated to such energies, one can explore the structure of atomic nuclei.

1.9. The wavelength and the frequency in a wave guide are related by

λ =−

c

n n2

02

(1.9.1)

Express the group velocity vg in terms of c and the phase velocity vp = λn.

SOLUTION

First, we find how the angular frequency w depends on the propagation number k. We have w = 2p n; using Eq. (1.9.1), we have

ω πλ

ππ

( )kc c k= + = +2 2

4

2

2 02

2 2

2 02

n n (1.9.2)

Hence, the group velocity is

vg

ddk c k

kc c k c= =

+

= =ω π

π

π ππλ

2

24

2

4 22

22 2

2 02

2

2

2 2

n

n ππ λn n= =c c

p

2 2

v (1.9.3)

SUPPLEMENTARY PROBLEMS

1.10. Derive an expression for the energy of a photon in eV when its wavelength is given in nanometers. Determine the energy of a 500-nm photon in eV.

Ans. E = (1239.8 eV⋅ nm)/l, 2.48 eV

1.11. An electron crashes into the metal mask of an old color television tube operating at 20.0 kV. Find the shortest wavelength x-rays that will be emitted.

Ans. lmin = 0.062 nm

1.12. Using Eq. (1.7) show that the energy levels of the hydrogen atom are given by

Enn = − 13 6

2. eV

(1.12.1)


1.13. How much energy would it take to raise a hydrogen atom from its ground state (n = 1) to its first excited state (n = 2)?

Ans. 10.2 eV

1.14. Refer to Problem 1.9 and find the group velocity for the following relations: (a) n = 23

πρλ

ϒ (water waves in

shallow water; ° is the surface tension and r the density). (b) n = g2πλ (water waves in deep water).

Ans. (a) v vg p= 32

; (b) v vg p= 12

1.15. Suppose that light of irradiance 10−10 W/m2 usually falls normally upon a metal surface. The atoms are approximately 3 Å or 3 × 10−10 m apart and it is given that there is one free electron per atom. The binding energy of an electron at the surface is 5 eV. Assume that the light is uniformly distributed over the surface and its energy absorbed by the surface electrons. If the incident radiation is treated classically (as waves), how long must one wait after the beam is switched on until an electron gains enough energy to be released as a photoelectron?

Ans. Approximately 2800 years.

1.16. Consider a monochromatic beam of light of irradiance I and frequency n striking a completely absorbing surface. Suppose that the light is incident along the normal to the surface. Using classical electromagnetic theory, one can show that on the surface a pressure (P) called the radiation pressure is acting, which is related to the irradiance by P = I/c. Is this relation also valid according to quantum theory?

Ans. Yes. Phc

N= n, where N is the flux of the photon beam.

1.17. Suppose that monochromatic light is scattered by an electron. Use Problem 1.3 to find the shift in the wavelength when the scattering angle is 90°. What is the fractional increase in the wavelength in the visible region (say, l = 4000 Å)? What is the fractional increase for x-ray photons of l = 1 Å = 0.1 nm?

Ans. Δλ θ= − =11 0 024 3

m ce( cos ) . .Å For l = 4000 Å, the fractional shift is 0.006 percent. For l = 1 Å

it is 2 percent.

1.18. We wish to show that wave properties of matter are irrelevant for the macroscopic world. Take as an example a tiny particle of diameter 1 μm and mass m = 10−15 kg. Calculate the de Broglie wavelength corresponding to this particle if its speed is 1 mm/s.

Ans. l = 6.6 × 10−7 nm.

1.19. Consider a virus of size 1.0 nm. Suppose that its density is equal to that of water and that the virus is located in a region that is approximately equal to its size. What is minimum speed of the virus?

Ans. vmin ≈ 1 m/s.

1.20. A beam of high-energy photons impinges on a target and some are backscattered by collisions with electrons that are essentially at rest. Determine the wavelength shift experienced by the scattered photons

Ans. 4.85 × 10−12 m.

1.21. If E is the energy of the incident photons in the previous problem, show that

Em c

m c Ese

e

=+

2

2 2/ (1.21.1)

is the energy of the scattered photons.

13

CHAPTER 2

Mathematical Background

2.1 The Complex Field CThe complex field, denoted by C, is the field generated by the complex numbers a + bi, where a and b are real numbers and i is the solution of the equation x2 + 1 = 0, i.e., i = −1. If z = a + bi, then a is called the real part of z and denoted Re(z); b is called the imaginary part of z and denoted Im (z). The complex conjugate of z a bi a bi= + −is and is denoted by z*. Summation and multiplication of complex numbers is performed in the following manner:

(a + bi) + (c + di) = (a + c) + (b + d )i (2.1)

(a + bi) (c + di) = (ac − bd ) + (bc + ad )i (2.2)

If z ≠ 0 we define z−1 and division by z as

zz

zz

a

a b

b

a bi− = =

++ −

+1

2 2 2 2

*

* (2.3)

wz

wz= −1 (2.4)

Figure 2.1 represents a geometric realization of the complex field as points in the plane.

Fig. 2.1

q

y

b

O aReal

z = a + ib

Imag

inar

y

x

The distance between the point z and O is denoted z a b zz= + =2 2 * and is called the modulus of z. The angle q is called the argument of z and denoted by arg(z). Since points in the plane can be characterized by polar coordinates, i.e., a pair (r, q) where r > ≤ ≤0 0 2and θ π , one can write a complex number in terms of its modulus and argument. As one can easily verify,

a = r cos q b = r sin q (2.5)

CHAPTER 2 Mathematical Background14

and

r a bba

= + = ⎛⎝⎜

⎞⎠⎟

−2 2 1θ tan (2.6)

and therefore z r i rei= + =(cos sin ) .θ θ θ

2.2 Vector Spaces over CA vector space over C is a collection of elements V that is closed under associative addition (+) of its elements (called vectors), and that satisfies the following conditions for each scalar a, b in C and vector v, u in V:

1. V contains a unique element denoted 0 that satisfies

v v v+ = + =0 0 (2.7)

0 is called the null vector.

2. a v is also in V.

3. a (v + u) = a v + a u.

4. (a + b )v = a v + b v.

5. (a ⋅ b )v = a (bv).

6. 0 ⋅ v = 0, a ⋅ 0 = 0, 1 ⋅ v = v.

An Important Example—Cn

Consider elements of the form (z1, z2, . . . , zn), where the zi are complex numbers. We define addition of such elements by

(z1, z2, . . . , zn) + (w1, w2, . . . , wn) = (z1 + w1, z2 + w2, . . . , zn + wn) (2.8)

and we define multiplication by a scalar (a complex number z) by

z(z1, z2, . . . , zn) = (zz1, zz2, . . . , zzn) (2.9)

It can be verified that the collection of these elements has all the properties of a vector space over C. This important vector space is denoted Cn.

Some Useful DefinitionsA collection of vectors u1, . . . , un in V, spans V if every element in V can be written as a linear combination of the u’s; that is,

v = + +a u a un n1 1 � (2.10)

where a1, . . . , an are complex numbers. The vectors u1, . . . , un are called linearly independent if a u a un n1 1 0+ + =� implies a a an1 2 0= = = =� . If u1, . . . , un are linearly independent and span V they are called a basis of V. The number n is unique and is called the dimension of V. Suppose that W is a collection of vectors from a vector space V. W is a subspace of V if: (1) for every v, w, in W, v + w is also in W; (2) for every w in W and every scalar a, a v is also in W.

2.3 Linear Operators and Matrices

Linear OperatorsLet V be a vector space over the complex field C. A map ˆ :T V V→ is an operator on V if it satisfies the following condition for every a, b in C and every, u, v in V:

ˆ( ) ˆ( ) ˆ( )T u T T uα β α βv v+ = + (2.11)


If T and S are linear operators, their sum, the linear operator ˆ ˆ,T S+ is defined by

( ˆ ˆ) ( ) ˆ( ) ˆ( )T S u T u S u+ = + (2.12)

for every u in V. Similarly, we define the product of two linear operators by

( ˆ ˆ)( ) ˆ[ ˆ( )]T S T S⋅ =v v (2.13)

for every v in V. The set of linear operators equipped with addition and multiplication is therefore an algebra over the complex field. For now, let us restrict ourselves to a finite dimensional vector space.

Assume e1, . . . , en is a basis of V and let T be a linear operator on V. Applying T to e1, . . . , en we get

ˆ( )

ˆ( )

T e a e a e

T e a e a e

n n

n n nn n

1 11 1 1

1 1

= + +

= + +

�

��

(2.14)

where aij are complex numbers. Now we define the matrix representation of T relative to the basis e by

[ ˆ] ( )T a

a a a

a a a

a a a

e ji

n

n

n n

= =

11 21 1

12 22 2

1 2

…

…

� … �

… nnn

⎛

⎝

⎜⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟

(2.15)

Note that the matrix representation of an operator is dependent on the choice of basis. For infinite matri-ces it is possible to sum and multiply infinite matrices like finite matrices, though one must pay attention to convergence whenever infinite sums are involved. Linear operators are of great importance in quantum mechanics, since, as we shall see in the next chapters, they represent physical quantities such as energy, momentum, etc.

Inner ProductAn inner product on V is a function ⟨ ⟩u, v from V × V to the complex field (i.e., taking every pair of vectors to a complex number), that satisfies the following conditions for every u, v, u′ in V and a in C:

(i)

ii

ii

⟨ ⟩ = ⟨ ⟩

⟨ + ′ ⟩ = ⟨ ⟩ + ⟨ ′ ⟩

u u

u u u u

, ,

( ) , , ,

(

*v v

v v v

ii

iv if

) , ,

( ) ,

⟨ ⟩ = ⋅ ⟨ ⟩

⟨ ⟩ > ≠

α αu u

u u u

v v

0 0

(2.16)

A vector space that has an inner product is called an inner product space.We can use the inner product to specify some useful definitions. The norm of a vector v is

v v v= ⟨ ⟩, (2.17)

If ⏐⏐v ⏐⏐ = 1, then v is called a unit vector and is said to be normalized.Two vectors u and v are orthogonal if

⟨ ⟩ =u, v 0 (2.18)


A set of vectors {ui} is orthogonal if any pair of two separate elements is orthogonal; this is, ⟨ ⟩ =u ui j, 0 for i ≠ j. In particular, the set is orthonormal if in addition each of its elements is a unit vector, or compactly,

⟨ ⟩ =u ui j ij, δ (2.19)

where dij is the Kronecker delta function, which is 0 for i ≠ j and 1 otherwise. An important result, used frequently in quantum mechanics, is the Cauchy–Schwartz inequality: For all vectors u and v,

⟨ ⟩ ≤ ⋅u u, v v (2.20)

Operators and Inner ProductsSuppose T is a linear operator on V and suppose V is an inner product space. It can be shown that there is a unique linear operator denoted ˆ†T that satisfies:

⟨ ⟩ = ⟨ ⟩ˆ , , ˆ†Tu u Tv v (2.21)

for every u, v in V. This operator is called the conjugate operator of ˆ.T If ˆ ( )A ij= α is a complex matrix, ˆ†A is defined as ˆ ( ) ,† *A ji= α which is found by swapping indices and taking the complex conjugate. If A represents an operator ˆ,T then ˆ†A represents ˆ ,†T which justifies the use of the same symbol † in each case. If ˆ ˆ ,†T T= then T is called a Hermitian operator or self-conjugate operator. If ˆ ˆ ,†T T= − then T is an anti-Hermitian operator. If T preserves the inner product, that is, ⟨ ⟩ = ⟨ ⟩ˆ , ˆ ,Tu T uv v for every u, v in V, then T is a unitary operator. If ˆ ˆ ˆ ˆ,† †TT T T= then T is a normal operator. Two vectors v and u are orthogonal if ⟨ ⟩ =v , .u 0

2.4 Eigenvectors and EigenvaluesLet T be a linear operator on V. A complex number l is called an eigenvalue (also known as a characteristic value) of T if it satisfies Tv v= λ for some v in V. The vector v is called the eigenvector of T corresponding to l. The same definition holds for matrices. Note that if V has a basis that consists of eigenvectors of ˆ,T then T is represented relative to that basis as a diagonal matrix. Diagonal matrices are not only easy to work with, but also reflect important characteristics of the physical system such as quanta of energy, and so forth.

Characteristic Polynomial Suppose that a given linear operator T is represented in some basis by the matrix ˆ.A The characteristic poly-nomial of T is defined by

Δ( ) det ( ˆ ˆ)t I A= −λ (2.22)

where l is the parameter (scalar) and I is the identity or unit matrix. The characteristic equation of T is defined by

Δ(t) = 0 (2.23)

These expressions are independent of the basis chosen.The following result provides a method for finding the eigenvalues of a matrix or operator: the scalar l is

an eigenvalue of an operator T if and only if it is a root of its characteristic polynomial, that is, Δ(t) = 0.If A is a Hermitian or unitary matrix, then there exists a unitary matrix U such that ˆ ˆ ÛAU −1 is a diagonal matrix

(this theorem will not be proved). Note also that if A and B are Hermitian matrices then a necessary and sufficient condition that they can be simultaneously diagonalized is that they commute, i.e., ˆ ˆ ˆ ÂB BA= (see Problem 2.16). These concepts have important physical meaning and will be discussed in greater detail in Chap. 4.

2.5 Fourier Series and the Fourier Transform

Fourier Series Consider a function f (x) over the interval 0 < <x l. The function is called square integrable if

f x dxl

( ) 2

0∫ (2.24)


is defined (i.e., convergent). It can be shown that the set of all such functions is an infinite dimensional vector space, denoted L l2 0( , ). We can define for L l2 0( , ) an inner product

⟨ ⟩ = ∫f g f x g x dxl

, ( ) ( )*

0

(2.25)

Every function f (x) in L2(0, l) can be expanded in a Fourier series,

f x f e kl

nnik x

n

nn( ) = =

=−∞

∞

∑ 2π (2.26)

According to this relation, we can consider the functions el

enik x

n= 1 as a “basis” of the infinite dimensional

space L2(0, l ): every function (vector) in this space can be expanded as a linear combination of the basis vectors. It can be shown that the {en} form an orthonormal basis, that is, ⟨ ⟩ =e ei j ij, .δ The coefficients fn in the expansion are called Fourier coefficients and are derived using the relation

fl

f t e dtnik t

ln= −∫1

0

( ) (2.27)

Since the functions en are periodic, of period l, it is not difficult to show that the Fourier expansion developed above holds also for periodic functions f (x) of period l.

Fourier Transform Now consider a function f (x) defined on (− ∞, ∞) that is not necessarily periodic. We can imagine f (x) to be an approximation of periodic functions whose period approaches ∞. The numbers kn become progressively denser until we have in the limit a continuous range of functions eikx. This is the intuitive basis of the fol-lowing result:

f x F k e dkikx( ) ( )=−∞

∞

∫1

2π (2.28)

where F(k) is given by

F k f x e dxikx( ) ( )= −

−∞

∞

∫1

2π (2.29)

F(k) and f x( ) are said to be Fourier transforms of each other. The Parseval–Plancherel formula states that a function and its Fourier transform have the same norm:

f x dx F k dk( ) ( )2 2

−∞

∞

−∞

∞

∫ ∫= (2.30)

2.6 The Dirac Delta FunctionIn Sec. 2.3 we used the Kronecker dmn function, which returns the value 1 whenever the integers n and m are equal, and 0 otherwise. There is a continuous analogue to Kronecker’s d-function—the Dirac delta function (Dirac d-function). Define the function de (x) as

δε

ε ε

εε ( )xx

x

=− < <

>

⎧

⎨⎪⎪

⎩⎪⎪

12 2

02

for

for

(2.31)


Consider the arbitrary function f (x), well defined for x = 0 with negligible variation over the interval [−e /2, e /2]. If e is sufficiently small, then we have

δ δε ε( ) ( ) ( ) ( ) ( )x f x dx f x dx f≅ =−∞

∞

−∞

∞

∫∫ 0 0 (2.32)

Taking the limit as e → 0 we define the d-function by

lim ( ) ( ) ( ) ( )ε εδ δ

→ −∞

∞

∫⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪= =

0x f x dx x f x dx ff ( )0

−∞

∞

∫ (2.33)

More generally, we can write

δ ( ) ( ) ( )x x f x dx f x− =−∞

∞

∫ 0 0 (2.34)

One can easily show that δ ( )x y dx− =−∞

∞

∫ 1 and that d (x − y) = 0 for x ≠ y. Although we use the term

d-function, it is not a function in the regular sense; it is really a more complicated object called a distribution (it is not defined at the point x = y). That is, we only consider it when it appears inside an integral:

f f x x y dy→ −−∞

∞

∫ ( ) ( )δ (2.35)

As this is a linear operation that maps a function to a number, the d-function can be viewed as a functional. The d-function is often used to describe a particle located at a point r0 = (x0, y0, z0) in a three-dimensional Euclidian space by defining a d (r − r0 ):

δ δ δ δ( ) ( ) ( ) ( )r r− = − − −0 0 0 0x x y y z z (2.36)

The integral of d over the whole space is 1, indicating the existence of the particle. On the other hand, d vanishes when r ≠ r0.

It is straightforward to demonstrate that the following results hold for the d-function:

1. d (−x) = d (x)

2. δ α α δ( ) ( )x x= 1

3. xd (x − x0) = x0d (x − x0)

4. δ δ δ( ) ( ) ( )x y y z dy y z− − = −−∞

∞

∫

The c -function and the Fourier Transform The Fourier transform of the d-function is

1

2

1

2πδ

π( )x y e dx eikx ikx− =−

−∞

∞−∫ (2.37)

The inverse Fourier transform then yields

δ π π( ) ( )x y e e dk e dkiky ikx ik x y− = =−

−∞

∞−

−∞

∞

∫ ∫12

12

(2.38)


SOLVED PROBLEMS

2.1. The complex conjugate of z a bi a bi= + −is , denoted by z*. Show that

(a) zz z* = 2

(b) z + z* is real

(c) ( )* * *z z z z1 2 1 2+ = +

(d ) ( )* * *z z z z1 2 1 2=

(e) z z z z1 2 1 2=

SOLUTION

(a) zz a bi a bi a b z* ( )( )= + − = + =2 2 2

(b) z z a bi a bi a+ = + + − =( ) ( ) ,2 which is real

(c) ( ) [( ) ( )] [( ) (* *z z a b i a b i a b b1 2 1 1 2 2 1 1 1+ = + + + = + + + bb i

a a b b i a b i a b i

2

1 2 1 2 1 1 2 2

) ]

( ) ( ) ( ) ( )

*

= + − + = − + − = zz z1 2* *+

(d) ( ) [( ) ( )] [( ) (* *z z a b i a b i a a b b a1 2 1 1 2 2 1 2 1 2= + + = − + 11 2 2 1

1 2 1 2 1 2 2 1 1 1

b a b i

a a b b a b a b i a b

+

= − − + = −

) ]

( ) (

*

ii a b i z z)( ) * *2 2 1 2− =

(e) z z z z z z z z z z z z z z z1 22

1 2 1 2 1 2 1 2 1 1 2 2= = = =( ) ( )* * * * *11

22

2z

2.2. Calculate 11

5+−

⎛⎝⎜

⎞⎠⎟

ii

.

SOLUTION

Method a: 11

1 11 1

5 5+−

⎛⎝⎜

⎞⎠⎟ = + +

− +⎡⎣⎢

⎤⎦⎥

=ii

i ii i

( )( )( )( )

(112

22

5 55+ +⎡

⎣⎢⎤⎦⎥

= ⎛⎝⎜

⎞⎠⎟ = =i i i

i i) )(1

(2.2.1)

Method b: 11

2 45 452 45 4

5+−

⎛⎝⎜

⎞⎠⎟ = +

−ii

(cos sin )(cos sin

� �

� 55

5 4

4

5

2 5

�)

( )

⎡

⎣⎢

⎤

⎦⎥ =

⎛

⎝⎜⎞

⎠⎟

= =

−e

e

e e

i

i

i i

π

π

π π

/

/

/ //2 90 90= + =cos sin� �i i

(2.2.2)

2.3. Here is a set of functions: f (z) = z2, g(z) = z3, and h(z) = z. Are they linearly independent on the real z axis?

SOLUTION

The condition for linear independence is that

a f z a g z a h z1 2 3 0( ) ( ) ( )+ + = (2.3.1)

implies

a1 = a2 = a3 = 0 (2.3.2)

Here that is

a z a z a z12

23

3 0+ + = (2.3.3)

If this is to be true for all z, it must indeed be that

a1 = a2 = a3 = 0 (2.3.4)

And the set of functions is linearly independent.

2.4. Consider the set of functions: f (y) = 4y2, g(y) = 2y, and h(y) = 10y. Are they linearly independent on the real y axis?

SOLUTION

We need to determine if

a f y a g y a h y1 2 3 0( ) ( ) ( )+ + = (2.4.1)


can only be true if

a a a1 2 3 0= = = (2.4.2)

Here,

a y a y a y12

2 34 2 10 0+ + = (2.4.3)

and if a1 = 0, a2 = −5, and a3 = 1, the sum will be zero for all y. Hence these functions are not independent:

h y g y f y( ) ( ) ( )= + ⋅5 0 (2.4.4)

2.5. Consider the set of vectors in ordinary three-dimensional space: A = (6, 0, 0), B = (0, − 4, 0), and C = (0, 0, 5) are they linearly independent?

SOLUTION

We have to examine

a a a1 2 3 0A B C+ + = (2.5.1)

and since A = 6i, B = − 4j, and C = 5k, it follows that

a a a1 2 36 4 5 0i j k+ − + =( ) (2.5.2)

Since each term is perpendicular to the others the composite vector can only be zero if 6a1 = 0, − 4a2 = 0, and 5a3 = 0. That is, a1 = a2 = a3 = 0. The vectors A, B, and C are linearly independent.

2.6. Show that the sum and product of two linear operators are linear operators.

SOLUTION

Suppose that T and S are linear operators, so

( ˆ ˆ) ) ˆ ) ˆ )

ˆ ) ˆ(

T S u T u S u

T u T

+ + ≡ + + +

= +

( ( (

(

α α α

α

v v v

v )) ˆ( ) ˆ( )

( ˆ ˆ) ) ( ˆ ˆ)

+ +

= + + +

S u S

T S u T S

α

α

v

v( ( )

(2.6.1)

and

( ˆ ˆ) ) ˆ[ ˆ( )] ˆ[ ˆ( ) ˆ( )]T S u T S u T S u S⋅ + ≡ + = +( α α αv v v

== + = ⋅ + ⋅ˆ[ ˆ( )] ˆ[ ˆ( )] ( ˆ ˆ) ( ) ( ˆ ˆ)T S u T S T S u T Sα αv (vv ) (2.6.2)

2.7. Let V be the space of infinitely differentiable functions in one variable. Prove that differentiation is a linear operator.

SOLUTION

We define the map ddx

as

ddx

f f x( ) ( )= ′ (2.7.1)

and using basic calculus we get

ddx

f g f g f x g xddx

fddx

( ) [ ] ( ) ( ) ( ) (+ = + ′ = ′ + ′ = +α α α α gg) (2.7.2)

2.8. Let V be Cn, i.e., the collection of n-tuples a = (a1, . . . , an), where the ai are complex numbers. Show

that ⟨ ⟩ ==

∑a b, *a bi i

i

n

1

is an inner product of V.


SOLUTION

We begin by checking the four conditions that an inner product on V must satisfy:

⟨ ⟩ = =⎡

⎣

⎢⎢

⎤

⎦

⎥⎥

= ⟨ ⟩= =

∑ ∑a b b a, ,* *

*

*a b a bi i

i

n

i i

i

n

1 1

(2.8.1)

⟨ + ′ ⟩ = + ′ = ⋅ + ′ ⋅= =

∑ ∑a a b, ( ) * *a a b a b a bi i i

i

n

i i i

i

n

1 1

ii

i

n

* , ,= ⟨ ⟩ + ⟨ ′ ⟩=

∑ a b a b

1

(2.8.2)

and

⟨ ⟩ = = = ⟨ ⟩= =

∑ ∑α α α αa b a b, ( ) ,* *a b a bi i

i

n

i i

i

n

1 1

(2.8.3)

Finally,

⟨ ⟩ = ===

∑∑a a, *a a ai i i

i

n

i

n2

11

(2.8.4)

and is greater than zero if one of the ai is different from zero.

2.9. If A and B are operators, prove

(a) that ( ˆ ) ˆ† †A A=

(b) that ( ˆ ˆ) ˆ ˆ† † †AB B A=

(c) that ˆ ˆ , ( ˆ ˆ ),† †A A i A A+ − and that ˆ ˆ†AA are Hermitian operators.

SOLUTION

(a) For every u and v in V,

⟨ ⟩ = ⟨ ⟩ = ⟨ ⟩ = ⟨ ⟩ = ⟨ˆ , , ˆ ˆ , , ( ˆ ) ( ˆ† † * † † *A u A u A u u Av v v v AA u† †) ,v ⟩

Thus we obtain ˆ ( ˆ ) .† †A A=

(b) For every u and v in V,

⟨ ⟩ = ⟨ ⟩ = ⟨ ⟩ = ⟨v v v v, ( ˆ ˆ) ˆ ˆ , ˆ , ˆ , ˆ ˆ† † † †AB u AB u B A u B A uu⟩ (2.9.1)

Hence, ˆ ˆ ( ˆ ˆ) .† † †B A AB=

(c) We write

( ˆ ˆ ) ˆ ( ˆ ) ˆ ˆ ˆ ˆ† † † † † † †A A A A A A A A+ = + = + = + (2.9.2)

Here we use the fact that the sum of conjugates is the conjugate of the sum, ( ˆ ˆ) ˆ ˆ ,† † †A B A B+ = + which can be easily verified, and we also use the result of part (a).

[ ( ˆ ˆ )] ( ˆ ˆ ) ( ˆ ˆ) ( ˆ ˆ† † * † † †i A A i A A i A A i A A− = − = − − = − ††) (2.9.3)

where we have used the fact that the conjugate of a complex number is the same as its conjugate as an operator. And finally,

( ˆ ˆ ) ( ˆ ) ˆ ˆ ˆ† † † † † †AA A A AA= = (2.9.4)

according to part (b).

2.10. Show that the eigenvalues of a Hermitian operator are real.

SOLUTION

Suppose l is an eigenvalue of ˆ,T and ˆ ˆ .†T T= For every v ≠ 0 in V,

λ λ λ

λ

⟨ ⟩ = ⟨ ⟩ = ⟨ ⟩ = ⟨ ⟩ = ⟨ ⟩

= ⟨ ⟩

v v v v v v v v v v

v v

, , ˆ , , ˆ ,

,

T T

** * ,= ⟨ ⟩λ v v (2.10.1)

Since ⟨ ⟩v v, is a positive real number (v ≠ 0), it follows that l = l∗, so l is a real number. The fact that the eigenvalues of Hermitian operators are real is of great importance, since these eigenvalues can represent physical quantities.


2.11. Show that eigenvectors that correspond to different eigenvalues of a Hermitian operator are orthogonal.

SOLUTION

Suppose Tv v= λ and ˆ ,Tu u= μ where μ λ≠ . Now,

λ λ μ⟨ ⟩ = ⟨ ⟩ = ⟨ ⟩ = ⟨ ⟩ = ⟨ ⟩ = ⟨v v v v v v, , ˆ , , ˆ , ˆ ,†u u T u T u Tu uu u⟩ = ⟨ ⟩μ* ,v (2.11.1)

so,

( ) , ( ) ,*λ μ λ μ− ⟨ ⟩ = − ⟨ ⟩ =v vu u 0 (2.11.2)

( ,*μ μ= since T is Hermitian). But l − m ≠ 0; therefore, ⟨ ⟩ =v , ,u 0 i.e., v and u are orthogonal.

2.12. Show that Hermitian, anti-Hermitian, and unitary operators are normal operators.

SOLUTION

If ˆ ˆ†T T= , then ˆ ˆ ˆ ˆ ˆ .† †T T T T T= = 2 Also, if ˆ ˆ†T T= − , then ˆ ˆ ˆ ˆ ˆ .† †T T T T T= = − 2 If T is unitary, then ⟨ ⟩ = ⟨ ⟩ˆ , ˆ ,Tu T uv v for every u, v in V. Using the definition of conjugate operator and taking u = v, we get

⟨ ⟩ = ⟨ ⟩ = ⟨ ⟩u u Tu Tu u T T u, ˆ , ˆ , ˆ ˆ† (2.12.1)

hence,

⟨ − ⟩ =u I T T u, ( ˆ ˆ ˆ )† 0 (2.12.2)

for every u in V. Since ˆ ˆ ˆ†I T T− is a Hermitian operator, it follows that ˆ ˆ ˆ .†I T T− = 0 This also completes the proof showing T is a normal operator.

2.13. Let V be the space of nonzero square integrable continuous complex functions in one variable. For every pair of functions, define

⟨ ⟩ =−∞

∞

∫f g f x g x dx, ( ) ( )* (2.13.1)

Show that with this definition, V is an inner product space.

SOLUTION

We must check the following conditions:

⟨ ⟩ = = = ⟨ ⟩−∞

∞

−∞

∞

∫f g f x g x dx g x f x dx g f, ( ) ( ) ( ) ( ) ,* * *∫∫ (2.13.2)

⟨ + ′ ⟩ = + ′

=

−∞

∞

∫f f g f x f x g x dx

f x g x d

, [ ( ) ( )] ( )

( ) ( )

*

* xx f x g x dx

f g f g

+ ′

= ⟨ ⟩ + ⟨ ′ ⟩−∞

∞

−∞

∞

∫∫ ( ) ( )

, ,

*

(2.13.3)

and

⟨ ⟩ = = = ⟨ ⟩−∞

∞

∫α α α αf g f x g x dx f x g x dx f g, ( ) ( ) ( ) ( ) ,* *

−−∞

∞

∫ (2.13.4)

⟨ ⟩ =−∞

∞

∫f f f x dx, ( ) 2 (2.13.5)

Since f is continuous and f ≠ 0 in a neighborhood, its integral also differs from zero; hence ⟨ ⟩ ≠f f, .0

2.14. (a) Show that if ⟨ ⟩ = ⟨ ⟩v v, ,u w for every v in V, then u w= .

(b) Show that if T and S are two linear operators in V that satisfy ⟨ ⟩ = ⟨ ⟩ˆ , ˆ ,T u S uv v for every u, v in V, then ˆ ˆ.S T=

SOLUTION

(a) The condition ⟨ ⟩ = ⟨ ⟩v v, ,u w implies that ⟨ − ⟩ =v , u w 0 for every v in V. In particular, if v = u − w we obtain

⟨ − − ⟩ =u w u w, 0 (2.14.1) Hence, u − w = 0, that is, u = w.


(b) According to part (a), ⟨ ⟩ = ⟨ ⟩ˆ , ˆ ,T u S uv v for every v, u in V which implies that ˆ ˆ ;T Sv v= i.e., ˆ ˆ.T S=

2.15. Let A and B be Hermitian matrices. Show that A and B can be simultaneously diagonalized (that is, with the same matrix ˆ )U if and only if ˆ ˆ ˆ ˆ.AB BA=

SOLUTION

Suppose ˆ ˆ ˆ ˆ ˆ ˆ ˆ ÛAU D UBU D− −= =11

12and where D1 and D2 are diagonal matrices. Hence,

ˆ ( ˆ ˆ) ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ Û AB U UAU UBU D D D D− − −= = =1 1 11 2 2 1 == =− − −ˆ ˆ ˆ ˆ ˆ ˆ ˆ ( ˆ ˆ) ÛBU UAU U BA U1 1 1

(2.15.1)

Multiplying on the right with U and on the left with U −1 we get ˆ ˆ ˆ ˆ.AB BA= We leave it to the reader to prove the other direction. This result is of great importance in quantum mechanics.

2.16. Show that the modulus of the eigenvalues of a unitary operator is equal to 1.

SOLUTION

Suppose T is a unitary operator, and let v ≠ 0 be an eigenvector with an eigenvalue l. Then,

⟨ ⟩ = ⟨ ⟩ = ⟨ ⟩ = ⟨ ⟩v v v v v v v v, ˆ , ˆ , ,*T T λ λ λλ (2.16.1)

Hence,

λλ λ* = =21 (2.16.2)

2.17. Suppose that f is an integrable function.

(a) If l ≠ 0 is a real number and g(x) = f (lx + y), prove that

G k e Fkiky( ) = ⎛

⎝⎜⎞⎠⎟

1λ λ

λ/ (2.17.1)

where F and G are the Fourier transforms of f and g, respectively.

(b) Prove that if x f is also integrable, then F(k) is a differentiable function and

F f x F ix f x[ ( )] [ ( )]′ = − (2.17.2)

SOLUTION

(a) By definition,

G k g x e dx f x y e dx

f

ikx ikx( ) ( ) ( )

(

= = +

=

−

−∞

∞−

−∞

∞

∫ ∫ λ

λxx y e e d x y

e

ik x y iky

i

+ +

=

− +

−∞

∞

∫ ) ( )( )( )/ /λ λ λλ λ1

1λ

kky i k s ikyf s e ds e Fk/ / /λ λ λ

λ λ( ) ( )−

−∞

∞

∫ = ⎛⎝⎜

⎞⎠⎟

1

(2.17.3)

(b) Consider the expression

F k h F kh

f x e eh

dikxihx( ) ( )

( )+ − = −⎛

⎝⎜⎞⎠⎟

−

−∞

∞

∫12

1π

xx (2.17.4)

Taking lim ,h

ihxeh

ix→

−⎛⎝⎜

⎞⎠⎟

= −0

1 we obtain

F f x ix f x e dx F ix f xikx[ ( )] ( ) [ ( )]′ = − = −−

−∞

∞

∫12π

(2.17.5)

2.18. Show that

(a) F x x F x eikx

[ ( )] [ ( )]δ δ− = −0

0

(b) F axa

F ka

[ ( )] .δ δ= ( )⎡⎣⎢

⎤⎦⎥

1


SOLUTION

(a) By definition,

F x x x x e dx z eikx i[ ( )] ( ) ( )δπ

δπ

δ− = − =−

−∞

∞−∫0 0

12

12

zzk ix k ikxe dz F x e

− −

−∞

∞

=∫ 0 0[ ( )]δ (2.18.1)

(b)

F ax ax e dxa

z eixk[ ( )] ( ) ( )δπ

δπ

δ= =−

−∞

∞

−∞

∞

∫ ∫12

12

1 −− = ⎛⎝⎜

⎞⎠⎟

⎡⎣⎢

⎤⎦⎥

ikz adza

Fka

/ 1 δ (2.18.2)


2.19. Prove the triangle inequality for complex numbers; that is, show that z z z z1 2 1 2+ ≤ + .

2.20. Are the two vectors

A = (4, −5, 0) and B = (−12, 15, 0)

in ordinary three-dimensional space, linearly independent?

Ans. No, −3a1 = a2 or −3A = B.

2.21. Show that the vectors (1, 1, 0), ( , ),0 0, 2 and (i, i, i) are linearly dependent over the complex field.

2.22. Find the eigenvalues and eigenvectors of the matrix ˆ .A = ⎛⎝⎜

⎞⎠⎟

01

10

Hint: If l is an eigenvalue, then ˆ ,Av v= λ or

( ˆ ˆ)A I− =λ v 0 for some v ≠ 0; this implies that det ( ˆ ˆ) .A I− =λ 0 Solve this equation for l, then substitute l and find v.

Ans. λ λ1 2111

111

= = ⎛⎝⎜

⎞⎠⎟ = − = −

⎛⎝⎜

⎞⎠⎟, , , .1 2v v

2.23. Show that the matrix

ˆ cos sinsin cos

T =−⎛

⎝⎜⎞⎠⎟

θ θθ θ (2.23.1)

is unitary. If uxy

= ⎛⎝⎜

⎞⎠⎟ is a vector in the plane, what is the geometric interpretation of u Tu→ ˆ ?

2.24. Demonstrate that the system 12

12

1 12

π π π π π,

1. . . . .sin , sin , , cos , cos ,k k k k .

⎧⎨⎩

⎫⎬⎭ is also orthonormal.

2.25. Consider the space of polynomials with degree less than or equal to n. We can think of each polynomial p x a a x a xn

n( ) = + + +0 1 � as a vector in the space Cn +1, (a0, a1, . . . , an). In fact, this is the representation of

p(x) relative to the basis {1, x, . . . , xn}. What is the matrix that represents the operator ddx

relative to the basis?

Ans.

0 1 0 00 0 2 0

0 00 0 0

��

� � � � ��

n

⎛

⎝

⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟

2.26. Find the Fourier transform of e x− 2 2/ .

Ans. F t e k( ) .= − 2 2/

2.27. (a) Find the Fourier series of f xx

x( ) , .= − ≤ ≤π π2

0 2

(b) Using part (a), show that π4

12 1

0

= −+

=

∞

∑ ( ).

n

nn

Ans. (a) π − = − =

= −∞

∞

=

∞

∑ ∑x ien

nxn

inx

n n2

0

sin( ).

25

CHAPTER 3

The Schrödinger Equation and Its Applications

3.1 Wavefunctions of a Single ParticleIn quantum mechanics, a particle is characterized by a wavefunction Ψ(r, t), which contains information about the spatial state of the particle at time t. The wavefunction Ψ(r, t) is a complex function of the three coordinates x, y, z and of the time t. The interpretation of the wavefunction is as follows: the probability dP(r, t) of the particle being at time t in a volume element d3r = dx dy dz located at the point r is

dP t C t d r( , ) ( , )r r= Ψ 2 3 (3.1)

where C is a normalization constant. The total probability of finding the particle anywhere in space, at time t, must be equal to unity; therefore,

dP t( , )r =∫ 1 (3.2)

According to Eqs. (3.1) and (3.2) we conclude:

(a) The wavefunction Ψ(r, t) has to be square-integrable; that is

Ψ( , )r t d r2 3∫ (3.3)

must be finite.

(b) The normalization constant is given by the relation

1 2 3

Ct d r= ∫ Ψ( , )r (3.4)

When C = 1 we say that the wavefunction is normalized. A wavefunction Ψ(r, t) must be defined and con-tinuous everywhere.

3.2 The Schrödinger EquationConsider a particle of mass m having a potential energy V(r, t). The time evolution of the wavefunction is governed by the Schrödinger equation:

it

t mt V t t�

�∂∂ = − ∇ +Ψ Ψ Ψ( , )

( , ) ( , ) ( , )r

r r r2

2

2 (3.5)

26

where ∇2 is the Laplacian operator, ∂2 / ∂x2 + ∂2 / ∂y2 + ∂2 / ∂z2. Pay attention to two important properties of the Schrödinger equation:

(a) The Schrödinger equation is a linear homogeneous differential equation in Ψ. Consequently, the superpo-sition principle holds; that is, if Ψ1(r, t), Ψ2(r, t), . . . , Ψn(r, t) are solutions of the Schrödinger equation,

then Ψ Ψ==∑α i ii n

n

t( , )r is also a solution.

(b) The Schrödinger equation is a first-order equation with respect to time; therefore, the state at time t0 determines its subsequent state at all times.

3.3 Particle in a Time-Independent PotentialThe wavefunction of a particle having a time-independent potential energy V(r) satisfies the Schrödinger equation:

it

t mt V t�

�∂∂ = − ∇ +Ψ Ψ Ψ( , )

( , ) ( ) ( , )r

r r r2

2

2 (3.6)

Performing a separation of variables Ψ(r, t) = y (r)f (t), we have f (t) = Ae−iw t (A and w are constant), where y (r) must satisfy the equation

− ∇ + =��

22

2mVψ ψ ωψ( ) ( ) ( ) ( )r r r r (3.7)

where �ω is the energy of the state E (see Problem 3.3). This is a stationary Schrödinger equation, where a wavefunction of the form

Ψ( , ) ( ) ( )r r rt e ei t iEt= =− −ψ ψω /� (3.8)

is called a stationary solution of the Schrödinger equation, since the probability density in this case does not depend on time [see Problem 3.3, part (b)]. Suppose that at time t = 0 we have

Ψ( , ) ( )r r0 = ∑ψ n

n

(3.9)

where yn(r) are the spatial parts of stationary states, Ψni t

t e n( , ) ( ) .r r= −ψ ω In this case, according to the

superposition principle, the time evolution of Ψ(r, 0) is described by

Ψ( , ) ( )r rt ei t

n

n= −∑ψ ω (3.10)

For a free particle we have V(r, t) ≡ 0, and the Schrödinger equation is satisfied by solutions of the form

Ψ( , ) ( )r k rt Aei t= −i ω (3.11)

where A is a constant; k and w satisfy the relation ω = �k2 2/ m. Solutions of this form are called plane waves.

Note that since the Ψ(r, t) are not square-integrable, they cannot rigorously represent a particle. On the other hand, a superposition of plane waves can yield an expression that is square-integrable and can therefore describe the dynamics of a particle,

Ψ( , )( )

( ) [ ( ) ]r k k rt g e d ki k t= −∫1

2 3 23

πω

/i (3.12)

CHAPTER 3 The Schrödinger Equation and Its Applications

27

A wavefunction of this form is called a wave packet. We often study the case of a one-dimensional wave packet traveling in the positive x direction,

Ψ( , ) ( ) [ ( ) ]x t g e dki kx k t= −

−∞

∞

∫1

2πωk (3.13)

See Sec. 1.3.

3.4 Scalar Product of Wavefunctions: OperatorsWith each pair of wavefunctions f (r) and y (r), we associate a complex number defined by

( , ) ( ) ( )*φ ψ φ ψ= ∫ r r d r3 (3.14)

where (f, y) is the scalar product of f (r) and y (r) (see Chap. 2).An operator A acting on a wavefunction y (r) creates another wavefunction y ′(r). An operator is called a

linear operator if this correspondence is linear, i.e., if for every complex number a1 and a2,

ˆ[ ( ) ( )] ˆ ( ) ˆ ( )A A Aα ψ α ψ α ψ α ψ1 1 2 2 1 1 2 2r r r r+ = + (3.15)

There are two sets of operators that are important:

(a) The spatial operators ˆ , ˆ,X Y and Z are defined by

ˆ ( , , , ) ( , , , )ˆ ( , , , ) ( ,

X x y z t x x y z t

Y x y z t y x y

Ψ ΨΨ Ψ

== ,, , )

ˆ ( , , , ) ( , , , )

z t

Z x y z t z x y z tΨ Ψ=

⎧

⎨⎪

⎩⎪

(3.16)

(b) The momentum operators ˆ , ˆp px y and pz are defined by

ˆ ( , , , ) ( , , , )

ˆ ( , , , )

p x y z ti x

x y z t

p x y z t

x

y

Ψ Ψ

Ψ

= ∂∂

=

�

��

�

i yx y z t

p x y z ti z

x y zz

∂∂

= ∂∂

Ψ

Ψ Ψ

( , , , )

ˆ ( , , , ) ( , , , tt)

⎧

⎨

⎪⎪⎪⎪

⎩

⎪⎪⎪⎪

(3.17)

The mean value of an operator A in the state y (r) is defined by

⟨ ⟩ = ∫ˆ ( )[ ˆ ( )]*A A d rψ ψr r 3 (3.18)

The root-mean-square deviation is defined by

ΔA A A= ⟨ ⟩ − ⟨ ⟩ˆ ˆ2 2 (3.19)

where A2 is the operator ˆ ˆ.A A⋅ Consider the operator called the Hamiltonian of the particle. It is defined by

ˆ ˆ( , )ˆ ˆ( , )H

mV t

mV t= − ∇ + ≡ +�

22

2

2 2r

pr (3.20)

CHAPTER 3 The Schrödinger Equation and Its Applications

CHAPTER 3 The Schrödinger Equation and Its Applications28

where p2 is condensed notation for the operator ˆ ˆ ˆ .p p px y z2 2 2+ + Using the operator formulation, the Schrödinger

equation is written in the form

it

tH t�

∂∂ =Ψ Ψ( , ) ˆ ( , )r

r (3.21)

If the potential energy is time-independent, a stationary solution must satisfy the equation

ˆ ( ) ( )H Eψ ψr r= (3.22)

where E is a real number called the energy of the state. Equation (3.22) is the eigenvalue equation of the operator ˆ ;H the application of H on the eigenfunction y (r) yields the same function, multiplied by the cor-responding eigenvalue E. The allowed energies are therefore the eigenvalues of the operator ˆ .H

3.5 Probability Density and Probability CurrentConsider a particle described by a normalized wavefunction Ψ(r, t). The probability density is defined by

ρ( , ) ( , )r rt t= Ψ 2 (3.23)

At time t, the probability dP(r, t) of finding the particle in an infinitesimal volume d3r located at r is equal to

dP t t d r( , ) ( , )r r= ρ 3 (3.24)

The integral of r (r, t) over all space remains constant at all times. Note that this does not mean that r (r, t) must be time-independent at every point r. Nevertheless, we can express a local conservation of probability in the form of a continuity equation,

∂

∂ + ∇ =ρ( , )( , )

rJ r

tt

ti 0 (3.25)

where J(r, t) is the probability current, defined by

J r( , ) ( ) ( ) Re* * *tmi m i

= ∇ − ∇⎡⎣ ⎤⎦ = ∇⎛⎝⎜

⎞� �2

1Ψ Ψ Ψ Ψ Ψ Ψ⎠⎠⎟

⎡⎣⎢

⎤⎦⎥

(3.26)

Consider two regions in a space separated by a potential energy step or barrier, see Fig. 3.1.

Fig. 3.1 (a) Potential step; (b) potential barrier.

I

(a)

II

x

V(x)

I

(b)

II

x

V(x)


We define transmission and reflection coefficients as follows. Suppose that a particle (or a stream of par-ticles) is moving from region I through the potential energy step (or barrier) to region II. In the general case, a stationary state describing this situation will contain three parts. In region I the state is composed of the incoming wave with probability current JI and a reflected wave of probability current JR. In region II there is a transmitted wave of probability current JT .

The reflection coefficient is defined by

RJJ

R

I= (3.27)

The transmission coefficient is defined by

TJJ

T

I= (3.28)

SOLVED PROBLEMS

3.1. Figure 3.2 depicts a plane passing through a point (x0, y0, z0), perpendicular to a propagation vector k. Using this diagram show that Eq. (3.11) represents a plane wave.

SOLUTION

The position vector of an arbitrary point (x, y, z) is

r = + +x y zi j k (3.1.1)

where i, j, and k are the usual unit basis vectors. The vector (r − r0) goes from (x0, y0, z0) to (x, y, z) and is given by

( ) ( ) ( ) ( )r r− = − + − + −0 0 0 0x x y y z zi j k (3.1.2)

Setting

( )r r k− =0 0i (3.1.3)

causes ( )r r− 0 to sweep out a plane perpendicular to the propagation vector k = + +k k kx y zi j k.

(x, y, z)

(x0, y0, z0)

y

x

z

r – r0

k

r

0r0

Fig. 3.2


Hence

k x x k y y k z zx y z( ) ( ) ( )− + − + − =0 0 0 0 (3.1.4)

and since x0, y0, and z0 are constants, as are kx, ky, and kz,

k x k y k zx y z+ + = constant (3.1.5)

Or more concisely

k ri = constant (3.1.6)

This is the equation of a plane perpendicular to k. Hence

ψ ( ) cosr k r= A ( )i (3.1.7)

or

ψ ( )r k r= A ei( )i (3.1.8)

is the equation for a stack of harmonically varying planes. To set the planes moving at a speed v = w/k we need only replace ( ) ( )k r k ri iby − ωt whereupon

Ψ( , ) expr k rt A i t= −[ ]( )i ω (3.1.9)

becomes the equation of a harmonic plane wave.

3.2. Starting with the Schrödinger equation in one dimension and using a de Broglie plane wave as a solution, show that when V = 0 this leads to the correct nonrelativistic relationship between energy and momentum.

SOLUTION

In one dimension with V = 0, Eq. (3.5) becomes

ix tt m

x t

x�

�∂∂ = − ∂

∂Ψ Ψ( , ) ( , )2 2

22 (3.2.1)

In general, a plane wave has the form

Ψ( , ) exp[ ( )]r k rt A i t= −i ω (3.2.2)

When k is pointing in the positive x direction k ri = k xx or, since k = kx under those circumstances,

Ψ( , ) exp[ ( )]x t A i kx t= − ω (3.2.3)

From de Broglie’s hypothesis p = h/l, and since k = 2p/l, p k= � . Similarly, E h= =n �ω. Thus

Ψ( , ) exp[ ( ) ]x t A i px Et= − /� (3.2.4)

Putting this into Schrödinger’s equation yields

i A i px Et iEm x

A i px� � ��

exp[ ( ) ]( ) exp[ (− − = − ∂∂/ /

2

2−− Et ip) ]( )/ /� � (3.2.5)

E x tm

ip x tΨ Ψ( , ) ( ) ( , )= − ��

22

2/ (3.2.6)

and

Epm

=2

2 (3.2.7)

which is the proper nonrelativistic formula.


3.3. Consider a particle subjected to a time-independent potential V(r). (a) Assume that a state of the particle is described by a wavefunction of the form Ψ(r, t) = y (r)f(t). Show that f(t) = Ae−iwt (A is constant) and that y (r) must satisfy the equation

− ∇ + =��

22

mVψ ψ ωψ( ) ( ) ( ) ( )r r r r (3.3.1)

where m is the mass of the particle. (b) Prove that the solutions of the Schrödinger equation of part (a) lead to a time-independent probability density.

SOLUTION

(a) We substitute Ψ(r, t) = y (r) f(t) in the Schrödinger equation:

idf t

dtf t

mf t V�

�ψ ψ( )( )

( ) ( ) ( )r r= − ∇⎡

⎣⎢⎢

⎤

⎦⎥⎥

+2

2

2(( ) ( )r rψ (3.3.2)

In the regions in which the wavefunction Ψ(r, t) does not vanish, we divide both sides of Eq. (3.3.2) by y (r) f(t) and obtain

if t

df tdt m

V� �( )

( )( )

( ) ( )= − ∇⎡

⎣⎢⎢

⎤

⎦⎥⎥

+12

22

ψ ψr r r (3.3.3)

The left-hand side of Eq. (3.3.3) is a function of t only, and does not depend on r. On the other hand, the right-hand side is a function of r only. Therefore, both sides of Eq. (3.3.3) depend neither on r nor on t, and are thus constants that we will set equal to �ω for convenience. Hence,

if t

df tdt

id f t

dt� � �

1( )

( ) [ln ( )]= = ω (3.3.4)

Therefore,

ln ( ) ( )f t i dt i t C f t Ae i t= − = − + = = −∫ ω ω ω (3.3.5)

where A is constant. Substituting in (3.3.3), we see that y (r) must satisfy the equation

− ∇ + =��

22

2mVψ ψ ωψ( ) ( ) ( ) ( )r r r r (3.3.6)

(b) For a function of the form Ψ(r, t) = y (r)e−iw t, the probability density is by definition

ρ ψ ψω ω( , ) ( , ) ( ) ( )*r r r rt t e ei t i t= = ⎡⎣ ⎤⎦ ⎡⎣ ⎤⎦ =− −Ψ 2 ψψ ψ ψω ω( ) ( ) ( )*r r re ei t i t− = 2 (3.3.7)

We see that the probability density does not depend on time. This is why this kind of solution is called “stationary.”

3.4. Consider the classical Hamiltonian for a one-dimensional system of two particles of masses m1 and m2 subjected to a potential that depends only on the distance between the particles x1 − x2,

Hpm

pm

V x x= + + −12

1

22

21 22 2

( ) (3.4.1)

(a) Write the Schrödinger equation using the new variables x and X, where

x x x Xm x m x

m m= − =

++1 2

1 1 2 2

1 2(relative distance) ((center of mass) (3.4.2)

(b) Use a separation of variables to find the equations governing the evolution of the center of mass and the relative distance of the particles. Interpret your results.


SOLUTION

(a) In terms of x1 and x2, the wavefunction of the two particles is governed by the Schrödinger equation:

ix x t

tH x x t

mx x

�

�

∂∂ =

= −∂

ΨΨ

Ψ

( , , ) ˆ ( , , )

( ,

1 21 2

2

1

21

222

12

2

2

21 2

22 1 22

, ) ( , , )( ) (

t

x mx x t

xV x x x

∂−

∂∂

+ −� ΨΨ 11 2, , )x t (3.4.3)

In order to transform to the variables x and X, we have to express the differentiations ∂ ∂212/ x and ∂ ∂2

22/ x

in terms of these new variables. We have

∂∂ = ∂

∂ = − ∂∂ = +

∂∂ = +

xx

xx

Xx

mm m

Xx

mm m1 2 1

1

1 2 2

2

1 21 1 (3.4.4)

Thus, for an arbitrary function g(x1, x2) we obtain

∂

∂ = ∂∂

∂∂ + ∂

∂∂∂ =

g x xx

g x Xx

xx

g x XX

Xx

( , ) ( , ) ( , )1 2

1 1 1

∂∂∂ + +

∂∂

g x Xx

mm m

g x XX

( , ) ( , )1

1 2 (3.4.5)

Similarly,

∂

∂ = ∂∂

∂∂ + ∂

∂∂∂ =

g x xx

g x Xx

xx

g x XX

Xx

( , ) ( , ) ( , )1 2

2 2 2−− ∂

∂ + +∂

∂g x X

xm

m mg x X

X( , ) ( , )2

1 2 (3.4.6)

or

∂

∂ = ∂∂ + +

∂∂

∂∂ = − ∂

∂ + +∂

∂x xm

m m X x xm

m m X1

1

1 2 2

2

1 2 (3.4.7)

For the second derivatives in x1 and x2 we have

∂∂

= ∂∂ + +

∂∂

⎛⎝⎜

⎞⎠⎟

∂∂ + +

∂∂

⎛⎝

2

12

1

1 2

1

1 2x xm

m m X xm

m m X⎜⎜⎞⎠⎟

= ∂∂

+ +∂∂

∂∂ + +

∂∂

∂∂ +

2

21

1 2

1

1 2

1

x

mm m x X

mm m X x

mm11 2

2 2

2+⎛⎝⎜

⎞⎠⎟

∂∂m X

(3.4.8)

The wavefunction must be a smooth function for both x1 and x2; so, we can interchange the order of differentiation and obtain

∂∂

= ∂∂

+ +⎛⎝⎜

⎞⎠⎟

∂∂

+ +∂

∂

2

12

2

21

1 2

2 2

21

1 2

2

x x

mm m X

mm m XX x

∂∂ (3.4.9)

For x2 we have

∂∂

= − ∂∂ + +

∂∂

⎛⎝⎜

⎞⎠⎟

− ∂∂ + +

∂∂

2

22

2

1 2

2

1 2x xm

m m X xm

m m X⎛⎛⎝⎜

⎞⎠⎟

= ∂∂

+ +⎛⎝⎜

⎞⎠⎟

∂∂

− +∂2

22

1 2

2 2

22

1 2

2

x

mm m X

mm m ∂∂

∂∂X x

(3.4.10)


Substituting Eqs. (3.4.9) and (3.4.10) in Eq. (3.4.3), we get

ix X t

t m x

mm m

��∂

∂ = − ∂∂

+ +⎛⎝⎜

⎞⎠⎟

∂∂

Ψ( , , ) 2

1

2

21

1 2

2 2

2 XX

mm m X x

x X t

m

21

1 2

2

2

2

2

2

+ +∂

∂∂∂

⎡

⎣⎢⎢

⎤

⎦⎥⎥

− ∂∂

Ψ( , , )

�

xx

mm m X

mm m X x2

2

1 2

2 2

22

1 2

2+ +

⎛⎝⎜

⎞⎠⎟

∂∂

− +∂

∂∂∂

⎡

⎣⎢⎢

⎤

⎦⎦⎥⎥

+

= − +⎛⎝⎜

⎞⎠⎟

Ψ Ψ( , , ) ( ) ( , , )x X t V x x X t

m m�

2

1 221 1 ∂∂

∂+ − +

⎛⎝⎜

⎞⎠⎟

2

2

2

1 221Ψ Ψ( , , )

( ) ( , , )x X t

xV x x X t

m m� ∂∂

∂

2

2Xx X tΨ( , , )

(3.4.11)

(b) Since the Hamiltonian is time-independent, Ψ(x, X, t) = y (x, X) f (t) (we separate the time and the spatial variables; see Problem 3.3). The equation governing the stationary part y (x, X) is ˆ ( , ) ( , ),H x X E x Xψ ψ= total where Etotal is the total energy. Substituting in Eq. (3.2.11) we arrive at

−+⎛

⎝⎜⎞⎠⎟

∂∂

+ −� �2

1 2

1 2

2

22m m

m mx X

xV x x X

ψ ψ( , )( ) ( , )

22

1 2

2

221

m mx X

XE x X+

⎛⎝⎜

⎞⎠⎟

∂∂

=ψ ψ( , )( , )total (3.4.12)

Performing a separation of the variables y (x, X) = x(x)h(X), Eq. (3.4.12) becomes

−+⎛

⎝⎜⎞⎠⎟

∂∂

+ =� �2

1 2

1 2

2

2

2

21

21

ξξ

( )( )

( )x

m mm m

x

xV x ηη

η( )

( )X m m

X

XE

1

1 2

2

2+∂

∂+ total (3.4.13)

The left-hand side of Eq. (3.4.13) depends only on x; on the other hand, the right-hand side is a func-tion only of X. Therefore, neither side can depend on x or X, and both are thus equal to a constant. We set

− +∂∂

=�2

1 2

2

221 1

ηη

( )( )

X m mx

XEcm (3.4.14)

By inspection, we conclude that Eq. (3.4.14) is the equation governing the stationary wavefunction of a free particle of mass m1 + m2, i.e.,

− +∂

∂=�

2

1 2

2

221

m mX

XE X

η η( )( )cm (3.4.15)

Note that the wavefunction corresponding to the center of mass of the two particles behaves as a free particle of mass m1 + m2 and energy Ecm. This result is completely analogous to the classical case. Returning to Eq. (3.4.13), the equation for the relative position of the two particles is

−+⎛

⎝⎜⎞⎠⎟

∂∂

+ =�2

1 2

1 2

2

22ξξ

( )( )

( )x

m mm m

x

xV x Etotall cm− E (3.4.16)

Equation (3.4.16) governs the stationary wavefunction of a particle of mass (m1 + m2)/m1m2 confined by a potential energy V(x) and having a total energy Etotal − Ecm. Thus the relative position of the two particles behaves as a particle with an effective mass (m1 + m2)/m1m2 and energy Etotal − Ecm confined by an effective potential energy V(x). This is also analogous to the classical case.


3.5. Consider a particle of mass m confined in a finite one-dimensional “potential well” V(x); see Fig. 3.3.

Prove that (a) d xdt

pm

⟨ ⟩ = ⟨ ⟩, and (b)

d pdt

dVdx

⟨ ⟩ = − where ⟨ ⟩x and ⟨ ⟩p are the mean values of the

coordinate and momentum of the particle, respectively, and − dVdx

is the mean value of the force

acting on the particle. This result can be generalized to other kinds of operators and is called Ehrenfest’s theorem.

SOLUTION

(a) Suppose that the wavefunction Ψ(x, t) refers to a particle. The Schrödinger equation is

∂

∂ = ∂∂

−Ψ Ψ Ψ( , ) ( , )( ) ( , )

x tt

im

x t

x

iV x x t

��2

2

2 (3.5.1)

and its conjugate equation is ∂

∂ = − ∂∂

+Ψ Ψ Ψ* *

*( , ) ( , )( ) ( , ).

x tt

im

x t

x

iV x x t

��2

2

2 [Notice that we assume

V(x) to be real.] The integral Ψ( , )x t dx2

−∞

∞

∫ must be finite; so we get

lim ( , ) lim ( , ) lim(

x x xx t x t

x→∞ →−∞ →∞

= = ∂Ψ Ψ Ψ2 2 0 and,, )

lim( , )t

xx txx∂ = ∂

∂ =→−∞

Ψ0 (3.5.2)

Hence, the time derivative of ⟨ ⟩x is

d xdt

ddt

x t x x t dxx tt

⟨ ⟩ = = ∂∂

−∞

∞

−∞∫ Ψ Ψ Ψ**

( , ) ( , )( , )

∞∞

−∞

∞

∫ ∫+ ∂∂x x t dx x t x

x tt

dxΨ Ψ Ψ( , ) ( , )

( , )* (3.5.3)

Substituting the Schrödinger equation and its conjugate gives

d xdt

im

x t

xx x t dx

i⟨ ⟩ = − ∂∂

+−∞

∞

∫��2

2

2Ψ Ψ Ψ

**( , )

( , ) (xx t V x x t dx

im

x t xx t

, ) ( ) ( , )

( , )( , )*

Ψ

Ψ Ψ

−∞

∞

∫+ ∂

∂�

2

2

xxdx

ix t V x x t dx2

−∞

∞

−∞

∞

∫ ∫⎡

⎣⎢⎢

⎤

⎦⎥⎥

−�

Ψ Ψ*( , ) ( ) ( , )

== − ∂∂

−→∞ −∫

im

x t

xx x t dx x

�2

2

2lim( , )

( , ) (*

*

ξ ξ

ξ Ψ Ψ Ψ ,, )( , )

t xx t

xdx

∂∂

⎡

⎣⎢⎢

⎤

⎦⎥⎥−∫

2

2Ψ

ξ

ξ

(3.5.4)

Fig. 3.3

I II

0

V(x)

V0

x


Integration by parts gives

d xdt

im

x tx

x x t⟨ ⟩ = − ∂

∂⎡

⎣⎢

⎤

⎦⎥

→∞−

�2 lim

( , ), )

*

ξξ

Ψ Ψ(

ξξ

ξ

ξ

− ∂∂

∂∂

⎧⎨⎪

⎩⎪

−

−∫Ψ Ψ

Ψ

*

*

( , )[ )]

x tx x

x x t dx

x t

( ,

( , )))

[ , ]*xx tx x

x t xx∂

∂⎡⎣⎢

⎤⎦⎥

+ ∂∂

∂

− −∫Ψ Ψ Ψ( ,

( )(

ξ

ξ

ξ

ξ,, )t

xdx∂

⎫⎬⎪

⎭⎪ (3.5.5)

Using Eq. (3.5.2), the first and third terms equal zero; so we have

d xdt

im

x tx

x t dx⟨ ⟩ = − − ∂

∂ −→∞ −∫

�2 lim

( , ))]

*

ξ ξ

ξ Ψ Ψ ( ,∂∂

∂∂

∂⎡

⎣⎢⎢

+ ∂∂

−

−

∫ Ψ Ψ

Ψ

*

*

( , ) ( , )

( , )

x tx

xx tx

dx

x tx

ξ

ξ

ξξ

ξ

ξ

ξ

∫ ∫∂∂ + ∂

∂⎤

⎦⎥⎥−

xx tx

dx x tx tx

dxΨ Ψ Ψ( ,

( , )) ( , )* (3.5.6)

Eventually, integration by parts of the first term gives

d xdt

im x t x t

⟨ ⟩ = − − ⎡⎣ ⎤⎦→∞ −−∫�

2 lim )*ξ ξ

ξ

ξ

ξΨ Ψ( , ( , ) ++ ∂

∂⎡

⎣⎢⎢

⎤

⎦⎥⎥

=

−∫2

1

Ψ Ψ

Ψ

*

*

( , )( , )

( , )

x tx tx

dx

mx t

ξ

ξ

��i

x tx

dxm

p∂

∂ = ⟨ ⟩−∞

∞

∫ Ψ( , ) 1 (3.5.7)

(b) Consider the time derivative of ⟨ ⟩p :

d pdt

ddt

x ti

x tx

dx

ix

⟨ ⟩ = ∂∂

= ∂

−∞

∞

∫ Ψ Ψ

Ψ

*

*

( , )( , )�

� ( , ttt

x tx

dxi

x tt

x t) ))*

∂∂

∂ + ∂∂

∂

−∞

∞

−∞

∞

∫ ∫Ψ Ψ Ψ( ,( ,

( ,� ))∂x

dx (3.5.8)

Since Ψ( , )x t has smooth derivatives, we can interchange the time and spatial derivatives in the second term. Using the Schrödinger equation, Eq. (3.3.8) becomes

d pdt m

x t

x

x tx

dx V x⟨ ⟩ = − ∂

∂∂

∂ +−∞

∞

∫�2 2

22Ψ Ψ*( , ) ( , )

( )) ( , )( , )

)

*

*

Ψ Ψ

Ψ Ψ

x tx tx

dx

mx t

∂∂

+ ∂

−∞

∞

−∞

∞

∫

∫�2 3

2 ( ,(( ,

( ,x t

xdx x t

xV x x t dx

)) [ ( ) ( , )]*

∂− ∂

∂−∞

∞

∫3 Ψ Ψ (3.5.9)

Integration by parts of the first term gives

Ix t

x

x tx

dxx t≡ ∂

∂∂

∂ = ∂

−∞

∞

→∞∫2

2Ψ Ψ Ψ* *( , ) ( , )

lim( ,

ξ

)) ( , ) ( , ) ( , )*

∂∂

∂⎡

⎣⎢

⎤

⎦⎥ − ∂

∂∂

∂−

xx tx

x tx

x t

x

Ψ Ψ Ψ

ξ

ξ 2

22−∫

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪ξ

ξ

dx (3.5.10)


Using Eq. (3.5.2), we arrive at

Ix t

xx t

xdx= − ∂

∂∂

∂

⎡

⎣⎢⎢

⎤

⎦⎥

→∞ −∫lim( , ) ( , )*

ξ ξ

ξ Ψ Ψ2

2 ⎥⎥ (3.5.11)

Again, integration by parts gives

I x tx t

xx t= − ∂

∂⎡

⎣⎢

⎤

⎦⎥ +

→∞−

lim ( , )( , )

( ,* *

ξξ

ξ

Ψ Ψ Ψ2

2 ))( , )

( , )( ,*

∂∂

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪

= ∂

−∫3

3

3

Ψ

Ψ Ψ

x t

xdx

x tx

ξ

ξ

tt

xdx

)

∂−∞

∞

∫ 3 (3.5.12)

Returning to Eq. (3.5.9), we finally have

d pdp

V x x tx tx

dx x tdV⟨ ⟩ = ∂

∂ −−∞

∞

∫( ) ( , )( , )

( , )* *Ψ Ψ Ψ (( )( , )

( , ) ( )( , )*

xdx

x t dx

x t V xx tx

Ψ

Ψ Ψ

−∞

∞

−∞

∞

∫

∫− ∂∂ ddx

dVdx

= − (3.5.13)

3.6. Consider a particle described by a wavefunction Ψ( ).r, t Calculate the time-derivative ∂

∂ρ( , )

,r tt

where

ρ( , )r t is the probability density, and show that the continuity equation ∂

∂ + ∇ =ρ( , )( , )

rJ r

tt

ti 0 is

valid, where J(r, t) is the probability current, equal to 1m

Re Ψ Ψ* .�i

∇⎛⎝⎜

⎞⎠⎟

⎡⎣⎢

⎤⎦⎥

SOLUTION

Using the Schrödinger equation,

it

t mt V t t�

�∂∂ = − ∇ +Ψ Ψ Ψ( , )

( ) ( ) ( )r

r r r2

2

2, , , (3.6.1)

Assuming V(x) is real, the conjugate expression is − ∂∂ = − ∇ +i

tt m

t V t t��Ψ Ψ Ψ

** *( , )( ) ( ) ( ).

rr r r

22

2 , , ,

According to the definition of ρ ρ( ) , ( ) ( ) (*r r r, that is, Eq. (3.23) , ,t t t= Ψ Ψ rr, t); hence,

∂

∂ = ∂∂ + ∂

∂ρ( , ) ( , )

( , ) ( , )( , )*

*r rr r

rtt

tt

t tt

tΨ Ψ Ψ Ψ

(3.6.2)

Using Eq. (3.6.1) and its conjugate, we arrive at

∂∂ = ∇⎡

⎣⎢⎤⎦⎥

−ρ( , )( ) ( ) (*rr r r

tt mi

t ti

V�

�212 Ψ Ψ, , , tt t t t

mit) ( ) ( ) ( ) ( , )* *Ψ Ψ Ψ Ψr r r r, , ,− ∇⎡

⎣⎢⎤⎦⎥

+

�2

2

112

2

it V t t

mit

��Ψ Ψ Ψ Ψ* *( , ) ( , ) ( , ) [ ( , ) ( ,r r r r r= − ∇ tt t t) ( , ) ( , )]*− ∇Ψ Ψr r2 (3.6.3)

We set

J r r( , ) Re [ ( , )* *tm i mi

t= ∇⎛⎝⎜

⎞⎠⎟

⎡⎣⎢

⎤⎦⎥

= ∇12Y Y Y� � YY Y Y( , ) ( , ) ( , )]*r r rt t t− ∇ (3.6.4)


Using the theorem ∇ = ∇ + ∇i i i( ) ( ) ( ),U U A UA A we have

∇ = ∇ ∇ + ∇ − ∇ ∇ −i i iJ r( , ) ( ) ( ) ( ) ( ) ( )* * *t

mi�

22Y Y Y Y Y Y ψψ ( )*

* *

∇⎡⎣ ⎤⎦

= ∇ − ∇⎡⎣ ⎤⎦

2

2 22

Y

Y Y Y Y�mi (3.6.5)

so

∂

∂ + ∇ =ρ( , )( , )

rJ r

tt

ti 0 (3.6.6)

3.7. Consider the wavefunction

Y( , ) [ ]/ / /x t Ae Be eipx ipx ip t m= + − −� � �2 2 (3.7.1)

Find the probability current corresponding to this wavefunction.

SOLUTION

The probability current is by definition

J x tmi x x

( , ) **

= ∂∂ − ∂

∂⎛

⎝⎜⎞

⎠⎟�

2 Y Y Y Y (3.7.2)

The complex conjugate of Y is Y* * / * / /( , ) ( ) ;x t A e B e eipx ipx ip t m= +− � � �2 2 so a direct calculation yields

J x tmi

A e B eip

Aeiipx ipx ipx( , ) ( )* / * / /= + −−�

��

2pp

Be

ipA e

ipB e

ipx

ipx ip

�

� �

�

�

−

−

⎛⎝⎜

⎞⎠⎟

⎡⎣⎢

− − +

/

* / * xx ipx ipxAe Be

pm

A A

/ / /

*

( )

(

� � �⎛⎝⎜

⎞⎠⎟ + ⎤

⎦⎥

= −

−

22⏐ ⏐ BBe AB e B A A Beipx ipx− −+ −⎡⎣ − − −2 2 2 2 2/ * / *) (� � ⏐ ⏐ ⏐ ⏐ iipx ipxAB e B

pm

A B

/ * / )

( )

� �+ + ⎤⎦

= −

2 2

2 2

⏐ ⏐

⏐ ⏐ ⏐ ⏐ (3.7.3)

Note that the wavefunction Y( , )x t expresses a superposition of two currents of particles moving in oppo-

site directions. Each of the currents is constant and time-independent in its magnitude. The term e ip t m− 2 2/ � implies that the particles are of energy p m2 2/ . The amplitudes of the currents are A and B.

3.8. Show that for a one-dimensional square-integrable wave packet,

J x dxpm

( )−∞

∞

∫ = ⟨ ⟩ (3.8.1)

where J(x) is the probability current.

SOLUTION

Consider the integral Y( , ) .x t dx2

−∞

∞

∫ This integral is finite, so we have lim ( , ) .x

x t→±∞

=Y 2 0 Hence,

J x dxim

x tx tx

x tx t

( ) ( , )( , )

( , )( , )*

*= ∂

∂ − ∂∂

�2 Y Y Y Y

xxdx

⎡

⎣⎢

⎤

⎦⎥

−∞

∞

−∞

∞

∫∫ (3.8.2)



Y Y Y Y( , )( , )

lim ( , ) ( , )*

*x tx tx

dx x t x t∂

∂ = ⎡−∞

∞

→∞∫ ξ ⎣⎣ ⎤⎦ − ∂∂

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪

= −

−−∫ξ

ξ

ξ

ξ Y Y

Y

( , )( , )*x t

xx t dx

**( , )( , )

x tx tx

dx∂

∂−∞

∞

∫ Y (3.8.3)

Therefore, we have

J x dxm

x ti x

x t dxpm

( ) ( , ) ( , )*= ∂∂ = ⟨ ⟩

−∞

∞

−∞

∞

∫ ∫1 Ψ Ψ� (3.8.4)

3.9. Consider a particle of mass m held in a one-dimensional potential V(x). Suppose that in some region V(x) is constant, V(x) = V. For this region, find the stationary states of the particle when (a) E > V, (b) E < V, and (c) E = V, where E is the energy of the particle.

SOLUTION

(a) The stationary states are the solutions of

− ∂∂

+ =�2 2

22mx

xV x E x

ψ ψ ψ( )( ) ( ) (3.9.1)

For E > V, we introduce the positive constant k defined by �2 2 2k m E V/ = − , so that

∂

∂+ =

2

22 0

ψ ψ( )( )

x

xk x (3.9.2)

The solution of this equation can be written in the form

ψ ( )x Ae A eikx ikx= + ′ − (3.9.3)

where A and A′ are arbitrary complex constants.

(b) We introduce the positive constant r defined by �2 2 2ρ / m V E= − ; so Eq. (3.9.1) can be written as

∂

∂− =

2

22 0

ψ ρ ψ( )( )

x

xx (3.9.4)

The general solution of Eq. (3.9.4) is ψ ρ ρ( )x Be B ex x= + ′ − where B and B′ are arbitrary complex constants.

(c) When E = V we have ∂

∂=

2

2 0ψ ( )

;x

x so ψ ( )x is a linear function of x, ψ ( )x Cx C= + ′ where C and C′ are

complex constants.

3.10. Consider a particle of mass m confined in an infinite one-dimensional potential well of width a:

V xa

xa

( ) = − ≤ ≤∞

⎧⎨⎪

⎩⎪

02 2

otherwise (3.10.1)

Find the eigenstates of the Hamiltonian (i.e., the stationary states) and the corresponding eigenenergies.

SOLUTION

For x > a/2 and x < −a/2 the potential is infinite, so there is no possibility of finding the particle outside the well. This means that

ψ ψxa

xa>⎛

⎝⎜⎞⎠⎟ = <⎛

⎝⎜⎞⎠⎟ =

20

20 (3.10.2)


Since the wavefunction must be continuous, we also have ψ ψ( ) ( ) .a a/ /2 2 0= − = For − ≤ ≤a x a/ /2 2 the potential energy is constant, V(x) = 0; therefore, we can rely on the results of Problem 3.9. We distinguish between three possibilities concerning the energy E. As in Problem 3.9, part (a), for E > 0 we define the positive constant k, �2 2 2k m E/ = ; so we obtain ψ ( ) .x Ae A eikx ikx= + ′ − Imposing the continuous conditions, we arrive at

I IIAe A e Ae A eika ika ika ika/ / / /2 2 2 20 0+ ′ = + ′ =− − (3.10.3)

Multiplying Eq. (3.10.3I) by eika / 2 we obtain ′ = −A Aeika. Substituting A′ into Eq. (3.10.3II) yields

Ae Ae eika ika ika− − =/ /2 2 0 (3.10.4)

Multiplying Eq. (3.10.4) by e ika− / 2 and dividing by A [if A = 0 then y (x) ≡ 0] we obtain e eika ika− − = 0. Using the relation e iiα α α= +cos sin we have −2i sin (ka) = 0. The last relation is valid only if ka = np, where n is an integer. Also, since k must be positive, n must also be positive. We see that the possible posi-tive eigenenergies of the particle are

Ekm m

na

n

mann= = ⎛

⎝⎜⎞⎠⎟ =

� � �2 2 2 2 2 2 2

22 2 2

π π (3.10.5)

The corresponding eigenfunctions are

ψ π πn

ik x ik a ik x in x a in a xx Ae Ae e Ae en n n( ) / (= − = −− − )) /

/ ( / / ) ( / /

a

in in x a in x aAe e e

C

= −[ ]

=

− − −π π π2 1 2 1 2

ssin ( , , )nxa

nπ −⎛⎝⎜

⎞⎠⎟

⎡⎣⎢

⎤⎦⎥

=12

1 2 … (3.10.6)

where C is a normalization constant obtained by

1 1

222

2

2

Cn

xa

dxa

a

= −⎛⎝⎜

⎞⎠⎟

⎡⎣⎢

⎤⎦⎥−∫ sin

/

/

π (3.10.7)

Defining yxa

= − 12

and dydxa

= , Eq. (3.8.7) becomes

1

21 22

2

1

0

1

0

Ca n y dy

any dy= = − =

− −∫ ∫sin ( ) [ cos( )]π π aay

nyn

a2

22 2

1

0

−⎡⎣⎢

⎤⎦⎥

=−

sin( )ππ (3.10.8)

Therefore, C a= 2/ . Finally,

ψ πn xa

nxa

( ) sin= −⎛⎝⎜

⎞⎠⎟

⎡⎣⎢

⎤⎦⎥

2 12

(3.10.9)

Consider now the case when E < 0. As in Problem 3.9, part (b), we introduce the positive constant r, �

2 2 2ρ / m E= − . Stationary states should be of the form ψ ρ ρ( ) .x Be B ex x= + ′ − Imposing the boundary con-ditions, we obtain

I IIBe B e Be B ea a a aρ ρ ρ ρ/ / / /2 2 2 20 0+ ′ = + ′ =− − (3.10.10)

Multiplying, Eq. (3.10.10I) by e aρ / 2 yields ′ = −B Be aρ , so Be Be ea a a− − =ρ ρ ρ/ / .2 2 0 Multiplying by e aρ / 2 and dividing by B, we obtain 1 02− =e aρ . Therefore, 2 0ρa = . Since r must be positive, there are no states with corresponding negative energy.


Finally, we consider the case when E = 0. According to Problem 3.9, part (c), we have ψ ( ) .x Cx C= + ′ Imposing the boundary conditions yields

Ca

C Ca

C2

02

0+ ′ = − + ′ = (3.10.11)

Solving these equations yields C C= ′ = 0; the conclusion is that there is no possible state with E = 0.

3.11. Refer to Problem 3.10. At t = 0 the particle is in a state described by a linear combination of the two lowest stationary states:

ψ αψ βψ α β( , ) ( ) ( )x x x0 11 22 2= + + =( )⏐ ⏐ ⏐ ⏐ (3.11.1)

(a) Calculate the wavefunction Ψ( , )x t and the mean value of x and px as a function of time.

(b) Verify the Ehrenfest theorem, md x dt px⟨ ⟩ = ⟨ ⟩/ .

SOLUTION

(a) Consider part (c) of Problem 3.3. The time-evolution of the stationary states is of the form

Ψn n nx t x iE t( , ) ( ) exp ( )= −ψ /� (3.11.2)

Consequently, using Eqs. (3.10.9) and (3.11.2) the superposition principle gives

Ψ Ψ Ψ( , ) ( , ) ( , )

sin

x t x t x t

axa

= +

= −⎛⎝⎜

⎞⎠⎟

α β

α π

1 2

2 12

⎡⎡⎣⎢

⎤⎦⎥

−⎛

⎝⎜⎞

⎠⎟⎡

⎣⎢⎢

⎤

⎦⎥⎥

+exp sinπ β π

2

22

22

i t

ma a� xx

ai t

ma−⎛

⎝⎜⎞⎠⎟

⎡⎣⎢

⎤⎦⎥

−⎛

⎝⎜⎞

⎠⎟⎡

⎣⎢⎢

⎤12

4

2

2

2expπ �

⎦⎦⎥⎥

(3.11.3)

We now calculate

⟨ ⟩ = = +−∫x x t x x t dx x t

a

a

Ψ Ψ Ψ Ψ*

/

/* * *( , ) ( , ) ( , )

2

2

1α β 22 1 22

2*

/

/

( , ) ( , ) ( , )x t x x t x t dxa

a

⎡⎣ ⎤⎦ +⎡⎣ ⎤⎦−

α βΨ Ψ∫∫∫= + +

− −α β2

12 2

2

2

22

2x x t dx x x t dxa

a

Ψ Ψ( , ) ( , ) Re/

/

aa

a

a

a

x x t x t dx/

/* *

/

/

( , ) ( , )2

2

1 22

2

∫ ∫−

⎡

⎣⎢⎢

⎤

⎦α β Ψ Ψ ⎥⎥

⎥

(3.11.4)

Consider each of the three elements separately:

I x x t dxa

xxa

a

a

1 12

2

22

212≡ = −⎛

⎝⎜⎞⎠⎟

⎡

−∫ Ψ ( , ) sin/

/

π⎣⎣⎢

⎤⎦⎥−∫ dx

a

a

/

/

2

2

(3.11.5)

Defining yxa

dydxa

= − =12

, , so

I a y y dy a y y dy12

1

02

1

0

2 1 2= + = +− −∫ ∫( ) sin ( ) sin ( )π π aa y dysin ( )2

1

0

π−∫ (3.11.6)

Solving these integrals yields

I ay y y y

1

2

21

0

24

24

2

8= − −

⎡

⎣⎢⎢

⎤

⎦⎥⎥ −

sin( ) cos( )ππ

ππ

++ −⎡⎣⎢

⎤⎦⎥

= − + =−

ay y a a2

24 2 2

01

0sin( )ππ (3.11.7)


One can repeat this procedure to show that

I x x t dxa

xxa

a

a

a

a

2 22

2

22

2

22

2≡ = −− −∫ ∫Ψ ( , ) sin

/

/

/

/

π 112

0⎛⎝⎜

⎞⎠⎟

⎡⎣⎢

⎤⎦⎥

=dx (3.11.8)

Note that this result can arise from different considerations. The function f xxa

( ) sin= −⎛⎝⎜

⎞⎠⎟

⎡⎣⎢

⎤⎦⎥

2 212

π is an even function of x:

f xxa

xa

( ) sin sin− = − −⎛⎝⎜

⎞⎠⎟

⎡⎣⎢

⎤⎦⎥

= − +⎛212

212

2

π π ⎝⎝⎜⎞⎠⎟

⎡⎣⎢

⎤⎦⎥

= − +⎛⎝⎜

⎞⎠⎟ +⎡

⎣⎢⎤⎦⎥

⎧⎨⎩

2

212

2sin π πxa

⎫⎫⎬⎭

= −⎛⎝⎜

⎞⎠⎟

⎡⎣⎢

⎤⎦⎥

=

2

2

212

sin ( )π xa

f x (3.11.9)

On the other hand, f (x) = x is an odd function of x; therefore, x x asin [ ( )]2 2 1 2π / /− is an even function of x, and its integral vanishes from −a/2 to a/2. Consider now the last term in Eq. (3.11.4):

I x x t x t dx

ax

a

a

a

a

3 1 22

2

2

22

≡

=

−

−

∫ Ψ Ψ*

/

/

/

/

( , ) ( , )

sin∫∫ −⎛⎝⎜

⎞⎠⎟

⎡⎣⎢

⎤⎦⎥

−⎛⎝⎜

⎞⎠⎟

⎡⎣⎢

⎤⎦⎥

π πxa

xa

12 2

12sin exxp −

⎛

⎝⎜⎞

⎠⎟3

2

2

2π i t

madx

� (3.11.10)

Defining y x d dy dx a= − =/ / /1 2, , and ω π= 3 22 2�/ ma , we obtain

I ae y y y dy aei t i t3

1

0

2 1 2= + =− −

−∫ω ωπ π( ) sin ( ) sin ( ) (( ) [cos ( ) cos ( )]2 112

3

16

9

1

0

2

y y y dy

ae i

+ −

=

−

−

∫ π π

πωωt (3.11.11)

Finally, returning to Eq. (3.11.4) we obtain

⟨ ⟩ = =−xa

ea

ti t16

92

32

92 2πα β

πα β ωωRe ( ) Re ( ) cos(* * )) Re ( )sin ( )*+⎡⎣ ⎤⎦i tα β ω (3.11.12)

Consider the mean value of the momentum:

⟨ ⟩ = ∂∂

= +

−∫pi x

x t dx

tx t

xa

a

Ψ Ψ

Ψ

*

/

/

* *

( , )

[ ( , )

2

2

1

�

� α ββ α β* *

/

/

( , )]( , ) ( , )

ΨΨ Ψ

22

21 2x t

x tx

x tx

a

a

−∫∂

∂ +∂

∂⎡⎡⎣⎢

⎤⎦⎥ dx (3.11.13)

We calculate separately each of the four terms in Eq. (3.11.13):

ΨΨ

12

21

2

22*

/

/

/

/( , )sin

− −∫ ∫∂∂ = −

a

a

a

ax tx

dxa a

xa

π π 112

12

⎛⎝⎜

⎞⎠⎟

⎡⎣⎢

⎤⎦⎥

−⎛⎝⎜

⎞⎠⎟

⎡⎣⎢

⎤⎦⎥

cos π xa

dx (3.11.14)


sin π xa

−⎛⎝⎜

⎞⎠⎟

⎡⎣⎢

⎤⎦⎥

12 is an even function of x and cos π x

a−⎛

⎝⎜⎞⎠⎟

⎡⎣⎢

⎤⎦⎥

12

is an odd function, so their product is an

odd function; therefore, the integral of the product between x a= − /2 and x a= /2 equals zero. Also,

ΨΨ

22

22

2

2 2*

/

/

/

/

( , )( , )

sin− −∫ ∂

∂ =a

a

a

a

x tx t

xdx

a aπ 22

212 2

12∫ −⎛

⎝⎜⎞⎠⎟

⎡⎣⎢

⎤⎦⎥

−⎛⎝⎜

⎞⎠⎟

⎡⎣⎢

⎤⎦

π πxa

xa

cos ⎥⎥ dx (3.11.15)

sin 212

π xa

−⎛⎝⎜

⎞⎠⎟

⎡⎣⎢

⎤⎦⎥ is an odd function of x and cos 2

12

π xa

−⎛⎝⎜

⎞⎠⎟

⎡⎣⎢

⎤⎦⎥ is an even once; therefore, their product

is an odd function, thus the integral between x a= − /2 and a = 2 vanishes. We have

I x tx t

xdx

aa

a

a

a

≡∂

∂ =− −∫ Ψ

Ψ1

2

22

22

4*

/

/

/

( , )( , )

sinπ //

cos2

12 2

12∫ −⎛

⎝⎜⎞⎠⎟

⎡⎣⎢

⎤⎦⎥

−⎛⎝⎜

⎞⎠⎟

⎡⎣⎢

⎤⎦

π πxa

xa ⎥⎥

−e dxi tω (3.11.16)

Defining yxa

= − 12

and dydxa

= , the integral I becomes

Ia

e y y dya

yi t= =−

−∫4

24

21

0π π π π πω sin ( ) cos ( )cos ( )

223

683

1

0

ππ

πω ω−⎡

⎣⎢⎤⎦⎥

=−

− −cos ( )ye

aei t i t

(3.11.17)

Finally,

Γ ≡∂

∂ =−− ∫Ψ

Ψ2

1

2

2

2

2*

/

/

/

/

( , )( , )

sinx tx t

xdx

aa

a

a

a π22

212

12∫ −⎛

⎝⎜⎞⎠⎟

⎡⎣⎢

⎤⎦⎥

−⎛⎝⎜

⎞⎠⎟

⎡⎣⎢

⎤⎦⎥

π πxa

xa

cos ee dxi tω (3.11.18)

Using the same definitions used above, we arrive at

Γ = = −−∫

22

2

1

0π π π πω ωa

e y y dya

ei t i tsin ( ) cos ( )cos (ππ

ππ

πωy y

aei t) cos ( )

23

683

1

0

−⎡⎣⎢

⎤⎦⎥

= −−

(3.11.19)

Substituting the results in equation Eq. (3.11.13), we finally reach

⟨ ⟩ = −⎡⎣ ⎤⎦−p

iae ex

i t i t83

� α β αβω ω* * (3.11.20)

(b) From Eq. (3.11.12)

⟨ ⟩ = −⎛

⎝⎜⎞

⎠⎟+x t

a i

mat( ) exp exp* *16

9

3

22

2

2πα β π α β� 33

2

2

2i

mat

π �⎛

⎝⎜⎞

⎠⎟⎡

⎣⎢⎢

⎤

⎦⎥⎥

(3.11.21)

Therefore, we have

md xdt

ma i

ma

i

mat

⟨ ⟩ = − −⎛16

9

3

2

3

22

2

2

2

2ππ α β π� �* exp

⎝⎝⎜⎞

⎠⎟+

⎛

⎝⎜⎞

⎠⎟⎡

⎣⎢⎢

⎤

⎦⎥⎥

=

α β π* exp3

2

83

2

2i

mat

ia

�

�αα β α βω ω* *e ei t i t− −⎡⎣ ⎤⎦ (3.11.22)

By inspection, the last expression is identical to ⟨ ⟩px . Thus, for this particular case Ehrenfest’s theorem is verified.


3.12. Refer again to Problem 3.10. Now suppose that the potential energy well is located between x = 0 and x = a:

V xx a

( ) =≤ ≤

∞{0 0otherwise

(3.12.1)

Find the stationary eigenstates and the corresponding eigenenergies.

SOLUTION

We begin by performing a formal shift of the potential energy well, �x x a= − /2, so the problem becomes identical to Problem 3.10:

V xa x a

( )��

= − ≤ ≤∞

⎧⎨⎩

0 2 2/ /otherwise

(3.12.2)

Using the solution of Problem 3.10, namely Eq. (3.10.5), the possible energies are

En

man = π 2 2 2

22

� (3.12.3)

where n is a positive integer. The corresponding eigenstates are given by Eq. (3.10.9)

ψ πn xa

nxa

( ) sin��

= −⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢

⎤

⎦⎥

2 12 (3.12.4)

Or, in terms of the original coordinate, we have

ψ π πn xa

n xa

n( ) sin= −⎛⎝⎜

⎞⎠⎟

2 (3.12.5)

3.13. Consider the step potential (Fig. 3.4):

V xV x

x( ) =

><

⎧⎨⎩

0 0

0 0 (3.13.1)

Consider a current of particles of energy E > V0 moving from x = −∞ to the right. (a) Write the stationary solutions for each of the regions. (b) Express the fact that there is no current coming back from x = +∞ to the left. (c) Use the matching conditions to express the reflected and transmitted amplitudes in terms of the incident amplitude. Note that since the potential is bounded, it can be shown that the derivative of the wavefunction is continuous for all x.

SOLUTION

(a) Referring to Problem 3.9 part (a), we define

kmE

km E V

1 2 2 22 2= = −� �

( ) (3.13.2)

Then the general solutions for the regions I (x < 0) and II (x > 0) are

ψ ψI II( ) ( )x A e A e x A e A eik x ik x ik x= + ′ = + ′− −

1 1 2 21 1 2 iik x2 (3.13.3)

Fig. 3.4

I II

0

V(x)

x

V0


(b) The wavefunction of form eikx represents particles coming from x = −∞ to the right, and e ikx− represents particles moving from x = +∞ to the left. ψ1( )x is the superposition of two waves. The first one is of incident particles propagating from left to right and is of amplitude A1; the second wave is of amplitude

′A1 and represents reflected particles moving from right to left. Since we consider incident particles com-ing from x = −∞ to the right, it is not possible to find in II a current that moves from x = + ∞ to the left. Therefore, we set ′ =A2 0. Thus, ψ II ( )x represents the current of transmitted particles with corresponding amplitude A2.

(c) First, we apply the continuity condition of ψ ( )x at x = 0, ψ ψI II( ) ( ).0 0= Substituting in Eq. (3.13.3) gives

A A A1 1 2+ ′ = (3.13.4)

Secondly, ∂

∂ψ ( )x

x should also be continuous at x = 0; we have

∂

∂ = − ′∂

∂ =ψ ψI II( ) ( )x

xik A e ik A e

xx

ikik x ik x

1 1 1 11 1

22 22A e

ik x (3.13.5)

Applying, ∂

∂ =∂

∂ψ ψI II( ) ( )

,0 0

x x we obtain

ik A A ik A1 1 1 2 2( )− ′ = (3.13.6)

Substituting A2 gives A A A A k k1 1 1 1 1 2+ ′ = − ′( ) ,/ which yields

′

=−+

AA

k kk k

1

1

1 2

1 2 (3.13.7)

Eventually, substituting Eq. (3.13.7) in Eq. (3.13.4) yields Ak kk k

A11 2

1 221 +

−+

⎛⎝⎜

⎞⎠⎟

= ; therefore,

AA

kk k

2

1

1

1 2

2= + (3.13.8)

3.14. Refer to Problem 3.13. (a) Compute the probability current in the regions I and II and interpret each term. (b) Find the reflection and transmission coefficients.

SOLUTION

(a) For a stationary state ψ ( ),x the probability current is time-independent and, as in Eq. (3.7.2), is equal to

J xmi

xx

xx

xx

( ) ( )( )

( )( )*

*= ∂

∂ − ∂∂

⎡

⎣⎢

⎤

⎦⎥

�2 ψ ψ ψ ψ (3.14.1)

Using Eq. (3.13.3) for region I, we have

J xmi A e A e ik A e i

ik x ik x ik xI ( ) * *= + ′( ) −−�

2 1 1 1 11 1 1 kk A e A e A e ik A e

ik x ik x ik x1 1 1 1 1 1

1 1 1′( ) − + ′( ) −− − −* iik x ik xik A e

km

A A

1 1

1 1

11

21

2

+ ′( )⎡⎣

⎤⎦

= − ′( )

*

�

(3.14.2)

Similarly, for region II we have

J xmi A e ik e A e ik

ik x ik x ik xII ( ) ( ) (*= − −−�

2 2 2 2 22 1 2 ))e

km

Aik x−⎡⎣ ⎤⎦ =2 22

2� (3.14.3)

The probability current in region I is the sum of two terms: �k A m1 12/ corresponds to the incoming cur-

rent moving from left to right, and − ′�k A m1 12/ corresponds to the reflected current (moving from right

to left). Note that the probability current in region II represents the transmitted wave.

(b) Using the definition of the reflection coefficient [refer to Eq. (3.27)], we have

RA k m

A k m

AA

=′

=′1

21

12

1

1

1

2�

�

/

/ (3.14.4)


Substituting Eq. (3.13.7), we arrive at

Rk k

k k

k k

k k=

−+

= −+

( )

( ) ( )1 2

2

1 22

1 2

1 221

4 (3.14.5)

The transmission coefficient is

TA k m

A k m

kk

AA

= =22

2

12

1

2

1

2

1

2�

�

/

/ (3.14.6)

Substituting Eq. (3.13.8), we arrive at

Tkk

kk k

k k

k k= +

⎛⎝⎜

⎞⎠⎟

=+

2

1

1

1 2

21 2

1 22

2 4

( ) (3.14.7)

3.15. Consider a free particle of mass m whose wavefunction at time t = 0 is given by

Ψ( , )( ) /

( )x

ae e dk

a k k ikx02 3 4

420

2

= − −

−∞

∞

∫π/ (3.15.1)

Calculate the time evolution of the wave packet Ψ( , )x t and the probability density Ψ( , ) .x t 2 Sketch qualitatively the probability density for t < 0, t = 0, and t > 0. You may use the following identity: For complex numbers a and b such that − < <π α π/ /4 4arg ( ) ,

e dyy− +

−∞

∞

=∫ α β πα

2 2( ) (3.15.2)

SOLUTION

The wave packet at t = 0 is a superposition of plane waves eikx with coefficients a

ea k k

( );/

( ) /

2 3 442

02

π− −

this is a

Gaussian curve centered at k = k0. The time evolution of a plane wave eikx has the form e e e eikx iE k t ikx i k t m− −=( ) / / .� �2 2

We set ω( ) .k k m= �2 2/ Using the superposition principle, the time evolution of the wave packet Ψ( , )x 0 is

Ψ( , )( ) /

( ) / [ ( ) ]x ta

e e dka k k i kx k t= − − −

−2 3 442

02

πω

∞∞

∞

∫ (3.15.3)

Our aim is to transform this integral into the form of Eq. (3.15.2). Therefore, we rearrange the terms in the exponent:

− − + − = − +⎛⎝⎜

⎞⎠⎟ +a

k k i kx k ta i t

mk

2

02

22

4 4 2( ) [ ( ) ]ω � aa

k ix ka

k

a i tm

k

a

2

0

2

02

2

2 4

4 2

+⎛⎝⎜

⎞⎠⎟ −

= − +⎛⎝⎜

⎞⎠⎟ −�

22

0

2

22

02

24 2

2k ix

a i tm

ak+

+⎛⎝⎜

⎞⎠⎟

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

+�

++⎛⎝⎜

⎞⎠⎟

+⎛⎝⎜

⎞⎠⎟

−ix

a i tm

ak

2

2

2

02

44 2

4� (3.15.4)

Substituting in Eq. (3.15.4) and using Eq. (3.15.2) yields

Ψ( , )

exp

ex/ /x ta

a k

a i tm

=

−⎛

⎝⎜

⎞

⎠⎟

+2

4

4 2

3 4 1 4

202

2π �pp

ak ix

ai tm

2

0

2

2

2

2

+⎛

⎝⎜⎞

⎠⎟

+

⎡

⎣

⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥

� (3.15.5)


The conjugate complex of Eq. (3.15.5) is

Ψ*/ /( , )

exp

ex ta

a k

a i tm

=

−⎛

⎝⎜

⎞

⎠⎟

−2

4

4 2

3 4 1 4

202

2π �xxp

ak ix

ai tm

2

0

2

2

2

2

−⎛

⎝⎜⎞

⎠⎟

−

⎡

⎣

⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥

� (3.15.6)

Hence,

� �Ψ( , )

exp

/x ta

a k

a i tm

23 2

202

22

2

4 2

=

−⎛

⎝⎜

⎞

⎠⎟

+⎛

⎝

π �⎜⎜

⎞

⎠⎟−

⎛

⎝⎜⎞

⎠⎟

⎛

⎝⎜

⎞

⎠⎟ − +

a i tm

a kx ia k

2

20

2

2 2

4 2

2

�exp

00

2

20

2

2 20

22

2

2

x

a i t m

a kx ia k x

a i t++

⎛

⎝⎜

⎞

⎠⎟ − −

−� �/ /mm

a t m a

a k

⎡

⎣

⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥

=+

−2 1

1 42 2 2 2 4

2

π � /exp

002

42 2

22

402

2

24

22

4a

t

ma

a kx

k+

⎛

⎝⎜⎞

⎠⎟+ −

⎛

⎝⎜

⎞

⎠⎟ +� � 00

2

4 2 2

2 2

4

2 1

1 4

am

xt

a t m

a t

+

⎡

⎣

⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥

=+

�

�

/

π 22 2 4

20

2

4 2 2 2

2

4/

/

/m a

a x k t m

a t mexp

( )−

−+

⎡

⎣⎢⎢

⎤

⎦⎥

�

� ⎥⎥

(3.15.7)

The probability density is a Gaussian curve for every time t entered at x k m tC = ( ) .� 0 / (That is, the wave packet moves with a speed v = �k m0 / .) The value of Ψ( , )x t 2 is maximal for t = 0 and tends to zero when t → ∞. The width of the wave packet is minimal for t = 0 and tends to ∞ when t → ∞; see Fig. 3.5.

Fig. 3.5

|Ψ(x, t)|2

t = 0

x

t > 0

2pa2

3.16. Consider a square potential barrier (Fig. 3.6):

V xx

V x l

l x

( ) =<< <<

⎧⎨⎪

⎩⎪

0 00

00 (3.16.1)

(a) Assume that incident particles of energy E > V0 are coming from x = −∞. (a) Find the stationary states. Apply the matching conditions at x = 0 and x = l. (b) Find the transmission and reflection coefficients. Sketch the transmission coefficient as a function of the barrier’s width l, and discuss the results.


SOLUTION

(a) Similar to Problem 3.9, part (a), we define

kmE

km E V

1 2 20

22 2

= =−

� �

( ) (3.16.2)

Thus, the stationary solutions for the three regions I (x < 0), II (0 < x < l), and III (x > l) are:

ψ

ψI

II

( )

( )

x A e A e

x A e A e

ik x ik x

ik x

= + ′

= + ′

−

−1 1

2 2

1 1

2 iik x

ik x ik xx A e A e

2

1 1

3 3ψ III ( ) = + ′

⎧

⎨⎪⎪

⎩⎪⎪

−

(3.16.3)

Each of the solutions describes a sum of terms representing movement from left to right, and from right to left. We consider incident particles from x = −∞, so there should be no particles in region III moving from x = ∞ to the left. Therefore, we set ′ =A3 0. The matching conditions at x = l enable us to express A2 and ′A2 in terms of A3. The continuity of ψ ( )x at x l= yields ψ ψII III so( ) ( ),l l=

A e A e A eik l ik l ik l

2 2 32 2 1+ ′ =− (3.16.4)

The continuity of ′ψ ( )x yields

ik A e ik A e ik A eik l ik l ik l

2 2 2 2 1 32 2 1− ′ =−

(3.16.5)

Equations (3.16.4) and (3.16.5) give

A

k kk

e A

Ak k

ke

i k k l

i

22 1

23

22 1

2

2

2

1 2=+⎡

⎣⎢⎤⎦⎥

′ =−

−( )

(( )k k lA1 2

3+⎡

⎣⎢⎤⎦⎥

⎧

⎨⎪⎪

⎩⎪⎪

(3.16.6)

The matching conditions at x = 0 yield

ψ ψI II( ) ( )0 0 1 1 2 2= ⇒ + ′ = + ′A A A A (3.16.7)

and

′ = ′ ⇒ − ′ = − ′ψ ψI II( ) ( )0 0 1 1 1 1 2 2 2 2ik A ik A ik A ik A (3.16.8)

so we obtain

Ak k

kA

k kk

A11 2

12

1 2

122 2

=+

+−

′ (3.16.9)

Fig. 3.6

II

0

IIII

V(x)

xl

V0


Using (3.16.6), we can express A1 in terms of A3:

Ak k

k ke

k kk k

ei k k l i

11 2

2

1 2

1 22

1 24 41 2=

+−

−−( ) ( )( ) (kk k lA

k k k kk k

1 2

3

1 22

1 22

1 24

+⎡

⎣⎢⎢

⎤

⎦⎥⎥

=+ − −

)

( ) ( )coss ( )

( ) ( )sin ( )k l i

k k k kk k

k l21 2

21 2

2

1 224

−+ + −⎡

⎣⎢⎢

⎤⎤

⎦⎥⎥

= −+

e A

k l ik k

k kk l

ik l1

3

212

22

1 222

cos ( ) sin ( )⎡⎡

⎣⎢⎢

⎤

⎦⎥⎥

e Aik l1

3 (3.16.10)

Similarly, we express ′A1 in terms of A3:

′ =−

++

′ =+ −

Ak k

kA

k kk

Ak k k k

k11 2

12

1 2

12

1 2 1 22 2 4

( )( )

11 2

1 2 2 1

2

1 2 1 2

4ke

k k k kk

ei k k l i k k l( ) ( )( )( )− ++

+ −⎡⎣⎣⎢

⎤⎦⎥

=− + −

+

A

k k k kk k

k l i

3

12

22

22

12

1 224

( ) ( )cos ( )

(( ) ( )sin ( )

k k k kk k

k l A22

12

12

22

1 22 34

− − −⎡

⎣⎢⎢

⎤

⎦⎥⎥

==−

ik k

k kk l e A

ik l22

12

1 22 32

1sin ( ) (3.16.11)

(b) The reflection coefficient is the ratio of squares of the amplitudes corresponding to the incident and reflection waves (compare to Problem 3.14):

RAA

=′11

2

(3.16.12)

Using the results of part (a), we obtain

R

k kk k

k l

k lk

=

−⎡

⎣⎢⎢

⎤

⎦⎥⎥

+

22

12

1 22

2

22

1

2sin ( )

cos ( )22

22

1 22

222

12 2 2

2+⎡

⎣⎢⎢

⎤

⎦⎥⎥

=−

kk k

k l

k k

sin ( )

( ) sin (( )

( ) sin ( )

k l

k k k k k l2

12

22

12

22 2 2

24 + − (3.16.13)

Finally, the transmission coefficient is

TAA

k lk k

k k

= =

++⎛

⎝⎜

⎞

⎠⎟

3

1

2

22

12

22

1 2

2

2

1

2cos ( ) sin (kk l

k k

k k k k k l

2

12

22

12

22

12

22 2 2

2

4

4)

( ) sin ( )=

+ − (3.16.14)

The dimensionless transmission coefficient oscillates periodically as a function of l (see Fig. 3.7) between its maximum value (one) and its minimum value [ ( )] .1 40

20

1+ − −V E E V/ When l is an inte-gral multiple of π /k2, there is no reflection from the barrier; this is called resonance scattering (see Chap. 15).


3.17. Consider the square potential barrier of Problem 3.16. Find the stationary states describing incident particles of energy E < V0. Compute the transmission coefficient and discuss the results.

SOLUTION

The method of solution is analogous to that of Problem 3.16. Referring to Problem 3-9, we define

kmE m V E

1 202

2 2= =

−� �

ρ( )

(3.17.1)

The stationary solutions for the three regions I (x < 0), II (0 < x < l), and III (x > l ) are

ψ

ψ ρ ρ

I

II

( )

( )

x A e A e

x A e A e

ik x ik x

x x

= + ′

= + ′

−

−

1 1

2 2

1 1

ψψ III ( )x A e A eik x ik x= + ′

⎧

⎨

⎪⎪

⎩

⎪⎪ −

3 31 1

(3.17.2)

We describe incident particles coming from x = − ∞, so we set ′ =A3 0. Applying the matching conditions in x = l gives

ψ ψ ρ ρII III( ) ( )l l A e A e A el l ik x= ⇒ + ′ =−

2 2 31 (3.17.3)

ψ ψ ρ ρ ρII III( ) ( )l l A e A e ik A el l ik l= ⇒ − ′ =−

2 2 1 31 (3.17.4)

From Eqs. (3.17.3) and (3.17.4) we obtain

Aik

e A Aik

eik l ik

21

3 21

2 21 1=

+⎡⎣⎢

⎤⎦⎥

′ =−−ρ

ρρ

ρρ( ) ( ++⎡

⎣⎢⎤⎦⎥

ρ)lA3 (3.17.5)

The matching conditions at x = 0 yield

ψ ψI II( ) ( )0 0 1 1 2 2= ⇒ + ′ = + ′A A A A (3.17.6)

′ = ′ ⇒ − ′ = − ′ψ ψ ρ ρI II( ) ( )0 0 1 1 1 1 2 2ik A ik A A A (3.17.7)


Aik

ikA

ikik

A11

12

1

122 2

=+

+−

′ρ ρ

(3.17.8)

Fig. 3.7

0

1

T

l

4k12k2

2

4k12k2

2 + (k12 – k2

2)2

p/k2 2p/k2


Using Eq. (3.17.5), we arrive at

Aik

ike

ikik

eik l ik

11

2

1

12

14 4=

+−

−−( ) ( )( ) (ρρ

ρρ

ρI 11

312 2

12+⎡

⎣⎢⎢

⎤

⎦⎥⎥

= −−

+ρ ρρ ρ)

sinh ( ) cosh (l

A ik

kl ρρl e A

ik l)

⎡

⎣⎢⎢

⎤

⎦⎥⎥

1

3 (3.17.9)

Finally, consider the transmission coefficient:

TAA

lk

k

= =

+−⎛

⎝⎜

⎞

⎠⎟

3

1

2

2 12 2

1

2

2

1

2cosh ( ) sinh (ρ

ρρ ρll

kk

l) sinh ( )

=

++⎛

⎝⎜

⎞

⎠⎟

1

12

12 2

1

2

2ρρ ρ

(3.17.10)

where we used the identity cosh2 a − sinh2 a = 1. Hence,

TE V E

E V E Vm V E

l

=−

− +−⎡

⎣⎢⎢

⎤

⎦

4

42

0

0 02 2 0

( )

( ) sinh( )

�⎥⎥⎥

(3.17.11)

We see that in contrast to the classical predictions, particles of energy E < V0 have a nonzero probability of crossing the potential barrier. This phenomenon is called the tunnel effect.

3.18. In this problem we study the bound states for a finite square potential well (see Fig. 3.8). Consider the one-dimensional potential defined by

V x Vx aa x a

a x( )

(

( )= −

< −− < <

<

⎧⎨⎪

⎩⎪

0

0

22 2

20

// /

/ (3.18.1)

I II

–a/2 a/2

V(x)

–V0

III

x

Fig. 3.8

(a) Write the stationary solutions for a particle of mass m and energy −V0 < E < 0 for each of the regions I (x < −a/2), II (−a/2 < x < a/2), and III (a/2 < x). (b) Apply the matching conditions at x = −a/2 and x = a/2. Obtain an equation for the possible energies. Draw a graphic representation of the equation in order to obtain qualitative properties of the solution.

SOLUTION

(a) Referring to Problem 3.9, we define

ρ = − =+2 2

20

2mE

km E V

� �

( ) (3.18.2)


Then the stationary solutions for each of the regions are

ψ

ψ

ψ

ρ ρI

II

III

( )

( )

(

x Ae A e

x Be B e

x

x x

ikx ikx

= + ′

= + ′

−

−

)) = ′ +

⎧

⎨

⎪⎪

⎩

⎪⎪ −C e Cex xρ ρ

(3.18.3)

Since y (x) must be bounded in regions I and III, we set A′ = C′ = 0; therefore,

ψ

ψ

ψ

ρ

ρ

I

II

III

( )

( )

( )

x Ae

x Be B e

x Ce

x

ikx ikx

x

=

= + ′

=

−

−

⎧⎧

⎨

⎪⎪

⎩

⎪⎪

(3.18.4)

(b) The continuity of y (x) and y ′(x) at x = −a/2 yields

Ae Be B e

Ae ikBe

a ika ika

a ika

− −

− −

= + ′

=

ρ

ρρ

/ / /

/ /

2 2 2

2 22 2− ′

⎧⎨⎪

⎩⎪ ikB eika /

(3.18.5)

Similarly, the matching conditions at x = a/2 yield

Ce Be B e

Ce ikBe

a ika ika

a ika

− −

−

= + ′

− =

ρ

ρρ

/

/ /

2 2 2

2

/ /

22 2− ′

⎧⎨⎪

⎩⎪−ikB e ika /

(3.18.6)

Hence, we can express B and B′ in terms of A:

Bik

ike A B

ikik

eik a= +⎛⎝⎜

⎞⎠⎟ ′ = − −− + − +ρ ρρ ρ

2 22( ) / ( iik a A) / 2⎛

⎝⎜⎞⎠⎟ (3.18.7)

We substitute Eq. (3.18.7) in Eq. (3.18.6) to obtain

C

ikik

eik

ike A

ikC

ik

ika ika= + − −⎛⎝⎜

⎞⎠⎟

− = +

−ρ ρ

ρ ρ

2 2

22 2ike

ikik

e Aika ika− −⎛⎝⎜

⎞⎠⎟

⎧

⎨⎪⎪

⎩⎪⎪

−ρ (3.18.8)

To obtain a nonvanishing solution of Eq. (3.18.8), we must have

− + − −⎛⎝⎜

⎞⎠⎟ = +−ρ ρ ρ ρ

ikik

ike

ikik

eik

ikeika ika

2 2 2iika ikaik

ike+ −⎛

⎝⎜⎞⎠⎟

−ρ2

(3.18.9)

which is equivalent to

ρρ

−+

⎛⎝⎜

⎞⎠⎟ =ik

ike ika

22 (3.18.10)

Equation (3.18.10) is an equation for E, since r and k depend only on E and on the constants of the problem. The solutions of Eq. (3.18.10) in terms of E are the energies corresponding to bound states of the well.

We shall transform Eq. (3.18.10) to express it in terms of k only. There are two possible cases. The first one is

I. ρρ

−+

⎛⎝⎜

⎞⎠⎟ = −ik

ikeika

2

(3.18.11)


The left-hand side of Eq. (3.18.11) is a complex number of modulus 1 and phase −2 tan−1 (k/r). Note that r + ik is the complex conjugate of r − ik. The right-hand side of Eq. (3.18.11) is also a complex

number of modulus 1, and its phase is π π π+ − = = +ka e e e eika i ika i ka; .( ) Therefore, we have

tan tan− ⎛⎝⎜

⎞⎠⎟ = − +⎛

⎝⎜⎞⎠⎟ ⇒ = − +⎛

⎝1

2 2 2 2k ka k kaρ

πρ

π⎜⎜

⎞⎠⎟

⎡⎣⎢

⎤⎦⎥

= − +⎛⎝⎜

⎞⎠⎟

= ⎛⎝⎜

⎞⎠⎟ =

tan

cot

π2 2

21

ka

kattan ( )ka/2

(3.18.12)

and

tanka

k2⎛⎝⎜

⎞⎠⎟ = ρ

(3.18.13)

We define kmV

k00

22 22

= = +�

ρ , where the parameter k0 is E-independent. Consider

1

21

222

2 2

20

cos ( )tan

ka

ka k

k

kk/

= + ⎛⎝⎜

⎞⎠⎟ = + = ⎛

⎝⎜⎞ρ⎠⎠⎟

2

(3.18.14)

Equation (3.18.11) is thus equivalent to the following system of equations:

cos

tan

ka kk

ka

2

20

0

⎛⎝⎜

⎞⎠⎟ =

⎛⎝⎜

⎞⎠⎟ >

⎧

⎨⎪⎪

⎩

⎪⎪

(3.18.15)

where we used Eqs. (3.18.13) and (3.18.14) together with the fact that both r and k are positive. We turn to the second possible case, i.e.,

II. ρρ

−+

⎛⎝⎜

⎞⎠⎟ =ik

ikeika

2 (3.18.16)

Similar arguments, as in case I, lead us to

− ⎛⎝⎜

⎞⎠⎟ = ⇒ = −−2

21tan tan

kka

ka kρ ρ (3.18.17)

Consider

sintan ( )

tan ( )2

2

2

2

222

1 2

ka ka

ka

k

k⎛⎝⎜

⎞⎠⎟ =

+=

+/

/ ρ22 (3.18.18)

Thus,

sin

tan

ka kk

ka

2

20

0

⎛⎝⎜

⎞⎠⎟ =

⎛⎝⎜

⎞⎠⎟ <

⎧

⎨⎪⎪

⎩

⎪⎪

(3.18.19)

In Fig. 3.9 we represent Eqs. (3.18.15) and (3.18.19) graphically. The straight line represents the func-

tion k/k0, and the sinusoidal arcs represent the functions sinka2

⎛⎝⎜

⎞⎠⎟ and cos .

ka2

⎛⎝⎜

⎞⎠⎟ The dotted parts

are the regions where the condition on tanka2

⎛⎝⎜

⎞⎠⎟ is not fulfilled.


The intersections marked with a point represent the solutions in terms of k. From these solutions it is possible to determine the possible energies. From Fig. 3.9 we see that if k0 ≤ p /a, that is, if

V Vma0 1

2 2

22≤ ≡ π �

(3.18.20)

then there exists only one bound state of the particle. Then, if V1 ≤ V0 < 4V1 there are two bound states, and so on. If V V0 1>> , the slope 1/k0 of the line is very small. For the lowest energy levels we have approximately

kna

n= =π( , , , . . .)1 2 3 (3.18.21)

and consequently,

En

maV= −π 2 2 2

2 02

� (3.18.22)

3.19. Consider a particle of mass m and energy E > 0 held in the one-dimensional potential −V0 d (x − a). (a) Integrate the stationary Schrödinger equation between a − e and a + e. Taking the limit e → 0, show that the derivative of the eigenfunction y (x) presents a discontinuity at x = a and determine it. (b) Relying on Problem 3.9, part (a), y (x) can be written

ψ

ψ

( )

( )

x A e A e x a

x A e A e

ikx ikx

ikx ikx

= + ′ <

= + ′

−

−

1 1

2 2 xx a>

⎧⎨⎪

⎩⎪ (3.19.1)

where k mE= 2 2/� . Calculate the matrix M defined by

A

AM

A

A2

2

1

1′⎛⎝⎜

⎞⎠⎟

= ′⎛⎝⎜

⎞⎠⎟

(3.19.2)

SOLUTION

(a) Using the Schrödinger equation,

− + − =�2 2

2 02md x

dxV x a x E x

ψ δ ψ ψ( )( ) ( ) ( ) (3.19.3)

Integrating between a − e and a + e yields

− + − =−

�2 2

2 02md x

dxdx V x a x dx E x dx

a

aψ δ ψ ψε

( )( ) ( ) ( )

++

−

+

−

+

∫∫∫ε

ε

ε

ε

ε

a

a

a

a

(3.19.4)

Fig. 3.9

0

1

kp /a 2p /a 3p /a 4p /a 5p /ak0


According to the definition of the d-function (see the Mathematical Appendix), the integration gives

− −⎛⎝⎜

⎞⎠⎟

+= + = −

�2

02md x

dxd x

dxV a

x a x a

ψ ψ ψε ε

( ) ( )( ) ==

−

+

∫E x dxa

a

ψε

ε

( ) (3.19.5)

Since y (x) is continuous and finite in the interval [a − e, a + e], in the limit e → 0,

− −⎡

⎣

⎢⎢

⎤

→>

→<

�2

2md x

dxd x

dxx ax a

x ax a

lim( )

lim( )ψ ψ

⎦⎦

⎥⎥

+ =V x0 0ψ ( ) (3.19.6)

We see that the derivative of y (x) presents a discontinuity at x = a that equals 2 02mV aψ ( ) ./�

(b) We have two matching conditions at x = a. The continuity of y (x) at x = a yields

A e A e A e A eika ika ika ika1 1 2 2+ ′ = + ′− − (3.19.7)

where the second matching condition is given in relation to Eq. (3.19.6) and yields

�

2

1 1 2 22m A ike A ike A ike A ikeika ika ika ika− ′ − + ′− −(( ) = − + ′( )−V A e A eika ika0 1 1

(3.19.8)

Equations (3.19.6) and (3.19.7) enable us to express A2 and ′A2 in terms of A1 and ′A1:

AmV

ikA

mV

ike A

Am

ika2

02 1

02

21

2

1= +⎛⎝⎜

⎞⎠⎟

+ ′

′ = −

−

� �

VV

ike A

mV

ikAika0

22

102 11

� �+ −

⎛⎝⎜

⎞⎠⎟

′

⎧

⎨

⎪⎪⎪

⎩

⎪⎪⎪

(3.19.9)

We therefore have

A

AM

A

A2

2

1

1′⎛⎝⎜

⎞⎠⎟

= ′⎛⎝⎜

⎞⎠⎟

(3.19.10)

where

M

mV

ik

mV

ike

mV

ike

mV

ika

ika=

+ +

− −

−1

1

02

02

2

02

2 0

� �

� iik�2

⎛

⎝

⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟

(3.19.11)

3.20. In this problem we study the possible energies (E > 0) of a particle of mass m held in a d-function periodic potential (see Fig. 3.10). We define a one-dimensional potential by

V xma

x na

n

( ) ( )= −=−∞

∞

∑�2

2λ δ (3.20.1)

Referring to Problem 3.9, part (a), for each of the regions Ωn [na < x < (n + 1) a], the stationary solution can be written in the form

ψ n ni k x na

ni k x nax B e C e( ) ( ) ( )= +− − − (3.20.2)

(a) Use Problem 3.19 to find the matrix T relating the regions Ωn + 1 and Ωn:

B

CT

B

Cn

n

n

n

+

+

⎛⎝⎜

⎞⎠⎟

=⎛⎝⎜

⎞⎠⎟

1

1

ˆ (3.20.3)


Prove that T is not a singular matrix. (b) Since T is a nonsingular matrix, we can find a basis (b1, b2) of C2 consisting of eigenvectors of the matrix ˆ.T We write

B

C0

01 1 2 2

⎛⎝⎜

⎞⎠⎟

= +β βb b (3.20.4)

where b1, b2 are complex numbers. Impose the condition that B Cn n

2 2+ does not diverge for n → ± ∞

to obtain a restriction on the eigenvalues of ˆ.T Express this restriction in terms of the possible energies E.

SOLUTION

(a) We compare the definitions of yn (x) and yn + 1(x) according to Eq. (3.20.2) and the definition of y (x) in Problem 3.19, part (b). The analogy is depicted in Table 3.1.

Fig. 3.10

Ω–2… …Ω–1 Ω0 Ω1 Ω2

–2a 2a x0–a

V(x)

a

Table 3.1

Problem 3.19 Problem 3.20

A1 Bn e−ikna

′A1 Cn eikna

A2 B enik n a

+− +

11( )

′A2 C enik n a

++

11( )

V0

�2

2m aλ

Also, the boundary between the two regions Ωn and Ωn + 1 is set in x = (n + 1)a, whereas in Problem 3.19 the boundary condition is imposed at x = a. Using this analogy we have

B e B eika

C enik n a

nik na

nik n

+− + −= −⎛

⎝⎜⎞⎠⎟

−11 1 2

( ) λ aa ik n a

nik n a

ni

ika

e

C e B e

λ2

2 1

11

⎛⎝⎜

⎞⎠⎟

=

− +

++ −

( )

( ) kk na ik n an

ik naika

e C eika

λ λ2 1 2

2 1⎛⎝⎜

⎞⎠⎟

+ +⎛⎝⎜

+( ) ⎞⎞⎠⎟

⎧

⎨⎪⎪

⎩

⎪⎪

(3.20.5)

We therefore have

B

CT

B

Cn

n

n

n

+

+

⎛⎝⎜

⎞⎠⎟

=⎛⎝⎜

⎞⎠⎟

1

1

ˆ (3.20.6)


where

T

ika

eika

e

ika

e

ika ika

ika=

−⎛⎝⎜

⎞⎠⎟

−

+ +

−12 2

21

λ λ

λ iika

e ikaλ2

⎛⎝⎜

⎞⎠⎟

⎛

⎝

⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟

− (3.20.7)

We see that T is not a singular matrix, since

det Tika

ika

ika

= +⎛⎝⎜

⎞⎠⎟

−⎛⎝⎜

⎞⎠⎟

+ ⎛⎝⎜

⎞⎠

12

12 2

λ λ λ⎟⎟ =

2

1 (3.20.8)

and therefore det ˆ .T = 0

(b) Since T is a nonsingular matrix, we can find a basis (b1, b2) of C2 consisting of eigenvectors of T with corresponding eigenvalues a1 and a2; these eigenvalues are the solutions of the cubic equation det ( ˆ ˆT I− =α ) 0. By definition,

ˆ

ˆ

T

T

b b

b b

1 1 1

2 2 2

=

=

⎧⎨⎪

⎩⎪

α

α (3.20.9)

Using Eq. (3.20.4), we have (for n = 1, 2, . . .)

B

CT T T

B

CTn

n n

⎛⎝⎜

⎞⎠⎟

=⎛⎝⎜

⎞⎠⎟

=ˆ ˆ ˆ ˆ�� times

0

0

nn n n( )β β β α β α1 1 2 2 1 1 1 2 2 2b b b b+ = + (3.20.10)

Consider

B CB

Cn nn

n

n2 22

1 1

2

1

2+ =

⎛⎝⎜

⎞⎠⎟

≥ β α b (3.20.11)

Therefore, α1 1≤ ; otherwise limn n nB C

→∞+( ) =2 2

∞. Similarly, we must have α2 1≤ . We apply a

similar consideration for n → −∞:

B

CT

B

Cnn n

n

0

01 2

⎛⎝⎜

⎞⎠⎟

=⎛⎝⎜

⎞⎠⎟

=−

−

ˆ , . . .for (3.20.12)

Hence,

B

CT

B

CTn

n

n n−

−

− −⎛⎝⎜

⎞⎠⎟

=⎛⎝⎜

⎞⎠⎟

= +ˆ ˆ ( )0

01 1 2 2β βb b == ( )⎡

⎣⎤⎦ + ( )⎡

⎣⎤⎦

− −βα

βα

α1

11 1

2

22n

n nn

n nT a Tˆ ˆb b2

== ′⎡⎣

⎤⎦ + =− −β

αβα

1

11

2

22n

n nn

n nT T T Tˆ ( ˆ ) [ ˆ ( ˆ )]b bββα

βα

nn n

n

11

2

22b b+ (3.20.13)

Therefore,

B CB

Cn nn

nn− −

−

−+ =

⎛⎝⎜

⎞⎠⎟

≥2 2

2

1

1

2

1

2βα

b (3.20.14)

so α1 1≥ ; otherwise ψ n x( )2 diverges for n → −∞, and similarly we must have α2 1≥ . Summing

our results, we must have α α1 2 1= = , i.e., the eigenvalues of T must be of modulus 1. Therefore, we can write

det ( ˆ ˆ)T e Ii− =φ 0 (3.20.15)


where f is a real constant. So,

12

12

−⎛⎝⎜

⎞⎠⎟

−⎡

⎣⎢

⎤

⎦⎥ +⎛

⎝⎜⎞⎠⎟

−ika

e eika

eika i iλ λφ kka ieka

−⎡

⎣⎢

⎤

⎦⎥ − =φ λ2

220

( ) (3.20.16)

A rearrangement of (3.20.16) gives

14

12

12

2

2 2+⎛

⎝⎜⎞

⎠⎟− −⎛

⎝⎜⎞⎠⎟

+ +⎛⎝

λ λ λk a

ika

eika

ika⎜⎜

⎞⎠⎟

⎡

⎣⎢

⎤

⎦⎥ + − =−e e e

kaika i iφ φ λ2

2

220

( ) (3.20.17)

or

1 22

02− +⎡⎣⎢

⎤⎦⎥

+ =cos ( ) sin ( )kaka

ka e ei iλ φ φ (3.20.18)

Consider the real part of (3.20.18):

1 22

2 0− +⎡⎣⎢

⎤⎦⎥

+ =cos ( ) sin ( ) cos cos ( )kaka

kaλ φ φ (3.20.19)

Using the relation cos (2f) = 2 cos2 f − 1, we arrive at

cos cos ( ) sin ( )φ λ= +kaka

ka2

(3.20.20)

Note that since k mE= 2 2/� , Eq. (3.20.20) is a constraint on the possible energies E:

cos ( ) sin ( )kaka

ka+ ≤λ2

1 (3.20.21)

We can represent this inequality schematically in the following manner. The function

f k kaka

ka( ) cos ( ) sin ( )= + λ2

(3.20.22)

behaves for k → ∞ as cos (ka) approximately. The schematic behavior of f(k) is depicted in Fig. 3.11.

We see that there are permitted bands of possible energies separated by domains where f k( ) ,≥ 1 and therefore the corresponding energy E does not correspond to a possible state. For E → ∞ the forbidden bands become very narrow, and the spectrum of the energy is almost continuous.

3.21. Consider a particle of mass m held in a three-dimensional potential written in the form

�V x y z V x U y W z( , , ) ( ) ( ) ( )= + + (3.21.1)

Derive the stationary Schrödinger equation for this case, and use a separation of variables in order to obtain three independent one-dimensional problems. Relate the energy of the three-dimensional state to the effective energies of the one-dimensional problem.

f (k)

ka

+1

permitted bands

–1

Fig. 3.11


SOLUTION

In our case the stationary Schrödinger equation is

− ∇ + + + =�2

2

2mV x U y W z EΨ Ψ Ψ( ) [ ( ) ( ) ( )] ( ( )r r r) (3.21.2)

where, Y (r) is the stationary three-dimensional state and E is the energy of the state. We assume thatY (r) can be written in the form Y (r) = f (x) c (y) y (z), so substituting in Eq. (3.21.2) gives

−

⎛

⎝⎜⎞

⎠⎟+

⎛�2 2

2

2

22md x

dxy z x

d y

dy

φ χ ψ φ χ( )( ) ( ) ( )

( )

⎝⎝⎜⎞

⎠⎟+

⎛

⎝⎜⎞

⎠⎟⎡

⎣⎢⎢

⎤

⎦⎥⎥

+

ψ φ χ ψ( ) ( ) ( )

( )z x y

d z

dz

2

2

[[ ( ) ( ) ( )] ( ) ( ) ( ) ( ) ( ) ( )V x U y W z x y z E x y z+ + =φ χ ψ φ χ ψ (3.21.3)

Dividing Eq. (3.21.4) by Y (r) and separating the x-dependent part, we get

− + = − + −� �2 2

2

2

21

21

m xd x

dxV x E U y W z

mφφ

χ( )( )

( ) ( ) ( )(( )

( )( )

( )y

d y

dy zd z

dz

2

2

2

21χ

ψψ+

⎛

⎝⎜⎞

⎠⎟⎡

⎣⎢⎢

⎤

⎦⎥⎥

(3.21.4)

The left-hand side of Eq. (3.21.4) is a function of x only, while the right-hand side is a function of y and z, but does not depend on x. Therefore, both sides cannot depend on x; thus they equal a constant, which we will denote by Ex. We have

− + =�2 2

22md x

dxV x x E xx

φ φ φ( )( ) ( ) ( ) (3.21.5)

We see that f (x) is governed by the equation describing a particle of mass m held in the one-dimensional potential V(x). Returning to Eq. (3.21.4), we can write

− + = − − −� �2 2

2

2

21

21

m yd y

dyU y E E W z

m zxχχ

ψ( )( )

( ) ( )( ))

( )d z

dz

2

2ψ⎡

⎣⎢⎢

⎤

⎦⎥⎥

(3.21.6.)

In Eq. (3.21.6) the left-hand side depends only on y, while the right-hand side depends only on z. Again, both sides must equal a constant, which we will denote by Ey. We have

− + =�2 2

22md y

dyU y y E yy

χ χ χ( )( ) ( ) ( ) (3.21.7)

Thus, c (y) is a stationary state of a fictitious particle held in the one-dimensional potential U( y). Finally, we have

− + =�2 2

22md z

dzW z z E zz

ψ ψ ψ( )( ) ( ) ( ) (3.21.8)

where we set Ez = E − Ex − Ey. Hence, the three-dimensional wavefunction Ψ(r) is divided into three parts. Each part is governed by a one-dimensional Schrödinger equation. The energy of the three-dimensional state equals the sum of energies corresponding to the three one-dimensional problems, E = Ex + Ey + Ez.


3.22. Solve Problems 3.13 and 3.14 for the case of particles with energy 0 < E < V0.

Ans. R = 1 and T = 0.

3.23. Consider the wavefunction

Ψ = −ei k z t( )ω (3.23.1)


The real part of Ψ corresponds to a harmonic plane wave traveling in the positive z direction. Determine the valve of the probability current.

Ans. p/m = v.

3.24. Consider a particle held in a one-dimensional complex potential V(x)(1 + ix ) where V(x) is a real function and x

is a real parameter. Use the Schrödinger equation to show that the probability current J mi x x= ∂

∂ − ∂∂

⎛

⎝⎜⎞

⎠⎟�

2 ψ ψ ψ ψ**

and the probability density r = y *y satisfy the corrected continuity equation ∂∂ + ∂

∂ =Jx t

V xρ ξ ρ2 ( ).

� (Hint:

Compare with Problem 3.6.)

3.25. Consider a particle of mass m held in a one-dimensional infinite potential energy well:

V xV x a

( ) =≤ ≤

∞⎧⎨⎩

0 0

otherwise (3.25.1)

Find the stationary states and the corresponding energies.

Ans. En

maV nn = + =π 2 2 2

2 021 2 3

�( , , , . . .). The corresponding states are the same as in Problem 3.12.

3.26. Consider an electron of energy 1 eV that encounters a potential barrier of width 0.1 nm and of energy-height 2 eV. What is the probability of the electron crossing the barrier? Repeat the same calculation for a proton.

Ans. For an electron T ≅ 0 78. ; for a proton T ≅ × −4 10 19.

3.27. (a) A particle of mass m and energy E > 0 encounters a potential energy well of width l and depth V0:

V xx

V x l

l x

( ) =<

− < <<

⎧⎨⎪

⎩⎪

0 00

00

Find the transmission coefficient. (Hint: Compare with Problem 3.16.) (b) For which values of l will the transmission be complete if the particle is an electron of energy 1 eV and V0 = 4 eV?

Ans. (a) TV

E E V

m E Vl

=

+ ++⎡

⎣⎢⎢

⎤

⎦⎥⎥

1

14

202

0

2 0

( )sin

( )

�

(b) l n m≅ × −( . ),2 7 10 10 where n is an integer.

3.28. An electron is held in a finite square potential energy well of width 1.0 × 10−10 m. For which values of the well’s depth V0 are there exactly two possible bound stationary states for the electron?

Ans. V1 ≤ V0 ≤ 4V1, where Vma1

2 2

2237 6= =π �

. eV.

3.29. Consider the wavefunction ψα

( ) .xN

x=

+2 2 (a) Calculate the normalization constant N where a is a real

constant. (b) Find the uncertainty Δx Δp (be careful calculating Δp!).

Ans. (a) N = 2 3απ ; (b) Δ Δx p = �

2.

3.30. Consider a particle of energy E > 0 confined in the potential energy well (Fig. 3.12)

V x

x aa x b

V b x b

b x aa x

( ) =

∞ < −− < < −− < <

< <∞ <

⎧

⎨

⎪⎪

⎩

⎪⎪

0

00


Show that for a stationary state with a nonvanishing probability of finding the particle to the right of the barrier (i.e., at b < x < a), there is also a nonvanishing probability of finding it to the left of the barrier (i.e., −a < x < −b). Note: For E < V0, this is another example of the tunnel effect of Problem 3.17.

3.31. Consider a particle of mass m confined in a one-dimensional infinite potential energy well:

V xx L

( ) =< <

∞{0 0otherwise

Suppose that the particle is in the stationary state, ψ πn x

Ln x

L( ) sin= ⎛

⎝⎜⎞⎠⎟

2 of energy E

n

mLn = π 2 2 2

22

�.

Calculate (a) ⟨ ⟩x and ⟨ ⟩p ; (b) ⟨ ⟩x2 and ⟨ ⟩p2 ; (c) Δx Δp.

Ans. (a) ⟨ ⟩ = ⟨ ⟩ =xL

p2

0, ; (b) ⟨ ⟩ = −⎛⎝⎜

⎞⎠⎟

⟨ ⟩ =x Ln

pn

L2 2

2 22

2 2 2

213

1

2ππ

, ;�

(c) Δ Δx p nn

= −ππ

�1

121

2 2 2 .

3.32. Consider a particle of mass m held in the potential

V x V x x l( ) [ ( ) ( )]= − + −0 δ δ

where l is a constant. Find the bound states of the particles. Show that the energies are given by the relation

e l− = ± −⎛⎝⎜

⎞⎠⎟

ρ ρα12

where E m= −�2 2 2ρ / and α = 2 0

2mV /� .

Fig. 3.12

V0

0–b b a x–a

V(x)

61

CHAPTER 4

The Foundations of Quantum Mechanics

4.1 Introduction

The State SpaceIn classical mechanics, the position of a particle is described by a vector having three real number elements. Though an analogous description exists in quantum mechanics, there are many significant differences. The state of a quantum mechanical system is described by an element of an abstract vector space called the state space and denoted as e. In Dirac notation, an element of this space is called a ket and is represented by the symbol ⏐ ⟩. A ket is a vector.

Observables In Chapter 2, the concept of a linear operator was introduced. The Hermitian operator is a linear operator that is equal to its adjoint (see Sec. 4.6). A fundamental concept of quantum mechanics is the observable. An observable is a Hermitian operator for which one can find an orthonormal basis of the state space that consists of the eigenvectors of the operator (e.g., linear momentum, mass, energy, angular momentum). If the state space is finite-dimensional, then any Hermitian operator is an observable. In Dirac notation, an operator is represented by a letter. Since the action of an operator on a vector yields another vector, an expression of the form A ⏐ψ⟩ also represents a ket.

The Dual Space Recall that a functional is a mapping from a vector space to the complex field. The dual space of the state space e consists of all linear functionals acting on e. It is designated by e*. In Dirac notation an element of e* is called a bra, and is designated by the symbol ⟨ ⏐. We can associate with any ket ⏐ φ ⟩ of e an element of e*, denoted by ⟨φ ⏐. In other words, for every ket � ψ ⟩ there is a bra ⟨ ψ ⏐ such that (⏐ ψ ⟩)* = ⟨ ψ ⏐ and (a ⏐ ψ ⟩)* = a*⟨ ψ ⏐ where a is a complex number. Moreover, ⏐ a ψ ⟩ = a ⏐ ψ ⟩ and so ⟨ a ψ ⏐ = a*⟨ ψ ⏐. The action of a bra ⟨ψ ⏐ on a ket ⏐ χ ⟩ is expressed by juxtaposing the two symbols, ⟨ψ ⏐ χ ⟩. By definition, this expression is a complex number. (The terms bra and ket come from “bracket.”) The correspondence between e and e* is closely related to the existence of a scalar product in e.

Scalar Product The basic properties of the scalar product are summarized below:

(a) ⟨ ⟩ = ⟨ ⟩φ ψ ψ φ⏐ ⏐ * (4.1)

(b) ⟨ + ⟩ = ⟨ ⟩ + ⟨ ⟩ψ φ φ ψ φ φ ψ⏐ ⏐ ⏐λ λ λ λ1 1 2 2 1 1 2 2 (4.2)

(c) ⟨ + ⟩ = ⟨ ⟩ + ⟨ ⟩λ λ λ λ1 1 2 2 1 1 2 2φ φ ψ φ ψ φ ψ⏐ ⏐ ⏐* * (4.3)

(d) ⟨ ⟩ψ ψ⏐ is real and positive; it is zero if and only if ⏐ψ⟩ = 0 (4.4)

CHAPTER 4 The Foundations of Quantum Mechanics62

Projector onto a Subspace of d Let ⏐ φ1 ⟩, ⏐ φ2 ⟩, . . . , ⏐ φm ⟩ be m normalized pairwise orthogonal vectors;

⟨ φi ⏐ φj ⟩ = dij i, j = 1, 2, . . . , m (4.5)

We denote by em the subspace of e spanned by these m vectors. The projector into the subspace em is defined by the linear projection operator

Pm i i

i

m

= ⟩ ⟨=

∑ ⏐ ⏐φ φ1

(4.6)

The operator P = ⟩ ⟨⏐ ⏐φ φ establishes the component of any other vector that lies in the direction of the vector ⏐ φ ⟩. Accordingly, ˆ .P ⏐ ⏐ ⏐ψ φ ψ φ⟩ = ⟨ ⟩ ⟩ Figure 4.1 presents a simple example of this concept. The set {⏐φ1 ⟩, ⏐φ2 ⟩, ⏐ φ3 ⟩} is an orthonormal set of vectors. The projection of an arbitrary vector ⏐ψ ⟩ into the plane spanned by ⏐ φ1⟩ and ⏐ φ2 ⟩ is given by ˆ ( ) ( ) .P2 1 1 2 2⏐ ⏐ ⏐ ⏐ ⏐ψ φ ψ φ φ ψ φ⟩ = ⟨ ⟩ ⟩ + ⟨ ⟩ ⟩

4.2 Postulates in Quantum MechanicsPostulate I The state of a physical system at time t0 is defined by specifying a ket ⏐ψ (t0)⟩ belonging to

the state space e.

Postulate II A measurable physical quantity A is described by an operator A acting on e.

Measurement of Physical Quantities: The validity of a physical theory is continuously investigated by comparing results calculated by the theory with measurements obtained in experiments. In the context of quantum mechanics the measurement of a physical quantity involves three principal questions:

(a) What are the possible results of the measurement?

(b) What is the probability of obtaining each of the possible results?

(c) What is the state of the system after the measurement?

The answers to these questions in the context of quantum mechanics are found in the following three postulates.

Postulate III The possible results of the measurement of a physical quantity are the eigenvalues of the corresponding observable A.

We can now answer the second question for the case of a discrete spectrum. The generalization to the case of a continuous spectrum is treated in Prob. 4.5.

|ψ⟩

|φ2⟩

|φ1⟩(⟨φ2|ψ⟩)|φ2

⟩ (⟨φ1|ψ⟩)|φ1⟩

P2|ψ⟩Ÿ

|φ3⟩

Fig. 4.1


Postulate IV Let A be a physical quantity with a corresponding observable A. Suppose that the system is in a normalized state ⏐ ψ ⟩, so ⟨ψ ⏐ ψ ⟩ = 1. When A is measured, the probability P(an) of obtaining the eigenvalue an of A is

P a un

i

g

ni

n

( ) = ⟨ ⟩=

∑1

2⏐ ψ (4.7)

where gn is the degeneracy of an and ⏐ ⏐ ⏐u u un n ng

n1 2⟩ ⟩ ⟩, , . . . , form an orthonormal basis of the subspace en that consists of eigenvectors of A with eigenvalues an.

In Prob. 4.3, we introduce a different (though equivalent) formulation of postulate IV. The subspace en of the state space defined in postulate IV is also called the eigenspace associated with an. The following postulate describes the state of the system after a measurement.

Postulate V If the measurement of a quantity A on a physical system in the state ⏐ ψ ⟩ gives the result an, then immediately after the measurement, the state is given by the normalized projection of ⏐ ψ ⟩ onto the

eigenspace en associated with an; that is1

⟨ ⟩⟩

ψ ψψ

⏐ ⏐⏐

ˆˆ ,

PP

n

nwhere P

nis the projector onto en.

4.3 Mean Value and Root-Mean-Square DeviationConsider a state described by a normalized ket, ⟨ψ ⏐ ψ ⟩ = 1. The mean value of an observable A in the state ⏐ψ ⟩ is defined by

⟨ ⟩ = ⟨ ⟩ˆ Â Aψ ψ ψ⏐ ⏐ (4.8)

The mean value of an observable has a clear physical meaning. Suppose the physical quantity represented by the operator A is measured a large number of times when the system is in the state ⏐ψ ⟩. Then ⟨ ⟩A ψ expresses the average of the results of the measurements (that is, the sum of each result multiplied by the probability of obtaining it). The derivation of this property is given in Problem 4.8.

The root-mean-square deviation of the observable A is defined by

ΔA A A= ⟨ ⟩ − ⟨ ⟩ˆ ˆ2 2ψ ψ (4.9)

The root-mean-square deviation has a direct physical interpretation. It characterizes the dispersion of the measurement results about ⟨ ⟩A ψ (see Problem 4.9).

4.4 Commuting ObservablesConsider two operators, A and ˆ.B In general, the expressions ˆ ÂB and ˆ ˆBA are not identical—multiplication of operators is not commutative. An important concept in quantum mechanics is the commutator [ ˆ , ˆ]A B of two operators defined by

[ ˆ , ˆ] ˆ ˆ ˆ Â B AB BA= − (4.10)

Some useful properties of a commutator are given in Problems 4.10, 4.11, and 4.12. If [ ˆ , ˆ] ,A B = 0 then A and B are called commuting operators. Given two observables A and B, they are said to be compatible when their associated operators A and B commute; [ ˆ , ˆ] .A B = 0 If A and B are not compatible they cannot be measured simultaneously. Consider the following theorem.

Theorem: Operator observables A and B commute if and only if there exists a basis of eigenvalues com-mon to both.

A set of operators ˆ , ˆ , ˆ , . . .A B C is called a complete set of commuting operator observables if all subpairs commute, and there exists a unique orthonormal basis of common eigenvectors. The uniqueness is within a multiplicative factor.


4.5 Function of an OperatorAssume that in a certain domain the function F of variable x can be expanded in a power seriesin x:

F x a xnn

n

( ) ==

∞

∑0

(4.11)

The corresponding function of the operator A is the operator ˆ ( ˆ)F A defined by a series that has the same coefficients an:

ˆ ( ˆ) ˆF A a Ann

n

==

∞

∑0

(4.12)

4.6 Hermitian ConjugationThe adjoint (or conjugate) of an operator A is denoted by ˆ .†A For every ⏐ φ ⟩ and ⏐ ψ ⟩ we have

⟨ ⟩ = ⟨ ⟩ψ φ φ ψ⏐ ⏐ ⏐ ⏐ˆ ˆ† *A A (4.13)

The basic properties of the adjoint of an operator are derived in Problems 4.13 and 4.17. An operator A is Hermitian if it is identical to its adjoint:

ˆ ˆ ˆ†A A Ais Hermitian ⇔ = (4.14)

An inspection of Eq. (4.13) shows that in order to obtain the Hermitian (or the adjoint) of any expression, it suffices to apply the following procedure:

Replace the constants by their complex conjugates.

Replace the kets by the bras associated with them.

Replace the bras with the kets associated with them.

Replace the operators by their adjoint operators.

Reverse the order of the factors (the position of the constants is of no importance). For example,

λ λ⟨ ⟩ → ⟨ ⟩φ ψ ψ φ⏐ ⏐ ⏐ ⏐ˆ ˆ ˆ ˆ* † †AB B A (4.15)

4.7 Discrete and Continuous State SpacesA discrete set of kets {⏐ui ⟩, i = 1, 2, . . . } is orthonormal if it satisfies the following relation:

⟨ui ⏐ uj ⟩ = dij (4.16)

For a continuous set of kets {⏐wa ⟩, l1 ≤ a ≤ l2}, the orthonormalization relation is written as

⟨ ⟩ = − ′′w wα α δ α α⏐ ( ) (4.17)

A set of kets constitutes a basis of the state space e if every ket ⏐ ψ ⟩ belonging to e has a unique expansion on these kets:

⏐ ⏐ψ⟩ = ⟩∑ C ui i

i

(4.18)


and for the continuous case:

⏐ ⏐ψ⟩ = ⟩∫C w d( )α αα (4.19)

It can be proved that an orthonormal set of kets constitutes a basis if and only if it satisfies the closure relation (see Problems 4.19 and 4.20):

⏐ ⏐ ⏐u u I w wi i⟩ ⟨ = ⟩ ⟨ˆ for the continuous case, α α ⏐⏐d I

i

α∫∑ =⎛⎝⎜

⎞⎠⎟

ˆ (4.20)

where I denotes the identity operator in e. Using the notion of the projector onto the space spanned by the set of kets, we can write these relations in an equivalent form:

P u I P w Ii{ } ˆ ( { } ˆ)= =or α (4.21)

4.8 RepresentationsThe validity of a physical theory is established by comparing experimentally obtained data with the calcu-lated results of the theory. When a basis is chosen in the abstract state space, each ket, bra, and operator can be characterized by specifying its coordinates for that basis. We say that the abstract object is represented by the corresponding set of numbers. Using these numbers, the theory-prescribed calculations are performed. Choosing a representation means choosing an orthonormal basis in the state space.

Representations of Kets and BrasIn a discrete basis {⏐ui ⟩}, a ket ⏐ ψ ⟩ is represented by the set of numbers Ci = ⟨ ui ⏐ ψ ⟩. These numbers can be arranged vertically to form a column matrix:

( )C

C

Ci =

⎛

⎝

⎜⎜⎜

⎞

⎠

⎟⎟⎟⋅⋅⋅

1

2 (4.22)

A bra ⟨φ ⏐ is represented by the sets of numbers b ui i* ,= ⟨ ⟩φ⏐ which are the complex conjugates of the com-

ponents of the ket ⏐ φ ⟩ associated with ⟨φ ⏐. These numbers can be arranged horizontally to form a row matrix,( , , . . .).* *b b1 2

In a continuous basis {⏐wa ⟩}, kets and bras are represented by a continuous infinity of numbers, that is, by a function of a. A ket ⏐ ψ ⟩ is represented by the set of numbers C (a) = ⟨wa ⏐ ψ ⟩, and a bra ⟨φ ⏐ is represented by b*(a) = ⟨φ ⏐wa ⟩. Once a representation is chosen, we can use the components of the ket and the bra to calculate their scalar product. In the discrete case,

⟨ ⟩ = ⟨ ⟩ =φ ψ φ ψ⏐ ⏐b C bi i* *in the continuous case, (( ) ( )α α αC d

i∫∑ ⎡

⎣⎢

⎤

⎦⎥ (4.23)

Representations of Operators In a discrete basis {⏐ui ⟩}, an operator is represented by the numbers

A u A uij i j= ⟨ ⟩⏐ ⏐ˆ (4.24)

These numbers can be arranged in a square matrix,

[ ]A

A A A

A A A

A Aij

j

j

i

=

⋅ ⋅ ⋅ ⋅ ⋅ ⋅⋅ ⋅ ⋅ ⋅ ⋅ ⋅

⋅⋅⋅

11 12 1

21 22 2

1 ii ijA2 ⋅ ⋅ ⋅ ⋅ ⋅ ⋅⋅⋅⋅

⎛

⎝

⎜⎜⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟⎟

(4.25)


For a continuous basis {⏐wa ⟩}, we associate with A a continuous function of two variables:

⟨ ′⟩ = ⟨ ⟩′α α α α, ˆw A w⏐ ⏐ (4.26)

As a consequence of Eq. (4.13),

( ˆ )† *A Aij i j= (4.27)

or

ˆ ( , ) ˆ ( , )† *A A′ = ′α α α α (4.28)

If A is an Hermitian operator ( ˆ ˆ),†A A= we have ˆ ( , ) ˆ ( , ).*A A′ = ′α α α α (Note that for the discrete case A Aij ji= * .) In particular, the diagonal elements of a Hermitian matrix are always real numbers.

Change of RepresentationWe provide a simple method to obtain the representation of a bra, ket, or operator in a given basis when its representation in another basis is known. For simplicity, assume that we perform a transformation from one discrete orthonormal basis {⏐ui ⟩} to another, {⏐vi ⟩}. Define the transformation matrix

Sik = ⟨ui ⏐vk ⟩ (4.29)

The Hermitian conjugate of Sik is given by

( ˆ ) ( )† *S S uki ik k i= = ⟨ ⟩v ⏐ (4.30)

To pass from the components of a ket ⏐ψ ⟩ represented in one basis to another, one applies the relation

⟨ ⟩ = ⟨ ⟩∑v k ki i

i

S u⏐ ⏐ψ ψ( ˆ )† (4.31)

or the inverse relation, ⟨ ⟩ = ⟨ ⟩∑u Si ik k

k

⏐ ⏐ψ ψv . For a bra ⟨φ ⏐ we have

⟨ ⟩ = ⟨ ⟩ ⟨ ⟩ = ⟨ ⟩⎡

⎣

⎢⎢

⎤

∑φ φ φ φ⏐ ⏐ ⏐ ⏐v vk i ik i k ki

k

u S u S( ˆ )†

⎦⎦

⎥⎥∑

i

(4.32)

Finally, the matrix elements of an operator A transform as

⟨ ⟩ = ⟨ ⟩ ⟨∑v vk l ki i j jl

i j

iA S u A u S u A u⏐ ⏐ ⏐ ⏐ ⏐ ⏐ˆ ( ˆ ) ˆ ˆ†

,

jj ik k l lj

k l

S A S⟩ = ⟨ ⟩∑ v v⏐ ⏐ˆ ( ˆ )†

,

(4.33)

� r Õ- and � p Õ-Representations In Sec. 4.1 we noted that to every ket ⏐ φ ⟩ there corresponds a bra ⟨φ ⏐. The converse is not necessarily true; there are bras with no corresponding kets. Nevertheless, in addition to the vectors belonging to e, we shall use generalized kets whose norm is not finite. At the same time, however, the scalar product of those kets with every ket is finite. The generalized kets do not represent physical states; they serve to help us analyze and interpret physical states represented by kets belonging to e.

Consider the physical system of a single particle. Together with the state space of the system we introduce another vector space, called the wavefuction space, denoted by F. This space consists of complex functions of the coordinates (x, y, z) having the following properties:

(a) The functions ψ (r) are defined everywhere as continuous and infinitely differentiable.

(b) The integral ⏐ ⏐ψ( )r 2 3∫ d r must be finite; i.e., ψ (r) must be square integrable.


To every function ψ (r) belonging to F there corresponds a ket ⏐ ψ ⟩ belonging to e. Using the wavefunctions φ (r) and ψ (r) corresponding to ⟨ φ ⏐ and ⏐ψ ⟩, we define the scalar product of ⟨ φ ⏐ and ⏐ψ ⟩:

⟨ ⟩ =∫φ ψ φ ψ⏐ *( ) ( )r r d r3 (4.34)

Consider two particular bases of F denoted { ( )}ξr r0

and { ( )}.np0r These bases are not composed of functions

belonging to F:

ξ δr r r r0

0( ) ( )= − (4.35)

and

npp rr

0

01

2 3 2( )( )

=π�

�

//

ei i (4.36)

To each ξr r0( ) we associate a generalized ket denoted by ⏐ r0 ⟩, and similarly for np r

0( ) we associate a general-

ized ket ⏐ p0 ⟩. The sets {⏐ r0 ⟩} and {⏐ p0 ⟩} constitute orthonormal bases in e :

⟨ ′⟩ = − ′ ⟩ ⟨ =∫r r r r r r d r I0 0 0 0 0 03⏐ ⏐ ⏐δ ( ) ˆ (4.37)

where we also have the following relations:

⟨ ′ ⟩ = − ′ ⟩ ⟨ =∫p p p p p p d p I0 0 0 0 0 03⏐ ⏐ ⏐δ ( ) ˆ (4.38)

We obtain two representations in the state space of a (spinless) particle: the {⏐r0 ⟩}- and {⏐ p0 ⟩}-representations. The correspondence between the ket ⏐ ψ ⟩ and the wavefunction associated with it is given by

ψ ψ( )r r0 0= ⟨ ⟩⏐ (4.39)

�ψ ψ( )p p0 0= ⟨ ⟩⏐ (4.40)

where �ψ ( )p is the Fourier transform of ψ (r). The value ψ (r)′ of the wavefunction at the point r is the compo-nent of the ket ⏐ψ ⟩ on the basis vector ⏐ r ⟩ of the ⏐ r ⟩-representation. Also, the value �ψ ( )p of the wavefunction in the momentum space at p is the component of the ket ⏐ ψ ⟩ on the basis vector ⏐ p ⟩ of the ⏐ p ⟩-representation.

Exchanging between the ⏐ r ⟩-representation and the ⏐ p ⟩-representation is accomplished analogously to the case of continuous bases. Note that

⟨ ⟩ = ⟨ ⟩ =r p p r p r⏐ ⏐ * /

( )

1

2 3 2π�

�/ ei i (4.41)

Now, we have

r ⏐ ⏐ ⏐ψ ψ⟩ = ⟨ ⟩ ⟨ ⟩∫ r p p d p3 (4.42)

and inversely,

⟨ ⟩ = ⟨ ⟩ ⟨ ⟩∫p ⏐ ⏐ ⏐ψ ψp r r d r3 (4.43)

Therefore, using Eq. (4.41), we obtain

ψπ

ψ( )( )

( )r pp r= ∫1

2 3 23

�

�/

/e d pi i � (4.44)


and

� iψπ

ψ( )( )

( )p rp r= −∫1

2 3 23

�

�/

/e d ri (4.45)

The Operators R and P Let ⏐ ψ ⟩ be a ket belonging to the state space and let ψ (x, y, z) = ⟨ r ⏐ ψ ⟩ = ψ (r) be its corresponding wavefunc-tion. The three operators ˆ , ˆ, ˆX Y Zand are defined by their action in the ⏐ r ⟩-representation:

⟨ ⟩ = ⟨ ⟩ ⟨ ⟩ = ⟨ ⟩ ⟨ ⟩ = ⟨r r r r r r⏐ ⏐ ⏐ ⏐ ⏐ ⏐ ⏐ ⏐ ⏐ˆ ˆ ˆX x Y y Z zψ ψ ψ ψ ψ ψψ ⟩ (4.46)

The operator X acting on ⏐ψ⟩ yields the ket ⏐ ′⟩ψ , which corresponds to the wavefunction ψ ′(x, y, z) = x ψ (x, y, z), and similarly for Y and ˆ.Z The operators ˆ , ˆ,X Y and Z are considered to be the components of a vector operator R. Similarly, the operators ˆ , ˆ ,P P

x y and Pz are defined by their action in the ⏐ p ⟩-representation:

⟨ ⟩ = ⟨ ⟩ ⟨ ⟩ = ⟨ ⟩ ⟨ ⟩p p p p p⏐ ⏐ ⏐ ⏐ ⏐ ⏐ ⏐ ⏐ˆ ˆ ˆP p P p Px x y y zψ ψ ψ ψ ψ == ⟨ ⟩pz p ⏐ ψ (4.47)

ˆ , ˆ ,P Px y and Pz are the components of the vector operator ˆ .P The operator observables R and P are of funda-

mental importance in quantum mechanics. Their commutation relations are called the canonical commutation relations:

[ ˆ , ˆ ] [ ˆ , ˆ ] [ ˆ , ˆ ]R P i R R P Pi j ij i j i j= = =�δ 0 0 (4.48)

Quantization Rules By quantization rules we mean the method for obtaining the quantum-mechanics analog of a classical quantity. Consider a system of a single particle. The operators ( ˆ , ˆ, ˆ)X Y Z are associated with the coordinates (x, y, z) of the particle, while ( ˆ , ˆ , ˆ )P P Px y z are associated with the momentum (px, py, pz). We shall often use the nota-

tion R for ( ˆ , ˆ, ˆ)X Y Z and P for ( ˆ , ˆ , ˆ ).P P Px y z

In classical mechanics, a physical quantity A related to a particle is expressed in terms of the particle’s position vector r and the momentum p. To obtain the corresponding quantum-mechanics observable, replace r R→ ˆ and p P→ ˆ . Since the expression obtained is not always Hermitian, we apply a symmetrization between R and P to obtain a Hermitian operator. In Problem 4.35 we demonstrate this method. Note that there exist quantum mechanical physical variables which have no classical equivalent (such as spin). These quantities are defined by the corresponding observables.

4.9 The Time EvolutionIn the previous sections we paid no attention to the time evolution of a system but rather considered a definite static state. We shall now present methods for treating the time evolution of a system. Consider the following postulate:

Postulate VI The time evolution of the state vector ⏐ ψ (t) ⟩ of a physical system is governed by the Schrödinger equation:

id t

dtH t t�

⏐⏐

ψψ

( ) ˆ ( ) ( )⟩

= ⟩ (4.49)

where ˆ ( )H t is the observable corresponding to the classical Hamiltonian of the system.

Some important implications of the Schrödinger equation must be noted:

(a) Since the Schrödinger equation is a first-order differential equation in t, it follows that if an initial state ⏐ψ (t0) ⟩ is given, the state ⏐ψ (t) ⟩ is determined; therefore, the time evolution is deterministic. Note that indeterminacy appears only when a physical quantity is measured.

(b) Let ⏐ψ1 (t) ⟩ and ⏐ψ2 (t)⟩ be two different solutions of the Schrödinger equation. If the initial state is ⏐ψ (t0)⟩ = a1⏐ψ1(t0) ⟩ + a2 ⏐ψ2 (t0) ⟩, where a1 and a2 complex numbers, then at time t the system is in the state ⏐ψ (t) ⟩ = a1⏐ψ1(t) ⟩ + a2 ⏐ψ2 (t) ⟩.


(c) At time t, the norm of the state vector remains constant:

ddt

t t⟨ ⟩ =ψ ψ( ) ( )⏐ 0 (4.50)

This means that the total probability of finding the particle is conserved (see Problem 4.40).

Time Evolution for a Conservative System A physical system is conservative if its Hamiltonian does not depend explicitly on time. In classical mechan-ics, the most important consequence of such an observation is the conservation of energy. Similarly, in quantum mechanics, a conservative system possesses important properties. Most of the problems in this book concern conservative systems.

The time evolution of a conservative system can be found rather simply. Suppose the Hamiltonian H does not depend explicitly on time. The time evolution of the system that was initially in the state ⏐ ψ (t0) ⟩ is found using the following procedure:

(a) Expand ⏐ψ (t0) ⟩ in the basis of eigenvectors of H:

⏐ ⏐ψ φ( ) ( )t a tnk nk

kn

0 0⟩ = ⟩∑∑ (4.51)

where ank (t0) = ⟨φn, k ⏐ψ (t0) ⟩.(b) To obtain ⏐ψ (t) ⟩ for t > t0, multiply each coefficient ank(t0) by e

iE t tn

− −( )0

/� where En is the eigenvalue of H associated with the state ⏐ φn, k ⟩:

⏐ ⏐ψ φ( ) ( )( )

,t a t enkiE t t

n k

kn

n⟩ = ⟩− −∑∑ 00

/ � (4.52)

This procedure can be generalized to the case of the continuous spectrum of ˆ .H So,

⏐ ⏐ψ φ( ) ( , )( )

,t a E t e dEkiE t t

E k

k

n⟩ = ⟩− −∫∑ 00

/ � (4.53)

The eigenstates of H are called stationary states.

Time Evolution of the Mean Value Let ⏐ψ (t) ⟩ be the normalized ket describing the time evolution of a physical system. The time evolution of the mean value of an observable A is governed by the equation

d A

dt iA H t

At

⟨ ⟩ = ⟨ ⟩ ∂∂

ˆ[ ˆ , ˆ ( )]

ˆ1�

(4.54)

If A does not depend explicitly on time, we have

d A

dt iA H t

⟨ ⟩ = ⟨ ⟩ˆ

[ ˆ , ˆ ( )]1�

(4.55)

By definition, a constant of motion is an observable A that does not depend explicitly on time and commutes with the Hamiltonian ˆ .H In this case,

d A

dt⟨ ⟩ =

ˆ0 (4.56)


4.10 Uncertainty RelationsAs we have seen in previous sections, the position or momentum of a particle in quantum mechanics is not characterized by a single number but rather by a continuous function. By the uncertainty of the position (or momentum) of a particle, we mean the degree of dispersion of the wavefunction relative to a central value. This quantity can be given a rigorous definition; however, that is beyond the scope of this volume.

The Heisenberg uncertainty relations give a lower limit for the product of the uncertainties of the position and the momentum of a particle:

Δ Δ Δ Δ Δ Δx p y p z px y z≥ ≥ ≥� � �/ / /2 2 2 (4.57)

Momentum and position are said to be conjugate variables. For the case of a conservative system, there is also a relation between the uncertainty of time Δt at which the system evolves to an appreciable extent, and the uncertainty of energy ΔE:

Δ Δt E ≥ �/2 (4.58)

Energy and time are conjugate variables; they are linked in a fundamental way.

4.11 The Schrödinger and Heisenberg Pictures In the formalism described in the previous sections, we considered the time-independent operators that cor-respond to the observables of the system. The time evolution is entirely contained in the state vector ⏐ψ (t) ⟩. This approach is called the Schrödinger picture. Nevertheless, since the physical predictions in quantum mechanics are expressed by scalar products of bras and kets of matrix elements of operators, it is possible to introduce a different formalism for the time evolution. This formalism is called the Heisenberg picture. Here the state of the system is described by a ket that does not vary over time, ⏐ ψH (t) ⟩ = ⏐ ψ (t0) ⟩. The observables corresponding to physical quantities evolve over time as

ˆ ( ) ˆ ( , ) ˆ ˆ ( , )†A t U t t A U t tH s= 0 0 (4.59)

where As is the observable in the Schrödinger picture and

ˆ ( , ) expˆ ( )U t t iH t ts

00= − −⎡

⎣⎢⎤⎦⎥�

(4.60)

The operator ˆ ( , )U t t0 is called the time evolution operator, and is a unitary operator. Note that this operator describes the time evolution of the state vector in the Schrödinger picture:

⏐ ⏐ψ ψs st U t t t( ) ˆ ( , ) ( )⟩ = ⟩0 0 (4.61)

SOLVED PROBLEMS

4.1. Two states of a physical system are given by

⏐ ψ1 ⟩ = (4i ⏐ φ1 ⟩ − 12i ⏐ φ2 ⟩) and ⏐ ψ2 ⟩ = (⏐ φ1 ⟩ − 6i ⏐ φ2 ⟩) (4.1.1)

Here ⏐ φ1 ⟩ and ⏐ φ2 ⟩ are orthonormal vectors. Determine both ⏐ψ1 + ψ2 ⟩ and ⟨ψ1 + ψ2 ⏐.

SOLUTION

⏐ ψ1 + ψ2 ⟩ = (4i ⏐ φ1 ⟩ − 12i ⏐ φ2 ⟩) + (⏐ φ1 ⟩ − 6i ⏐ φ2 ⟩) (4.1.2)


and

⏐ ψ1 + ψ2 ⟩ = (1 + 4i) ⏐ φ1 ⟩ − 18i ⏐ φ2 ⟩ (4.1.3)

Taking the complex conjugate of this leads to

(⏐ ψ1 + ψ2 ⟩)* = ⟨ψ1 + ψ2 ⏐ = (1 + 4i)* ⟨φ1 ⏐ − (18i)* ⟨φ2 ⏐ (4.1.4)

and so

⟨ψ1 + ψ2 ⏐ = (1 − 4i) ⟨φ1 ⏐ + 18i ⟨φ2 ⏐ (4.1.5)

4.2. Two states of a physical system are represented by

⏐ ψ1 ⟩ = (6i ⏐ φ1 ⟩ − 3i ⏐ φ2 ⟩) and ⏐ ψ2 ⟩ = (−2 ⏐ φ1 ⟩ + 4i ⏐ φ2 ⟩) (4.2.1)

Moreover, ⏐ φ1 ⟩ and ⏐ φ2 ⟩ are orthonormal. Determine the salar product ⟨ψ1 ⏐ ψ2 ⟩.

SOLUTION

We first find the bra ⟨ψ1 ⏐ corresponding to the ket ⏐ψ1 ⟩ = (6i ⏐ φ1 ⟩ − 3i ⏐ φ2 ⟩) using the fact that (⏐ ψ ⟩)* = ⟨ψ ⏐. Hence,

⟨ψ1 ⏐ = (6i ⏐ φ1 ⟩)* − (3i ⏐ φ2 ⟩)

*

= − 6i ( ⏐ φ1 ⟩)* + 3i (⏐ φ2 ⟩)

* (4.2.2)

and

⟨ψ1 ⏐ = −6i ⟨φ1 ⏐ + 3i ⟨φ2 ⏐ (4.2.3)

The scalar product is

⟨ψ1 ⏐ ψ2 ⟩ = (−6i ⟨ φ1 ⏐ + 3i ⟨ φ2 ⏐)(−2 ⏐ φ1 ⟩ + 4i ⏐ φ2 ⟩) (4.2.4)

Recall that ⏐ φ1 ⟩ and ⏐ φ2 ⟩ are orthonormal, which means that ⟨φ1 ⏐ φ1 ⟩ = ⟨φ2 ⏐ φ2 ⟩ = 1 and ⟨φ1 ⏐ φ2 ⟩ = ⟨φ2 ⏐ φ1 ⟩ = 0. We will use this after we multiply the scalar product,

⟨ψ1 ⏐ ψ2 ⟩ = (−6i ⟨φ1 ⏐)(−2 ⏐ φ1 ⟩) + (−6i ⟨φ1 ⏐)(4i ⏐ φ2 ⟩) + (3i ⟨φ2 ⏐)(−2 ⏐ φ1 ⟩) + (3i ⟨φ2 ⏐)(4i ⏐ φ2 ⟩)

= 12i ⟨φ1 ⏐ φ1 ⟩ + 24 ⟨φ1⏐ φ2 ⟩ − 6i ⟨φ2 ⏐ φ1 ⟩ − 12 ⟨φ2 ⏐ φ2 ⟩ (4.2.5)

Using the orthonormality of ⏐ φ1 ⟩ and ⏐ φ2 ⟩,

⟨ψ1 ⏐ ψ2 ⟩ = 12i − 12 (4.2.6)

4.3. With the previous problem in mind, determine the scalar product ⟨ψ2 ⏐ ψ1 ⟩. Verify that ⟨ψ1 ⏐ ψ2 ⟩* = ⟨ψ2⏐ ψ1 ⟩,

which is the equivalent of Eq. (4.1).

SOLUTION

Now find the bra ⟨ψ2 ⏐ corresponding to the ket ⏐ ψ2 ⟩ = (−2 ⏐ φ1 ⟩ + 4i ⏐ φ2 ⟩) using the fact that (⏐ ψ ⟩)* = ⟨ψ ⏐. Accordingly,

⟨ψ2 ⏐ = −2(⏐ φ1 ⟩)* + (4i)*(⏐ φ2 ⟩)

* (4.3.1)

and

⟨ψ2 ⏐ = −2 ⟨φ1 ⏐ − 4i ⟨φ2 ⏐ (4.3.2)

The scalar product is

⟨ψ2 ⏐ ψ1 ⟩ = (−2 ⟨φ1 ⏐ − 4i ⟨φ2 ⏐)(6i ⏐ φ1 ⟩ − 3i ⏐ φ2 ⟩)

= −12i ⟨φ1 ⏐ φ1 ⟩ + 6i ⟨φ1 ⏐ φ2 ⟩ + 24 ⟨φ2 ⏐ φ1 ⟩ − 12⟨φ2 ⏐ φ2 ⟩ (4.3.3)


and so,

⟨ψ2 ⏐ ψ1 ⟩ = −12i − 12 (4.3.4)

Since ⟨ψ1 ⏐ ψ2 ⟩ = 12i − 12, we see that

⟨ψ1 ⏐ ψ2 ⟩* = −12i − 2 = ⟨ψ2 ⏐ ψ1 ⟩ (4.3.5)

4.4. Let ⏐ ψ1 ⟩ and ⏐ ψ2 ⟩ be two orthogonal normalized states of a physical system:

⟨ψ1 ⏐ ψ2 ⟩ = 0 and ⟨ψ1 ⏐ ψ1 ⟩ = ⟨ψ2 ⏐ ψ2 ⟩ = 1 (4.4.1)

and let A be an observable of the system. Consider a nondegenerate eigenvalue of A2

denoted by αn

to

which the normalized state ⏐ φn ⟩ corresponds. We define P n n1 12( )α = ⏐ ⏐ ⏐⟨ ⟩φ ψ and P n n2 2

2( ) .α = ⏐ ⏐ ⏐⟨ ⟩φ ψ (a) What is the interpretation of P1(an) and P2(an)? (b) A given particle is in the state 3⏐ ψ1 ⟩ − 4i ⏐ ψ2 ⟩. What is the probability of getting an when A is measured?

SOLUTION

(a) According to the postulates of quantum mechanics, P1(an) is the probability of obtaining an when A is measured while the system is in the state ⏐ ψ1 ⟩. The same is the case with P2(an) in the state ⏐ ψ2 ⟩.

(b) The normalized state of the particle is

⏐⏐ ⏐

⏐ + ⏐ ⏐ ⏐ψ

ψ ψψ ψ ψ ψ

⟩ =⟩ − ⟩

⟨ ⟨ ⟩ − ⟩3 4

3 4 3 41 2

1 2 1 2

i

i i( )( ))( )=

⟩ − ⟩+

= ⟩ − ⟩3 4

9 1615 3 41 2

1 2

⏐ ⏐⏐ ⏐

ψ ψψ ψ

ii (4.4.2)

Using the postulates of quantum mechanics (for a summary of the theory, see Sec. 4.2), the probability of measuring an is

P in n n n( )α = ⟨ ⟩ = ⟨ ⟩ − ⟨ ⟩

=

⏐ ⏐ ⏐ ⏐ ⏐ ⏐ ⏐φ ψ φ ψ φ ψ21 2

2125

3 4

1255

3 4 3 41 2 1 2( )( * *⟨ ⟩ − ⟨ ⟩ ⟨ ⟩ + ⟨ ⟩φ ψ φ ψ φ ψ φ ψn n n ni i⏐ ⏐ ⏐ ⏐ ))

[= ⟨ ⟩ + ⟨ ⟩ + ⟨ ⟩ ⟨125

9 16 1212

22

1⏐ ⏐ ⏐ ⏐ ⏐ ⏐ ⏐φ ψ φ ψ φ ψn n ni φφ ψ φ ψ φ ψn n n

n

i

P

⏐ ⏐ ⏐2 2 1

1

12

125

9 16

⟩ − ⟨ ⟩ ⟨ ⟩

= +

* *]

{ ( )α PP in n n2 1 22 12( ) [ ]}*α + ⟨ ⟩ ⟨ ⟩Re φ ψ φ ψ⏐ ⏐

(4.4.3)

4.5. Consider postulate IV in Sec. 4.2, and generalize for the case of a continuous spectrum.

SOLUTION

Consider a physical observable A. Suppose that the system is in a normalized state ⏐ ψ ⟩; ⟨ψ ⏐ ψ ⟩ = 1. Let ⏐vαβ ⟩

form an orthonormal basis of the state space consisting of eigenvectors of A:

A⏐ ⏐v vαβ

αβα⟩ = ⟩ (4.5.1)

The index b distinguishes between eigenvectors corresponding to the same degenerate eigenvalue a of ˆ.A This index can be either discrete or continuous, and we assume that it is continuous and varies in the domain B(a). Since the spectrum of A is continuous, it is meaningless to speak about the probability of obtaining an eigenvalue a. Alternatively, we should speak about the differential probability dP(a) of obtaining a result between a and a + da. We then have an analogy to postulate IV for the discrete case,

dPd

B

d( )

( )

α β

ααα

β= ⟨ ⟩

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪∫ ⏐ ⏐ ⏐v ψ 2

(4.5.2)


4.6. Consider postulate IV for the case of a discrete spectrum. Show that an equivalent form for the probability of obtaining the eigenvalue an of the operator A is

P a P Pn n n( ) ˆ ˆ†= ⟨ ⟩ψ ψ⏐ ⏐ (4.6.1)

where Pn is the projector onto the eigensubspace of Â associated with an.

SOLUTION

Assume that ⏐ ⏐u un n1 2⟩ ⟩, , . . . , and ⏐un

gn ⟩ form an orthonormal basis of the eigensubspace associated with an.

By definition, Eq. (4.6) becomes,

P u un ni

i

g

ni

n

= ⟩ ⟨=

∑⏐ ⏐1

(4.6.2)

Therefore,

⟨ ⟩ = ⟨ ⟩ ⟨ ⟩ ⟨ ⟩=

∑ψ ψ ψ ψ⏐ ⏐ ⏐ ⏐ ⏐ˆ ˆ†P P u u u un n ni

ni

nj

nj

j

gn

1ii

g

ni

nj

ij

j

g

i

g

ni

n nn

u u

u

= ==∑ ∑∑= ⟨ ⟩ ⟨ ⟩

= ⟨

1 11

ψ ψ⏐ ⏐ δ

⏐⏐ψ⟩=

∑ 2

1j

gn

(4.6.3)

The two formulations are equivalent.

4.7. Consider two kets ⏐ ψ ⟩ and ⏐ ψ ′ ⟩ such that ⏐ ψ ′ ⟩ = eiq ⏐ ψ ⟩ where q is a real number. (a) Prove that if ⏐ ψ ⟩ is normalized, then so is ⏐ ψ ′ ⟩. (b) Demonstrate that the predicted probabilities for an arbitrary measurement are the same for ⏐ ψ ⟩ and ⏐ ψ ′ ⟩; therefore, ⏐ ψ ⟩ and ⏐ ψ ′ ⟩ represent the same physical state.

SOLUTION

(a) We assume that ⏐ ψ ⟩ is normalized, or ⟨ψ ⏐ ψ ⟩ = 1. Then

⟨ ′ ′⟩ = ⟨ ⟩ = ⟨ ⟩ =−ψ ψ ψ ψ ψ ψθ θ⏐ ⏐ ⏐ ⏐e ei i 1 (4.7.1)

(b) According to postulate IV (see Sec. 4.2), the probabilities predicted for a measurement depend on terms of the form ⏐ ⏐ ⏐⟨ ⟩un

i ψ 2 or ⏐ ⏐ ⏐⟨ ′⟩uni ψ 2. We have

⏐ ⏐ ⏐ ⏐ ⏐ ⏐⟨ ′⟩ ⟨ ′⟩ ⟨ ′⟩ ⟨ ⟩ −u u u e u eni

ni

ni i

ni iψ ψ ψ ψθ2 = =* θθ ψ ψ⟨ ⟩ = ⟨ ⟩u un

ini⏐ ⏐ ⏐ ⏐* 2 (4.7.2)

Therefore, the predicted probabilities for the states ⏐ ψ ⟩ and ⏐ ψ ′ ⟩ are the same.

4.8. Consider a large number of measurements of an observable performed on the system. Show that the mean value of an observable expresses the average of the results. Assume that the spectrum of the operator consists of both a discrete and a continuous part, but for simplicity, assume it to be nondegenerate.

SOLUTION

Consider first an eigenvalue an belonging to the discrete part of the spectrum. From a quantity of N mea-surements of A (the system being in the normalized state ⏐ ψ ⟩) the eigenvalue an will be obtained N(an) times with

N a

NP an

nn

( )( )⎛

⎝⎞⎠ →

→∞ (4.8.1)


where P(an) is the probability of obtaining an in a measurement. Similarly, if dN(α) expresses the number of experiments that yield a result between α and α + dα in the continuous part of the spectrum, we have

dNN

dPn

( ) ( )α α( ) →→∞

(4.8.2)

The average of the results of the N measurements is the sum of the values divided by N. It is therefore equal to

Average (NN

a N aN

dNn n

n

) ( ) ( )= + ∫∑1 1 α α (4.8.3)

For N → ∞, we obtain

Average (N a P a dPn n

n

→ ∞ = + ∫∑) ( ) ( )α α (4.8.4)

Suppose now that ⏐ un ⟩ for n = 1, 2, . . . , together with ⏐va ⟩, where a is a continuous index, form an ortho-normal basis of the state space consisting of eigenvalues of ˆ :A

ˆ Â u a u An n n⏐ ⏐ ⏐ ⏐⟩ = ⟩ ⟩ = ⟩v vα αα (4.8.5)

The closure relation of this basis is

⏐ ⏐ ⏐ ⏐u u dn n

n

⟩ ⟨ + ⟩ ⟨ =∫∑ v vα α α I (4.8.6)

So, using Eq. (4.8.4), we arrive at

Average ( +N a u d

a

n n

n

→ ∞ = ⟨ ⟩ ⟨ ⟩

=

∫∑) ⏐ ⏐ ⏐ ⏐ ⏐ ⏐ψ ψ2 2α ααv

nn

n

n nu u d∑ ∫⟨ ⟩ ⟨ ⟩ ⟨ ⟩ ⟨ ⟩ψ ψ ψ ψ⏐ ⏐ ⏐ ⏐+ α αα αv v

Using Eq. (4.8.5), we obtain

Average (N A u u An n→ ∞ = ⟨ ⟩ ⟨ ⟩ + ⟨ ⟩ ⟨) ˆ ˆψ ψ ψ⏐ ⏐ ⏐ ⏐ ⏐ ⏐v vα α ψψ

ψ

⟩

= ⟨ ⟩ ⟨ + ⟩ ⟨⎡

⎣

⎢⎢

⎤

⎦

⎥

∫∑

∫∑

d

A u u d

n

n n

n

α

αα α⏐ ⏐ ⏐ ⏐ ⏐ˆ v v⎥⎥

⟩⏐ ψ

Substituting the closure relation, we finally get

Average (N A→ ∞ = ⟨ ⟩) ˆψ ψ⏐ ⏐

4.9. Consider another formulation for the root-mean-square deviation of the operator A (in the normalized state ⏐ ψ ⟩):

Δ A A A= ⟨ − ⟨ ⟩ ⟩( ˆ ˆ )2 (4.9.1)

(a) Show that this definition is equivalent to that given in Eq. (4.9). (b) Use the formulation Eq. (4.9.1) to interpret the term root-mean-square deviation.

SOLUTION

(a) By the given definition we have

⟨ − ⟨ ⟩ ⟩ = ⟨ − ⟨ ⟩ ⟩( ˆ ˆ ) ( ˆ ˆ )A A A A2 2ψ ψ⏐ ⏐ (4.9.2)


Note that in this equation the term ⟨ ⟩A is actually a shortened form of ⟨ ⟩ˆ Â I , where I is the identity operator;⟨ ⟩A is a scalar. Hence,

⟨ − ⟨ ⟩ ⟩ = ⟨ − ⟨ ⟩ + ⟨ ⟩ ⟩

=

ψ ψ ψ ψ⏐ ⏐ ⏐ ⏐( ˆ ˆ ) ( ˆ ˆ ˆ ˆ )A A A A A A2 2 22

⟨⟨ ⟩ − ⟨ ⟩ ⟨ ⟩ + ⟨ ⟩ ⟨ ⟩ψ ψ ψ ψ ψ ψ⏐ ⏐ ⏐ ⏐ ⏐ˆ ˆ ˆ Â A A A2 22 (4.9.3)

Using the known definition of mean value, we have

⟨ ⟩ − ⟨ ⟩ ⟨ ⟩ + ⟨ ⟩ = ⟨ ⟩ − ⟨ ⟩ˆ ˆ ˆ ˆ ˆ Â A A A A A2 2 2 22 (4.9.4)

So, the two definitions coincide.

(b) The root-mean-square deviation expresses the average of the square of the deviations of A from its mean value ⟨ ⟩ˆ .A It therefore characterizes the dispersion of the measurement results about ⟨ ⟩A . For example, if the spectrum of A is continuous and the probability has a Gaussian shape, then ⟨ ⟩A characterizes the peak of the curve (the value of maximal probability), and ΔA characterizes the width of the Gaussian curve.

4.10. Prove that for the operators ˆ , ˆ ,A B and ˆ,C the following identities are valid:

(a) [ ˆ , ˆ] [ ˆ , ˆ]B A A B= −

(b) [ ˆ ˆ, ˆ] [ ˆ , ˆ] [ ˆ, ˆ]A B C A C B C+ = +

(c) [ ˆ , ˆ ˆ] [ ˆ , ˆ] ˆ ˆ [ ˆ , ˆ]A BC A B C B A C= +

SOLUTION

(a) By definition,

[ ˆ , ˆ] ˆ ˆ ˆ ˆ ( ˆ ˆ ˆ ˆ) [ ˆ , ˆ]B A BA AB AB BA A B= − = − − = − (4.10.1)

(b) By definition,

[ ˆ ˆ , ˆ] ( ˆ ˆ) ˆ ˆ( ˆ ˆ) ˆ ˆ ˆ ˆ Â B C A B C C A B AC BC C+ = + − + = + − ˆ ˆ ˆ

( ˆ ˆ ˆ ˆ) ( ˆ ˆ ˆ ˆ) [ ˆ , ˆ] [ ˆ

A CB

AC CA BC CB A C

−

= − + − = + BB C, ˆ ] (4.10.2)

(c) We write

[ ˆ , ˆ ˆ] ˆ( ˆ ˆ) ( ˆ ˆ) ˆ ( ˆ ˆ ˆ ˆ ˆ Â BC A BC BC A ABC BAC= − = − )) ( ˆ ˆ ˆ ˆ ˆ ˆ) [ ˆ , ˆ] ˆ ˆ [ ˆ , ˆ]+ − = +BAC BCA A B C B A C (4.10.3)

4.11. Suppose the operators A and B commute with their commutator, i.e., [ ˆ , [ ˆ , ˆ]] [ ˆ , [ ˆ , ˆ]] .B A B A A B= = 0 Show that (a) [ ˆ , ˆ ] ˆ [ ˆ , ˆ];A B nB A Bn n= −1 (b) [ ˆ , ˆ] ˆ [ ˆ , ˆ].A B nA A Bn n= −1

SOLUTION

(a) Consider the following procedure:

[ ˆ , ˆ ] ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ Â B AB B A ABB BAB Bn n n n n= + = − +− −1 1 (( ˆ ˆ) ˆ ˆ( ˆ ˆ) ˆ ˆ ˆ ˆ ÂB B B BA B B AB Bn n n− − −− + + −⋅ ⋅ ⋅2 3 1 nn

n n

BA

A B B B A B B B

−

− −= + + +⋅ ⋅ ⋅

1

1 2

ˆ ˆ

[ ˆ , ˆ] ˆ ˆ[ ˆ , ˆ] ˆ ˆ nn A B−1[ ˆ , ˆ] (4.11.1)

Using the fact that B commutes with [ ˆ , ˆ],A B we obtain

[ ˆ , ˆ ] ˆ [ ˆ , ˆ] ˆ [ ˆ , ˆ] Â B B A B B A B Bn n n n= + + +− − −⋅ ⋅ ⋅1 1 11 1[ ˆ , ˆ] ˆ [ ˆ , ˆ]A B nB A Bn= − (4.11.2)

(b) According to Problem 4.10, part (a), [ ˆ , ˆ] [ ˆ , ˆ ].A B B An n= − Using part (a) of the solution, we obtain

[ ˆ , ˆ] ˆ [ ˆ , ˆ] ˆ [ ˆ , ˆ]A B nA B A nA A Bn n n= − =− −1 1 (4.11.3)


4.12. Consider the operators A and B presented in Problem 4.11. Prove that (a) for every analytic function F(x)

we have [ ˆ , ( ˆ)] [ ˆ , ˆ] ( ˆ),A F B A B F B= ′ where F ′(x) denotes the derivative of F(x). (b) e e e eA B A B A Bˆ ˆ ˆ ˆ [ ˆ , ˆ ] .= + /2

SOLUTION

(a) First we prove, using induction, that for every n = 1, 2, . . . we have

[ ˆ , ˆ ] [ ˆ , ˆ] Â B n A B Bn n= −1 (4.12.1)

Proof: For n = 1, Eq. (4.12.1) is clearly true. Suppose that this equation is verified for n. Then, using part (c) in Problem 4.10 for n + 1, we have

[ ˆ , ˆ ] [ ˆ , ˆ ˆ ] [ ˆ , ˆ] ˆ ˆ [ ˆ , ˆ ]A B A BB A B B B A Bn n n n+ = = +1 == + −[ ˆ , ˆ] ˆ ˆ [ ˆ , ˆ] Â B B Bn A B Bn n 1 (4.12.2)

B and [ ˆ , ˆ]A B commute, so we finally have

[ ˆ , ˆ ] [ ˆ , ˆ] ˆ [ ˆ , ˆ] ˆ ( )[ ˆ ,A B A B B n A B B n An n n+ = + = +1 1 ˆ] ˆB Bn (4.12.3)

Equation (4.12.1) is therefore established. Consider now the expansion of the function F(x) in a power

series, F x a xnn

n

( ) .= ∑ Using (4.12.1) we obtain

[ ˆ , ( ˆ)] ˆ , ˆ [ ˆ , ˆ ]A F B A a B a A Bnn

n

n

n

n=⎡

⎣

⎢⎢

⎤

⎦

⎥⎥

= =∑ ∑ [[ ˆ , ˆ] Â B na Bnn

n

−∑ 1 (4.12.4)

The power series expansion of the derivative of F(x) is ′ = −∑F x na xnn

n

( ) .1 Therefore, by inspection we can conclude that

[ ˆ , ( ˆ)] [ ˆ , ˆ] ( ˆ)A F B A B F B= ′ (4.12.5)

(b) Consider an operator ˆ ( )F s depending on the real parameter s:

ˆ ( )ˆ ˆ

F s e eAs Bs= (4.12.6)

The derivative of F with respect to s is

dFds

dds

e e edds

eAs Bs As Bsˆ

ˆˆ ˆ ˆ ˆ= ⎛

⎝⎜⎞⎠⎟ + ⎛

⎝⎜⎞⎠⎟ = AAe e e Be

Ae e e Be

As Bs As Bs

As Bs As A

ˆ ˆ ˆ ˆ

ˆ ˆ ˆ ˆ

ˆ

ˆ ˆ

+

= + − ss As Bs As Ase e A e Be F sˆ ˆ ˆ ˆ

( ˆ ˆ ) ˆ ( )= + − (4.12.7)

Using part (a), we can write

[ , ˆ] [ ˆ , ] [ ˆ , ˆ] [ ˆ , ˆ]ˆ ˆ ˆ

e B B e s B A e s A BAs As As= − = − = eeAsˆ (4.12.8)

Therefore, e B Be s A B eAs As Asˆ ˆ ˆˆ ˆ [ ˆ , ˆ]= + and e Be B s A BAs Asˆ ˆˆ ˆ [ ˆ , ˆ].− = + Substituting in Eq. (4.12.7) we obtain

dFds

A B s A B F sˆ

( ˆ ˆ [ ˆ , ˆ]) ˆ ( )= + + (4.12.9)

Since ˆ Â B+ and [ ˆ , ˆ]A B commute, we can integrate this differential equation. This yields

ˆ ( ) ˆ ( ) ( ˆ ˆ ) [ ˆ , ˆ ]F s F e A B s A B s+ + +02 2/ (4.12.10)

Setting s = 0, we obtain F e eA B( ) ˆ ˆ ˆˆ ˆ0 0 0= = =⋅ ⋅ I I I⋅ . Finally, substituting ˆ ( )F 0 and s = 1 in Eq. (4.12.10),

we obtain e e e eA B A B A Bˆ ˆ ˆ ˆ [ ˆ , ˆ ] .= + /2


4.13. Let ⟨ψ ⏐ be the corresponding bra of the ket ⏐ ψ ⟩. We designate by ⏐ ψ ′ ⟩ the result of the action of the operator A on ⏐ ψ ⟩, so ⏐ ⏐′⟩ = ⟩ψ ψˆ .A Let ⟨ψ ′ ⏐ be the bra corresponding to ⏐ ψ ′ ⟩. Prove that

⟨ ′ = ⟨ψ ψ⏐ ⏐ ˆ†A (4.13.1)

SOLUTION

Recall the basic definition of a bra as a functional acting on the state space. The two functionals ⟨ψ ′ ⏐ and ⟨ψ ⏐ ˆ†A are identical if their action on an arbitrary ket ⏐ φ ⟩ yields the same result; i.e., we have to show that

⟨ ′ ⟩ = ⟨ ⟩ψ φ ψ φ⏐ ⏐ ⏐ˆ†A (4.13.2)

Now, using Eq. (4.13) we have

⟨ ⟩ = ⟨ ⟩ = ⟨ ′⟩ψ φ φ ψ φ ψ⏐ ⏐ ⏐ ⏐ ⏐ˆ ˆ† * *A A (4.13.3)

and according to the basic property of the scalar product [see Eq. (4.1)], we have

⟨ ⟩ = ⟨ ′ ⟩ψ φ ψ φ⏐ ⏐ ⏐ˆ†A (4.13.4)

4.14. Given that ( ˆ) ˆ† * †aA a A= and ( ˆ ) ˆ† †A A= and ( ˆ ˆ) ˆ ˆ† † †A B A B+ = + and ( ˆ ˆ ˆ) ˆ ˆ ˆ ,† † † †ABC C B A= where a is a complex number and ˆ , ˆ ,A B and C are operators, determine the Hermitian conjugate of ( ˆ ˆ ).4 2 6 2+ +i A B Once you finish this problem go on to study Problem 4.17.

SOLUTION

Call ( ˆ ˆ )4 2 6 2+ +i A B the operator ˆ.C Then

ˆ ( ˆ ˆ ) ( ) ( ˆ) ( ˆ )† † † † †C i A B i A B= + + = + +4 2 6 4 2 62 2 (4.14.1)

Since ( )†4 4= and ( ˆ) ˆ† †2 2i A i A= − and ( ˆ ) ( ˆ ) ˆ ˆ ( ˆ )† † † † †6 6 6 62 2 2B B B B B= = = , it follows that

ˆ ˆ ( ˆ )† † †C iA B= − +4 2 6 2 (4.14.2)

4.15. Given operators A and B where ˆ ˆ ˆ.†A B B= + Show that A is Hermitian, independent of whether B is or is not Hermitian.

SOLUTION

ˆ ˆ ( ˆ ˆ) ( ˆ ) ˆ† † † † † †A A B B B B= = + = + (4.15.1)

From Problem 2.6 ( ˆ ) ˆ† †A A= , hence ˆ ˆ ˆ ˆ† †A B B A= + = regardless of whether B is Hermitian or not.

4.16. An operator A is said to be skew-Hermitian or anti-Hermitian if ˆ ˆ.†A A= − Show that the operator i B B( ˆ ˆ )†+ is anti-Hermitian. See Sec. 2.3.

SOLUTIONˆ ( ˆ ˆ )†A i B B= + , hence ˆ [ ( ˆ ˆ )]† † †A i B B= + but in general, ( ˆ) ˆ† * †bB b B= where b is a complex number. Therefore,

ˆ ( ) ( ˆ ˆ) ( ˆ ˆ)† * † †A i B B i B B= + = − + (4.16.1)

and

ˆ ˆ†A A= −

4.17. Derive the following properties of the adjoint of an operator:

(a) ( ˆ ) ˆ† †A A=

(b) ( ˆ) ˆ ,† * †λ λA A= where l is a complex number


(c) ( ˆ ˆ) ˆ ˆ† † †A B A B+ = +

(d) ( ˆ ˆ) ˆ ˆ† † †AB B A=

SOLUTION

First, recall that two operators are identical if their matrix elements in a basis of the state space are the same. Therefore, if for arbitrary ⏐ φ ⟩ and ⏐ ψ ⟩ we have ⟨ ⟩ = ⟨ ⟩φ ψ φ ψ⏐ ⏐ ⏐ ⏐ˆ ˆ ,A A1 2 then A

1 and A2

are identical. In the following derivations we also use some of the basic properties of conjugation of complex numbers that were given in Chap. 2.

(a) Using Eq. (4.13) we have

⟨ ⟩ = ⟨ ⟩ψ φ φ ψ⏐ ⏐ ⏐ ⏐( ˆ ) ˆ† † † *A A (4.17.1)

and using Eq. (4.13) again, we have

⟨ ⟩ = ⟨ ⟩φ ψ ψ φ⏐ ⏐ ⏐ ⏐ˆ ˆ† *A A (4.17.2)

Therefore,

⟨ ⟩ = ⟨ ⟩ = ⟨ ⟩ = ⟨ψ φ φ ψ ψ φ ψ⏐ ⏐ ⏐ ⏐ ⏐ ⏐ ⏐ ⏐( ˆ ) ˆ ( ˆ ) ˆ† † † * * *A A A A φφ⟩ (4.17.3)

(b) We write

⟨ ⟩ = ⟨ ⟩ = ⟨ ⟩

= ⟨

ψ φ φ ψ φ ψ

φ

⏐ ⏐ ⏐ ⏐ ⏐ ⏐

⏐

( ˆ) ˆ [ ˆ ]

ˆ

† * *

*

λ λ λ

λ

A A A

AA A A⏐ ⏐ ⏐ ⏐ ⏐ψ ψ φ ψ φ⟩ = ⟨ ⟩ = ⟨ ⟩* * † * †ˆ ˆλ λ (4.17.4)

(c) We write

⟨ + ⟩ = ⟨ + ⟩ = ⟨ ⟩ + ⟨ψ φ φ ψ φ ψ φ⏐ ⏐ ⏐ ⏐ ⏐ ⏐ ⏐( ˆ ˆ) ˆ ˆ) [ ˆ ˆ† *A B A B A BB

A B A

⏐

⏐ ⏐ ⏐ ⏐ ⏐ ⏐

ψ

φ ψ φ ψ ψ φ ψ

⟩

= ⟨ ⟩ + ⟨ ⟩ = ⟨ ⟩ + ⟨

]

ˆ ˆ ˆ

*

* * † ⏐⏐ ⏐ ⏐ ⏐ˆ ( ˆ ˆ )† † †B A Bφ ψ φ⟩ = ⟨ + ⟩ (4.17.5)

(d) Let us define ⏐ ⏐χ⟩ = ⟩ˆ .B ψ Using the results of Problem 4.13, we have ⟨ = ⟨χ ⏐ ⏐ψ ˆ .†B Now,

⟨ ⟩ = ⟨ ⟩ ≡ ⟨ ⟩ = ⟨ψ φ φ ψ φ⏐ ⏐ ⏐ ⏐ ⏐ ⏐ ⏐ ⏐( ˆ ˆ) ˆ ˆ ˆ ˆ† * * †AB AB A Aχ χ φφ ψ φ⟩ = ⟨ ⟩⏐ ⏐ˆ ˆ† †B A (4.17.6)

4.18. Consider a Hermitian operator A that has the property ˆ ˆ.A I3 = Show that ˆ ˆ.A I=

SOLUTION

First we find the possible eigenvalues of ˆ.A Suppose ˆ ,A ⏐ ⏐ψ ψ⟩ = ⟩α so we have

⏐ ⏐ ⏐ ⏐ ⏐ ⏐ψ ψ ψ ψ ψ ψ⟩ = ⟩ = ⟩ = ⟩ = ⟩ = ⟩ˆ ˆ ( ) Â A A A3 2 2 2 3α α α α (4.18.1)

Therefore, a3 = 1. The possible values of a are then

α = − + − −12

32

12

32

1i i, , (4.18.2)

Since A is Hermitian its eigenvalues are real; therefore, the only possible eigenvalue of A is a = 1. We can choose an orthonormal basis of the state space consisting of eigenvalues of ˆ ,A so ˆ .A u u

i i⏐ ⏐⟩ = ⟩ Every state

⏐ φ⟩ can be expanded as

⏐ ⏐ ⏐ ⏐φ φ⟩ = ⟩ ⟩ = ⟩∫u u dsi sor if the basis has a ccontinuous index⎛⎝⎜

⎞⎠⎟∑

i

(4.18.3)

Finally,

ˆ ˆ Â A u A u ui i i

iii

⏐ ⏐ ⏐ ⏐ ⏐φ φ⟩ = ⟩ = ⟩ = ⟩ = ⟩∑∑∑ (4.18.4)

which implies ˆ ˆ.A I=


4.19. Prove that if an orthonormal discrete set of kets { , , , . . .}⏐u ii ⟩ = 1 2 constitutes a basis, then it follows that

⏐ ⏐u u Ii i

i

⟩ ⟨ =∑ ˆ (4.19.1)

SOLUTION

Let ⏐ψ⟩ be an arbitrary ket belonging to the state space. Since { }⏐ui ⟩ is a basis, there exists, by definition, a

unique expansion ⏐ ⏐ψ⟩ = ⟩∑C ui i

i

. We use the orthonormalization relation Eq. (4.16) to obtain

⟨ ⟩ = ⟨ ⟩ = =∑∑u C u u C Cj i j i i ij j

jj

⏐ ⏐ψ δ (4.19.2)

So,

⏐ ⏐ ⏐ ⏐ ⏐ ⏐ ⏐ψ ψ ψ⟩ = ⟩ = ⟨ ⟩ ⟩ = ⟩ ⟨⎡

⎣⎢⎢

⎤

⎦⎥⎥

⟩∑C u u u u ui i i i i i

iiii∑∑ (4.19.3)

Note that since ⟨ ⟩ui ⏐ ψ is a scalar we could change the place of this expression. We see that for any ket ⏐ ψ⟩

the action of the operator ˆ({ })P u u ui i i

i

⏐ ⏐ ⏐⟩ = ⟩ ⟨∑ on that ket yields the same ket ⏐ ψ⟩. Therefore, it is, by

definition, the identity operator, ˆ({ }) ˆ.P u Ii

⏐ ⟩ =

4.20. Show that if the closure relation is valid for an orthonormal continuous set { },⏐wα ⟩ then this set constitutes a basis.

Let ⏐ ψ ⟩ be an arbitrary ket belonging to the state space. Using the closure relation we have

⏐ ⏐ ⏐ ⏐ψ ψ ψ⟩ = ⟩ = ⟩ ⟨ ⟩∫I w w dα α α (4.20.1)

SOLUTION

Defining C(a) = ⟨wa ⏐ ψ ⟩ we have ⏐ ⏐ψ⟩ = ⟩∫C w d( ) .α αα We see that any ket ⏐ψ ⟩ has an expansion on the

⏐ wa ⟩. To show that this expansion is unique we assume that we have two expansions:

⏐ ⏐ ⏐ ⏐ψ ψ⟩ = ⟩ ⟩ = ′ ⟩∫∫C w d C w d( ) ( )α α α αα α (4.20.2)

Subtracting we obtain

[ ( ) ( )]C C w dα α αα− ′ ⟩ =∫ ⏐ 0 (4.20.3)

Applying ⟨ ′wα ⏐ on this ket, [ ( ) ( )]C C w w dα α αα α− ′ ⟨ ⟩ =′∫ ⏐ 0 and using the orthonormalization relation we obtain

[ ( ) ( )] ( )C C dα α δ α α α− ′ ′ − =∫ 0 (4.20.4)

Equation (4.20.4) is valid only if C C( ) ( ) .′ − ′ ′ =α α 0 Therefore, for any a ′ we have C C( ) ( ),′ = ′ ′α α and the expansion of any ket ⏐ ψ ⟩ on {⏐ wa ⟩} is unique.

4.21. Suppose that in a certain basis {⏐ ui ⟩}, the operators A and B are represented by the matrices (Ai j) and (Bi j), respectively; the ket ⏐ ψ ⟩ is represented by ci; and the bra ⟨φ⏐ by bi

*.

(a) Obtain the matrix representation of the operator ˆ ˆ.AB

(b) Find the representation of the ket ˆ .A ⏐ ψ⟩

(c) Obtain an expression for the scalar ⟨ ⟩φ ψ⏐ ⏐A in terms of the various representations.


SOLUTION

(a) Consider the matrix element of ˆ ˆ :AB

( ˆ ˆ) ˆ ˆ ˆ ˆ ÂB u AB u u AIB uij i j i j= ⟨ ⟩ = ⟨ ⟩⏐ ⏐ ⏐ ⏐ (4.21.1)

Using the closure relation we obtain

( ˆ ˆ) ˆ ÂB u A u u B u A Bij i k k j i k k j

kk

= ⟨ ⟩ ⟨ ⟩ = ∑∑ ⏐ ⏐ ⏐ ⏐ (4.21.2)

(b) By definition, the ket A ⏐ ψ⟩ is represented by the numbers ′ = ⟨ ⟩c u Ai i ⏐ ⏐ˆ .ψ Using the closure relation between A and ⏐ ψ ⟩, we can write

′ = ⟨ ⟩ = ⟨ ⟩ ⟨ ⟩ = ∑∑c u AI u A u u A ci i i j j i j j

jj

⏐ ⏐ ⏐ ⏐ ⏐ˆ ˆ ˆψ ψ (4.21.3)

and in matrix form,

′′

⎛

⎝

⎜⎜⎜

⎞

⎠

⎟⎟⎟

=

⎛

⎝

⎜⎜

⋅ ⋅ ⋅⋅ ⋅ ⋅

c

c

A A

A A1

2

11 12

21 22

::

::

⎜⎜

⎞

⎠

⎟⎟⎟

⎛

⎝

⎜⎜⎜

⎞

⎠

⎟⎟⎟

c

c1

2

::

(4.21.4)

(c) We write

⟨ ⟩ = ⟨ ⟩ ⟨ ⟩ ⟨ ⟩ =φ ψ φ ψ⏐ ⏐ ⏐ ⏐ ⏐ ⏐ˆ ˆ *

,

A u u A u u b A ci i j j i i j j

i j∑∑∑

i j,

(4.21.5)

or in matrix form,

⟨ ⟩ =

⋅ ⋅ ⋅ ⋅ ⋅ ⋅

φ ψ⏐ ⏐ˆ ( . . . . . .)* * *A b b b

A A A

A

i

j

1 2

11 12 1

211 22

1 2

2A

A A A

c

c

ci i ij

i

j

:

:

⋅ ⋅ ⋅

⎛

⎝

⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟

⎛

⎝

⎜

�⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟

(4.21.6)

4.22. Suppose that ⏐φn ⟩, where n = 1, 2, . . . , form an orthonormal basis for the state space of a physical system. Let A be an operator with matrix elements A Amn m n= ⟨ ⟩φ φ⏐ ⏐ˆ . Show that the operator A can be written as

ˆ

,

A Amn m n

m n

= ⟩ ⟨=

∞

∑ ⏐ ⏐φ φ1

(4.22.1)

SOLUTION

Recall that two operators are identical if and only if their matrix elements in a certain basis are identical. Therefore, we write the matrix elements of the expression in Eq. (4.22.1) as

⟨ ⟩ ⟨⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

⟩ = ⟨=

∞

∑φ φ φ φ φ φk mn m n

m n

l k mA⏐ ⏐ ⏐ ⏐ ⏐, 1

⟩⟩ ⟨ ⟩ ⟨ ⟩

= ⟨ ⟩

=

∞

∑ φ φ φ φ

φ φ

m n n l

m n

km m n nl

A

A

⏐ ⏐ ⏐

⏐ ⏐

ˆ

ˆ

, 1

δ δ == ⟨ ⟩=

∞

∑ φ φk

m n

A⏐ ⏐ˆ

,

1

1

(4.22.2)

where we used the orthonormalization relations ⟨ ⟩ =φ φi j i j⏐ δ .


4.23. Consider a two-dimensional physical system. The kets ⏐ ψ1 ⟩ and ⏐ ψ2 ⟩ form an orthonormal basis of the state space. We define a new basis ⏐ φ1 ⟩ and ⏐ φ2 ⟩ by

⏐ ⏐ ⏐ ⏐ ⏐ ⏐φ ψ ψ φ ψ ψ1 1 2 2 1 2

1

2

1

2⟩ = ⟩ + ⟩ ⟩ = ⟩ − ⟩( ) ( ) (4.23.1)

An operator P is represented in the ⏐ ψi ⟩-basis by the matrix

( )aij = ⎛⎝⎜

⎞⎠⎟

11ε

ε

Find the representation of P in the basis ⏐ φi ⟩, i.e., find the matrix �a Pij i j= ⟨ ⟩φ φ⏐ ⏐ˆ .

SOLUTION

Method 1: We define the transformation matrix Tij i j= ⟨ ⟩ψ ϕ⏐ . We calculate its elements; for example,

T11 1 1 1 1 2

12

12

1 012

= ⟨ ⟩ = ⟨ ⟩ + ⟩ = + =ψ φ ψ ψ ψ⏐ ⏐ ⏐ ⏐( ) ( )

and

T22 2 2 2 1 2

12

12

0 112

= ⟨ ⟩ = ⟨ ⟩ − ⟩ = − = −ψ φ ψ ψ ψ⏐ ⏐ ⏐ ⏐( ) ( )

and so on. Then we find

T = −⎛⎝⎜

⎞⎠⎟

12

1 11 1

(4.23.2)

The adjoint matrix is ˆ .†T = −⎛⎝⎜

⎞⎠⎟

12

1 11 1

Using the closure relation ⏐ ⏐ψ ψi i

i

I⟩ ⟨ ==

∑1

2

ˆ, we obtain

�a P Pkl k l k i i j j l

i j

= ⟨ ⟩ = ⟨ ⟩ ⟨ ⟩ ⟨ ⟩φ φ φ ψ ψ ψ ψ φ⏐ ⏐ ⏐ ⏐ ⏐ ⏐ˆ ˆ

, == =∑ ∑=

1

2

1

2

ˆ ˆ†

,

T a Tki i j jl

i j

We can accomplish the calculation in matrix form:

( )�akl = −⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟ −

⎛⎝⎜

⎞⎠⎟

12

1 11 1

11

12

1 11 1

εε == −

⎛⎝⎜

⎞⎠⎟

+ −+ − +

⎛⎝⎜

⎞⎠⎟

=+

12

1 11 1

1 11 1

12

2 2 00

ε εε ε

ε22 2

1 00 1+

⎛⎝⎜

⎞⎠⎟ =

+−

⎛⎝⎜

⎞⎠⎟ε

εε (4.23.3)

Method 2: Observing that ⏐φi ⟩ are actually eigenvectors of P,

1

112

11

12

11

112

εε

εε ε⎛

⎝⎜⎞⎠⎟

⎛⎝⎜

⎞⎠⎟ =

++

⎛⎝⎜

⎞⎠⎟ = +( )

111

⎛⎝⎜

⎞⎠⎟

and

1

112

11

12

11

1ε

εεε ε⎛

⎝⎜⎞⎠⎟ −

⎛⎝⎜

⎞⎠⎟ =

−− +

⎛⎝⎜

⎞⎠⎟ = −( )

112

11−

⎛⎝⎜

⎞⎠⎟

Therefore,

ˆ ( ) ˆ ( )P P⏐ ⏐ ⏐ ⏐φ φ φ φ1 1 2 21 1⟩ = + ⟩ ⟩ = − ⟩ε ε


This implies that in the ⏐φi⟩-representation P is diagonal:

( )aij =+

−⎛⎝⎜

⎞⎠⎟

1 00 1

εε (4.23.4)

4.24. Refer to Problem 4.23 and obtain the representation of the ket eP ⏐ψ1⟩ in the ⏐ψ i ⟩-basis.

SOLUTION

Since P is diagonal in the basis, it is easier to work in this basis. Hence,

e e e eP Pˆ ˆ⏐ ⏐ ⏐ ⏐φ φ φ φ1

11 2

12⟩ = ⟩ ⟩ = ⟩+ −ε ε

so we obtain

e e e eP Pˆ ˆ⏐ ⏐ ⏐ ⏐ψ φ φ ψ1 1 2

11

12

12

12⟩ = ⟩ + ⟩⎛

⎝⎜⎞⎠⎟

= ⟩ ++ε 112

11

12

1 112

+ − −

+ −

⟩ + ⟩ − ⟩⎡⎣ ⎤⎦

= +

ε ε ε

ε ε

⏐ ⏐ ⏐ψ ψ ψe e

e e( )) ( )⏐ ⏐ψ ψ11 1

2⟩ + − ⟩⎡⎣ ⎤⎦+ −e eε ε

Therefore, eP ⏐ψ1⟩ is represented in the ⏐ ψ i ⟩-basis as

ee e e

e e

P ⏐ψ1 2⟩ =+−

⎛

⎝⎜⎞

⎠⎟−

−

ε ε

ε ε (4.24.1)

4.25. (a) Show that the ket ⏐ r⟩, where r = ( , , ),x y z is an eigenvector of the operator observable X with an eigenvalue x. (b) Show that ⏐ p⟩, where p = ( , , ),p p px y z is an eigenvector of Px with an eigenvalue px .

SOLUTION

(a) Using the r-representation we have ⟨ ′ ⟩ = ′⟨ ′ ⟩r r r r⏐ ⏐ ⏐ˆ .X x Substituting the representation for ⟨ ′ ⟩r r⏐ we obtain

⟨ ′ ⟩ = ′ ′ − = ′ −r r r r r r⏐ ⏐ˆ ( ) ( )X x xδ δ (4.25.1)

where ′ = ′ ′ ′r ( , , ).x y z Therefore, we have ⟨ ′ ⟩ = ⟨ ′ ⟩r r r r⏐ ⏐ ⏐ˆ .X x Since this holds for all ′r we have

X x⏐ ⏐ ⏐r r= ⟩ (4.25.2)

(b) In the p-representation, we apply the same method as in part (a), so

⟨ ′ ⟩ = ′ ⟨ ′ ⟩ = ′ ′ − = ′ − =p p p p p p p p⏐ ⏐ ⏐ˆ ( ) ( )P p p p px x x xδ δ ⟨⟨ ′ ⟩p p⏐ (4.25.3)

Therefore, ˆ .P px⏐ ⏐p p⟩ = ⟩ In conclusion, since analogous arguments can be applied to the y- and z-components, one can write

ˆ

ˆ

ˆ

ˆX x

Y y

Z z

Px⏐ ⏐

⏐ ⏐

⏐ ⏐

⏐r r

r r

r r

r⟩ = ⟩

⟩ = ⟩

⟩ = ⟩

⎧

⎨⎪⎪

⎩⎪⎪

⟩ == ⟩

⟩ = ⟩

⟩ = ⟩

⎧

⎨⎪⎪

⎩⎪⎪

p

P p

P p

x

y y

z z

⏐

⏐ ⏐

⏐ ⏐

r

r r

r r

ˆ

ˆ

(4.25.4)

4.26. Prove that ⟨ ⟩ = ∇⟨ ⟩r P r⏐ ⏐ ⏐ˆ .ψ ψ�i (b) Write an expression for ⟨ ⟩φ ψ⏐ ⏐px using the wavefunctions

corresponding to ⏐φ⟩ and ⏐ψ⟩.

SOLUTION

(a) Consider, for example, the x-component (the y- and z-components can be treated in a completely analo-gous manner). We have

⟨ ⟩ = ⟨ ⟩ ⟨ ⟩∫r r p p⏐ ⏐ ⏐ ⏐ ⏐ˆ ˆP p d px xψ ψ 3 (4.26.1)


where we use the closure relation of the p-representation. Using Eqs. (4.41) and (4.47) in Sec. 4.8 we obtain

⟨ ⟩ = ∫r pp r⏐ ⏐ˆ( )

ˆ ( )//P e p d px

ixψ

πψ1

2 3 23

�

�i � (4.26.2)

This expression is the Fourier transform of ˆ ( ),px�ψ p which is

�i x

∂∂ψ( )

.r

We therefore have

⟨ ⟩ = ∂∂r r⏐ ⏐ˆ ( )p

i xx ψ ψ� (4.26.3)

(b) Suppose that φ( )r and ψ( )r are the wavefunctions corresponding, respectively, to ⏐ φ⟩ and ⏐ ψ⟩; so,

φ φ ψ ψ( ) ( )r r r r= ⟨ ⟩ = ⟨ ⟩⏐ ⏐ (4.26.4)

Using the closure relation of the r-representation together with the result of part (a) we obtain

⟨ ⟩ = ⟨ ⟩ ⟨ ⟩ = ∂∂∫ ∫φ ψ φ ψ φ ψ

⏐ ⏐ ⏐ ⏐ ⏐ˆ ˆ ( )( )*p p d r

ix xr r rr3 �

xxd r3 (4.26.5)

4.27. Show

(a) that [ ˆ, ˆ]x y = 0

(b) that [ ˆ , ˆ ]p px y = 0

(c) that [ ˆ, ˆ ]x p ix = �

(d ) that [ ˆ, ˆ ]x py = 0

SOLUTION

(a) Using the r-representation we obtain the action of [ ˆ, ˆ]x y on an arbitrary ket ⏐ψ⟩:

⟨ ⟩ = ⟨ ⟩ − ⟨ ⟩r r r⏐ ⏐ ⏐ ⏐ ⏐ ⏐[ ˆ, ˆ] ˆˆ ˆ ˆx y xy yxψ ψ ψ

Using Eq. (4.46) in Sec. 4.8, we arrive at ⟨ ⟩ = ⟨ ⟩ − ⟨ ⟩r r r⏐ ⏐ ⏐ ⏐ ⏐ ⏐[ ˆ, ˆ] ˆ ˆ .x y x y y xψ ψ ψ So,

⟨ ⟩ = ⟨ ⟩ − ⟨ ⟩ =r r r⏐ ⏐ ⏐ ⏐ˆ, ˆx y xy yxψ ψ ψ 0 (4.27.1)

Since this is valid for any ⟨r ⏐ and arbitrary ⏐ψ⟩, we have [ ˆ, ˆ] .x y = 0

(b) We apply the same method in the p-representation:

⟨ ⟩ = ⟨ ⟩ − ⟨ ⟩

=

p p p⏐ ⏐ ⏐ ⏐ ⏐ ⏐[ ˆ , ˆ ] ˆ ˆ ˆ ˆp p p p p p

p

x y x y y xψ ψ ψ

xx y y x x y y xp p p p p p p⟨ ⟩ − ⟨ ⟩ = ⟨ ⟩ − ⟨ ⟩p p p p⏐ ⏐ ⏐ ⏐ ⏐ ⏐ˆ ˆψ ψ ψ ψ == 0 (4.27.2)

(c) We write ⟨ ⟩ = ⟨ ⟩ − ⟨ ⟩r r r⏐ ⏐ ⏐ ⏐ ⏐ ⏐[ ˆ, ˆ ] ˆˆ ˆ ˆ ;x p xp p xx x xψ ψ ψ so,

⟨ ⟩ = ⟨ ⟩ − ∂∂ ⟨ ⟩ =r r r⏐ ⏐ ⏐ ⏐ ⏐ ⏐[ ˆ, ˆ ] ˆ ˆx p x p

i xx

ixx xψ ψ ψ� � ∂∂

∂ ⟨ ⟩ − ∂∂ ⟨ ⟩

x i xxr r⏐ ⏐ψ ψ�

( ) (4.27.3)

If ψ( )r is the wavefunction corresponding to ⏐ψ⟩, we have

⟨ ⟩ = ∂∂ − ∂

∂⎡⎣⎢

⎤⎦⎥

=rr r

⏐ ⏐[ ˆ, ˆ ]( ) ( ( ))

x pi

xx

xxx ψ ψ ψ� ��

� �i

xx

xx

i i∂

∂ − − ∂∂

⎡⎣⎢

⎤⎦⎥

= = ⟨ψ ψ ψ ψ( )( )

( )( )

rr

rr r ⏐⏐ψ⟩

Since the calculation is valid for all ⏐ψ⟩ and for any ⏐ r⟩, we obtain [ ˆ, ˆ ] .x p ix = �


(d) Again applying the method used in part (c), we obtain

⟨ ⟩ = ⟨ ⟩ − ∂∂ ⟨ ⟩

=

r r r⏐ ⏐ ⏐ ⏐ ⏐ ⏐[ ˆ, ˆ ] ˆ ˆx p x pi y

x

ix

y yψ ψ ψ�

� ∂∂∂ − ∂

∂ = ∂∂ − ∂

∂y i yx

ix

yx

yψ ψ ψ ψ ψ

( )( )

( )( ) ( )

rr

rr r� � ⎡⎡

⎣⎢⎤⎦⎥

= 0 (4.27.4)

4.28. Consider the following operators:

ˆ ( ) ( ) ˆ ( )( )

O x x x O x xd x

dx13

2ψ ψ ψ ψ= = (4.28.1)

Find the commutation relation [ ˆ , ˆ ].O O1 2

SOLUTION

Method 1: Substituting the operators O1 and O2 in the commutation relation we obtain

[ ˆ , ˆ ] ˆ ( ˆ ( )) ˆ ( ˆ ( ))(

O O O O x O O x x xd

1 2 1 2 2 13ψ ψ ψ ψ= − = xx

dxx

ddx

x x

xd x

dxx x x

)[ ( )]

( )( )

⎡⎣⎢

⎤⎦⎥

−

= −

3

4 23

ψ

ψ ψ ++⎡⎣⎢

⎤⎦⎥

= −xd x

dxx x3 33

ψ ψ( )( ) (4.28.2)

Method 2: According to the action of x and p in the x-representation, we have ˆ Ô x13= and ˆ ˆ Ô ix p2 = /�.

Therefore,

[ ˆ , ˆ ] [ ˆ , ˆ ˆ]O Oi

x xp1 23=

� (4.28.3)

and using Problem 4.12

[ ˆ , ˆ ] [ ˆ, ˆ ˆ] ([ ˆ, ˆ] [ Ô Oi

x x xpi

x x x p x1 22 23 3= = +

� �xx p x, ˆ]) = −3 3 (4.28.4)

Or equivalently, [ ˆ , ˆ ] ( ) ( ).O O x x x1 233ψ ψ= −

4.29. The angular momentum is defined by L r p= × (for example, L yp zpx z y= − ). Use the commutation relations between R and P and the properties of the commutator derived in Problem 4.10 to find the following commutation relations: (a) [ ˆ , ˆ ];L Lx y (b) [ ˆ , ˆ ]L Ly x

2 and [ ˆ , ˆ ];L Lz x2 (c) [ ˆ , ˆ ].L Lx

2

SOLUTION

(a) By definition,

[ , ] [ ˆ ˆ ˆ ˆ , ˆ ˆ ˆ ˆ ] [ ˆ ˆ , ˆL L YP ZP ZP XP YP Zx y z y x z z= − − = ˆ ] [ ˆ ˆ , ˆ ˆ ]P ZP XPx y z+ (4.29.1)

where we used the fact that ˆ ˆYPz commutes with ˆ ˆXPz and ˆ ˆZPy commutes with ˆ ˆ .ZPx Using the relation derived in Problem 4.4, part (c), we then have

ˆ[ ˆ , ˆ ] ˆ ˆ[ ˆ , ˆ ] ˆ ˆ ˆ ˆ ˆY P Z P X Z P P i YP i XPz x z y x+ = − +� � yy zi L= � ˆ (4.29.2)

(b) We write

[ ˆ , ˆ ] ˆ [ ˆ , ˆ ] [ ˆ , ˆ ] ˆ ˆL L L L L L L L i Ly x y y x y x y y2 = + = − � ˆ ˆ ˆL i L Lz z y− � (4.29.3)

Similarly,

[ ˆ , ˆ ] ˆ [ ˆ , ˆ ] [ ˆ , ˆ ] ˆ ˆ ˆL L L L L L L L i Lz x z z x z x z z2 = + = � LL i L Ly y z+ � ˆ ˆ (4.29.4)


(c) We write

[ ˆ , ˆ ] [ ˆ , ˆ ] [ ˆ , ˆ ] [ ˆ , ˆ ]L L L L L L L Lx x x y x z x

2 2 2 2= + +

= 00 0− − + + =i L L i L L i L L i L Ly z z y z y y z� � � �ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ (4.29.5)

This result also holds for [ ˆ , ˆ ]L Ly2 and [ ˆ , ˆ ].L Lz

2

4.30. A particle is described by the wavefunction

ψ( )/

/xa

e ax= ⎛⎝

⎞⎠

−−π 1 4

22

(4.30.1)

Calculate Δx and Δp, and verify the uncertainty relation.

SOLUTION

We begin by considering the matrix element of x:

⟨ ⟩ = ⟨ ⟩ = = =−

−∞

∞

−∞

∞

∫∫x x x x dxa

xe dxaxψ ψ ψ π⏐ ⏐ ⏐ ⏐( ) 2 2

0 (4.30.2)

where we used the fact that xe ax− 2

is an odd function. Also,

⟨ ⟩ = = =−∞

∞−

−∞

∞ ∞

∫ ∫ ∫x x x dxa

x e dxaax2 2 2 2

0

2

2⏐ ⏐ψ π π( ) xx e dxa

a aax2

3 2

2

21 2

2

12

− = =πΓ( )

//

(4.30.3)

so,

Δx x xa

= ⟨ ⟩ − ⟨ ⟩ =2 2 12 (4.30.4)

In order to find Δp we calculate the wavefunction in the momentum representation:

�ψ ψ( ) ( )//

p e x dxa

ipx= = ⎛⎝⎜

⎞⎠⎟

−

−∞

∞ −

∫1

2

1

2

1

π ππ

� �

�44

2

1 4

2

1

2

2

e e dx

a ae

ipx ax−

−∞

∞−

−

∫= ⎛

⎝⎜⎞⎠⎟

/ /

/

�

�ππ π −− −= ⎛

⎝⎜⎞⎠⎟

p a p a

ae

2 2 2 221 4

21 1//

/� �

� π (4.30.5)

Since �ψ( )p is an odd function, we obtain ⟨ ⟩ =p 0, and

⟨ ⟩ = =−

−∞

∞−

∞

∫pa

p e dpa

p ep a p a2 2 2

0

1 1 22 2 2 2

� �

� �

π π/ /∫∫ = =dp

a a

a2 2

2 1 22 3 2

2

� �

�

ππ /

/( ) /

so, we obtain

Δp p pa= ⟨ ⟩ − ⟨ ⟩ =2 2

2� (4.30.6)

Eventually, the uncertainty relation will be Δ Δx p = �/2. This example demonstrates the basic nature of the uncertainty relation. If we choose a wavefunction with smaller dispersion around the central position ⟨ ⟩x , we obtain a higher dispersion of the momentum around ⟨ ⟩x .

4.31. A particle is in the state ⏐ψ⟩ and its wavefunction is ψ ψ( ) .r r= ⟨ ⟩⏐ (a) Find the mean value of the operator ˆ .A = ⟩ ⟨⏐ ⏐r r (b) Calculate ⟨ ⟩r p⏐ ⏐ˆ .ψ (c) Find the mean value of the operator ˆ [ ˆ ˆ ] ,k mr r r p p r r= ⟩ ⟨ + ⟩ ⟨⏐ ⏐ ⏐ ⏐ /2 where p is the momentum operator and m is the mass of the particle.


SOLUTION

(a) By definition,

⟨ ⟩ = ⟨ ⟩ = ⟨ ⟩ ⟨ ⟩ = =ˆ ˆ ( ) ( ) ( )*A Aψ ψ ψ ψ ψ ψ ψ⏐ ⏐ ⏐ ⏐ ⏐ ⏐r r r r r 2 (4.31.1)

(b) The x-component of ⟨ ⟩r p⏐ ⏐ˆ ψ equals

⟨ ⟩ = ⟨ ⟩ = ∂∂r p r

r⏐ ⏐ ⏐ ⏐ˆ ˆ ( )ψ ψ ψ

x xpi x�

(4.31.2)

Therefore, ⟨ ⟩ = ∇⎡⎣⎢

⎤⎦⎥

r p r⏐ ⏐ˆ ( ) .ψ ψxx

i�

Similarly for y and z, we obtain ⟨ ⟩ = ∇r p⏐ ⏐ˆ .ψ ψ�i

(c) By definition,

⟨ ⟩ = ⟨ ⟩ ⟨ ⟩ + ⟨ ⟩ ⟨ ⟩ =ψ ψ ψ ψ ψ ψ⏐ ⏐ ⏐ ⏐ ⏐ ⏐ ⏐ ⏐ˆ [ ˆ ˆ ]k

mr r r p p r r1

211

2

1

m i i

m

ψ ψ ψ ψ

ψ

* *

*

( ) ( ) ( ) ( )

Re

r r r r� �

�

∇ + ∇⎡⎣⎢

⎤⎦⎥

=ii

∇⎛⎝⎜

⎞⎠⎟

⎡⎣⎢

⎤⎦⎥

ψ

This example demonstrates the basic nature of the uncertainty relation: If we choose a wavefunction with smaller dispersion around the central position ⟨ ⟩x , we get a higher dispersion of the momentum around ⟨ ⟩p .

4.32. The parity operator Π is defined by

Π ⏐ ⏐r r⟩ = − ⟩ (4.32.1)

(a) Let ⏐ψ⟩ be an arbitrary ket with corresponding wavefunction ψ( ).r Find the wavefunction corresponding to ˆ .Π ⏐ψ⟩

(b) Show that Π is a Hermitian operator.

(c) Find the operator ˆ .Π2 What are the possible eigenvalues of ˆ ?Π

(d) We define the operators

ˆ ( ˆ ˆ ) ˆ ( ˆ ˆ )p I p I+ −= + = −12

12Π Π (4.32.2)

For an arbitrary ket ⏐ψ⟩ we also define

⏐ ⏐ ⏐ ⏐ψ ψ ψ ψ+ + − −⟩ = ⟩ ⟩ = ⟩ˆ ˆp p (4.32.3)

Show that ⏐ψ+ ⟩ and ⏐ψ− ⟩ are eigenvectors of ˆ .Π

(e) Prove that the wavefunctions corresponding to ⏐ψ+ ⟩ and ⏐ψ− ⟩ are even and odd functions, respectively.

SOLUTION

(a) We begin by considering the ket ⏐ ⏐ψ ψ⟩ = ⟩∫ ( ) ,r r d r3 hence

ˆ ( )[ ˆ ] ( )Π Π⏐ ⏐ ⏐ψ ψ ψ⟩ = ⟩ = − ⟩∫ ∫r r r rd r d r3 3 (4.32.4)

Changing the integration variable to ′ = −r r, the wavefunction corresponding to Π ⏐ψ⟩ is

⟨ ⟩ = − ′ ⟨ ′⟩ ′ = − ′ − ′∫ ∫r r r r r r r⏐ ⏐ ⏐ˆ ( ) ( ) ( )Π ψ ψ ψd r d3 3δ ′′ = −r ψ( )r (4.32.5)

(b) Using part (a) we have ⟨ ⟩ = ⟨ − ⟩r r⏐ ⏐ ⏐ˆ .Π ψ ψ Therefore, ⟨ = ⟨ −r r⏐ ⏐ˆ .Π On the other hand, taking the Hermitian conjugate of Eq. (4.32.1) yields ⟨ = ⟨ −r r⏐ ⏐ˆ .†Π Since this is valid for any ⟨r ⏐ it follows that ˆ ˆ .†Π Π=


(c) We have

ˆ ˆ ˆ ˆΠ Π Π Π2 ⏐ ⏐ ⏐ ⏐r r r r⟩ = ⟩ = − ⟩ = ⟩ (4.32.6)

Since this is valid for any ⏐ r⟩, we have ˆ ˆΠ2 = I . Suppose that ⏐φ⟩ is an eigenvector of Π with an eigenvalue p, ˆ .Π ⏐ ⏐φ φ⟩ = ⟩p So, on the one hand we have

ˆ ˆΠ2 ⏐ ⏐ ⏐φ φ φ⟩ = ⟩ = ⟩I (4.32.7)

and, on the other hand, we have,

Π Π Π2 2⏐ ⏐ ⏐ ⏐φ φ φ φ⟩ = ⟩ = ⟩ = ⟩ˆ ( ) ˆp p p (4.32.8)

Therefore, p2 1= . But since Π is a Hermitian operator, its eigenvalues must be real. Therefore, the possible eigenvalues are +1 and −1.

(d) We have

ˆ ˆ ˆ ( ˆ ˆ ) ( ˆ ˆ )Π Π Π Π Π Π⏐ ⏐ ⏐ ⏐ψ ψ ψ ψ+ +⟩ = ⟩ = + ⟩ = + ⟩p I12

12

2 (4.32.9)

Using part (c) we arrive at

ˆ ( ˆ ˆ) ˆΠ Π⏐ ⏐ ⏐ ⏐ψ ψ ψ ψ+ + +⟩ = + ⟩ = ⟩ = ⟩12 I p (4.32.10)

Hence, ⏐ψ+ ⟩ is an eigenvector of Π with an eivenvalue +1. Similarly, we can conclude that ⏐ψ− ⟩ is an eigenvector of Π with eigenvalue −1.

(e) Using part (a) we have ⟨ ⟩ = −+ +r r⏐ ⏐ˆ ( ).Π ψ ψ On the other hand, relying on part (d),

⟨ ⟩ = ⟨ ⟩ = ++ + +r r r⏐ ⏐ ⏐ˆ ( )Π ψ ψ ψ (4.32.11)

Therefore, ψ ψ+ +− = +( ) ( ),r r and ψ+ is an even function. Similarly, ⟨ ⟩ = −− −r r⏐ ⏐ˆ ( )Π ψ ψ and

⟨ ⟩ = − ⟨ ⟩ = −− − −r r r⏐ ⏐ ⏐ˆ ( )Π ψ ψ ψ (4.32.12)

Therefore ψ ψ− −= −( ) ( ),r r and ψ− is an odd function. Note that we can write any ⏐ψ⟩ as ⏐ ⏐ ⏐ψ ψ ψ⟩ = ⟩ + ⟩+ − . Thus we have obtained a method for separating a wavefunction into even and odd parts.

4.33. Consider a one-dimensional physical system described by the classical Hamiltonian

Hpm

V x= +2

2( ) (4.33.1)

(a) Show that [ ˆ , ˆ] ˆ .H x i p m= − � / (b) For a stationary state find ⟨ ⟩p (consider only square integrable states).

SOLUTION

(a) Using Eq. (4.11.3) and considering the commutation relation,

[ ˆ , ˆ] [ ˆ , ˆ] [ ˆ( ), ˆ] ˆ [ ˆ, ˆH xm

p x V x xm

p p x= + =12

12 22 ]] ˆ+ = −0

im

p�

(4.33.2)

(b) In a stationary state we have ˆ ,H ⏐ ⏐ψ ψ⟩ = ⟩λ where l is the eigenvalue. Since H is a Hermitian operator, we also have ⟨ = ⟩ψ ψ⏐ ⏐ˆ .H λ Using part (a) we finally obtain

⟨ ⟩ = ⟨ ⟩ = ⟨ − ⟩ = ⟨ˆ ˆ ˆ ˆ ˆ ˆ [ ˆp pim

Hx xHim

xψ ψ ψ ψ ψ⏐ ⏐ ⏐ ⏐ ⏐� �

λ ⏐⏐ ⏐ ⏐ψ ψ ψ⟩ − ⟨ ⟩ =λ ˆ ]x 0 (4.33.3)

4.34. Consider a free particle in one dimension whose wavefunction at t = 0 is given by

ψ( , )| | /

x N e e dkk k ikx0 0= −

−∞

∞

∫ (4.34.1)

where N is a normalization constant and k0 is a real number. In a measurement of the momentum at time t, find the probability P(p, t) of getting a result between −p1 and p1.


SOLUTION

First note that the relation between the wavefunction of the particle ψ( , )x t and its wavefunction in the momentum representation �ψ( , )p t is

ψ ψ( , ) ( , )/x t e p t dpipx= −

−∞

∞

∫1

2π�

� � (4.34.2)

(This is a Fourier transform.) Substituting k p= /� in ψ( , )x 0 we obtain

ψ( , )/ /x

Ne e dp

p k ipx0 0= −

−∞

∞

∫�� (4.34.3)

Therefore,

�ψ( , )/

pN

ep k

0 2 0= −�

��π (4.34.4)

From the normalization condition of �ψ( , )p 0 we can find the constant N:

⏐ ⏐�ψ( , )/

p dpN

e dpNp k

02 2

222

22

0

−∞

∞

−∞

∞−∫ ∫= =π π

� �� −−⎛

⎝⎜⎞⎠⎟

⎡

⎣⎢

⎤

⎦⎥ = =− ∞� �k

e k Np k0 2

0 02

2 2 10/ π (4.34.5)

Therefore, Nk

= 12 0π

, and

�ψ( , )/

pk

ep k

01

0

2 0= −

�

� (4.34.6)

The Hamiltonian of a free particle is H p m= 2 2/ . The basis ⏐ p⟩ of the state space consists of eigenvectors of H:

ˆ ˆH p

pm

p E pp⏐ ⏐ ⏐⟩ = ⟩ = ⟩2

2 (4.34.7)

Note that for every p p t, ( , )�ψ is actually the coefficient of ⏐ p⟩ in the expansion of the state of the particle ⏐ψ( )t ⟩ in the basis ⏐ p⟩:

⏐ ⏐ψ ψ( ) ( , )t p t p dp⟩ = ⟩−∞

∞

∫ � (4.34.8)

where �ψ ψ( , ) .p t p= ⟨ ⟩⏐ The time evolution of ⏐ψ( )t ⟩ is described by

⏐ ⏐ψ ψ( ) ( , )/ /

t p e p dpk

eiE t p kp⟩ = ⟩ =

−∞

∞− −∫ � 0

1

0

0� �

� −−∞

∞−∫ ⟩e p dpip t m2 2/ � ⏐ (4.34.9)

Or equivalently,

�ψ( , )/ /p t

ke e

p k ip t m= − −1

0

202

�

� � (4.34.10)

So finally we obtain

P p t p t dpk

p

p

p

p

p

( , ) ( , ) exp12

0

212

1

1

1

1

= =−− ∫∫ ⏐ ⏐�ψ

� 222

0 0 0

21

mt

pk

dpk

tpm

p

� � � �−

⎛

⎝⎜⎞

⎠⎟⎡

⎣⎢⎢

⎤

⎦⎥⎥

= ∫ exp −−⎛

⎝⎜⎞

⎠⎟2

0

pk

dp�

4.35. Consider a classical quantity f expressed in terms of the dynamic variables r and p, so that f (r, p). Suppose that in f (r, p) there appears a term of the form r pi . Using the quantization rules, find the quantum mechanical operator corresponding to the term r pi .


SOLUTION

Let the operator R correspond to the classical coordinate r, and the operator P correspond to the classical momentum p. Note that ˆ ˆR Pi is not a Hermitian operator:

( ˆ ˆ ) ( ˆ ˆ ˆ ˆ ˆ ˆ ) ˆ ˆ ˆ ˆ ˆ† †R Pi = + + = + +XP YP ZP P X P Yx y z x y PP Zzˆ ˆ ˆ= P Ri (4.35.1)

In order to obtain the Hermitian operator corresponding to r pi , we must perform a symmetrization of the operator ˆ ˆR Pi :

12

12[ ˆ ˆ ( ˆ ˆ ) ] ( ˆ ˆ ˆ ˆ )†R P R P R P P Ri i i i+ = + (4.35.2)

As an exercise, prove that this operator is indeed a Hermitian operator.

4.36. Consider a physical system with a three-dimensional state space. An orthonormal basis of the state space is chosen; in this basis the Hamiltonian is represented by the matrix

H =⎛

⎝⎜⎜

⎞

⎠⎟⎟

2 1 01 2 00 0 3

(4.36.1)

(a) What are the possible results when the energy of the system is measured? (b) A particle is in the

state ⏐ψ⟩, represented in this basis as 13

ii

i−

⎛

⎝⎜⎜

⎞

⎠⎟⎟

. Find ⟨ ⟩ ⟨ ⟩ˆ , ˆ ,H H 2 and ΔH.

SOLUTION

(a) The possible energies are the eigenvalues of H that are found by solving the equation det ( ˆ ˆ) ,H I− =λ 0 or

2 1 01 2 00 0 3

2 1 3 4 3 32 2−

−−

= − − − = − +λ

λλ

λ λ λ λ[( ) ]( ) ( )( −−

= − −

λ

λ λ

)

( ) ( )3 12 (4.36.2)

Therefore, E1 = 1 and E2 = 3. Note that E1 is a nondegenerate eigenvalue where E2 is degenerate, so a two-dimensional subspace corresponds to it.

(b) Method 1: We write

⟨ ⟩ = − −⎛

⎝⎜⎜

⎞

⎠⎟⎟

−⎛

⎝ψ ψ⏐ ⏐ˆ ( )H i i i

ii

i

13

13

2 1 01 2 00 0 3

⎜⎜⎜

⎞

⎠⎟⎟

= − − −⎛

⎝⎜⎜

⎞

⎠⎟⎟

= + + =13

3

13 1 1 3

53( ) ( )i i i

iii

(4.36.3)

Also,

⟨ ⟩ = ⟨ ⟩ = − −⎛

⎝⎜⎜

⎞

⎠⎟⎟

ˆ ˆ ( )H H i i i2 2

2

13

2 1 01 2 00 0 3

ψ ψ⏐ ⏐iii

i

i i iii

−⎛

⎝⎜⎜

⎞

⎠⎟⎟

= − −⎛

⎝⎜⎜

⎞

⎠⎟⎟

−13

2 1 01 2 00 0 3

( )33

13

9

13 1 1 9

ii i i

iii

⎛

⎝⎜⎜

⎞

⎠⎟⎟

= − − −⎛

⎝⎜⎜

⎞

⎠⎟⎟

= + +( ) ( )) = 113 (4.36.4)

and

ΔH H H= ⟨ ⟩ − ⟨ ⟩ = − =ˆ ˆ2 2 113

259

2 23

(4.36.5)


Method 2: We define

⏐ ⏐uii u

i1 2

12 0

00⟩ = −

⎛

⎝⎜⎜

⎞

⎠⎟⎟

⟩ =⎛

⎝⎜⎜

⎞

⎠⎟⎟

(4.36.6)

Thus, ⏐ ⏐ ⏐ψ⟩ = ⟩ + ⟩23

131 2u u . Note that ⏐ u1⟩ and ⏐ u2 ⟩ are eigenvectors of H:

H uii

i⏐ 1

12

2 1 01 2 00 0 3 0

12

⟩ =⎛

⎝⎜⎜

⎞

⎠⎟⎟

−⎛

⎝⎜⎜

⎞

⎠⎟⎟

= −ii u E u0

1 1 1

⎛

⎝⎜⎜

⎞

⎠⎟⎟

= ⟩ = ⟩⏐ ⏐ (4.36.7)

Similarly, ˆ .H u u E u⏐ ⏐ ⏐2 2 2 23⟩ = ⟩ = ⟩ The eigenvectors ⏐ u1⟩ and ⏐ u2 ⟩ are orthogonal since they cor-respond to different eigenvalues of H. So we obtain

⟨ ⟩ = ⟨ + ⟨⎛⎝⎜

⎞⎠⎟

⟩ + ⟩⎛⎝⎜

⎞ˆ ˆH u u H u u23

13

23

131 2 1 2⏐ ⏐ ⏐ ⏐

⎠⎠⎟= ⟨ ⟩ + ⟨ ⟩ = + =2

323

23 1

531 1 1 2 2 2E u u E u u⏐ ⏐ (4.36.8)

Also,

⟨ ⟩ = ⟨ ⟩ = ⟨ + ⟨⎛⎝⎜

⎞⎠⎟

ˆ ˆH H u u H E u2 21 2 1 1

23

13

23ψ ψ⏐ ⏐ ⏐ ⏐ ⏐ ⟩⟩ + ⟩

⎛⎝⎜

⎞⎠⎟

= + =13

23

13

1132 2 1

222E u E E⏐ (4.36.9)

and ΔH H H= ⟨ ⟩ − ⟨ ⟩ =ˆ ˆ .2 2 2 2 3/

4.37. Refer to Problem 4.36. Suppose that the energy of the system was measured and a value of E = 1 was found. Subsequently we perform a measurement of a variable A described in the same basis by

A ii

=−

⎛

⎝⎜⎜

⎞

⎠⎟⎟

5 0 00 20 2

(4.37.1)

(a) Find the possible results of ˆ.A (b) What are the probabilities of obtaining each of the results found in part (a)?

SOLUTION

(a) The possible results are the eigenvalues of A obtained by solving the secular equation

det ( ˆ ˆ) ( )[( ) ] ( )( )( )A I− = − − − = − − −λ λ λ λ λ λ5 2 1 5 3 12 (4.37.2)

Therefore, a1 = 1, a2 = 3, and a3 = 5.

(b) The energy E = 1 is a nondegenerate eigenvalue of the Hamiltonian, so after the energy measurement the state of the system is well defined by the eigenvector

ψ = −⎛

⎝⎜⎜

⎞

⎠⎟⎟

12

11

0 (4.37.3)

Now we can find the eigenvectors of A corresponding to each of the eigenvalues obtained in part (a). This can be accomplished directly by solving the equation

5 0 00 20 2

ii

a j−

⎛

⎝⎜⎜

⎞

⎠⎟⎟

⎛

⎝⎜⎜

⎞

⎠⎟⎟

=⎛

⎝⎜⎜

⎞

⎠⎟⎟

αβγ

αβγ

(4.37.4)

for each j. For example, for a1 we have

5

22

α αβ γ ββ γ γ

=+ =

− + =

⎧⎨⎪

⎩⎪i

i (4.37.5)


Therefore, α = 0. Choosing arbitrarily β = 1 we obtain γ = i, so after normalization we get

ξ1

12

01=

⎛

⎝⎜⎜

⎞

⎠⎟⎟i

(4.37.6)

In the same manner, we obtain the eigenvectors of A corresponding to a2 and a3:

ξ ξ2 3

12

0

1

12

100

=⎛

⎝⎜⎜

⎞

⎠⎟⎟

=⎛

⎝⎜⎜

⎞

⎠⎟⎟

i (4.37.7)

Finally, the probability P(aj) of a measurement yielding aj is P aj( ) .= ⟨ ⟩⏐ ⏐ ⏐ξ12ψ Thus,

P a i( ) ( )1

2

212

0 112

11

0

14 1

14= − −

⎛

⎝⎜⎜

⎞

⎠⎟⎟

= − = (4.37.8)

Similarly, we obtain

P a i( ) ( )2

2

14 0 1

11

0

14= − −

⎛

⎝⎜⎜

⎞

⎠⎟⎟

= (4.37.9)

and

P a( ) ( )3

2

12 1 0 0

11

0

12= −

⎛

⎝⎜⎜

⎞

⎠⎟⎟

= (4.37.10)

4.38. A particle of mass m is confined within an infinite one-dimensional well, between x = 0 and x = L. The stationary states ⏐φn ⟩ of the particle correspond to the energies

En

mLnn = =π 2 2 2

221 2

�, , . . . (4.38.1)

and to the wavefunctions φn xL

nxL

( ) sin .= ⎛⎝

⎞⎠

2 π Consider the case in which at time t = 0 the

particle is in the state ⏐ ⏐ ⏐ψ φ φ( ) [ ] .0 21 2⟩ = ⟩ + ⟩ / (a) Find the time-dependent ⏐ψ( ) .t ⟩ (b) Calculate

the wavefunction ψ( , ).x t

SOLUTION

(a) Since E mL12 2 22= π � / and E mL2

2 2 22= π � / , we have,

⏐ ⏐ ⏐ψ φ φ( )/ /

t e e eiE t iE t⟩ = ⟩ + ⟩⎡

⎣⎤⎦ =− − −1

212

1 2

1 2� � ii t mL i t mLeπ π2 2 2 22

12

2� �/ /⏐ ⏐φ φ⟩ + ⟩⎡

⎣⎢⎤⎦⎥

− (4.38.2)

(b) The wavefunction ψ( , )x t is obtained by ⟨ ⟩x t⏐ ψ( ) ; that is

ψ ψ φ( , ) ( ) expx t x t xi

mLt= ⟨ ⟩ = ⟨ ⟩ −

⎛

⎝⎜⎞

⎠⎟⏐ ⏐

12 21

2

2π � ++ ⟨ ⟩ −

⎛

⎝⎜⎞

⎠⎟⎡

⎣⎢⎢

⎤

⎦⎥⎥

= −

xi

mLt

L

i

⏐ φ2

2

2

1

exp

exp

π

π

�

22

2

2

22

1 2� �t

mL

xL L

i t

mL

⎛

⎝⎜⎞

⎠⎟⎛⎝⎜

⎞⎠⎟ + −

⎛sin exp

π π⎝⎝⎜

⎞

⎠⎟⎛⎝⎜

⎞⎠⎟sin

2π xL

(4.38.3)


4.39. Show that the norm of the state vector evolving from the Schrödinger equation remains constant.

SOLUTION

Consider the Schrödinger equation:

ddt

ti

H t t⏐ ⏐ψ ψ( ) ˆ ( ) ( )⟩ = ⟩1�

(4.39.1)

Taking the Hermitian conjugates of both sides of Eq. (4.39.1) we obtain

ddt

ti

t H ti

t H t⟨ = − ⟨ = − ⟨ψ ψ ψ( ) ( ) ˆ ( ) ( ) ˆ ( )†⏐ ⏐ ⏐1 1� �

(4.39.2)

since H(t) is an observable and it must therefore be a Hermitian operator. So we get

ddt

t td t

dtt t

d t⟨ ⟩ = ⟨ ⟩ ⟩ + ⟨ψ ψ ψ ψ ψ ψ( ) ( )

( )( ) ( )

( )⏐ ⏐ ⏐

⏐ ⟩⟩

= − ⟨⎡⎣⎢

⎤⎦⎥

⟩ + ⟨

dt

it H t t t

i1 1� �

ψ ψ ψ( ) ˆ ( ) ( ) ( ) ˆ⏐ ⏐ ⏐ HH t t( ) ( )⏐ψ ⟩⎡⎣⎢

⎤⎦⎥

= 0 (4.39.3)

4.40. The Hamiltonian of a particle in a potential V(r) is

ˆ ˆ ˆ( ˆ )Hm

V= +12

2P R (4.40.1)

(a) Write the Schrödinger equation in the r-representation. (b) Repeat part (a) in the p-representation.

SOLUTION

(a) Consider the Schrödinger equation:

iddt

t H t� ⏐ ⏐ψ ψ( ) ˆ ( )⟩ = ⟩ (4.40.2)

Projecting this equation into the r-basis, we obtain

it

tm

t V t�∂∂ ⟨ ⟩ = ⟨ ⟩ + ⟨r r P r R⏐ ⏐ ⏐ ⏐ ⏐ψ ψ ψ( ) ˆ ( ) ˆ( ˆ ) (

12

2 ))⟩ (4.40.3)

The wavefunction corresponding to ⏐ψ( )t ⟩ is ψ ψ( , ) ( ) .r rt t= ⟨ ⟩⏐ We also have

⟨ ⟩ = ⟨ + + ⟩

= − ∂

r P r⏐ ⏐ ⏐ ⏐ˆ ( ) ( ˆ ˆ ˆ ) ( )2 2 2 2

2

ψ ψt P P P tx y z

�22

2 2

2

22 2

∂+ ∂

∂+ ∂

∂⎛

⎝⎜⎞

⎠⎟= − ∇

x y zx y z t tψ ψ( , , , ) ( ,� r )) (4.40.4)

and we have ⟨ ⟩ =r R r r⏐ ⏐ˆ( ˆ ) ( ) ˆ( ) ( , ).V t V tψ ψ Therefore,

it

tm

V t��∂

∂ = − ∇ +⎡

⎣⎢

⎤

⎦⎥ψ ψ( , ) ( ) ( , )r r r

22

2 (4.40.5)

(b) We begin by projecting the Schrödinger equation onto the p-basis:

it

tm

t V t�∂∂ ⟨ ⟩ = ⟨ ⟩ + ⟨p p P p R⏐ ⏐ ⏐ ⏐ ⏐ψ ψ ψ( ) ˆ ( ) ˆ( ˆ ) (

12

2 ))⟩ (4.40.6)

The wavefunction in the momentum representation is defined by �ψ ψ( , ) ( ) .p pt t= ⟨ ⟩⏐ So we have

⟨ ⟩ =p P p⏐ ⏐ˆ ( ) ( , )2 2ψ ψt p t� (4.40.7)


In order to calculate the term ⟨ ⟩p R⏐ ⏐ˆ( ˆ ) ( )V tψ in Eq. (4.40.6), we insert the closure relation in the p-basis between ˆ( ˆ )V R and ⏐ψ( ) ,t ⟩ and obtain

⟨ ⟩ = ⟨ ′⟩ ⟨ ′ ⟩ ′∫p R p R p p⏐ ⏐ ⏐ ⏐ ⏐ˆ( ˆ ) ( ) ˆ( ˆ ) ( )V t V t d pψ ψ 3 (4.40.8)

Using the closure relation in the r-basis we have

⟨ ′⟩ = ⟨ ⟩ ⟨ ′⟩ =∫p R p p r r R p⏐ ⏐ ⏐ ⏐ ⏐ˆ( ˆ ) ˆ( ˆ )( )

V V d r3 1

2π�33 2

3/

/ ˆ ( ˆ )e V d ri−∫ ⟨ ′⟩r p r R pi � ⏐ ⏐ (4.40.9)

We also have

⟨ ′⟩ = ⟨ ′⟩ = ′r R p r r p r r p⏐ ⏐ ⏐ˆ( ˆ ) ( ) ( ) /V V V ei i � (4.40.10)

So, using Eqs. (4.40.8) to (4.40.10) we see that

⟨ ⟩ = − ′ ′p R p p p⏐ ⏐ˆ( ˆ ) ( )( )

( ) ( , )/V t V t dψπ

ψ1

2 3 23

�� 33 ′∫ p (4.40.11)

where

� iV V e d ri( )( )

( )/( )/p p r r p p− ′ = − − ′∫1

2 3 23

π�

� (4.40.12)

Note that �V ( )p is the Fourier transform of V(r). Finally, we have

it

tpm

t V��

∂∂ = + − ′

�� ψ ψ

π( , )

( , )( )

( )/p

p p p2

3 221

2ψψ( , )′ ′∫ p t d p3 (4.40.13)

4.41. Show that the operator exp ( ˆ )−ilpx /� describes a displacement of a distance l along the x-direction.

SOLUTION

Consider the problem in the x-representation. We search for an operator A acting on a wavefunction ψ( ),x with

ˆ ( ) ( )A x x lψ ψ= − (4.41.1)

Using the Taylor expansion, we can write

ψ ψ ψ ψ ψ( ) ( ) ( ) ! ( )( )

!(x l x l x

lx

ln

n

− = − ′ + ′′ + + −⋅ ⋅ ⋅2

2nn x)( ) + ⋅ ⋅ ⋅ (4.41.2)

In the x-representation, the momentum operator acts as ˆ ( ) ( ) .p x i x xxψ ψ= − ∂ ∂� / Therefore,

ψ ψ ψ ψ( ) ( ) ˆ ( ) !ˆ (x l x

ilp x

ilp xx x− = − + ⎛

⎝⎜⎞⎠⎟� �

12

22 )) !

ˆ ( )

expˆ

+ + −⎛⎝⎜

⎞⎠⎟ +

= −

⋅ ⋅ ⋅ ⋅ ⋅ ⋅1n

ilp x

ilp

n

xn

x

�ψ

��

⎛⎝⎜

⎞⎠⎟

ψ( )x (4.41.3)

4.42. Assume the validity of all the postulates given in Sec. 4.2 except postulate II; i.e., we introduce a system whose Hamiltonian is not Hermitian. Consider a system whose state space is two-dimensional. Suppose ⏐φ1⟩ and ⏐φ⟩2 form an orthonormal basis of the state space and are eigenvectors of the Hamiltonian with eigenvalues E1 5= � and E i2 4= −( ) ,� respectively. (a) Suppose that at time t = 0 the system is in the state ⏐φ1⟩. What is the probability of finding the system at time t in the state ⏐φ1⟩? (b) Repeat part (a) for ⏐φ2⟩. (c) Interpret the results of parts (a) and (b).

SOLUTION

(a) Using the postulates of quantum mechanics, the state vector at time t is

⏐ ⏐ ⏐ψ φ φ( )/

t e eiE t it⟩ = ⟩ = ⟩− −1

15

1�

(4.42.1)

The probability of finding the system in the state ⏐φ1⟩ at time t is, then, P t e it1

5 2 1( ) .= =−⏐ ⏐


(b) In this case, we have

⏐ ⏐ ⏐ψ φ φ( )/ ( )t e e

iE t i i t⟩ = ⟩ = ⟩− − −2

24 2

2�

(4.42.2)

The probability of finding the system in ⏐φ2 ⟩ is P t e ei i t it2

4 2 2 2( ) .( )= =− − −⏐ ⏐

(c) By inspection, we see that the state ⏐φ2 ⟩ is unstable. The probability of finding the system in this state decreases exponentially. This is not the case for the state ⏐φ1⟩, which is stable and remains in the initial state permanently. This means that the Hamiltonian is not a Hermitian, and therefore cannot represent rigorously an independent physical system. Nevertheless, the system could have been a part of a larger system, and then, phenomenologically, the notion of complex energies proves to be useful for taking into account the instability of states.

4.43. Consider a particle in a stationary potential V(r). Show that

( )ˆ ˆ

( )ˆ

ˆ( ˆ )a bd

dt md

dtV

⟨ ⟩ = ⟨ ⟩ ⟨ ⟩ = − ⟨∇ ⟩R P PR (4.43.1)

(a) and (b) are known as the Ehrenfest equations and are analogous to the classical Hamilton-Jacobi equations.

SOLUTION

We begin by considering the Hamiltonian of the system:

ˆˆ

ˆ( ˆ )Hm

V= +PR

2

2 (4.43.2)

Since the observables p and V ( ˆ )R do not depend explicitly on time, we have, according to Eq. (4.55),

d

dt iH

i m⟨ ⟩ = ⟨ ⟩ =

⎡

⎣⎢⎢

⎤

⎦⎥⎥

ˆ[ ˆ , ˆ ] ˆ ,

ˆRR R

P1 12

2

� � (4.43.3)

where we used the fact that R and V ( ˆ )R commute. Using the canonical commutation relations we can obtain

ˆ ,ˆ

ˆRP

P2

2mim

⎡

⎣⎢⎢

⎤

⎦⎥⎥

= ⟨ ⟩� (4.43.4)

Hence, d dt m⟨ ⟩ = ⟨ ⟩ˆ ˆ .R P/ / Also, using Eq. (4.55) for P and Problem 4.12,

d

dt iH

iV

i⟨ ⟩ = ⟨ ⟩ = ⟨ ⟩ = ⟨ˆ

[ ˆ , ˆ ] [ ˆ , ˆ ( ˆ )] [P

P P R1 1 1� � �

−− ∇ ⟩ = − ⟨ ∇ ⟩i V V� ˆ( ˆ )] ˆ( ˆ )R R (4.43.5)

Compare with Problem 3.5.

4.44. Assume that in the Schrödinger picture all the operators are time-independent. (a) Work in the Heisenberg picture and derive an equation expressing the time evolution of an operator ˆ ( ).A tH (b) Show that Eq. (4.55) is also valid in the Heisenberg picture.

SOLUTION

(a) In the Schrödinger picture, combining the Schrödinger equation and Eq. (4.61), we have

itU t t t H U t t ts s s�

∂∂ ⟩ = ⟩ˆ ( , ) ( ) ˆ ˆ ( , ) ( )0 0 0 0⏐ ⏐ψ ψ (4.44.1)

Since this is valid for any ⏐ψ s t( )0 ⟩, we obtain i U t t t H U t ts�∂ ∂ =ˆ ( , ) ˆ ˆ ( , ).0 0/ Hs is a Hermitian operator, so

we also have −∂∂ =it

U t t H U t ts� ˆ ( , ) ˆ ˆ ( , ).† †0 0 We differentiate Eq. (4.59) with respect to time and obtain

dA t

dt tU t t A U t t UH

s

ˆ ( ) ˆ ( , ) ˆ ˆ ( , ) ˆ† †= ∂∂

⎡⎣⎢

⎤⎦⎥

+0 0 (( , ) ˆ ˆ ( , )t t AtU t ts0 0

∂∂

⎡⎣⎢

⎤⎦⎥

(4.44.2)


Substituting the time derivatives we arrive at

dA t

dt iU t t H A U t t

iH

s s

ˆ ( )[ ˆ ( , ) ˆ ] ˆ ˆ ( , )†= − +1 1

0 0� �ˆ ( , ) ˆ ˆ ˆ ( , )†U t t A H U t ts s0 0 (4.44.3)

Since ˆ ( , ) ˆ ( , )†U t t U t t0 0 is equal to the identity operator, we insert this product between As and Hs and obtain

dA tdt i

U t t H U t t U tHs

ˆ ( )[ ˆ ( , ) ˆ ˆ ( , )][ ˆ ( ,† †= − 1

0 0�tt A U t t

iU t t A U t t

s

s

0 0

0 0

1

) ˆ ˆ ( , )]

[ ˆ ( , ) ˆ ˆ ( , )][†+�

ˆ ( , ) ˆ ˆ ( , )]†U t t H U t ts0 0

Using Eq. (4.59) we finally obtain idA t

dtA t H tH

H H�ˆ ( )

[ ˆ ( ), ˆ ( )].=

(b) The mean value of an operator in the Heisenberg picture is

⟨ ⟩ = ⟨ ⟩ˆ( ) ˆ ( )A t A tH H Hψ ψ⏐ ⏐ (4.44.4)

On the right-hand side of Eq. (4.44.4), only ˆ ( )A tH depends on time. Therefore,

d A

dtdA t

dtHH

H

⟨ ⟩ = ⟨⎡

⎣⎢⎢

⎤

⎦⎥⎥

⟩ˆ ˆ ( )

ψ ψ (4.44.5)

We assume that A is time-independent in the Schrödinger equation, so using the result of part (a) we obtain

d A t

dt iA H tH

H H

⟨ ⟩= ⟨ ⟩

ˆ ( )[ ˆ , ˆ ( )]

1�

(4.44.6)

4.45. In this problem we show that for a conservative system the greater the energy’s uncertainty, the faster the time evolution. Consider a Hamiltonian with a continuous spectrum, and assume that the spectrum is nondegenerate. Consider a state ⏐ψ( )t0 ⟩ with an uncertainty energy ΔE and show that if Δt is the time interval at the end of which the system evolves to an appreciable extent, then

Δ Δt E � � (4.45.1)

SOLUTION

A state ⏐ψ( )t0 ⟩ can be written in the form

⏐ ⏐ψ φ( ) ( )t E dEE0 ⟩ = ⟩∫α (4.45.2)

where ⏐φE ⟩ is an eigenstate of H with an eigenvalue E. We define a state for which ⏐ ⏐α( )E 2 has the form depicted in Fig. 4.2.

Fig. 4.2

ΔE

E00

a(E) 2

E


In this case, ΔE represents the uncertainty of the energy of the system. Using Eq. (4.53), the state ⏐ψ( )t0 ⟩ evolves to

⏐ ⏐ψ φ( ) ( )( )/

t E e dEiE t t

E⟩ = ⟩− −∫α 0 � (4.45.3)

In order to estimate the time interval during which the system evolves to an appreciable extent, we calculate the probability of finding the system in a state ⏐ χ⟩. This probability is

P t t E e diE t t

E( , ) ( ) ( )( )/χ χ α χ= ⟨ ⟩ = ⟨ ⟩− −∫⏐ ⏐ ⏐ ⏐ψ φ2 0 �

EE2

(4.45.4)

If ΔE is sufficiently small, we can neglect the variation of ⟨ ⟩χ ⏐ φE relative to the variation of α( );E therefore, replacing ⟨ ⟩χ ⏐ φE by ⟨ ⟩χ ⏐ φE0

, we obtain

P t E e dEEiE t t

( , ) ( )( )/χ χ α≅ ⟨ ⟩ − −∫⏐ ⏐ ⏐φ

0

022

� (4.45.5)

Thus, P t( , )χ is approximately the square of the modulus of the Fourier transform of α( )E and using the properties of the Fourier transform, the width Δt of P t( , )χ is related to ΔE by

Δ Δt

E�

� 1 (4.45.6)

where Δt is the time period during which there is an appreciable probability of finding the system in ⏐χ⟩, and therefore it can serve as an estimation of the time during which the system evolves to an appreciable extent.


4.46. Consider the projector onto a subspace εm of e (see Sec. 4.1). Verify that ˆ ˆ .P Pm m2 =

4.47. Repeat Problem 4.19 for the case of a continuous set of kets.

4.48. Repeat Problem 4.20 for the case of a discrete set of kets.

4.49. Consider the following four expressions (A is an operator):

(i) ii) iii⟨ ⟩ ⟨ ⟩ ⟨ ⟩ ⟨ ⟨ ⟩ψ φ ψ φ ψ φ ψ ψ φ⏐ ⏐ ⏐ ⏐ ⏐ ⏐ ⏐ˆ ( ˆ ( ) Â A A φφ ψ ψ φ ψ⟩ ⟨ ⟩ ⟨ ⟩⏐ ⏐ ⏐ ⏐( ) ˆ îv A A

(a) For each of the expressions, find whether it is a scalar, operator, ket, or bra.

(b) Obtain the Hermitian conjugate of each expression.

Ans. (a) (i) scalar; (ii) bra; (iii) operator; (iv) ket.

(b) (i) ⟨ ⟩ ⟨ ⟩ ⟨ ⟩ ⟨ ⟩φ ψ φ ψ ψ φ ψ φ⏐ ⏐ ⏐ ⏐ ⏐ ⏐ˆ ˆ ;† * *A Aor (ii) ⟨ ⟩ ⟩ψ φ ψ⏐ ⏐* †ˆ ;A (iii) ⟨ ⟩ ⟩ ⟨ψ φ ψ φ⏐ ⏐ ⏐* †ˆ ;A

(iv) ⟨ ⟩ ⟨φ ψ ψ⏐ ⏐ ⏐ˆ ˆ .* †A A

4.50. Derive the expression of the scalar product

⟨ ⟩ = ⟨ ⟩∑ ∫φ ψ φ ψ⏐ ⏐b C b C di i

i

* *( ) ( )and = α α α (4.50.1)

in terms of components of the ket and the bra in a given representation. (Hint: Use the closure relations.)

4.51. Show that e ix a2π / and eiap /� commute for every real number a. [Hint: Use Problem 4.12, part (b).]

4.52. Show that the transformation matrix between two orthonormal bases [Eq. (4.29)] is a unitary transformation,

i.e., ˆ ˆ ˆ ˆ ˆ.† †SS S S I= =


4.53. Derive Eqs. (4.31), (4.32), and (4.33) using the orthonormality and closure relations for the two bases { }⏐ ui ⟩ and { }.⏐vk ⟩

4.54. Refer to Problem 4.34. (a) What is the form of the wave packet at time t = 0? (b) Calculate the product Δ Δx p at t = 0.

Ans. (a) ψ( , ) ;xk

k x0

2 1

10

02 2=

+π (b) Δ Δx p = �/ 2.

4.55. Using the Schrödinger equation, derive Eq. (4.54).

4.56. Derive Eqs. (4.52) and (4.53) of postulate VI. [Hint: First find the time evolution of an eigenvector of the Hamiltonian and then use property (b) of the Schrödinger equation; see Sec. 4.9.]

4.57. Find the operator describing a shift of p0 in the x-direction momentum. (Hint: Compare to Problem 4.42.)

Ans. eip x0 /

.�

98

Harmonic Oscillator

5.1 IntroductionIn this chapter we consider a particle moving as a harmonic oscillator with a potential energy of

V x kx k( ) (= =12

2 constant) (5.1)

The general differential equation for the oscillator potential can be solved using a technique that is frequently exploited in solving quantum mechanics problems. Many problems in physics can be reduced to a har-monic oscillator with appropriate conditions. In classical mechanics, for example, when expanding potentials around a classical equilibrium point, to the second order, we obtain the harmonic potential kx2 2/ .

Schrödinger Equation The classical Hamiltonian for the one-dimensional harmonic oscillator is

Hpm

kx= +2 2

2 2 (5.2)

where k m= ω 2. The variables w and m are, respectively, the angular frequency and the mass of the oscillator. We have

ˆˆ ˆ

ˆHPm

m Xm

d

dx

mX= + = − +

2 2 2 2 2

2

22

2 2 2 2ω ω�

(5.3)

Thus, the stationary Schrödinger equation is

− + =�2 2

2

22

2 2md x

dx

mx x E x

ψ ψ ψ( )( ) ( )

ω (5.4)

The eigenfunctions that are the solutions of the Schrödinger equation are

ψn n nxx

nH

xe( )

!

//= ⎛

⎝⎜⎞⎠⎟ ( ) −1 1

22

1 422 2

πλ λλ (5.5)

where λ = �/mω and Hn( )ς are the Hermite polynomials. The eigenvalues of the harmonic oscillator that

are the eigenenergies are

E n nn = +( ) =12

0 1 2�ω , , , . . . (5.6)

CHAPTER 5


5.2 The Hermite PolynomialsThe Hermite polynomial H

n( )ς is a polynomial of degree n that is symmetric for even n and antisymmetric

for odd n. The Hermite polynomial is a solution of the differential equation

d H

d

dH

d

EHn n n

n

2

22

21 0

( ) ( )( )

ς

ςς

ςς ω ς+ − −

⎛

⎝⎜⎞

⎠⎟=

� (5.7)

This equation can be reduced to

d H

d

dH

dnHn n

n

2

22 2 0

( ) ( )( )

ς

ςς

ςς ς− + = (5.8)

The Hermite polynomials also satisfy the following relations:

dH

dnHn

n

( )( )

ςς ς= −2

1 (5.9)

and

H H nHn n n+ −= −1 12 2( ) ( ) ( )ς ς ς ς (5.10)

The generating function of the Hermite polynomials is

S t eH

ntt t n

n

n( , )( )!

ςςς= =− +

=

∞

∑2 2

0

(5.11)

where

Hd

dtS tn

n

n t( ) [ ( , )]ς ς= =0

(5.12)

More information on Hermite polynomials is given in the Mathematical Appendix.

5.3 Two- and Three-Dimensional Harmonic OscillatorsSimilar to the one-dimensional case, the classical Hamiltonian in the two-dimensional case is

Hp p

m

m x m yx y x y

2

2 2 2 2 2 2

2 2 2=

++ +

ω ω (5.13)

In this case the Hamiltonian is separable in x and y, so the problem is reduced to two one-dimensional har-monic oscillators, one in x and the other in y. The eigenfunctions in this case are

ψ ψ ψn n n nx y x y

x y x y( , ) ( ) ( )= (5.14)

where ψn i

i

x( ) is the eigenfunction of the one-dimensional harmonic oscillator. The eigenvalue corresponding to ψ

n nx y

x y( , ) is

E n nn n x x y y

x y

= +⎛⎝⎜

⎞⎠⎟

+ +⎛⎝⎜

⎞⎠⎟

� �ω ω12

12

(5.15)

The generalization to the three-dimensional case is straightforward.

5.4 Operator Methods for a Harmonic OscillatorEigenfunctions can be thought of as an orthonormal basis of unit vectors in an n-dimensional vector space that is obtained by solving the Schrödinger equation. Here we will go a step further. We will find the eigenvalue

CHAPTER 5 Harmonic Oscillator100

spectrum and eigenfunctions using operators alone. The lowering and raising or annihilation and creation operators, a and ˆ ,†a are defined by

ˆ ˆˆ

ˆ ˆˆ

†am

XiPm

am

XiPm

= +⎛

⎝⎜⎞

⎠⎟= +

⎛

⎝⎜⎞

⎠⎟ω

ωω

ω2 2� � (5.16)

These operators are very useful tools for representing the eigenfunctions of the harmonic oscillator. Note that the Hamiltonian of the harmonic oscillator can be written as

ˆ ˆ ˆ†H a a= +⎛⎝⎜

⎞⎠⎟

�ω 12

(5.17)

or

ˆ ˆ ˆ†H aa= −⎛⎝⎜

⎞⎠⎟

�ω 12

(5.18)

It can be proved that the commutation relations for these operators are

[ ˆ, ˆ ] [ ˆ , ˆ] ˆ [ ˆ , ˆ ] ˆ† † †a a H a a H a a= = − =1 � �ω ω (5.19)

Let us denote the nth state of the harmonic oscillator ψn

x( ) as ⏐n⟩, so a and ˆ†a satisfy (see Problem 5.10)

ˆ

ˆ†

a n n n

a n n n

⏐ ⏐

⏐ ⏐

⟩ = − ⟩⟩ = + + ⟩

⎧⎨⎪

⎩⎪

1

1 1 (5.20)

Now we can justify the names lowering and raising operators for a and ˆ ,†a respectively. Thus, one can build the state ⏐n⟩ as

⏐ ⏐nn

a n⟩ = ⟩10

!( ˆ )†

(5.21)

where ⏐0⟩ is the vacuum state (n = 0).

SOLVED PROBLEMS

5.1. A one-dimensional harmonic oscillator is characterized by the potential

V x kx( ) = 12

2 (5.1.1)

where k is a real positive constant. It can be shown that the angular frequency is ω = k m/ , where m is the mass of the oscillator. (a) Solve the stationary Schrödinger equation for this potential and find the stationary eigenstates for this system. (b) Refer to part (a), and find the energy eigenvalues of the oscillator. What is the minimal energy eigenvalue? Explain.

SOLUTION

(a) The Hamiltonian of this system can be written as

Hpm

kx= +2

2

212

(5.1.2)

or

Hm

d

dx

mx= − +�

2 2

2

22

2 2ω

(5.1.3)


Thus, the eigenvalue equation is

− + =�2 2

2

22

2 2md x

dx

mx x E x

ψ ψ ψ( )( ) ( )

ω (5.1.4)

We define ε ω= 2E

�, and we change the variable to ζ ω= m

x�

; hence, we have

d

dx

ddx

dd

ddx

d

d

ddx

2

2

2

2

2ψ ψ ψ= ⎛⎝⎜

⎞⎠⎟

= ⎛⎝⎜

⎞⎠⎟

=ζζ

ζζ mm d

d

ωζ�

2

2ψ

(5.1.5)

Therefore,

� �ω ζ

ζζ ω ζ ζ2 2 0

2

22d

dE

ψ ψ ψ( )( ) ( )+ − = (5.1.6)

or

d

d

2

22 0

ψ ψζ

ε ζ+ − =( ) (5.1.7)

For large z (large x), the dominant part of the differential equation, Eq. (5.1.7), is

d

d

2

22 0

ψ ψζ

ζ− = (5.1.8)

The solution for this equation points to the asymptotic behavior of the wavefunction for large z :

ψ( ) ~ /ζ ξe− 2 2 (5.1.9)

So, we can assume

ψ( ) ( ) /ζ ζ ξ= −H e2 2 (5.1.10)

Substituting into Eq. (5.1.8) yields

d

d

dd

H e H e

H e

2

22 22 2ψ

ζ ζ ζ ζ ζ

ζ

ζ ξ= ′ −

= ′′

− −[ ( ) ( ) ]

( )

/ /

−− − − −− ′ − +ζ ζ ζ ζζ ζ ζ ζ ζ2 2 2 22 2 2 22/ / /( ) ( ) ( )H e H e H e/ 22

(5.1.11)

or

d

dH H H e

2

22 22 1

2ψζ

ζ ζ ζ= ′′ − ′ + − −[ ( ) ] / (5.1.12)

Thus, we have

[ ( ) ] ( ) /′′ − ′ + − + − =− −H H H e He2 1 02 2 2 22 2

ζ ζ ε ζζ ζ/ (5.1.13)

We obtain the Hermite polynomials differential equation,

d H

d

dHd

H2

22 1 0

( ) ( )( ) ( )

ζζ

ζ ζζ

ε ζ− + − = (5.1.14)

The wavefunction’s behavior around ζ = =0 0( )x is accounted for by these polynomials. In order to

solve this equation we substitute H ann

n

( ) ,ζ ζ==

∞

∑0

so that

d H

da n n a n nn

n

n

n

n

2

22

0

2

0

1 2 1ζ

ζ= − = + +−

=

∞

+

=

∞

∑ ∑( ) ( ) ( ))ζ n (5.1.15)


and

− = −=

∞

∑2 2

0

ζ ζ ζdHd

nan

n

n (5.1.16)

Hence,

[ ( ) ( ) ( ) ]a n n na an n nn

n

+

=

∞

+ + − + − =∑ 2

0

2 1 2 1 0ε ζ (5.1.17)

Therefore, all the coefficients of this series must vanish:

a n n n an n+ + + + − − =

22 1 2 1 0( )( ) ( )ε (5.1.18)

or

an

n na

n n+ = + −+ +2

2 12 1

ε( ) ( )

(5.1.19)

We set a0

0≠ and a1

0= to obtain the value of a a a m2 4 2, , . . . (m = positive integer); similarly, we set a

00≠ and a1

0= to obtain the values of a a a m3 5 2 1, , . . . , − (m = positive integer). The a0 or a1 values are computed using a normalization condition for the wavefunction.

(b) As in part (a), we wish the wavefunction to asymptotically approach e−ζ 2 2/ for large z. To begin, set the values of the coefficients of H(z ) to zero for some value n. For that n, we obtain

2 1 0n + − =ε (5.1.20)

That is, ε = +2 1n , or

E nn

= +⎛⎝⎜

⎞⎠⎟

12

�ω (5.1.21)

Hence, we obtain the quantization condition for the energy eigenvalues. Without an energy source, the system reaches its minimal energy eigenvalue E0 2= /�ω at the temperature T = 0 K. This value is imposed by the uncertainty relation

Δ Δx p = �2 (5.1.22)

and is the minimal energy eigenvalue the system can have.

5.2. A particle with energy E = �ω /2 moves under the potential of a harmonic oscillator. Compute the probability that the particle is found in the classically forbidden region. Compare this result to the probability of finding the particle in higher energy levels.

SOLUTION

For the classical harmonic oscillator, we have

x A t p m A tn n= = −cos( ) sin ( )ω ω ω (5.2.1)

Hence, the energy is

Epm

m xm A

nn= + =

22 2

2 2

212 2

ωω

(5.2.2)


which yields AE

mnn=

22ω

. The classically forbidden region is x An> or xE

mn>

22ω

. Thus, the prob-

ability of finding the particle in the classically forbidden region is

P x x dx x x dxn n n

A

n nA

n

n

= + =−∞

− ∞

∫ ∫ψ ψ ψ ψ ψ* *( ) ( ) ( ) ( ) 2 nn nA

n n

A

x x dx

x x dx

n

n

*

*

( ) ( )

( ) ( )

ψ

ψ ψ

∞

∞

∫

∫= −1 2 (5.2.3)

Considering the ground state, we have

P x x dx e dxA

x

A0 0 0 22 2

1

0

2 2

0

= =∞

−∞

∫ ∫ψ ψ* /( ) ( )π λ

λ (5.2.4)

Changing integration variables η λ= x / , we obtain

P e d e dA

A

00

21

22

0

20

= = −−∞

−∫ ∫πη

πηη

λ

ηλ

/

/

(5.2.5)

We have A0

1/λ = ; hence,

P e d0

0

1

12 2

= − −∫πηη (5.2.6)

Solving this numerically, we obtain P0 = 0.157 8 (see Problem 12.8). For excited states the probability for being in the classically forbidden region is

Pn

Hx

e dxn n

A

nx

n

= −⎛⎝⎜

⎞⎠⎟

= −∫ −1 21

21

12

0

2 2 2

π λ λ πλ

!

/

22 12

0

2 2

n n

Ax

nH

xe d

xn

−−∫ ⎛

⎝⎜⎞⎠⎟

⎛⎝⎜

⎞⎠⎟!

/

λ λλ (5.2.7)

Putting η λ= x / , we arrive at

Pn

H e dn n n

An

= − −−∫1

1

2 12

0

2

πη η

λη

!( )

/

(5.2.8)

Using the known Hermite polynomials H H H0 1 2

21 2 4 2( ) , ( ) , ( ) ,η η η η η= = = − and A13/λ = , we obtain:

P e d1

2

0

3

14 2

= − −∫πη ηη (5.2.9)

The numerical solution is P1 = 0.111 6. We also find

P e d24 2

0

5

4 2

11

416 16 4

11

4 4

2

= − − +

= − −

−∫πη η η

πη η

η( )

( ++ =−∫ 1 0 095 12

0

5

) .e dη η (5.2.10)

Thus, we have seen that P0 = 0.157 3, P1 = 0.111 6, and P2 = 0.095 1. Note that the value of Pn is smaller for higher energy levels. The reason for this is that particles with high energy are “more classical” than those with lower energies, and hence the probability for particles in higher energy levels to be in the classically forbidden region is less.


5.3. Using the uncertainty relation Δ Δp x ≥ �/2, estimate the energy ground state of the harmonic oscillator.

SOLUTION

The classical Hamiltonian of the harmonic oscillator is

Hpm

mx= +

2 22

2 2ω

(5.3.1)

The expectation value of the energy is

⟨ ⟩ = = ⟨ ⟩ + ⟨ ⟩ˆ ˆˆH E

pm

mx

2 22

2 2ω

(5.3.2)

We can write

Δ Δp p p x x x2 2 2 2 2 2= ⟨ ⟩ − ⟨ ⟩ = ⟨ ⟩ − ⟨ ⟩ˆ ˆ ˆ ˆ (5.3.3)

for the harmonic oscillator ⟨ ⟩ = ⟨ ⟩ =ˆ ˆ .p x 0 The proof for these results is as follows:

⟨ ⟩ = =−∞

∞

−∞

∞

∫ ∫ˆ ( ) ˆ ( ) ˆ ( )*x x x x dx x x dxn n nψ ψ ψ 2 (5.3.4)

The integral of the antisymmetric function x xnψ ( ) 2 over a symmetric interval is zero; hence, ⟨ ⟩ =ˆ .x 0 Similarly,

⟨ ⟩ = −∂

∂−∞

∞

∫ˆ ( )( )

*p i xx

xdxn

n� ψψ

(5.3.5)

Changing the variables to ζλ

= x and λ ω

= �

m, we have

⟨ ⟩ = −∂

∂−∞

∞

∫ˆ ( )( )

*p i dnn� ψ

ψζ

ζζ ζ (5.3.6)

so,

∂

∂=

∂∂

+−ψ

ψn n

n n

H e

n

( ) ( )

!( )

/ζζ

ζζ π λ

ζ ζζ 2 2

2 (5.3.7)

Thus, we obtain

⟨ ⟩ = −∂∂ −

−∞

∞−∫ˆ

!( ) (* / *p

i

n

He d i

n nn�

�πλ

ζ ζ ζζ

2

2 2ψ ψ ζζ ζ ζ ζ) ( )ψ d−∞

∞

∫ (5.3.8)

Notice that

⟨ ⟩ =∫ˆ ~ ( ) ( )*x dψ ψζ ζ ζ ζ 0 (5.3.9)

As the Hermite polynomials are either symmetric or antisymmetric, the multiple HH

nn( )( )

ζτ

ζ∂

∂ is always

antisymmetric, and for the same reason that ⟨ ⟩x vanishes, ⟨ ⟩p also vanishes. Thus,

Epm

mx= +Δ Δ

2 22

2 2ω

(5.3.10)

According to the uncertainty relation, the minimal value of Δp is Δ Δpx

= �2 ; hence,

Em x

mx= +�

2

2

22

8 2ΔΔω (5.3.11)


Finally, the minimal value of E x( )Δ is obtained by

dEd x m x

m x( ) ( )Δ ΔΔ= − + =�

2

32

40ω (5.3.12)

So, Δxm0 2

= �ω . Also,

d E

d x m xm

x x

2

2

2

42

0

3

40

( ) ( )Δ ΔΔ Δ=

= + >� ω (5.3.13)

Hence, the minimal value is

Em x

mxmin ( )

( )= + = + =� � � �2

02

2

02

4 2 4 4 2ΔΔω ω ω ω (5.3.14)

as we expected. Here we obtained the exact solution by relying on the lower bound of the uncertainty rela-tion Δ Δx p = �/2. This follows from the result that in the ground state, we have a Gaussian form of the eigenfunction:

ψ( ) ( ) / / ( ) /x e eipx x x= − − −

2 1 4 40

2 2

πσ σ� (5.3.15)

Though the uncertainty relation is normally used to estimate the ground state energy eigenvalue, for the case given above we can evaluate it exactly.

5.4. Find the eigenfunctions and eigenvalues of a two-dimensional isotropic harmonic oscillator; find the degeneracy of the energy levels. The classical Hamiltonian of this system is

Hp

m

p

mm x yx y= + + +

2 22 2 2

2 212

ω ( ) (5.4.1)

SOLUTION

The Hamiltonian of the system can be separated into two parts, H H Hx y

= + , where

Hp

mm x

Hp

mm y

xx

yy= + = +

2 2 2 2 2 2

2 2 2 2ω ω

(5.4.2)

Thus, the wavefunction can be written as a multiple of two functions, ψx

x( ) (the eigenfunction of ˆ )Hx

and ψ

yy( ) (the eigenfunction of ˆ )H

y with eigenvalues E n

x x= +�ω ( )1 2/ and E n

y y= +�ω ( ),1 2/ respectively.

So, we have ˆ ,H Eψ ψ= where ψ ψ ψ( , ) ( ) ( );x y x yx y

= hence,

ˆ ( , ) ( ˆ ˆ ) ( ) ( ) ˆ ( ) ( )H x y H H x y H x yx y x y x x yψ ψ ψ ψ ψ= + = + ψψ ψ

ψ ψ ψ ψ ψ ψ

x y y

x x y y x y x y x y

x H y

E E E E

( ) ˆ ( )

( )= + = + (5.4.3)

Therefore,

E E E n n nx y x y= + = + + ≡ +( ) ( )1 1� �ω ω (5.4.4)

The degeneracy of each state E n nx y

( , ) is computed as follows: (n + 1) is an integer that assumes all values from 0 to ∞. We can see from Fig. 5.1 that (n + 1) = constant, defines a line in the nx ny space. One can also see that the degeneracy of the state n is n + 1.


5.5. Consider a particle with charge + e moving under a three-dimensional isotropic harmonic potential:

V r m r( ) = 12

2 2ω (5.5.1)

in an electric field E = E0i. Find the eigenstates and the energy eigenvalues of the particle.

SOLUTION

The classical Hamiltonian of the system is

Hm

meE x= + −p

r2 2

202 2

ω (5.5.2)

We separate the Hamiltonian into three parts: H H H Hx y z

= + + , where

Hpm

mx eE x

Hp

mm

y

Hpm

xx

yy

zz

= + −

= +

=

2 22

0

2 22

2

2 2

2 2

2

ω

ω

++

⎧

⎨

⎪⎪⎪⎪

⎩

⎪⎪⎪⎪

mz

ω 22

2

(5.5.3)

Notice that Hy and Hz are identical to the Hamiltonian of the one-dimensional harmonic oscillator, so we can write the wavefunction as ψ ψ ψ ψ( , , ) ( ) ( ) ( ),x y z x y z= 1 2 3 where ψ2

( )y and ψ3( )z are the wavefunctions

of the one-dimensional harmonic oscillator:

ψ

ψ

2

2

2

3

3

1

2

1

2

2 2

2 2

3

( )!

( )

( )

/yn

H y e

zn

n ny

n

=

=

−

π λ

π λ

λ

!!( ) /H z en

z

3

2 22−

⎧

⎨

⎪⎪

⎩

⎪⎪

λ

(5.5.4)

with λ ω= �

m. The equation for ψ1

( )x is

ˆ ( )H xm x

mx eE x Exψ

ψψ ψ ψ1

2 21

2

22

1 0 1 1 12 2= −∂∂

+ − =� ω (5.5.5)

Fig. 5.1

1

1

2

3

4

0 2 3 4 nxn + 1 = nx + ny + 1 = 4

n + 1 = nx + ny + 1 = 2

ny


Changing the variables to ζλ ω

= −x eE

m0

� yields

d

d

E eE

m

21

21 0

2

3 12

1

20

ψψ ψ

ζ ω ωζ+ +

⎛

⎝⎜⎜

⎞

⎠⎟⎟

− =� �

( ) (5.5.6)

We obtain the differential equation for a one-dimensional harmonic oscillator with the solution

ψ1

1

21

2 1 1

2

( )!

( ) /ζπ λ

ζ ζ= −n n

nH e (5.5.7)

or

ψ1

1

03

1

2

121 1

( )!

( ) expxn

H xx eE

mn n= − −⎛⎝⎜

⎞⎠⎟π λ λ ω�

22⎡

⎣⎢⎢

⎤

⎦⎥⎥

(5.5.8)

The quantization condition in this case is

2

2 11 02

3 1

E eE

mn

� �ω ω+ = +

( ) (5.5.9)

so the energy eigenvalues are

( )( )

E neE

mn1 10

2

21

12 2

= +⎛⎝⎜

⎞⎠⎟

−�ωω

(5.5.10)

In conclusion, the wavefunctions are

ψ ψ ψ ψ( , , ) ( ) ( ) ( )x y z x y z= 1 2 3 (5.5.11)

and the energy eigenvalues are

E E E E n n neE

n n n n n n1 2 3 1 2 3 1 2 3

32

= + + = + + +⎛⎝⎜

⎞⎠⎟ −�ω

( 002

22

)

mω (5.5.12)

5.6. Consider a particle with mass m in a one-dimensional harmonic potential. At t = 0 the normalized wavefunction is

ψ( )/

/x e x=⎛⎝⎜

⎞⎠⎟

−12

1 422 2

πσσ (5.6.1)

where σ ω2 ≠ �

m is a constant. Find the probability that the momentum of the particle at t > 0 is

positive.

SOLUTION

We denote by �ψ( , )p t the wavefunction of the particle in the momentum space at time t. The probability P for a positive momentum is

P p t dp=∞

∫ ⏐ ⏐�ψ( , ) 2

0

(5.6.2)

We can write �ψ( , )p t as a linear combination of the eigenfunctions in the momentum space:

� �ψ( , ) ( ) ( / )p t C p en n

n

i n t==

∞− +∑ φ

0

1 2 ω (5.6.3)


where �φn p( ) are the stationary eigenfunctions in the momentum space and the coefficients are

C x xn n= ⟨ ⟩φ ψ( ) ( ) .⏐ � Note that, here, φn x( ) are the eigenfunctions in the coordinate space. ψ( , )x t can also be written as

ψ φ( , ) ( ) ( / )x t C x en ni n t

n

= − +∑ 1 2 ω (5.6.4)

The functions φn x( ) are either symmetric or antisymmetric, as are their Fourier transforms �φn p( ). This attribute is conserved for every t; thus ψ( , )p 0 is symmetric, � �ψ ψ( , ) ( , ),p p0 0= − as is � �ψ ψ( , ) ( , ).p t p t= − Hence,

⏐ ⏐ ⏐ ⏐ ⏐� � �ψ ψ ψ( , ) ( , ) ( , )p t dp p t dp p t2

0

2

0

∞ ∞

∫ ∫= − = − + ⏐⏐ ⏐ ⏐2

0

20−∞

−∞∫ ∫=dp p t dp�ψ( , ) (5.6.5)

Using the fact that �ψ( , )p t is normalized, that is,

⏐ ⏐ ⏐ ⏐ ⏐ ⏐� � �ψ ψ ψ( , ) ( , ) ( , )p t dp p t dp p t2 2

0

2

−∞

∞ ∞

∫ ∫= +−−∞∫ =0

1dp (5.6.6)

we obtain

P p t dp p t dp= = =−∞

∞

∫∫ ⏐ ⏐ ⏐ ⏐� �ψ ψ( , ) ( , )2 20

0

12 (5.6.7)

5.7. (a) Refer to the initial condition in Problem 5.6 and calculate ψ( , ).x t (b) Given that at t = 0 the particle is in state

ψ φ φ( ) [ ( ) ( )]x x x= +12 0 1

(5.7.1)

where φn x( ) are the eigenfunctions of a one-dimensional harmonic oscillator. Compute the expectation value of x at t > 0.

SOLUTION

(a) First, note that the given ψ( )x is not ψ0( )x (the eigenfunction) since σ ω

2 ≠ �

m, so to find ψ( , )x t we

must write ψ( )x as a linear combination of the eigenfunctions φn x( ):

ψ φ( ) ( )x C xn n

n

==

∞

∑0

(5.7.2)

and

ψ φ( , ) ( ) ( / )x t C x en ni n t

n

= − +∑ 1 2 ω (5.7.3)

where

C x x x x dxn n n= ⟨ ⟩ =−∞

∞

∫φ ψ φ ψ( ) ( ) ( ) ( )*⏐ (5.7.4)

Now, writing λω

2 = �

m, we have

φn n nxn

Hx x

( )!

exp/

= ⎛⎝⎜

⎞⎠⎟ − ⎛

⎝⎜⎞⎠⎟

⎡

⎣

1

2

121 2

2

π λ λ λ⎢⎢

⎤

⎦⎥ (5.7.5)

So,

Cn

Hx

xn n n= ⎛⎝⎜

⎞⎠⎟ −1

2

1 12

11 2 1 2

2

( !) ( )exp/ /π λ πσ λ λ22 2

1+⎛⎝⎜

⎞⎠⎟

⎡⎣⎢

⎤⎦⎥

−∞

∞

∫ σdx (5.7.6)


Recall that H xn( )/λ are either symmetric (for even n) or antisymmetric (for odd n); hence, since

H xn( )/λ is antisymmetric and exp − +⎛⎝⎜

⎞⎠⎟

⎡⎣⎢

⎤⎦⎥

12

1 122x λ λ

is symmetric, Cn vanishes for odd n. Thus we

need only compute

Cm

Hx

xm m m2 1 2 221

4 2

12= ⎛

⎝⎜⎞⎠⎟ −

−∞

∞

∫[ ( )! ]exp/π σλ λ

σσ λλ σ

2 2

2 2+⎛

⎝⎜⎞

⎠⎟⎡

⎣⎢⎢

⎤

⎦⎥⎥

dx (5.7.7)

Substituting variables η σ λλ σ

= +2 2

2 22 and x =

+2 2 2

2 2

λ σλ σ

η we obtain,

Cm

H em m m2 2

2

2 21

4 2

2 2

=+

⎛

⎝⎜⎜

⎞

⎠⎟⎟−∞

∞−∫π λσ

σλ σ

η η

( )!

22

2

4 2

2

2 2

2 2

2 2 2

2

λ σλ σ

η

λσπ λ σ

σλ

+

=+ −∞

∞

∫

d

mHm m( )!( ) 22 2

2

+

⎛

⎝⎜⎜

⎞

⎠⎟⎟

−

ση ηe dn (5.7.8)

Using the identity

H ax e dxm

mam

x m2

22 21( )

( )!!

( )−

−∞

∞

∫ = −π (5.7.9)

we get

Cm

mn m

m

2 2 2 2

2 2

2 2

2 2

4=

+−+

⎛

⎝⎜

⎞

⎠⎟

λσλ σ

σ λσ λ

( )!

( !) ( ) (5.7.10)

or

ψ ψ( , ) ( ) ( / )x t x en nn t

n

= − +

=

∞

∑σ ω2 2

2 1 2

0

(5.7.11)

(b) It is given that at t = 0 we have

ψ φ φ( , ) [ ( ) ( )]x x x012 0 1= + (5.7.12)

Thus, for t > 0,

ψ φ φ( , ) ( ) ( )/ /x t x e x ei t i t= +⎡⎣

⎤⎦

− −12 0

21

3 2ω ω (5.7.13)

By definition, the expectation value of x is

⟨ ⟩ = ⟨ ⟩ = ⟨ ⟩ + ⟨ˆ ( , ) ˆ ( , ) ( ) ˆ ( ) (x x t x x t x x xψ ψ φ φ φ1

2 0 0 1 xx x x

e x x x e xi t i t

) ˆ ( )

( ) ˆ ( ) (

φ

φ φ φ

1

0 1 1

⟩⎡⎣

+ ⟨ ⟩ + ⟨− ω ω )) ˆ ( )x xφ0 ⟩⎤⎦ (5.7.14)

Let us compute each part separately:

⟨ ⟩ = =−∞

∞

∫φ φ φ φ φ0 0 0 0 02( ) ˆ ( ) ( ) ( ) ( )*x x x x x dx x⏐ ⏐ ⏐ ⏐ ddx

−∞

∞

∫ (5.7.15)


Since φ02( )x is symmetric and x is an antisymmetric function, the integration vanishes on a symmetric

interval, ⟨ ⟩ =φ φ0 0 0( ) ˆ ( ) ,x x x and also on ⟨ ⟩ =φ φn nx x x( ) ˆ ( ) .0 We turn now to compute

⟨ ⟩ =

=

−∞

∞

∫φ φ φ φ0 1 0 1

1 2

1 1

2

( ) ˆ ( ) ( ) ( )*

/

x x x x x dx⏐ ⏐

π λ ππ λ λ λλ

1 2 0 1

2 2

/

/Hx

Hx

xe dxx⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

−∞

∞−∫ (5.7.16)

We have H x0

1( )/λ = and H x x1

2( )/ /λ λ= (see the Mathematical Appendix). Therefore,

⟨ ⟩ = =−

−∞

∞

∫φ φ0 1 22

22 1 1

2

2 2

( ) ˆ ( ) /x x x x e dxx⏐ ⏐ π λ λλ

(5.7.17)

or ⟨ ⟩ =φ φ0 1 2( ) ˆ ( )x x xm

⏐ ⏐�

ω and

⟨ ⟩ = ⟨ ⟩ =φ φ φ φ1 0 0 1 2( ) ˆ ( ) ( ) ˆ ( ) *x x x x x xm

⏐ ⏐ ⏐ ⏐�

ω (5.7.18)

So, we finally obtain

⟨ ⟩ =ˆ cos ( )xm

t�

2 ω ω (5.7.19)

5.8. Consider the one-dimensional harmonic oscillator with the Hamiltonian

Hpm

m x= +2

2 2

212

ω (5.8.1)

We define new operators

ˆˆ

ˆ ˆpP

mq X

m= =ω

ω� �

and (5.8.2)

so ˆ ˆ ˆ .H q p= +( )�ω2

2 2 (a) compute the commutation relation [ ˆ, ˆ].p q (b) For the operators a and ˆ†a

defined as

ˆ ( ˆ ˆ) ˆ â q ipm

Xi

mP= + = +1

2 2ω

ω� (5.8.3)

ˆ ( ˆ ˆ) ˆ ˆ†a q ipm

Xi

mP= − = −1

2 2ω

ω� (5.8.4)

compute a n⏐ ⟩ and ˆ ,†a n⏐ ⟩ where ⏐ n⟩ is the eigenfunction of the oscillator for the nth energy state.

SOLUTION

(a) We use the known commutation relation [ ˆ , ˆ] ,X P i= � so

[ ˆ, ˆ]ˆ

, ˆ [ ˆ , ˆ ]p qP

mX

mP X i=

⎡

⎣⎢⎢

⎤

⎦⎥⎥

= = −ω

ω� � �

1 (5.8.5)

(b) Using the result obtained in part (a), we can write

ˆ ˆ ( ˆ ˆ) ( ˆ ˆ) [ ˆ ˆ ( ˆ ˆ†a a q ip q ip q p i pq= − + = + − −12

12

2 2 ˆ ˆ)]

( ˆ ˆ [ ˆ, ˆ]) ( ˆ ˆ )

qp

q p i p q q p= + − = + −12

12

12 2 2 2 (5.8.6)


So, substituting in Eq. (5.8.1), we have

ˆ ˆ ˆ†H a a= +⎛⎝⎜

⎞⎠⎟

�ω 12

(5.8.7)

Now we turn to compute the commutation relation a and ˆ :†a

[ ˆ , ˆ] [ ˆ ˆ, ˆ ˆ] [ ˆ, ˆ]†a a q ip q ip i q p= − + = = −12

1 (5.8.8)

Thus, ˆ ˆ ˆ ˆ .† †a a aa− = −1 Therefore, we obtain

ˆ ˆ ˆ†H aa= −⎛⎝⎜

⎞⎠⎟

�ω 12

(5.8.9)

We also need to compute the commutation relation of a and ˆ†a with ˆ ,H

[ ˆ, ˆ ] [ ˆ, ˆ ˆ] [ ˆ, ˆ ] ˆ ˆ† †a H a a a a a a a= = =� � �ω ω ω (5.8.10)

Similarly,

[ ˆ , ˆ ] [ ˆ , ˆ ˆ ] [ ˆ , ˆ] ˆ ˆ† † † † † †a H a aa a a a a= = = −� � �ω ω ω (5.8.11)

Thus, using the eigenvalue equation of the energy ˆ ( ) ,H n n n⏐ ⏐⟩ = + ⟩�ω 1 2/ we can write

ˆ ˆ ˆ†H n a a n⏐ ⏐⟩ = +⎛⎝⎜

⎞⎠⎟

⟩�ω 12

(5.8.12)

Therefore, ˆ ˆ .†a a n n n⏐ ⏐⟩ = ⟩ Similarly,

ˆ ˆ ˆ†H n aa n⏐ ⏐⟩ = −⎛⎝⎜

⎞⎠⎟

⟩�ω 12

(5.8.13)

so ˆ ˆ ( ) .†aa n n n⏐ ⏐⟩ = + ⟩1 We apply ˆ [ ˆ , ˆ ]† †a a H= − 1�ω

on the state ⏐n⟩, so

ˆˆ ˆ ˆ ˆ

ˆ†† †

a na H

nHa

n n a⏐ ⏐ ⏐⟩ = − ⟩ + ⟩ = − +⎛⎝⎜

⎞⎠⎟� �ω ω

12

†††ˆ ˆ

⏐ ⏐nHa

n⟩ + ⟩�ω

(5.8.14)

or

ˆ ( ˆ ) ( ˆ )† †H a n n a n⏐ ⏐⟩ = +⎛⎝⎜

⎞⎠⎟

⟩�ω 32

(5.8.15)

Hence, we conclude that ˆ†a n⏐ ⟩ is a state that is proportional to ⏐ n + ⟩1 , i.e.,

⏐ ⏐ ⏐ψ+ +⟩ ≡ ⟩ = + ⟩ˆ†a n nα 1 (5.8.16)

where α+ is a constant given by

α+ + += ⟨ ⟩ = ⟨ ⟩2 ψ ψ⏐ ⏐ ⏐n aa nˆ ˆ† (5.8.17)

We have already seen that ˆ ˆ ( ) ;†aa n n n⏐ ⏐⟩ = + ⟩1 thus α+ = +2 1( ).n Choosing α+ = +n 1, we finally get

ˆ†a n n n⏐ ⏐⟩ = + + ⟩1 1 (5.8.18)

Similarly, we apply ˆ [ ˆ, ˆ ]a a H= 1�ω

on the state ⏐ n⟩ and find

ˆˆ ˆ ˆ ˆ

( ˆ )a naH

nHa

n n a n⏐ ⏐ ⏐ ⏐⟩ = ⟩ − ⟩ = +⎛⎝⎜

⎞⎠⎟

⟩� �ω ω

12

−− ⟩ˆ

( ˆ )H

a nω�

⏐ (5.8.19)

or

H a n n a n⏐ ⏐⟩( ) = −⎛⎝⎜

⎞⎠⎟

⟩( )�ω 12

(5.8.20)


So, we conclude that a n⏐ ⟩ is a state that is proportional to ⏐ n − ⟩1 , i.e.,

⏐ ⏐ ⏐ψ− −⟩ ≡ ⟩ = − ⟩a n nα 1 (5.8.21)

where α− is also a constant

α− − −= ⟨ ⟩ = ⟨ ⟩2 ψ ψ⏐ ⏐ ⏐n a a nˆ ˆ† (5.8.22)

We have seen that ˆ ˆ ;†a a n n n⏐ ⏐⟩ = ⟩ therefore α− =2 n. Choosing α− = n , we get

a n n n⏐ ⏐⟩ = − ⟩1 (5.8.23)

Note that if we apply a to the ground state ⏐ 0⟩, we get

a ⏐ 0 0⟩ = (5.8.24)

Thus, we introduce the lowering and raising operators a and ˆ†a , defined above, that satisfy

ˆ

ˆ†

a n n n

a n n n

⏐ ⏐

⏐ ⏐

⟩ = − ⟩

⟩ = + + ⟩

⎧⎨⎪

⎩⎪

1

1 1 (5.8.25)

5.9. Compute the matrix elements of the operators X and P for the one-dimensional harmonic oscillator,

X n X k x X x dxnk n k= ⟨ ⟩ =−∞

∞

∫⏐ ⏐ˆ ( ) ˆ ( )*φ φ (5.9.1)

P n P k x P x dxnk n k= ⟨ ⟩ =−∞

∞

∫⏐ ⏐ˆ ( ) ˆ ( )*φ φ (5.9.2)

where φn x( ) are the eigenfunctions of the harmonic oscillator.

SOLUTION

Let us write X and P using the lowering and raising operators a and ˆ†a (see Problem 5.8):

ˆ ( ˆ ˆ ) ( ˆ ˆ )† †Xm

a am

a a= + = +12

22

� �

ω ω (5.9.3)

Similarly,

ˆ ( ˆ ˆ ) ( ˆ ˆ)† †Pm

i ma a i

ma a= − = −ω

ωω

22

2� �

(5.9.4)

from which we can now compute

⟨ ⟩ = ⟨ + ⟩ = ⟨ ⟩ + ⟨n X km

n a a km

n a k⏐ ⏐ ⏐ ⏐ ⏐ ⏐ˆ ( ˆ ˆ ) ( ˆ†� �2 2ω ω nn a k⏐ ⏐ˆ )† ⟩ (5.9.5)

We have seen that

ˆ

ˆ†

a k k k

a k k k

⏐ ⏐

⏐ ⏐

⟩ = − ⟩

⟩ = + + ⟩

⎧⎨⎪

⎩⎪

1

1 1 (5.9.6)

Therefore, we have

⟨ ⟩ = ⟨ − ⟩ + + ⟨ + ⟩( ) =n X km k n k k n k m

k n⏐ ⏐ ⏐ ⏐ˆ,

� �2 1 1 1 2ω ω δ kk n kk− ++ +( )1 11δ , (5.9.7)

where

δnm

n mn m

= =≠

⎧⎨⎩

10

(5.9.8)


Hence,

⟨ ⟩ =

+ = +

= −n X k

nm

k n

nm

k n⏐ ⏐ˆ

( )�

�

12

1

21

0

ω

ωotherwise

⎧⎧

⎨

⎪⎪

⎩

⎪⎪

(5.9.9)

In the same way, we can compute

⟨ ⟩ = ⟨ − ⟩ = ⟨ ⟩n P k im

n a a k im

n a k⏐ ⏐ ⏐ ⏐ ⏐ ⏐ˆ ( ˆ ˆ) ˆ† †ω ω� �2 2 −− ⟨ ⟩( )n a k⏐ ⏐ˆ (5.9.10)

Now, using the relation in Eq. (5.9.6), we have

⟨ ⟩ = + ⟨ + ⟩ − ⟨ − ⟩( )

= +

n P k im

k n k k n k

im

k

⏐ ⏐ ⏐ ⏐ˆ ω

ω

�

�

2 1 1 1

211 1 1δ δn k n kk, ,+ −−( ) (5.9.11)

So, we obtain

⟨ ⟩ =

= −

− + = +n P k

im n

k n

im n

k n⏐ ⏐ˆ ( )

ω

ω

�

�

2 1

12 1

0 otherwiise

⎧

⎨

⎪⎪⎪

⎩

⎪⎪⎪

(5.9.12)

We can express ⟨ ⟩n X k⏐ ⏐ˆ and ⟨ ⟩n P k⏐ ⏐ˆ in a matrix form as

⟨ ⟩ =

⋅ ⋅ ⋅⋅ ⋅ ⋅⋅ ⋅ ⋅⋅ ⋅ ⋅

⋅⋅⋅⋅

n X km

⏐ ⏐ˆ �2

0 1 0

1 0 2

0 2 0 3

0 0 3 0ω

⋅⋅⋅

⎛

⎝

⎜⎜⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟⎟

(5.9.13)

and

⟨ ⟩ =

−−

−

⋅ ⋅ ⋅⋅ ⋅ ⋅⋅ ⋅ ⋅⋅ ⋅ ⋅

n P k im

⏐ ⏐ˆ ω�2

0 1 0

1 0 2

0 2 0 3

0 0 3 0

⋅⋅⋅⋅⋅⋅⋅

⎛

⎝

⎜⎜⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟⎟

(5.9.14)

As expected, X and P are represented by Hermitian matrices.

5.10. Consider a one-dimensional oscillator in the nth energy level. Compute the expectation values

⟨ ⟩ ⟨ ⟩ ⟨ ⟩ ⟨ ⟩ˆ , ˆ , ˆ , ˆX X P P2 2

What can you say about the uncertainty relation Δ x Δ p?

SOLUTION

Using the operators a and ˆ ,†a one can find that

⟨ ⟩ = +ˆ ( )Xm

n2

22 1

�

ω (5.10.1)

⟨ ⟩ = +ˆ ( )Pm

n2

22 1

ω� (5.10.2)


and ⟨ ⟩ =ˆ ,P 0 ⟨ ⟩ =ˆ .X 0 Therefore,

ΔP P Pm

n= ⟨ ⟩ − ⟨ ⟩ = +ˆ ˆ ( )2 2

22 1

ω�

ΔX X Xm

n= ⟨ ⟩ − ⟨ ⟩ = +ˆ ˆ ( )2 2

22 1

�ω

so

Δ Δx p n= +�2 2 1( ) (5.10.3)

Hence, the ground state satisfies the minimum of the uncertainty relation:

Δ Δx p = �2 (5.10.4)

5.11. The simplest molecular crystals are formed from noble gasses such as neon, argon, krypton, and xenon. The interaction between the ions in such a molecular crystal is approximated by the Lennard–Jones potential:

V r Vr r

( ) =⎛⎝⎜

⎞⎠⎟

−⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢⎢

⎤

⎦⎥⎥

40

12 6σ σ (5.11.1)

The values of V0 and s for the noble gasses are listed in Table 5.1.

Table 5.1

Ne Ar Kr Xe

V0(eV) 0.003 1 0.010 4 0.014 0 0.020 0

s (Å) 2.74 3.40 3.65 3.98

Find the approximate ground state energy of a single ion in such a crystal. Hint: The ion near the minimal value of V(r) can be treated as a harmonic oscillator.

SOLUTION

We begin by approximating the potential V(r) near the minima by a polynomial of the form

V r Vk

r r O r rm m m( ) ( ) [( ) ]≈ + − + −2

2 3 (5.11.2)

where Vm is the value of V(rm) and rm are the minima. Hence,

dV r

drV

r rr

r r m mm

m

( )

== − +

⎛

⎝⎜

⎞

⎠⎟ = ⇒4 12 6 00

12

13

6

7σ σ == 21 6/ σ (5.11.3)

and thus V(rm) = −V0. Similarly,

kd V r

drV

r rr r

m

= = −⎛

⎝⎜⎞

⎠⎟=

=

2

2 0

12

14

6

84 156 42 3( ) σ σ

66 22 3 02i / V

σ (5.11.4)

Now we can approximate the behavior of an ion in the crystal to the behavior of a harmonic oscillator. The

ground state of a harmonic oscillator with potential V r U k r r( ) ( )= + ( ) −0 02

2 is

E Ukm

U0 0 02 2

= + = +� �ω (5.11.5)


where m is the mass of the ion. Therefore,

E Vkm

Vm

Vm0

1 30

023 2≈ + = −� � i /

σ (5.11.6)


5.12. Show that the eigenfunctions of the harmonic oscillator in the ground state and in the first excited state have inflection points wherever the condition V(x) = E is satisfied, i.e.,

m

x nω ω

22

212

= +( )� (5.12.1)

5.13. Find the eigenenergies and eigenfunctions for a particle moving under the potential

V xm

x x

x

( ) =>

∞ ≤

⎧

⎨⎪

⎩⎪

ω 22

20

0

(5.13.1)

Hint: It is easy to solve the Schrödinger equation for x > 0 and for x < 0 separately, and then demand that the eigenfunction for all values of x will be continuous.

Ans. The eigenfunctions are fn for n odd where fn are the eigenfunctions of the harmonic oscillator. The

corresponding eigenenergies are E nn = +⎛⎝⎜

⎞⎠⎟�ω 1

2 .

5.14. Consider an isotropic three-dimensional harmonic oscillator. (a) Perform a separation of variables and find the eigenstates of the system. (b) Find the eigenenergies and determine the degeneracy of the levels.

Ans. (a) ψ( , , )( )

( ) ( ) ( )/ (

x y zH x H y H zn n n

n n=

+ +

1

23 2

1 2 3

1 2πλ nn

x y z

n n ne

3

2 2 2 2

1 2 3

2

)

( ) /

! ! !

− + + λ

(b) gn

nn n

n= − +

−= + +( )!

!( )!( )( )3 1

3 11 2

2

5.15. The wavefunction of a harmonic oscillator at time t = 0 is

ψ φ φ φ( , )x A A A0 2121 2 3= + + (5.15.1)

where φn is the stationary eigenfunction of the oscillator for the nth state and A is a normalization constant. (a) Compute the constant A. (b) Compute the eigenfunciton ψ (x, t) for all values of t. (c) Calculate the average ⟨ E ⟩ at times t = 0, t = p /w, and t = 2p /w.

Ans. (a) A = 27

; (b) ψ φ φ φ( , )x t e e ei t i t i t= + +− − −27 2

121

3 22

5 23

7ω ω ω/ / /22⎛⎝⎜

⎞⎠⎟

;

(c) ⟨ ⟩ = ⟨ ⟩ = ⟨ ⟩ == = =E E Et t t0

23114

πω

πω

ω�

5.16. Consider an isotropic two-dimensional harmonic oscillator. (a) Write the stationary Schrödinger equation for the oscillator. Solve the equation in cartesian coordinates. (b) Write the stationary Schrödinger equation in polar coordinates and solve it for the ground state. Is this state degenerate?

Ans. (a) Schrödinger equation:

1

2 2

2

2

2

2

22 2

m x yx y

mx y x y

∂∂

+ ∂∂

⎛

⎝⎜⎞

⎠⎟+ +ψ ψ( , ) ( ) ( ,

ω)) ( , )= E x yψ (5.16.1)

ψ00

2 2

4( , ) exp ( )x y

m mx y= − +⎡

⎣⎢

⎤⎦⎥

ωπ

ω�

(5.16.2)


(b) Schrödinger equation:

1

21 1

22

2

2

22

m r rr r

r

r mr r

∂∂ + ∂

∂+ψ ψ ψ( , )

( , )( , )θ θ

θω θ == E rψ( , )θ (5.16.3)

ψ002

4( , ) exprm m

rθ ωπ

ω= −⎛⎝⎜

⎞⎠⎟�

(5.16.4)

and the state is not degenerate (ground state).

5.17. Compute the matrix elements ⟨ ⟩n X m⏐ ⏐ˆ 2 and ⟨ ⟩n P m⏐ ⏐ˆ 2 for the one-dimensional harmonic oscillator.

Ans. ⟨ ⟩ =

−++ +

= −

n X mm

m m

m

m m

n m

⏐ ⏐ˆ

( )

( )

( )( )2

2

1

2 1

1 2

0

�ω

22

2

n m

n m

== +

⎧

⎨⎪⎪

⎩⎪⎪ otherwise

(5.17.1)

⟨ ⟩ = −

− = −− + =

+n P m

mm m n m

m n m

m⏐ ⏐ˆ

( )( )

( )(2

2

1 22 1

1

�ωmm n m+ = +

⎧

⎨⎪⎪

⎩⎪⎪

2 20

)otherwise

(5.17.2)

5.18. Compute ⟨ ⟩n PX m⏐ ⏐ˆ ˆ for the one-dimensional harmonic oscillator.

Ans. ⟨ ⟩ =

=

− = −

+ +

n PX m

im n

in n m n

in n

⏐ ⏐ˆ ˆ ( )

( )(

�

�

�

2

2 1 2

2 2 1)) m n= +

⎧

⎨

⎪⎪⎪

⎩

⎪⎪⎪ 2

(5.18.1)

5.19. Compute the matrix elements ⟨ ⟩n X m⏐ ⏐ˆ 3 and ⟨ ⟩n X m⏐ ⏐ˆ 4 for the one-dimensional harmonic oscillator.

Ans. ⟨ ⟩ =

⎛⎝⎜

⎞⎠⎟ + + + = +

n X m

mn n n m n

⏐ ⏐ˆ

( ) ) )

3

3 2

2 3 2 1 3�

ω

/

( (

331

2 1

3 2

3 2

3 2

�

�

( )nm

m n

nm

m n

+⎛⎝⎜

⎞⎠⎟ = +

⎛⎝⎜

⎞⎠⎟ =

ω

ω

/

/

−−

⎛⎝⎜

⎞⎠⎟ − − = −

⎧

⎨

⎪⎪⎪⎪⎪

⎩

⎪⎪⎪

1

2 1 2 33 2

�m

n n n m nω

/

(( ) )⎪⎪⎪

(5.19.1)

⟨ ⟩ =

⎛⎝⎜

⎞⎠⎟ + + + + =

n X m

mn n n n m

⏐ ⏐ˆ

( ) ( ) ( ) )

4

2

2 1 2 3 4�

ω ( nn

mn n n m n

m

+

⎛⎝⎜

⎞⎠⎟ + + = +

⎛⎝⎜

⎞⎠⎟

4

4 2 1 2

2 2

2�

�

ω

ω

( ) ( 2)

222

2

3 2 1

4 2 2 1 2

( )

( ) ( )

n n m n

nm

n n m n

+ + =

− ⎛⎝⎜

⎞⎠⎟ − = −�

ω

��2 1 2 3 4

2

mn n n n m nω

⎛⎝⎜

⎞⎠⎟ − − − = −

⎧

⎨

⎪⎪⎪⎪

( ) ) )( (

⎪⎪⎪⎪

⎩

⎪⎪⎪⎪⎪⎪⎪

(5.19.2)

117

Angular Momentum

6.1 IntroductionIn classical mechanics the angular momentum of a particle with linear momentum p and position r is defined as

L r p= × (6.1)

where, in cartesian coordinates,

L p r= = =( , , ) ( , , ) ( , , )L L L p p p x y zx y z x y z (6.2)

By contrast, in quantum mechanics these observables are replaced by their corresponding quantum operators. Nowadays, there are three different notational schemes in common use. For example, the linear momentum operator is often written as P, p, or ˆ .p Here we will use the latter representation, which is slightly less prone to misinterpretation. Thus, the angular momentum operator is

ˆ ˆ ˆ ˆ ˆ

ˆ ˆ ˆ

L yp zp i yz

zy

L zp

x z y

y x

= − = − ∂∂ − ∂

∂⎛⎝⎜

⎞⎠⎟

=

�

−− = − ∂∂ − ∂

∂⎛⎝⎜

⎞⎠⎟

= − = −

ˆ ˆ

ˆ ˆ ˆ ˆ ˆ

xp i zx

xz

L xp yp

z

z y x

�

ii xy

yx

�∂

∂ − ∂∂

⎛⎝⎜

⎞⎠⎟

⎧

⎨

⎪⎪⎪⎪

⎩

⎪⎪⎪⎪

(6.3)

and

ˆ ˆ ˆ ˆL L L Lx y z2 2 2 2= + + (6.4)

In cartesian coordinates, the commutation relations between ˆ ( , , )L j x y zj = are

[ ˆ , ˆ ] ˆL L i Lx y z= � (6.5)

[ ˆ , ˆ ] ˆL L i Ly z x= � (6.6)

[ ˆ , ˆ ] ˆL L i Lz x y= � (6.7)

CHAPTER 6

CHAPTER 6 Angular Momentum118

6.2 Commutation RelationsUsing the commutation relations in Sec. 6.1, one can also find another useful commutation relation:

[ ˆ , ˆ ] [ ˆ , ˆ ] [ ˆ , ˆ ] [ ˆ , ˆ ]L L L L L L Lz x y2 2 2 20 0L = ⇒ = = = (6.8)

[ ˆ , ˆ ] ˆL r i Li j ijk k

k

= ∑� ε (6.9)

[ ˆ , ˆ ] ˆL p i pi j ijk k

k

= ∑� ε (6.10)

[ ˆ , ˆ ] [ ˆ , ˆ ] [ ˆ , ˆ ˆL p L r Li i i2 2 0= = =r p⋅ ] (6.11)

where

εijk

ijkijk= −

1 have cyclic permutation1 have anticyclic permutation0 otherwise

⎧⎨⎪

⎩⎪

6.3 Lowering and Raising OperatorsWe define the raising operator as

ˆ ˆ ˆL L iLx y+ = + (6.12)

Similarly, the lowering operator is defined as

ˆ ˆ ˆL L iLx y− = − (6.13)

so we can write

ˆˆ ˆ

ˆˆ ˆ

LL L

LL L

ix y=+

=−+ − + −

2 2 (6.14)

L+ and L− are not Hermitian operators since it can be proved that

ˆ ˆ†L L+ −= (6.15)

Moreover,

ˆ ˆ ( ˆ ˆ ˆ ˆ )L L L L L Lz2 2 1

2= + ++ − − + (6.16)

and

ˆ ˆ ˆ ˆ ˆL L L L Lz z+ − = − +2 2� (6.17)

ˆ ˆ ˆ ˆ ˆL L L L Lz z− + = − +2 2� (6.18)

Thus, we have the commutation relations:

[ ˆ , ˆ ]L L2 0± = (6.19)

[ ˆ , ˆ ] ˆL L Lz ± ±= ± � (6.20)

[ ˆ , ˆ ] ˆL L Lz+ − = 2� (6.21)

The operators L− and L+ enable us to represent all the eigenfunctions of L2 and Lz using only one eigenfunction and the operators L+ and ˆ .L−


6.4 Algebra of Angular MomentumThe operators L2 and Lz describe physical quantities and therefore, they must be Hermitian operators; that is,

( ˆ ) ˆ ( ˆ ) ˆ† †L L L Li i= ⇒ =2 2 (6.22)

One can verify that L2 and Lz are commutative operators, [ ˆ , ˆ ]L Lz2 0= [see Problem 6.2, part (a)]; it is thus

possible to find the simulation eigenfunctions of both L2 and ˆ ( ),L lmz | ⟩ which constitute a complete ortho-normal basis:

ˆ ( )L lm l l lm2 21| |⟩ = + ⟩� (6.23)

L lm m lmz | |⟩ = ⟩� (6.24)

Applying the raising and lowering operators to ⏐ lm⟩ gives

ˆ ( ) ( ) , ( ) )L lm l l m m l m l m l m+ ⟩ = + − + + ⟩ = − + +| |1 1 1 1� �( || l m, + ⟩1 (6.25)

ˆ ( ) ( ) , ( ) )L lm l l m m l m l m l m− ⟩ = + − − − ⟩ = + − +| |1 1 1 1� �( || l m, − ⟩1 (6.26)

Note that if ⏐ lm⟩ is an eigenvector of L2 with eigenvalue l (l + 1), then for a fixed l there are (2l + 1) possible eigenvalues for ˆ :Lz

m l l l l= − − + −, , . . . , , . . . , ,1 0 1 (6.27)

Thus,

ˆ ,L l l+ ⟩ =| 0 (6.28)

ˆ ,L l l− − ⟩ =| 0 (6.29)

The basis ⏐ lm⟩ is orthonormal, i.e,

⟨ ⟩ =l m l m l l m m1 1 2 21 2 1 2

| δ δ (6.30)

This basis is called the standard basis. The closure relation for the standard basis is

| |lm lm I

m l

l

l

⟩ ⟨ == −=

∞

∑∑ ˆ

0

(6.31)

6.5 Differential RepresentationsThe representation of eigenvectors and eigenvalues is often more convenient using spherical coordinates:

x r y r z r= = =sin cos sin sin cosθ ϕ θ ϕ θ (6.32)


The representation of the angular momentum operators in spherical coordinates is

ˆ sincostan

ˆ cos

L i

L i

x

y

=∂

∂ +∂

∂⎛⎝⎜

⎞⎠⎟

= −∂

�

�

ϕ θϕθ ϕ

ϕ ∂∂ +∂

∂⎛⎝⎜

⎞⎠⎟

= −∂

∂

⎧

⎨

⎪⎪⎪⎪

⎩

⎪⎪⎪⎪

θϕθ ϕ

ϕ

sintan

L iz �

(6.33)

which yields

ˆtan sin

L2 22

2 2

2

2

1 1= − ∂∂

+ ∂∂ + ∂

∂

⎛

⎝⎜⎞

⎠⎟�

θ θ θ θ ϕ (6.34)

ˆ cotL e ii+ =

∂∂ +

∂∂

⎛⎝⎜

⎞⎠⎟

�ϕ

θ θ ϕ (6.35)

ˆ cotL e ii−

−= −∂

∂ +∂

∂⎛⎝⎜

⎞⎠⎟

�ϕ

θ θ ϕ (6.36)

Thus, the eigenvectors of L2 and Lz are functions that depend on the angles q and j only; hence, we can represent the wavefunction as

ψ( , , ) ( ) ( , )r R r Ylmθ ϕ θ ϕ= (6.37)

For a central potential V(r) = V(r), we find that Ylm ( , )θ ϕ are the spherical harmonics, where

|lm Ylm⟩ = ( , )θ ϕ (6.38)

The algebraic representation of Ylm ( , )θ ϕ for m > 0 is

Yl l m

l mP el

m mlm i( , ) ( )

( )!( )!

(cos )θ ϕ π θ= −+ −

+12 1

4mmϕ (6.39)

and for m < 0,

Yl l m

l mPl

m ml

m( , ) ( )( )!( )!

(cos| | | |θ ϕ π= − + −+1

2 14

θθ ϕ)eim (6.40)

P xlm ( ) are the associated Legendre functions defined by

P x xd

dxP xl

m mm

m l( ) ( ) ( )= −1 2 (6.41)

where P xl ( ) are the Legendre polynomials,

P xl

d

dxxl

l

l

l

ll( )

( )

!( )= − −1

212 (6.42)

This is known as the Rodrigues formula.


Note that the Ylm ( , )θ ϕ are uniquely defined except for sign, which is changeable. The spherical harmonic

functions, the associated Legendre functions, and Legendre polynomials are described in detail in the Math-ematical Appendix.

6.6 Matrix Representation of an Angular MomentumWe have already mentioned in Chap. 4 that an operator can be represented in matrix form; this representation depends on the basis vectors (eigenvectors) that we choose. For an angular momentum operator, we usually use the standard basis ⏐ lm⟩, so every matrix element Aij that represents the operator A satisfies

A li A lji j = ⟨ ⟩| |ˆ (6.43)

Thus, for every l = constant, we can write a (2l + 1) × (2l + 1) matrix for ˆ ,L2 ˆ ,Lx ˆ ,Ly and ˆ ;Lz that is,

( ˆ ) ˆ ( )L li L lj l li j i j2 2 21= ⟨ ⟩ = +| | � δ (6.44)

( ˆ ) ˆL li L lj jz i j z i j= ⟨ ⟩ =| | �δ (6.45)

( ˆ ) ˆ ( ) ) (,L li L lj l m l m lx ij x i j= ⟨ ⟩ = − + + ++| |

�2

1 1( δ ++ − +⎡⎣ ⎤⎦−m l m i j) ) ,( 1 1δ (6.46)

( ˆ ) ˆ ( ) ) (,L li L lj l m l m ly i j y i j= ⟨ ⟩ = − + + −+| |

�2

1 1( δ ++ − +⎡⎣ ⎤⎦−m l m i j) ) ,( 1 1δ (6.47)

For the case when l = 1, we have

L2 2

11 10 1 1

21 0 00 1 00 0 1

11

1=⎛

⎝⎜⎜

⎞

⎠⎟⎟

⟩ ⟩ − ⟩⟩

�

| | |

|

| 00

1 1

⟩

− ⟩|

(6.48)

and

Lx =⎛

⎝⎜⎜

⎞

⎠⎟⎟

⟩ ⟩ − ⟩⟩

�

2

0 1 01 0 10 1 0

11 10 1 1

11

10

| | |

|

| ⟩⟩

− ⟩

=−

−⎛

⎝⎜⎜

⎞

⎠⎟⎟

⟩ ⟩ −

|

| | |

1 12

0 00

0 0

11 10 1

Li

i ii

y

�

11

11

10

1 1

⟩⟩

⟩

− ⟩

|

|

|

Lz =−

⎛

⎝⎜⎜

⎞

⎠⎟⎟

⟩ ⟩ − ⟩⟩

�

1 0 00 0 00 0 1

11 10 1 1

11

10

| | |

|

| ⟩⟩

− ⟩|1 1

(6.49)

6.7 Spherical Symmetry PotentialsFrom classical mechanics we know that when a spherical symmetry potential V(x, y, z) = V(r) acts on a particle, its angular momentum is a constant of motion. In terms of quantum mechanics, this means that the angular momentum operator L2 commutes with the Hamiltonian:

ˆ ˆ ˆ( )ˆ

Hpm

V rm r r

rr

L

m= + = − ∂

∂∂∂

⎛⎝⎜

⎞⎠⎟

+2 2

22

2

2 21

2

�

rrV r2 + ˆ( ) (6.50)

where the angular dependence of the Hamiltonian is found only in ˆ .L2 We can thus split the wavefunction in two: an angular part depending only on q and j, and a radial part depending only on r (see Problems 6.16 and 6.18).


6.8 Angular Momentum and RotationsLet ⏐ψ ⟩ be a state vector of a system in a certain coordinate system O. To represent the state vector in another coordinate system O′ we define the rotation operator ˆ ,UR such that the state vector in O′ is given by

| |′⟩ = ⟩ψ ψUR (6.51)

For a system O′ obtained by the rotation of O around an axis in the direction of the unit vector n through an angle q, UR is given as

ˆ ( , ) exp Ûi

R θ θn n L= −⎛⎝⎜

⎞⎠⎟� ⋅ (6.52)

where L is the angular momentum operator. L is said to be a generator of rotation. One can conclude from the definition that

⟨ ′ = ⟨ψ ψ| | ˆ †UR (6.53)

Note that to obtain UR we usually use the infinitesimal rotation operator:

ˆ ( , ) Û di

dR θ θn L n= −1� ⋅ (6.54)

Also note that

ˆ ( , ) ˆ ( , ) Û U IR R2 0π n n= = (6.55)

UR can be used as a rotation operator not only for state vectors, but also for other operators or observables. Thus, an observable A in the system O is transformed to A in the system O′ such that

ˆ ˆ ˆ ˆ †′ =A U AUR R (6.56)

Or similarly,

ˆ ˆ ˆ ˆ†A U A UR R= ′ (6.57)

SOLVED PROBLEMS

6.1. Using the definition of angular momentum, L = r × p, prove the following commutation relations:

(a) [ ˆ , ˆ ] ˆ ;L r i ri j ijk k

k

= ∑� ε (b) [ ˆ , ˆ ] ˆ ( , , , , ).L L i L i j k x y zi j ijk k

k

= =∑� ε Note that if A and B are vector

operators, then the kth component of the vector operator ˆ Â B× is

( ˆ ˆ ) ˆ Â B× = ∑k ijk i j

i j

A Bε (6.1.1)

Also use the identity ε ε δ δ δ δijk mnk im jn

k

in jm= −∑ .

SOLUTION

(a) Using the definition L = r × p we obtain ˆ ˆ ˆ ;L r pi kli k l

k l

= ∑ε thus,

[ ˆ , ˆ ] [ ˆ ˆ , ˆ ] ( ˆ [ ˆ , ˆL r r p r r p ri j kli k l j kli k l j= =ε ε ]] [ ˆ , ˆ ] ˆ )+∑∑ r r pk j l

k lk l

(6.1.2)


Using the commutation relations [ ˆ , ˆ ]r rk j = 0 and [ ˆ , ˆ ] ,p r il j l j= − �δ where

δlj

l j=

={10 otherwise

(6.1.3)

we obtain

[ ˆ , ˆ ] ( ) ˆ ˆL r i r i r ii j kji lj k kji k ik= − = − = −ε δ ε ε� � � jj k

k

ijk k

kkk l

r i rˆ ˆ∑ ∑∑∑ = � ε (6.1.4)

(b) We decompose the commutation relation [ ˆ , ˆ ] ˆL L i Li j ijk k

k

= ∑� ε into the following three commutation

relations: [ ˆ , ˆ ] ˆ ; [ ˆ , ˆ ] ˆ ;L L i L L L i Lx y z y z x= =� � and [ ˆ , ˆ ] ˆ .L L i Lz x y= � Note that

ˆ (ˆ ˆ ) ˆ ˆ ˆ ˆ ˆ (ˆ ˆ ) ˆ ˆL r p r p L rx x y z z y y y z= × = − = × =r p r p pp r px x z− ˆ ˆ (6.1.5)

Thus,

[ ˆ , ˆ ] [ ˆ ˆ ˆ ˆ , ˆ ˆ ˆ ˆ ]

[ ˆ

L L r p r p r p r p

r

x y y z z y z x x z= − −

= yy z z x y z x z z y zp r p r p r p r p r pˆ , ˆ ˆ ] [ ˆ ˆ , ˆ ˆ ] [ ˆ ˆ , ˆ ˆ− − xx z y x yr p r p] [ ˆ ˆ , ˆ ˆ ]+ (6.1.6)

We compute each part separately:

[ ˆ ˆ , ˆ ˆ ] ˆ [ ˆ , ˆ ˆ ] [ ˆ , ˆ ˆr p r p r p r p r r py z z x y z z x y z x= + ]] ˆ

ˆ ( ˆ [ ˆ , ˆ ] [ ˆ , ˆ ] ˆ ) ( ˆ [ ˆ

p

r r p p p r p r

z

y z z x z z x z= + + rr p r r p py x y z x z, ˆ ] [ ˆ , ˆ ] ˆ ) ˆ+ (6.1.7)

Now, using the known relations

[ ˆ , ˆ ] [ ˆ , ˆ ] [ ˆ , ˆ ] [ ˆ , ˆp p p r i r p r rz x z z y x y z= = − =0 0� ]] = 0 (6.1.8)

we obtain [ ˆ ˆ , ˆ ˆ ] ˆ ˆ .r p r p i r py z z x y x= − � Similarly,

[ ˆ ˆ , ˆ ˆ ] ˆ [ ˆ , ˆ ˆ ] [ ˆ , ˆ ˆr p r p r p r p r r py z x z y z x z y x z= + ]] ˆ

ˆ ( ˆ [ ˆ , ˆ ] [ ˆ , ˆ ] ˆ ) ([ ˆ ,

p

r r p p p r p r

z

y x z z z x z y= + + ˆ ] ˆ ˆ [ ˆ , ˆ ]) ˆr p r r p px z x y z z+ = 0 (6.1.9)

and

[ ˆ ˆ , ˆ ˆ ] ˆ [ ˆ , ˆ ˆ ] [ ˆ , ˆ ˆr p r p r p r p r r pz y z x z y z x z z x= + ]] ˆ

ˆ ([ ˆ , ˆ ] ˆ ˆ [ ˆ , ˆ ]) ([ ˆ ,

p

r p r p r p p r

y

z y z x z y x z= + + ˆ ] ˆ ˆ [ ˆ , ˆ ]) ˆr p r r p pz x z z x y+ = 0 (6.1.10)

Also,

[ ˆ ˆ , ˆ ˆ ] ˆ [ ˆ , ˆ ˆ ] [ ˆ , ˆ ˆr p r p r p r p r r pz y x z z y x z z x z= + ]] ˆ

ˆ ([ ˆ , ˆ ] ˆ ˆ [ ˆ , ˆ ]) ([ ˆ ,

p

r p r p r p p r

y

z y x z x y z z= + + ˆ ] ˆ ˆ [ ˆ , ˆ ]) ˆ ˆ ˆr p r r p p i r px z x z z y x y+ = � (6.1.11)

Thus, we obtain

[ ˆ , ˆ ] ( ˆ ˆ ˆ ˆ ) (ˆ ˆ ) ˆL L i r p r p i i Lx y x y y x z= − = × =� � �r p zz (6.1.12)

We leave it to the reader to prove the other two relations.

6.2. Prove the following relations for the angular momentum operator: (a) [ ˆ , ˆ ] ;L Lz2 0= (b) ˆ ˆ ˆL L L× = i� .

SOLUTION

(a) The operator L2 can be written as ˆ ˆ ˆ ˆ ,L L L Lx y z2 2 2 2= + + and hence

[ ˆ , ˆ ] [ ˆ ˆ ˆ , ˆ ] [ ˆ , ˆ ] [L L L L L L L Lz x y z z x z2 2 2 2 2= + + = + ˆ , ˆ ] [ ˆ , ˆ ]L L L Ly z z z

2 2+ (6.2.1)


We compute each part separately:

[ ˆ , ˆ ] ˆ [ ˆ , ˆ ] [ ˆ , ˆ ] ˆL L L L L L L Lx z x x z x z x2 = + (6.2.2)

We have shown in Problem 6.1 that [ ˆ , ˆ ] [ ˆ , ˆ ] ˆ .L L L L i Lx z z x y= − = − � Therefore,

[ ˆ , ˆ ] ( ˆ ˆ ˆ ˆ )L L i L L L Lx z x y y x2 = − +� (6.2.3)

Similarly, using the commutation relation [ ˆ , ˆ ] ˆ ,L L i Ly z x= � we have

[ ˆ , ˆ ] ˆ [ ˆ , ˆ ] [ ˆ , ˆ ] ˆ ( ˆL L L L L L L L i Ly z y y z y z y y2 = + = � ˆ ˆ ˆ )L L Lx x y+ (6.2.4)

Since Lz commutes with itself, [ ˆ , ˆ ] ,L Lz z2 0= we arrive at

[ ˆ , ˆ ] ( ˆ ˆ ˆ ˆ ) ( ˆ ˆ ˆ ˆL L i L L L L i L L Lz x y y z y x x2 = − + + +� � LLy) = 0 (6.2.5)

(b) We will compute separately the components of ˆ ˆ :L L×

( ˆ ˆ ) ˆ ˆ ˆ ˆ [ ˆ , ˆ ] ˆ

( ˆ ˆ

L L

L

× = − = =

×

x y z z y y z xL L L L L L i L�

LL

L L

) ˆ ˆ ˆ ˆ [ ˆ , ˆ ] ˆ

( ˆ ˆ )

y z x x z z x y

z

L L L L L L i L= − = =

× =

�

ˆ ˆ ˆ ˆ [ ˆ , ˆ ] ˆL L L L L L i Lx y y x x y z− = =

⎧

⎨

⎪⎪⎪

⎩

⎪⎪⎪ �

(6.2.6)

Thus, summing over the three components we obtain ˆ ˆ ˆ .L L L× = i�

6.3. Consider a system of two particles; each particle has its own angular momentum operator, L1 and ˆ .L2 Show that ˆ ˆ ˆL L L= +1 2 is an angular momentum operator; in other words, show that L satisfies the relation in part (b) of Problem 6.2.

SOLUTION

As L1 and L2 are both angular momentum operators, for the sum ˆ ˆ ˆL L L= +1 2 we have

ˆ ˆ ( ˆ ˆ ) ( ˆ ˆ ) ( ˆ ˆ ) ( ˆ ˆL L L L L L L L L L× = + × + = × + ×1 2 1 2 1 1 2 22 1 2 2 1) ( ˆ ˆ ) ( ˆ ˆ )+ × + ×L L L L (6.3.1)

In Problem 6.2, part (b), we saw that if L is an angular momentum operator, then ˆ ˆ ˆL L L× = i� . Thus,

ˆ ˆ ˆ ˆ ( ˆ ˆ ) ( ˆ ˆ ) ( ˆL L L L L L L L L× = + + × + × =i i i� � �1 2 1 2 2 1 11 2 1 2 2 1

1 2

+ + × + ×

= + × +

ˆ ) ( ˆ ˆ ) ( ˆ ˆ )

ˆ ( ˆ ˆ )

L L L L L

L L Li� (( ˆ ˆ )L L2 1× (6.3.2)

We will now compute the term ˆ ˆL L1 2× :

ˆ ˆ ( ˆ ˆ ˆ ˆ ) ( ˆ ˆ ˆL L1 2 1 2 1 2 1 2 1× = − + −L L L L L L Ly z z y z x xi ˆ ) ( ˆ ˆ ˆ ˆ )L L L L Lz x y y x2 1 2 1 2j k+ − (6.3.3)

Similarly,

ˆ ˆ ( ˆ ˆ ˆ ˆ ) ( ˆ ˆ ˆL L2 1 2 1 2 1 2 1 2× = − + −L L L L L L Ly z z y z x xi ˆ ) ( ˆ ˆ ˆ ˆ )L L L L Lz x y y x1 2 1 2 1j k+ − (6.3.4)

Since L1 and L2 are different operators, their components commutate; hence, we obtain

( ˆ ˆ ) ( ˆ ˆ )L L L L1 2 2 1 0× + × = (6.3.5)

So finally,

ˆ ˆ ( ˆ ˆ ) ( ˆ ˆ ) ( ˆ ˆ ) ˆL L L L L L L L L× = + × + = + =2 1 1 2 1 2i i� � (6.3.6)


6.4. Consider the following relations:

ˆ ˆ ˆ ˆ ˆ ˆL L iL L L iLx y x y+ −= + = − (6.4.1)

ˆ ( ) ( ) ,L lm l l m m l m+ ⟩ = + − + + ⟩| |� 1 1 1 (6.4.2)

ˆ ( ) ( ) ,L lm l l m m l m− ⟩ = + − − − ⟩| |� 1 1 1 (6.4.3)

L lm m lmz | |⟩ = ⟩� (6.4.4)

ˆ ( )L lm l l lm2 21| |⟩ = + ⟩� (6.4.5)

Consider a system of l = 1, and find the matrix representations of ˆ , ˆ , ˆ ,L L Lx y z and L2 in the basis of eigenvectors of Lz and ˆ .L2

SOLUTION

First, we note that the ˆ , ˆ , ˆ ,L L Lx y z and L2 are Hermitian operators, as are their matrix representations; for each component of the matrix aij we have a aij ji= * . For a system that has an angular momentum l = 1, the eigenvectors of Lz are

1 1 1

0

corresponding to

corresponding t

l m= =,

oo

corresponding to

l m

l m

= =

− = = −

⎧

⎨⎪⎪

⎩

1 0

1 1 1

,

,⎪⎪⎪

(6.4.6)

To find the matrix representation of Lx we need to compute the following relations:

ˆ ( ˆ ˆ ) ˆ

ˆ

L L L L

L

x

x

| | | |

|

112

112

12

0

0

⟩ = + ⟩ = ⟩ = ⟩

⟩ =

+ − −�

112

02

1 1

112

( ˆ ˆ )

ˆ ( ˆ

L L

L Lx

+ −

+

+ ⟩ = ⟩ + − ⟩( )

− ⟩ = +

| | |

|

�

ˆ ) ˆL L− +− ⟩ = − ⟩ = ⟩

⎧

⎨

⎪⎪⎪⎪

⎩

⎪⎪⎪⎪

| | |112

12

0�

(6.4.7)

If we choose the standard basis

1100

0010

1001

≡⎛

⎝⎜⎜

⎞

⎠⎟⎟

≡⎛

⎝⎜⎜

⎞

⎠⎟⎟

− ≡⎛

⎝⎜⎜

⎞

⎠⎟⎟

(6.4.8)

then the matrix representation of Lx is

Lx =⎛

⎝⎜⎜

⎞

⎠⎟⎟

�

2

0 1 01 0 10 1 0

(6.4.9)

Similarly, for Ly we have

ˆ ( ˆ ˆ )

ˆ ( ˆ ˆ )

Li

L Li

Li

L L

y

y

112

12

0

012

0

= − = ⟩

= −

+ −

+ −

�|

== − −

− ⟩ = − − = −

⎧

⎨

⎪⎪

+ −

i

Li

L Li

y

�

�

21 1

112

12

0

( )

ˆ ( ˆ ˆ )|

⎪⎪⎪

⎩

⎪⎪⎪⎪

(6.4.10)


Hence,

Li

i ii

y =−

−⎛

⎝⎜⎜

⎞

⎠⎟⎟

�

2

0 00

0 0 (6.4.11)

Also, for Lz we have ˆ , ˆ ,L Lz z| | |1 1 0 0⟩ = ⟩ ⟩ =� and ˆ ;Lz − = − −1 1� thus,

Lz =−

⎛

⎝⎜⎜

⎞

⎠⎟⎟

�

1 0 00 0 00 0 1

(6.4.12)

For L2 we have ˆ , ˆ ,L L2 2 2 21 2 1 0 2 0| | | |⟩ = ⟩ ⟩ = ⟩� � and ˆ ;L2 21 2 1| |− ⟩ = − ⟩� thus,

L2 221 0 00 1 00 0 1

=⎛

⎝⎜⎜

⎞

⎠⎟⎟

� (6.4.13)

6.5. What is the probability that a measurement of Lx will equal zero for a system that has an angular

momentum of one and is in the state 1

14

123

⎛

⎝⎜⎜

⎞

⎠⎟⎟

?

SOLUTION

First, we will find the eigenvectors of Lx for l = 1 in the basis of ˆ ;Lz i.e., we want to find the eigenvectors and eigenvalues of

Lx =⎛

⎝⎜⎜

⎞

⎠⎟⎟

�

2

0 1 01 0 10 1 0

(6.5.1)

Assuming that the eigenvalues of Lx are �λ / 2 , the secular equation of Lx is

det ( )−

−−

⎛

⎝⎜⎜

⎞

⎠⎟⎟

= − − + = − =λ

λλ

λ λ λ λ λ1 0

1 10 1

1 2 02 3 (6.5.2)

Hence, λ = ±0 2, and thus the eigenvalues of Lx are ±� or 0. The eigenvector corresponding to the eigenvalue � is

| | | |1 1 0 1⟩ =⎛

⎝⎜⎜

⎞

⎠⎟⎟

= ⟩ + ⟩ + − ⟩x

abc

a b c (6.5.3)

where ⏐ a ⏐2 + ⏐ b ⏐2 + ⏐ c ⏐2 = 1 is the normalization condition. Therefore,

�

�2

0 1 01 0 10 1 0

⎛

⎝⎜⎜

⎞

⎠⎟⎟

⎛

⎝⎜⎜

⎞

⎠⎟⎟

=⎛

⎝⎜⎜

⎞

⎠⎟⎟

abc

abc

(6.5.4)

or

b a= 2 (6.5.5)

a c a b+ = +2( ) (6.5.6)

b c= 2 (6.5.7)


From Eqs. (6.5.5) and (6.5.7) we obtain b a c= =2 2 ; thus, using the normalization condition, we have

a a a a2 2 22 112

+ + = ⇒ = (6.5.8)

Hence, the eigenvector ⏐ 1⟩x is

| | | |112

1

21

12

1 2 0 1⟩ =⎛

⎝⎜⎜

⎞

⎠⎟⎟

= ⟩ + ⟩ + − ⟩x ( ) (6.5.9)

Similarly, the eigenvector corresponding to the eigenvalue zero is

| | | |012 1 0 1⟩ =

⎛

⎝⎜⎜

⎞

⎠⎟⎟

= ⟩ + ⟩ + − ⟩x

abc

a b c (6.5.10)

where a, b, and c satisfy the normalization condition and

�2

0 1 01 0 10 1 0

0⎛

⎝⎜⎜

⎞

⎠⎟⎟

⎛

⎝⎜⎜

⎞

⎠⎟⎟

=abc

(6.5.11)

or

b a c= + =0 0 (6.5.12)

Therefore, a a a2 20 1 1 2+ + = ⇒ = / . Finally, the eigenvector ⏐ 0⟩x is

| | |012

101

12

1 1⟩ =−

⎛

⎝⎜⎜

⎞

⎠⎟⎟

= ⟩ − − ⟩x ( ) (6.5.13)

Also, the eigenvector corresponding to the eigenvalue −� is

| | | |− ⟩ =⎛

⎝⎜⎜

⎞

⎠⎟⎟

= ⟩ + ⟩ + − ⟩1 1 0 1x

abc

a b c (6.5.14)

where a, b, and c satisfy the normalization condition, and

�

�2

0 1 01 0 10 1 0

⎛

⎝⎜⎜

⎞

⎠⎟⎟

⎛

⎝⎜⎜

⎞

⎠⎟⎟

= −⎛

⎝⎜⎜

⎞

⎠⎟

abc

abc⎟⎟

(6.5.15)

or

b a a c b b c= − + = − = −2 2 2 (6.5.16)

Thus, b a c= − = −2 2 ; using the normalization condition, we obtain a a a a2 2 22 1 1 2+ + = ⇒ = / . Hence,

| | | |− ⟩ = −⎛

⎝⎜⎜

⎞

⎠⎟⎟

= ⟩ − ⟩ + − ⟩112

1

21

12 1 2 0 1x ( ) (6.5.17)


So, we can write

| | | |α⟩ ≡⎛

⎝⎜⎜

⎞

⎠⎟⎟

= ⟩ + ⟩ + − ⟩114

123

114

1 2 0 3 1( ) (6.5.18)

In the basis of the eigenvectors of Lx we have

| | | 1 | | | | −1α α α α⟩ = ⟨ ⟩ ⟩ + ⟨ ⟩ ⟩ + ⟨− ⟩ ⟩x x x x x x x1 0 0 1 (6.5.19)

We compute the terms separately:

x ⟨ ⟩ = + + = +1

12 14

1 2 2 32 2

14| α ( ) (6.5.20)

x ⟨ ⟩ = − = −0128

1 317

| α ( ) (6.5.21)

and

x ⟨− ⟩ = − + =

−1

1

2 141 2 2 3

2 2

14| α ( ) (6.5.22)

The probability that a measurement of Lx yields zero is therefore

Px x( )0 017

2= ⟨ ⟩ =| | |α (6.5.23)

6.6. Apply the operators ˆ ˆ ˆL L iLx y+ ≡ + and ˆ ˆ ˆL L iLx y− ≡ − on the eigenstates of L2 and ˆ ( )L lmz | ⟩ and interpret the physical meaning of the results. Use the following steps: (a) Find the Hermitian conjugate of ˆ .L+ (b) Calculate the norm of L lm+ ⟩| and ˆ .L lm− ⟩| (c) Calculate the eigenvalues of L2 and Lz for the state L lm+ ⟩| and ˆ .L lm− ⟩|

SOLUTION

(a) The Hermitian conjugate of L+ is ˆ ˆ ˆ† † †L L iLx y+ = − ; but, since ˆ ˆ†L Lx x= and ˆ ˆ†L Lx y= , we have ˆ ˆ .†L L+ −= (b) The norm of L lm+ ⟩| is

ˆ ( ˆ ) ˆ ) ( ˆ ˆ )† †L lm L lm L lm lm L L lm+ + + + += ⟩ ⟩ = ⟨2

| | | |( ⟩⟩ = ⟨ ⟩− +lm L L lm| |( ˆ ˆ ) (6.6.1)

We see that

ˆ ˆ ( ˆ ˆ )( ˆ ˆ ) ˆ ˆ ˆ ˆL L L iL L iL L L iLx y x y x y y− + = − + = + −2 2 LL iL L

L L i L L L L

x x y

x y x y z

+

= + + = − −

ˆ ˆ

ˆ ˆ [ ˆ , ˆ ] ˆ ˆ2 2 2 2� ˆLz

(6.6.2)

Thus, substituting ˆ ˆ ,L L− + we obtain

ˆ ( ˆ ˆ ) ( ˆ ˆ ˆ )L lm lm L L lm lm L L Lz z+ − += ⟨ ⟩ = ⟨ − −2 2 2| | | � || lm

l l m m l l m m

⟩

= + − − = + − +� �2 2 21 1 1[ ( ) ] [ ( ) ( )] (6.6.3)

The norm of L lm− ⟩| is || | || | |ˆ ˆ ˆ .L lm lm L L lm− + −⟩ = ⟨ ⟩2 Again,

ˆ ˆ ( ˆ ˆ ) ˆ ˆ ) ˆ ˆ ˆ ˆL L L iL L iL L L iLx y x y x y y+ − = + − = + +( 2 2 LL iL L L L i L L

L L

x x y z x y

z

− = − −

= − +

ˆ ˆ ˆ ˆ [ ˆ , ˆ ]

ˆ ˆ ˆ

2 2

2 2�LLz (6.6.4)


Hence, we obtain

ˆ ( ˆ ˆ ˆ ) [ ( ) ]L lm lm L L L lm l l m mz z− = − + = + − +

=

2 2 2 2 21� �

��2 1 1[ ( ) ( )]l l m m+ − − (6.6.5)

(c) First, consider the commutation relations:

[ ˆ , ˆ ] [ ˆ , ˆ ˆ ] [ ˆ , ˆ ] [ ˆ , ˆL L L L iL L L i L Lx y x2 2 2 2

+ = + = + yy] = 0 (6.6.6)

and

[ ˆ , ˆ ] [ ˆ , ˆ ˆ ] [ ˆ , ˆ ] [ ˆ , ˆL L L L iL L L i L Lx y x2 2 2 2

− = − = − yy] = 0 (6.6.7)

This means that ˆ ˆ ˆ ˆL L L L2 2+ += and ˆ ˆ ˆ ˆ .L L L L2 2

− −= The eigenvalues of L2 for L lm+ ⟩| and L lm− ⟩| are

ˆ ( ˆ ) ˆ ( ˆ ) ( ) ˆ

ˆ

L L lm L L lm l l L lm

L

2 2 2 1+ + +⟩ = ⟩ = + ⟩| | |�

22 2 2 1( ˆ ) ˆ ( ˆ ) ( ) ˆL lm L L lm l l L lm− − −⟩ = ⟩ = + ⟩

⎧⎨⎪

| | |�⎩⎩⎪ (6.6.8)

That is, L lm+ ⟩| and L lm− ⟩| are eigenstates of L2 with eigenvalues �2 1l l( ).+ Before we continue to calculate the eigenvalues of Lz note that

[ ˆ , ˆ ] [ ˆ ˆ , ˆ ] [ ˆ , ˆ ] [ ˆ , ˆL L L iL L L L i L Lz x y z x z y+ = + = + zz y xi L L L] ˆ ˆ ˆ= − − = − +� � � (6.6.9)

Hence, ˆ ˆ ˆ ˆ ˆL L L L Lz z+ + +− = −� and ˆ ˆ ˆ ˆ ˆ .L L L L Lz z+ + += + � Similarly,

[ ˆ , ˆ ] [ ˆ ˆ , ˆ ] [ ˆ , ˆ ] [ ˆ , ˆL L L iL L L L i L Lz x y z x z y− = − = + zz y xi L L L] ˆ ˆ ˆ= − + = −� � � (6.6.10)

Therefore, ˆ ˆ ˆ ˆ ˆL L L L Lz z− − −− = � and ˆ ˆ ˆ ˆ ˆ .L L L L Lz z− − −= − � Thus, we can calculate

ˆ ˆ ( ˆ ˆ ˆ ) ˆ ˆ ˆL L lm L L L lm L L lm Lz z z+ + + +⟩ = + ⟩ = ⟩ +| | |� � ++

+ + +

⟩

= ⟩ + ⟩ = + ⟩

|

| | |

lm

m L lm L lm m L lm� � �ˆ ˆ ( ) ˆ1 (6.6.11)

and also

ˆ ˆ ( ˆ ˆ ˆ ) ˆ ˆ ˆL L lm L L L lm L L lm Lz z z− − − −⟩ = − ⟩ = ⟩ −| | |� � −−

− − −

⟩

= ⟩ − ⟩ = − ⟩

|

| | |

lm

m L lm L lm m L lm� � �ˆ ( ) ˆ1 (6.6.12)

We see that L lm+ ⟩| and L lm− ⟩| are eigenstates of Lz with eigenvalues (m + 1) � and (m − 1) �, respec-tively. To conclude:

ˆ ( ) ( )

ˆ ( ) ( )

L lm l l m m

L lm l l m m

+

−

= + − +

= + − −

⎧

⎨⎪

�

�

1 1

1 1⎩⎩⎪

(6.6.13)

ˆ ( ˆ ) ( ) ˆ )

ˆ ( ˆ )

L L lm l l L lm

L L lm

2 2

2

1+ +

−

⟩ = + ⟩

⟩ =

| |

|

� (

��2 1l l L lm( ) ˆ )+ ⟩

⎧⎨⎪

⎩⎪ −( |

(6.6.14)

and,

ˆ ( ˆ ) ( ) ˆ )

ˆ ( ˆ ) (

L L lm m L lm

L L lm

z

z

+ +

+

⟩ = + ⟩

⟩ =

| |

|

�

�

1 (

mm L lm− ⟩

⎧⎨⎪

⎩⎪ −1) ˆ )( |

(6.6.15)


From Eq. (6.6.14) we see that L lm+ ⟩| and L lm− ⟩| are proportional to | lm′⟩ (note that m′ is distinct from m). From Eq. (6.6.15) we conclude that L lm+ ⟩| is proportional to ⏐l′, m + 1⟩ and that L lm− ⟩| is proportional to | ′ − ⟩l m, ;1 thus,

ˆ ~ , ; ˆ ~ ,L lm l m L lm l m+ −⟩ + ⟩ ⟩ − ⟩| | | |1 1 (6.6.16)

Recall that the norm of ⏐ l′, m + 1⟩ and L lm− ⟩| is 1; hence, from Eq. (6.6.13) we get

ˆ ( ) ( ) ,L lm l l m m l m+ ⟩ = + − + + ⟩| |� 1 1 1 (6.6.17)

ˆ ( ) ( ) ,L lm l l m m l m− ⟩ = + − − − ⟩| |� 1 1 1 (6.6.18)

So, we see that the operators L+ and L− allow us to “travel” between the eigenvalues of L2 and ˆ .Lz Also note that ˆ ,L l l+ ⟩ =| 0 and ˆ , .L l l− − ⟩ =| 0

6.7. Compute the expressions ⟨ ⟩lm L lmx| |ˆ 2 and ⟨ ⟩lm L L lmx y| |( ˆ ˆ ) in the standard angular momentum basis.

SOLUTION

We begin by representing Lx and ˆ ,Ly using L+ and ˆ :L−

ˆˆ ˆ

ˆˆ ˆ

LL L

LL L

ix y=+

=−+ − + −

2 2and (6.7.1)

Keeping in mind that

ˆ ( ) ( ) ,L lm l l m m l m+ ⟩ = + − + + ⟩| |� 1 1 1 (6.7.2)

and

ˆ ( ) ( ) ,L lm l l m m l m− ⟩ = + − − − ⟩| |� 1 1 1 (6.7.3)

the operator Lx2 can be written as

ˆ ( ˆ ˆ ) ( ˆ ˆ ˆ ˆ ˆL L L L L L L Lx2 2 2 21

414

2 2= − = − + ++ − + − + − −ˆ )L+ (6.7.4)

The terms L+2 and L−

2 do not contribute to the expression ⟨ ⟩lm L lmx| |ˆ 2 since

⟨ ⟩ ⟨ + ⟩ =

⟨ ⟩ ⟨

+

−

lm L lm lm l m

lm L lm lm l

| | |

| | |

ˆ ~ ,

ˆ ~

2

2

2 0

,, m − ⟩ =

⎧⎨⎪

⎩⎪ 2 0

(6.7.5)

Thus, to compute ⟨ ⟩lm L lmx| |ˆ 2 we consider only the contributions of ˆ ˆL L− + and ˆ ˆ ;L L+ − that is,

⟨ ⟩ = ⟨ + ⟩

= ⟨

+ − − +lm L lm lm L L L L lmx| | | |ˆ ( ˆ ˆ ˆ ˆ )

[

2 12

12

llm L L lm lm L L lm

l l

| | | |( ˆ ˆ ) ( ˆ ˆ ) ]

[ (

+ − − +⟩ + ⟨ ⟩

= +�2

1)) ( ) ˆ , ( ) ( )( ˆ− − ⟨ − ⟩ + + − + ⟨+m m lm L l m l l m m lm1 1 1 1| | | LL l m− + ⟩| , )]1

= + − − + − − ⟨ ⟩

+ +

�2

21 1 1 1[ ( ) ( ) ( ) ( )

(

l l m m l l m m lm lm

l l

|

11 1 1 1

21

2

) ( ) ( ) ( ) ]

[ ( )

− + + − + ⟨ ⟩

= + −

m m l l m m lm lm

l l

|

�mm m l l m m

l l m

( ) ( ) ( )]

[ ( ) ]

− + + − +

= + −

1 1 1

12 2� (6.7.6)


Next, we compute ⟨ ⟩lm L L lmx y| |( ˆ ˆ ) . Using the operators L+ and ˆ ,L− we obtain

ˆ ˆ ( ˆ ˆ )( ˆ ˆ ) ( ˆ ˆL Li

L L L Li

L Lx y = + − = − −+ − + − + −14

14

2 2 ˆ ˆ ˆ ˆ )L L L L+ − − ++ (6.7.7)

Once again the terms of L+2 and L−

2 do not contribute to ⟨ ⟩lm L L lmx y| |( ˆ ˆ ) ; thus,

⟨ ⟩ = ⟨ − + −+ − − +lm L L lmi lm L L L Lx y| | |( ˆ ˆ ) ( ˆ ˆ ˆ ˆ ˆ1

42 2 LL L lm i

lm L L L L lm+ − − + + −⟩⎡⎣ ⎤⎦ = ⟨ − ⟩⎡ˆ ) ( ˆ ˆ ˆ ˆ )| | |14 ⎣⎣

⎤⎦

= ⟨ ⟩ − ⟨ ⟩⎡⎣ − + + −14i lm L L lm lm L L lm| | | |( ˆ ˆ ) ( ˆ ˆ ) ⎤⎤⎦

= + − + ⟨ + ⟩ − + −−�4 1 1 1 1i l l m m lm L l m l l m( ) ( ) ˆ , ( )| | (( ) ˆ ,m lm L l m− ⟨ − ⟩⎡⎣ ⎤⎦+1 1| |

= + − + + − + ⟨ ⟩⎡⎣

−

�2

41 1 1 1

il l m m l l m m lm lm

l

( ) ( ) ( ) ( )

(

|

ll m m l l m m lm lmi m+ − − + − − ⟨ ⟩⎤⎦ =1 1 1 1

2

2

) ( ) ( ) ( ) |�

(6.7.8)

6.8. Consider a particle with a wavefunction

ψ( , , ) ( ) [( ) / ]x y z N x y z e x y z= + + − + +2 2 2 2α (6.8.1)

where N is a normalization constant and a is a parameter. We measure the values of L2 and ˆ .Lz Find the probabilities that the measurements yield: (a) ˆ , ˆ ;L Lz

2 22 0= =� (b) ˆ , ˆ ;L Lz2 22= =� �

(c) ˆ , ˆ .L Lz2 22= = −� � Use the known relations

Y e Y Yi11

10

113

83

4( , ) sin ( , ) cos (θ ϕ π θ θ ϕ π θ θϕ= − = − − ,, ) sinϕ π θ ϕ= − −3

8e i (6.8.2)

SOLUTION

First, we will express ψ (x, y, z) in spherical coordinates:

x r y r z r= = =sin cos sin sin cosθ ϕ θ ϕ θ (6.8.3)

where r2 = x2 + y2 + z2. So,

ψ( , , ) [sin (cos sin ) cos ]r N re rθ ϕ θ ϕ ϕ θ α= + + − 2 2/ (6.8.4)

We write ψ (r, q, j) as a multiple of two functions ψ (r, q, j) = R(r)T(q, j) where R r Nre r( ) = − 2 2/α and

T a Ylm lm

l m

( , ) ( , ) sin cos sin sin cosθ ϕ θ ϕ θ ϕ θ ϕ θ= = + +∑∑ (6.8.5)

The coefficients alm are determined by

a lm T Y T d dlm lm= ⟨ ⟩ = ∫| ( , ) ( ) ( , )*θ ϕ θ ϕ θ ϕ (6.8.6)

Using the properties of spherical harmonics, one can prove that

T Y Y i Y Y( , )θ ϕ π= −( ) − +( )⎡

⎣⎢⎤⎦⎥

+− −83

12

121

111

11

11 44

3

23 1 1 2

10

11

11

10

π

π

Y

i Y i Y Y= + − − +⎡⎣ ⎤⎦−( ) ( ) (6.8.7)


To compute the probabilities, we must normalize the function T(q, j); we denote the normalized function by T ′(q, j) = b T (q, j), where

β θ ϕ θ ϕ θ ϕ β π π β2 2 223

2 2 2 4 1T T d d*( , ) ( , ) ( )∫ = + + = = (6.8.8)

or β π= 1 4/ . Hence we have

′ = + − − +⎡⎣ ⎤⎦−T i Y i Y Y( , ) ( ) ( )θ ϕ 1

61 1 21

111

10 (6.8.9)

Thus, the probabilities are computed as follows:

(a) For L2 22= � and Lz = 0, we have

P T= ⟨ ′⟩ = =| | |1 016

213

22

, ˆ (6.8.10)

(b) For L2 22= � and Lz = �, we have

P Ti= ⟨ ′⟩ = − − =| | |1 1

16

13

22

, (6.8.11)

(c) For L2 22= � and Lz = −�, we have

P Ti= ⟨ − ′⟩ = + =| | |1 1

16

13

22

, (6.8.12)

6.9. A symmetrical top with moments of inertia Ix = Iy and Iz in the body axes frame is described by the Hamiltonian

ˆ ( ˆ ˆ ) ˆHI

L LI

Lx

x yz

z= + +12

12

2 2 2 (6.9.1)

Note that moments of inertia are parameters and not operators. ˆ , ˆ ,L Lx y and Lz are the angular momentum operators in the body axes frame. (a) Calculate the eigenvalues and the eigenstates of the Hamiltonian. (b) What values are expected for a measurement of ˆ ˆ ˆL L Lx y z+ + for any state? (c) The state of the top at time t = 0 is ⏐ l = 3, m = 0⟩. What is the probability that for a measurement of Lx at t Ix= 4π /� we will obtain the value �?

SOLUTION

(a) We begin by writing the Hamiltonian as

ˆ ( ˆ ˆ ˆ ) ˆHI

L L LI I

Lx

x y zz z

z= + + + −⎛⎝⎜

⎞⎠⎟

12

12

12

2 2 2 2 == + −⎛⎝⎜

⎞⎠⎟

12

12

12

2 2

IL

I IL

x z xz

ˆ ˆ (6.9.2)

where L is the total angular momentum. Recall that if A is an operator that has the eigenvalues li (i = 1, . . . , n), then the eigenvalues of f A( ˆ) (where f A( ˆ) is a function of ˆ)A are f i( ).λ Therefore, the eigenvalues of the energy are

EI

l lI I

mlmx z x

= + + −⎛⎝⎜

⎞⎠⎟

��

22 2

21

12

12

( ) (6.9.3)

So, the eigenstates of the Hamiltonian are those of L2 and ˆ ,Lz i.e., the spherical harmonics Ylm ( , )θ ϕ with

the eigenenergies Elm .


(b) Measuring ˆ ˆ ˆL L Lx y z+ + for the top, we find the top at eigenstate Ylm ( , );θ ϕ that is, a measurement of

ˆ ˆ ˆL L Lx y z+ + yields

⟨ + + ⟩ = ⟨Y L L L Y Yl

mx y z l

mlm( , ) ( ˆ ˆ ˆ ) ( , ) ( , )θ ϕ θ ϕ θ ϕ| | || |

ˆ ˆ ˆ ˆˆ ( , )

L L L Li

L Y

Y

z lm+ − + −+

+−

+⎛

⎝⎜

⎞

⎠⎟ ⟩

= ⟨

2 2θ ϕ

llm

z lmL Y m( , ) ( , )θ ϕ θ ϕ| | ⟩ = � (6.9.4)

(c) The state of the top at t = 0 is ψ( , , ) ( , ),t Y= =0 30θ ϕ θ ϕ which is an eigenstate of the Hamiltonian. A

measurement of Lz for this state yields zero, and since it is an eigenstate of ˆ ,H the top will always remain in this state. Therefore, the probability of the measurement of � is zero.

6.10. The spherical harmonic functions are defined by

Y C P emlm

lm im

1 ( , ) (cos )θ ϕ θ ϕ= (6.10.1)

where Clm is a normalization constant and P xl

m ( ) are the associated Legendre functions defined by

P x xd

dxP x P xl

m mm

m l lm( ) ( ) ( ) ( )| | /

| |

| |= − = −1 2 2 (6.10.2)

Compute the function Y m1 ( , )θ ϕ for m = 0, ±1.

SOLUTION

Consider the Legendre polynomial P x x1( ) ;= so, ddx

P x( ( )) .1 1= Therefore, relying on Eq. (6.10.2) (see the Mathematical Appendix), we have

P x P x x1 11 21′ = = −( ) ( ) (6.10.3)

Similarly, P x x10 ( ) ;= thus, using Eq. (6.10.1) we obtain

Y C P e C ei i11

11

11

11( , ) (cos ) sinθ ϕ θ θϕ ϕ= = (6.10.4)

Also,

Y C e Y Ci1

11

110

10− − −= =( , ) sin ( , ) cosθ ϕ θ θ ϕ θϕ (6.10.5)

Using the normalization condition we arrive at

d Y Y d d Cm

llmϕ θ ϕ θ ϕ θ θ ϕ

π π

0

2

1 101∫ ∫ = ⇒( ) ( , ) ( , ) sin (* )) cos2 2

00

2

1θ θππ

d =∫∫ (6.10.6)

or, − = =2 13

410 2 2

10

0

π θ θ π

π

( ) cos (cos ) ;C d Cthat is,∫∫ . Similarly,

C C d e e di i11

11

0

2

0

= =⎛

⎝⎜

⎞− −∫ ∫ϕ θ θ θ θπ

ϕ ϕπ

sin sin sin⎠⎠⎟ =

⎛

⎝⎜

⎞

⎠⎟ =

− −

∫2

3

0

2

23

8π θ θ π

π

sin d (6.10.7)

Finally, we have

Y Y e Yi10

11

113

43

8( , ) cos ( , ) sin ( ,θ ϕ π θ θ ϕ π θ θ ϕϕ= = − )) sin= −3

8π θ ϕe i (6.10.8)


6.11. Solve the eigenvalue equation ˆ ( , ) ( , ),L Y Y2 2θ ϕ λ θ ϕ= � and find the eigenvalues of ˆ .L2 Use the expression for L2 in spherical coordinates.

ˆsin sin

sinL2 22

2

2

1 1= − ∂∂

+ ∂∂

∂∂

⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢�

θ ϕ θ θ θ θ⎤⎤

⎦⎥ (6.11.1)

SOLUTION

We begin by substituting the expression for L2 in the eigenvalue equation, so we obtain

1 12

2

2sin sinsin (

θ ϕ θ θ θ θ∂

∂+ ∂

∂∂

∂⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢⎢

⎤

⎦⎥⎥Y θθ ϕ λ θ ϕ, ) ( , )= − Y (6.11.2)

We solve this equation using the variables separation method; thus, we substitute Y ( , ) ( ) ( )′ =θ ϕ ϕ θΦ Θ and get

Θ Φ Φ Θ

Φ Θsin sin

sin ( )2

2

2θ ϕ θ θ θ θ λ ϕd

d

dd

dd

+ ⎛⎝⎜

⎞⎠⎟ = − (( )θ (6.11.3)

Dividing Eq. (6.11.2) by Θ Φ( ) ( )

sin

θ ϕθ2 we obtain

1

02

22

ΦΦ

ΘΘd

d

dd

ddϕ

θθ θ θ λ θ+ ⎛

⎝⎜⎞⎠⎟ + =

sinsin sin (6.11.4)

We now have two parts: The first, 1 2

2ΦΦd

dϕ, is a function of j only, and the second,

sinsin

θθ θ θΘ

Θdd

dd

⎛⎝⎜

⎞⎠⎟ +

λ θsin ,2 is a function of q only; the sum of these parts yields zero. Therefore, each of them must be a constant. We set

1 2

22

ΦΦd

dm

ϕ= − (6.11.5)

and

sin

( )sin sin

θθ θ θ θ λ θΘ

Θdd

dd

m⎛⎝⎜

⎞⎠⎟ + =2 2

(6.11.6)

The solution of Eq. (6.11.5) is

Φ( )ϕ ϕ= eim (6.11.7)

To qualify as a periodic function, Φ( )ϕ must satisfy the condition Φ Φ( ) ( );ϕ π ϕ+ =2 that is e im2 1π = , thus, m must be an integer number, m = 0, ±1, ±2, . . . . Now Eq. (6.13.5) can be expressed in terms of x = cos q, where

d

ddxd

ddx

ddx

xddxθ θ θ= = − = − −sin 1 2 (6.11.8)

Substituting into Eq. (6.11.6), we now have

1

1 1 02

2 2 2−−⎡

⎣⎢⎤⎦⎥

+ − − =x d

dxx

ddx

x mΘΘ

( ) ( )λ (6.11.9)

We rearrange Eq. (6.11.9) in order to obtain the usual form of the generalized Legendre equation:

ddx

xddx

m

x( )1

102

2

2−⎡⎣⎢

⎤⎦⎥

+ −−

⎛

⎝⎜⎞

⎠⎟=

ΘΘλ (6.11.10)


Note that under the transformation x → −x, Eq. (6.11.10) is unchanged. This means that the solutions of the generalized Legendre equation are either symmetric or antisymmetric in x. Consider the equation for m2 0= :

ddx

xddx

( )1 02−⎡⎣⎢

⎤⎦⎥

+ =Θ

Θλ (6.11.11)

Assume that the solution can be represented by a power series; so, Θ( ) .x x a xsn

n

n

==

∞

∑0

We leave it for the reader to show that by substituting we obtain

(( )( ) [( )( ) ](s n s n a x s n s n ans n

n+ + + + − + + + −++2 1 12 λ xxs n

n

+

=

∞

=∑ )) 0

0

(6.11.12)

Hence, each coefficient must vanish, and we have

( ) ) [( ) ) ]s n s n a n s n an n+ + + + = + + + −+2 1 1 12( ( λ (6.11.13)

or

as n s ns n s n

an n+ = + + + −+ + + +2

12 1

( )( )( )( )

λ (6.11.14)

The function Θ(x) is bounded at x = 1 (q = 0), so the condition (s + n)(s + n + 1) − l = 0 must hold for l. That is, l must be of the form l = l (l + 1), where l is an integer number. Hence, the eigenvalues of L2 are �

2 1l l( ).+ The solution of Eq. (6.11.11) can be represented as

Θl l

l

llx

l

d

dxx( )

!( )= −

1

212 (6.11.15)

Similarly, the general solutions of Eq. (6.11.10) are

Θlm

m

lm

l m

l mxl

xd

dx( )

( )

!( )| | /

( | |)

( | |= − −+

+1

21 2 2

))| | /( ) ( ) ( ) ( )x x

d

dxP xl m m

m

m l2 2 21 1 1− = − − (6.11.16)

6.12. Consider a particle in a central potential. Given that ⏐ lm⟩ is an eigenstate of L2 and ˆ :Lz (a) Compute the sum Δ ΔL Lx y

2 2+ . (b) For which values of l and m does the sum in part (a) vanish?

SOLUTION

(a) The uncertainties ΔLx2 and ΔLx

2 are defined as

Δ ΔL L L L L Lx x x y y y2 2 2 2 2 2= ⟨ ⟩ − ⟨ ⟩ = ⟨ ⟩ − ⟨ ⟩ˆ ˆ ˆ ˆ (6.12.1)

Using the raising and lowering operators L+ and ˆ ,L− we write ˆˆ ˆ

LL L

x =++ −2

and ˆˆ ˆ

.LL L

iy =−+ −2

There-fore, we have

ˆ ( ˆ ˆ ˆ ˆ ˆ ˆ ) ˆ ( ˆL L L L L L L L Lx y2 2 2 21

414

= + + + = −+ − + − − + ++ − + − − ++ − −2 2ˆ ˆ ˆ ˆ ˆ )L L L L L (6.12.2)

So,

⟨ ⟩ = ⟨ ⟩ = ⟨+⎛

⎝⎜

⎞

⎠⎟ ⟩ =

⟨

+ −ˆ ˆˆ ˆ

L lm L lm lmL L

lmx x| | | |2

0

ˆ ˆˆ ˆ

L lm L lm lmL L

ilmy y⟩ = ⟨ ⟩ = ⟨

−⎛

⎝⎜

⎞

⎠⎟ ⟩ =+ −| | | |

20

⎧⎧

⎨

⎪⎪⎪

⎩

⎪⎪⎪

(6.12.3)


since

ˆ ( ) ( ) ,

ˆ (

L lm l l m m l m

L lm l l

+

−

⟩ = + − + + ⟩

⟩ = +

| |

|

�

�

1 1 1

1)) ( ) ,− − − ⟩

⎧⎨⎪

⎩⎪ m m l m1 1| (6.12.4)

Similarly, we can compute

⟨ ⟩ = ⟨ ⟩ = ⟨ + ++ − +

ˆ ˆ ( ( ˆ ˆ ˆ ˆL lm L lm lm L L L Lx x2 2 2 21

4| | | −− − +

+ −

+ ⟩

= ⟨ + ⟩ + ⟨

ˆ ˆ ) )

( ( ˆ ˆ )

L L lm

lm L L lm lm

|

| | |14

2 2 (( ˆ ˆ ˆ ˆ ) )L L L L lm+ − − ++ ⟩| (6.12.5)

Relying on the properties of the raising and lowering operators we have

ˆ ~ , ˆ ~ ,L lm l m L lm l m+ −⟩ + ⟩ ⟩ − ⟩2 22 2| | | | (6.12.6)

We also have

ˆ ˆ ˆ ( ( ) ( ) , ) [ (L L lm L l l m m l m l+ − +⟩ = + − − − ⟩ =| |� �1 1 1 2 ll m m lm+ − − ⟩1 1) ( )] | (6.12.7)

and

ˆ ˆ ˆ ( ( ) ( ) , ) [ (L L lm L l l m m l m l− + −⟩ = + − + + ⟩ =| |� �1 1 1 2 ll m m lm+ − + ⟩1 1) ( )] | (6.12.8)

Thus, we obtain

⟨ ⟩ = + − − + + − + =ˆ [ ( ) ( ) ( ) ( )] [L l l m m l l m mx2 2 21 1 1 1 2� � ll l m( ) ]+ −1 2 (6.12.9)

Similarly,

⟨ ⟩ = ⟨ ⟩ = + −ˆ ˆ [ ( ) ]L L l l my x2 2 2 22 1� (6.12.10)

Finally, we have

Δ ΔL L L L L L lx y x x y y2 2 2 2 2 2 24+ = ⟨ ⟩ − ⟨ ⟩ + ⟨ ⟩ − ⟨ ⟩ =ˆ ˆ ˆ ˆ [� (( ) ]l m+ −1 2 (6.12.11)

(b) Using the result of part (a), we see that Δ ΔL Lx y2 2+ vanishes when l (l + 1) − m2 = 0; that is,

m2 = l (l + 1). Using the fact that m and l must be integers, we conclude that this condition is satisfied only when l = m = 0.

6.13. Consider a system with a state function

ψ( , ) expr t Nr

r= = −

⎛

⎝⎜

⎞

⎠⎟0

2

2

02ξ (6.13.1)

where x = x + iy; N is a normalization constant and r0 is a given parameter. It is also given that the eigenfunctions of L2 and Lz are the spherical harmonic functions

Y x y zr

Y x y zzr

Y x y z11

10

113

83

4( , , ) ( , , ) ( , ,= − = −

πξ

π ))*

= − 38π

ξr (6.13.2)

(a) What are the values obtained from a measurement of L2 and ˆ ?Lz Also find the probability for each measurement. (b) Write the three eigenfunctions of L2 and Lx corresponding to the given spherical harmonics. (c) Find the values that are expected from a measurement of ˆ .Lx What is the probability for each value?


SOLUTION

(a) Consider the operators L2 and ˆ .Lz They operate only on the part of the function that depends on the angles j and q. Note that we can write ψ as

ψ( , ) exp ( , , )r t Nrr

rY x y z= = −

−⎛

⎝⎜

⎞

⎠⎟0

83 2

2

02 1

1π (6.13.3)

Hence, we see that the possible values in a measurement of L2 and Lz are 2 2� and �, respectively, with

a probability of 100 percent (since L2 and Lz operate only on Y x y z11( , , ), which is an eigenfunction of

these operators with these eigenvalues).

(b) Consider a system K′ in which the x′, y′, and z′ axes are parallel to the x, y, and z axes of our system. In this system the operator Lz ′ is similar to Lx in K; thus the eigenfunction of Lz ′ is also the eigenfunction of Lx with the following substitutions: x′ → y; y′ → z; z′ → x. The eigenfunction of Lz ′ is

( ( , , ))Y x y zr

x iy

xL

z11

2

38

38′ ′ ′ = − ′

′ = − ′ + ′

′ + ′′ πξ

π yy z2 2+ ′ (6.13.4)

Therefore, the eigenfunction of Lx is

( ( , , ))Y x y zy iz

x y z

y izrL

x11

2 2 2

38

38

= −+

+ += −

+π π (6.13.5)

Since L2 commutes with Lz and ˆ , ( )L Yx Lx

11 is an eigenfunction of ˆ ;Lx it is also an eigenfunction of ˆ .L2

Similarly,

( ) ( )Yxr

Yy iz

rL Lx x

01

113

43

8= =

−−π π (6.13.6)

(c) Following parts (a) and (b), we use the expansion theorem to write (see Chap. 4)

ψ( , ) exp ( ) (r t Nrr

rY

iY

Lx

= = −⎛

⎝⎜

⎞

⎠⎟ +0

2

43 2

2

02 0

11

π −− +⎡⎣

⎤⎦

⎛⎝⎜

⎞⎠⎟

111) ( )L L

x xY (6.13.7)

Consider only the part of ψ that is an eigenfunction of Lx and ˆ :L2

P x y z Yi

Y YL L L

x x x( , , ) ( ) ( ) ( )= + −⎡

⎣⎤⎦

⎛⎝⎜

−α 01

11

11

2⎞⎞⎠⎟

(6.13.8)

where a is a normalization constant, ⟨ ⟩ = + +⎛⎝⎜

⎞⎠⎟ = ⇒ =P P| α α2 1

12

12

112

. Therefore,

P x y z Yi

Y YL L L

x x x( , , ) ( ) ( ) ( )= + −⎡

⎣⎤⎦

⎛⎝

−12 20

11

111

⎜⎜⎞⎠⎟

(6.13.9)

The values expected from the measurements of Lx and L2 are therefore as follows: For L2 22= � and

ˆ ,Lx = 0 the probability is ⟨ ⟩ =( ) .Y PLx

01

2 12

| For L2 22= � and ˆ ,Lx = � the probability is ⟨ ⟩ =( ) .Y PLx

11

2 14

|

Finally, for L2 22= � and ˆ ,Lx = −� the probability is ⟨ ⟩ =( ) .Y PLx

11

2 14

|

6.14. Consider a particle in a spherical and infinite potential energy well:

V rr a

a r( ) =

≤ ≤

∞ <

⎧⎨⎪

⎩⎪

0 0 (6.14.1)

(a) Write the differential equations of the radial and angular parts, and solve the angular equation. (b) Compute the energy levels and the stationary wave equation for l = 0.


SOLUTION

(a) We begin by writing the Hamiltonian of the system:

ˆ ˆ( ) ( )H

pm

V rm

V r= + = − ∇ +2 2

2

2 2�

(6.14.2)

where ∇2 in spherical coordinates is

∇ = ∂∂

+ ∂∂

∂∂

⎛⎝⎜

⎞⎠⎟

+22

2 21 1 1 1r r

rr

( )sin

sinsinθ θ θ θ 22

2

2 2

2

2 21

θ ϕ∂

∂⎡

⎣⎢⎢

⎤

⎦⎥⎥

= ∂∂

−2

r rr

L

r( )

ˆ

� (6.14.3)

Thus,

ˆ ( )ˆ

( )Hm r r

rL

mrV r= − ∂

∂+ +�

2 2

2

2

221

2 (6.14.4)

The differential equation for the stationary wavefunction ψ (r, q, j) is

ˆ ( )ˆ

( )Hm r r

rL

mrV r Eψ ψ ψ ψ ψ= − ∂

∂+ + =�

2 2

2

2

221

2 (6.14.5)

It is evident that [ ˆ , ˆ ] ;H L2 0= hence, we write ψ( , , ) ( ) ( , )r R r Ynl lmθ ϕ θ ϕ= and obtain

− ∂∂

+�2 2

2

2

2mY

r rrR r

R r L Ylm

nlnl l

m( , )[ ( )]

( ) ( ,θ ϕ θ ϕϕθ ϕ θ ϕ

)( ) ( ) ( , ) ( ) ( , )

2 2mrR r V r Y ER r Ynl l

mnl l

m+ = (6.14.6)

Since Ylm( , )θ ϕ is the eigenfunction of ˆ ,L2

ˆ ( , ) ( ) ( , )L Y l l Ylm

lm2 2 1θ ϕ θ ϕ= +�

Hence, the radial equation is

− ∂∂

+ + +⎡

⎣⎢⎢

� �2 2

2

2

221

21

m r rrR r

mrl l V rnl[ ( )] ( ) ( )

⎤⎤

⎦⎥⎥

=R r ER rnl nl( ) ( ) (6.14.7)

(b) For l = 0, we have

− ∂∂

+ =�2 2

2 0 0 021

m r rrR r V r R r ER rn n n[ ( )] ( ) ( ) ( ) (6.14.8)

We denote R r R rn0 ( ) ( ).= For r > a, the function must vanish [because V(r) is infinite]; therefore, we have for 0 ≤ r ≤ a:

− ∂∂

=�2 2

221

m r rrR r ER r( ( )) ( ) (6.14.9)

We substitute U(r) = rR(r); hence, − ∂∂

=�2 2

22mU

rEU r( ), or

∂∂

+ =2

2 22

0U

r

mEU r

�( ) (6.14.10)

The solution of Eq. (6.14.10) is

U r A kr B kr( ) cos ( ) sin )= + ( (6.14.11)

where k mE= 2 2/� . A and B are constants that can be determined using the boundary conditions:

1. The value of U vanishes on r = 0: U (r = 0) = [rR(r)]⏐r = 0 = 0. 2. The value of U vanishes on r = a: U (r = a) = [rR(r)]⏐r = a = 0.

Thus, from the first condition we have U(0) = A = 0, and using the second condition,

U (a) = B sin (ka) = 0 ⇒ ka = np (6.14.12)


we obtain

Ema

nn = π 2 2

22

2

� (6.14.13)

Finally, to compute the value of B we use the normalization condition of the wavefunction R(r):

R rU r

r

Bkr

rr a

( )( )

sin( )

= =≤ ≤⎧

⎨⎪

⎩⎪

0

0 otherwise

(6.14.14)

Hence,

| |R r r dr Bkr

rr dr B

a

( )sin ( )2 2 2

0

2

22 2

0

4 4 4π π π= =∫∫∞

ssin( )

cos sin

kr dr

Bk

x x x

a

ka

0

2

0

4 12

12

∫= − +⎡

⎣⎢⎤⎦⎥

π==

= =n

n Bn a

πππ

21

2 2

/ (6.14.15)

so Ba

=1

2π. Thus, for l = 0 we have

ψ( , , ) ( ) sinr R ra r

mErθ ϕ

π= =

⎛⎝⎜

⎞⎠⎟

1

2

1 22

� (6.14.16)

6.15. Consider the classical Hamiltonian of a three-dimensional isotropic harmonic oscillator:

Hm

p p pm

x y zx y z= + + + + +1

2 22 2 2

22 2 2( ) ( )

ω (6.15.1)

(a) Write the Hamiltonian in spherical coordinates. (b) Find the eigenfunctions of the Hamiltonian in spherical coordinates. (c) Find the energy eigenvalues.

SOLUTION

(a) We begin by writing

ˆ ˆ ˆp p px y zx x x

2 2 2 22

2

2

2

2

2+ + = − ∂∂

+ ∂∂

+ ∂∂

⎛

⎝⎜⎞

⎠⎟= −� ��

2 2∇ (6.15.2)

which, in spherical coordinates, becomes

− ∇ = −∂∂

+∂

∂∂

∂⎛⎝⎜

⎞⎠⎟

� �2 2 2

2

2 2

1 1r

r

r r

( )

sinsin

θ θ θ θ ++∂

∂

⎡

⎣⎢⎢

⎤

⎦⎥⎥

12 2

2

2r sin θ ϕ (6.15.3)

Using ˆsin

sinsin

L2 22

2

21 1= − ∂

∂∂

∂⎛⎝⎜

⎞⎠⎟

+ ∂∂

⎡

⎣⎢� θ θ θ θ θ ϕ⎢⎢

⎤

⎦⎥⎥

, we arrive at

− ∇ = − ∂∂

+��2 2

2 2

2

2

2rr

r

L

r

( ) ˆ (6.15.4)

In spherical coordinates, the Hamiltonian is therefore

ˆ ( ) ˆH

m rr

r

L

mr

mr= − ∂

∂+ +�

2 2

2

2

2

22

21

2 2ω

(6.15.5)


(b) The angular dependence of the Hamiltonian comes only from ˆ ;L2 therefore, writing the eigenfunction in the form ψ( , , ) ( ) ( , ),r R r Yl

mθ ϕ θ ϕ= we have

ˆ ( , )( ( ))

( ) ˆHm

Yr

d

drrR r

R r

mrL Yl

m

ψ = − +�2 2

2 22

2 2

θ ϕllm

lmm

r R r Y E( , ) ( )θ ϕ ω+ =2

2

2ψ (6.15.6)

or

ˆ ( ) ( ( ))( )

HR r Ym

Yr

d

drrR r

l l

mrlm l

m

= − + +� �2 2

2

2

21

2 22

2 22

2 2R r Y

mR r Y

mr R r Y

ER r Y

lm

lm

lm

l

( ) ( ) ( )

( )

+ +

=

ω ω

mm (6.15.7)

We get the radial equation

− + + +⎡

⎣⎢⎢

� �2 2

2

2

2

22

21 1

2 2m rd

drrR r

l l

mr

mr( ( ))

( ) ω ⎤⎤

⎦⎥⎥

=R r ER r( ) ( ) (6.15.8)

By substituting u(r) = rR(r), Eq. (6.15.8) becomes

− + + +⎛

⎝⎜⎞

⎠⎟=� �

2 2

2

2

3

2

21 1

2 2m rd u

dr

l l

mr

mr u r

( )( )

ωEE

u rr( )

or

d

dr

l l

r

mr

mEu r

2

2 2

2 2

22

21 2

0− + − +⎡

⎣⎢⎢

⎤

⎦⎥⎥

=( )( )

ω� �

(6.15.9)

We denote β ω22 2

2≡ m

� and ε ≡ 2

2mE

�; so, we obtain

d

dr

l l

rr u r

2

2 22 21

0−+

− +⎡

⎣⎢⎢

⎤

⎦⎥⎥

=( )

( )β ε (6.15.10)

Note that for large r, the dominant part of Eq. (6.15.10) is d

drr u r

2

22 2 0−

⎛

⎝⎜⎞

⎠⎟=β ( ) . Therefore, for

larger r,

u r g r e r( ) ~ ( ) −β 2 2/ (6.15.11)

Let us compute

d u

dr

ddr

g e rge

g e

r r

r

2

22 22 2

2

= ′ −

= ′′

− −

−

( )

(

β β

β

β/ /

/22 2 2 2 2 22 2 2

2− − ′ +

= ′′

− − −β β ββ β βge rg e r ger r r/ / / )

(gg g rg r g e r− − ′ + −β β β β2 2 22

) / (6.15.12)

Hence, we have

′′ − − ′ + − + − −⎡⎣⎢

⎤⎦⎥

g g rg r gl l g

rr g gβ β β β ε2

12 22

2 2( )ee r− =β 2 2 0/ (6.15.13)

The differential equation for g(r) is

′′ − ′ + − − + =g rg gl l

rg2

102β ε β( )

( ) (6.15.14)


We substitute g r r a rsn

n

n

( ) ==

∞

∑0

(for a0 ≠ 0), so ′ = + + −

=

∞

∑g a n s rns n

n

( ) ,1

0

and

′′ = + + − = ++ −+

= −

∞

=

∞

∑∑g a n s n s r a nns n

n

nn

( )( ) (1 22

20

ss n s rs n+ + + +2 1)( ) (6.15.15)

Note that g

ra r a rn

s nn

s n

nn

22

2

20

= =+ −+

+

= −

∞

=

∞

∑∑ , so Eq. (6.15.4) becomes

a n s n s l l r ans n

n

n++

= −

∞

+ + + + − + + −∑ 2

2

2 1 1 2[( )( ) ( )] [ ββ ε β( ) ]n s rs n

n

+ + − =+

=

∞

∑ 0

0

(6.15.16)

For n = −2, we have [s(s − 1) − l (l + 1)] a0 = 0. Since a0 ≠ 0, it follows that s = l + 1. For n = −1, we have [(s + 1) s − l (l + 1)] a1 = 0. Since s = l + 1, we obtain a1 = 0; so,

an l

n l n l l lan n+ = − − +

+ + + + − +2

3 23 2 1

ε β β ( )( )( ) ( ) (6.15.17)

(c) The eigenfunction must be bounded for large r, so we must demand that g(r) be a polynomial of a finite degree; i.e., we set an0

0= for a certain n0:

ε β β− − +

+ + + + − + =3 2

3 2 100

0 0

( )( ) ) ( )

n ln l n l l l( (6.15.18)

or ε β β= + + =3 2 202

0( ) .n l mEn /� Thus, the energy eigenvalues are

Em

n lm

mn ln l′ = + ′ + = + ′ + =� �

�

2 2

23 2

23 2

3[ ( )] [ ( )]β β ω

22+ ′ +⎛

⎝⎜⎞⎠⎟n l �ω (6.15.19)

6.16. Consider the infinitesimal rotation operator:

ˆ ( , ) ˆ Û d I dR θ θn L n= − ⋅ (6.16.1)

where n is a unit vector. Find the rotation operator for a finite angle q. Hint: Define dq = q/N for N→ ∞.

SOLUTION

Let ⏐ψ ⟩ be a state vector in a coordinate system O. The state vector in coordinate system O′ that rotates around n by an angle q (relative to O) is

| |′⟩ = ⟩ψ ψ[ ˆ ( , )]U dRNθ n (6.16.2)

Hence, the rotation operator for a finite angle q is ˆ ( , ) [ ˆ ( , )] .U U dR RNθ θn n= Defining dq = q/N, we arrive at

ˆ ( , ) lim ˆ Û Ii

NR N

N

θ θn L n= −⎛

⎝⎜⎞⎠⎟→∞ � ⋅ (6.16.3)

Recall that lim ;N

N

Ne

→∞+⎛

⎝⎜⎞⎠⎟ =1

α α using this identity we finally obtain

ˆ ( , ) lim ˆˆ

U IN

iR N

N

θ θnL n= + −

⎛

⎝⎜⎞

⎠⎟⎡

⎣⎢⎢

⎤

⎦⎥⎥→∞

1 ⋅�

== −⎛⎝⎜

⎞⎠⎟exp î

�θ L n⋅ (6.16.4)


6.17. (a) Refer to Problem 6.16 and compute the rotation operator around n = j, which is a unit vector in the y-direction, for l = 1. (b) Use the rotation operator obtained in part (a), and find the representation of the eigenvectors of Lx in the standard basis of ˆ .Lz

SOLUTION

(a) Consider the rotation operator ˆ exp ˆ .Ui

R = −⎛⎝⎜

⎞⎠⎟

θ�

L n⋅ For n = j, we obtain

ˆ exp ˆ!

Ûi

nR y

n

y

n

= −⎛⎝⎜

⎞⎠⎟ = −⎛

⎝⎜⎞⎠⎟

=

∞

∑� �θ θL L

1 1

0

(6.17.1)

Let us compute

L i

i ii

iy

�=

−−

⎛

⎝⎜⎜

⎞

⎠⎟⎟

=−

−⎛

⎝⎜1

2

0 00

0 0 2

0 1 01 0 10 1 0⎜⎜

⎞

⎠⎟⎟

(6.17.2)

Ly

�

⎛⎝⎜

⎞⎠⎟ = −

−−

−

⎛

⎝⎜⎜

⎞

⎠⎟⎟

2

12

1 0 10 2 01 0 1

(6.17.3)

and

L i iy

�

⎛⎝⎜

⎞⎠⎟ = − −

−

⎛

⎝⎜⎜

⎞

⎠⎟⎟

=−3

8

0 2 02 0 2

0 2 0 2

0 1 01 0 −−

⎛

⎝⎜⎜

⎞

⎠⎟⎟

=10 1 0

Ly

� (6.17.4)

so we obtain

ˆ ˆ ( )( )!

ˆ (U I

in

L iR

n

n

y= + −+

⎛⎝⎜

⎞⎠⎟ + −+

=

∞

∑ θ θ2 1

02 1 �

))( )!

ˆ

ˆˆ ( )

2

1

2

2

2

1

n

n

y

yn n

n

L

I iL

=

∞

+

∑ ⎛⎝⎜

⎞⎠⎟

= − −

�

�θ 11

0

22

12 1

12( )!

ˆ ( )( )!n

L

nn

yn n

n+ +

⎛⎝⎜

⎞⎠⎟

−

=

∞

=

∞

∑ �θ∑∑ (6.17.5)

Note that

sin( )( )!

cos( )

(θ θ θ θ= −

+ − = −+

=

∞

∑ 12 1

112 1

0

2n n

n

n n

n 221

nn

)!=

∞

∑ (6.17.6)

therefore,

ˆ ˆ sinˆ

(cos )ˆ

U I iL L

Ry y= − + −

⎛⎝⎜

⎞⎠⎟θ θ

� �1

2

(6.17.7)

or

ˆ sinUR =

⎛

⎝⎜⎜

⎞

⎠⎟⎟

+−

−⎛

⎝⎜⎜

⎞1 0 00 1 00 0 1 2

0 1 01 0 10 1 0

θ

⎠⎠⎟⎟

− −−

−−

⎛

⎝⎜⎜

⎞

⎠⎟⎟

=

+−

(cos )

cos s

θ

θ

11 0 1

0 2 01 0 1

12

iin cos

sincos

sin

cos sin c

θ θ

θ θ θ

θ θ

2

12

2 21

2 2

1

−

−

−− + oosθ

2

⎛

⎝

⎜⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟

(6.17.8)


(b) To obtain the eigenvectors of Lx by using the eigenvectors of ˆ ,Lz we must rotate the eigenvectors of Lz by q = p /2; hence, in this case, where j is a unit vector in the y direction, we have

ˆ ( , )UR π /

/ / /

1/ 2 0 1/ 2

1/2 1/ 2 1/2

2

1 2 1 2 1 2

j =−

−⎛

⎝

⎜⎜⎜

⎞⎞

⎠

⎟⎟⎟

(6.17.9)

Thus,

| | | | |1 2 1 0 2 0⟩ = ⎛⎝⎜

⎞⎠⎟ ⟩ ⟩ = ⎛

⎝⎜⎞⎠⎟ ⟩x R x RU Uˆ , ˆ ,

π πj j −− ⟩ = ⎛

⎝⎜⎞⎠⎟ − ⟩1 2 1x RU ,

πj | (6.17.10)

where

| | |1100

0010

1001

⟩ =⎛

⎝⎜⎜

⎞

⎠⎟⎟

⟩ =⎛

⎝⎜⎜

⎞

⎠⎟⎟

− ⟩ =⎛

⎝⎜⎜

⎞

⎠⎠⎟⎟

(6.17.11)

are the standard basis. Therefore,

| | | |11 2

1 2

12 1

12

012 1⟩ =

⎛

⎝⎜⎜

⎞

⎠⎟⎟

= ⟩ + ⟩ + − ⟩x

/

1/ 2/

(6.17.12)

| | |01 2

1 2

12

1 1⟩ =−⎛

⎝

⎜⎜

⎞

⎠

⎟⎟

= − ⟩ − − ⟩x

/0

/

( ) (6.17.13)

and

| | | |− ⟩ = −⎛

⎝⎜⎜

⎞

⎠⎟⎟

= ⟩ − ⟩ + − ⟩11 2

1 2

12 1

12

012 1x

/

1/ 2/

(6.17.14)


6.18. Prove the following relations: (a) [ ˆ , ˆ ] ˆ .L p i pi j ijk k

k

= ∑� ε (b) [ ˆ , ˆ ] [ ˆ , ˆ ] [ ˆ , ˆ ˆ ] .L p L r Li i i2 2 0= = =r pi

Recall that i, j, and k can assume the values x, y, and z, and that eijk is

εij k

ijk xyzijk= −

11

cyclic permutation ofantticyclic permutation of

otherwisexyz

0

⎧⎨⎪

⎩⎪ (6.18.1)

Hint: By definition, L = r × p and use ˆ (ˆ ˆ ) ˆ ˆ .

,

L r pk k ijk i j

i j

= × = ∑r p ε

6.19. Prove the following relations for the angular momentum operator: (a) ˆ ˆ ˆ ˆ ˆ ˆ ;L L L p L Lz z2 2= − ++ − �

(b) [ ˆ , ˆ ] ˆ .L L Lz ± ±= ± �

6.20. Show that if the matrices of Lx and Ly are real, i.e.,

⟨ ′ ′⟩ = ⟨ ′ ′⟩ ⟨ ′ ′⟩ =lm L l m lm L l m lm L l mx x y| | | | | |ˆ ˆ ˆ* * ⟨⟨ ′ ′⟩lm L l my| |ˆ (6.20.1)

then the matrix of Lz is imaginary, ⟨ ′ ′⟩ = ⟨ ′ ′⟩lm L l m lm L l mz z| | | |ˆ ˆ .* Hint: Recall that [ ˆ , ˆ ] ˆ .L L i Lx y z= �

6.21. For a system with an angular momentum l = 1, find the eigenvalues and eigenvectors of ˆ ˆ ˆ ˆ .L L L Lx y y x+

Ans. | | | | | |v v v1 2 31 012

1 1 1 112

⟩ = ⟩ ⟩ = ⟩ + − ⟩ ⟩ = −, ; ( , , ); (i ii | |1 1 1 1, , )⟩ + − ⟩


6.22. In a system with an angular momentum l = 1, the eigenvalues of Lz are given by ⏐ + 1⟩, ⏐ 0⟩, and ⏐ −1⟩, where

ˆ ˆ ˆL L Lz z z| | | | |+ ⟩ = + ⟩ − ⟩ = − − ⟩ = ⟩1 1 1 1 0� � (6.22.1)

The Hamiltonian is ˆ ( ˆ ˆ ),H L Lx y= −ω0 2 2

� where w0 is a constant. Find (a) The matrix representation of H in the

basis ⏐ + 1⟩, ⏐ 0⟩, and ⏐ −1⟩; (b) the eigenvalues and the eigenvectors.

Ans. (a) ⏐ + 1⟩ ⏐ 0⟩ ⏐ −1⟩

H =⎛

⎝⎜⎜

⎞

⎠⎟⎟

+ ⟩⟩

− ⟩�ω0

1 0 00 2 00 0 1

10

1

|||

(b) The eigenvalues and eigenstates are ω ω0 01 2� � ( 0 ),( ,| |+ ⟩ ⟩ and ω0 1�( ).|− ⟩

6.23. Prove that, in spherical coordinates, the operators ˆ , ˆ ,L Lx y and Lz are written as

ˆ sin cos cot

ˆ cos

Li

Li

x

y

= − ∂∂ + ∂

∂⎛⎝⎜

⎞⎠⎟

=

�

�

ϕ ϕ ϕ θ ϕ

ϕ ∂∂∂ − ∂

∂⎛⎝⎜

⎞⎠⎟

= ∂∂

⎧

⎨

⎪⎪⎪⎪

⎩

⎪⎪⎪⎪

θ ϕ θ ϕ

ϕ

sin cot

Liz

�

(6.23.1)

6.24. The Hamiltonian of a three-dimensional isotropic harmonic oscillator is

Hm

p p p m x y zx y z= + + + + +1

212

2 2 2 2 2 2 2( ) ( )ω (6.24.1)

Calculate the following commutation relations: (a) [ ˆ , ˆ ];H Lz (b) [ ˆ , ˆ ];H Hz

(c) [ ˆ, ˆ ],L Hz where Hz =

12

12

2 2 2

mp m zzˆ .+ ω

Ans. (a) [ ˆ , ˆ ] ;H Lz = 0 (b) [ ˆ , ˆ ] ;H Hz = 0 (c) [ ˆ, ˆ ] .L Hz = 0

6.25. Prove that the time derivative of the mean value of the angular momentum operator L is given by

d

dtV

⟨ ⟩= − ⟨ × ∇ ⟩

ˆˆLr (6.25.1)

where V is the potential. What can you say about the time derivative of L for a central potential?

Ans. For a central potential, ∇ ∝ ⇒ × ∇ =V Vˆ ˆ ,r r 0 and the time derivative of L vanishes; thus, the eigenvalues of L2 are time-independent.

6.26. Use the following data to compute P x4( ): (a) P x4 ( ) is a polynomial of the fourth degree; (b) P4 1 1( ) ;= (c) P x4 ( )

is orthogonal to 1, x, x2, and x3; i.e., x P x dxk4

1

1

0( ) =−∫ for k = 0, 1, 2, 3. Hint: Choose P x4 ( ) to be of the form

P x C xkk

k

4

0

4

( ) .==

∑ Ans. P x x x4

4 218

35 30 3( ) ( ).= + +

6.27. Let ⏐ψ ⟩ be a state function of a certain system and ˆ ( , )UR θ n be a rotation operator with angle q around n (n is a unit vector), so that | |′⟩ = ⟩ψ ψUR is the state function rotated by angle q about n. Using a matrix representation,

show that for l = 1, ˆ ( , ) exp Ûi

R θ θn n L= −⎛⎝⎜

⎞⎠⎟� ⋅ (this operator is the rotation operator for all values of l).

145

CHAPTER 7

Spin

7.1 DefinitionsSpin is an intrinsic property of particles. This property was deduced from the Stern-Gerlach experiment. The formal definition of the spin operator S is analogous to the angular momentum operator (see Chap. 6),

ˆ ( )S s s2 21α α= + � (7.1)

α being an eigenfunction of S2 and s (s + 1) the corresponding eigenvalues. We also define

ˆ ˆ ˆ ˆS S S Sx y z2 2 2 2= + + (7.2)

where ˆ , ˆS Sx y, and Sz obey the following commutation relations:

[ ˆ , ˆ ] ˆ [ ˆ , ˆ ] ˆ [ ˆ , ˆ ] ˆS S i S S S i S S S ix y z y z x z x= = =� � �SSy (7.3)

Analogous to angular momentum, the quantum number of spin in the z direction is ms = −s, −s + 1, . . . , + s, and

S mz sα α= � (7.4)

7.2 Spin 1/2For particles (an electron, for example) with spin 1/2 we have ms = ± 1/2 and two distinct eigenvectors of S2 and Sz denoted by + 1

2 and − 1

2. These eigenvectors are called the standard basis, where

ˆ ˆS Sz2 1

22 1

212

12

34 2

± = ± ± = ± ±��

(7.5)

As its name implies, it is this basis that is usually used, though alternative bases are available. Any wavefunc-tion in the spin space can be written as a linear combination of the standard basis.

7.3 Pauli MatricesThe Pauli matrices ˆ ( ˆ , ˆ , ˆ )σ σ σ σ= x y z are defined using

ˆ ˆS = �2

σ (7.6)

where

ˆ ˆ ˆσ σ σx y z

ii

= ⎛⎝⎜

⎞⎠⎟ =

−⎛⎝⎜

⎞⎠⎟ = −

⎛⎝⎜

⎞⎠

0 11 0

00

1 00 1⎟⎟ (7.7)

CHAPTER 7 Spin146

S is written in the standard basis. The commutation relations of the Pauli matrices are

[ ˆ , ˆ ] ˆ [ ˆ , ˆ ] ˆ [ ˆ , ˆ ] ˆσ σ σ σ σ σ σ σx y z y z x z xi i i= = =2 2 2 σσ y (7.8)

Other useful relations for the Pauli matrices are

ˆ ˆ ˆ ˆσ σ σx y z I2 2 2= = = (7.9)

and also

( ˆ )( ˆ ) ( ) ˆ ˆ ( )σ σ σ⋅ ⋅ ⋅ ⋅= + ×A B A B A BI i (7.10)

where A and B are two spatial vectors.

7.4 Lowering and Raising OperatorsAnalogously to the angular momentum, we define the raising and lowering spin operators:

ˆ ˆ ˆ ˆ ˆ ˆS S iS S S iSx y x y+ −= + = − (7.11)

where

ˆ ˆS S+ −+ = + = −12

12

120 � (7.12)

ˆ ˆS S+ −− = + − =12

12

12 0� (7.13)

7.5 Rotations in the Spin SpaceTo find the representation of a state α in a given coordinate system that is rotated by an angle q around an axis in the direction of the unit vector u (see Fig. 7.1), we compute

α θ α' = −⎛⎝⎜

⎞⎠⎟⋅exp î

�u S (7.14)

z z′

y′

x′

x y

u

Fig. 7.1

CHAPTER 7 Spin 147

Thus, the rotation matrix is

ˆ exp ˆ cos ( ) sin ( )

Ui e

R

i

= −⎛⎝⎜

⎞⎠⎟ =

−⋅−

�θ θ θ ϕ

u S/ /2 2

ssin ( ) cos ( )θ θϕ/ /2 2ei

⎛

⎝⎜⎞

⎠⎟ (7.15)

Notice that for j = 0 (rotation around the z axis) we have

ˆ cos ( ) sin ( )sin ( ) cos ( )

UR =−⎛

⎝⎜⎞⎠⎟

θ θθ θ

/ // /2 22 2

(7.16)

which is a rotation of q /2 around the z axis. The rotation of a spin vector differs from that of a spatial vector. This result is unique to the spin vector and can thus be used to define a spin vector. A spin vector is called a spinor.

7.6 Interaction with a Magnetic FieldConsider a system consisting of particles with a spin S. Applying a magnetic field B will introduce an addi-tional term to the free Hamiltonian H0, so that

ˆ ˆ ˆ ˆ întH H H H

emc

= + = + ⋅0 0

BS (7.17)

SOLVED PROBLEMS

7.1. Calculate the commutation relation [ ˆ , ˆ ]σ σi j , where j = x, y, z and σ i are the Pauli matrices.

SOLUTION

We begin by considering the Pauli matrices:


ii

= ⎛⎝⎜

⎞⎠⎟ =

−⎛⎝⎜

⎞⎠⎟ = −

⎛⎝⎜

⎞⎠

0 11 0

00

1 00 1⎟⎟

Therefore, we see that

[ ˆ , ˆ ] ˆ ˆ ˆ ˆσ σ σ σ σ σx y x y y x

ii

= − = ⎛⎝⎜

⎞⎠⎟

−⎛⎝⎜

⎞0 11 0

00 ⎠⎠⎟ −

−⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

= −⎛⎝⎜

⎞⎠⎟ −

−

00

0 11 0

00

00

ii

ii

ii

⎛⎛⎝⎜

⎞⎠⎟ = −

⎛⎝⎜

⎞⎠⎟ =2

1 00 1

2i i zσ (7.1.1)

Also,

[ ˆ , ˆ ] ˆ ˆ ˆ ˆσ σ σ σ σ σy z y z z y

ii

= − =−⎛

⎝⎜⎞⎠⎟ −

⎛⎝⎜

00

1 00 1

⎞⎞⎠⎟ − −

⎛⎝⎜

⎞⎠⎟

−⎛⎝⎜

⎞⎠⎟

= ⎛⎝⎜

⎞⎠⎟ −

−−

1 00 1

00

00

0

ii

ii

iii

i x02⎛

⎝⎜⎞⎠⎟ = σ (7.1.2)

and, finally,

[ ˆ , ˆ ] ˆ ˆ ˆ ˆσ σ σ σ σ σz x z x x z= − = −⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞1 00 1

0 11 0⎠⎠⎟ − ⎛

⎝⎜⎞⎠⎟ −

⎛⎝⎜

⎞⎠⎟

= −⎛⎝⎜

⎞⎠⎟ −

−

0 11 0

1 00 1

0 11 0

0 11 0

⎛⎛⎝⎜

⎞⎠⎟ = 2i yσ (7.1.3)

So, we conclude that

[ ˆ , ˆ ] ˆσ σ ε σi j ijk ki= 2 (7.1.4)

CHAPTER 7 Spin148

where

εijk

ijkijk= −

11

have cyclic permutationhave anticyclic permutation

otherwise0

⎧⎨⎪

⎩⎪ (7.1.5)

7.2. Using the basis vectors of Sz eigenvectors, calculate Si + 12 and ˆ ( , , )S i x y zi − =1

2 , where + 12 and

− 12

are the eigenvectors of Sz with eigenvalues + �/2 and −�/2, respectively.

SOLUTION

The basis vectors of Sz eigenvectors are (see Sec. 4.2)

ˆ ˆ ˆS Si

iSx y z= ⎛

⎝⎜⎞⎠⎟ =

−⎛⎝⎜

⎞⎠⎟ = −

� � �2

0 11 0 2

00 2

1 00 11

⎛⎝⎜

⎞⎠⎟ (7.2.1)

and ˆ ˆS = �2

σ . Denoting + ≡ ⎛⎝⎜

⎞⎠⎟

12

10

, − ≡ ⎛⎝⎜

⎞⎠⎟

12

01

, we write

Sx + = ⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟ = ⎛

⎝⎜⎞⎠⎟ = −1

21

20 11 0

10 2

01 2

� � �22 (7.2.2)

Sx − = ⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟ = ⎛

⎝⎜⎞⎠⎟ = +1

21

20 11 0

01 2

10 2

� � �22 (7.2.3)

Note that Sx produces a transition between the eigenstates of Sz, so that when Sx operates on one eigenstate it produces a multiple of the other. Similarly, for Sy:

Si

i ii

y + =−⎛

⎝⎜⎞⎠⎟

⎛⎝⎜

⎞⎠⎟ = ⎛

⎝⎜⎞⎠⎟ =1

2 20

010 2

02

� � � −− 12 (7.2.4)

Si

ii

y − =−⎛

⎝⎜⎞⎠⎟

⎛⎝⎜

⎞⎠⎟ =

−⎛⎝⎜

⎞⎠⎟ = −1

2 20

001 2 0

� � �22

12+ (7.2.5)

And so, as expected,

Sz + = −⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟ = ⎛

⎝⎜⎞⎠⎟ = +1

2 21 00 1

10 2

10 2

� � � 112 (7.2.6)

Sz − = ⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟ = − ⎛

⎝⎜⎞⎠⎟ = −1

2 20 11 0

01 2

01 2

� � � −− 12 (7.2.7)

7.3. (a) If the z component of an electron spin is + �/2, what is the probability that its component along a direction z′ that forms an angle q with the z axis equals + �/2 or −�/2 (see Fig. 7.2)? (b) What is the average value of the spin along z′?

n

ϕ

θ

z

z'

yx

Fig. 7.2

CHAPTER 7 Spin 149

SOLUTION

(a) The present state of the electron is + 12 ; the spin operator component along z′ is

ˆ ˆ ˆSz ′ = =⋅ ⋅S n n�2

σ (7.3.1)

where n is a unit vector along z′. In our case n = i sinq cosj + j sinq sinj + k cosq and therefore,

ˆ ˆ sin cos ˆ sin sin ˆ cosS S S Sz x y z′ = + +θ ϕ θ ϕ θ (7.3.2)

The eigenvalues of Sz ′ are + �/2 or −�/2, and the eigenvectors of Sz ′ with the basis eigenvectors of Sz are

+ = + + −12

12

12

� a b (7.3.3)

Sz′ + = + +12

122

� � � (7.3.4)

and

− = + + −12

12

12

� c d (7.3.5)

Sz′ − = − −12

122

� � � (7.3.6)

where a, b, c, and d are complex constants. By substituting Eq. (7.3.2) and Eq. (7.3.3) into Eq. (7.3.4) we obtain

( ˆ sin cos ˆ sin sin ˆ cos )S S S a bx y zθ ϕ θ ϕ θ+ + + + −( 12

12 )) = + + −( )�

212

12a b (7.3.7)

Using the known relations

ˆ

ˆ

ˆ

S

Si

S

x

y

z

+ = −

+ = −

+ = +

⎧

⎨

⎪⎪⎪

⎩

12

12

12

12

12

12

2

2

2

�

�

�⎪⎪⎪⎪

− = +

− = − −

− = − −

ˆ

ˆ

ˆ

S

Si

S

x

y

z

12

12

12

12

12

12

2

2

2

�

�

�

⎧⎧

⎨

⎪⎪⎪

⎩

⎪⎪⎪

(7.3.8)

Eq. (7.3.7) turns into the form

�

�

ai

b

2

2

12

12

12sin cos sin sin cos

si

θ ϕ θ ϕ θ− + − + +{ }+ nn cos sin sin cosθ ϕ θ ϕ θ+ − + − −{ } = + + −1

212

12

122

i a b� 11

2( ) (7.3.9)

Hence, we obtain

a ia b ba b

sin cos sin sin coscos sin cos

θ ϕ θ ϕ θθ θ ϕ

+ − =+ −− ={ ib asin sinθ ϕ (7.3.10)

or ab

i= +

+ +( cos )sin (cos sin )

1 12

θθ ϕ ϕ must be a unit vector; thus, ⏐ a ⏐2 + ⏐ b ⏐2 = 1 and b

22

211

1+ +⎛

⎝⎜⎞

⎠⎟=( cos )

sin,

θθ

so

b2

2 2

2

2

2 24

2

42

= + =⎛⎝⎜

⎞⎠⎟

=

⎛sin

cossin

cos

sinθθ

θθ

θ⎝⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

= ⎛⎝⎜

⎞⎠

cos

cossin

2

2

22

42

2

θ

θθ

⎟⎟ (7.3.11)

CHAPTER 7 Spin150

We choose b = eij sin(q/2); hence,

ae

eii= + ⎛

⎝⎜⎞⎠⎟ =

⎛⎝⎜

⎞( cos )

sinsin

cos1

2

22

2

θθ

θθ

ϕϕ ⎠⎠⎟

⎛⎝⎜

⎞⎠⎟

= ⎛⎝⎜

⎞⎠⎟

sin

sincos

θ

θθ22 (7.3.12)

so we obtain

+ = ⎛⎝⎜

⎞⎠⎟ + + ⎛

⎝⎜⎞⎠⎟ −1

212

122 2

� cos sinθ θ ϕei (7.3.13)

Since − 12

� is orthogonal to + 12

� we have

+ − = ⎛⎝⎜

⎞⎠⎟ + ⎛

⎝⎜⎞⎠⎟ = ⇒ = −−1

212 2 2

0c d e cicos sin tθ θ ϕ aan

θ ϕ2

⎛⎝⎜

⎞⎠⎟

−e di (7.3.14)

Note that − 12

� is also a unit vector, so ⏐ c ⏐2 + ⏐ d ⏐2 = 1. Substituting c, we obtain [tan2(q/2) + 1] ⏐ d ⏐2 = 1,

or ⏐ d ⏐2 = cos2 (q /2). We choose d = −cos(q /2), and so c = −e−ij sin(q /2). Therefore,

− = − ⎛⎝⎜

⎞⎠⎟ + + ⎛

⎝⎜⎞⎠⎟ −−1

212

122 2

� sin cosθ θϕe i (7.3.15)

The present state of the electron represented by the basis eigenvectors of Sz ′ is

+ = + + + + + − − = ⎛⎝⎜

⎞⎠⎟ +1

212

12

12

12

12

12

122

� �� cosθ ++ ⎛

⎝⎜⎞⎠⎟ −−sin

θ ϕ2

12e i � (7.3.16)

Therefore, the probability that the spin component along z′ is + �/2:

P+⎛

⎝⎜⎞⎠⎟ = + + = ⎛

⎝⎜⎞⎠⎟

�2 2

12

12

2 2� cosθ

(7.3.17)

and the probability that it is −�/2:

P −⎛⎝⎜

⎞⎠⎟ = − + = ⎛

⎝⎜⎞⎠⎟

�2 2

12

12

2 2� sinθ

(7.3.18)

(b) The average value of the spin along z′ is ˆ ˆ .S Sz z′ ′= + +12

12 Using the relation

ˆ ˆ cos sinS S ez zi

′ ′−+ = ⎛

⎝⎜⎞⎠⎟ + + ⎛

⎝⎜⎞⎠⎟

12

122 2

θ θ ϕ� −−⎛⎝⎜

⎞⎠⎟

= ⎛⎝⎜

⎞⎠⎟ + − ⎛

⎝⎜⎞⎠⎟

−12

122 2 2

� � �cos sinθ θ

e iiϕ −⎛⎝⎜

⎞⎠⎟

12

� (7.3.19)

we obtain

ˆ ˆ cos sinS Sz z′ ′= + + = + ⎛⎝⎜

⎞⎠⎟ + − ⎛1

212

12

122 2 2

� �θ θ⎝⎝⎜

⎞⎠⎟ −⎛

⎝⎜⎞⎠⎟

= ⎛⎝⎜

⎞⎠⎟ + + −

−e iϕ

θ

12

12

122 2

�

� �cos siin

cos c

θ

θ

ϕ2

2 2

12

12

⎛⎝⎜

⎞⎠⎟ + −⎡

⎣⎢⎤⎦⎥

= ⎛⎝⎜

⎞⎠⎟

−e i �

�oos sin sin

θ θ θϕ ϕ2 2 2

⎛⎝⎜

⎞⎠⎟ − ⎛

⎝⎜⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

⎡⎣

−e ei i⎢⎢

⎤⎦⎥

= ⎛⎝⎜

⎞⎠⎟ − ⎛

⎝⎜⎞⎠⎟

⎡⎣⎢

⎤⎦⎥

=� �2 2 2

2 2cos sincoθ θ ssθ2 (7.3.20)

CHAPTER 7 Spin 151

7.4. Consider a particle with spin 1/2. (a) Find the eigenvalues and eigenfunctions of the operator Sx + Sy where Si is the spin operator in the i direction (i = x, y, z). (b) Assume that α designates the eigenfunction of S x + S y that belongs to the maximal eigenvalue, and that the particle is in state α . If we measure the spin in the z direction, what are the values and their probabilities? (c) The particle is in state α . Find, if possible, the direction n in which the spin measurement will, with certainty, yield the value ˆ .Sn = �/2

SOLUTION

(a) We begin by writing the matrices

ˆ ˆ ˆS Si

iSx y z= ⎛

⎝⎜⎞⎠⎟ =

−⎛⎝⎜

⎞⎠⎟ = −

� � �2

0 11 0 2

00 2

1 00 11

⎛⎝⎜

⎞⎠⎟ (7.4.1)

thus,

ˆ ˆ Â S Si

ix y= + =−

+⎛⎝⎜

⎞⎠⎟

�2

0 11 0

(7.4.2)

To find the eigenvalues of this operator (l�/2), we must solve the equation det [A − (l�/2) Î ] = 0; that is,

det [� �2

11

02

42− −

+ −⎛⎝⎜

⎞⎠⎟

⎧⎨⎩

⎫⎬⎭

= ⇒ ⎛⎝⎜

⎞⎠⎟ −λ

λ λii

(( )( )]1 1 0− + =i i (7.4.3)

So, l2 − 2 = 0, which yields λ = ± 2 , and the eigenvalues of A are ± �/ 2 . Recall that the kernel of a matrix A is the set of vectors u for which Au = 0. The eigenfunction of A corresponding to the eigenvalue +�/ 2 is

ker ker�2

2 1

1 2

2 1

1

− −+ −

⎛⎝⎜

⎞⎠⎟

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪=

− −+

i

i

i

i −−⎛⎝⎜

⎞⎠⎟2

(7.4.4)

That is the state a b+ + −12

12 , where

− −

+ −⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟ = ⎛

⎝⎜⎞⎠⎟ ⇒

− + −2 1

1 2

00

2 1i

i

ab

a i( )bb

i a b

=+ − =

⎧⎨⎩

0

1 2 0( ) (7.4.5)

Thus, ai

b= +2

1. For a b+ + −1

212 to be normalized we must satisfy the condition ⏐ a ⏐2 + ⏐ b ⏐2 = 1;

hence,

2

11 12

+ +⎛

⎝⎜

⎞

⎠⎟ =

ib (7.4.6)

which yields b = 1 2/ and ai

i e i

= + = − =−1

11

2 2

4π /

. Therefore, the first eigenstate v112

12= + + −a b

is found to be v1

412

122

12

= + + −−e iπ /

. Simiarly, for the second eigenfunction of A corresponding

to the eigenvalue −�/ 2 we obtain

ker ker�2

2 1

1 2

2 1

1 2

−+

⎛⎝⎜

⎞⎠⎟

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪=

−+

⎛⎝

i

i

i

i⎜⎜⎞⎠⎟

(7.4.7)

Or v 212

12= + + −c d , where

2 1

1 2

00

2 1 0−+

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟ = ⎛

⎝⎜⎞⎠⎟ ⇒

+ − =i

i

cd

c i d( )

(( )1 2 0+ + =⎧⎨⎩ i c d

(7.4.8)

CHAPTER 7 Spin152

so ci

d= − +2

1. The normalization condition of v2 yields d = 1 2/ and c

ie= − + =1

123 4π / / , and

therefore, v2

3 412

122

12

= + + −e π /

; so, finally

( ˆ ˆ ) ( ˆ ˆ )S S S Sx y x y+ = − + =v v v v1 1 2 22 2� �

(7.4.9)

(b) The maximal eigenvalue of Sx + Sy is +�/ 2; thus,

απ

= = + + −−

v1

412

122

12

e i /

(7.4.10)

The values that can be obtained from a measurement of Sz are ±�/2. The probability for Sz = �/2 is

Pe i

�2 2

12

12

2 4 2⎛⎝⎜

⎞⎠⎟ = + = =

−

απ /

(7.4.11)

Therefore the probability for Sz = −�/2 is

P P−⎛⎝⎜

⎞⎠⎟ = − ⎛

⎝⎜⎞⎠⎟ =� �

21

212 (7.4.12)

(c) If the measurement of an observable gives only one result, then the state of the system is an eigenstate of that observable; thus, the state α is the eigenstate of a spin operator in a certain direction (the one we wish to find). As we have seen in part (a),

απ

= = + + −−

v1

412

122

12

e i /

(7.4.13)

v1 is also an eigenstate of Sx + Sy with the eigenvalue �/ 2 , that is,

( ˆ ˆ ) ( ˆ ˆ )S S S Sx y x y+ = ⇒ + =α α α α� �

212 2

(7.4.14)

Hence, α is the eigenstate of ( ˆ ˆ )S Sx y+ / 2 and the measurement of ( ˆ ˆ )S Sx y+ / 2 always yields the result �/2. Note that ( ˆ ˆ )S Sx y+ / 2 is the spin operator in the direction of the spatial unit vector n = i + j where i and j are unit vectors in the x and y directions, respectively.

7.5. Consider a particle with spin 1/2. (a) What are the eigenvalues and eigenvectors of Sx, Sy, and Sz? (b) Consider a particle in eigenstate Sx. What are the possible results and their probabilities if we measure the z component of the spin? (c) At t = 0, the particle is in the eigenstate Sx, which corresponds

to the eigenvalue −�/2. The particle is in a magnetic field and its Hamiltonian is ˆ ˆ .Hemc

Sz= B Find the

state at t > 0. (d) If we measure Sx at t = t1, what is the result? What is the result for a measurement of Sz at t = t1? Explain the difference in t1-dependence. (e) Calculate the expectation values of Sx and Sz at t = t1.

SOLUTION

(a) Consider the matrices Sx, Sy, and Sz written in the basis eigenvectors of Sz,

ˆ ˆ ˆS Si

iSx y z= ⎛

⎝⎜⎞⎠⎟ =

−⎛⎝⎜

⎞⎠⎟ = −

� � �2

0 11 0 2

00 2

1 00 11

⎛⎝⎜

⎞⎠⎟ (7.5.1)

First, we shall determine the eigenvectors of Sz. For eigenvalue + �/2 we have + = ⎛⎝⎜

⎞⎠⎟

12

10z

and for eigen-

value −�/2 we have − = ⎛⎝⎜

⎞⎠⎟

12

01z

. The eigenvalues of Sx are �l /2, where det (Sx − (�l /2)Î ) = 0; that is,

det� � �2

0 11 0 2

00 2

⎛⎝⎜

⎞⎠⎟ − ⎛

⎝⎜⎞⎠⎟

⎧⎨⎩

⎫⎬⎭

= ⎛⎝⎜

⎞⎠⎟

λλ

221

10

−− =λ

λ (7.5.2)

CHAPTER 7 Spin 153

or l2 − 1 = 0. Therefore, we obtain the eigenvalues ±�/2. The eigenvector corresponding to the eigenvalue

+�/2 is + = + + − ≡ ⎛⎝⎜

⎞⎠⎟

12

12

12x z z

a bab

, hence

Sab

abx x x

+ = + ⇒ ⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟ = ⎛

⎝⎜12

122 2

0 11 0 2

� � � ⎞⎞⎠⎟ (7.5.3)

Solving Eq. (7.5.3), we obtain b = a. Now, + 12 x

must be normalized, so we set the condition ⏐ a ⏐2 + ⏐ b ⏐2 = 1. Substituting for a, we obtain

2 112

2b a b= ⇒ = = (7.5.4)

Thus, the eigenvector of Sx with eigenvalue +�/2 is

+ = ⎛⎝⎜

⎞⎠⎟ = + + −( )1

212

12

12

11

12x z z

(7.5.5)

The other eigenvector − 12 x

(with eigenvalue −�/2) is obtained either from orthogonality and normaliza-

tion conditions (since the two eigenvectors belong to different eigenvalues), or in the same manner in which the first eigenvector was obtained. We will follow the former course:

− = + + − ≡ ⎛⎝⎜

⎞⎠⎟

12

12

12x z z

c bcd

(7.5.6)

and

x xc d

c d− + =⎛⎝⎜

⎞⎠⎟

= + =12

12

1 2

1 2 2 20( )

/

/ (7.5.7)

giving c = −d. Using the normalization condition ⏐ c ⏐2 + ⏐ d ⏐2 = 1, we can choose c d= − = 1 2/ and obtain

− = −⎛⎝⎜

⎞⎠⎟ = + − −( )1

212

12

12

11

12x z z

(7.5.8)

Similarly, the eigenvalues of Sy are (�/2)l, where

det ˆ ˆS Ii

iy −⎛⎝⎜

⎞⎠⎟ = ⎛

⎝⎜⎞⎠⎟

− −−

⎛⎝⎜

⎞⎠⎟ =� �

2 20

2

λ λλ (7.5.9)

or l2 − 1 = 0; the eigenvalues of Sy are also ±�/2, and the eigenvector corresponding to the eigenvalue, +�/2 is

+ = + + − ≡ ⎛⎝⎜

⎞⎠⎟

12

12

12y z z

a bab

(7.5.10)

where

Si

iab

aby y

+ =−⎛

⎝⎜⎞⎠⎟

⎛⎝⎜

⎞⎠⎟ = ⎛

⎝⎜⎞⎠⎟ =1

2 20

0 2 2� � � ++ 1

2 y (7.5.11)

and ia = b. Using the normalization condition ⏐ a ⏐2 + ⏐ b ⏐2 = 1, we obtain 2⏐ b ⏐2 = 1, so we can choose b = 1 2/ and a i= − / 2 . And finally, we obtain

+ = − + + −( )12

12

12

12y z z

i (7.5.12)

Using the orthogonality relation of − 12 y

, to + 12 y

, we have − = + + −12

12

12y z z

c d and

y yc d

i ic d− + =−⎛

⎝⎜⎞⎠⎟

= − + =12

12

2

1 2 2 20( )

/

/ (7.5.13)

CHAPTER 7 Spin154

so d = ic, and from the normalization condition we get ⏐ c ⏐2 + ⏐ d ⏐2 = 2⏐ c ⏐2 = 1. Thus, c = 1 2/ and d i= / 2; therefore,

− = + + −( )12

12

12

12y z z

i (7.5.14)

(b) As we found in part (a), the eigenstates of Sx are

+ = + + −( ) − = + − −( )12

12

12

12

12

12

12

12x z z x z z

(7.5.15)

If we measure the spin component in the z direction, the state of the particle will be either + 12 z

, resulting

in Sz = �/2, or − 12 z

, resulting in Sz = −�/2. The probability for Sz = �/2 is

P z x

�2

12

12

12

2⎛⎝⎜

⎞⎠⎟ = + ± = (7.5.16)

and for Sz = −�/2 is

P z x−⎛

⎝⎜⎞⎠⎟ = + ± =�

212

12

12

2 (7.5.17)

Note that if the initial state is either + 12 x

or − 12 x

, we obtain the same results.

(c) At t = 0, the particle is in initial state:

+ = + + −( )12

12

12

12x z z

(7.5.18)

We want to find the time evolution of this state, therefore we use the Schrödinger equation, it

H�∂∂ =ψ ψ.ˆ

In as much as the Hamiltonian is time-independent, we write ψ (r, s, t) = φ1(r, s)φ2(t); substituting in

the Schrödinger equation gives

φφ

φ φ12

2 1( )( )

( ) ˆ ( )r s r s, ,∂

∂ =t

tt H (7.5.19)

Assuming that φ2(t) is of the form φ2(t) = e−iEt /�, where E is a constant, we obtain

E e t H E tiE tφ φ φ φ φ1 2 1 1 2( ) ( ) ˆ ( ) ( ) (/r s r s r s, , ,− = ⇒� )) ( ) ˆ ( )= φ φ2 1t H r s, (7.5.20)

and we must require that φ1(r, s) = Eφ1(r, s). In other words, φ1(r, s) must be an eigenfunction of the Hamiltonian H. Note that

ˆ ˆ ( .) ˆHemc

S Sz z= =Bconst (7.5.21)

Thus, the eigenstates of H are similar to the eigenstates of Sz, where the eigenvalues of H are the eigenvalues of Sz multiplied by the constant eB/mc. Therefore,

ψ112 2

( )r s, = + =z

Ee

mcB�

(7.5.22)

and

ψ12 1

2( ) /r s, , t = +−e ie t mc

z

B (7.5.23)

Also,

ψ112 2

( )r s, = − = −z

Ee

mcB�

(7.5.24)

which gives

ψ12 1

2( ) /r s, , t = −eie t mc

z

B (7.5.25)

CHAPTER 7 Spin 155

Therefore, each state of the particle can be written as

ψ ψ ψ( ) ( ) ( ) /r s r s r s, , , , , ,t t t e ie t mc= + = +−α β α1 22B 11

22 1

2z

i mc

ze+ −β / (7.5.26)

For our system, the initial condition is

ψ( )r s, , tz z z z

= = + + −( ) = + + −012

12

12

12

12α β (7.5.27)

Hence, α β= = 1 2/ , giving

ψ( ) / /r s, , t e eie t mc

z

ie t mc

z= + + −( )−1

22 1

22 1

2B B (7.5.28)

(d) A measurement of Sx or S z will give either +�/2 or −�/2. The probability for a measurement Sx = +�/2 is

P t e ex xie t mc+⎛

⎝⎜⎞⎠⎟ = + = −−�

212

12 1

2 21ψ( ) (/r s, ,

B iie t mc e tmc

B B1 2

22 1

2/

) cos= ⎛⎝⎜

⎞⎠⎟ (7.5.29)

and for Sx = −� /2 we have

P t e ex xie t mc−⎛

⎝⎜⎞⎠⎟ = − = −−�

212

12 1

2 21ψ( ) (/r s, ,

B iie t mc e tmc

B B1 2

22 1

2/

) sin= ⎛⎝⎜

⎞⎠⎟

(7.5.30)

Similarly, the probability for Sz = +�/2 is

P t ez zie t mc+⎛

⎝⎜⎞⎠⎟ = + = =−�

212

112 1

2 22

1ψ( )/r s, ,

B

22 (7.5.31)

and for Sz = −� /2,

P t ez zie t mc−⎛

⎝⎜⎞⎠⎟ = − = =�

212

12

12 1

2 22

1ψ( )/r s, ,

B (7.5.32)

(e) We can calculate the expectation value of Si in two ways: the first by calculating ⟨ ⟩ψ ψ( ) ˆ ( )r s r s, , , ,t S ti1 1⏐ ⏐ and the second by summing over the products of the possible values multiplied by their probability. Using the second method,

⟨ ⟩ = + +⎛⎝⎜

⎞⎠⎟ − −⎛

⎝⎜⎞⎠⎟ =ˆ cosS P P

ex x x

� � � � �2 2 2 2 2

2 B ttmc

e tmc

e1 2 12 2 2

⎛⎝⎜

⎞⎠⎟

− ⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢

⎤

⎦⎥ =sin cos

B � BB tmc

1⎛⎝⎜

⎞⎠⎟

(7.5.33)

Similarly,

⟨ ⟩ = + +⎛⎝⎜

⎞⎠⎟ − −⎛

⎝⎜⎞⎠⎟ = −⎛S P Pz z z

� � � � �2 2 2 2 2

12

12⎝⎝⎜

⎞⎠⎟ = 0 (7.5.34)

Note that ⟨ ⟩Sx is not conserved in time; this is because [ ˆ , ˆ ] [ ˆ , ˆ ]H Semc

S Sx z x= ≠B0 , while ⟨ ⟩Sz is conserved

since

[ ˆ , ˆ ] [ ˆ , ˆ ]H Semc

S Sz z z= =B0 (7.5.35)

7.6. (a) Prove that [S2, Sz] = 0 where ˆ ˆ ˆ ˆS S S Sx y z2 2 2 2= + + . (b) Show that the eigenvectors’ basis of Sz

diagonalizes S2. Find the eigenvalues of S2.

SOLUTION

(a) In Problem 7.1, we found that [ ˆ , ˆ ] ˆ ; [ ˆ , ˆ ] ˆσ σ σ σ σ σx y z y z xi i= =2 2 ; and [ ˆ , ˆ ] ˆσ σ σz x yi= 2 . Therefore, recalling that ˆ ˆS = �σ /2 we write

[ ˆ , ˆ ] ˆ [ ˆ , ˆ ] ˆ [ ˆ , ˆ ] ˆS S i S S S i S S S ix y z y z x z x= = =� � �SSy (7.6.1)

CHAPTER 7 Spin156

Hence,

[ ˆ , ˆ ] [ ˆ ˆ ˆ , ˆ ] [ ˆ , ˆ ]S S S S S S S Sz x y z z i z

i

2 2 2 2 2= + + = ∑ (7.6.2)

where i = x, y, z. We see that

[ ˆ , ˆ ] ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆS S S S S S S S S S Sz i z z i i z i i z

2 2 2= − + − SS

S S S S S S S S S S

i

i i z z i i z z i i= − + − =ˆ ( ˆ ˆ ˆ ˆ ) ( ˆ ˆ ˆ ˆ ) ˆ SS S S S S Si i z i z i[ ˆ , ˆ ] [ ˆ , ˆ ] ˆ+ (7.6.3)

so [ ˆ , ˆ ]S Sz z2 0= . Also,

[ ˆ , ˆ ] ˆ [ ˆ , ˆ ] [ ˆ , ˆ ] ˆ [ ˆS S S S S S S S i Sx z x x z x z x2 = + = − � xx y y xS S Sˆ ˆ ˆ ]+ (7.6.4)

and

[ ˆ , ˆ ] ˆ [ ˆ , ˆ ] [ ˆ , ˆ ] ˆ ( ˆS S S S S S S S i Sy z y y z y z y2 = + = − � yy x x yS S Sˆ ˆ ˆ )+ (7.6.5)

Finally,

[ ˆ , ˆ ] [ ˆ , ˆ ] ( ˆ ˆ ˆ ˆ ) (S S S S i S S S S iz i z y x x y2 2 2= = + −� � ˆ ˆ ˆ ˆ )S S S Sx y y x

i

+ =∑ 0 (7.6.6)

(b) To obtain the matrix representation of S2 we calculate it using the matrices of Sx, Sy, and Sz in the basis

of the eigenvectors of Sz; that is,

ˆ ˆ ˆS Si

iSx y z= ⎛

⎝⎜⎞⎠⎟ =

−⎛⎝⎜

⎞⎠⎟ = −

� � �2

0 11 0 2

00 2

1 00 11

⎛⎝⎜

⎞⎠⎟ (7.6.7)

Hence,

ˆ ˆ ˆ ˆ ( ˆ ˆ ˆS S S Sx y z x y z2 2 2 2

22 2 2

2= + + = ⎛

⎝⎜⎞⎠⎟ + +� σ σ σ )) (7.6.8)

Using the known result that ˆ ˆσ i I2 = , we obtain

ˆ ˆS I22 2

32

34

1 00 1

= ⎛⎝⎜

⎞⎠⎟ = ⎛

⎝⎜⎞⎠⎟

� � (7.6.9)

We see that the S2 is diagonalized (in the basis of the eigenvectors of Sz). From linear algebra, we know that if a vector basis diagonalizes the matrix of an operator, then the basis is comprised of the operator’s eigenvectors, i.e., + 1

2 and − 12 are also the eigenvectors of S2. In other words, we conclude that if the

commutation relation of two operators is zero, then we can find similar eigenvectors for both of them. To find the eigenvalue of S2 for the eigenvector + 1

2 we calculate

S2 12

2 234

1 00 1

10

34

10

+ = ⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟ = ⎛

⎝⎜⎞⎠⎟ =� � 33

4

212

� + (7.6.10)

So, the eigenvalue of + 12 is 3�

2/4, and the eigenvalue of − 12 is

S2 12

2 234

1 00 1

01

34

01

− = ⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟ = ⎛

⎝⎜⎞⎠⎟ =� � 33

4

212

� − (7.6.11)

Thus the eigenvalue of − 12 is also 3�

2/4. Note that if we set s = 1/2 to be the quantum number of the total spin, then (like the angular momentum theory) the eigenvalue 3�

2/4 can be written as �2s (s + 1).

7.7. Find the result of applying the operators Sx + iSy and Sx − iSy on the eigenvectors + 12

and − 12

of Sz. What is the importance of these operators?

SOLUTION

We begin with the operator Sx + iSy and calculate

( ˆ ˆ ) ˆ ˆS iS S iS ix y x y+ + = + + + = − + ⎛⎝⎜

⎞⎠

12

12

12

122 2

� �⎟⎟ − =i 1

2 0 (7.7.1)

CHAPTER 7 Spin 157

and

( ˆ ˆ ) ˆ ˆS iS S iS ix y x y+ − = − + − = + + ⎛⎝⎜

⎞⎠

12

12

12

122 2

� �⎟⎟ + = +i 1

212� (7.7.2)

For the operator Sx − iSy, we have

( ˆ ˆ ) ˆ ˆS iS S iS ix y x y− + = + − + = − − ⎛⎝⎜

⎞⎠

12

12

12

122 2

� �⎟⎟ − = −i 1

212� (7.7.3)

and

( ˆ ˆ ) ˆ ˆS iS S iS ix y x y− − = − − − = + − ⎛⎝⎜

⎞⎠

12

12

12

122 2

� �⎟⎟ − + =( )i 1

2 0 (7.7.4)

To conclude, we have

ˆ ˆ ˆ ˆS S S S+ − + −+ = + = − − = + − =12

12

12

12

12

120 0� � (7.7.5)

where, from Eq. (7.11), S+ ≡ Sx + iSy and S− ≡ Sx − iSy. The latter relations justify calling S+ a spin-raising operator, since it increases the spin in z direction from −� /2 to +� /2. Similarly, we call S− a spin-lowering operator, since it lowers the z component of the spin from +� /2 to −� /2. S+ and S− allow us to jump from one eigenstate of Sz to the other. These operators are very useful in spin calculations.

7.8. Using the operators S+ and S−, compute the matrices Sx and Sy; show that ˆ ˆ ˆ ˆS S S Sx y z2 2 2 2= + + is

diagonalized in the basis of eigenvectors of Sz.

SOLUTION

The spin-raising S+ operator and the spin-lowering S− operator are defined as

ˆ ˆ ˆ ˆ ˆ ˆS S iS S S iSx y x y+ −= + = − (7.8.1)

Hence, we can write

ˆ ( ˆ ˆ ) ˆ ( ˆ ˆ )S S S Si

S Sx y= + = −+ − + −12

12

(7.8.2)

Therefore,

ˆ ˆ ˆ ˆ ˆ ( ˆ ˆ ) ( ˆS S S S S S S Sz x y z2 2 2 2 2 21

414

= + + = + + −+ − + −−

= + + + + −

−

+ + − − + −

ˆ )

ˆ ( ˆ ˆ ˆ ˆ ˆ )ˆ

S

S S S S S Sz S

2

2 2 214

14

(( ˆ ˆ ˆ ˆ ˆ ˆ )

ˆ ( ˆ ˆ ˆ

S S S S S S

S S Sz

+ + − − + −

+ −

− − +

= + +

2 2

2 12

SS S− +ˆ ) (7.8.3)

To find the matrix representation of S2 we compute

ˆ ˆ ( ˆ ˆ ˆ ˆ ) ˆS S S S S S Sz z2 1

22 1

221

2+ = + +⎛

⎝⎜⎞⎠⎟ + =+ − − + ++ + + + +

= ⎛⎝⎜

⎞⎠⎟ + +

+ − − +12

12

12

212

12

12

2

ˆ ˆ ˆ ˆS S S S

� �22

04 2

34

12

2 212

212S+ − + = +

⎛

⎝⎜⎞

⎠⎟+ = +� � �

(7.8.4)

And also

ˆ ˆ ( ˆ ˆ ˆ ˆ ) ˆS S S S S S Sz z2 1

22 1

221

2− = + +⎛

⎝⎜⎞⎠⎟ − =+ − − + −− + − + −

= ⎛⎝⎜

⎞⎠⎟ − +

+ − − +12

12

12

212

12

12

2

ˆ ˆ ˆ ˆS S S S

� �22

04 2

34

12

2 212

2

S− + + = +⎛

⎝⎜⎞

⎠⎟− =� � �

(7.8.5)

CHAPTER 7 Spin158

Therefore,

+ −

=+

−

⎛

⎝

⎜⎜⎜

⎞

⎠

⎟⎟⎟

=

12

12

212

12

2

2

23

40

03

4

34

[ ˆ ]S

�

�

� 11 00 1

⎛⎝⎜

⎞⎠⎟

(7.8.6)

which is diagonalized.

7.9. For a particle with spin 1/2, compute, in two ways, the expectation value of iSx Sy Sx, where the particle

wavefunction is 1

212

12+ + −( ): (a) using S+ and S− operators, where S+ = Sx + iSy and S− = Sx − iSy;

(b) in a direct way.

SOLUTION

(a) Consider the matrices S+ and S−:

ˆ ( ˆ ˆ ) ˆ ( ˆ ˆ )S S S Si

S Sx y= + = ++ − + −12

12

(7.9.1)

Therefore,

ˆ ˆ ˆ ˆ ( ˆ ˆ )( ˆ ˆ )( ˆ Â iS S Sii

S S S S S Sx y x≡ = + − ++ − + − + −8)) ( ˆ ˆ ˆ ˆ ˆ ˆ ) ( ˆ ˆ )

(

= − + − +

=

+ + − − + − + −18

18

2 2S S S S S S S S

ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆS S S S S S S S S S S+ + − + − + − + + − +− + − + −3 2 2 2 ˆ ˆ ˆ ˆ ˆ )S S S S S− − + − −+ −2 3 (7.9.2)

Recall that

ˆ ˆ ˆ ˆS S S S+ + − −+ = − = + + = − − =12

12

12

12

12

120 0� � (7.9.3)

Hence,

ˆ ˆS S+ −− = + =2 12

2 120 0 (7.9.4)

Therefore, all the expressions in A that contain S+2 or S−

2 do not contribute to the expectation value, that is,

⟨ ⟩ = + + −( ) + + −( )⎡⎣

⎤⎦

=

ˆ ˆ ˆ Â iS S Sx y x

12

116

12

12

12

12

++ + −( ) + + + −( )⎡⎣

⎤⎦− + − + − +

12

12

12

12

ˆ ˆ ˆ ˆ ˆ ˆS S S S S S (7.9.5)

It can be seen that

ˆ ˆ ˆ ˆ ˆ ˆS S S S S S− + − − + −+ = − − =12

3 12

12 0� (7.9.6)

and also,

ˆ ˆ ˆ ˆ ˆ ˆS S S S S S+ − + + − ++ = + = +12

12

3 120 � (7.9.7)

Substituting into Eq. (7.9.5), we obtain

⟨ ⟩ = + + −( ) + + −( ) − + + −( ) − + +A�

312

12

12

12

12

12

12

116 22 0( ){ } = (7.9.8)

(b) The matrix representation of iSxSySx in the standard basis is

ˆ ˆ ˆ Â iS S Si i

ix y x= = ⎛⎝⎜

⎞⎠⎟

−⎛⎝⎜

⎞⎠⎟

�3

80 11 0

00

0 11 0

⎛⎛⎝⎜

⎞⎠⎟ = ⎛

⎝⎜⎞⎠⎟ −

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

i�3

80 11 0

0 11 0

0 11 0

== ⎛⎝⎜

⎞⎠⎟ −

⎛⎝⎜

⎞⎠⎟ = −

⎛⎝⎜

⎞⎠⎟

0 11 0

0 11 0 8

0 11 0

3i� (7.9.9)

CHAPTER 7 Spin 159

The particle wavefunction in the standard basis is 12

12

11

12

12+ + −( ) ≡ ⎛

⎝⎜⎞⎠⎟ and, therefore,

⟨ ⟩ =−⎛

⎝⎜⎞⎠⎟

⎛⎝⎜

⎞⎠⎟ =ˆ ( ) ( )A

i i� �3 3

161 1

0 11 0

11 16

1 1−−⎛

⎝⎜⎞⎠⎟ =

11

0 (7.9.10)

7.10. Consider the commutation relations:

[ ˆ , ˆ ] ˆS S i Sx y z= � (7.10.1)

[ ˆ , ˆ ] ˆS S i Sz y x= � (7.10.2)

[ ˆ , ˆ ] ˆS S i Sz x y= � (7.10.3)

Given that Sx, Sy, and Sz are Hermitian operators with eigenvalues ± �/2, find the matrix representation of Sx, Sy, and Sz in a basis where Sz is diagonalized.

SOLUTION

Note that Sx, Sy, and Sz each have two eigenvectors and that they are Hermitian operators; thus, we conclude that their matrix representation is 2 × 2. So,

ˆ ˆ ˆSa b

c dS

a b

c dS

a bx y z=

⎛⎝⎜

⎞⎠⎟

=⎛⎝⎜

⎞⎠⎟

=1 1

1 1

2 2

2 2

3 33

3 3c d⎛⎝⎜

⎞⎠⎟

(7.10.4)

We want to express the matrices in a basis in which Sz is diagonalized; thus, we write

Sz = −⎛⎝⎜

⎞⎠⎟ = −

⎛⎝⎜

⎞⎠⎟

�

�

�//

2 00 2 2

1 00 1

(7.10.5)

Substituting Sx and Sz in Eq. (7.10.3) gives

� �2

1 00 1 2

1 1

1 1

1 1

1 1−⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

−⎛⎝⎜

⎞⎠

a b

c d

a b

c d ⎟⎟ −⎛⎝⎜

⎞⎠⎟ =

⎛⎝⎜

⎞⎠⎟

1 00 1

2 2

2 2( )i

a b

c d� (7.10.6)

or

�2

1 1

1 1

1 1

1 1

a b

c d

a b

c d− −⎛⎝⎜

⎞⎠⎟

−−−

⎛⎝⎜

⎞⎠⎟

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎭⎪=

⎛⎝⎜

⎞⎠⎟

ia b

c d� 2 2

2 2 (7.10.7)

Thus, we obtain

0

01

1

2 2

2 2

−⎛⎝⎜

⎞⎠⎟

=⎛⎝⎜

⎞⎠⎟

=ib

ic

a b

c dSyˆ (7.10.8)

Sy is a Hermitian matrix; i.e., ˆ ˆ†S Sy y= , or 0

00

01

1

1

1

−⎛⎝⎜

⎞⎠⎟

=−

−

⎛

⎝⎜

⎞

⎠⎟

ib

ic

ic

ib

*

*. Therefore, b1 = c1 ≡ a. Hence,

ˆ ˆ* *S

i

iS

a

dy x=−⎛

⎝⎜⎞⎠⎟

=⎛

⎝⎜⎞

⎠⎟0

01

1

αα

α

α (7.10.9)

Substituting Sz and Sy in Eq. (7.10.2) gives

�2

0

01 00 1

1 00 1

0−⎛⎝⎜

⎞⎠⎟ −

⎛⎝⎜

⎞⎠⎟ − −

⎛⎝⎜

⎞⎠⎟

−i

i

iαα

α* ii

ia

dα

α

α* *01

1

⎛⎝⎜

⎞⎠⎟

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪=

⎛

⎝⎜⎞

⎠⎟� (7.10.10)

or

12

0

0

0

0ii

i

i

i

αα

αα* *

⎛⎝⎜

⎞⎠⎟

−−

−⎛⎝⎜

⎞⎠⎟

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪=

aa

d

a

d1

1

1

1

0

0

α

αα

α

α

α* * *

⎛

⎝⎜⎞

⎠⎟⎛⎝⎜

⎞⎠⎟

=⎛

⎝⎜⎞

⎠⎟ (7.10.11)

Thus, we obtain

ˆ ˆ* *S

i

iSy x=

−⎛⎝⎜

⎞⎠⎟

=⎛⎝⎜

⎞⎠⎟

0

0

0

0

αα

αα

(7.10.12)

CHAPTER 7 Spin160

Finally, we substitute Sx and Sy in Eq. (7.10.1) and obtain

0

0

0

0

0

0

0αα

αα

αα

αα* * *

⎛⎝⎜

⎞⎠⎟

−⎛⎝⎜

⎞⎠⎟

−−⎛

⎝⎜⎞⎠⎟

i

i

i

i ** 0 21 00 1

2⎛⎝⎜

⎞⎠⎟

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪= −

⎛⎝⎜

⎞⎠⎟

i� (7.10.13)

or

−

−

⎛

⎝⎜⎜

⎞

⎠⎟⎟

−−⎛

⎝⎜⎜

⎞

⎠⎟⎟

=i

i

i

i

iα

α

α

α

2

2

2

2

20

0

0

0 21 0�

00 10

0 41 00 1

2

2

2

−⎛⎝⎜

⎞⎠⎟ ⇒

−

⎛

⎝⎜⎜

⎞

⎠⎟⎟

= −⎛⎝⎜

⎞⎠⎟

α

α�

(7.10.14)

Thus, ⏐a ⏐2 = �2/4. If we choose a to be a real positive number (a = � /2), we obtain the standard representa-tion of Sx, Sy, and Sz:

ˆ ˆ ˆS Si

iSx y x= ⎛

⎝⎜⎞⎠⎟ =

−⎛⎝⎜

⎞⎠⎟ = −

� � �2

0 11 0 2

00 2

1 00 11

⎛⎝⎜

⎞⎠⎟ (7.10.15)

7.11. Using the Pauli matrices, prove: (a) ( ˆ ˆ ( ˆ ˆ ( ˆ ˆ ˆ ˆ ( ˆ ˆ )σ σ σi i i iA B A B A B) ) )= + ×I i , where ˆ ( ˆ , ˆ , ˆ )σ σ σ σ= x y z and

I is a 2 × 2 matrix, ˆ ( ˆ , ˆ , ˆ )A = A A Ax y z , ˆ ( ˆ , ˆ , ˆ )B = B B Bx y z ; (b) exp ˆ cos( ) ˆ ˆ sin( ).−⎛

⎝⎜⎞⎠⎟ = −i

I iθ σ θ σ θ

22 2n ni i/ /

Recall that we can expand an operator A in a Taylor series, e nAA n

n

ˆ

!( ˆ)= ∑ 1

(see Chap. 4).

SOLUTION

(a) We begin by considering the Pauli matrices:

ˆ ˆ ˆσ σ σx x x

ii

= ⎛⎝⎜

⎞⎠⎟ =

−⎛⎝⎜

⎞⎠⎟ = −

⎛⎝⎜

⎞⎠

0 11 0

00

1 00 1⎟⎟ (7.11.1)

so

ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆˆ

ˆσ σ σ σi A = + + =

⎛

⎝⎜⎜

⎞

⎠⎟⎟x x y y z z

x

x

A A AA

A

0

0++

−⎛

⎝⎜⎜

⎞

⎠⎟⎟

+−

⎛

⎝⎜⎜

⎞

⎠⎟⎟

=

0

0

0

0

iA

iA

A

A

A

y

y

z

z

z

ˆ

ˆ

ˆ

ˆ

ˆ ˆ ˆ

ˆ ˆ ˆ

A iA

A iA A

x y

x y z

−

+ −

⎛

⎝⎜⎜

⎞

⎠⎟⎟

(7.11.2)

Similarly, ˆ ˆˆ ˆ ˆ

ˆ ˆ ˆ.σ i B =

B B iB

B iB B

z x y

x y z

−

+ −

⎛

⎝⎜⎜

⎞

⎠⎟⎟

Thus, we obtain

( ˆ ˆ )( ˆ ˆ )ˆ ˆ ˆ

ˆ ˆ ˆσ σi iA B =

−

+ −

⎛

⎝⎜⎜

⎞A A iA

A iA A

z x y

x y z ⎠⎠⎟⎟

−

+ −

⎛

⎝⎜⎜

⎞

⎠⎟⎟

=+

ˆ ˆ ˆ

ˆ ˆ ˆ

ˆ ˆ

B B iB

B iB B

A B

z x y

x y z

z zˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ Â B A B iA B iA B A B iA Bx x y y x y y x z x z+ + − − yy x z y z

x z y z z x z

A B iA B

A B iA B A B iA

− +

+ − −

ˆ ˆ ˆ ˆ

ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆB A B A B iA B iA B A By x x z z y x x y y y+ + − +

⎛

⎝⎜⎜⎜

⎞

⎠⎟⎟

= +− −

( ˆ ˆ ) ˆ( ˆ ˆ ˆ ˆ ) ( ˆ ˆ ˆ

A Bi Ii A B A B A B Ax y y x z x xx z y z z y

x z z x

B i A B A B

A B A B i A

ˆ ) ( ˆ ˆ ˆ ˆ )

( ˆ ˆ ˆ ˆ ) ( ˆ

+ −

− + yy z z y y x x yB A B i A B A Bˆ ˆ ˆ ) ( ˆ ˆ ˆ ˆ )

( ˆ ˆ

− −

⎛

⎝⎜⎜

⎞

⎠⎟⎟

= A i BB) ˆ ( ˆ ˆ ˆ ˆ ) ( ˆ ˆ Î A B A Bi

iA B Ax y y x z x+ − −

⎛⎝⎜

⎞⎠⎟ + −

00 xx z y z z yB A B A B

ii

ˆ ) ( ˆ ˆ ˆ ˆ )0 11 0

00−

⎛⎝⎜

⎞⎠⎟ + − ⎛

⎝⎜⎞⎠⎟

(7.11.3)

CHAPTER 7 Spin 161

Note that

ˆ ˆ ˆ ˆ ˆ

ˆ ˆ ˆ

( Â B× =

⎛

⎝

⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟

=i j k

A A A

B B B

Ax y z

x y z

yy z z y z x x z x y yB A B A B A B A B Aˆ ˆ ˆ ) ( ˆ ˆ ˆ ˆ ) ( ˆ ˆ ˆ− + − + −i j ˆ )Bx k (7.11.4)

so that

( ˆ ˆ )( ˆ ˆ ) ( ˆ ˆ ) ˆ ( ˆ ˆ ) ˆ ( ˆ ˆσ σ σi iA B A B A B A⋅ = + × + ×I iz z BB A B

A B

) ˆ ( ˆ ˆ ) ˆ

( ˆ ˆ ) ˆ ˆ ( ˆ ˆ )

y y x xi i

I i

σ σ

σ

+ ×

= + ×A Bi i (7.11.5)

(b) We expand the exponent as

exp ˆ!

ˆ−⎛⎝⎜

⎞⎠⎟ = −⎛

⎝⎜⎞⎠⎟

=

∞

∑in

in

n

θ σ θ σ2

12

0

n ni i (7.11.6)

Note that

( ˆ )ˆ

ˆ( )n

ni

iσ

σn nI n

ni=

⎧⎨⎩⎪

− =for evenfor odd

11

1 1 2

for even

for odd

n

i nn( )( )( ) /− −⎧⎨⎩

−

Thus, we obtain

exp ˆ( )!

în

Inθ σ θ

21

2 2

2

n i⎛⎝⎜

⎞⎠⎟ = ⎛

⎝⎜⎞⎠⎟

⎡

⎣⎢⎢

⎤

⎦⎥⎥

nn

n

nn

=

∞ +

=∑ − +

⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢⎢

⎤

⎦⎥⎥

0

2 112 1 2( )!

( ˆ )θ σn i

00

2

0

12 2

12

∞

=

∞

∑

∑= ⎛⎝⎜

⎞⎠⎟ − −

+ˆ

( )!ˆ ( )

(I

ni

n

n

n

nθ σn i11 2

0

2 1

)!n

n

=

∞ +

∑ ⎛⎝⎜

⎞⎠⎟

θ (7.11.7)

Using the known expansions of

cos( )!

sin( )

( )!α α α α= = −

+=

∞+

=∑

2

0

2 1

02

12 1

n

n

nn

nn n

∞∞

∑

we eventually obtain

exp ˆ cos ˆ ˆ sin−⎛⎝⎜

⎞⎠⎟ = ⎛

⎝⎜⎞⎠⎟ − ⎛i

I iθ σ θ σ θ2 2 2

n ni i ⎝⎝⎜⎞⎠⎟ (7.11.8)

7.12. Consider the eigenvectors of Sn, the spin component in the n direction, where n is a unit vector:

n = + +i j ksin cos sin sin cosθ ϕ θ ϕ θ (7.12.1)

Find the rotation operator UR, where

ˆ Û UR R+ = + ′ − = − ′12

12

12

12 (7.12.2)

+ 12

and − 12

are the standard bases of Sz eigenvectors; + ′12 and − ′1

2 are the eigenvectors of Sn with eigenvalues +�/2 and −�/2, respectively. Recall that

+ ′ = ⎛

⎝⎜⎞⎠⎟ + + ⎛

⎝⎜⎞⎠⎟ −

− ′ = −

12

12

12

12

2 2cos sin

θ θ ϕei

ssin cosθ θϕ2 2

12

12

⎛⎝⎜

⎞⎠⎟ + + ⎛

⎝⎜⎞⎠⎟ −

⎧

⎨⎪⎪

⎩⎪⎪

−e i (7.12.3)

CHAPTER 7 Spin162

SOLUTION

We choose + = ⎛⎝⎜

⎞⎠⎟ − = ⎛

⎝⎜⎞⎠⎟

12

12

10

01

and , so that

+ ′ =⎛⎝⎜

⎞⎠⎟

− ′ = −12

12

2

22cos ( )

sin ( )sin (θ

θθ

ϕ/

//

ei))

cos ( )e i−⎛

⎝⎜⎞⎠⎟

ϕ

θ /2 (7.12.4)

Assume that the matrix representation of UR is Ua bc dR = ⎛

⎝⎜⎞⎠⎟ ; then the condition from Eq. (7.12.2)

UR − = − ′12

12 can be written as

a bc d ei

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟ =

⎛⎝⎜

⎞⎠

10

2

2

cos ( )

sin ( )

θθ ϕ

/

/ ⎟⎟ ⇒ ⎛⎝⎜

⎞⎠⎟ =

⎛⎝⎜

⎞⎠⎟

ab ei

cos ( )

sin ( )

θθ ϕ

/

/

2

2 (7.12.5)

Similarly, for UR − = − ′12

12 we obtain b

de i⎛

⎝⎜⎞⎠⎟ = −⎛

⎝⎜⎞⎠⎟

−sin ( )cos ( )

θθ

ϕ//

22

; so finally, we get

ˆ cos ( ) sin ( )

sin (U

a bc d

eR

i

= ⎛⎝⎜

⎞⎠⎟ =

− −θ θθ

ϕ/ /

/

2 2

2)) cos ( )

cos ( )cos ( )

eiϕ θ

θθ

/

//

2

2 00 2

⎛

⎝⎜⎞

⎠⎟

= ⎛⎝⎜

⎞⎠⎟⎟ +

−⎛⎝⎜

⎞⎠⎟ +

+0 22 0

0sin ( ) cossin ( ) cos

sin (θ ϕθ ϕ

//

θθ ϕθ ϕ

θ

//

22 0

2

) sinsin ( ) sin

cos ˆ

⎛⎝⎜

⎞⎠⎟

= ⎛⎝⎜

⎞⎠⎟ −I ii iysin (cos ) ˆ sin ( sin ) ˆθ ϕ σ θ ϕ

2 2⎛⎝⎜

⎞⎠⎟ − ⎛

⎝⎜⎞⎠⎟ − σσ

θ θ σ

x

I i= ⎛⎝⎜

⎞⎠⎟ − ⎛

⎝⎜⎞⎠⎟cos ˆ sin ( ˆ )

2 2u i

(7.12.6)

where u = −i sinj + j cosj is a unit vector (see Fig. 7.3). Note that unn

= ××

kk

, so

ki j k

i j× = = − +n 0 0 1sin cos sin sin cos

sin sinθ ϕ θ ϕ θ

θ ϕ ssin cos sinθ ϕ θ⇒ × =k n (7.12.7)

k

u

n

ϕϕ

sin θ

z

sin ϕ

sin θ cos ϕ

θ

yx

cos ϕ

sin θsin ϕ

Fig. 7.3

CHAPTER 7 Spin 163

In Prob. 7.11, part (b), we obtain the result

cos ˆ sin ( ˆ ) exp ˆθ θ θ2 2

⎛⎝⎜

⎞⎠⎟ − ⎛

⎝⎜⎞⎠⎟ = −I i

iu S ui i

�SS⎛

⎝⎜⎞⎠⎟ (7.12.8)

In conclusion, the rotation operator is

ˆ exp Ûi

R = −⎛⎝⎜

⎞⎠⎟

θ�

u i S (7.12.9)

where u is a unit vector in the direction of the axis around which we want to rotate the system, u = ××

kk

nn

.

Furthermore, n is a unit vector in the direction of the new z axis, and q is the angle between the new and old z axes.


7.13. Prove that ˆ ˆ ˆ ˆσ σ σx y z I2 2 2= = = , where I is a 2 × 2 unit matrix.

7.14. Calculate the anticommutation relation [ ˆ , ˆ ]σ σi j + where we defined [ ˆ , ˆ] ˆ ˆ ˆ ˆ.A B AB BA+ = +

Ans. [ ˆ , ˆ ] .σ σi j + = 0

7.15. Show that the matrix of ˆ ˆ ˆ ˆS S S Sx y z2 2 2 2= + + is diagonalized in the basis of eigenvectors of both Sx and Sy.

7.16. Calculate the value of ⟨ ⟩Si and ΔSi (i = x, y, z) for the spinor 12

2 12

2 12e ei iϕ ϕ/ / .+ + −( )−

Ans. ⟨ ⟩ = Δ = ⟨ ⟩ = − Δ =ˆ cos , sin ; ˆ cos ,S S S Sx x y y

� � � �2 2 2

ϕ ϕ ϕ22

02

cos ; ˆ , .ϕ ⟨ ⟩ = Δ =S Sz z

�

7.17. The matrix representation of Sx in a certain basis is Sx = −⎛⎝⎜

⎞⎠⎟

�2

1 00 1

. Find the basis and the matrix representation of Sy and Sz.

Ans. + = + + −( ) − = + − −( ) =12

12

12

12

12

12

12

12x x x x x x yS; ; ˆ �

220 11 0 2

00

⎛⎝⎜

⎞⎠⎟ =

−⎛⎝⎜

⎞⎠⎟; S

iiz

�.

7.18. Consider the rotation operator

ˆ ( , ) exp ˆ exp Ûi i

R θ θ θ σu u S u= ⎛⎝⎜

⎞⎠⎟ = ⎛

⎝⎜⎞⎠⎟�

i i2 (7.18.1)

By rotating the eigenvectors of Sz, find the eigenvectors of Sx and Sy in the standard basis.

Ans. + = = =⎛⎝⎜

⎞⎠⎟ + = + + −( )

−

12

12

12

12

12

212x R z z z

U ,θ πu j

xx R z z z

y

U= = =⎛⎝⎜

⎞⎠⎟ − = + − −( )

+ =

ˆ ,θ π2

12

12

12

12

12

u j

ˆ ,U iR z z z

y

θ π= = −⎛⎝⎜

⎞⎠⎟ + = − + + −( )

−

212

12

12

12

12

u i

== = = −⎛⎝⎜

⎞⎠⎟ − = + + −( )ˆ ,U iR z z z

θ π2

12

12

12

12u i

164

Hydrogen-like Atoms

8.1 A Particle in a Central PotentialThe Hamiltonian of a particle of mass M placed in a central potential V(r) is

ˆ ˆ ˆ( ) ˆ( )HM

V rM

V r= + = − ∇ +p2 22

2 2�

(8.1)

where the Laplacian ∇2 in spherical coordinates is

∇ = ∂∂

+ ∂∂

+ ∂∂

+ ∂∂

⎛

⎝⎜

22

2 2

2

2 2

2

2

1 1 1 1r r r θ θ θ θ ϕtan sin

⎞⎞

⎠⎟ (8.2)

Comparing Eq. (8.2) with the expression for the operator L2 obtained in Chap. 6, we see that H can be written as

ˆ ˆ ˆ( )HM r r

rMr

V r= − ∂∂

+ +�2 2

2 22

21 1

2L (8.3)

The three components of L commute with L2, and therefore, according to Eq. (8.3), they also commute with H:

[ ˆ , ˆ ] [ ˆ , ˆ ] [ ˆ , ˆ ]H L H L H Lx y z

= = = 0 (8.4)

We can now solve the three eigenvalue equations:

ˆ ( , , ) ( , , )H r E rψ ψθ ϕ θ ϕ= (8.5)

ˆ ( , , ) ( ) ( , , )L2 21ψ ψr l l rθ ϕ θ ϕ= + � (8.6)

ˆ ( , , ) ( , , )L r m rzψ ψθ ϕ θ ϕ= � (8.7)

to determine those states that are eigenfunctions of H, L2, and Lz (we used the notations of Chap. 6). Using

separation of variables (see Problem 8.1), we get

ψ( , , ) ( ) ( , )r R r Ynl lmθ ϕ θ ϕ= (8.8)

CHAPTER 8


where Ylm is the spherical harmonic function and Rnl(r) is the radial function (which does not depend on the

quantum number m). Since the Ylm( , )θ ϕ are normalized, by definition,

( ) ( ) sin*Y Y d dlm

lm

ll mm′

′ ′=∫∫ θ θ ϕ δ δππ

00

2

(8.9)

The normalization condition is

r R r dr2 2

0

1( ) =∞

∫ (8.10)

According to Problem 8.1, the radial equation for Rnl(r) is

− + + +⎡

⎣⎢

⎤

⎦⎥ =� �

2 2 2

221 1

2M rddr

rl l

MrV r R rnl

( )( ) ( ) EER rnl ( ) (8.11)

We can simplify this equation by writing

R rr

U rnl nl

( ) ( )= 1 (8.12)

from which we have

− + + +⎡

⎣⎢

⎤

⎦⎥ =� �

2 2 2

221

2Mddr

rl l

MrV r U r EUnl

( )( ) ( ) nnl r( ) (8.13)

Equation (8.13) is analogous to the one-dimensional problem of a particle of mass M moving in an effective potential Veff (r), where

V r V rl l

Mreff ( ) ( )( )= + + 1

2

2

2

� (8.14)

For the angular part, we have the equations:

− ∂∂ =i Y mYl

mlm

ϕ θ ϕ θ ϕ( , ) ( , ) (8.15)

− ∂∂

∂∂

⎛⎝⎜

⎞⎠⎟

+ ∂∂

⎡

⎣⎢

⎤

⎦⎥

1 12

2

2sinsin

sinθ θ θ θ θ ϕYl

m (( , ) ( ) ( , )θ ϕ θ ϕ= +l l Ylm1 (8.16)

8.2 Two Interacting ParticlesConsider a system of two spinless particles of mass m1 and m2 and positions r1 and r2. We assume that the potential energy depends only on the distance between the particles, V ( ).r r1 2− The study of the motion of the two particles is simplified if we adopt the coordinates of the center of mass:

rr r

cm=

++

m m

m m1 1 2 2

1 2

(8.17)

and the relative coordinates:

r r r= −1 2

(8.18)

We can then derive the equations (see Problem 8.2):

− + ∇ =�2

1 2

2

2 ( )( ) ( )

m mEφ φr rcm cm cm

(8.19)

CHAPTER 8 Hydrogen-like Atoms166

and

− ∇ +⎡

⎣⎢

⎤

⎦⎥ =�

22

2μ V E( ) ( ) ( )r r rχ χcm (8.20)

where m is the reduced mass of the two particles:

μ =+

m m

m m1 2

1 2

(8.21)

From Eq. (8.19) we conclude that the center of mass behaves like a free particle of mass m m1 2+ and energy Ecm. The relative motion of the two particles is determined by Eq. (8.20) and is analogous to the motion of a particle of mass m placed in a potential V(r).

8.3 The Hydrogen AtomThe hydrogen atom consists of a proton of mass mp = × −1 67 10 27. kg and charge e = × −1 6 10 19. C, and an electron of mass me = × −0 91 10 30. kg and charge − e. The interaction between these two particles is essentially electrostatic, and the potential energy in MKS units is

V rk e

r( ) = − 0

2

Some authors still use CGS, or Gaussian, units when dealing with atoms in which case

V rer

( ) = −2

(8.22)

where r is the distance between the two particles. Since mp is much greater than me, the reduced mass m of the system is very close to me:

μ =+

≅ −⎛

⎝⎜

⎞

⎠⎟

m m

m mm

m

me p

e pe

e

p

1 (8.23)

This means that the center of mass of the system is practically in the same place as the proton; the relative motion can be identified, to a good approximation, with the electron.

According to Eqs. (8.8) and (8.12), we may write the states of the system in the form

ψnlm nl lmr

rU r Y( , , ) ( ) ( , )θ ϕ θ ϕ= 1

(8.24)

We introduce the Bohr radius a0, which characterizes atomic dimensions:

ae0

2

20 52= ≅�

μ. Å (8.25)

and the ionization energy of the hydrogen atom:

Ee

1

4

2213 6= ≅μ

�. eV (8.26)

To solve the radial equation for the hydrogen atom, we define r = r / a0 and λk l k lE E= − / 1 . The radial equation, Eq. (8.13), then becomes

d

d

l lUkl k l

2

2 221 2

0ρ ρ ρ λ ρ− + + −

⎡

⎣⎢

⎤

⎦⎥ =( )

( ) (8.27)


where we use the index k instead of n n k l; .= + The radial equation is solved by performing a change of function (see Problem 8.1):

U ekl k lk l( ) ( )ρ ξ ρρλ

=−

(8.28)

and expanding xkl in powers of r:

ξ ρ ρ ρk ls

qq

q

C( ) ==

∞

∑0

(8.29)

The coefficients Cq can be obtained from the recursion relation (see Problem 8.1):

Ck

kk q

lqq

qq

= − +⎛⎝⎜

⎞⎠⎟

−− −

+( )

( )!( )!

( )!!(

12

11

12 1qq l

C+ +2 1 0) (8.30)

The solution for Rnl (r) can be written in the form

Rna

n l

n nenl ( )

( )!

[( )!]/ρ ρ= − ⎛

⎝⎜⎞⎠⎟

− −+

−2 1

2 10

3

32ρρ ρl

n llL ++2 1( ) (8.31)

where Lpq ( )ρ are the associated Laguerre polynomials (for detailed information, see the Mathematical

Appendix). Some examples of the radial functions are

R r a en lr a

= =− −=1 0 0

3 22 0

,/ /

( ) ( ) (8.32)

R r ara

en lr a

= =− −= −⎛

⎝⎜⎞⎠⎟2 0 0

3 2

0

22 2 1

20

,/ /

( ) ( ) (8.33)

R r ara

en lr a

= =− −=2 1 0

3 2

0

22

1

30

,/ /

( ) ( ) (8.34)

8.4 Energy Levels of the Hydrogen AtomFor fixed l, there exists an infinite number of possible energy values:

EE

k lkkl

n= −+

=( )

, , , . . .2 1 2 3 (8.35)

Each of them is at least (2l + 1)-fold degenerate. This essential degeneracy results from the radial equation being independent of the quantum number m. Some of the energy values manifest accidental degeneracy. Here the Ekl do not depend on k and l separately but only on their sum. We set n = k + l, and then

En

Ee

n nn= − = − = − ×1

2

113 6

2 1

4

2 2 2

μ�

. eV (8.36)

The shell characterized by n is said to contain n subshells, each corresponding to one of the values of l:

l n= −0 1 2 1, , , . . . , (8.37)


Each subshell contains 2l + 1 distinct states corresponding to the possible values of m,

m l l l l= − − + −, , . . . , ,1 1 (8.38)

The total degeneracy of the energy level En is

g ln n

n nn

l

n

= + = − + ==

−

∑ ( )( )

2 12 1

22

0

1

(8.39)

If one takes into account the electron’s spin (which can be in one of two possible orientations) then the number gn should be multiplied by 2.

For historical reasons (from the period in which the study of atomic spectra resulted in empirical classifica-tion of the lines observed), the various values of l are associated with letters of the Latin alphabet, as follows:

( )

( )

( )

( )

( )

l s

l p

l d

l f

l g

= ↔

= ↔

= ↔

= ↔

= ↔

0

1

2

3

4

� �

in alphhabetical order

(8.40)

8.5 Mean Value ExpressionsThe following list includes some mean value expressions of rk that are useful in many problems:

⟨ ⟩ ≡ +∞

∫r r R r drk knl

2 2

0

[ ( )] (8.41)

⟨ ⟩ = − +ra

n l l0 2

23 1[ ( )] (8.42)

⟨ ⟩ = + − +ra n

n l l2 02 2

2

25 1 3 1[ ( )] (8.43)

1 1

02r a n

= (8.44)

and

1 1

1 2202 3r a n l

=+( )/

(8.45)

8.6 Hydrogen-like AtomsThe results obtained in Sec. 8.5 originate in calculations for systems of two particles with mutual attraction energy inversely proportional to the distance between them. There are many physical systems that satisfy this condition: deuterium, tritium, ions that contain only one electron, muonic atoms, positronium, etc. The results are applicable to these systems, provided that we properly select the constants introduced in the cal-culations. For example, if the charge of a nucleus is Z, then e2 → Ze2 in all the calculations.


SOLVED PROBLEMS

8.1. (a) Write the eigenvalue equation for a particle in a central potential V(r), and perform the separation of variables in the wavefunction. Obtain the radial equation and the two angular equations. (b) Solve the radial equation for the potential of the hydrogen atom V r e r( ) .= − 2 /

SOLUTION

(a) Consider the Hamiltonian of the system:

ˆ ( )ˆ

( )Hr r

rL

rV r= − ∂

∂+ +�

2

2

2

221

2μ μ (8.1.1)

We have the following eigenvalue equation:

− ∂∂

+ +⎡

⎣⎢⎢

⎤

⎦⎥⎥

�2 2

2

2

221

2μ μθ ϕ

r rr

L

rV r r( )

ˆ( ) ( , ,ψ )) ( , , )= E rψ θ ϕ (8.1.2)

The three observables H , L2, and Lz commute. Thus, we can look for functions ψ (r, q, j) that are

eigenfunctions of L2 and Lz. We have the following system of differential equations:

ˆ ( , , ) ( , , )H r E rψ ψθ ϕ θ ϕ= (8.1.3)

ˆ ( , , ) ( ) ( , , )L r l l r2 21ψ ψθ ϕ θ ϕ= + � (8.1.4)

and

ˆ ( , , ) ( , , )L r m rzψ ψθ ϕ θ ϕ= � (8.1.5)

Note that we have three differential equations for ψ (r, q, j), which is a function of three variables. Since

ˆtan sin

L2 22 2

2

21 1= − ∂

∂+ ∂

∂ + ∂∂

⎛

⎝⎜⎞

⎠⎟�

θ θ θ θ ϕ (8.1.6)

and L iz

= − ∂∂

�ϕ

(see Chap. 6), Eqs. (8.1.4) and (8.1.5) can be replaced by

− ∂∂

+ ∂∂ + ∂

∂

⎛

⎝⎜⎞

⎠⎟=

θ θ θ θ ϕθ ϕ2 2

2

21 1

tan sin( , , ) (ψ r l ll r+ 1) ( , , )ψ θ ϕ (8.1.7)

and

− ∂∂

=ir

m rψ ψ( , , )

( , , )θ ϕ

ϕθ ϕ (8.1.8)

The solutions ψ (r, q, j) to these equations corresponding to fixed values of l and m must be products of a function of r and the spherical harmonic Y

lm ( , ):θ ϕ

ψ( , , ) ( ) ( , )r R r Ylmθ ϕ θ ϕ= (8.1.9)

Substituting Eq. (8.1.9) in Eqs. (8.1.2), (8.1.8), and (8.1.9), we obtain

− + + +⎡

⎣⎢⎢

⎤

⎦⎥⎥

� �2 2

2

2

221 1

2μ μrd

drr

l l

rV r R( )

( )( ) (rr ER r) ( )= (8.1.10)

− ∂∂

+ ∂∂ + ∂

∂⎡

⎣⎢

⎤

⎦⎥ =

θ θ θ θ ϕθ ϕ2 2 2

1 1tan sin

( , ) (Y l llm ++ 1) ( , )Yl

m θ ϕ (8.1.11)


and

− ∂∂

=i Y mYlm

lm

φθ ϕ θ ϕ( , ) ( , ) (8.1.12)

Equation (8.1.10) is the radial equation; Eqs. (8.1.11) and (8.1.12) are the angular equations. From Eq. (8.1.12) we can conclude that the j-dependence of Y

lm ( , )θ ϕ is of the form eimj. Thus,

Y G elm

lm lm( , ) ( ) ,θ ϕ θ ϕ= where G

lm ( )θ is a function of q only.

(b) We write the radial equation in the form

− + + +⎡

⎣⎢⎢

⎤

⎦⎥⎥

� �2 2

2

2

221 1

2μ μrd

drr

l l

rV r Rk( )

( )( ) ll k l k lr E R r( ) ( )= (8.1.13)

Introducing the function ukl(r) = rRkl(r) we arrive at

− + + +⎡

⎣⎢⎢

⎤

⎦⎥⎥

=� �2 2

2

2

221

2μ μd

dr

l l

rV r u rk l

( )( ) ( ) EE u rk l k l ( ) (8.1.14)

We define an effective potential:

V V rl l

reff = + +( )

( )1

2

2

2�

μ (8.1.15)

We may view Eq. (8.1.14) as a one-dimensional problem, i.e., a particle of mass m moving in the effective potential Veff, the one difference being that r assumes nonnegative values only. To express Eq. (8.1.14) in dimensionless form, we define

Ee

ae

E

EraI k l

k l

l= = =

−=μ

μλ ρ

4

2 0

2

202�

� (8.1.16)

Equation (8.1.14) becomes

d

d

l luk l k l

2

2 221 2

0ρ ρ ρ λ ρ− + + −

⎡

⎣⎢⎢

⎤

⎦⎥⎥

=( )( ) (8.1.17)

Let us define u ek l k lk l( ) ( );ρ ξ ρρλ

=−

we now obtain

d

d

dd

l lk l k l

2

2 222 1

ρλ ρ ρ ρ

ξ− + − +⎛

⎝⎜⎞⎠⎟

⎡

⎣⎢⎢

⎤

⎦⎥⎥

( )(ρρ) = 0 (8.1.18)

with the boundary condition xkl(0) = 0. An expansion of xkl(r) in a power series of r yields

ξ ρ ρ ρk ls

qq

q

C( ) ,==

∞

∑0

where C0 is the first nonzero coefficient. Thus,

d

dq s Ck l

q

qq sξ ρ

ρ ρ( )

( )= +=

∞+ −∑

0

1 (8.1.19)

and

d

dq s q s Ck l

q

qq s

2

2

0

21ξ ρ

ρρ

( )( )( )= + + −

=

∞+ −∑ (8.1.20)

Substituting Eqs. (8.1.19) and (8.1.20) into Eq. (8.1.18), we obtain a power series on the left-hand side and zero on the right-hand side; thus, the coefficients of the powers of r equal zero. We assume that the solution of Eq. (8.1.13) behaves at the origin as rs:

R r Crk l r

s( ) ~→0

(8.1.21)


Substituting Eq. (8.1.21) in to Eq. (8.1.13) we obtain

l l s s( ) ( )+ − + =1 1 0 (8.1.22)

which is satisfied if s = l or s = −(l + 1). Therefore, for a given value of Ekl, there are two linearly indepen-dent solutions of Eq. (8.1.13). The solutions behave at the origin as rl and 1/rl + 1, respectively. The latter solution must be rejected, as it can be shown that ( ) ( , )1 1/r Yl

lm+ θ ϕ is not a solution of the eigenvalue

equation, Eq. (8.1.2), for r = 0. It follows that the solutions of Eq. (8.1.13) go to zero at the origin for all l, since u r Crk l r

l( ) ~ .→

+

0

1 Therefore, the condition ukl(0) = 0 should be added to Eq. (8.1.13). In the power

series that we obtain we now take the lowest term and equate its coefficient to zero. It follows that

[ ( ) ( )]− + + − =l l s s C1 1 00 (8.1.23)

Since C0 ≠ 0, we have s = −l or s = l + 1. Next, we set the coefficient of the general term rq + s − 2 equal to zero (for s = l + 1) and obtain the following recurrence relation:

q q l C q l Cq k l q( ) [( ) ]+ + = + − −2 1 2 1 1λ (8.1.24)

Hence, assuming that C0 is known, we can calculate C1, C2, . . . . Since Cq /Cq −1 → 0 when q → ∞, the series is convergent for all r. One can show that

Ck l

kk q

lqq

qq

= − +⎛⎝⎜

⎞⎠⎟

−− −

++( )

( )!( )!

( )!(

12 1

12 1

22 1 0lC+ )!

(8.1.25)

where C0 can be determined from the normalization condition:

r R r dr u r drk l kl2 2

0

2

0

( ) ( )=∞ ∞

∫ ∫ (8.1.26)

8.2. A hydrogen atom can be viewed as two point-charged particles—a proton and an electron with Coulomb’s interacting potential between them. Write the Schrödinger equation for such a system and separate it into two parts: one describing the motion of the center of mass, and another describing the relative motion of the proton and the electron.

SOLUTION

The Schrödinger equation for the proton and the electron is

−∇

+∇⎡

⎣⎢⎢

⎤

⎦⎥⎥

+⎛

⎝⎜⎜

⎞

⎠⎟⎟

=�

212

22

m mV r E

p e

( ) ψ ψ (8.2.1)

where me and mp denote the mass of the proton and the electron, respectively. The indices 1 and 2 refer to

the proton and the electron, respectively. The potential between the particles is

V r V r r Zex x y y z z

( ) ( )( ) ( ) (

= − = −− + − + −

1 22

1 22

1 22

1 2

1

))2

2

= − Zer

(8.2.2)

Define the relative coordinates:

x x x y y y z z zr r r

= − = − = −2 1 2 1 2 1 (8.2.3)

and the center of mass coordinates rm r m r

m mp e

p ecm

=++

1 2 . For the differential operators we have

∂∂

= +⎛

⎝⎜⎞

⎠⎟∂

∂− +

∂∂

2

12

2 2

2

22

x

m

m m x

m

m m xp

p e

p

p ecm cm∂∂ + ∂∂x xr r

2

2 (8.2.4)


and

∂

∂= +

⎛

⎝⎜⎞

⎠⎟∂

∂+ +

∂∂ ∂

2

22

2 2

2

22

x

mm m x

mm m x

e

p e

e

p e rcmxx xrcm

+ ∂∂

2

2 (8.2.5)

Similar relations hold for the operators ∂

∂∂

∂∂

∂

2

12

2

22

2

12y y z

, , , and ∂

∂

2

22z

. Substituting the operators into Eq. (8.2.1), we obtain

− +∂

∂+ ∂

∂+ ∂

∂

⎛

⎝⎜

⎞

⎠⎟ + ∂�

2 2

2

2

2

2

221 1

m m x y zp e cm cm cmμ

22

2

2

2

2

2

2

∂+ ∂

∂+ ∂

∂

⎛

⎝⎜

⎞

⎠⎟

⎡

⎣⎢⎢

⎤

⎦⎥⎥

−⎛

⎝⎜⎜ x y z

Zer

r r r

⎞⎞

⎠⎟⎟

=ψ ψE (8.2.6)

where m is the reduced mass, μ =+

m m

m mp e

p e

. We separate the wavefunction ψ into two parts. The first part

depends only on the center-of-mass coordinates, while the second part depends only on the relative coordi-nates, ψ φ χ= ( ) ( ).r rrcm Substituting into Eq. (8.2.6) We arrive at

− + ∇⎡

⎣⎢

⎤

⎦⎥ =� �

22

2

21

21

φ φ χ( )( )

( )r m mr

rp e rcmcm cm μ ∇∇ + +

⎡

⎣⎢⎢

⎤

⎦⎥⎥r r

Zer

E r22

χ( ) (8.2.7)

For Eq. (8.2.7) to be valid for all values of rcm and rr, each side of the equation must be equal to a constant. Therefore we obtain two separate equations:

�2

2

20

( )( )

m mE r

p e+ ∇ +⎡

⎣⎢⎢

⎤

⎦⎥⎥

=cm cm cmφ (8.2.8)

and

�

22

2

20μ ∇ + +

⎛

⎝⎜⎞

⎠⎟=r r r

Zer

E rχ( ) (8.2.9)

Ecm is the translational kinetic energy of the center-of-mass frame and Er is the relative energy. Clearly, we have E = Ecm + Er. To obtain the wavefunction of a hydrogen atom’s electron one must solve Eq. (8.2.9) (see Problem 8.1).

8.3. The wavefunction of an electron in a hydrogen-like atom is ψ (r) = Ce−r/a, where a = a0 / Z; a0 ≈ 0.5 Å is the Bohr radius (the nucleus charge is Ze and the atom contains only one electron). (a) Compute the normalization constant. (b) If the nucleus number is A = 173 and Z = 70, what is the probability that the electron is in the nucleus? Assume that the radius of the nucleus is 1.2 × A1/3 fm. (c) What is the probability that the electron is in the region x, y, z > 0?

SOLUTION

(a) The normalization condition is ψ ψ* .d r3 1=∫∫∫ Substituting for ψ we have

C r e dr d d C r e dr a r a2 2 2

0

2 2 2

00

4−∞

−∞

∫ ∫∫ =/ /sinϕ θ θ ππ

rr =∫ 10

2π

(8.3.1)

The integral in Eq. (8.3.1) is

r e dra a ar a2 2

0

3 3 3

23

22

4−

∞

∫ =⎛⎝⎜

⎞⎠⎟

=⎛⎝⎜

⎞⎠⎟

=/ ( ) !Γ (8.3.2)

Therefore, Ca a

=⎛⎝⎜

⎞⎠⎟

=14

4 13

1 2

3π π

/

.


(b) Denoting by R the radius of the nucleus, the probability that the electron is found in the nucleus is

P r r dr d d C r e r aR

= =∫ ∫ ∫ −2 2

0

22

0

2 2

00

4ψ( ) sin /ϕ θ θ ππ πRR

dr∫ (8.3.3)

Since R is small compared to a (R ~ 1 fm = 10−5 Å and a ~ 1 Å), we can consider | ψ |2 as a constant in

the nucleus, i.e., e−2r/a ~ e−2R/a ~ 1. Thus, we have

Pa

r drRa

Zra A= = ⎛

⎝⎜⎞⎠⎟ = ⎛

⎝⎜⎞⎠⎟

= ×4 43

43

1 1 132

30

0

3

. 00 1 26

00

−∫ =R

r( . fm) (8.3.4)

(c) The wavefunction is independent of both q and j (it is a symmetrical function). Thus the probability that the electron is found in 1/8 of the space (i.e., in x, y, z > 0) is simply 1/8.

8.4. Compute the normalized momentum distribution of a hydrogen atom electron in states 1s, 2s, and 2p.

SOLUTION

The normalized momentum distribution is |ψ (p)|2, where ψ (p) is the wavefunction in the momentum rep-resentation. In order to find ψ (p), we perform a Fourier transform of the wavefunction ψ (r).

ψ ψ( )( )

( )/p rp r= −∫1

2 3

3

π�

�e d ri ⋅ (8.4.1)

We then substitute into Eq. (8.4.1) the explicit forms of ψ1s(r), ψ2s(r), and ψ2p(r), and obtain

ψ

ψ

1

3 2

2 2 2 2

1

2

1 2 1

1s

s

pa

p a

p

( )[( )]

( )

/

= ⎛⎝⎜

⎞⎠⎟ +π � �/

== ⎛⎝⎜

⎞⎠⎟ +

⎧

⎨

⎪⎪

⎩

⎪⎪

1 2 1

12 2 2 2 4πa

p a� �[( )]/

(8.4.2)

and

ψ2

3 2

2 2 2 3

2 212

2 1

1 4s pa

p a

p a( )

[( )]

/

= ⎛⎝⎜

⎞⎠⎟ +π � �/ / ��

�

2

2

2

2

3

2 2

14

1

2

2 1

−⎛

⎝⎜⎞

⎠⎟

= ⎛⎝⎜

⎞⎠⎟ψ s p

a

p a( )

( ) [(π // /� �2 6

2 2

2

2

1 4

14+

−⎛

⎝⎜⎞

⎠⎟

⎧

⎨

⎪⎪⎪

⎩

⎪⎪⎪ )]

p a

(8.4.3)

There are three different eigenfunctions for the state 2p: m = − 1, 0, 1. Thus,

m

p ia

aap

p apz

=

= − ⎛⎝⎜

⎞⎠⎟ +

0

1

12

3 2

2 2 2

:

( )[(

/

ψ π � � �/ /44

1

3

2

2

2

3 2

2 2 2 2

)]

( )( )

[(ψ p

zpa ap

p a= ⎛

⎝⎜⎞⎠⎟ +π � � �/ 11 4 6/ )]

⎧

⎨

⎪⎪

⎩

⎪⎪

(8.4.4)

and

m

p ia a p ip

p apx y

= ±

= − ⎛⎝⎜

⎞⎠⎟

±

1

122

3 2

2

:

( )( )

[(

/

ψπ � �

22 2 3

2

2

2

3 2

1 4

1

2

/ /�

�

+

= ⎛⎝⎜

⎞⎠⎟

±

)]

( )( )

ψ px yp

a a p ip

π

22

2 2 2 2 61 4� �[( )]p a / /+

⎧

⎨

⎪⎪⎪

⎩

⎪⎪⎪

(8.4.5)

8.5. Consider a wavefunction for a hydrogen-like atom:

ψ( , ) ( ) cos/ /r Z Zr Zre Z rθ π θ= − −181

263 2 3 (8.5.1)


where r is expressed in units of a0. (a) Find the corresponding values of the quantum numbers n, l, and m. (b) Construct from ψ (r, q) another wavefunction with the same values of n and l, but with a different magnetic quantum number, m + 1. (c) Calculate the most probable value of r for an electron in the state corresponding to ψ and with Z = 1.

SOLUTION

(a) Consider the exponential factor in ψ (r, q ); it has the form exp( ).− −Er Since E = − Z 2/n2, we conclude that n = 3. The angular quantum number l can be determined either by exploiting the factor rl, which multiplies the Laguerre polynomial in hydrogen-like wavefunctions, or by carrying out the following operation:

ˆ ( , ) ˆ ( ) cos ( )sin

sin cL2 2 1ψ r f r f rθ θ θ θ θ θ= = − ∂∂

∂∂L oos

( )sin

(sin )

θ

θ θ θ

⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢

⎤

⎦⎥

= ⎡⎣⎢

⎤⎦⎥

=f rd

d1 2 22 1f r l l r( ) cos ( ) ( , )θ θ= + ψ (8.5.2)

Thus, l = 1. To find the magnetic quantum number, we use the operator ˆ :Lz

ˆ ( , ) [ ( ) cos ] ( , )Lz

r i f r m rψ ψθϕ

θ θ= − ∂∂

= =0 (8.5.3)

It follows then that m = 0.

(b) In order to generate a new hydrogen-like wavefunction with a magnetic quantum number m + 1, we use the raising operator L+ (see Chap. 6). Since l = 1 and m = 0, we have

ˆ ( )( )L l m l mm m m+ + += − + + =ψ ψ ψ1 21 1 (8.5.4)

We use the differential representation of L+:

ˆ ˆ ˆ (sin cos ) (cos sin )L L iL i i i ix y+ = + = − ∂∂ + +ϕ ϕ θ ϕ ϕ ccotθ ϕ

∂∂ (8.5.5)

and obtain

ˆ ( ) cos ( ) sinL e f r e f rmi i

+ =+= ∂

∂ = −ψ 0ϕ ϕ

θ θ θ (8.5.6)

Combining Eqs. (8.5.4) and (8.5.6) we obtain

ψmi Z rf r e Z Zr Zr e+

−= − = − −13 2 31

2

1

816( ) sin ( )/ /θ

πϕ ssinθ ϕei (8.5.7)

(c) The most probable value of r occurs when (r ψ )2 assumes its maximum value. For Z = 1 we have

∂

∂ = = ∂∂ − = − +

⎛− −( )( ) / /r

r rr r e e

rr rr rψ

0 63

5 122 3 33

2

⎝⎝⎜⎞

⎠⎟ (8.5.8)

We obtain the quadratic equation r2 − 15r + 36 = 0; its roots are r = 12 and r = 3. Evaluating | r ψ | we find that it is maximal for r = 12. Therefore, the most probable value of r is 12a0.

8.6. Consider a particle in a central field and assume that the system has a discrete spectrum. Each orbital quantum number l has a minimum energy value. Show that this minimum value increases as l increases.

SOLUTION

We begin by writing the Hamiltonian of the system:

ˆ ( )( )H

mr rr

r ml l

rV r= ∂

∂∂∂

⎛⎝⎜

⎞⎠⎟

+ + +� �2

22

2

22 21

(8.6.1)


Using ˆ ( ),Hmr r

rr

V r1

2

22

2= − ∂

∂∂∂

⎛⎝⎜

⎞⎠⎟

+� we have

ˆ ˆ ( )H H

ml l

r= + +

1

2

221�

(8.6.2)

The minimum value of the energy in the state l is

E Hm

l l

rd rl

l lmin* ˆ ( )= + +⎡

⎣⎢⎢

⎤

⎦⎥⎥∫ ψ ψ1

2

23

21�

(8.6.3)

The minimum value of the energy in the state l + 1 is given by

E Hm

l l l

rl

lmin* ˆ ( )( )+

+= + + +⎡

⎣⎢⎢

⎤

⎦⎥⎥

11 1

2

221 2ψ ψ�

ll d r+∫ 13 (8.6.4)

Equation (8.6.4) can be written in the form

Em

l

rd r H

ml

l l lmin* * ˆ+

+ + += + + +∫11

2

2 13

1 1

212

ψ ψ ψ� � ll l

rd rl

( )+⎡

⎣⎢⎢

⎤

⎦⎥⎥ +∫ 1

2 13ψ (8.6.5)

Since |ψl + 1|2 and �2

2

1m

l

r

+ are positive, the second term in Eq. (8.6.5) is always positive. Consider now the

first term of Eq. (8.6.5). The minimum eigenvalue of the Hamiltonian ˆ ˆ ( )H H

ml l

r= + +

1

2

221�

corresponds to the eigenfunction ψl . Thus,

ψ ψ ψi l lHm

l l

rd r H* *ˆ ( ) ˆ

0

2

23

1 021+ +⎡

⎣⎢⎢

⎤

⎦⎥⎥

< +∫ +� ��

2

2 13

21

ml l

rd rl

( )+⎡

⎣⎢⎢

⎤

⎦⎥⎥ +∫ ψ (8.6.6)

This proves that E El lmin min< + 1.

8.7. Write the Schrödinger equation for a two-dimensional hydrogen atom. Suppose that the potential

energy is − e2/r, where r x y= +2 2 . Using separation of variables, find the radial and the angular equations. Solve the angular equation. Describe the quantum numbers that characterize the bound states and the degeneracies of the system.

SOLUTION

Consider the Schrödinger equation in two dimensions:

− ∂∂

∂∂

⎛⎝⎜

⎞⎠⎟

+ ∂∂

⎡

⎣⎢⎢

⎤

⎦⎥⎥

−�2

2

2

2

2

21 1

m r rr

r r

er

ψ ψϕ

ψψ ψ= E (8.7.1)

Performing a separation of variables ψ = R(r)Φ(j), we obtain the angular equation

∂

∂= −

2

22Φ Φ( )

( )ϕ

ϕϕm (8.7.2)

The constant m must be an integer number, so the solution of Eq. (8.7.2) is

Φm

ime( )ϕπ

ϕ= 1

2 (8.7.3)

Consider the radial equation:

− +⎛

⎝⎜

⎞

⎠⎟ + −� �

2 2

2

2 2

2

2

21

2md R

dr rdRdr

m

mrR r

er

R( ) (rr ER r) ( )= (8.7.4)

Every state Rn |m |(r) is characterized by the principal quantum numbers n and the absolute value of the angu-lar quantum number m. The energies of the system are En |m |. Every state with m ≠ 0 is twofold degenerate, and the states with m = 0 are not degenerate.


8.8. The muon is a particle with fundamental properties similar to those of the electron, with the exception of mass.

m meμ = 207 (8.8.1)

The physical system formed by a m+ and an electron is called muonium. Muonium behaves like a light isotope of hydrogen, and its electrostatic attraction is the same as that of a proton and an electron. Determine the ionization energy and Bohr radius of muonium.

SOLUTION

The reduced mass of muonium is

μμμ

μ=

+= = −

⎛⎝⎜

⎞⎠⎟

m m

m mm me

ee e

207208

11

208 (8.8.2)

The Bohr radius is

ae

a0

2

2 0 11

200( ) ( )muonium H= ≅ +⎛

⎝⎜⎞⎠⎟

�

μ (8.8.3)

where a0(H) is the Bohr radius of the hydrogen atom. The ionization energy is

Ee

E1

4

2 121

1200

( ) ( )muonium H= ≅ −⎛⎝⎜

⎞⎠⎟

μ�

(8.8.4)

where E1(H) = 13.6 eV is the ionization energy of the hydrogen atom. The study of the muon is of great interest. The two particles that comprise muonium are not subject to strong nuclear interactions, thus enabling energy levels to be calculated with great precision.

8.9. Prove the following relation between the spherical harmonic functions:

Y Ylm lm

m l

m l

* ( , ) ( , ) .θ ϕ θ ϕ ==−

=+

∑ const (8.9.1)

Use the expansion of the Legendre polynomials (see the Mathematical Appendix):

Pl ml m

P Pl lm

lm(cos )

( )!( )!

(cos ) (cosγ θ θ=−+

⏐ ⏐⏐ ⏐ 1 22 1 2) ( )eim

m l

m l

ϕ ϕ−=−

=+

∑ (8.9.2)

where g is the angle between two directions given by q1, j1 and q2, j2.

SOLUTION

We write the spherical harmonic functions in the form

Yl l m

l mPlm

m m

lm( , )

( ) ( )( )!( )!

( ) /

θ ϕ = − + −+

+14

2 12

π((cos )θ ϕeim (8.9.3)

Then,

Y Yl l m

l mPlm lm l

m* ( , ) ( , )( )!( )!

(cosθ ϕ θ ϕ π= + −+

2 14

θθ)2

m l

m l

m l

m l

= −

= +

= −

= +

∑∑ (8.9.4)

We substitute q1 = q2 = q, and j1 = j2 = j into Eq. (8.9.2) and obtain

Pl ml m

P Pl lm

l

m l

(cos )( )!( )!

(cos ) ( )γ θ= −+ = =

= −

20 1

mm l= +

∑ (8.9.5)


Substituting Eq. (8.9.5) into Eq. (8.9.4) we arrive at

Y Yl

lm lm

m l

m l

* ( , ) ( , )θ ϕ θ ϕ π= +

= −

= +

∑ 2 14

(8.9.6)

Since (2l + 1)/4p is a constant, we have established the proof.

8.10. The parity operator is defined by the replacement r → − r (see Chap. 4). How does the parity operator affect the electron’s wavefunction in a hydrogen atom?

SOLUTION

In a hydrogen atom we can express the wavefunctions using the spherical coordinates (r, q, j); we determine how the parity operation affects these coordinates (see Fig. 8.1).

Fig. 8.1

θ P

z

ϕ

x

y

r

We see that under the parity operator r → r, q → p − q and φ π ϕ→ + . Since the radial part of the hydrogen atom’s eigenfunctions depends only on r, we conclude that the parity operator affects only the spherical

harmonics part. For spherical harmonics, we have Y a ell

ll il( , ) (sin ) ;θ ϕ θ ϕ= thus,

Y Yll l

ll( , ) ( ) ( , )π θ π ϕ θ ϕ− + = −1 (8.10.1)

Therefore, under the parity operator,

Y Yll l

ll( , ) ( ) ( , )θ ϕ θ ϕ→ −1 (8.10.2)

Moreover, since ∂

∂ → − ∂∂θ θ and

∂∂ → ∂

∂ϕ ϕ , it follows that the operators L± are not affected by the parity

operation. Since we have obtained the explicit form of Ylm ( , )θ ϕ by applying the operator L− on Y

ll , we can

conclude, without any further calculation, that

Y Ylm l

lm( , ) ( ) ( , )π θ π ϕ θ ϕ− + = −1 (8.10.3)

In other words, under the parity operation

Y Ylm l

lm( , ) ( ) ( , )θ ϕ θ ϕ→ −1 (8.10.4)



8.11. Consider a hydrogen atom in a state n = 2, l = 0, and m = 0. Find the probability that an electron has a value r that is smaller than the Bohr radius.

Ans. 0.176.

8.12. For an electron in the state n and l = n − 1 in a hydrogen-like atom, find the most probable value of r.

Ans. r = n2/Z in units of a0.

8.13. Show that the degeneracy of the nth shell in a hydrogen atom equals 2n2. Take into account the spin of the electron but not the spin of the proton.

8.14. The six wavefunctions of the state 2p for the hydrogen atom are

m m Are

ae

m m

l s

r ai

l s

= + = ± =

= =

+

−

112

0

1

2

0

0

, , sin

,

/

ψ θ ϕ

±± =

= − = ± =

−

−

12

112

0

2

0

1

0

, cos

, ,

/

ψ

ψ

Bre

a

m m Cre

r a

l s

θ

−−−

r ai

ae

/

sin2

0

0

θ ϕ

(8.14.1)

where a0 is the Bohr radius and A, B, and C are the normalization constants. (a) Compute the constants A, B, and C. (b) Show that the sum ψm

l

2 is a function of r only. (c) Compute ⟨ ⟩r for ml = 0.

Ans. ( ) , , ; ( ) .a Aa

Ba

Ca

c r a= − = = ⟨ ⟩ =1

8

1

4 2

1

85

03

03

03 0

π π π

8.15. Consider a hydrogen atom in the state with the quantum numbers n and l. Calculate the dispersion of the distance

of the electron from the nucleus. Note that the dispersion is defined by ⟨ ⟩ −r r2 2 .

Ans. n n l l2 2 2 22 1

2( ) ( )

.+ − +

8.16. In a hydrogen atom the wavefunction ψ(r) describes the relative motion of a proton and an electron. If the coordinates of the center of mass of this system are x = 0, y = 0, and z = 0, show that the probability density of

the proton equals m M

mm M

m+⎛

⎝⎜⎞⎠⎟

+⎛⎝⎜

⎞⎠⎟

3 2

ψ r .

8.17. For a two-dimensional hydrogen-like atom, the Schrödinger equation is ( )−∇ − =2 2Z r E/ ψ ψ (in atomic units). Use cylindrical coordinates to find the equations for R(r) and Φ(j).

Ans. d

dm

rddr

rdRdr

Zr

m

r

2

22

2

21 2Φ Φ

ϕϕ= − ⎛

⎝⎜⎞⎠⎟ + − +( ) and EE R r

⎛

⎝⎜⎞

⎠⎟=( ) .0

8.18. Consider a particle in a spherical well, V rV r a

r a( ) .=

− <>

⎧⎨⎩

0

0 Assuming that the angular momentum is zero,

find the particle’s energy spectrums.

Ans. The energy spectrums are given by ka nk

mV= − ⎛

⎝⎜⎞⎠⎟

π arcsin�

2 0

and Ekm

= �2 2

2. These equations can be

solved either graphically or numerically (see Chap. 12).

179

Particle Motion in an Electromagnetic Field

9.1 The Electromagnetic Field and Its Associated PotentialsConsider an electromagnetic field, characterized by the values of the electric field E( , )r t and of the magnetic field B( , )r t . The fields E( , )r t and B( , )r t are not independent; they must satisfy Maxwell’s equations. It is possible to introduce a scalar potential φ (r, t) and a vector potential A r( , )t such that

E = −∇ − ∂∂φ 1

c tA

(9.1)

and

B = ∇ × A (9.2)

Using Maxwell’s equations, it can be shown that φ and A can always be found. However, when E and B are given, φ and A are not uniquely determined. When we choose a particular set of potentials, we say that we choose a gauge. From one set of potentials (φ, A) we can obtain another set, (φ′, A′) by performing a gauge transformation:

′ = − ∂∂φ φ 1

cf t

t( , )r

(9.3)

and

′ = + ∇A A rf t( , ) (9.4)

where f (r, t) is an arbitrary function of r and t (see Problem 9.2). The equations describing the physical sys-tem involve the potentials φ and A, but we shall see that in quantum mechanics, as in classical physics, the predictions of the theory do not depend on the gauge chosen (that is, the particular set of φ and A describing the electromagnetic field). This important property is called the gauge invariance (see Problem 9.5).

Let us consider two examples of gauges describing a constant magnetic field in the z direction, B = B0k. First we have the symmetric gauge,

A r= − × = −12

12

0 0 0

Bi j kx y z

B (9.5)

or A = −B0

20( , , )y x . Another gauge is the Landau gauge:

A = −( , , )B0 0 0y (9.6)

CHAPTER 9

CHAPTER 9 Particle Motion in an Electromagnetic Field180

9.2 The Hamiltonian of a Particle in the Electromagnetic FieldConsider a particle of mass m and charge q. The classical equation of motion (in CGS units) in the presence of electric and magnetic fields E and B is

md

dtq

qc

2

2

rv= + ×E B (9.7)

The classical Hamiltonian that leads to this equation of motion is

Hm

qc

qc

q= −⎛⎝⎜

⎞⎠⎟ −⎛

⎝⎜⎞⎠⎟ +1

2p A p Ai φ (9.8)

where φ and A are the potentials relating to E and B according to Eqs. (9.1) and (9.2) (see Problem 9.1).In this chapter we use a semiclassical theory for particle motion in an electromagnetic field. In this theory

the field is analogous to a classical field, while the system is treated according to the postulates of quantum mechanics. Thus, the particle is described by a wavefunction Ψ(r, t), and the Hamiltonian is written as in Eq. (9.8), but now ˆ , ˆp A, and φ represent the corresponding operators (see Problem 9.3).

When we perform a gauge transformation according to Eqs. (9.3) and (9.4), the wavefunction describing the particle transforms (see Problem 9.4) as

� ′ = ⎡⎣⎢

⎤⎦⎥

Ψ Ψ( , ) exp ( , ) ( , )r r rtiqc

f t t�

(9.9)

9.3 Probability Density and Probability CurrentGiven a wavefunction Ψ (r, t), the probability density is

ρ = Ψ( , )r0

2t (9.10)

where ρ expresses the probability of finding the particle at time t at the point r0. For particles with mass m and charge q (without a magnetic moment), the probability current density is

j A= ∇ − ∇ −⎡⎣⎢

⎤⎦⎥

12

2m i

qc

�( )Ψ Ψ Ψ Ψ Ψ Ψ∗ ∗ ∗ (9.11)

If we consider a particle with spin s and a magnetic moment ms, we have

j A S= ∇ ∇ −⎡⎣⎢

⎤⎦⎥

+ ∇ ×12

2m i

qc

css�

( ) (Ψ Ψ Ψ Ψ Ψ Ψ Ψ∗ ∗ ∗ ∗−μ

ΨΨ) (9.12)

The continuity equation

∂∂ + ∇ =⋅ρ

tj 0 (9.13)

relates the probability density and the probability current (see Problem 9.3). Both ρ and j do not depend on the gauge chosen, and they are said to be gauge-invariant; see Problem 9.5. The “real” current corresponding to a particle of charge q is defined by

I j= q (9.14)


9.4 The Magnetic MomentFor a particle with a magnetic moment ls in a magnetic field B, the interaction Hamiltonian is

H sint = −l iB (9.15)

This term should be added to the Hamiltonian. An electron of spin S has a magnetic moment

l = − egmc2

S (9.16)

where g, the gyromagnetic relation constant is very close to 2:

g = + +⎛⎝

⎞⎠ =⋅ ⋅ ⋅2 1 2 2 002 319

απ . (9.17)

9.5 UnitsIn discussing electromagnetic phenomena, it is customary to adopt one of the many possible systems of units. The MKS system is popular in solving practical or engineering problems. In the study of the interaction of electromagnetic radiation with the fundamental constituents of matter, it is sometimes more convenient to adopt the Gaussian system of units. Here we will use the latter system.

SOLVED PROBLEMS

9.1. The classical equation of motion for a particle with mass m and charge q in the presence of electric and magnetic fields E and B is

m qqc

a v= + ×E B (9.1.1)

where a is the acceleration of the particle and v is its velocity vr

r av

r= ≡ = ≡( )ddt

ddt

and . E and B

must satisfy Maxwell’s equations, so it is possible to define the vector potential A (r, t) and the scalar potential φ (r, t) such that

E = −∇ − ∂∂φ 1

c tA

(9.1.2)

and

B = ∇ × A (9.1.3)

Show that the Hamiltonian

Hm

qc

qc

q= −⎛⎝⎜

⎞⎠⎟ −⎛

⎝⎜⎞⎠⎟ +⋅1

2p A p A φ (9.1.4)

leads to the equation of motion. Use the Hamilton equations:

rp

=∂∂H

(9.1.5)

and

pr

= − ∂∂H

(9.1.6)


Proceed with the following steps: (a) Write r as a function of p and A. (b) Write r as a function of p and A. (c) Use Eq. (9.1.6) to write p as a function of v and A. (d) Use the vector “chain rule,”

ddt t

ddt t

A A rA

Av A= ∂

∂ + ∇⎛⎝⎜

⎞⎠⎟ = ∂

∂ + ∇⋅ ⋅( ) (9.1.7)

and the vector identity

( ) ( ) ( )v A v A v A⋅ ⋅∇ = − × ∇ × + ∇ (9.1.8)

to find ddtA

. (e) Combine parts (a) to (d ) to get the equation of motion.

SOLUTION

(a) Using Eqs. (9.1.5) and (9.1.4) we get

rp

p A p A= ∂∂ −⎛

⎝⎜⎞⎠⎟ −⎛

⎝⎜⎞⎠⎟ +⎡

⎣⎢⎤⎦⎥

=⋅12

1m

qc

qc

qφmm

qc

p A v−⎛⎝⎜

⎞⎠⎟ = (9.1.9)

(b) As in part (a) we obtain

rr

p Ap= = −⎛

⎝⎜⎞⎠⎟

⎡⎣⎢

⎤⎦⎥

= −ddt

ddt m

qc m

ddt

qc

1 1 dddt m

qc

Ap A⎡

⎣⎢⎤⎦⎥

= −⎡⎣⎢

⎤⎦⎥

1 (9.1.10)

(c) From Eqs. (9.1.6) and (9.1.4) we arrive at

pr

p A p A= − ∂∂ = −∇ = −∇ −⎛

⎝⎜⎞⎠⎟ −⎛

⎝⎜⎞⎠⎟ +⋅H

Hm

qc

qc

q1

2φφ⎡

⎣⎢⎤⎦⎥

(9.1.11)

Recall that r and p are independent phase space variables in Hamilton’s approach, so ∇ =⋅ p 0. Using ∇ =⋅( )p p 0, we write Eq. (9.1.11) as

p p A A= ∇ −⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

⎡⎣⎢

⎤⎦⎥

− ∇⋅1m

qc

qc

q φ (9.1.12)

From Eq. (9.1.9) and from Eq. (9.1.12) we have

p v A= ∇ − ∇⋅qc

q( ) φ (9.1.13)

(d) From Eqs. (9.1.7) and (9.1.8) we obtain

ddt tA A

v A v A= ∂∂ − × ∇ × + ∇ ⋅( ) ( ) (9.1.14)

Finally, using Eq. (9.1.3) we have

AA A

v v A= = ∂∂ − × + ∇ ⋅d

dt tB ( ) (9.1.15)

(e) Combining Eqs. (9.1.10), (9.1.13), and (9.1.15) we obtain

r vA= × − ∂

∂ + ∇⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢

⎤

⎦⎥

1 1m

qc

qc t

( )B φ (9.1.16)

Multiplying Eq. (9.1.16) by m and using Eq. (9.1.2) we arrive at

mqc

qr v= × +( )B E (9.1.17)

which is the equation of motion.


9.2. Let A (r, t) and φ (r, t) satisfy Eqs. (9.1.2) and (9.1.3). For electric and magnetic fields, E and B, are the potentials A and φ determined uniquely? If not, explain.

SOLUTION

Assume that A1 and A2, φ1 and φ2 satisfy Eqs. (9.1.2) and (9.1.3) with the same E and B, namely,

E = −∇ −∂∂ = − ∇ +

∂∂

⎛⎝⎜

⎞⎠⎟

φ φ11

221 1

c t c tA A

(9.2.1)

and

B = ∇ × = ∇ ×A A1 2 (9.2.2)

Now, if A and φ are determined uniquely, then we must have A1 = A2 and φ1 = φ2. We define a ≡ A1 − A2 and φ = φ1 − φ2 and investigate whether a = 0 and φ = 0. From Eq. (9.2.2) we obtain

∇ × =a 0 (9.2.3)

Since the gradient of any function f (r, t) satisfies ∇ × ∇ =( )f 0, one can show that a = ∇f for some function f (r, t). If we use Eq. (9.2.1), we obtain

∇ + ∂∂ =φ 1

0c t

a (9.2.4)

From Eq. (9.2.4) we get ∇ + ∇ ∂∂

⎛⎝⎜

⎞⎠⎟

=φ 10

cft

or

φ = − ∂∂ +1

cft

C t( ) (9.2.5)

where C(t) is a function of t. Without loss of generality, choose C = 0, since this corresponds to shifting the energy by a constant. From Eq. (9.2.5) we therefore obtain

a = ∇ = − ∂∂f

cft

φ 1 (9.2.6)

where f (r, t) is any function of r and t. We see that a and φ are not necessarily zero. The potentials A and φ are not determined uniquely since f is arbitrary. The nonuniqueness in Eq. (9.2.6) is called gauge freedom. This means that if A and φ satisfy Eqs. (9.1.2) and (9.1.3), then A′ and φ′ obtained by the transformation equations

′ = + ∇ ′ = − ∂∂A A f

cft

φ φ 1 (9.2.7)

are also potentials.

9.3. (a) Write the quantum Hamiltonian for a particle with mass m and charge q in the presence of an electromagnetic field. (b) What is the probability density for finding the particle in r = r0 at t = t0? (c) Obtain the equation of conservation of probability and find the probability current density.

SOLUTION

(a) From the classical Hamiltonian (9.1.4) we reach the quantum Hamiltonian by replacing r and p with the operators r and p. Remember, however, that A(r, t) and φ (r, t) are functions of r, so we must also replace r with r in these functions. Thus, we obtain

ˆ ˆ ˆ (ˆ, ) ˆ(ˆ, )Hm

qc

t q t= −⎛⎝⎜

⎞⎠⎟ +1

2

2

p A r rφ (9.3.1)

(b) Let Ψ(r, t) be the wavefunction of the particle. Then the probability density of finding the particle in r = r0 at t = t0 is

ρ ( , ) ( , ) ( , ) ( , )r r r r0 0 0 0

2

0 0 0 0t t t t= =Ψ Ψ Ψ∗ (9.3.2)

(c) First calculate ∂∂ρt

:

∂∂ = ∂

∂ = ∂∂ + ∂

∂ρ ∗

∗∗

t t t t( )Ψ Ψ Ψ Ψ Ψ Ψ

(9.3.3)


Using the Schrödinger equation and its complex conjugate − ∂∂ =i

tH�

Ψ Ψ∗

∗( ˆ )

∂∂ = − −ρ ∗ ∗

t iH H

1�

[( ˆ ) ( ˆ )]Ψ Ψ Ψ Ψ (9.3.4)

Use the representations

ˆ ˆr r p= = − ∇i� (9.3.5)

In a coordinate representation, ˆ (ˆ, )A r t becomes a vector function, so

ˆ (ˆ, ) ( , )A r A rt t= (9.3.6)

and the quantum Hamiltonian is

Hm

iqc

iqc

q= ∇ +⎛⎝⎜

⎞⎠⎟ ∇ +⎛

⎝⎜⎞⎠⎟ +⋅1

2� �A A φ (9.3.7)

Equation (9.3.4) then gives

∂∂ = − − ∇ +⎛

⎝⎜⎞⎠⎟ − ∇ +⎛

⎝⎜⎞⎠⎟⋅ρ

t i mi

qc

iqc

1 12�

� �Ψ ΨA A ∗∗

∗

⎡⎣⎢

⎤⎦⎥

⎧⎨⎩

− − ∇ +⎛⎝⎜

⎞⎠⎟ ∇ +⎛

⎝⎜⎞⋅Ψ 1

2mi

qc

iqc

� �A A⎠⎠⎟⎡⎣⎢

⎤⎦⎥⎫⎬⎭

Ψ (9.3.8)

which can be written as

∂∂ = −∇ ∇ − ∇ −⎡

⎣⎢⎤⎦⎥

⎧⎨⎩

⎫⎬⎭

ρ ∗ ∗t m i

qc

12

2�( )Ψ Ψ Ψ Ψ Ψ ΨA (9.3.9)

The equation describing probability conservation is

∂∂ + ∇ =⋅ρ

tj 0 (9.3.10)

where j is the probability current density. From Eqs. (9.3.9) and (9.3.10) we conclude

j A= ∇ − ∇ −⎡⎣⎢

⎤⎦⎥

12

2m i

qc

�( ) ˆΨ Ψ Ψ Ψ Ψ Ψ∗ ∗ ∗ (9.3.11)

which is the probability current density for a particle moving in a region with an electromagnetic field. In a vacuum in which there is no electromagnetic field, A = 0, and Eq. (9.3.11) is reduced to the known probability current density described in Chap. 3.

9.4. According to the postulates of quantum mechanics, a given physical system is characterized by a state vector Ψ . Consider a particle of mass m and charge q influenced by an electric field E and a magnetic field B. In Problem 9.2 we have shown how different pairs of potentials A and φ can describe the same E and B. In this problem, we study how the state vector Ψ depends on the choice of gauge (A and φ). Follow these steps: (a) Write the Hamiltonian with A and φ; then with A′ and φ′ relate A and φ by Eq. (9.2.7). (b) Write the Schrödinger equation for the two cases. (c) Show that if Ψ is the solution of the first Schrödinger equation, then

�Ψ Ψ( , ) ( , )( , ) /r rrt e tiq f t c= � (9.4.1)

is the solution of the second equation [where f is the same as in Eq. (9.2.7)]. (d) Discuss the results.


SOLUTION

(a) According to Eq. (9.1.4), the classical Hamiltonian for A and φ is

Hm

qc

qc

q= −⎛⎝⎜

⎞⎠⎟ −⎛

⎝⎜⎞⎠⎟ +⋅1

2p A p A φ (9.4.2)

Similarly, for A′ and φ′ we have

�Hm

qc

qc

q= − ′⎛⎝⎜

⎞⎠⎟ − ′⎛

⎝⎜⎞⎠⎟ + ′⋅1

2p A p A φ (9.4.3)

Using Eq. (9.2.7) we obtain

�Hm

qc

qc

fqc

qc

f q= − − ∇⎛⎝⎜

⎞⎠⎟ − − ∇⎛

⎝⎜⎞⎠⎟ + −⋅1

2p A p A φ qq

cft

∂∂ (9.4.4)

(b) The Schrödinger equation for the first case is

H id

dtΨ Ψ= � (9.4.5)

We can use Eq. (9.4.2) to write the Schrödinger equation, in the coordinates representation, by replacing p with − ∇i� , and obtain

1

2

2

mi

qc

q t i− ∇ −⎛⎝⎜

⎞⎠⎟ +

⎡

⎣⎢⎢

⎤

⎦⎥⎥

= ∂� �ˆ ( , )

(A r

rφ Ψ Ψ ,, )tt∂ (9.4.6)

For the second case we have

H id

dt�Ψ Ψ= � (9.4.7)

Using Eq. (9.4.4) we have, in the coordinates representation,

1

2

2

mi

qc

qc

f qqc

ft

− ∇ − − ∇⎛⎝⎜

⎞⎠⎟ + − ∂

∂⎡

⎣⎢⎢

⎤

⎦⎥⎥

� A φ �ΨΨ Ψ( , )

( , )r

rt i

tt

= ∂∂�

� (9.4.8)

(c) Suppose that Ψ(r, t) is a solution of Eq. (9.4.6). Define

�Ψ Ψ( , ) ( , )( , ) /r rrt e tiq f t c= � (9.4.9)

We wish to show that �Ψ is the solution of Eq. (9.4.8). Using Eqs. (9.4.6) and (9.4.9)

it

tqc

f tt

e tiq f t c�

�∂∂ = − ∂

∂�Ψ Ψ( , ) ( , )

( , )( , ) /r rrr ++ ∂

∂⎛⎝⎜

⎞⎠⎟

= − ∂∂

e it

t

qc

f t

iq f t c( , ) / ( , )

( , )

r r

r

��

Ψ

ttt e

mi

qc

qiq f t c�Ψ( , ) ˆ( , ) /r Ar+ − ∇ −⎛⎝⎜

⎞⎠⎟ +�

�1

2

2

φφ⎡

⎣⎢⎢

⎤

⎦⎥⎥

e tiq f t c( , ) / ( , )r r� �Ψ (9.4.10)

So,

it

tqc

f tt

q t e�∂

∂ = − ∂∂ +⎡

⎣⎢⎤⎦⎥

+�

�Ψ Ψ( , ) ( , )( , )

r rrφ iiq f t c iq

mi

qc

e( , ) / ˆr A��

12

2

− ∇ −⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢⎢

⎤

⎦⎥⎥

− ff t c t( , ) / ( , )r r� �Ψ (9.4.11)


We calculate the last term in the right-hand side of Eq. (9.4.11):

− ∇ −⎛⎝⎜

⎞⎠⎟ − ∇ −⎛

⎝⎜⎞⎠⎟

⎡⎣⎢

⎤⎦⎥

⋅ −iqc

iqc

e iq f� �ˆ ˆ (A A rr

r

r

A

, ) /

( , ) /

( , )

ˆ

t c

iq f t

t

iqc

e

�

�

�Ψ

= − ∇ −⎛⎝⎜

⎞⎠⎟ ⋅ − cc q

cf t i

qc

t

e

��− ∇ − ∇ −⎛

⎝⎜⎞⎠⎟

⎡⎣⎢

⎤⎦⎥

=

( , ) ˆ ( , )r A r�Ψ

−− − ∇ − ∇ −⎛⎝⎜

⎞⎠⎟ − ∇ − ∇⋅iq f t c q

cf i

qc

qc

f i( , ) / ˆr A�� −−⎛

⎝⎜⎞⎠⎟

qc

tˆ ( , )A r�Ψ (9.4.12)

hence,

it

tqc

f tt

qm

qc

f iq

� �∂

∂ = − ∂∂ + + − ∇ − ∇ −

�Ψ( , ) ( , )r r φ 12 cc

tˆ ( , )A r⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢⎢

⎤

⎦⎥⎥

2�Ψ (9.4.13)

So, �Ψ( , )r t is indeed the solution of the Schrödinger equation (9.4.8).

(d) When we pass from one gauge to another, the state vector describing the system is transformed by the

unitary transformation e−iq f (r, t)/c�, where f (r, t) is the function relating the two gauges. For the wavefunc-tion, the gauge transformation corresponds to a phase change that varies from one point to another and is therefore not a global phase factor. However, the physical predictions obtained by using the wavefunc-tions ψ and �ψ are the same, since the operators that describe the physical quantities are also transformed when we change between the gauges (see Problem 9.5).

9.5. In Problem 9.4 it was shown that when we perform a gauge transformation

A A A→ ′ = + ∇

→ ′ = − ∂∂

⎧⎨⎪

⎩⎪

f

cft

φ φ φ 1 (9.5.1)

The wavefunction describing a particle of mass m and charge q transforms according to

Ψ Ψ Ψ( , ) ( , ) ( , )( , ) /r r rrt t e tiq f t c→ ′ = � (9.5.2)

(a) Do the probability density and the probability current change when we pass from one gauge to another? (b) Suppose that at time t we want to measure a physical quantity Q. Does the probability of obtaining an eigenvalue q of Q depend on the gauge? (Assume for simplicity that q is nondegenerate.)

SOLUTION

(a) The probability density in the first gauge is

ρ ∗( , ) ( , ) ( , ) ( , )r r r rt t t t= =Ψ Ψ Ψ2 (9.5.3)

After the gauge transformation, and according to Eq. (9.5.2),

′ = ′ = ′ ′ =ρ ∗( , ) ( , ) ( , ) ( , ) ( , ) /r r r r rt t t t eiq f tΨ Ψ Ψ2 cc iq f t ct e t t t� �Ψ Ψ Ψ Ψ( , ) ( , ) ( , ) ( , )( , ) /r r r rr− =∗ ∗ (9.5.4)

We see that the probability density is gauge-invariant. Now, the probability current density in the first gauge is

j A=1

2m iqc

�( )Ψ Ψ Ψ Ψ Ψ Ψ∗ ∗ ∗∇ − ∇ −{ }2

(9.5.5)


When we perform the gauge transformation, Eq. (9.5.1), we have

′ = ∇ −−j r r12m i

e e eiq f t c iq f t c i� � �[ ( )( , ) / ( , ) /Ψ Ψ∗ qq f t c iq f t ce

qc

f

( , ) / ( , ) /( )]

( )(

r r

A

� �Ψ Ψ∇{− + ∇

− ∗

2ee e

m iiqc

iq f t c iq f t c− }=

( , ) / ( , ) /)( )r r� �

��

Ψ Ψ∗

12

ΨΨ Ψ Ψ Ψ Ψ Ψ Ψ Ψ Ψ∗ ∗ ∗ ∗∇ + ∇ + ∇ − ∇⎡⎣⎢

⎤⎦⎥

− + ∇fiqc

fqc

f�

2( )A ∗∗

∗ ∗ ∗

Ψ

Ψ Ψ Ψ Ψ Ψ Ψ

⎧⎨⎩

⎫⎬⎭

= ∇ − ∇⎡⎣ ⎤⎦ −{ }12

2m i

qc

�A

We see that the probability current density is gauge-invariant.

(b) Suppose that Φ(r, t) is the eigenfunction of Q corresponding to the eigenvalue q:

ˆ ( , ) ( , )Q t q tΦ Φr r= (9.5.6)

According to the postulates of quantum mechanics (see Chap. 4), the probability of obtaining q when the system is in the state Ψ(r, t) is

P t tq = =φ φ∗Ψ Ψ( , ) ( , )r r (9.5.7)

When we make the gauge transformation, Eq. (9.5.1), the wavefunction φ (r, t) will transform to

φ φ φ( , ) ( , ) ( , )( , ) /r r rrt t e tiq f t c→ ′ = � (9.5.8)

The probability of obtaining q will be determined according to Eqs. (9.5.2) and (9.5.9):

′ = ′ ′ = −P t t e t eqiq f t c i' φ φ∗ ∗( , ) ( , ) ( , )( , ) /r r rrΨ � qq f t c

qt t t P( , ) / ( , ) ( , ) ( , )r r r r� Ψ Ψ= =φ∗ (9.5.9)

We can conclude that all the physical predictions do not depend on the gauge that has been chosen.

9.6. A one-dimensional harmonic oscillator consists of a particle with mass m and potential energy

V x m x( ) = 12

2 2ω (9.6.1)

This particle has a charge q and is placed in a uniform electric field E parallel to the x axis, E = E i. (a) Find a suitable potential field φ (x) corresponding to the electric field. (b) Write the Hamiltonian of the particle. (c) Perform a coordinate transformation y = ax + b (a and b are constants), such that in the y coordinate the Hamiltonian is similar to that of a one-dimensional harmonic oscillator (with no charge). What are a and b? (d ) Find the energy eigenvalues and eigenstates of the system.

SOLUTION

(a) We have E = E i and we seek φ(x, t) such that

E = −∇φ (9.6.2)

Since B = 0, we seek a gauge in which A = 0. Integrating Eq. (9.6.2) we obtain φ (x) = −ex + c, where c is a constant of integration. Let us choose c = 0; then

φ ( )x x= −ε (9.6.3)

(b) The total classical Hamiltonian is

Hpm

m x x= + −2

2 2

212

ω ε (9.6.4)

The first term on the right-hand side of Eq. (9.6.4) is the standard kinetic term, the second term is the harmonic oscillator potential energy, and the third term is the electrical potential energy.


(c) Write Eq. (9.6.4) in the following form:

Hp

mm y Hy

yy= + +

22 2

0212

ω (9.6.5)

where H0 is a constant and y = ax + b. Consider the kinetic term. We see that py = px, so a = 1. Now we can substitute y = x + b into Eq. (9.6.5) and obtain

Hpm

m x b Hpm

m x m bxyx x= + + + = + +2

2 20

22 2 2

212 2

12

ω ω ω( ) ++ +12

2 20m b Hω (9.6.6)

From Eqs. (9.6.4) and (9.6.6) we see that Hx = Hy only if b = −e /mw 2 and H0 = −e 2/ 2mw2. To conclude, if we perform the coordinate transformation y = x − e /mw2, we get a one-dimensional harmonic oscil-lator with no charge, and the energy shifted by −e 2/ 2mw2.

(d) The energy eigenvalues of a one-dimensional harmonic oscillator are

E nn = +⎛⎝⎜

⎞⎠⎟

12

12

�ω (9.6.7)

corresponding to the eigenstate ψn We have a shifted harmonic oscillator; thus, the energy eigenvalues are now,

E nmn = +⎛

⎝⎜⎞⎠⎟ −1

212

12

2

2�ω εω

(9.6.8)

Its eigenfunctions are

ψ ψn ny xm

( ) = −⎛⎝⎜

⎞⎠⎟

εω 2 (9.6.9)

As a function of y, Eq. (9.6.9) expresses the standard one-dimensional harmonic oscillators’ eigenfunctions. Note that as a function of x, however, those eigenfunctions are different.

9.7. Consider the constant magnetic field B = B0k. (a) Find the potential A corresponding to the symmetric

gauge A r= ×12

B. (b) Find the potential A corresponding to a nonsymmetric gauge. (c) Compute the

gauge function f (r, t) relating the two gauges used in parts (a) and (b).

SOLUTION

(a) In the symmetric gauge A r= − ×12

B we get

A = −⎛

⎝

⎜⎜

⎞

⎠

⎟⎟

= − +12

0 0

12

12

i j ki jx y z y x

BB B

0

0 0 (9.7.1)

so

A = −B02

0( , , )y x (9.7.2)

(b) We can use any other gauge and find a different A. As an example, we can try to find A only in the x direction, A = �Ax i. In that case,

∇ × = ∂∂

∂∂

∂∂

⎛

⎝

⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟

= +∂∂

⎛

⎝A

i j k

x y z

A

Az

x

x

�

�

0 0

⎜⎜⎞

⎠⎟ −

∂∂

⎛

⎝⎜

⎞

⎠⎟ =j k k

�Ayx B0

(9.7.3)

By integrating Eq. (9.7.3) we obtain �A y cx = − +B0 . We can choose c = 0, so

� � �A y A Ax y z= − = =B0 0 (9.7.4)


(c) We want to find the gauge function f (r) such that A A= + ∇� f (see Problem 9.2). From Eqs. (9.7.2) and (9.7.4), we find that

A

A

= −

= −

⎧⎨⎪

⎩⎪

B

B

0

0

20

0 0

( , , )

( , , )

y x

y� (9.7.5)

or, explicitly,

A y A f y f

A x Ay f

x x x x

y y y

= − = + ∂ = − + ∂

= = + ∂ = ∂

BB

B

00

0

2

2

�

� ff

⎧

⎨⎪

⎩⎪

(9.7.6)

Hence,

∂∂ = ∂

∂ =fx

yfy

xB B0 02 2

(9.7.7)

By integrating Eq. (9.7.7) we obtain

f x y xy( , ) .= +B02

const (9.7.8)

9.8. A particle with mass m and charge q is in a region of a constant magnetic field B. Assume that B is in the k direction and use the Landau gauge; i.e., A = −( , , )B y 0 0 . (a) What is the Hamiltonian of the particle? (b) Show that the Hamiltonian commutes with px and pz. (c) Work with the basis of the eigenstates of px and pz and use a separation of variables to show that for the y component, the Schrödinger equation reduces to a Schrödinger equation of a harmonic oscillator (see Problem 9.6). (d) Find the eigenstates and eigenenergies of the Hamiltonian.

SOLUTION

(a) The classical Hamiltonian is

Hm

qc

pqc m

qc

y= −⎛⎝⎜

⎞⎠⎟ −⎛

⎝⎜⎞⎠⎟ = +⎛

⎝⎜⎞1

21

2p A A pi B i⎠⎠⎟ +⎛

⎝⎜⎞⎠⎟i p

qc

yB i (9.8.1)

where i is a unit vector in the x direction. The Hamiltonian operator is therefore

ˆ ( ˆ ˆ ) ˆ ˆ ˆHm

p pm

pqc

ym

py z x= + + +⎛⎝⎜

⎞⎠⎟ =1

21

21

22 2

2

B xx y z xp pqc

ypqc

y2 2 22

22+ + + + ⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢⎢

⎤

⎦ˆ ˆ ˆ ˆ ˆB B ⎥⎥

⎥ (9.8.2)

(b) To find the commutation relations between H and px or pz, we use the known relations

[ ˆ , ˆ ] [ ˆ , ˆ ] [ ˆ , ˆ] [ ˆ , ˆ] [ ˆ ,p p p p p y p z px y x z x x z= = = = ˆ]y = 0 (9.8.3)

and obtain

[ ˆ , ˆ ] [ ˆ , ˆ ] ˆ [ ˆ , ˆ ]H pm

p pqc

y p px x x x x= +⎛⎝⎜

⎞⎠

12

22 B⎟⎟ (9.8.4)

By definition, [ ˆ , ˆ ] [ ˆ , ˆ ]p p p px x z z= = 0, so we easily find that [ ˆ , ˆ ]H px = 0, and also for pz

[ ˆ , ˆ ] [ ˆ , ˆ ]H pm

p pz z z= =12

02 (9.8.5)


(c) Since H commutes with px and pz, we can find eigenstates of H that are also eigenstates of px and pz (recall also that [ ˆ , ˆ ]p px z = 0). We use a separation of variables; namely, ψ (x, y, z) = ψx(x)ψy(y)ψz(z). For ψx(x) and ψz(z) we choose the eigenstates of px and pz, respectively:

ψ ψ

ψ ψ

x pip x

z pip z

x x e

z z e

x

z

x

z

( ) ( )

( ) ( )

/

/

≡ =

≡ =

⎧⎨⎪

�

�

⎩⎩⎪ (9.8.6)

so

ψ ψ( , , ) ( )/ /

x y z e e yip x ip z

yx z= � �

(9.8.7)

where px and pz are now constant numbers (these are the eigenvalues). Using Eqs. (9.8.2) and (9.8.7) we get the Schrödinger equation:

ˆ ˆ ˆ ˆHm

p p pq p

cy

qc

yx z yxψ = + + + + ⎛

⎝⎜⎞⎠⎟

⎡12

22 2 22

2BB

⎣⎣⎢⎢

⎤

⎦⎥⎥

=ψ ψ( , , ) ( , , )x y z E x y z (9.8.8)

Note that in Eq. (9.8.8), px and pz are constant numbers and only py and y are operators. Let us denote 1

22 2

mp p ax z( ) ;+ = then Eq. (9.8.8) can be written as

1

21

22

22

mp

q pmc

ym

qc

yyxˆ ˆ ˆ+ ⎛

⎝⎜⎞⎠⎟

+ ⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢⎢

B B ⎤⎤

⎦⎥⎥

= −ψ ψ( , , ) ( ) ( , , )x y z E a x y z (9.8.9)

We see now that the y component of the Schrödinger equation is similar to the Hamiltonian of Problem 9.6 [see, for example, Eq. (9.6.4)]. In order to show that the y component is identical to the Hamiltonian of a harmonic oscillator we make a transformation similar to the one in Problem 9.6; that is,

ˆ ˆ ˆ

ˆ ˆ ˆ

y y ycpq

p p p

x

y y y

→ = +

→ =

⎧⎨⎪

⎩⎪

�

�

B (9.8.10)

The Schrödinger equation, Eq. (9.8.9), then becomes

1

21

2 22

22

2

mp

mqc

ypmyxˆ ˆ (� �+ ⎛

⎝⎜⎞⎠⎟ −

⎡

⎣⎢⎢

⎤

⎦⎥⎥

=B ψ EE a− ) ψ (9.8.11)

or

1

21

2 22

22

2

mp

mqc

y Epmyzˆ ˆ

� �+ ⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢⎢

⎤

⎦⎥⎥

= −B ψ⎛⎛

⎝⎜

⎞

⎠⎟ ψ (9.8.12)

If we denote �E E p mz= − 2 2/ , Eq. (9.8.12) becomes

1

212

2 2 2

mp m y x y z E x yy Bˆ ˆ ( , , ) ( ,� � � � �+⎡

⎣⎢⎤⎦⎥

=ω ψ ψ ,, )z (9.8.13)

where ωB

qcm

22

= ( )B . We see that Eq. (9.8.13) is indeed a Schrödinger equation for a one-dimensional

harmonic oscillator.

(d) Since Eq. (9.8.13) is the Schrödinger equation of a harmonic oscillator, we know its eigenvalues and eigenstates:

�E nqmc

nn B= +⎛⎝⎜

⎞⎠⎟ = +⎛

⎝⎜⎞⎠⎟� �ω 1

212

B (9.8.14)

and

ψ π��

yB m y

nym

e H y( ) ( )/

/= ⎛⎝⎜

⎞⎠⎟

−ω ω�

�1 4

22B (9.8.15)


where Hn(x) are Hermite polynomials. The eigenvalues of the original Hamiltonian, Eq. (9.8.2), are E [see Eq. (9.8.8)]. Hence,

E Epm

qmc

npmn n

z z= + = +⎛⎝⎜

⎞⎠⎟ +�

2 2

212 2

�B

(9.8.16)

where the eigenfunctions ψn(x, y, z) are

ψnB ip x ip z

x y zm

e em

x z( , , ) exp/

/ /= ⎛⎝⎜

⎞⎠⎟

−ω

π��

1 4 ωωB xn

xycpq

H ycpq2

2

�+⎛

⎝⎜⎞⎠⎟

⎡

⎣⎢⎢

⎤

⎦⎥⎥

+⎛⎝⎜

⎞⎠⎟B B

(9.8.17)

9.9. Solve Problem 9.8 for a particle of spin 1/2 (an electron, for example) and with a magnetic moment l = msS.

SOLUTION

(a) We add to the Hamiltonian, Eq. (9.8.1), the interaction energy between the spin and the magnetic field,

H = −l iB (9.9.1)

and obtain the total classical Hamiltonian:

Hm

qc

= −⎛⎝⎜

⎞⎠⎟ −1

2

2

p A l iB (9.9.2)

The magnetic field is B = B k , and we use the gauge A = −( , , )B y 0 0 to obtain the Hamiltonian operator:

ˆ ˆ ˆ ˆˆ

ˆ ˆHm

p p pq p

cy

qc

yx z yx= + + + + ⎛

⎝⎜⎞⎠⎟

12

22 2 22B B 22⎡

⎣⎢⎢

⎤

⎦⎥⎥

−μs

zsS

B ˆ (9.9.3)

(b) One can easily see that the Hamiltonian, Eq. (9.9.3), commutes with px and pz. The only term that we

need to check (after using the results of Problem 9.8) is μs

zsS

B ˆ . Since the degrees of freedom of the

spin are free from the spatial ones, we have [ ˆ , ˆ ]p S = 0. Specifically,

ˆ , ˆ ˆ , ˆps

S ps

Sxs

z zs

z

μ μB B⎡⎣⎢

⎤⎦⎥

= ⎡⎣⎢

⎤⎦⎥

= 0 (9.9.4)

(c) Including the spin states, we use the basis of the eigenstates of px and pz as well as of S2 and Sz; namely, our wavefunction is

ψ χ ψ χ( , , ) ( ) ( ,/ /

x y z e e y s Sip x ip z

y zx z

spin /= =� �1 2 )) (9.9.5)

where χ (s = 1/2, Sz) is the spin state of the electron that is an eigenstate of S2 and Sz:

ˆ ( , ) ( ) ( , ˆ ) (S s S s s s S sz z2 2 21 2 1 1 2

43

χ χ χ= = + = =/ /� � == 1 2/ , )Sz (9.9.6)

ˆ ( , ) ( , ) (S s S S s S sz z z zχ χ χ= = = = ±⎛⎝⎜

⎞⎠⎟1 2 1 2

12

/ /� � == 1 2/ , )Sz (9.9.7)

We will represent the operator S using the Pauli matrices ˆ ˆS = �2

σ . The states χ 12

12, ±( ) can be written as

χ χ12

12

12

12

10

01

, ,+( ) = ⎛⎝⎜

⎞⎠⎟ −( ) = ⎛

⎝⎜⎞⎠⎟ (9.9.8)

see Chap. 7.


(d) In order to find the eigenfunctions and eigenvalues, we follow Problem 9.8, part (d), and write the Schrödinger equation:

1

212

12

2 2 2 2

mp m y

mp

sS xy B z

sz

ˆ ˆ ˆ (� �+ + −⎛⎝⎜

⎞⎠⎟

ωμB ψ ,, , ) ( , , )y z E x y zχ ψ χspin spin= (9.9.9)

where [following Problem 9.8, part (d); see Eqs. (9.8.10) and (9.8.13)]

� ˆy y

cpq

qcm

x

B

= +

=

⎧

⎨⎪

⎩⎪

BBω

(9.9.10)

and pz, s = 1/2, and Sz = ±1/2 are constants. Defining

E Eqcm s

Ssz= − +B B

μ (9.9.11)

we obtain, from Eq. (9.9.9), a standard one-dimensional harmonic oscillator Schrödinger equation,

1

212

2 2 2

mp m y Eyˆ ˆ

� �+⎛⎝⎜

⎞⎠⎟ =ωB ψ ψ (9.9.12)

with the eigenvalues E n= +�ωB ( )1 2/ and the eigenfunctions ψ (x, y, z) χspin, where ψ (x, y, z) is as given in Eq. (9.8.17). Hence, the eigenvalues of our Schrödinger equation, Eq. (9.9.9), are

Eqmc

npm s

Sz sz= +⎛

⎝⎜⎞⎠⎟ + −�

BB

12 2

2 μ (9.9.13)

These eigenvalues are known as the Landau levels.

9.10. Consider the particle of Problem 9.8. (a) Assume that the particle is in a very large, but finite, box: 0 ≤ x ≤ Lx, − Ly ≤ y ≤ Ly, and 0 ≤ z ≤ Lz. Write the eigenfunctions for that case. (b) Find the number of states per unit area (in the xy plane).

SOLUTION

(a) Consider the Schrödinger equation

ˆ ( , , ) ( , , )H x y z E x y zψ ψ= (9.10.1)

where H is given by Eq. (9.8.2). We also have the boundary conditions

I (ψ ψx x Lx= = = =0 0) ( ) (9.10.2)

II (ψ ψy L y Ly y= − = = =) ( ) 0 (9.10.3)

III ψ ψ( ) ( )z z Lz= = = =0 0 (9.10.4)

Using the separation of variables of Problem 9.8 and Eqs. (9.10.2) and (9.10.4), we replace Eq. (9.8.6) with

ψ

ψ

xx

x

zz

z

xL

p x

zL

p z

( ) sin ( )

( ) sin ( )

=

=

⎧

⎨⎪⎪

⎩⎪⎪

12

12

(9.10.5)

where

p

Ln n

pL

n n

xx

x x

zx

z z

= =

= =

⎧ π

π

�

�

0 1 2

0 1 2

, , , . . .

, , , . . .⎨⎨⎪⎪

⎩⎪⎪

(9.10.6)


Assuming that Ly is very large such that q

cLy

B2

12

�>> , the y-part of the wavefunction, Eq. (9.8.15), will

hardly be affected by the boundary condition Eq. (9.10.3), as is the case for the ψ( )�y wavefunction. The eigenstates are therefore [see Eq. (9.8.17)]

ψ( , , ) sin( ) sin (/

x y zm

L Lp x pB

x zx z= ⎛

⎝⎜⎞⎠⎟

ωπ�

1 41

2zz

my

cpq

H ycpq

B xn

x) exp−

+⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢⎢

⎤

⎦⎥⎥

+ω

2

2

� B B⎛⎛⎝⎜

⎞⎠⎟

(9.10.7)

The eigenenergies are [see Eq. (9.8.16)]

Eqmc

nm L

nn n yz

zy z

= +⎛⎝⎜

⎞⎠⎟ +

⎛⎝⎜

⎞⎠⎟

��B 1

21

2

22π

(9.10.8)

where we used pz = p �nz /L [see Eq. (9.10.6)]. Note that Eq. (9.10.8) does not depend on nx, so we have a degeneracy.

(b) The number of states in the xy plane is the number of different possible nx and ny, such that the particle

is inside the region 0 ≤ x ≤ Lx, − Ly ≤ y ≤ Ly. We note that in the y direction we have a harmonic oscil-

lator centered at y cp qx0 = − / B [see Eqs. (9.8.10) and (9.8.11)]. Assuming that the deviations from the equilibrium point y = y0 are small, we need only to demand that − Ly ≤ y0 ≤ Ly. Hence,

− ≤ − ≤Lcpq

Lyx

yB (9.10.9)

Using Eq. (9.10.6) we get − ≤ −⎛⎝⎜

⎞⎠⎟

≤Lc

q Ln Ly

xx yB

π�, or

− ⎛⎝⎜

⎞⎠⎟ ≤ ≤ ⎛

⎝⎜⎞⎠⎟

qc

L L nq

cL Lx y x x y

B Bπ π� �

(9.10.10)

The number of different states in the region 0 ≤ x ≤ Lx and − Ly ≤ y ≤ Ly is the number of different nx in Eq. (9.10.10), namely,

nq

cL Lx x y= B

� (9.10.11)

Including the two spin states for each nx, we finally find the total number of states:

Nq

cL Lx y= 2

B�

(9.10.12)

Therefore, the number of states per unit area is

nN

qc L L

L Lq

cx y

x y= = =

area

2

2

BB�

� (9.10.13)

9.11. Refer to Problem 9.10. In the case px = 0, show that the current I is indeed zero.

SOLUTION

Using the definition of the probability current density, see Eq. (9.3.11), we obtain the probability current:

J = = ∇ − ∇( ) −⎡⎣⎢

⎤⎦⎥

qqm i

qc

j A2

2� ψ ψ ψ ψ ψ ψ∗ ∗ ∗ (9.11.1)

Since ψ is real, ψ ψ ψ ψ∗ ∗∇ − ∇ = 0, and

J = − qmc

2

2Aψ ψ∗ (9.11.2)

It was shown in Problem 9.5 that the probability current is gauge-invarient. We can choose, for example, the vector potential A = −( , , )B y 0 0 (see Problem 9.8). Hence,

J J

Jqmc

y

y z

x

= =

=

⎧⎨⎪

⎩⎪

0

2

2

B ψ ψ∗ (9.11.3)


Using Eq. (9.10.7) px = 0, we see that ψ*ψ is an even function of y. The current I is

I J= ∫ dx dy dz (9.11.4)

We have Iy = Iz = 0, and

Ix x z

LL

L

qmc

y y dy x dx z dzzx

= ∫∫−

22 2 2

002B ψ ψ ψ( ) ( ) ( )

yy

yL

∫ (9.11.5)

Since ψ( )y2 is an even function (only) in the case where px = 0), we finally get ψ( )y y dy

L

L

y

y2

0=−∫ and

Ix = 0. The classical motion of the particle is a circle and so the total current in the x or y directions is zero.

9.12. For the particle in Problem 9.10 and electric field E = E j: (a) Find the eigenstates and eigenvalues of the particle. (b) If px = 0, show that Ix ≠ 0 even though E is only in the y direction. What is the drift velocity?

SOLUTION

(a) We add to the Hamiltonian, Eq. (9.8.1), the potential energy:

H qelectric = φ (9.12.1)

where E = −∇φ. Since E = E j, we have φ = −E y, and the total classical Hamiltonian is

Hm

qc

qm

p p pq p

cyx z y

x= −⎛⎝⎜

⎞⎠⎟ + = + + +1

21

222

2 2 2p A φB

−− + ⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢⎢

⎤

⎦⎥⎥

22

2mq yqc

yEB

(9.12.2)

Working in a coordinate representation, we get the Schrödinger equation:

1

22

222

2

2

2

2

2m x y z

q pc

y mq yx− ∂∂

+ ∂∂

+ ∂∂

⎛

⎝⎜⎞

⎠⎟+ −�B

E ++ ⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢⎢

⎤

⎦⎥⎥

=qc

y x y z E x y zB 2

2 ψ ψ( , , ) ( , , ) (9.12.3)

where we use the fact that Helectric commutes with px and pz. The equation for ψ (y) is

1

22

222

2

22

m y

q pc

m q yqc

yx�∂

∂+ −⎛

⎝⎜⎞⎠⎟

+ ⎛⎝⎜

⎞⎠⎟

⎡ BE

B

⎣⎣⎢⎢

⎤

⎦⎥⎥

=ψ ψ( ) ( )y yε (9.12.4)

where ε = − −Epm

pm

x z2 2

2 2. Defining

y ycpq

x

B= + −B

vDω (9.12.5)

where vD

c= EB and ωB = q

cmB

, and using Eq. (9.12.3), we get

1

212

22

22 2

m ym y y E yB− ∂

∂+

⎛

⎝⎜⎞

⎠⎟=� ω ψ ψ( ) ( ) (9.12.6)

where

E Epm

p mzx D D= − + −

22

212

v v (9.12.7)


The eigenstates of Eq. (9.12.6) are the standard harmonic oscillator eigenfunctions, and the energy spectrum is

E Epm

p m

n

n n n nz

x D D

B y

x y z y= + − +

= +⎛⎝⎜

⎞⎠

22

212

12

v v

�ω ⎟⎟ +⎛

⎝⎜

⎞

⎠⎟ − +1

212

2 2

22 2

m Ln

Ln m

zz

D

xx D

π π� �vv (9.12.8)

Note that, unlike Eq. (9.10.8), Eq. (9.12.8) depends on nx and the degeneracy is removed (due to the electric field).

(b) The current, Eq. (9.11.4), is I J= ∫ dx dy dz. Using Eq. (9.11.3) we have Iy = Iz = 0, and

IxL

Lq

mcy y dy

y

y

=−∫

22

2B ψ( ) (9.12.9)

Notice that in contrast to Problem 9.11, here, even in the case where px = 0, the function ψ( )y2 is not

even since from Eq. (9.12.5) we can conclude that for px = 0,

y y D

B= −

vω (9.12.10)

Note that ψ( )y2 is even in y but not in y. If we make the coordinate transformation y y→ in

Eq. (9.12.9) we obtain

Iq

mcy yx

D

BL

L

y D B

y D B

= +⎛⎝⎜

⎞⎠⎟− +

−

∫2

2B v

v

v

ωω

ω

/

/

( )ψ 22d y (9.12.11)

Now using LyD

B>>

vω , we obtain

Iq

mcy y d yx

D

B≈ +⎛

⎝⎜⎞⎠⎟−∞

∞

∫2

2

2B v

ω ψ( ) (9.12.12)

The first term (linear with y ) will give zero since the integrand is antisymmetric. The second term will give, as we expected,

Iq

mcy d y

qmc

qxD

B

D

BD= = =

−∞

∞

∫2

22

2 2B Bv v

vω ωψ( ) (9.12.13)

where vD is the drift velocity ( )vD c= E B/ .

9.13. Consider a spinless particle of mass m and charge q, subjected simultaneously to a scalar potential V(r)

and a magnetic field B = B0k. Use the symmetric gauge A r= − ×12

B and find the Hamiltonian of

the particle. Write it as a sum of H0, corresponding to the case of no magnetic field, and the additional term H1.

SOLUTION

We have

Hm

qc

V= −⎛⎝⎜

⎞⎠⎟ +1

2

2

p A r( ) (9.13.1)


Using Eq. (9.5), calculate

p A p r v p−⎛⎝⎜

⎞⎠⎟ = + × + × +q

cp

qc

q

c

22

2

22 4[ ( ) ( ) ] (i iB B rr ×

= + − + − + +

B)

( ) (

2

2 02

02

22 4p

qc

p y p x yp xpq

cx y x y

B Bxx y

pq

cxp yp

q

cx y py x

2 2

2 02

02

22 2 2

4

+

= + − + + =

)

( ) ( )B B

++ + +q

cL

q

cx yz

B B02

02

22 2

4( ) (9.13.2)

Substituting Eq. (9.13.2) into Eq. (9.13.1), we obtain

Hm

pqmc

Lq

mcx y Vz= + + + +1

2 2 82 0

202

22 2B B

( ) ( )r (9.13.3)

We see that H = H0 + H1, where

Hm

p V021

2= + ( )r (9.13.4)

and

HL q

mx yz

10

202

2 2

8= − + +

μB B�

( ) (9.13.5)

and where m denotes the Bohr magneton, μ = qmc�

2.

9.14. Polarized electrons, with a spin polarization (+) in the z direction, enter a region of constant magnetic field B = B0i. The electrons move in the y direction. After time T, the electrons reach a Stern–Gerlach apparatus in which the magnetic field is in the z direction. (a) Write the interaction Hamiltonian in the region of a constant magnetic field. (b) In a detector D we can detect only electrons with spin polarization (−) in the z direction. Find the values of B0 such that all the electrons will reach the detector D. (c) For the smallest value of B0 [found in part (b)], what percent of the electrons will reach D if the traveling time in the constant magnetic field region is T/2 (not T)?

SOLUTION

(a) The interaction between the electron and the magnetic field is due to the magnetic moment of the electron,

lee

em c

= 2S, and the external magnetic field B = B0 i. The interaction Hamiltonian is

Hem c

em c

See e

xint = = =l i iB2 20 0B B

S i (9.14.1)

We can use the two-vector representation of the ± z spin states (see Chap. 7),

+ → ⎛

⎝⎜⎞⎠⎟

− → ⎛⎝⎜

⎞⎠⎟

⎧

⎨⎪⎪

⎩⎪⎪

z

z

10

01

(9.14.2)

In this representation, the electron spin operator can be described by the Pauli matrices:

ˆ ˆS = �2

σ (9.14.3)

where


ii

= ⎛⎝⎜

⎞⎠⎟ = −

⎛⎝⎜

⎞⎠⎟ = ⎛

⎝⎜⎞⎠⎟

0 11 0

00

1 00 1

(9.14.4)


Using Eq. (9.14.4), we can write Eq. (9.14.1) as

întH

em ce

= ⎛⎝⎜

⎞⎠⎟

� B0 0 11 0

(9.14.5)

(b) In order to find the state of the electrons at time t we need to solve the time-dependent Schrödinger equation:

it

H�∂

∂ =ψ

ψˆ (9.14.6)

The state ψ can be written as

ψ( ) ( ) ( )t t z t z= + + −+ −α α (9.14.7)

where α α+ −+ =2 2 1, or in the two-vector representation,

ψ( ) ( ) ( )( )

( )t t t

t

t= ⎛

⎝⎜⎞⎠⎟ + ⎛

⎝⎜⎞⎠⎟ =+ −

+

−α α

αα

10

01

⎛⎛⎝⎜

⎞⎠⎟

(9.14.8)

Using Eqs. (9.14.5) and (9.14.8), the Schrödinger equation, Eq. (9.14.6), becomes

it

t

tem c

t

e�

�∂∂

⎛⎝⎜

⎞⎠⎟

= ⎛⎝⎜

⎞⎠⎟

+

−

+αα

α( )

( )

(B0 0 11 0

))

( )

( )

( )ααα−

+

−

⎛⎝⎜

⎞⎠⎟

=⎛⎝⎜

⎞⎠⎟t

em c

t

te

� B0 (9.14.9)

Equation (9.14.9) is equivalent to the following two equations:

id t

dtt

αω α+

−=( )

( )0 (9.14.10)

id t

dtt

αω α−

+=( )

( )0 (9.14.11)

where ω0 0= e m ceB / . Making another derivative of Eq. (9.14.11) we get

id t

dt

d tdt

2

2 0

αω

α− +=( ) ( )

(9.14.12)


d t

dtt

2

2 02α

ω α−−= −

( )( ) (9.14.13)

and similarly,

d t

dtt

2

2 02α

ω α++= −

( )( ) (9.14.14)

The solutions of Eqs. (9.14.13) and (9.14.14) are

α ω ωα ω

+ + +

− −

= +=

( ) cos ( ) sin ( )

( ) cos ( )

t a t b t

t a t0 0

0 ++⎧⎨⎩ −b tsin ( )ω0

(9.14.15)

where a± and b± are constants determined by the initial condition. The initial condition is

ψ( )t z= = + = ⎛⎝⎜

⎞⎠⎟0

10

(9.14.16)

So, a+ = 1 and a− = 0. From α α+ −+ =2 2 1 we get b+ = 0 and b− = 1. Thus, the solutions of Eq. (9.14.15) are

α ωα ω

+

−

==

⎧⎨⎩

( ) cos ( )

( ) sin ( )

t t

t t0

0

(9.14.17)


and the quantum state, Eq. (9.14.8), is

ψ( )cos ( )

sin ( )t

t

t=

⎛⎝⎜

⎞⎠⎟

ωω

0

0

(9.14.18)

After a time T, the state of the electrons is

ψ( )cos ( )

sin ( )t

T

T=

⎛⎝⎜

⎞⎠⎟

ωω

0

0

(9.14.19)

If we want all the electrons to reach the detector D, we must demand that

ψ( )T z= − = ⎛⎝⎜

⎞⎠⎟

01

(9.14.20)

since the detector D detects only electrons with polarization −z. From Eqs. (9.14.18) and (9.14.19) we obtain cos( )ω0 0T = and sin( )ω0 1T = , or, equivalently,

ω π π0 20 1 2T n n= + = ± ±, , , . . . (9.14.21)

Using ω0 0= e m ceB / , we finally get

B0 2= +⎛⎝⎜

⎞⎠⎟

m ceT

ne π π (9.14.22)

(c) The minimum positive value for B0 satisfying Eq. (9.14.21) is, for n = 1,

( )minB0 2=πm c

eTe (9.14.23)

Assuming that B0 equals Eq. (9.14.22), the quantum state ψ( )t after time T / 2 is

ψ( )cos ( )

sin ( )T

T

T/

/

/2

2

20

0=

⎛⎝⎜

⎞⎠⎟

ωω (9.14.24)

Now, using Eq. (9.14.22), we have

ω π0

02= =

em c Te

( )minB (9.14.25)

Hence, from Eqs. (9.14.23) and (9.14.24),

ψ( )cos ( )

sin ( )T

T

T/

/

/2

2

212

11

0

0=

⎛⎝⎜

⎞⎠⎟

= ⎛⎝⎜

⎞⎠

ωω ⎟⎟ (9.14.26)

The probability of finding the electron in the detector D is

P z TD = − = ⎛⎝⎜

⎞⎠⎟ =ψ( ) ( )/2

12

0 101

12

22

(9.14.27)

9.15. In this problem we examine how the energy levels of the hydrogen atom are modified in the presence of a static magnetic field; this effect is called the Zeeman effect. We shall ignore the effects of spin (the “normal” Zeeman effect). Suppose that the mass of the electron is m and its charge is q. (a) We denote by H0 the Hamiltonian of the electron in the hydrogen atom (without a magnetic field). Write the eigenstates of H0 that are also eigenstates of L2 and Lz. What are the corresponding eigenvalues? (b) Suppose that the atom is placed in a uniform magnetic field B0 along the k axis. Write the new Hamiltonian. Are the states of part (a) also eigenstates of the new Hamiltonian? How are the energy

levels modified? Assume that the term q

mx y

2 22 2

8B

( )+ is negligible compared to μB

zL�B0

ˆ (this can

be shown by a detailed calculation).


SOLUTION

(a) The eigenstates of the Hamiltonian of the hydrogen atom can be written in the form

φnl m nl lmr R r Y( , , ) ( ) ( , )θ ϕ θ ϕ= (9.15.1)

The number n determines the energy level, En = −E1/n2. The energy levels in a hydrogen atom are degen-erate; for each n the number l can assume one of the values l = 0, 1, 2, . . . , n − 1, and m is an integer between −l and l. The total degeneracy of the energy level En is n2 (without spin). The wavefunction φnl m is an eigenfunction of L2 with an eigenvalue l (l + 1) �2. It is also an eigenfunction of Lz with an eigenvalue m�.

(b) According to Problem 9.13, the classical Hamiltonian is the sum of H0 and

H Lq

mx yz1 0

2 22 2

8= − + +μ

�B

B( ) (9.15.2)

Now we assume (without a detailed proof) that the second term in Eq. (9.15.2) is negligible when com-pared to the first term. Since φnl m( )r is an eigenstate of Lz, we have

( ˆ ˆ ) ( ) ˆ ( ) ˆ ( )H H H Lnlm nlm z nlm0 1 0 0+ = − =φ φ φr r rμ�B (( ) ( )E mn nlm− μB0 φ r (9.15.3)

We see that φnlm( )r are also eigenstates of the new Hamiltonian, but the energies are shifted by mμB0.Also, because of the presence of the magnetic field, the degeneracy is removed.

9.16. An electron is constrained to move on a one-dimensional ring of radius R, see Fig. 9.1. At the center of the ring there is a constant magnetic flux F in the z direction. (a) Find the vector potential A on the ring in the gauge in which it is independent of j. (b) Write the Schrödinger equation for the constrained electron. (c) What are the general boundary conditions on the wavefunctions of the electron? (d) Find the eigenstates and eigenenergies of the electron. Use functions of the form eikj.

Magnetic flux

j

y

x

e

R

Fig. 9.1

SOLUTION

(a) The magnetic field is B = Bk. The magnetic flux through the surface bounded by the ring is

Φ = dx dy

insidethe ring

∫ ∫ B ik (9.16.1)


We would like to find A on r = R, such that B = ∇ × A and A does not depend on f. From Eq. (9.16.1) we obtain

Φ = ∇ ×∫∫ ( )A ik dS

S

(9.16.2)

where S is the surface bounded inside the ring. Using Stokes’s theorem we can write Eq. (9.16.2) as

Φ = ∫ A li d

C

(9.16.3)

where C is the boundary of S, which is the ring r = R, and d l is along the curve C. Now,

d Rdl = ( )ϕ i (9.16.4)

where i is a unit vector tangential to the ring. From Eqs. (9.16.3) and (9.16.4) it follows that

Φ = ∫ A R dϕ

π

ϕ0

2

(9.16.5)

Using the gauge in which A does not depend on j, we get, from Eq. (9.16.5), F = 2pRAj. Finally,

A A

AR

r z= =

=

⎧⎨⎪

⎩⎪

0

2ϕ πΦ (9.16.6)

(b) Given the symmetry of the problem, it is more convenient to use cylindrical coordinates. To write the Schrödinger equation, we have to express the gradient ∇ in cylindrical coordinates as follows:

∇ = ∂∂ + ∂

∂ + ∂∂q iρ ρ ϕ

1k

z (9.16.7)

where q, i, and k are unit vectors in the r, j, and z directions, respectively. Since the electron is constrained to move on the ring, j, R, and z are all constant. Thus, the only nonvanishing part of ∇ in

Eq. (9.16.7) is i 1ρ ϕ

∂∂ . Applying Eqs. (9.16.6) and (9.16.7) on the ring we get

Hm

iec m

iR

ec R

= − ∇ −⎛⎝⎜

⎞⎠⎟ = − ∂

∂ −⎛⎝⎜

12

12

12

2

� �A ϕ πΦ ⎞⎞

⎠⎟= − ∂

∂ −⎛⎝⎜

⎞⎠⎟

2

2

21

2 2mRi

ec

� ϕ πΦ

(9.16.8)

and the Schrödinger equation is

1

2 22

2

mRi

ec

E− ∂∂ −⎛

⎝⎜⎞⎠⎟

=� ϕ π ϕ ϕΦ ψ ψ( ) ( ) (9.16.9)

(c) Since j is defined over 2p, the general boundary condition for any function of j determines that the

function will be periodic in 2p ; so, we have ψ ψ( ) ( )ϕ π ϕ+ =2 and similarly, for ∂∂ψϕ . We consider

only absolute values—in quantum mechanics it is only ⏐ψ ⏐2 that has real physical meaning.

(d) Check whether ψ( ) ( )ϕ ϕ= =1N

e kik constant are solutions of Eq. (9.16.9). First, we find the normaliza-tion constant N:

R d RN

ψ( )ϕ ϕ ππ

22

0

2

21

1= =∫ (9.16.10)


So, NR

= 12π

. Next, we use ψ φ( ) = 1N

eikϕ in Eq. (9.16.9) and obtain

1

2 222 2

2

mRk k

ec

ec

� �− ⎛⎝⎜

⎞⎠⎟ + ⎛

⎝⎜⎞⎠⎟

⎡

⎣⎢⎢

⎤

⎦⎥⎥

Φ Φπ π == E (9.16.11)

or, equivalently,

�ke

cmR E−⎛

⎝⎜⎞⎠⎟ =Φ

22

22

π (9.16.12)

We define Φ0 ≡ ce�

and write Eq. (9.16.12) as

kmR

E−⎛⎝⎜

⎞⎠⎟

=ΦΦ0

2 2

22

� (9.16.13)

From the boundary condition and the wavefunction ψ( )ϕ ϕ= 1N

eik , we have

2 2 0 1 2π πk n n= = ± ±, , , . . . (9.16.14)

From Eqs. (9.16.13) and (9.16.14) we get the eigenenergies:

EmR

nn = −( )�2

2 0

2

2Φ Φ/ (9.16.15)

and the eigenstates:

ψnik

Re( )ϕ

πϕ= 1

2 (9.16.16)

9.17. Refer to Problem 9.16, Eqs. (9.16.15) and (9.16.16). The magnetic field is zero on the ring (recall that the flux is inside the ring but not on the ring). (a) In classical mechanics, a particle (electron) constrained to move on the ring will not be affected by the magnetic flux. Is this also the case in quantum mechanics? Is the energy of the electron a function of the flux F ? (b) Plot a graph describing the ground state of the electron as a function of F (or F /F0). (c) The current on the ring can be defined by

I = cdHdΦ (9.17.1)

where H is the Hamiltonian and F the flux. Write the current operator I in the coordinates representation. (d ) Calculate the expectation value of I in state ψn. Find the relation between the energy and the current of the state ψn.

SOLUTION

(a) Using Eq. (9.16.15), we can easily see that the energy’s eigenvalues for the electron depend on F ; thus, in contrast to classical mechanics, in quantum mechanics a particle can be affected by a magnetic field even when the magnetic field is zero in the region in which the particle moves. This surprising phenomenon is known as the Aharonov–Bohm effect.

(b) The energy eigenvalues are

EmR

nn = −⎛⎝⎜

⎞⎠⎟

�2

20

2

2

ΦΦ (9.17.2)

The ground states depend on F (or F /F0). For −1/2 < F /F0 < 1/2, the minimum energy in Eq. (9.17.2)

corresponds with n = 0 (Fig. 9.2). For F /F0 > 1/2, the value n = 0 is no longer the minimum energy

(the ground state). For 12

320< <Φ Φ/ , the minimum energy in Eq. (9.17.2) corresponds to n = 1. For

32

520< <Φ Φ/ , ψn = 2 is the ground state, and so on. For

n n− < < +12

120Φ Φ/ the ground state is ψn.

So, the ground state is periodic in F /F0 with period 1, as shown in Fig. 9.2.


(c) Using Eqs. (9.17.1) and (9.16.8) we have

I = ∂∂ − ∂

∂ −⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢⎢

⎤

⎦⎥⎥

=cmR

iec

c

mRΦΦ1

2 22

2

� ϕ π 22

2 2

2 2

4

−⎛⎝⎜

⎞⎠⎟ − ∂

∂ −⎛⎝⎜

⎞⎠⎟

= ∂∂

ec

iec

eh

mRi

π ϕ π

π

�Φ

ϕϕ −⎛⎝⎜

⎞⎠⎟

ΦΦ0

(9.17.3)

(d) The expectation value of I is

⏐ ˆ ( )[ ˆ ( )]( )*I ⟩ = =∫ −

n n ninI R d e

eh

mRψ ψϕ ϕ ϕ

π

πϕ

0

2

24 220

2

0⎛⎝⎜

⎞⎠⎟

∂∂ −⎛

⎝⎜⎞⎠⎟

⎡

⎣⎢

⎤

⎦⎥

=

∫π

ϕϕ ϕi e R d

eh

inΦ Φ/

22 0 0πmRn

ehmR

n( ) ( )− = − −Φ Φ Φ Φ/ / (9.17.4)


Em

enn=

⟩⎛⎝⎜

⎞⎠⎟2

2⏐ ˆ

( )I (9.17.5)


9.18. Consider an electron in a region of a constant magnetic field of 1 gauss in the z direction. Assume that the electron is in a very large box, 0 ≤ x ≤ Lx, −Ly ≤ y ≤ Ly, and 0 ≤ z ≤ Lz. What is the number of state per unit area (in the xy plane)?

Ans. According to Eq. (9.10.11), nN= ≅

area m80

12 .

9.19. Solve Problem 9.8 using the symmetric gauge, A = −⎛⎝⎜

⎞⎠⎟

B B2 2

0y x, , . Show that the eigenvalues in Eq. (9.8.16) are the same (as they must be).

Ans. Hm

pq

cy p

qc

x px y z= +⎛⎝⎜

⎞⎠⎟ + −⎛

⎝⎜⎞⎠⎟ +

⎡

⎣⎢

12 2 2

2 22B B

⎢⎢

⎤

⎦⎥⎥

.

–5/2 –2

n = –2ground state

n = –1ground state

n = 0ground state

n = 1ground state

Eground state

–3/2 –1 –1/2 0 1/2 1 3/2 2 5/2 F/F0

n = 2ground state

14 2mR2

�2

Fig. 9.2


9.20. Using Eq. (9.9.2), solve Problem 9.3 for a charged particle with spin and a magnetic moment ls.

Ans. (a) H mi

qc s= − ∇ −⎛

⎝⎜⎞⎠⎟ − ⋅1

2

2

� A l B. (b) ρ ψ ψ∗( ) ( ( ).r r r0 0 0= )

(c) j A S= ∇ − ∇( ) − + ∇ ×�2mi

qmc

cssψ ψ ψ ψ ψ ψ ψ ψ∗ ∗ ∗ ∗μ

( ˆ ). (9.20.1)

9.21. Conductivity is defined by

σ =iVtot (9.21.1)

where itot is the total current per unit length and V is the electric potential. Consider Problem 9.12. For this case, E = E j, and φ = −E y, so V Ly= 2E . The total current in the x direction is ( )i Nix xtot = , where N is the number

of states in a complete Landau level, which is given in Eq. (9.10.12). Find s for this case.

Ans. s = e2/�.

9.22. Consider the following classical harmonic oscillator Hamiltonian:

H p p p m x yx z z02 2 2

02 2 21

212

= + + − +( ) ( )ω (9.22.1)

(a) Is it possible to find a basis of eigenstates that is common to H0 and Lz? (b) Assume that the oscillator has a

charge of q and is placed in a region of constant magnetic field B = B0k. Use the gauge A r= − ×12

B and find the corresponding Hamiltonian of the system.

Ans. (a) Yes, since [ ˆ , ˆ ]H Lz0 0= .

(b)

ˆ ( ˆ ˆ ˆ )Hm

p p pm

qc

mx z z= + + + ⎛⎝⎜

⎞⎠⎟

+12

12 2

12

2 2 2 02

2Bωω ω0

2 2 02

202 2

212

⎛

⎝⎜

⎞

⎠⎟ + ⎛

⎝⎜⎞⎠⎟

+⎛

⎝⎜

⎞

⎠⎟ˆ ˆx

qc

m yB⎡⎡

⎣⎢⎢

⎤

⎦⎥⎥

+ −qmc

p y p xx y

B02

( ˆ ˆ ˆ ˆ).

9.23. Refer to Problem 9.22. (a) Is it possible to find a basis of eigenstates that is common to Lz and the Hamiltonian of Problem 9.22, part (b)? (b) Are the eigenstates of part (b) also the eigenstates of past (a)?

Ans. (a) Yes, since [ ˆ , ˆ ]H Lz = 0. (b) No, since [H , H0] ≠ 0.

9.24. Consider a hydrogen atom placed in a constant magnetic field of 104 gauss. Calculate the wavelengths corresponding to the three transitions between the levels 3d and 2p.

Ans. E E E Eemc

E Eemc1 = Δ = Δ + = Δ − =32 2 32 3 32 12 2

65; ; ;� �B B λ 000 6500 0 22 3Å Å.; .,λ = ±

204

Solution Methods in Quantum Mechanics—Part A

10.1 Time-Independent Perturbation TheoryThe quantum mechanical study of a conservative physical system (whose Hamiltonian is not explicitly time-dependent) is based on the eigenvalue equation of the Hamiltonian operator. Some systems (e.g., the har-monic oscillator) are simple enough to be solved exactly. In general, the equation is not amenable to analytic solutions and an approximate solution is sought, usually using computer-based numerical methods.

In this section, we present the widely used time-independent perturbation theory. We use a method that is often encountered in physics: we study the primary factors that produce the main properties of the system, then we attempt to explain the secondary effects neglected in the first approximation.

Perturbation theory is appropriate when the Hamiltonian H of the system can be put in the form

ˆ ˆ ˆH H W= +0 λ (10.1)

where the eigenstates and eigenvalues of H0 are known and l is a parameter. The operator lW must be “much smaller” than H0; that is, the relation lW << H0, i.e., l << 1, must hold and the matrix elements of W are comparable in magnitude to those of H0. More precisely, the matrix elements of W are of the same magnitude as the difference between the eigenvalues of H0.

The Unperturbed StateWe assume that the unperturbed energies (that is, the eigenvalues of H0) form a discrete spectrum Ep

0, where

p is an integral index. We denote the corresponding eigenstates by | φpi ⟩, where the additional index i distin-

guishes between the different linearly independent eigenvectors corresponding to the same eigenvalue in the case of a degenerate eigenvalue. We have

H Epi

p pi

00| |φ φ⟩ = ⟩ (10.2)

where | φpi ⟩ form an orthonormal basis of the state space,

⟨ ⟩ =

⟩ ⟨ =

⎧

⎨⎪⎪

⎩⎪⎪

∑∑φ φ

φ φ

pi

qj

pq ij

pi

pi

ip

I

|

| |

δ δ

ˆ (10.3)

Possible Effects of the PerturbationWhen the parameter l is equal to zero, H(l) is equal to the unperturbed Hamiltonian H0. The eigenvalues E(l) of H(l) generally depend on l. Figure 10.1 represents possible forms of the variation of energy levels with respect to l.

CHAPTER 10


In the case of a nondegenerate energy level, the perturbation may either affect the energy level (E10

in

Fig. 10.1) or not affect it (as in case of E20). For a degenerate energy level, it is possible that the perturba-

tion “splits” it into distinct energy levels, as in the case of E30 in Fig. 10.1. We say then that the perturbation

removes the degeneracy of the corresponding eigenvalue of H0. The perturbation may also leave the degen-eracy of an energy level, as in the case of E4

0 in Fig. 10.1.

Approximate Solution for the Eigenvalue EquationWe are looking for the eigenstates |ψ( )λ ⟩ and eigenvalues E (l) of the Hamiltonian H(l):

ˆ ( ) ( ) ( ) ( )H Eλ λ λ λ| |ψ ψ⟩ = ⟩ (10.4)

We shall assume that E (l) and |ψ( )λ ⟩ can be expanded in a power series of l in the form

Eq

q( )λ ε λε λ ε= + + +0 1 � (10.5)

| | | |ψ( )λ λ λ⟩ = ⟩ + ⟩ + + ⟩0 1 �q

q (10.6)

When the parameter is equal to zero, we have the energy level and eigenstate of the unperturbed Hamiltonian. When l << 1, each element in the series expansions, Eqs. (10.5) and (10.6), is much smaller (in general) then the previous one; in practice, it usually suffices to consider only the first few elements. The element contain-ing l is called the first-order correction, the one containing l2 is called the second-order correction, etc.

10.2 Perturbation of a Nondegenerate LevelConsider a particular nondegenerate eigenvalue En

0 of the unperturbed Hamiltonian, with eigenvector | φn ⟩ (this eigenvector is unique to within a constant factor). We now give first- and second-order corrections for the energy level and corresponding eigenvector (the derivation is given in Problem 10.1).

E E WW

E En n n n

pi

n

n p

( ) ˆˆ

λ λ λ= + ⟨ ⟩ +⟨ ⟩

−0 2

2

0 0φ φφ φ

| || |

++∑∑≠

O

ip n

( )λ3 (10.7)

l1 l0

E01

E 02

E03

E04

E(l)

Fig. 10.1

CHAPTER 10 Solution Methods in Quantum Mechanics—Part A206

| || |

|ψ φφ φ

φn ( )ˆ

λ λ⟩ = ⟩ +⟨ ⟩

−⟩∑∑

≠

npi

p

n ppi

ip n

W

E E0 0 ++ −⟨ ⟩ ⟨ ⟩

−

⎡

⎣⎢⎢∑λ2

0 0 2

φ φ φ φpi

p pi

p

n pi

W W

E E

| | | |ˆ ˆ

( )pp n

pi

qj

qj

n

n p n

W W

E E E

≠∑

+⟨ ⟩ ⟨ ⟩

− −

φ φ φ φ| | | |ˆ ˆ

( )(0 0 0 EEO

qjq n

pi

03

)( )

⎤

⎦⎥⎥

⟩ +∑∑≠

| φ λ (10.8)

Note that the first-order correction for the energy level is simply the mean value of the perturbation term lW in the unperturbed state | φn ⟩.

10.3 Perturbation of a Degenerate StateAssume that the level En

0 is gn-fold degenerate. We present a method for calculating the first-order correction for the energies and the zero-order correction for the eigenstates. The derivation is given in Problem 10.8.

Arrange the numbers ⟨ ⟩′φ φni

niW| |ˆ in a gn × gn matrix (i is the row index and i′ the column index). This

matrix, which we denote W (n), is “cut” out of the matrix that represents W in the { }|φpi ⟩ basis. Note that W (n)

is not identical to W; it is an operator in the gn-dimensional space corresponding to the energy level En0.

The first-order corrections ε1j of the energy level En

0 are eigenvalues of the matrix W(n). The zero-order

eigenstates corresponding to En0 are the eigenvectors of W (n). Let ε1

11 2jnj f( , , )( )= … be the roots of the char-

acteristic equation of W(n) (that is, the eigenvalues of W(n) ). The degenerate energy level splits, to the first order, into fn

( )1 distinct sublevels:

E E j f gn k nj

n n,( )( ) , , ,λ λε= + = ≤0

111 2 … (10.9)

When f gn n( )1 = we say that, to the first order, the perturbation W completely removes the degeneracy of the level

En0 . When f gn n

( )1 < the degeneracy is only partially removed, or not at all removed if fn( ) .1 1=

Suppose that a specific sublevel E En j nj

, ( )λ λε= +01 is q-fold degenerate, in the sense that there are q

linearly independent eigenvectors of W (n) corresponding to it. We distinguish between two completely different situations:

1. Suppose that there is only one exact energy level E(l) that is equal to the first order to En, j. This energy is

q-fold degenerate. [In Fig. 10.1 for example, the energy E(l) that approaches E40 when l → 0 is twofold

degenerate.] In this case, the zero-order eigenvector |0⟩ of H (l) cannot be completely specified, since the only condition is that this vector belongs to the q-dimensional eigensubspace of H(l) corresponding to E(l). This situation often arises when the H0 and l W possess common symmetry properties, implying an essential degeneracy of H(l).

2. A second possibility arises when several different energies E(l) are equal, to the first order, to En, j. The difference between these energies appears in the calculation of the second order or higher. In this case, an eigenvector of H (l) corresponding to one of these energies approaches an eigenvector of En, j for l → 0; the inverse however, does not hold.

10.4 Time-Dependent Perturbation TheoryConsider a physical system with Hamiltonian H0. We assume the spectrum of H0 to be discrete and nonde-generate (the formulas can be generalized to other situations). We have

H En n n0 | |φ φ⟩ = ⟩ (10.10)

Suppose that H0 is time-independent, but, at t = 0, a time-dependent perturbation is applied to the system

ˆ ( ) ˆ ˆ ( )H t H W t= +0 λ (10.11)


where l is a parameter, l << 1, and W (t) is an operator of the same magnitude as H0, and is zero for t < 0. Suppose that the system is initially in the state |φi ⟩, which is an eigenstate of H0 with eigenvalue Ei . We present an expression for calculating the first-order approximation of the probability Pi f (t) of finding the sys-tem in another eigenstate | φf ⟩ of H0 at time t. The derivation of this expression is given in Problem 10.12.

P t e t dtif

i t

f i

tf i( ) ( )= ′∫λ2

20

2

�

ω ω ′ ′ (10.12)

where wi f is the Bohr angular frequency, defined by

ωifi fE E

=−�

(10.13)

and wi f (t) is the matrix element of W(t):

W t W tfi f i( ) ˆ ( )= ⟨ ⟩φ φ| | (10.14)

Consider now the case of transition between a state |φi ⟩ and a state |φf ⟩ of energy Ef belonging to a continu-ous part of the spectrum of H0. In this case, the probability of transition at time t, | | |⟨ ⟩φ ψf t( ) 2, is actually a probability density. That is, we must integrate the probability density over a range of final states in order to give a physical prediction.

The time-dependent perturbation theory can be applied to this situation. One very important result is Fermi’s golden rule. This formula relates to the case of a constant perturbation. It can be demonstrated that, in this case, transitions can occur only between states of equal energies. The probability density Pf i of transi-tion from |φi ⟩ to |φf ⟩ increases linearly with time, and

WdP t

dtW t Efi

fif i f= = ⟨ ⟩

( )ˆ ( ) ( )

2 2π ρ�

| | | |ψ ψ (10.15)

where r (Ef ) is the density of the final states.

SOLVED PROBLEMS

10.1. Derive the formulas for the first- and second-order energy corrections for a time-independent perturbation. Also, derive the first-order corrections to the eigenstates. Assume that there is no degeneracy.

SOLUTION

We write the Hamiltonian in the form H = H0 + lW , where H0 is the Hamiltonian of the unperturbed system and W is the perturbation (l << 1). We assume that the eigenstates |ψ( )λ ⟩ and the eigenenergies E (l) of the perturbed system can be expanded in a power series of l:

| | | |ψ +( )λ λ λ⟩ = ⟩ + ⟩ + ⟩0 1 �q

q (10.1.1)

and

Eq

q( )λ ε λε λ ε= + + +0 1 � (10.1.2)

Substituting into the Schrödinger equation, we obtain

( ˆ ˆ )H W qq

q

q

q

q

0

0 0

+ ⟩⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

=⎡

=

∞ ∞

∑ ∑λ λ λ ε|′

′

′ =⎣⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

⟩⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥=

∞

∑λq

q

q|0

(10.1.3)


Then, by equating the coefficients of successive powers of l, we obtain

H0 00 0| |⟩ = ⟩ε (10.1.4)

( ˆ ) ( ˆ )H W0 0 11 0 0− ⟩ + − ⟩ =ε ε| | (10.1.5)

and

( ˆ ) ( ˆ )H W0 0 1 22 1 0 0− ⟩ + − ⟩ − ⟩ =ε ε ε| | | (10.1.6)

For the nth order we obtain

( ˆ ) ( ˆ )H n W n n n0 0 1 21 2 0 0− ⟩ + − − ⟩ − − ⟩ + − ⟩ =ε ε ε ε| | | |� (10.1.7)

Note that we are free to choose the norm and the phase of |ψ( )λ ⟩, so we require that |ψ( )λ ⟩ is normalized and that its phase is such that the inner product ⟨ ⟩0 | ψ( )λ is a real number. This implies that

⟨ ⟩ = ⟨ ⟩ = ⟨ ⟩ = ⟨ ⟩ = ⟨ ⟩ = − ⟨ ⟩0 0 1 0 1 1 0 0 0 2 2 0 1 112| | | | | | (10.1.8)

For the nth order we obtain

⟨ ⟩ = ⟨ ⟩ = − ⟨ − ⟩ + ⟨ − ⟩ + + ⟨ − ⟩ + ⟨0 0 1 1 2 2 2 212| | | | |n n n n n( � 11 1| n − ⟩) (10.1.9)

Note that when l → 0, we have ε00= En

( ). Using Eq. (10.1.4), we conclude that |φn ⟩ is proportional to | 0⟩; therefore, we choose | |φn ⟩ = ⟩0 . Multiply Eq. (10.1.5) on the left by ⟨φn |:

⟨ − ⟩ + ⟨ − ⟩ =φ φn nH W| | | |( ˆ ) ( ˆ )0 0 11 0 0ε ε (10.1.10)

The first term in Eq. (10.1.10) is zero; therefore,

ε1 0= ⟨ ⟩ = ⟨ ⟩φ φ φn n nW W| | | |ˆ ˆ (10.1.11)

For the first order we have

E E W On n n n( ) ˆ ( )( )λ λ λ= + ⟨ ⟩ +0 2φ φ| | (10.1.12)

We see that the first-order correction to the energy is simply equal to the mean value of the perturbation term W in the unperturbed state |φn ⟩. Multiplying Eq. (10.1.5) by the basis vectors ⟨φp | we obtain

⟨ − ⟩ + ⟨ − ⟩ = ≠φ φ φp n p nH E W p n| | | |( ˆ ) ( ˆ ) ( )( )0

011 0ε (10.1.13)

This leads to the equation

( ) ˆ( ) ( )E E Wp n p p n0 0 1 0− ⟨ ⟩ + ⟨ ⟩ =φ φ φ| | | (10.1.14)

where we used the orthogonality of the basis vectors. Then,

⟨ ⟩ =−

⟨ ⟩ ≠φ φ φpp n

p nE EW p n| | |1

10 0( ) ( )

ˆ ( ) (10.1.15)

Since ⟨ ⟩ = ⟨ ⟩ =φn | |1 0 1 0, we arrive at

|| |

|1 0 0⟩ =⟨ ⟩

−⟩

≠∑ φ φ

φp n

p np n

W

E E

ˆ

( ) ( ) p (10.1.16)


Therefore, to the first order, the eigenvectors |φn( )λ ⟩ of H that correspond to the unperturbed state |φn ⟩ can be written as

| || |

|ψ φφ φ

φn np n

n pp

p n

W

E E( )

ˆ

( ) ( )λ λ⟩ = ⟩ +⟨ ⟩

−⟩

≠∑ 0 0 ++ O( )λ 2 (10.1.17)

To obtain the second-order correction of the energy we multiply Eq. (10.1.6) by ⟨φn |:

⟨ − ⟩ + ⟨ − ⟩ − ⟨ ⟩φ φ φ φn n n n nH E W| | | | |( ˆ ) ( ˆ )( )0

01 22 1ε ε == 0 (10.1.18)

This leads to ε2 1= ⟨ ⟩φn W| |ˆ . Substituting Eq. (10.1.16) for |ψ( )λ ⟩, we arrive at

ε2 0 0=⟨ ⟩

−≠

∑ φ φp n

n pp n

W

E E

| |ˆ

( ) ( ) (10.1.19)

Therefore, to the second order, the energy is given by

E E WW

En n n np n

n

( ) ˆˆ

( )(λ λ λ= + ⟨ ⟩ +

⟨ ⟩0 22

0φ φφ φ

| || | | |

)) ( ) ( )−

+≠

∑ EO

pp n

03λ (10.1.20)

10.2. Consider a particle in the two-dimensional, symmetrical, infinite potential well. The particle is subject to the perturbation W = Cxy, where C is a constant. (a) What are the eigenenergies and eigenfunctions of the unperturbed system? (b) Compute the first-order energy correction. (c) Find the wavefunction of the first excited level.

SOLUTION

(a) For the unperturbed system, the wavefunctions and eigenenergies are (see Chap. 3)

ψn n x yL

n xL

n yL1 2

0 1 22,

( ) ( , ) sin sin= ⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞π π⎠⎠⎟

(10.2.1)

EmL

n nn n1 2

02 2

2 12

22

2,( ) ( )= +π �

(10.2.2)

(b) The first-order correction to the energy is given by

ΔE Wn n n n n n1 2 1 2 1 2

1 0 0,

( ),

( ),

( )ˆ= ⟨ ⟩ψ ψ| | (10.2.3)

Thus,

Δ = ⎛⎝⎜

⎞⎠⎟

⎛⎝

EC

Lx

nL

x dx ynLn n1 2

12

12

14,

( ) sin sinπ π

⎜⎜⎞⎠⎟

=∫∫2 2

004

dyL C

LL

(10.2.4)

(c) In order to find the wavefunction of the first excited level, we compute the following matrix elements:

⟨ ⟩ = ⟨ ⟩ =ψ ψ ψ ψ1 20

1 20

2 10

2 10 1

,( )

,( )

,( )

,( )ˆ ˆ| | | |W W

442 2L C (10.2.5)

and

⟨ ⟩ = ⟨ ⟩ =ψ ψ ψ ψ1 20

2 10

2 10

1 20 2

,( )

,( )

,( )

,( )ˆ ˆ| | | |W W

556

81 42 2

πL C (10.2.6)

Thus, the eigenvalue equation can be written as

L Cu

u2 2

4

4

1

2

14

256

81256

81

14

π

π

⎛

⎝

⎜⎜⎜

⎞

⎠

⎟⎟⎟

⎛⎝⎜

⎞⎠⎟

= λλu

u1

2

⎛⎝⎜

⎞⎠⎟

(10.2.7)


where λπ1 2 4

2 214

256

81, = ±⎛⎝⎜

⎞⎠⎟

L C and u1

12

= , u2

12

= or u1

12

= , u2

12

= − . Note that as the first

excited level is twofold degenerate, there are two solutions for the wavefunctions:

ψ ψ ψ ψ ψ ψ( ), ,

( ),

( ),

( )( ) (11 20

2 10 1

1 20

2 101

212

= + = − )) (10.2.8)

10.3. Consider a harmonic oscillator with a force constant k and a reduced mass m. The small perturbation ˆ ˆW ax= 3 is applied to the oscillator. Compute the first-order correction to the wavefunctions and the

first nonvanishing correction to the eigenenergies.

SOLUTION

The Hamiltonian of the system is given by

ˆ ˆ ˆ ˆ ˆH H Wm

ddx

kx axo= + = − + +�2 2

2 3

212

(10.3.1)

The eigenenergies for the unperturbed Hamiltonian are E nn( ) ( )0 1 2= + / �ω , and the eigenfunctions are

given by

φn nx

nxn

e H x( ) /( )!

( )0 21

2

2

= −απ αα (10.3.2)

where a ≡ mw /� and Hn are the Hermite polynomials. Note that when we compute the first-order correction En

( )1 , we obtain an integral with an integrand of an odd function; the integral therefore vanishes and we have the result E n ax kn

( ) ˆ1 3 0= ⟨ ⟩ =| | . For the second-order correction we obtain

En W k

E E

n ax nn

n kk n

( )( ) ( )

ˆ ˆ22

0 0

3

= ⟨ ⟩−

= ⟨

≠∑ | | | | | | | ++ ⟩

−+ ⟨ + ⟩

−+

3 12

03

0

3 2

0| | | | |

E E

n ax n

E En n n n( ) ( ) ( )

ˆ

++ −

+ ⟨ − ⟩−

+ ⟨

10

3 2

01

01

( ) ( ) ( )

ˆ ˆ| | | | | |n ax n

E E

n ax

n n

33 2

03

03| |n

E En n

− ⟩− −

( ) ( )

(10.3.3)

Note that this result can be obtained by using the relation | | | | | | | |⟨ ⟩ = ⟨ ⟩ ⟨ ⟩∑n ax m a n x k k x m

k

ˆ ˆ ˆ3 2 2 . The required matrix elements are

⟨ + ⟩ = ⟨ + ⟩ ⟨ + + ⟩ = +n ax n a n x n n x n a

n| | | | | |ˆ ˆ ˆ (3 23 2 2 3

1))( )( )

( )

n n+ +2 3

2 3α (10.3.4)

⟨ + ⟩ = ⟨ + ⟩ ⟨ + + ⟩ + ⟨n ax n a n x n n x n n| | | | | | |ˆ [ ˆ ˆ ˆ3 21 2 2 1 xx n n x n an2

3

31 31

8| | |⟩ ⟨ + ⟩ = +ˆ ]

( )

α (10.3.5)

⟨ − ⟩ = ⟨ − ⟩ ⟨ − − ⟩ + ⟨n ax n a n x n n x n n x| | | | | | |ˆ ˆ ˆ ˆ3 21 2 2 1 223

31 38

| | |n n x n an⟩ ⟨ − ⟩ =ˆα

(10.3.6)

and

⟨ − ⟩ = ⟨ − ⟩⟨ − − ⟩ = −n ax n a n x n n x n a

n n| | | | | |ˆ ˆ ˆ (3 23 2 2 3

11 2

8 3)( )n −α

(10.3.7)

Substituting into Eq. (10.3.3) yields

Ea

n nn( )2

2

3215

4

1130

= − + +⎛⎝⎜

⎞⎠⎟�ωα

(10.3.8)


The same matrix elements are required to calculate the first-order correction to the wavefunctions; hence, we obtain

φ φ φ φn nn k

k n

k W n

E EO a= + ⟨ ⟩

−+ =( )

( )(

ˆ( )0

2

0 00 2 0| | | | )) ( )( )( )

( )+ − − +⎡

⎣⎢ −

≠∑ a n n n

O an

k n2

13

1 22 3

0 2

�ωα α φ

++ − + + − +− +3

23 1

12

13

11

01

0nn

nn n n

n nα αφ φ( ) ( )( )( ) ( ++ + ⎤

⎦⎥ ++

2 32 3

0 2)( )( )( )n

O anα φ (10.3.9)

10.4. Consider a particle of mass m in a one-dimensional infinite potential well of width a:

V xx a

( ) =≤ ≤

∞{0 0otherwise

(10.4.1)

The particle is subject to perturbation of the form

W x a x a( ) ( )= −ω δ0 2/ (10.4.2)

where a is a real constant with dimension of energy. (a) Calculate the changes in the energy level of the particle in the first order of w0. (b) This problem can be solved without using perturbation theory;

find the exact solution. Defining k mE= 2 2/� , show that the possible levels of energy are given by one of the following equations:

sin tanka ka k

ma20

2

2

0

⎛⎝⎜

⎞⎠⎟ = ⎛

⎝⎜⎞⎠⎟ = −or

�ω

(10.4.3)

How do these results depend on the absolute value and sign of w0? Show that for w0 → 0 one obtains the results of part (a).

SOLUTION

(a) For the unperturbed system the energy eigenvalues and eigenfunctions are given by

ψn nxa

nxa

En

ma0 0

2 2 2

22

2( ) sin ( )= ⎛

⎝⎜⎞⎠⎟ =π π �

(10.4.4)

The first-order corrections of the energy eigenvalues are given by

Δ = ⟨ ⟩ = ⎛⎝⎜

⎞⎠⎟∫E W

anxan n n

a( ) ( ) ( )ˆ sin1 0 0 2

0

2ψ ψ π| | aa x

adx

n

nω δ

ω0

02

2

0−⎛

⎝⎜⎞⎠⎟ = ⎧

⎨⎩

odd

even (10.4.5)

(b) Turning now to the exact solution, we divide the well potential into two regions: I and II, as shown in Fig. 10.2. The wavefunction for region I is ψI(x) = A sin (kx), and for region II, ψII(x) = B sin [k (a − x)].

I

0 a/2 a

II

x

V(x)

Fig. 10.2


From the boundary condition ψI(x = a/2) = ψII(x = a/2) we have A = B. Using the normalization condition

| |ψ( )x dxa

2

0

1=∫ , we obtain A B n a= = 2 2π / . Hence, from the discontinuity relation between the deriva-

tives ′ =ψ I /( )x a 2 and ′ =ψ II /( )x a 2 , we obtain

′ = = ′ = −→ −

ψ ψI II/ /( ) ( ) lim ( )/

x a x am

W xa

a

2 22

2 02� ε ε

//

sin ( )

( ) sin (

2

2 0

2

22

2

+

∫= ′ = −

ε

ω

akx dx

x am

a kψ II /�

aa/2) (10.4.6)

Therefore, k ka k kama

kacos ( ) cos ( ) sin ( )/ / /2 22

202= − −ω

�, so

2 22

2 202

2

k kama

ka kacos ( ) sin ( ) tan ( )/ / /= − ⇒ = −ω

�

� kkmaω0

(10.4.7)

For sin (ka/2) = 0, we obtain the unperturbed solution corresponding to k = p n/2, where n is an even number. As w0 → 0 we get −�

2/maw0 → ± ∞, which, from Eq. (10.4.7), occurs when ka/2 = p /2 + n p, or k = p n/a for odd n. We introduce z ≡ −ka/2 + p n /2, where n is an odd number. In this case, tan (ka /2) = cot z. Using the expansion of cot x in the vicinity of zero we can write

cot tanzz ka

ka kma

≈ = − = ( ) = −1 12 2

2

0π ωn/2 /�

(10.4.8)

Note that the last equality comes from Eq. (10.4.7). Therefore,

kn

ak

m2 02

20− ⎛

⎝⎜⎞⎠⎟ − =π ω

� (10.4.9)

and kn

an

am

1 2

20

212

8, = ± ⎛

⎝⎜⎞⎠⎟ +

⎛

⎝⎜⎜

⎞

⎠⎟⎟

π π ω�

. Using the expression

kn

an

am

= + ⎛⎝⎜

⎞⎠⎟ +

⎛

⎝⎜⎜

⎞

⎠⎟⎟

12

820

2π π ω

� (10.4.10)

and the expansion 1 12

1+ = + + <<ε ε ε�( ) we obtain

kn

an

aan

m na

= + + ⎛⎝⎜

⎞⎠⎟

⎛

⎝⎜⎜

⎞

⎠⎟⎟

≈ +12

18 22

02

π ππ

ω π�

mm

n

ωπ

02

� (10.4.11)

The energy eigenvalues are therefore

Ekm

n

man = ≈ +� �2 2 2 2 2

2 02 22

π ω (10.4.12)

The first term on the right-hand side of Eq. (10.4.12) corresponds to the unperturbed energy eigenvalues, and the second term is the first-order correction that we obtained in part (a).

10.5. Consider a particle with mass m in a two-dimensional square box of length L. There is a weak potential in the box given by

V x y V L x x y y( , ) ( ) ( )= − −02

0 0δ δ (10.5.1)


(a) Evaluate the first-order correction to the energy of the ground state. (b) Write the expressions for the second-order correction to the energy, and the first-order correction to the wavefunction of the ground state. Explain how you would calculate the expressions for (x0, y0) = (L /2, L /2). (c) Find an expression for the energy of the first excited state to the first order in V0. What is the difference between the energy sublevels for (x0, y0) = (L /4, L /4)? (d) For the first excited state, find the points (x0, y0) defining a potential V(x, y) that do not remove the degeneracy. Explain your result in terms of the symmetry of the problem.

SOLUTION

(a) The eigenfunctions and eigenenergies of the unperturbed state are (see Chap. 3)

ψ ( ) (( , ) sin sin0 02x y

Ln

Lx

kL

y Enk= ⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

π π )) ( )= +π 2 2

22 2

2

�

mLn k (10.5.2)

The ground state is nondegenerate, but since E E120

210( ) ( )= , the first excited state is degenerate. For the

ground state, the first-order correction of the energy is

E VL

xL11

1110

110

224( ) ( ) ( )ˆ sin s= ⟨ ⟩ = ⎛

⎝⎜⎞⎠⎟ψ ψ| |

πiin ( ) ( )2

02

0 000

4

π δ δyL

V L x x y y dx dyLL

⎛⎝⎜

⎞⎠⎟ − −

=

∫∫VV

xL

yL0

2 0 2 0sin sinπ π⎛

⎝⎜⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

(10.5.3)

(b) For the second-order correction of the energy of the ground state we have

EV

E Enk

n kn

112

0110 2

110 0

( )( ) ( )

( ),

( )

ˆ=

⟨ ⟩−

| | | |ψ ψ

,,( , ) ( , )

sin sin

kn k

L

nxL

kyL

≠

∑

=

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞

1 1

24 π π

⎠⎠⎟ − − ⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞V L x x y yx

Ly

L02

0 0δ δ π π( ) ( ) sin sin ⎠⎠⎟

− −

=

∫∫∑≠

dx dy

mLn kn k

n k

2

2 2

22 2

1 1 22

π �( ),

( , ) ( , )

44 00 0 0V

nxL

kyL

xL

sin sin sinπ π π⎛

⎝⎜⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎠⎟

⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢

⎤

⎦⎥

− −

sin

( ),

π

π

yL

mLn kn k

0

2 2

22 2

2

22

�

(( , ) ( , )n k ≠

∑1 1

(10.5.4)

When (x0, y0) = (L /2, L /2) we obtain

EV

n k

112

02

22

2 2

162 2

2

( )sin sin

=

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

π π

π �

mmLn k

V mL

n kn k

22 2

1 1

02 2

2 2

2

32 2 1

( ),( , ) ( , )

− −

=

≠

∑

π � (( ),

( , ) ( , )

2 2 2

1 1

− −

⎛

⎝

⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟

≠

∑ n kn k

n kodd

(10.5.5)


For the first-order correction of the ground state we have

|| |

ψψ ψ

111

0110

110 0

( )( ) ( )

( ) ( )

,

ˆ⟩ =

⟨ ⟩−

nk

nkn k

V

E E

(( , ) ( , )

( )

( )

n k

nk

mLn k

L

≠

∑ ⟩

=− −

1 1

0

2 2

22 2

21

22

4

| ψ

π �ssin sin ( ) ( )

π π δ δxL

yL

V L x x y y⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟ − −0

20 0∫∫∫∑ ⎡

⎣⎢

× ⎛⎝⎜

⎞⎠⎟

⎛

≠n k

n k

n xL

kyL

,( , ) ( , )

sin sin

1 1

π π⎝⎝⎜

⎞⎠⎟

⎤⎦⎥

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

=

dx dyL

nxL

kyL

2

4

sin sinπ π

VVxL

yL

nxL0

0 0 0sin sin sinπ π π⎛

⎝⎜⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

ssin

( ),

( , )

π

π

kyL

n k

mL

Ln k

n k

0

2 2 2 2

22

2

2⎛⎝⎜

⎞⎠⎟

− −≠

�

(( , )

sin sin

1 1

∑ ⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

π πnxL

kyL (10.5.6)

We turn to the case where x0 = y0 = L /2. We substitute n = 2p + 1 and k = 2q + 1, and obtain

ψ111 0

2 2 2 2

41

2 1( )

( )( ) sin

( )=+ + +

− ++mLV

p q p q

pp q

π �

ππ πxL

q yL

p q

⎡⎣⎢

⎤⎦⎥

+⎡⎣⎢

⎤⎦⎥

=

∞

∑,

sin( )

0

2 1 (10.5.7)

(c) The first excited state is degenerate, E E120

210( ) ( )= ; according to Sec. 10.3, the secular equation will be

V E V

V V E

12 12 121

12 12

21 12 21 21 121 0

,( )

,

, ,( )

−

−= (10.5.8)

Thus, we obtain

E V V V V V121

12 12 21 21 12 12 21 2121

2 4( ), , , ,( )= + ± − + | 112 21

2, |⎡

⎣⎢⎤⎦⎥ (10.5.9)

where V n V x y mnm = ⟨ ⟩| |( , ) and

V VnxL

kyL

lnk lm, sin sin sin= ⎛

⎝⎜⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

4 00 0π π π xx

LmyL

0⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟sin

π (10.5.10)

For x0 = y0 = L /4 we obtain V12, 12 = V12, 21 = V21, 12 = V21, 21 = 2V0, whereupon

E V V V V VV12

10 0 0 0

20

212 2 2 2 2 4 2

0

4( )

( ) ( )= + − +⎡⎣

⎤⎦ =±

00

⎧⎨⎪

⎩⎪ (10.5.11)

(d ) The degeneracy will not be removed if

( ), , ,V V V12 12 21 212

12 2124 0− + =| | (10.5.12)

Thus,

sin sin sin2 0 2 0 2 02 2π π πxL

yL

xL

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

− ⎛⎝⎜

⎞⎠⎠⎟

⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢

⎤

⎦⎥

+ ⎛⎝⎜

⎞⎠⎟

sin

sin sin

2 02

02

4

π

π

yL

xL

22 0 2 0 2 02 2π π πxL

yL

yL

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

=sin sin 00


so

sin sin sin2 0 2 0 2 02 2π π πxL

yL

xL

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

+ ⎛⎝⎜

⎞⎠⎠⎟

⎛⎝⎜

⎞⎠⎟

= ⇒ ⎛⎝⎜

⎞⎠⎟

=sin sin sin2 0 0 002 2π π πy

Lx

Ly

L⎛⎛⎝⎜

⎞⎠⎟

= 0

(10.5.13)

Hence, each of the variables x0 and y0 can assume the values 0, L, and L /2; altogether we attain nine points in the two-dimensional box. These nine points are the only points where the symmetry of the system is not removed when the perturbation is applied.

10.6. Consider a particle of mass m subjected to the Hamiltonian

H

pm

m rr a

pm

r a

=+ ≤ ≤

>

⎧

⎨⎪⎪

⎩⎪⎪

2 2 2

22 2

0

2

ω (10.6.1)

where r x y= +2 2 . Use the second-order perturbation to find the corrections to the ground state energy.

SOLUTION

One can write the Hamiltonian in the following form:

Hpm

m r V r H V r= + + = +2

2 202

12

ω ( ) ( ) (10.6.2)

where the perturbation is

Vr a

m r r a=

≤ ≤− >

⎧⎨⎩

0 0

22 2ω / (10.6.3)

The wavefunction of H0 for the ground state is φ00 2

2

22

2( ) expr r

r= −⎛

⎝⎜⎞

⎠⎟λ λ, where λ ω= �/m

and E00( ) = �ω . (In the function φ00 one of the zeros corresponds to an eigenfunction of the unperturbed

Hamiltonian, and the other zero corresponds to the ground state.) For the first order in V we have

E m rr

r dra

0 22 2

2

22 1

2= − ⎛

⎝⎜⎞⎠⎟ −

⎛

⎝⎜⎞

⎠⎟=

∞

∫�ωλ

ωλ

exp ��ω ω

λ λ− −

⎛

⎝⎜⎞

⎠⎟+

⎛

⎝⎜⎞

⎠⎟⎡

⎣

⎢⎢

⎤

⎦

⎥⎥

∞

21

2

2

2

2expr r

a

(10.6.4)

or E m a m a0

2 2

2 1= − +⎡⎣⎢

⎤⎦⎥ −

⎛⎝⎜

⎞⎠⎟�

�

� �ω ω ω ωexp . This result is valid for �w << mw 2a2. In the second order,

the first state that contributes to the energy correction is φ10(r) at energy 3�w, yielding the contribution

− −⎛

⎝⎜⎞

⎠⎟⎛

⎝⎜

⎞

⎠⎟

∞

∫12 210

2 2

00

2

�ωωφ φ( ) ( )r

m rr r dr

a

== − − −⎛

⎝⎜⎞

⎠⎟+ +

⎛

⎝⎜⎞

⎠⎟1

2 21

2

2

2

2

4

4��

ωω

λ λ λexp

r r r

a

∞∞⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

= − −⎛

⎝⎜⎞

⎠⎟

2

2

8�

�ω ω

expm a

112 2 2 4

2

2

+ +⎛

⎝⎜⎞

⎠⎟⎡

⎣⎢⎢

⎤

⎦⎥⎥

m a m aω ω� �

(10.6.5)


Hence, we obtain

Em a m a

0

2 2

21

81≅ − +

⎛

⎝⎜⎞

⎠⎟−

⎛

⎝⎜⎞

⎠⎟− +�

��

�ω ω ω ω ωexp

mm a m a m aω ω ω2 2 2 4

2

2 22� � �

+⎛

⎝⎜⎞

⎠⎟−

⎛

⎝⎜⎞

⎠⎟exp (10.6.6)

Note that this analysis is incorrect only if E > mw2a2 / 2.

10.7. Consider now a three-dimensional problem. In a given orthonormal basis, the Hamiltonian is represented by the matrix

�HC

CC

=−

⎛

⎝⎜⎜

⎞

⎠⎟⎟

+⎛

⎝⎜⎜

⎞

⎠⎟⎟

1 0 00 3 00 0 2

0 00 0

0 0

(10.7.1)

Here � � �ˆ ˆ ˆH H H= +0 1 and C is a constant, C << 1. (a) Find the exact eigenvalues of �H . (b) Use the

second-order perturbation to determine the eigenvalues. (c) Compare the results of parts (a) and (b).

SOLUTION

(a) The eigenvalues of �H are the roots of the equation det(ˆ ˆ)�H I− =λ 0,

01 0

3 00 0 2

21

32=

−−

− −= − − −

− = − −λ

λλ

λ λλ λ

CC

CC

CC

C( ) ( )[[ ]λ λ2 24 3− + − C (10.7.2)

Thus, l = C − 2, 2 1 2± + C .

(b) The second-order correction to the energy may be written as E E E En n n n= + +( ) ( ) ( ) ,0 1 2 or

( ) (ˆ

) (ˆ

) ( ) ( )� � �E H H

H H

E Ei ii iiik ki

i k

= + +−

0 11 1

0 0

kk i≠∑ (10.7.3)

It can be seen that (H0)ii = 1, 3, and −2. The first-order energy correction is given by H111 0= , H22

1 0= , and H C33

1 = . For the second correction we have

EH H

E E

H H

E E12 12

1211

10

20

131

311

10

3

( )( ) ( ) ( ) (=

−+

− 00

2 2

203 2) = − + = −C C

(10.7.4)

EH H

E E

H H

E E22 21

1121

20

10

231

321

20

3

( )( ) ( ) ( ) (=

−+

− 00

2 2

3 10 0

3 2) = − + ⋅ =C C (10.7.5)

and

EH H

E E

H H

E E32 31

1131

30

10

321

231

30

2

( )( ) ( ) ( ) (=

−+

− 00) (10.7.6)

Thus,

EC

1

2

12

= − (10.7.7)

EC

2

2

32

= + (10.7.8)

and

E C3 2= − + (10.7.9)


(c) We expand 2 1 2± + C in a binomial series:

2 1 2 112

312

112

12 2 2 2 2± + = ± + +⎛⎝⎜

⎞⎠⎟ = + − <<C C C C C� , ( )) (10.7.10)

This gives the same result as the second-order corrections [Eqs. (10.7.7) and (10.7.8)].

10.8. Derive the first-order correction of a degenerate state according to perturbation theory (Sec. 10.3).

SOLUTION

We assume that the energy level Ep is g-fold degenerate, so we have g orthonormal vectors |φpk ⟩ such that

H Epk

p pk

0 | |φ φ⟩ = ⟩ (10.8.1)

We add a perturbation lW to the Hamiltonian H0, and we seek the possible energy levels e corresponding to the first-order correction state |0⟩ :

[ ˆ ˆ ] ( )H W Ep0 0 0+ ⟩ = + ⟩λ λε| | (10.8.2)

We have

⟨ ⟩ = ⟨ ⟩φ φpk

pkW| | |ˆ 0 0ε (10.8.3)

Using the closure relation for the basis { }|φpk ⟩ we obtain

⟨ ⟩ ⟨ ⟩ = ⟨ ⟩′′

′′

′′∑∑ φ φ φ φp

kpk

pk

k

pk

p

W| | | |ˆ 0 0ε (10.8.4)

Since |0⟩ is an eigenvector of H0 with the eigenvalue Ep, it is orthogonal to every |φ ′′ ⟩p

k for p′ ≠ p, so

⟨ ⟩ ⟨ ⟩ = ⟨ ⟩′ ′

′ ≠∑ φ φ φ φp

kpk

pk

k

g

pkW| | | |ˆ 0 0

1

ε (10.8.5)

We define the matrix {W p} by

W Wijp

pi

pj= ⟨ ⟩φ φ| |ˆ (10.8.6)

Equation (10.8.5) is equivalent to the vector equation W p | |0 0⟩ = ⟩ε . Therefore, the possible values of e are the solutions of

det ( ˆ ˆ)W Ip − =λ 0 (10.8.7)

10.9. Consider an electron of mass m in a three dimensional box with energy 3p 2�2/ ma2. A weak electric field in the z direction and of strength E is applied to the system; the perturbation is then W = eEz. Compute the first-order correction to the electron’s energy.

SOLUTION

A free electron in the three-dimensional box has energy p2�

2n2/2ma2; and so n n nx y z2 2 2 6+ + = . Three vec-

tors satisfy this condition:

( , , ) ( , , ) ( , , ) ( , , ) ( , ,n n n n n n n nx y z x y z x y= =1 1 2 1 2 1 nnz ) ( , , )= 2 1 1 (10.9.1)

The state is therefore threefold degenerate. The wavefunctions for these three possibilities are

φ112 38 2= ⎛

⎝⎜⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞a

xa

ya

za

sin sin sinπ π π

⎠⎠⎟ (10.9.2)


φ121 38 2= ⎛

⎝⎜⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞a

xa

ya

za

sin sin sinπ π π

⎠⎠⎟ (10.9.3)

and

φ 211 38 2= ( ) ⎛

⎝⎜⎞⎠⎟

⎛⎝⎜

⎞⎠⎟a

xa

ya

za

sin sin sinπ π π (10.9.4)

Note that ⟨ ⟩ = ⟨ ⟩ = ⟨2 1 1 2 1 1 1 2 1 1 2 1 1 1 2, , ˆ , , , , ˆ , , , , ˆ| | | | |z z z ||1 1 2, , ⟩ and

⟨ ⟩ = ⎛⎝⎜

⎞⎠⎟∫2 1 1 2 1 1

8 23

2

0

2, , ˆ , , sin sin| |za

xa

dxa π

00

2

0

2

0

2

a ay

ady z

za

dz

az

∫ ∫⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

=

π πsin

sinaa

za

dza∫ ⎛

⎝⎜⎞⎠⎟ =π

2 (10.9.5)

It can be similarly shown that ⟨ ⟩ = ⟨ ⟩ = ⟨2 1 1 1 2 1 2 1 1 1 1 2 1 2 1, , ˆ , , , , ˆ , , , , ˆ| | | | |z z z ||1 1 2 0, , ⟩ = . Thus, all the off-diagonal matrix elements vanish and the energy is given by

Ema

e a= +32

2 2

2π � E

(10.9.6)

10.10. Consider a hydrogen atom placed in a uniform static electric field E that points along the k direction. The term that corresponds to this interaction in the Hamiltonian is

W e z= − E (10.10.1)

Note that for the electric fields typically produced in a laboratory, the condition W << H0 is satisfied. The appearance of the perturbation removes the degeneracy from some of the hydrogen states. This phenomenon is called the Stark effect. Calculate the Stark effect for n = 2 in a hydrogen atom.

SOLUTION

Before we explicitly calculate the matrix elements of the perturbation, we note that the perturbation has nonzero matrix elements only between states of opposite parity; as we are considering level n = 2, the rel-evant states are those with l = 0 and l = 1. Using symmetry, the m-values of the two states must be equal. Therefore,

2 2 0 2 1 2 1

0 2 2 0 0 0

s p m p m p m

W

s W p m

s

, , ,

ˆ

ˆ ,

= = = −

=

⟨ = ⟩⟨

| |

22 0 2 0 0 00 0 0 00 0 0 0

p m W s, ˆ= ⟩

⎛

⎝

⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟

| | (10.10.2)

An explicit calculation gives ⟨ = ⟩ =2 0 2 3 0p m W s ea, ˆ| | E , where a0 is the Bohr radius. Note that the matrix element is linear in E, so this correction is called the linear Stark effect. We transform to the basis that diagonalizes the perturbation; that basis is

{ , , , , ( , , ),| | | |2 1 2 112

2 0 2 01

p m p m s m p m= − ⟩ = ⟩ = ⟩ + = ⟩22

2 0 2 0( , , )}s m p m| |= ⟩ − = ⟩ (10.10.3)

Schematically, Fig.10.3 depicts how the linear Stark effect removes some of the degeneracy of the n = 2 level.


10.11. Consider a planar molecule consisting of four atoms: one atom is of type A and the three other atoms are of type B; see Fig. 10.4. An electron in the molecule can be found in the vicinity of each atom. If the electron is close to atom A, it has energy E1

0( ); if it is close to any of the B atoms, it has energy E2

0( ), where E E10

20( ) ( )< . We denote the states by

| | | |1 1000 2 0100 3 0010 4 0001⟩ ⟩ ≡ ⟩ ≡ ⟩ ≡≡ ( ) ( ) ( ) ( ) (10.11.1)

(a) For the first approximation, the electron cannot move from one atom to another. Using the basis { , , , }| | | |1 2 3 4⟩ ⟩ ⟩ ⟩ , write the Hamiltonian H0 for this approximation. (b) For the case in which an electron can move from an atom B to atom A and back, but cannot move from one B to another, we denote by a the energy associated with the transition from atom A to an atom B, where a << E1. Write the perturbation in this case. (c) Using perturbation theory, calculate the second-order correction to the energy of the state |1⟩ and the first-order correction to the states |2⟩ , | 3⟩, and | 4⟩. (d) Calculate exactly the corrections to the energies of the states. Show that when a << E1, one obtains the result of part (c).

Fig. 10.3

3ea0E

3ea0E

12

1 (|2s, m = 0⟩ – |2p, m = 0⟩)2

(|2s, m = 0⟩ + |2p, m = 0⟩)

|2p, m = –1⟩ and |2p, m = 1⟩

B

A

B B

(1)

(3)(4)

(2)

Fig. 10.4

SOLUTION

(a) In the basis { , , , }| | | |1 2 3 4⟩ ⟩ ⟩ ⟩ the Hamiltonian H0 is represented by the following matrix:

ˆ

( )

( )

( )

( )

H

E

E

E

E

0

10

20

20

20

0 0 0

0 0 0

0 0 0

0 0 0

=

⎛

⎝

⎜⎜⎜⎜⎜

⎞⎞

⎠

⎟⎟⎟⎟⎟

(10.11.2)


(b) The perturbation matrix representing the transitions between the state |1⟩ and each of the states | 2⟩, | 3⟩, or | 4⟩ is

W

a a aaaa

=

⎛

⎝

⎜⎜⎜

⎞

⎠

⎟⎟⎟

00 0 00 0 00 0 0

(10.11.3)

(c) The energy level E10( ) is nondegenerate. For the perturbation W the second-order correction is, in

accordance with Eq. (10.6),

E E Wi W

E Ei1

21

02

10 01 1

1( ) ( )( ) ( )

ˆˆ

= + ⟨ ⟩ − ⟨ ⟩−

| || | | | == +

−=

∑ Ea

E E10

2

10

20

2

43( )

( ) ( )

i

(10.11.4)

The energy level E20( ) is threefold degenerate, so we need to use perturbation theory for a degenerate state.

Since the matrix elements of W between the states | 2⟩, | 3⟩, and | 4⟩ are zero, the secular equation is

det0 0 0

0 0 00 0 0

0−

−−

⎛

⎝⎜⎜

⎞

⎠⎟⎟

=ε

εε

(10.11.5)

and therefore − =ε2 0, and the first-order correction to energy level E20( ) is zero:

E E E E E E21

20

31

20

41

20( ) ( ) ( ) ( ) ( ) ( )= = = (10.11.6)

We see that, to the first order, the degeneracy is not removed.

(d ) The total Hamiltonian is

ˆ

( )

( )

( )

( )

H

E a a a

a E

a E

a E

=

⎛

⎝

⎜⎜⎜⎜⎜

⎞1

0

20

20

20

0 0

0 0

0 0 ⎠⎠

⎟⎟⎟⎟⎟

(10.11.7)

To find the eigenenergies of H , we must solve the quadratic equation det (H − l I ) = 0. An explicit calculation gives

( )( ) ( )( ) ( ) ( )E E a E10

20 3 2

20 23 0− − − − =λ λ λ (10.11.8)

or

( ) [( )( ) ]( ) ( ) ( )E E E a20 2

10

20 23 0− − − − =λ λ λ (10.11.9)

Thus, we obtain

λ1 2 10

20

10

20 2 21

212,

( ) ( ) ( ) ( )( ) ( )= + ± − +⎡⎣⎢

E E E E a ⎤⎤⎦⎥

=λ 3 4 20

,( )E (10.11.10)

We see that the degeneracy level E20( ) is not completely removed by the perturbation, but it is partly

removed. For a E<< 10( ) we have

λ1 2 10

20

10

20

2

1

12

112

,( ) ( ) ( ) ( )( ) ( )

(= + ± − +E E E E

a

E(( ) ( )

( ) ( ) ( )

)

( ) (

020 2

10

20

101

2

−

⎡

⎣⎢⎢

⎤

⎦⎥⎥

≅ + ±

E

E E E −− +−

⎛

⎝⎜

⎞

⎠⎟

⎡

⎣⎢⎢

⎤

⎦⎥⎥

Ea

E E20

2

10

20 21

6( )( ) ( ))

( ) (10.11.11)


So,

λ λ1 10

2

10

20 2 2

02

10

2

3 3= +−

= −−

Ea

E EE

a

E E( )

( ) ( )( )

( ) (( )0 (10.11.12)

This is in accordance with the second-order correction for the E10( ), see Eq. (10.11.4), and the first-

order correction for the level E20( ).

10.12. Derive the transition probability equation for the first-order, time-dependent perturbation theory.

SOLUTION

Let cn(t) be the components of the vector |ψ( )t ⟩ in the { }|φn ⟩ basis:

| | |ψ φ φ ψ( ) ( ) ( ) ( )t c t c t tn

n

n n n⟩ = ⟩ = ⟨ ⟩∑ (10.12.1)

We define W W tnk n k= ⟨ ⟩φ φ| |ˆ ( ) . The Schrödinger equation is

iddt

t H W t t� | |ψ ψ( ) [ ˆ ˆ ( )] ( )⟩ = + ⟩0 λ (10.12.2)

By multiplying Eq. (10.12.2) by ⟨φn | and using Eq. (10.12.1) we obtain

idc t

dtE c t W t c tn

n n nk k

k

�( )

( ) ( ) ( )= + ∑λ (10.12.3)

Using the Bohr angular frequency wnk = (En − Ek)/� and the substitution c t a t en niE t

n( ) ( )/= − �

, Eq. (10.12.3) becomes

ida t

dte W t a tn i t

nk k

k

nk�( )

( ) ( )= ∑λ ω (10.12.4)

We write bn(t) in the form of a power series expansion in l:

a t a t a t a tn n n n( ) ( ) ( ) ( )( ) ( ) ( )= + + +0 1 2 2λ λ � (10.12.5)

We seek the solution to the first order in l. For t < 0 we assume the system to be in the state |φi⟩, so accord-ing to Eq. (10.12.1) and the relation between an(t) and cn(t) we have

a tn ni( )= =0 δ (10.12.6)

If we substitute Eq. (10.12.5) into Eq. (10.12.4) and equate the coefficient of lr on both sides of the equa-tion, we obtain [by using Eq. (10.12.6)]

idb

dte W t e W tn i t

nk kii t

ni

k

nk n i�

( )

( ) ( )1

= =∑ ω ωδ (10.12.7)

Equation (10.7.7) can be integrated to obtain

a ti

e W t dtni t

nk

tn k( )( ) ( )1

0

1= ′ ′′∫�ω

(10.12.8)

Finally, the transition probability Pif (t) between the states |φi ⟩ and |φf ⟩ is, according to Eq. (10.12.1), equal to | |c tf ( ) 2. Note that af (t) and cf (t) have the same modulus, and, to the first order,

a t a t a tf f f( ) ( ) ( )( ) ( )≅ +0 1λ (10.12.9)


Since the transition is between two different stationary states, we have b tf( )( )0 0= , and consequently

P t a t e W t dtif f

i t

fif i( ) ( ) ( )( )= = ′ ′

′λ λ ω2 1 22

20

| |�

tt

∫⎡

⎣⎢⎢

⎤

⎦⎥⎥

2

(10.12.10)

where we used Eq. (10.12.8).

10.13. Consider a one-dimensional harmonic oscillator with angular frequency w0 and electric charge q. At time t = 0 the oscillator is in the ground state. An electric field is applied for time t, so the perturbation is

W tq x t

( ) = − ≤ ≤{ E 00

τotherwise

(10.13.1)

where E is a field strength and x is a position operator. (a) Using first-order perturbation theory, calculate the probability of transition to the state n = 1. (b) Using first-order perturbation theory, show that a transition to n = 2 is impossible.

SOLUTION

(a) We denote by P01 the probability of transition from the ground state to n = 1; then, according to the first-order time-dependent perturbation theory,

P e W dti t

011

20

21

1 01

10( )( ) ˆτ ωτ

= ⟨ ⟩⎡

⎣⎢⎢

⎤

⎦⎥⎥

=∫�| |

��2

01 0

10e dt x q x x dxi tω

τ

∫ ∫ −⎡

⎣⎢⎢

⎤

⎦⎥

−∞

∞

φ φ( )( ) ( )E⎥⎥

2

(10.13.2)

where φ0(x) and φ1(x) are the energy eigenfunctions in the coordinate representation for n = 0 and n = 1, respectively. Using results for a harmonic oscillator we know that (see Chap. 5)

φ φ0 1 40 0

2

1

1 12

( ) exp ( )/xx

xx

x= −⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢⎢

⎤

⎦⎥⎥

=π

11 121 4

03 2

0

2

π / / expx

xx

−⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢⎢

⎤

⎦⎥⎥

(10.13.3)

where xm0 ≡ �

ω . Substituting these into Eq. (10.13.2) we obtain

P e qm

dti t

011

200

12

0( )( )τ ωω

τ

= −⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢⎢

⎤

⎦∫�

�E ⎥⎥

⎥=

⎡

⎣⎢⎢

⎤

⎦⎥⎥

=

∫2

2

0 0

2

2

2

2

10( )

( )

qm

e dt

qm

i tE

E

�ωω

τ

��ωω τ

ω0

0

0

222

sin ( )//

⎡⎣⎢

⎤⎦⎥

(10.13.4)

(b) To the first order, for the transition n = 0 → n = 2, we can write

P e W dti t

021

20

21

2 020( )( ) ˆτ ωτ

= ⟨ ⟩⎡

⎣⎢⎢

⎤

⎦⎥⎥∫�

| | (10.13.5)

We have

⟨ ⟩ = ⟨ − ⟩ = − ⟨ + ⟩ =2 0 2 02

2 0 0| | | | | |ˆ ( ) †W q x qm

a aE E�

ω ( ) (10.13.6)

where we used the relation xm

a a= +�2 ω ( )† (see Chap. 5). Therefore, P02

1 0( )( )τ = .


10.14. Consider a one-dimensional harmonic oscillator embedded in a uniform electric field. The field can be considered as a small perturbation that depends on time according to

E( ) exptA t= − ⎛

⎝⎜⎞⎠⎟

⎡

⎣⎢

⎤

⎦⎥πτ τ

2

(10.14.1)

where A is a constant. If the oscillator was in the ground state until the field was turned on at t = 0, compute, in the first approximation, the probability of its excitation as a result of the action of the perturbation.

SOLUTION

Consider the total momentum p imparted to the oscillator by the field over the duration of the perturbation:

p e t dteA

e dt eAt= = = =−∞

∞−

−∞

∞

∫ ∫E( ) .( / )

πττ 2

const (10.14.2)

We see that p does not depend on the time constant t of the perturbation. This means that the areas under the curves of Fig. 10.5 are equal, for every t.

0 t

E(t)

Fig. 10.5

The probability of a transition from the state n to the state k is given by

P V e dtnk kni t

kn=⎡

⎣⎢⎢

⎤

⎦⎥⎥−∞

∞

∫12

2

�

ω (10.14.3)

where V V dxk n k n=−∞

∞

∫ ψ ψ( )* ( )0 0 is the matrix element of the perturbation and ωkn k nE E= −| |( ) ( )0 0 /�. Let e,

m, and w denote the charge, mass, and natural frequency of the oscillator, respectively, where x denotes its deviation from its equilibrium position. In the case of a uniform field, the perturbation is given by

V x t ex t x( , ) ( )= − E ∼ (10.14.4)

The oscillator is the ground state (n = 0), so the nonvanishing elements of the perturbation matrix are

V Vp

mt

01 10

2

2= = − − ⎛

⎝⎜⎞⎠⎟

⎡

⎣⎢⎢

⎤

⎦⎥⎥πτ ω τ

�exp (10.14.5)

Thus, in the first approximation, a uniform field can produce a transition of the oscillator only to the first excited

state. If we substitute Eq. (10.14.5) and ω ω ωkn k nE E= = − =100 0| |( ) ( ) /� into Eq. (10.14.3) we obtain

Pp

mti t

tdt01

2

2

2

2= − ⎛

⎝⎜⎞⎠⎟

⎡

⎣⎢⎢

⎤

⎦⎥⎥−∞

∞

∫πτ ωω τexp

⎡⎡

⎣⎢⎢

⎤

⎦⎥⎥

2

(10.14.6)


Using the identity e dx ei x xβ α β απα

−

−∞

∞−∫ =

2 2 4/ we arrive at

Pp

m01

22

212

= −⎡⎣⎢

⎤⎦⎥�ω ωτexp ( ) (10.14.7)

We conclude that for a given classically imparted momentum p, the probability of the excitation decreases with the increase of t ; so, for t >> 1/w this probability is extremely small. This is the case of a so-called adiabatic perturbation. On the contrary, for a rapid perturbation t << 1/w the probability of excitation is constant. Note that in the limit t → 0,

lim ( ) ( ) ( )τ

δ δ→

= =0E t A t

pe

t (10.14.8)

and we have a sudden perturbation. In this case, the probability assumes the value

limτ ω→

=0 01

2

2P

pm�

(10.14.9)

which is equal to the ratio of the classically imparted energy p2/2m to the difference between the energy levels of the oscillator, �w. The criterion for the applicability of perturbation theory is that the probabil-ity of excitation must be small compared to the probability that the oscillator will remain in the ground state:

P P P01 01 011 1<< − <<( ) or (10.14.10)

It is apparent from Eq. (10.14.7) that

pm

eAm

2 2

2 2= <<( )

�ω (10.14.11)

is a sufficient condition for Eq. (10.14.10). However, if the field’s change is sufficiently adiabatic, that is, t >> 1/w, then the condition of Eq. (10.14.11) is too rigorous and perturbation theory can be applied. (This is if p2/2m is of the order of �w.)

10.15. Consider a linear oscillator in its ground state. Suppose that the equilibrium point begins to move slowly and uniformly at time t = 0, and that it stops at time t = T. Using the adiabatic approximation, find the probability that the oscillator will be excited. What is the validity of this approximation? (In an adiabatic approximation one assumes that the perturbation changes very slowly with time. It turns out that for adiabatic perturbation, the probability of excitation is very small.)

SOLUTION

The classical Hamiltonian of the oscillator at t ≥ 0 has the form

H tm

p m x a t( ) [ ( )]= + −12

12

2 2 2ω (10.15.1)

where a(t), the position of the equilibrium point, is v0 t according to the given condition, with v0 = constant being the velocity of the equilibrium point. The instantaneous eigenfunctions of the Hamiltonian, Eq. (10.16.1), have the form

ψn n

m

n

mx a t= ⎛

⎝⎜⎞⎠⎟ − −⎡

⎣⎢⎤⎦

ωπ

ω� �

1 421

2 2

/

!( ( ))exp ⎥⎥ − −

⎡

⎣⎢

⎤

⎦⎥H

mx a tn

ω�

[ ( )] (10.15.2)

The matrix element of the operator ∂∂ = − −

ˆ[ ( )]

Ht

m x a tω 20v computed from these functions, is nonzero

only for the transition n = 0 → n = 1 (recall that the initial state is the ground state),

∂∂

⎛

⎝⎜⎞

⎠⎟= −H

tm

m10

20 2

ω ωv�

(10.15.3)


Evidently, the spectrum of the energy levels of the oscillator does not change during the motion of the equilibrium point; i.e., all the w are constant. The probability amplitude of the first excited state is obtained by substituting w10 = w, so

C ti

mm

e im

ei t i t1 2

20 0

12

12

( ) ( ) (≈ − − = −�

��ω

ω ω ωω ωv v 11) (10.15.4)

Therefore, the probability that at time t the oscillator will be in the first excited state is

P t C tm

t1 12 0

2

1( ) ( ) ( cos ( ))= = −| |v

�ω ω (10.15.5)

Note that this probability oscillates with time. Thus, the probability of excitation for t ≥ T is

P Tm

t102

1( ) ( cos ( ))= −v

�ω ω (10.15.6)

For the adiabatic approximation to be valid, the inequality P1(T ) << 1 must hold for all t. This is equivalent to the condition

v0 << �ωm

(10.15.7)

10.16. Consider a hydrogen atom in its ground state at time t = 0. At the same time a uniform periodic electric field is applied to the atom. (a) Find the minimum frequency that the field needs in order to ionize the atom. (b) Using perturbation theory, find the probability of ionization per unit time. Assume that when the atom becomes ionized, its electron becomes free.

SOLUTION

(a) The equation for the transition probability per unit time from a state in a discrete spectrum to a state in a continuous spectrum, under the action of a periodic perturbation, has the form

dP V E E dn n nn n nn= − −2 2 0π δ ω

��| | ( )( ) (10.16.1)

where w is the frequency of the periodic perturbation, n represents the set of quantum numbers that characterizes the states of the discrete spectrum, dn is the corresponding infinitesimal energy interval of the continuous spectrum, En

( )0 is the unperturbed energy level in the discrete spectrum, En is the

unperturbed energy level in the continuous spectrum, and Vn n is the matrix element of the perturbation

operator for the considered transition. The perturbation operator has the form

ˆ ( ( ) ) ( ) sin ( ) ˆ exp( ) ˆ*W e t e t V i t V= = = − +E E⋅ ⋅r r0 ω ω eexp ( )i tω (10.16.2)

where E( )t is the electric field, | |E0 its amplitude, and V is given by

Vie=2 0E ⋅ r (10.16.3)

Note that the d-function in Eq. (10.16.1) assures that the transition takes place only when E Enn

− − =( )0 0�ω ; therefore,

ωmin( )

min( )= −1 0

�E Enn

(10.16.4)

Since the hydrogen atom is in its ground state, we have

( )( )minE E

menn

− =04

32� (10.16.5)

which gives us the minimum frequency of the electric field needed to ionize the atom.


(b) Consider the matrix element V V d rn nn n= ∫ ψ ψ* ˆ ( )0 3 . For ψn

( )0 we take the ground-state wavefunction of the hydrogen atom:

ψna

ra

ae

( ) exp0

3

2

21= −⎛

⎝⎜⎞⎠⎟ =

⎛

⎝⎜⎞

⎠⎟π μ�

(10.16.6)

For ψn we approximate

ψn

≈ 1

8 3πexp ( )i k r⋅ (10.16.7)

where n = �k2/4p m. Note that ψn( )0 is normalized to unity and ψ

n is normalized to d (n − n ′ ). Substi-

tuting all this into Vn n, we obtain

Vie

ai

ra

d rnn = − −⎛⎝⎜

⎞⎠⎟∫2

1

8

13 3 0

3

π πexp k r r⋅ ⋅E (10.16.8)

To calculate the integral, we use the spherical coordinates (r, q, j). We assume that the wave or propagation vector k is directed along the polar axis, and denote the angle between k and E0 by c. The scalar product E0 ⋅ r is

E0 0 0⋅ r r= + −E [cos cos sin sin cos ( )]χ θ χ θ ϕ ϕ (10.16.9)

where j0 is the corresponding coordinate of E0. Substituting Eq. (10.16.9) into Eq. (10.16.8) (denoting z ≡ cos q) we obtain

Vie

aikr z

ra

r drnn = − −⎛⎝⎜

⎞⎠⎟

⎡ ∞

∫E03

3

02πχ

( )cos exp

⎣⎣⎢⎢

⎤

⎦⎥⎥

=+−∫ z dz

e

a

a k

k a

E03

1

1 5

2 2 32

16

1

cos

( ) ( )

χ

π (10.16.10)

Let us now turn to dn:

d d k d k d kdk

dEd dE

mkd dEk k kn

nn n

= = = =3 2 22k Ω Ω Ω

� (10.16.11)

where we have used the relation En = �2k2/2m and dWk is an element of the solid angle with the axis

along k. Substituting Eqs. (10.16.10) and (10.16.11) into Eq. (10.16.1), we obtain

dPk

k aE E dn n kn n

| |E02 3 2

2 2 60

1

cos

( )( )( )χ

δ ω+

× − − � Ω ddEn

(10.16.12)

The probability of ionization when the electron makes a transition with a final wave vector k within the element d Wk is obtained by integrating Eq. (10.16.12) over dE

n. Using the properties of d-function

in Eq. (10.16.12), we have to consider only the point where E Enn= +( )0

�ω ; thus it follows that

kmE m2

2

2 2= =

−n

� �

( )minω ω (10.16.13)

and 1 2 2+ =k aω

ωmin. We now have

dPa

Ek = ⎛⎝⎜

⎞⎠⎟

−⎛⎝⎜

⎞⎠⎟

641

3 6 3 2

0πω

ωω

ω�min

min

/

cos22χ d kΩ (10.16.14)

[The probability is denoted now by dPk since Eq. (10.16.14) depends only on k.] We use the fact that cos2c = 1/3, and integrate Eq. (10.16.14) over the angles. We finally obtain the total probability of the atom ionization per unit time:

Pa

E= ⎛⎝⎜

⎞⎠⎟

−⎛⎝⎜

⎞⎠⎟

2563

13 6

0

3 2

02

�

ωω

ωω

min/

(10.16.15)


10.17. Consider a quantum system with two stationary eigenstates |1⟩ and |2⟩ . The difference between their eigenvalues is given by E2 − E1 = �w21. At time t = 0, when the system is in state |1⟩, a small perturbation that does not change in time and is equal to H′ is applied. The following matrix elements are given:

⟨ ′ ⟩ = ⟨ ′ ⟩ = ⟨ ′ ⟩ = −1 1 0 2 1 2 20| | | | | |ˆ ˆ ˆH H H� �ω ω (10.17.1)

(a) Using the first-order time-dependent perturbation theory, calculate the probability of finding the system at time t in state |1⟩, and the probability of finding it in state |2⟩ . (b) Solve the Schrödinger equation and find |ψ( )t ⟩. (c) What is the probability that at time t the system is in state |2⟩? When is the approximation used in part (a) correct? At what time (for the first order) will the system have probability 1 in state |2⟩?

SOLUTION

(a) From first-order time-dependent perturbation theory we have P| |1 2⟩ → ⟩, and since there are only two eigenstates for the system, we obtain

P P| | | |1 1 1 21⟩ → ⟩ ⟩ → ⟩= − (10.17.2)

Using the formula P e W t dtif

i tt

f if i( ) ( )1

20

21= ′ ′

′∫�

ω we arrive at

P e H dti t

t

| | | |1 2 20

2

22

0

22 1

121

⟩ → ⟩′≅ ⟨ ′ ⟩ ′ =∫� �

�ω ωˆ 22

0

2

02

21

2

02

21 211

1e dti

e

e

i tt

i t

i

ω ω

ω

ω ω

ω

′∫ ′ = −

=

( )

221 21 21 212 2 2

21

2

02

2t i t i t i te e

ie/ / / /

( )ω ω ω

ω ω− =− 22 2

2

21

21

2

02 2 2

2121

i ti

eti t

sin( )

sin( )/

ωω

ωωω

/

/= ωω ω

ωω21

2

02 21

21

2

22

2//

/=

sin( )t (10.17.3)

Since ω21 1t << , P t| |1 2 02 2 1⟩ → ⟩ ≅ <<ω and thus ω0 1t << . The two inequalities ω21 1t << and ω0 1t << are the

conditions for applicability of the first-order approximation. Consequently, P P| | | |1 1 1 21 1⟩ → ⟩ ⟩ − ⟩= − ≅ .

(b) Method 1: Diagonalization of the Hamiltonian H = H0 + H ′. First, we express the Hamiltonian explicitly in the basis of the eigenstates |1⟩ and | 2⟩:

HE

E

E

E= −

⎛⎝⎜

⎞⎠⎟

=⎛⎝⎜

⎞⎠⎟

1 0

0 2 21

1 0

0 1

�

� �

�

�

ωω ω

ωω (10.17.4)

where we used the relation E2 − E1 = �w2. The eigenvalue equation is Hv = lv, then (E1 − l)2 − (�w0)2 = 0 and λ ω1 2 1 0,

ˆ= ±E � . For two eigenvectors, v1 and v2, we obtain

v v1 2

12

11

12

1 212

11

12

= ⎛⎝⎜

⎞⎠⎟ = ⟩ + ⟩ = −

⎛⎝⎜

⎞⎠⎟ =( ) (| | || |1 2⟩ − ⟩) (10.17.5)

So, for the state |ψ( )t ⟩ we have

| | |ψ( )t a e a ei t i t⟩ = ⟩ + ⟩− −

1 1 2 21 2λ λ/� v v (10.17.6)

Using the initial condition | |ψ( )t = ⟩ = ⟩0 1 we get a a1 2 1 2= = / , and eventually

| ( ) ( )( ) ( )ψ t e e

i E t i E t⟩ = + ⟩− + − −12

11 0 1 0� � � �ω ω/ / | ++ − ⟩

=

− + − −

−

12

21 0 1 0( )( ) ( )

e e

e

i E t i E t

iE

� � � �ω ω/ / |

11

0 01 2t

t t/�

[cos ( ) sin ( ) ]ω ω| |⟩ − ⟩ (10.17.7)


Method 2: Explicit solution of the Schrödinger equation.

We write |ψ( )t ⟩ in the form

| | |ψ(t)⟩ = ⟩ + ⟩− −C t e C t e

iE t iE t1 2

1 21 2( ) ( )/ /� � (10.17.8)

Substituting this into the Schrödinger equation it

tH H t�

∂ ⟩∂ = + ′ ⟩|

|ψ ψ( )

( ˆ ˆ ) ( )0 , we obtain

i C t eiE

C t eiE t iE t

��

� �

11

11 1 1

i( ) ( )

− −+⎡⎣⎢

⎤⎦⎥

⟩/ / | ++ −⎡⎣⎢

⎤⎦⎥

⟩− −i C t e

iEC t e

iE t iE t�

��

22

22 1 2

i( )

/ / |

== + ′ ⟩ +− −( ˆ ˆ )[ ( ) ( )H H C t e C t e

iE t iE t0 1 2

1 21/ /� �| | 22⟩] (10.17.9)

Multiplying Eq. (10.17.9) by ⟨1 | yields

i C t eiE

C t e CiE t iE t

��

� �

11

1 11 1

i( ) ( )

/− −−⎡⎣⎢

⎤⎦⎥

=/(( ) ( ˆ ˆ )

( )

/t e H H

C t e

iE t

iE

−

−

⟨ ⟩ + ⟨ ′ ⟩

+

1 1 1 1 10

2

� | | | |

22

1

1 2 1 20

1 1

t

iE t

H H

E C t e

/( ˆ ˆ )

( )

�

�

⟨ ⟩ + ⟨ ′ ⟩

= +−

| | | |

/CC t e

iE t2 0

2( )��ω − / (10.17.10)

which leads to

i C t e C ti t

� �1 2 021

i( ) ( )= − ω ω (10.17.11)

where ω21 2 1

1= −�

( )E E . In the same way, multiplying Eq. (10.17.9) on the left by ⟨2 | results in

i C C t e C ti t

� � �2 1 0 2 2121

i( ) ( ) ( )ψ = −ω ω ω (10.17.12)

Equations (10.17.11) and (10.17.12) give a system of differential equations with coefficients C1(t) and C2(t):

i C t e C t

i C t C t e

i t

i

� �

�

1 2 0

2 1

21

21

i

i

( ) ( )

( ) ( )

=

=

− ω

ω

ωtt

C t� �ω ω0 2 21−

⎧⎨⎪

⎩⎪ ( ) (10.17.13)

Extracting C2(t) from the first equation, C ti

C t ei t

20

121( ) ( )= ω

ωi, and differentiating:

C ti

C t e i C t ei t i t

20

1 21 121 21

i ii i( ) ( ) ( )= +⎡

⎣⎢ω ωω ω ⎤⎤⎦⎥

(10.17.14)

Substituting these two expressions into the second equation in (10.17.13) we get

C t C t1 02

1 0i i

( ) ( )+ =ω (10.17.15)

Using the initial condition C1(t = 0) = 0, Eq. (10.17.15) gives C1(t) = cos (w0t). Thus, we can calculate the coefficient C2(t) and find C t i t e

i t2 0

21( ) sin( )= − ω ω , so eventually,

| | |ψ( ) [cos ( ) sin ( ) ]t e t i tiE t⟩ = ⟩ − ⟩− 1

0 01 2/� ω ω (10.17.16)

Note that the results of the two methods coincide.

(c) The probability of finding the system in state | 2⟩ is given by

P t t| | | | |1 22 2

02⟩ → ⟩ = ⟨ ⟩ =ψ ( ) sin ( )ω (10.17.17)

Therefore, the approximation used in part (a) is correct when w 0 t = 1. The system will be in state | 2⟩ when

P = 0 or ωπ π0 2

T k= ± + , k N∈ . At times Tk= ± +π

ωπω2 0 0

, k N∈ , the system will be in state | 2⟩.



10.18. Repeat Problem 10.3, using the raising and upper operators, a and a†, respectively.

10.19. Consider a one-dimensional oscillator with a linear perturbation lx. For the ground state energy, compute the first two orders of the perturbation (use dimensionless units).

Ans. E(l) = 1 − l2/4

10.20. A small perturbation, W = ax4, is applied to a harmonic oscillator with force constant k and reduced mass m. Compute the first-order correction to the eigenenergies and first nonvanishing correction to the wavefunctions.

Ans. Ea

n na

n( ) ( );1

22

0 00

23

2

12

3 2

4= + +⎛

⎝⎜⎞⎠⎟ = −

α ωαφ φ φ

� 220( ) + �

10.21. Consider the Hamiltonian ˆ ˆ ˆ( , )H H V x y= +0 , where ˆ ( ˆ ˆ )H m x y02 2 2 2= +ω / is the free Hamiltonian, and

ˆ( , ) ˆ ˆV x y m x y= λ ω 2 is a perturbation. (a) Find the exact ground state. (b) Using the second-order perturbation, calculate the ground state energy.

Ans. (a) ψ π02 1 8

2

11 1

4( , ) ( ) exp

( )/x ym m x= − − − + +⎛

⎝⎜ω λ ω λ λ� �

⎞⎞

⎠⎟

× − − + +⎛

⎝⎜⎞

⎠⎟− − +

exp( )

exp(m y mω λ λ ω λ λ1 1

41 12

�)) xy

2�

⎛

⎝⎜⎞

⎠⎟

(b) E02

2

8( ) = − λ ω�

, where the exact result is E0

212

1 18

= − + + ≅ − +� ��ω λ λ ω λ ω

( ) �

10.22. A particle with mass m and electric charge e moves in a one-dimensional harmonic potential, subjected to a weak electric field E. (a) Calculate the corrections to the energy levels and to the eigenstates of the first nonvanishing order. (b) Calculate the electric dipole moment of the particle. (c) Solve parts (a) and (b) exactly, and compare the results to the approximate solutions.

Ans. (a) Δ = − ⟩ = ⟩ − ⟩ − +−E

e

me

m

n nn n n

( );

EE

2

2 3 12 21

2ω ω| | |ψ φ φ�

|| φn b Pe

m+ ⟩⎡

⎣⎢

⎤

⎦⎥ =1

2

22; ( )

E

ω

10.23. A plane rotator with electric dipole moment d and an inertia moment I is subject to a uniform electric field E that lies in the plane of rotation. Calculate the first nonvanishing corrections to the energy levels of the rotator. Consider the field E as a small perturbation. Hint: The perturbation is W = −d ⋅ E.

Ans. E E EnI

I d

mn n n= + = +−

( ) ( ) ( )

( )0 2

2 2 2

2 22 4 1

�

�

E

10.24. Consider the classical Hamiltonian

HPm

Pm

m x x V x x= + + + + −12

22

02

12

22

1 22 212

ω ( ) ( )rel (10.24.1)

where V x x m x xrel ( ) ( )1 2 12

1 221

4− = −ω . (a) Find the exact energy of this system. (b) Assuming that

W = Vrel(x1 − x2), use the second-order perturbation to compute the energy of the ground state.

Ans. (a) E N n N nNn = + + + + =( ) ( ) ( , , , )1 2 1 2 0 10 02

12/ /� �ω ω ω … , where

E00 0 02

12

012

0

14

03

12

12 4 16

= + + ≅ + − +� � �� ω ω ω ω

ωω

ωω

�� ; (b) E0 012

0

14

034 16

= + −�� ω

ωω

ωω

10.25. In the first approximation, compute the energy of the ground state of a two-electron atom or ion having a nuclear charge Z. Consider the interaction between the electrons as a small perturbation.

Ans. E E E Z Zme≈ + = − −⎛

⎝⎜⎞⎠⎟

( ) ( )0 1 24

258 �


10.26. Consider a quantum system that has an orthonormal basis of three unperturbed states. The perturbed Hamiltonian is represented by the matrix:

ˆ

* *

H

E a

E b

a b E

=

⎛

⎝

⎜⎜⎜

⎞

⎠

⎟⎟⎟

1

1

2

00 (10.26.1)

where E2 > E1. Use a second-order nondegenerate perturbation to find the perturbed eigenvalues. Diagonalize the matrix and find the exact eigenvalues. Repeat using a second-order degenerate perturbation. Explain the inconsistencies arising from the different approaches.

Ans. Denote | | |1100

2010

3001

⟩ =⎛

⎝⎜⎜

⎞

⎠⎟⎟

⟩ =⎛

⎝⎜⎜

⎞

⎠⎟⎟

⟩ =⎛

, , and⎝⎝⎜⎜

⎞

⎠⎟⎟

. Using a nondegenerate perturbation, ΔE (1) = 0,

Δ = − Δ = − Δ⟩ ⟩ ⟩Ea

E EE

bE E

E| | |(| | | |

12

2

2 122

2

2 132( ) ( ) )) = − +

−| | | |a b

E E

2 2

2 1 (10.26.2)

The exact solution is

Δ = Δ = +− Δ = − +

⟩ ⟩ ⟩E Ea b

E EE

a b| | |

| | | | | | | |1 2

2 2

2 13

2 2

0EE E2 1− (10.26.3)

Using a degenerate perturbation, Δ = Δ = +−⟩ ⟩E E

a bE E| |

| | | |1 2

2 2

2 10;

10.27. Consider a molecule consisting of three atoms arranged on a line; see Fig. 10.6.

B A B

(1)(3) (2)

Fig. 10.6

If an electron is in the vicinity of atom A, its energy is E1; if it is in the vicinity of either of the atoms B, its energy is E2 (E1 < E2). (a) To the first approximation, assume that there is no transition of an electron between atoms. Find the Hamiltonian H0. (b) A small perturbation is applied and the electron can move from one atom to another. The energy associated with a transition is a, where a << E1. To the first order, compute the corrections to the energies and eigenstates (for E1 apply to the second order). (c) Suppose that at t = 0 the electron is near atom B. Using the approximation of part (b), calculate the probability that at t > 0 the electron will be in the vicinity of atom A. (d) Find the exact eigenenergies of the electron. (e) An electron can now move from an atom B to another atom B, and the energy associated with the transition is b, where b << a. Using perturbation theory, find the energies and eigenstates of H when the unperturbed Hamiltonian is H0 + W , where W is the perturbation introduced in part (b).

Ans. (a) ˆ ;H

E

E

E0

1

2

2

0 00 00 0

=⎛

⎝

⎜⎜

⎞

⎠

⎟⎟ (b) ˆ ; ˆ( )W

a aaa

E E W=⎛

⎝⎜⎜

⎞

⎠⎟⎟

= + ⟨ ⟩ + ⟨00 00 0

1 1 22

1 10 | |

| || | |ˆ;( )W

E EE

aE E

1 22

2 11

02

2 1

⟩− = + −

| || |

|| |

|ψ12 1 3 1

12 1

23 1

3⟩ = ⟩ − ⟨ ⟩− ⟩ − ⟨ ⟩

− ⟩ =ˆ ˆW

E EW

E E|| | | |1 2 3

1

2 11 2 1

2 1

⟩ − − ⟩ ⟩( ) ⟩ =− −

− −

⎛

aE E

aE E

aE E

+ ψor

⎝⎝

⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟

. The

states |ψ2⟩ and |ψ3⟩ are degenerate; therefore, to the first order, there are no corrections to the eigenstates.


(c) P| |2 1 0⟩→ ⟩ = ; (d ) E E E E E E a E1 3 10

20

10

20 2 2

2

12

8= = + ± − −( )( ) ( ) ( ) ( )( ) , == E20( ); (e) �E E b2 2= − ,

� �E E E E b E E b a1 3 1 2 1 22 21

22= = + + ± − − −( )( )

10.28. In the first approximation, compute the shift in the energy level of the ground state of a hydrogen-like atom resulting from the fact that the nucleus is not a charge point. Regard the nucleus as a sphere of radius R throughout the volume of which the charge Ze is distributed evenly. Hint: The potential energy of an electron is the field of a nucleus that has an evenly distributed charge,

V r

ZeR

r

Rr R

Zer

( ) =− −

⎛

⎝⎜⎞

⎠⎟≤ ≤

−

2 2

2

2

32

12

0for

for rr R≥

⎧

⎨

⎪⎪

⎩

⎪⎪

(10.28.1)

Ans. E E E ERa0 0

001

00

0

2

145

≈ + = −⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢⎢

⎤

⎦⎥⎥

( ) ( ) ( )

10.29. Consider a harmonic oscillator described by

Hm

p m t xx= +12

12

2 2 2ω ( ) (10.29.1)

where ω ω δω( ) cos ( )t at= +0 and δω ω<< 0 (a is a constant). Assume that at t = 0 the system is in the ground state. Using perturbation theory, find the transition probability from the ground state to a final state f. You may use the result ⟨ ⟩ =n x m| |ˆ2 0 2�ω / for n = 2 and zero otherwise.

Ans. P ta t

aff i

f i0

2

028

2

2→ ≈−

−⎡

⎣⎢

⎤( )

( ) sin ( )

( )δω

ω

ωω

/

/ ⎦⎦⎥ +( ), ,2 2 00

δ δωa a

232

Solution Methods in Quantum Mechanics—Part B

11.1 The Variational MethodThe perturbation theory studied in Chap. 10 is not the only approximation method in quantum mechanics. In this section, we present another method applicable to conservative system. Consider a physical system with time-independent Hamiltonian H. We assume for simplicity that the entire spectrum of H is discrete and nondegenerate:

( ˆ ) , , , . . .H E nn n n| |φ φ⟩ = ⟩ = 1 2 3 (11.1)

We denote by E0 the smallest eigenvalue of H (that is, the smallest energy of the system). An arbitrary state | ψ⟩ can be written in the form

| |ψ φ⟩ = ⟩∑ cn n

n

(11.2)

Then

⟨ ⟩ = ≥∑ ∑ψ ψ| |H c E E cn

n

n n

n

2

0

2 (11.3)

On the other hand,

⟨ ⟩ = ∑ψ ψ| cn

n

2 (11.4)

Thus, we can conclude that for every ket,

⟨ ⟩ =⟨ ⟩

⟨ ⟩ ≥ˆˆ

HH

Eψ ψ

ψ ψ| |

| 0 (11.5)

Equation (11.5) is the basis of the variational method. A family of kets |ψ (a)⟩ called trial kets, is chosen. The mean value of H in the states |ψ (a)⟩ is calculated, and the expression ⟨ ⟩H ( )α is minimized with respect to the parameter a. The minimal value obtained is an approximation of the ground state energy E0.

Equation (11.5) is actually a part of a more general result called the Ritz theorem: the mean value of the Hamiltonian H is stationary in the neighborhood of its discrete eigenvalues (see Problem 11.1).

The variational method can therefore be generalized and provide estimations for other energy levels besides the ground state. If the function ⟨ ⟩H ( )α obtained from the trial kets |ψ (a)⟩ has several extrema, they give approximate values of some of the energies En.

CHAPTER 11


11.2 Semiclassical Approximation (The WKB Approximation)Apart from the perturbation and variational methods described earlier, there is another method that is suitable for obtaining solutions to the one-dimensional Schrödinger equation. This is the so-called semiclassical, or WKB approximation (named after Wentzel, Kramers, and Brillouin). The WKB method can also be applied to three-dimensional problems if the potential is spherically symmetric and a radial differential equation can be separated.

The WKB method introduces an expansion in powers of � in which terms of order greater than �2 are neglected. Thus, one replaces the Schrödinger equation by its classical limit ( ).� → 0 However, the method can be applied even in regions in which the classical interpretation is meaningless (regions inaccessible to classical particles).

Consider the Schrödinger equation in one dimension:

d

dx

mE V x x

2

2 2

20

ψ ψ+ − =�

[ ( )] ( ) (11.6)

We consider only stationary states, and write the wavefunction in the form ψ( ) .( )x eiu x= We shall use the abbreviations

k x m E V x E V x

k x i xi

( ) [ ( )] ( )

( ) ( ) ( )

= − >

= − = −

12

�for

χ��

2m V x E E V x[ ( ) ] ( )− <

⎧

⎨⎪

⎩⎪ for

(11.7)

Substituting ψ(x) into Eq. (11.7) one finds that u(x) satisfies the equation

id u

dx

dudx

k x2

2

22 0− ⎛

⎝⎜⎞⎠⎟ + =[ ( )] (11.8)

In the WKB approximation, we expand u(x) in a power series of �:

u x ui

ui

u( ) = + + ⎛⎝⎜

⎞⎠⎟ +0 1

2

2

� �� (11.9)

and we consider only u0 and u1. Then we obtain the approximate wavefunction according to the WKB method:

ψ( )( )

exp ( )(

xC

k xi k x dx

C

k

x

= ′ ′⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪+∫1 2

| | | xxi k x dx

x

)exp ( )

|− ′ ′

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪∫ (11.10)

A region in which E > V(x) is called a classically allowed region of motion, while a region in whichE < V(x) is called classically inaccessible. The points in the boundary between these two kinds of regions are called turning points [where E = V(x)].

Applicability ConditionThe WKB approximation is based on the condition

12

2| | | |′ <<k x k x( ) ( ) (11.11)

This condition can be expressed in a number of equivalent forms. Using the de Broglie wavelength λ π= 2k

we can write Eq. (11.11) as

λπ4

dkdx

k<< (11.12)

CHAPTER 11 Solution Methods in Quantum Mechanics—Part B234

Adjacent to the turning points, for which k(x0) = 0, we have

kdkdx

x xx

≈ −0

0( ) (11.13)

Thus, the semiclassical approximation is applicable for a distance from the turning point satisfying the condition

| | x x− >>0 4λπ (11.14)

The Connection FormulasConsider a turning point. Assume that, except in its immediate neighborhood, the WKB approximation is applicable. The match between the WKB approximations on each side of the turning points depends on whether the classical region is to the left of the point (Fig. 11.1) or to the right of it (Fig. 11.2).

Fig. 11.1

V(x)

x = b

E

Fig. 11.2

V(x)

x = a

E

In the first case we have, x > b:

ψ11

1( ) cos ( )xA

kk x dx B

b

x

= ′ ′ −⎛

⎝⎜

⎞

⎠⎟∫ π (11.15)

while in the second case, for x < a,

ψ22

2( ) cos ( )xAk

k x dx Bx

a

= ′ ′ −⎛

⎝⎜

⎞

⎠⎟∫ π (11.16)

Application to the Bound StateThe WKB approximation can be applied to derive an equation for the energies of a bound state. Using the connection formulas in each side of the potential well one obtains (see Problem 11.13)

k x dx n na

b

( ) , , , . . .= +⎛⎝⎜

⎞⎠⎟ =∫ 1

20 1 2π (11.17)

which may be written

○ p x dx n n( ) , , , . . .= +⎛⎝⎜

⎞⎠⎟ =∫ 2

12

0 1 2π� (11.18)

This equation is called the Bohr-Sommerfeld quantization rule.


Barrier PotentialIf one considers a potential barrier of the form V(x) between x = a and x = b and a particle with energy E, the transmission coefficient in the WKB approximation is given by

T m V x E dxa

b

≈ − −⎡

⎣⎢⎢

⎤

⎦⎥⎥∫exp [ ( ) ]

22

� (11.19)

SOLVED PROBLEMS

11.1. We define ⟨ ⟩H by ⟨ ⟩ =⟨ ⟩

⟨ ⟩ˆ

ˆ,H

Hψ ψψ ψ| |

| where |ψ⟩ is any vector in the state space. Show that ⟨ ⟩H

is stationary (that is, δ ⟨ ⟩ =ˆ ),H 0 if, and only if, |ψ⟩ is an eigenvector of H with eigenvalue ⟨ ⟩H .

SOLUTION

We write the equation ⟨ ⟩ = ⟨ ⟩ ⟨ ⟩ˆ ˆH Hψ ψ ψ ψ| | |/ in the more convenient form ⟨ ⟩ ⟨ ⟩ = ⟨ ⟩ψ ψ ψ ψ| | |ˆ ˆ .H H Differentiating both sides gives

( ) ˆ ˆ ˆ⟨ ⟩ + ⟨ ⟩ ⟨ ⟩ + ⟨ ⟩ ⟨ ⟩ = ⟨ ⟩ + ⟨δ δ δ δψ ψ ψ ψ ψ ψ ψ ψ ψ| | | | |H H H || |H δ ψ⟩ (11.1.1)

As ⟨ ⟩H is a scalar value, we may rewrite Eq. (11.1.1) as

⟨ ⟩ ⟨ ⟩ = ⟨ − ⟨ ⟩ ⟩ + ⟨ − ⟨ ⟩ψ ψ ψ ψ ψ ψ| | | | |δ δ δˆ ( ˆ ˆ ) ( ˆ ˆ )H H H H H ⟩⟩ (11.1.2)

Defining | |φ ψ⟩ = − ⟨ ⟩ ⟩( ˆ ˆ ) ,H H we may reformulate Eq. (11.1.2) in the simpler form

⟨ ⟩ ⟨ ⟩ = ⟨ ⟩ + ⟨ ⟩ψ ψ ψ φ φ ψ| | |δ δ δH (11.1.3)

Equation (11.1.3) holds for any |d ψ⟩, in particular for |d ψ⟩ = |φ⟩d l, where d l is an infinitesimally small real number. Substituting into Eq. (11.1.3) we arrive at

⟨ ⟩ ⟨ ⟩ = ⟨ ⟩ψ ψ φ φ| |δ δ λH 2 (11.1.4)

Now, if δ ⟨ ⟩ =ˆ ,H 0 then according to Eq. (11.1.4) we must have |φ⟩ = 0, so that ˆ ˆ ,H H| |ψ ψ⟩ = ⟨ ⟩ ⟩ and we see that | ψ⟩ is an eigenvector of H with eigenvalue ⟨ ⟩ˆ .H On the other hand, if ˆ ˆ ,H H| |ψ ψ⟩ = ⟨ ⟩ ⟩ then accord-ing to Eq. (11.1.4) we must have δ ⟨ ⟩ =ˆ ,H 0 therefore ⟨ ⟩H is stationary.

11.2. Consider a one-dimensional harmonic oscillator:

Hm

d

dxm x= − +�

2 2

22 2

212

ω (11.2.1)

(a) For the one-parameter family of wavefunctions ψαα α( ) ( ),x e x= >− 2

0 find a wavefunction that minimizes ⟨ ⟩ˆ .H What is the value of ⟨ ⟩ˆ ?minH (b) For another one-parameter family of wavefunctions

ψββ β( ) ( ),x xe x= >− 2

0 find a wavefunction that minimizes ⟨ ⟩H and compute the value of ⟨ ⟩ˆ .minH (c) Repeat the same procedure for

ψγ γγ( ) ( )x

x=

+>1

02 (11.2.2)

SOLUTION

(a) We begin by considering ⟨ ⟩ˆ :H

⟨ ⟩ =− +

⎡

⎣⎢⎢

⎤

⎦⎥⎥−∞

∞

∫ˆ

( )*

H

xm

d

dxm xψ ψα αω�

2 2

22 2

212

(( )

( ) ( )*

x dx

x x dxm

m

ψ ψα α

α ω α

−∞

∞

∫= +�

22

218

1 (11.2.3)


We differentiate ⟨ ⟩H with respect to a :

d Hd m

m⟨ ⟩ = −

ˆ

α ωα

�2

222

18

1 (11.2.4)

From the condition d Hd⟨ ⟩ =

=

ˆ,α

α α0

0 we have �

22

022

18

10

mm− =ω

α and α ω

0 2= m

�; thus, α0 gives the mini-

mum value of ⟨ ⟩H (as can be easily verified). The wavefunction that minimizes ⟨ ⟩H is ψαω

0

2 2( ) ,x e m x= − / � and

⟨ ⟩ = + =ˆminH

mm

��

2

02

0218

1 12

α ω α ω (11.2.5)

Thus, ⟨ ⟩ˆminH coincides with the energy of the n = 0 level of a one-dimensional harmonic oscillator.

Note that the family of functions we are studying coincides with the ground-state wavefunction of the harmonic oscillator.

(b) We proceed with the same method as in part (a):

⟨ ⟩ =− +

⎡

⎣⎢⎢

⎤

⎦⎥⎥ˆ

( ) ( )*

H

xm

d

dxm x x dψ ψβ βω�

2 2

22 2

212

xx

x x dxm

m−∞

∞

−∞

∞

∫∫

= +ψ ψβ β

β ωβ

* ( ) ( )

32

38

12 2�

(11.2.6)

and

d Hd m

m⟨ ⟩ = − =ˆ

βω

β32

38

10

2 2

2�

(11.2.7)

We obtain β ω0

12

= m�

and ψβ

ω0

2 2( ) ;x xe m x= − / � so, ⟨ ⟩ =ˆ .minH32

�ω Thus, ⟨ ⟩ˆminH equals the energy of the

n = 1 level of the one-dimensional harmonic oscillator. (Try to explain this result.)

(c) Applying the procedure of parts (a) and (b), we obtain

⟨ ⟩ =− +

⎡

⎣⎢⎢

⎤

⎦⎥⎥ˆ

( ) ( )*

H

xm

d

dxm x x dψ ψγ γω�

2 2

22 2

212

xx

x x dxm

m−∞

∞

−∞

∞

∫∫

= +ψ ψγ γ

γ ω γ* ( ) ( )

�2

2

21 1

2 (11.2.8)

and

ψγ ωγ ω0

1

2

122 2 0( )x

x m m=

+=

�

�

/ (11.2.9)

hence, ⟨ ⟩ =ˆ .minH 212

�ω We see that ⟨ ⟩ˆminH is equal to 2 times the ground-state energy.

11.3. (a) Using the variational method, estimate the ground-state energy of an atom of hydrogen. Choose as trial functions the spherically symmetrical functions φa (r) whose r-dependence is given by

φα α α

α( )r

Cr

r

r= −⎛

⎝⎜⎞⎠⎟ ≤

>

⎧⎨⎪

⎩⎪

1

0

for

for (11.3.1)

where C is a normalization constant and a is the variational parameter. (b) Find the extremum value of a. Compare this value with the Bohr radius a0.


SOLUTION

(a) First we compute the normalization constant. This gives C2 315= /πα . The kinetic energy is given by

⟨ ⟩ = −⎡

⎣⎢⎢

⎤

⎦⎥⎥

Em

r rr

d r

drdrk

22

122

2

20

πα

αα

� φφ

( )( )∫∫ (11.3.2)


⟨ ⟩ = − ⎛⎝⎜

⎞⎠⎟

+Em

r rd r

dr mddr

rk

π πα

αα

� �2

0

2

φφ

φ( )( )

( αα

α

( ))r dr⎡⎣⎢

⎤⎦⎥∫

2

0

(11.3.3)

But since ( ( )) ( ( )) ,r r r rr rφ φα α α| |= == =0 0 the first term vanishes and we have

⟨ ⟩ = ⎡⎣⎢

⎤⎦⎥

=∫ −Em

ddr

r r drmk

π αα

α� �

2 2

0

2315

21( ( ))φ −−⎛

⎝⎜⎞⎠⎟

= − +⎡⎣⎢

⎤⎦⎥

=

∫−

2

152

43

5

2

0

23

rdr

m

α

α α α α

α

� �22

21

m α

(11.3.4)

The potential energy is

⟨ ⟩ = =

∞

∫V r r V r r dr ke r r d2 22

0

2 2π πα α αφ φ φ( ) ( ) ( ) ( )| | rr

ke rr r

drke

0

22 3

2

2

0

302 15

61

α

α

α α α

∫= − − +

⎛

⎝⎜⎞

⎠⎟= −∫∫

(11.3.5)

Thus, the total energy as a function of a is given by

⟨ ⟩ = ⟨ ⟩ + ⟨ ⟩ = −⎛

⎝⎜⎞

⎠⎟E E V

mke

k( )αα α51

212

2

2� (11.3.6)

(b) The extremum condition d ⟨E⟩ / da = 0 leads to

2 12

42

03 2

02

0

2

2� �m

kekme

α α α− −= ⇒ = (11.3.7)

Note that the Bohr radius is a kme02 2= � / ; thus, a0 = 4a0.

11.4. (a) Write the Schrödinger equation for the helium atom. What are the solutions for the ground state if one neglects the interaction between the two electrons? (b) Assume that the electrons perform an electric screening of each other and define Z as a variational parameter. Use the variation method to find ⟨ ⟩H and the screening charge.

SOLUTION

(a) We begin by considering the classical Hamiltonian of the helium atom:

Hpm

pm

Zer r

er

= + − +⎛⎝⎜

⎞⎠⎟

+12

22

2

1 2

2

122 21 1

(11.4.1)

where r12 = |r1 − r2|. We transform the Hamiltonian to units in which e m= = =� 1. In these units the Schrödinger equation becomes

− ∇ − ∇ − +⎛⎝⎜

⎞⎠⎟

+⎡

⎣⎢

⎤

⎦⎥

12

12

1 1 112

22

1 2 121Z

r r rrψ( , rr E r r2 1 2) ( , )= ψ (11.4.2)


If one neglects the term e2/r12 (the interaction term), the solutions are obtained by separation of variables:

ψ0 1 2 1 1 2 2

3 2

1 2

3 2

11( , ) ( ) ( )

/

r r u r u rZ

eZZ r= = −

π π/

/

/22

32 1 2e

Ze

Z r Z r r− − += π( )

(11.4.3)

Note that the factors Z

eZ

3 2

1 21

/

/π− τ

and Z

eZ

3 2

1 22

/

/π− τ

are the ground-state functions of a hydrogen-like atom.

(b) In the presence of another electron, each of the electrons is influenced by a decreased charge from the nucleus. We define Zeff = Z − s, where s is the screening charge. We choose the trial function to be

ψσσ

π σ( , )( )

exp[ ( )( )]r rZ

Z r r1 2

3

1 2= − − − + (11.4.4)

Since

ˆ ( )H Zr r r

ψ ψσ σσ σ σ= − − − − +⎡⎣⎢

⎤⎦⎥

2

1 2 12

1 (11.4.5)

then

⟨ ⟩ = = − −∫∫∫∫

ˆˆ

(

*

*

H

H d r d r

d r d r

Z

ψ ψ

ψ ψ

σ σ

σ σ

σ

31

32

31

32

))

( )

2

1 2 12

2 31

32

2

1− − +⎡⎣⎢

⎤⎦⎥

= − −

∫∫ σ σ

σ

σr r rd r d r

Z

ψ

−− −

+ −

− −∞

−

∫2 43

2

0

6

2

2

( )

( )

( )

(

Zr

r e dr

Z e

Z r

Z

σπ

σ π

σπ

σ

−− +

∫∫σ )( )r r

rd r d r

1 2

12

31

32

(11.4.6)

or,

⟨ ⟩ = − − − −−

+ − −ˆ ( )

( )

( )

( )(

H ZZ

Z

Z eZ

σ σπ

πσσ

σπ

23

2

6

2

22

−− +

∫∫σ )( )r r

rd r d r

1 2

12

31

32 (11.4.7)

We solve the last integral using the expansion of 1/r12 by Legendre polynomials (see the Mathematical Appendix):

1

10

112

1

1

20

1 2

1

r

rrr

P r r

rr

n

n

n

=

⎛⎝⎜

⎞⎠⎟

≤ ≤=

∞

∑ (cos )θ

22

10

2 1rP r r

n

n

n

⎛⎝⎜

⎞⎠⎟

≤ < ∞

⎧

⎨

⎪⎪⎪

⎩

⎪⎪⎪ =

∞

∑ (cos )θ

(11.4.8)

The only terms that contribute to the integral are the ones with n = 0 (since the exponent that enters the integral depends only on the values of r1 and r2, and not on the angle q); thus,

er

d r d r er

Z r rZ r

− − +− −∫∫ =

2

12

31

32

21 2

2 41

( ) ( )( )

σσ π

2212 2

10

12

11

2

2

2

r e dr r e drZ r

rZ r

r

− − − −∞

∫ ∫+⎛

( ) ( )σ σ

⎝⎝⎜⎜

⎞

⎠⎟⎟

⎡

⎣

⎢⎢

⎤

⎦

⎥⎥

=−

∫ d r

Z

32

2

55

8

πσ( )

(11.4.9)


Thus, the expression for ⟨ ⟩H is

⟨ ⟩ = − − − − + −ˆ ( ) ( ) ( )H Z Z Zσ σ σ σ2 258 (11.4.10)

Using the condition d Hd⟨ ⟩ =

ˆ

σ 0, we find σ0

516

= , and then Z eeff = 2716

.

11.5. Consider a one-dimensional attraction potential V(x) such that V(x) < 0 for all x. Using the variational principle, show that such a potential has at least one bound state.

SOLUTION

For a particle moving in this potential we may choose the following trial wavefunction:

ψ = −24 2aaxπ exp ( ) (11.5.1)

Note that the function is normalized to unity. Thus, for the ground-state energy we have

Em

d

dxV x dx0

2 2

22≤ − +

⎛

⎝⎜⎞

⎠⎟−∞

∞

∫ ψ ψ* ( )�

(11.5.2)

Since V(x) < 0 for all x, it remains to prove that E0 < 0. Substituting the trial function, we find

Ea

axm

d

dxV x ax0

22 2

22

2≤ − − +

⎛

⎝⎜⎞

⎠⎟−π exp ( ) ( ) exp (

� 22

22 2

22

2

)

exp ( ) exp (

dx

aax

md

dx

−∞

∞

∫= − −

⎛

⎝⎜⎞

⎠⎟−π

�aax V x ax dx

a am

2 2

2

2

22

) ( ) exp ( )+ −⎡

⎣⎢⎢

⎤

⎦⎥⎥

=

−∞

∞

∫π

�[[ ] exp ( ) ( ) exp( )1 2 2 22 2 2− − + −

⎡

⎣⎢⎢

⎤

⎦⎥⎥−

ax ax V x ax∞∞

∞

∫= − − +

dx

am

am

aax ax dx

aV

� �2 2

2 2

2 22

2 22

π πexp ( ) (xx ax dx) exp ( )−−∞

∞

−∞

∞

∫∫ 2 2

(11.5.3)

We define

′ = + −−∞

∞

∫Ea

ma

V x ax dx0

22

22

2�

π ( ) exp ( ) (11.5.4)

Thus, since the integral 2

2 22 2aax ax dxπ exp ( )−

−∞

∞

∫ has a positive value, E0 < E0′. Consider now the

minimum value of E0′:

∂ ′∂ = + − −

−∞

∞

∫Ea m a

V x ax dxa

x02

2 2

21

22

22

�π π( ) exp( ) VV x ax dx( ) exp ( )

−∞

∞

∫ − =2 02 (11.5.5)

Combining Eqs. (11.5.4) and (11.5.5) we obtain

( ) exp( ) ) ( )min′ = − +−∞

∞

∫Ea

ax ax V x dx02 22

2π (1 4 (11.5.6)

since exp (−2ax2) and (1 + 4ax2) are positive functions and V(x) is a negative function for all x, ( )min′ <E0 0 and so is E0.


11.6. Consider a particle in a one-dimensional potential V(x) = l x4. Using the variational method, find an

approximate value for the energy of the ground state. Compare it to the exact value Em

k0

21 31 06

2= . ,

� /

where k m= 2 2λ /� . Choose as a trial function ψ = −( ) ./2 1 4 2

α π α/ e x

SOLUTION

First, note that the trial function ψ( ) ( ) /x e x= −2 1 4 2

α π α/ is normalized to unity; that is, | |ψ 2 1dx =−∞

∞

∫ . The

Hamiltonian is ˆ ;Hm

d

dxx= − +�

2 2

24

2λ

thus,

⟨ ⟩ =− +

⎛

⎝⎜⎞

⎠⎟−∞

∞

∫ˆ

( ) ( )

(

*

*

H

xm

d

dxx x dxψ ψ

ψ

�2 2

24

2λ

xx x dx) ( )ψ−∞

∞

∫ (11.6.1)

The denominator equals one as [ψ(x) is normalized]; thus,

⟨ ⟩ = ⎛⎝⎜

⎞⎠⎟ −

⎛

⎝⎜⎞

⎠⎟−

−∞

∞

∫ˆ/

H em

d

dxx2

221 4 2 2

2

2απ

α � ααπ

απ λα α⎛

⎝⎜⎞⎠⎟ + ⎛

⎝⎜⎞⎠⎟

−

−∞

∞−∫

1 4 1 42 22/

e dx e xx x

/44

1 4

22 2

2

22

2 2

2

2

απ

απ α α

α

α

⎛⎝⎜

⎞⎠⎟

= −

−

−

/

e dx

me x

x

x�[ −− +

= −

−

−∞

∞

−∞

∞

−

∫∫ 12

22

4

2 4

22

2

]dx e x dx

me

xλ απ

απ α

α

� 22 22

2 22 2 2

22

22α α αα

π α λ απ

x x xx dxm

e dx e x−∞

∞− −∫ + +� 44dx

−∞

∞

−∞

∞

∫∫

(11.6.2)

The first integral is

Im

e x dxm

x1

22 2 2

22

22

42

24

14 2

2

≡ − = − =−� �απ α α

π α απα

α −−−∞

∞

∫ �2

2αm (11.6.3)

The second integral is

Im

e dxm m

x2

22

2 2

22

22

22

2

2

≡ = =−

−∞

∞

∫� � �απ α α

π α πα

αα (11.6.4)

and the third integral is

I e x dxax3

2 42 2

2 2 3

4 2 23

16

2

= = =−

−∞

∞

∫λ απ λ α

π απα

λα( )

(11.6.5)

Substituting these integrals, we obtain

⟨ ⟩ = − + + = +Hm m m

� � �2 2

2

2

223

16 23

16α α λ

αα λ

α (11.6.6)

Hence, d Hd m⟨ ⟩ = −

ˆ.α

λα

�2

032

38

Since d Hd⟨ ⟩ =

=

ˆ,α

α α0

0 we obtain

��

2

03 0 2

1 3

238 0 3

4mm− = → = ⎛

⎝⎜⎞⎠⎟

λα

α λ/


In terms of k m= 2 2λ /� we have α01 3 1 33 8= ( / ) ./ /k Substituting this value into Eq. (11.6.6), we obtain

⟨ ⟩ = =ˆ .minHm

km

k34

32

1 0822

1 32

1 32

1 3/ / /� � (11.6.7)

Comparing the last result to the exact value of E0, we see that we have quite a good approximation. The error is approximately 2 percent.

11.7. Consider a particle moving in an arbitrary potential. Assuming that the potential V(r) satisfies the semiclassical condition, estimate the number of discrete energy levels that the particle can occupy.

SOLUTION

The number of states that belong to a volume V in the phase space, and that correspond to a momentum in the

range 0 ≤ ≤p pmax, and that correspond to particle coordinates in the volume dV equals 43 2

33π

πp

dVmax ( )

.�

For

fixed r, the particle may assume a momentum that satisfies the condition E p m V= + ≤2 2 0/ ( ) .r Thus, the

maximal momentum is p mVmax ( ).= −2 r Substituting pmax, we obtain the number of states in volume dV:

dN mVdV= −4

32

23 2

3ππ

[ ( )]( )

/r�

(11.7.1)

So, the total number of states of the discrete spectrum is

Nm

V d r= −∫2

3

3 2

2 33 2 3

//[ ( )]

π �r (11.7.2)

The integration is carried over the region of space where V ( ) .r < 0 Note that the integral diverges if V(r) decreases as r n− , where n < 2.

11.8. (a) Find the condition for applicability of the WKB approximation to the case of the attracting Coulomb potential. (b) What are the implications of this condition for the Bohr model of the hydrogen atom?

SOLUTION

(a) We may write the applicability condition in the form

| |dxdp

p>> �

2 2 (11.8.1)

Omitting the factor 1/2, we obtain

�

p

dpdx2 1<< (11.8.2)

Note that

dpdx

ddx

m E Vmp

dVdx

mFp

= − = − =2 ( ) (11.8.3)

where FdV x

dx= − ( )

is the classical force. Substituting Eq. (11.8.3) into Eq. (11.8.2), we obtain the

following condition:

m F

p

� | |3 1<< (11.8.4)

For the attracting Coulomb field F r= −α / 2, we can roughly estimate the momentum by writing

p m Vmr

~ ~2 | |α

(11.8.5)


Thus, Eq. (11.8.4) becomes m r

m r

r

m

� �( )/ / /

/

/ /α

α α/

/

2

3 2 3 2 3 2

1 2

1 2 1 2 1= <<−

and finally,

rm

>> �2

α (11.8.6)

(b) The Bohr radius of a hydrogen atom is given by a mBohr /= �2 α; thus, Eq. (11.8.6) becomes r a>> Bohr .

For the Bohr model we know that the nth-level distance of an electron from a proton is given by r n an = 2

Bohr , and so the WKB approximation is applicable for the levels n >> 1.

11.9. Using the WKB approximation find the bounded states for a one-dimensional infinite potential well. Compare your results with the exact solution.

SOLUTION

Suppose that the boundaries of the potential well are at x a= ± . At the boundaries the wavefunction has a value of zero. From Eqs. (11.15) and (11.16) we have

00

1

2

= −= −

⎧⎨⎩

cos ( )cos ( )

B

B

ππ

(11.9.1)

and therefore, B B1 2 1 2= = / . Thus we get, according to Eq. (11.11),

k x dx ak nna

a

n( ) ( )′ ′ = = +−

+

∫ 2 1 �π (11.9.2)

We get

Ek

mn

mann= = +1

21

8

2 2 2 2 2

2

� �π ( ) (11.9.3)

Recall that the exact result is En

man = π 2 2 2

28

�.

11.10. Use a WKB approximation to obtain the energy levels of a linear harmonic oscillator.

SOLUTION

Consider the Bohr-Sommerfeld quantization rule:

p x dx n na

b

( ) ( ) ( , , , . . .)∫ = + =�π 1 2 0 1 2/ (11.10.1)

where p x m E V x( ) [ ( )]= −2 is the momentum of the oscillator, E is its energy, and V(x) is its potential

energy. Since ○ p dx p dxa

b

∫ ∫= 2 holds for a linear harmonic oscillator, we may write the Bohr–Sommerfeld

quantization rule in the form of Eq. (11.10.1). For the harmonic oscillator we have V m x= 12

2 2ω . The

points a and b are the turning points that are determined by the condition p a p b( ) ( )= = 0 or E V− = 0;

thus, E m x− =12

02 2ω . So, we have

aE

mb

E

m= − =2 2

2 2ω ω (11.10.2)

We introduce the new variable z xm

E= ω 2

2, and obtain

p x dxE

z dzE

a

b

( )∫ ∫= − =−

21 2

1

1

ωπω (11.10.3)


Comparing this result to Eq. (11.10.1), we obtain

E nn = +⎛⎝⎜

⎞⎠⎟�ω 1

2 (11.10.4)

Thus, in the case of the semiclassical approximation the result is identical to the exact solution.

11.11. Using the semiclassical approximation, calculate the transmission coefficient of a potential barrier

V xV

x

aa x a

( ) = −⎛

⎝⎜⎞

⎠⎟− ≤ ≤

⎧

⎨⎪

⎩⎪

0

2

21

0 otherwise

(11.11.1)

See Fig. 11.3.

SOLUTION

Let E be the energy of the particle and m its mass. The transmission coefficient in the semiclassical approximation is given by

T m V x E dxx

x

≈ − −⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪∫exp [ ( ) ]2

21

2

� (11.11.2)

where x1 and x2 are the turning points computed using the condition V(x) = E. Hence,

x aEV

x aEV1

02

01 1= − − = + − (11.11.3)

Thus, Eq. (11.11.2) becomes

T m Vx

aE

a E V

≈ − −⎛

⎝⎜⎞

⎠⎟−

⎡

⎣⎢⎢

⎤

⎦⎥⎥− −

exp[ ( /

22 10

2

21

�00

1 2

01 21

)]

[ ( / )]

/

/+ −

∫⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪

a E V

dx (11.11.4)

Computing the integral gives

Tm

Va V E

≈ −−⎡

⎣⎢

⎤

⎦⎥exp

( )π 2

0

0�

(11.11.5)

Note that the expression of T is valid if the exponent in Eq. (11.11.5) is large; that is,

π 21

0

0mV

a V E( )−>>

� (11.11.6)

0 x2

V0

E

x1–a xa

V(x)

Fig. 11.3


11.12. The limit � → 0 corresponds to the transition from quantum mechanics to classical mechanics. Assume that the wavefuncton can be written in the form ψ( , ) ,( , ) /r rt eiS t= � and that the system is in a stationary state, i.e., we can write S t Et( , ) ( ) .r r= −σ Derive the following conditions for the applicability of the semiclassical approximation: (a) ( )∇ >> ∇σ σ2 2

�| | and p2 >> ∇�| |i p ; (b) in a particular case of

one-dimensional motion, λ λ λπ>> d

dx2, where λ is the wavelength according to the de Broglie relation

λ = h p/ . (c) p mdVdx

3 >> � .

SOLUTION

(a) We begin by substituting the wavefunction Ψ( , ) /r t eiS= � into the Schrödinger equation:

− ∇ + = ∂∂

��

22

2mV i

tΨ Ψ Ψ

( )r (11.12.1)

Hence,

1

2 22

mS S

im

S VSt

( ) ( )∇ ∇ − ∇ + = − ∂∂i

�r (11.12.2)

Using the assumption that the system is in a stationary state, we substitute S t Et( , ) ( ) ,r r= −σ and arrive at

1

2 22 2

mim

V E( ) ( )∇ − ∇ + =σ σ�r (11.12.3)

To achieve the transition from quantum mechanics to classical mechanics we must take the limit

� → 0; then, the term − ∇im�

22σ in Eq. (11.12.3) can be neglected, and we obtain

1

22

mV E( ) ( )∇ + =σ r (11.12.4)

This can be considered an equation of classical mechanics, provided that ∇ =σ0 p. However, the essence of the semiclassical approach is to arrive at equations that lead to classical mechanics equa-tions, even for purely quantum systems where the transition � → 0 is not justified. Looking again at Eq. (11.12.3) we note that the transition from Eq. (11.12.3) to Eq. (11.12.4) can be achieved not only by taking the limit � → 0, but also by assuming that

( )∇ >> ∇σ σ02 2

0� | | (11.12.5)

Therefore, Eq. (11.12.5) is a condition for the applicability of the semiclassical approximation. Equa-tion (11.12.5) can be rewritten as ( ) :p = ∇σ

p2 >> ∇� | |i p (11.12.6)

(b) In the case of one-dimensional motion, ∇ =i pdpdx

; using Eq. (11.12.6) we have

1 2>> � | |dp dx

p

/ (11.12.7)

Differentiating the de Broglie relation λ = h p/ with respect to x, we obtain

ddx

hdp

p dx

dp

p dx

λ π= =2 22 � (11.12.8)

Then, according to Eq. (11.12.7), we have d

dxλ

π21<< , from which it follows that

λ λ λπ>> d

dx2 (11.12.9)

The condition specified by Eq. (11.12.9) can be interpreted as follows: along the distance of λ π/2 the change in the wavelength must be much less than the wavelength itself.


(c) From classical mechanics we know that p m E V= −2 ( ). Thus,

dpdx

dpdV

dVdx

m

m E V

dVdx

mp

dVdx

= = −−

= −2 ( )

(11.12.10)

so dpdx

mp

dVdx

= . Substituting into Eq. (11.12.7), we obtain

p mdVdx

3 >> � (11.12.11)

11.13. Using the WKB approximation, derive the Bohr–Sommerfeld quantization rule.

SOLUTION

Consider a one-dimensional case where E V x> [ ( )]min (see Fig. 11.4).

V(x)

0 x1ax2 x3bx1 x

E

Fig. 11.4

For any value of E there are only two turning points V a V b E( ) ( ) .= = The oscillating solution between two turning points is

ψosc = ′ ′ +⎡

⎣⎢⎢

⎤

⎦⎥⎥∫C

pk x dx

a

x

sin ( ) β (11.13.1)

where C and b are constants. In small vicinities x a x x b x1 2 3 4≤ ≤ ≤ ≤, ; the WKB approximation is not applicable where the wavefunctions in these vicinities are given by

ψ

χ

a

x

a

x

A

px dx x

A

p

( )

exp ( )

s

=

− ′ ′⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪∫2 1| |at

iin ( )k x dx xa

x

′ ′⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪

⎧

⎨

⎪⎪⎪

⎩

⎪⎪⎪ ∫ at 2

(11.13.2)

and

ψ

χ

b

b

x

x

B

px dx x

B

p

( )

exp ( )

s

=

− ′ ′⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪∫2 4| |at

iin ( )k x dx xx

b

′ ′ +⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪

⎧

⎨

⎪⎪⎪

⎩

⎪⎪⎪ ∫ π

4 3at

(11.13.3)

where k x m E V x( ) [ ( )]= −12

� and χ( ) [ ( ) ].x m V x E= −1

2�

We require a smooth transition from the oscil-

lating solution to the solutions in the vicinities of a and b and so the following conditions must be satisfied:

B Cn= − +( )1 1 (11.13.4)


k x dx nb

a

( )′ ′ = +∫ π π2

(11.13.5)

where n = 0 1 2, , , . . . . Recall that p k= � and introduces the loop integral

○ p x dx p x dxa

b

( ) ( )=∫ ∫2 (11.13.6)

This integral can be interpreted as integrating along the line from a to b and then back from b to a. Thus, substituting in Eq. (11.13.5), we arrive at

○ p x dx n n( ) , , , . . .= +⎛⎝⎜

⎞⎠⎟

⎞⎠⎟

=⎛

⎝⎜∫ 2

12

0 1 2π� (11.13.7)

which is the Bohr–Sommerfeld quantization rule.

11.14. Use the semiclassical approximation in order to find the radial part of the wavefunction for a particle moving in a central potential field.

SOLUTION

From the theory of a particle in a central potential, we know that the radial part of the corresponding wave-function can be written in the form R r u r r( ) ( ) ,= / where u(r) satisfies the following equation:

du r

dr

mE V r

l l

ru r

2

2 2 22 1

0( )

( ( ))( )

( )+ − − +⎡⎣⎢

⎤⎦⎥

=�

(11.14.1)

We will write u(r) in the form

u r C r iS r

( ) ( ) exp( )= ⎛

⎝⎜⎞⎠⎟

⎡⎣⎢

⎤⎦⎥�

(11.14.2)

where C(r) and S(r) are real functions. Substituting Eq. (11.14.2) into Eq. (11.14.1) we obtain

d C r

dri

S i dC rdr

dS rd

2

2( )

exp( ) ( )

� �⎛⎝⎜

⎞⎠⎟

⎡⎣⎢

⎤⎦⎥

+rr

iS dC rdr

iS i dS rdr

exp( )

exp( )

� � �⎛⎝⎜

⎞⎠⎟ + ⎛

⎝⎜⎞⎠⎟

⎛⎛⎝⎜

⎞⎠⎟

+ ⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟ +C r

iS i dS rdr

C( ) exp( )

(� �

2

rriS i d S r

dr

mE V r

) exp( )

( (

� �

�

⎛⎝⎜

⎞⎠⎟

⎛

⎝⎜⎞

⎠⎟

+ −

2

2

22

)))( )

( ) exp− +⎡⎣⎢

⎤⎦⎥

⎛⎝⎜

⎞⎠⎟ =l l

rC r

iS102 �

(11.14.3)

Setting the real and imaginary parts of the left-hand side of Eq. (11.14.3) to zero we arrive at

2 02dC r

drdS r

drC r

d S rdr

( ) ( )( )

( )+ = (11.14.4)

and

dS r

dr C rd C r

drm E V r

( )( )

( )[ ( )]⎛

⎝⎜⎞⎠⎟ − = − −

2 2 2

2 2� �

22

21l l

r

( )+ (11.14.5)

Integrating Eq. (11.14.4), we obtain C rdS r

dr( ) ( .)

( ).

/

= × ⎛⎝⎜

⎞⎠⎟

−

const1 2

Since �2 is assumed to be a small quantity,

we can solve Eq. (11.14.5) approximately. For small values of r, when the dominant term on the right-hand side of

Eq. (11.14.5) is �

2

21l l

r

( ),

+ we have

dS rdr

i l lr

( ) ( )≈

+� 1 with C r r( ) ~ , and we arrive at the approximation

� �

2 2

2

2

24C rd C r

dr r( )( ) ≈ (11.14.6)


We now substitute Eq. (11.14.6) into Eq. (11.14.5) and thereby obtain a better approximation:

S r m E V rl

rdr( ) [ ( )]

( )= − − +∫ 21 22 2

2� /

(11.14.7)

and

C r

m E V rl

r

( ).

[ ( )]( )

=

− − +

const

/4 2

1 22 2

2�

(11.14.8)

Substituting Eqs. (11.14.7) and (11.14.8) into Eq. (11.14.2) we obtain u(r), and then R(r) = u(r)/r.


11.15. Using the trial function ψ = −N rexp ( ),α 2 compute a variational upper limit for the ground state of a hydrogen atom and compare it to the exact value.

Ans. ⟨ ⟩ ≈ −ˆ .H 11 5 eV. The exact value is −13.6 eV.

11.16. Using the variational method, compute the energy of the ground state of a hydrogen atom. Use the following

trial functions: (a) ψ1 10= −

A ebr a/

, (b) ψ1 22

2

02

1

= +⎛

⎝⎜

⎞

⎠⎟

−

A br

a, and (c) ψ3 3

0

0= −A

ra

ebr a/

, where a0 is the Bohr

radius. Compare your results with the exact result and discuss the causes for the differences. Hint: Compare the behavior of ψ ψ1 2, , and ψ3 with the true wavefunction.

Ans. (a) b Hea

EH= ⟨ ⟩ = − = −12

2

0, ˆ ;min (b) b H EH= ⟨ ⟩ = −π

40 81, ˆ . ;min (c) b H E= ⟨ ⟩ = −3

20 75, ˆ . ;min H

where EH is the energy of the ground state of a hydrogen atom.

11.17. Using variational calculus, give an estimate for the binding energy of the deuteron. Assume that the potential between a proton and a neutron is V r Ae

r r( ) ,

/= − 0 and use ψ( )r Ce r= −β as a trial function, where A and C are

normalizaton constants and r0 is a characteristic length of the potential.

Ans. E = −2.1 MeV.

11.18. Show that for motion in a central field, the condition for applicability of the WKB approximation is l >> 1, where l is an angular momentum quantum number. Explain why the term “semiclassical approximation” is justified in this case.

Ans. Since an angular momentum equals L l= �, we obtain relatively large values of an angular momentum, so L is “almost classical.”

11.19. Consider the Hamiltonian of a nonharmonic oscillator ˆ .Hd

dxx x= − + +

2

22 4 Use the WKB approximation to

find the ground state for x → ∞.

Ans. ψ ~ exp1

3

3

xx

x±⎛

⎝⎜⎞

⎠⎟→ ∞as

11.20. Use the WKB approximation to compute the transmission coefficient of an electron going through the potential barrier depicted in Fig. 11.5.

V x V k x x V k( ) = − <⎧⎨⎪

⎩⎪0

2 20

12

2

0

/

otherwise (11.20.1)

Ans. T m V kx E dxV E k

V

= − − −⎛⎝⎜

⎞⎠⎟

− +

+

exp( ) /

(2

2120

2

2

2

0

0

�

EE k) /

∫⎡

⎣⎢⎢

⎤

⎦⎥⎥


0 x

V0

V(x)

V(x) = V0 – Dx

E

Fig. 11.6

11.21. Use the WKB approximation to find the transmission coefficient for the potential

V xx

V kx x( ) =<

− >⎧⎨⎩

0 000

(11.21.1)

0 x2V0

V(x)

V0

E

–k

2V0k

Fig. 11.5

where V0 and k are constant.

Ans. (a) T m E V k x dxV E k

= − − +⎡

⎣⎢⎢

⎤

⎦⎥⎥

=−

∫exp ( ) exp( ) /

22 0

0

0

�−− −

⎡

⎣⎢

⎤

⎦⎥

4 23 0

3 2mk

V E�

( ) /

11.22. What is the probability of a particle with zero angular momentum escaping from a central potential

V rV r a

rr a

( ) =− <

>

⎧⎨⎪

⎩⎪

0

α (11.22.1)

Ans. P mr

E dra

E

= − −⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢⎢

⎤

⎦⎥⎥

= −∫exp exp/

22

2� �

α αα22

11mE

Ea Ea Eacos− ⎛

⎝⎜⎞⎠⎟

− −⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢⎢

⎤

⎦⎥⎥

⎧⎨ α α α⎪⎪

⎩⎪

⎫⎬⎪

⎭⎪

249

CHAPTER 12

Numerical Methods in Quantum Mechanics

12.1 Numerical QuadratureThe numerical quadrature of the definite integral of a function f (x) between two limits a and b is accom-plished by dividing the interval [a, b] into N small intervals, between N + 1 points denoted by

a x x x bN= =0 1, , . . . , (12.1)

The points xi are equally spaced using a constant step h = (b − a)/N:

x x ih i Ni = + =0 0 1, , . . . , (12.2)

The basic idea behind quadrature is to write the integrals as the sum of integrals over small intervals:

f x dx f x dx f x dx fa

b

a

a h

a h

a h

( ) ( ) ( ) (∫ ∫ ∫= + + ⋅ ⋅ ⋅ ++

+

+2

xx dxb h

b

)−∫ (12.3)

and in these small intervals, approximate f (x) by a function that can be integrated exactly. We will dem-onstrate two methods of quadrature. The first method is called the trapezoidal method; it is based on the approximation of f (x) to a linear function, as shown in Fig. 12.1.

x0 x1 x2 xxN0

f (x)

Fig. 12.1

250

In this case, the integral f x dx f x f xh

i ix

x

i

i

( ) [ ( ) ( )] ,= ++

+

∫ 1 2

1

so if we denote f (xi) = fi ,we obtain

f x dx h f f f f fN Na

b

( ) ≈ + + + ⋅ ⋅ ⋅ + +⎡⎣⎢

⎤⎦⎥−∫ 1

2120 1 2 1 (12.4)

The second method is called Simpson’s method. It is based on the approximation of f (x) via a second-degree

polynomial at three points. In this case, the integral f x dx h f f fi i ix

x

i

i

( ) ,≈ + +⎡⎣⎢

⎤⎦⎥+ +

+

∫ 13

43

131 2

2

so

f x dx f x dx f x dxa

b

a

a h

a h

a h

( ) ( ) ( )= + + ⋅ ⋅ ⋅∫ ∫ ∫+

+

+2

2

4

++ ≈ + + + + ⋅ ⋅ ⋅ +−∫ f x dx

hf f f f fN

b h

b

( ) [ ]3

4 2 40 1 2 32

(12.5)

One should be aware of the fact that these methods are only an approximation of the exact integral. The approximation is improved as we consider larger N. In the trapezoidal method, the approximation error is proportional to 1/N 2, while in Simpson’s method, it is proportional to 1/N 4; i.e., in general, the Simpson method is more accurate than the trapezoidal method.

12.2 RootsIn order to determine the roots of a function f (x) we must solve the equation f (x) = 0. All numerical methods for finding roots depend on one or more initial guesses. In each algorithm, the root is approximated after a given number of iterations. Note that by initial guess we do not necessarily mean a close guess for the root, though the better the guess is, the faster the convergence will be (and less iterations will be needed). Thus, to obtain an efficient initial guess for a given root for the function f (x), it is helpful to first plot the function.

We describe three methods for finding roots. The first is called the bisection method. This method is useful when we know that the root is found in a specific interval, say, [x1, x2], as shown in Fig. 12.2.

CHAPTER 12 Numerical Methods in Quantum Mechanics

xx1 x20

f (x)

x0

Fig. 12.2

We know that the signs of f (x1) and f (x2) are opposite. In the first iteration we evaluate f (x) at the midpoint between x1 and x2; then we use the midpoint to replace the limit with the same sign. In each successive itera-tion, the interval containing the root gets smaller by a factor of 1/2, so the maximal error in our estimation (if we assume that the midpoint is the root we are searching for) is simply half of the interval between the new limits x1 and x2. Thus, we need n = log2 (e0 /e) iterations to obtain the root with a maximal error of e / 2. Note that e0 is the initial interval, ε0 2 1= −| |x x . The bisection method will always converge if the initial interval [x1, x2] contains a root (or singularity points).

251

The second algorithm, the Newton–Raphson method, uses the derivative f ′(x) at an arbitrary point x. We begin with an initial guess x1. Each new approximation for the root depends on the previous one:

x xf x

f xi i

i

i+ = −

′1 ( )

( ) (12.6)

We stop when the value of | |x xi i+ −1 is less than the tolerance we have selected. To understand how the method works, we write Eq. (12.6) in the form

f x f x x xi i i i( ) ( )( )+ ′ − =+1 0 (12.7)

Notice that the left-hand side of Eq. (12.7) is a linear extrapolation to the value of f (x i + 1), which should be zero.

The third method, called the secant method, is similar to the Newton–Raphson method. Here we do not evaluate the derivative but use the approximation,

′ ≈ ′ − ′−

−

−f xf x f x

x xi

i i

i i( )( ) ( )1

1 (12.8)

Hence we obtain

x xx x

f x f xf xi i

i i

i ii+

−

−= − −−

11

1( ) ( )( ) (12.9)

12.3 Integration of Ordinary Differential EquationsSolving differential equations is of paramount importance in physics. Many key results are formulated in terms of differential equations. We introduce several methods for solving differential equations of the form

dydx

f x y= ( , ) (12.10)

The methods differ in their accuracy, and in the time it takes to obtain the required accuracy. One should decide which method to use according to these criteria. Note that higher-order differential equations such as

d y

dxF x y

2

2 = ( , ) (12.11)

can be written as

dzdx

F x y zdydx

= =( , ) (12.12)

Thus, they can be solved using the same methods.The first method, the Euler method, is the simplest and least accurate method. Write Eq. (12.10) approxi-

mately as a difference equation:

ΔΔ

yx

f x y= ( , ) (12.13)

or

Δ Δy f x y x= ( , ) (12.14)

Iterate the value of y(x) from a starting point y0 = y(x0) by

y y f x y x xn n n n n n+ += + −1 1( , ) ( ) (12.15)

CHAPTER 12 Numerical Methods in Quantum Mechanics

CHAPTER 12 Numerical Methods in Quantum Mechanics252

then set Δ x = xn + 1 − xn = h (constant); thus,

y y f x y hn n n n+ = +1 ( , ) (12.16)

The point (xn + 1, yn + 1) depends only on the previous point (xn, yn). The accuracy of the iteration depends chiefly on the choice of h; a smaller h gives higher accuracy. The error in the approximation of yn + 1 is pro-portional to h2.

The second method, the Runge–Kutta method, is based on the Euler method using an approximation of f (x, y) by a given order to the Taylor series expansion. The higher the order of the Taylor series (i.e., the higher the order of the Runge–Kutta method), the better the accuracy. Consider the second-order Runge–Kutta method:

y y kn n+ = +1 2 (12.17)

where

k hf x y

k hf x h y k

n n

n n

1

2 12 2

=

= + +

⎧⎨⎪

⎩⎪

( , )

( / , )/ (12.18)

with an error proportional to h3. Similarly, the third order of the Runge–Kutta method is

y y k k kn n+ = + + +1 1 2 3

16

4( ) (12.19)

where

k hf x y

k hf x h y k

k hf x h

n n

n n

n

1

2 1

3

2 2

=

= + +

= +

( , )

( , )

(

/ /

,, )y k kn − +

⎧

⎨⎪⎪

⎩⎪⎪ 1 22

(12.20)

with an error proportional to h4. The fourth order of the Runge–Kutta method is

y y k k k kn n+ = + + + +1 1 2 3 4

16

2 2( ) (12.21)

where

k hf x y

k hf x h y k

k hf x h

n n

n n

n

1

2 1

3

2 2

=

= + +

= +

( , )

( , )

(

/ /

// /2 22

4 3

, )

( , )

y k

k hf x h y k

n

n n

+

= + +

⎧

⎨

⎪⎪⎪

⎩

⎪⎪⎪

(12.22)

with an error proportional to h5, and so on.The Schrödinger equation is a second-order differential equation. Thus, the methods described above need,

as an initial condition, the value of the wavefunction and its derivative at a given point. Since the value of the derivative of the wavefunction is usually not given, we are left only with the value of the wavefunction at two points (the boundaries). We demonstrate here an algorithm to solve second-order differential equations with two boundary conditions—the Numerov algorithm.

Numerov’s method is used to solve a differential equation of the form

d y

dxk x y S x

2

22+ =( ) ( ) (12.23)


We approximate the second derivative by the three-point difference formula:

y y y

hy

hyn n n

n n+ −− +

= ′′ + ′′′′1 12

2212

(12.24)

where ′′yn and ′′′′yn are the second and fourth derivatives at point xn, respectively. Using Eq. (12.23) we arrive at

′′′′ = − +=

yd

dxk x y S xn

x xn

2

22[ ( ) ( )] (12.25)

Denoting k(xn) = kn and S(xn) = Sn yields

′′′′ = − − +⎡⎣

⎤⎦ ++ + − −y

hk y k y k yn n n n n n n

12

12 1

21

21

21 hh

S S Sn n n2 1 12[ ]+ −− + (12.26)

Substituting Eq. (12.26) into Eq. (12.23) we obtain

112

2 1512

12

12

1

22+

⎛

⎝⎜⎞

⎠⎟− −

⎛

⎝⎜⎞

⎠⎟++ +

hk y

hk yn n n n ++

⎛

⎝⎜⎞

⎠⎟= + +− − + −

hk y

hS S Sn n n n n

2

12

1

2

1 112 1210( ) (12.27)

where the error is proportional to h6. This error can be shown to be better than that for the fourth order of the Runge–Kutta method.

Comment: All the following programs were written in standard FORTRAN 77 and were compiled on an IBM AIX RS/6000 workstation. The precision used was the default precision REAL *4.

SOLVED PROBLEMS

12.1. Write a FORTRAN subroutine:

Subroutine Simpson (FUNC, N, A, B, S)INTEGER NREAL FUNC (0:10000), A, B, S

This program computes the value of the integral of FUNC from A to B, using N iterations of the Simpson method. FUNC(0 : N) is an array of N + 1 values of the integral at N + 1 points separated by h = (B − A)/N. The value of the integral is updated in S.

SOLUTION

Consider the Simpson rule and note that

Sh= + + + +3

[FUNC(0) 4*FUNC(1) 2*FUNC(2) 4*FUNC(3) �� + FUNC(N)] (12.1.1)

The summation is slightly different for an odd and even N. One way to perform this summation is as follows:

S = FUNC(0) + FUNC(N)Do loop i from 1 to N – 1if i is evenS = S + 2*FUNC(i)elseS = S + 4*FUNC(i)end ifend doS = S * (B − A)/(3*N)


This algorithm can be written in FORTRAN 77 as follows:

C** Subroutine to compute the value of a definite integral.Subroutine Simpson (func, n, a, b, s)integer nreal func (0:1000) a, b, ss = 0s = func(0) + func(n)do 1 i=1, n–1

C** (1– mod(i, 2)) equals 0 if i even and equals 1 if i odd.s = s+2*2** (1– mod(i, 2))*func (i)

1 continues = s*(b-a)/(3*n)returnend

12.2. Write a program to compute the integral

e dxx

a

b2∫ (12.2.1)

using the Simpson method. The program should use as input the boundaries a, b, and N described in Sec. 12.1. Use different values of N for a = 0 and b = 1 to obtain an accuracy of 1 × 10−2.

SOLUTION

Consider the following program:

Program Problem 12.2integer nreal func (0:1000), a, b, sreal x, h

C** Get the boundaries of the interval.write (*,*)’Enter the interval bounds a and b:’read (*,*) a,b

C** Prepare file of results.open (unit=1, file=’results.txt’)write (1,*)’The value of the integral of the function exp (x**2)’

10 format (’from’, f4.2, ’to’, f4.2)write (1, 10) a,bwrite (1,*)’N S The integral’

C** Get the number of points N. 2 write (*,*)’Enter the number of points N (0<N<1001):’

write (*,*)’Enter N<0 to stop’read (*,*) nif (n.gt.1000.or.n.lt.1) goto 3

C** The step value between points.h=(b-a)/n

C** Compute the value of the function on the N points.do 1 i=0, n

x = a+h*float(i) func(i)=exp (x*x)

1 continue

C** Compute the value of the integral.call Simpson (func, n, a, b, s)


C** Print results.write (1,*) n, swrite (*,*) n, sgoto 2

3 stopend

C** Subroutine to compute the value of a definite integral.Subroutine Simpson (func, n, a, b, s)integer nreal func (0:1000), a, b, ss=0.s=func(0) + func(n)do 1 i=1, n–1

C** (1-mod(i, 2)) equals 0 if i even and equals 1 if i odd.s=s+2*2** (1–mod(i,2)) *func(i)

1 continues=s*(b–a)/(3*n)returnend

Running this program gives the following:

N S The Integral N S The Integral10 1.347725272 100 1.45034778120 1.402942777 110 1.45145905030 1.422343731 120 1.45238649840 1.432232141 130 1.45317101550 1.438225508 140 1.45384490560 1.442246556 150 1.45442903070 1.445130706 200 1.45647740480 1.447300673 500 1.46017658790 1.448992372 1000 1.461413145

Notice that we used the subroutine that we have written in Problem 12.1. Using subroutines makes the program easier to read, though it often slows the program. The output results show the values of N and the corresponding values of S for A = 0 and B = 1. From these results we see that different N values correspond to different S values, though large values of N the value of S is more stable, i.e., the changes in its value are small. This is mostly true for well-behaved functions like the one we are dealing with in this problem.

12.3. Write two different FORTRAN programs to solve the equation cos x = x. Consider an accuracy of five digits after the decimal point. Use the bisection method with x1 = 0 and x2 = 1, and the Newton–Raphson method with x1 = 1.

SOLUTION

Consider the graph of the functions y = cos x and y = x shown in Fig. 12.3. The solution of cos x − x = 0 is the value of x, where y = cos x and y = x intersect. We conclude from Fig. 12.3 that this happens in the interval [0, 1]; hence, a good starting guess for the bisection method would be x1 = 0 and x2 = 1. For each iteration we will get a new value XM = (x1 + x2)/2, and we will compare it with the value of XM in the previous iteration XMOLD. If the difference between XM and XMOLD is consistently less than 1 × 10−5, then we have an accuracy of five digits. Consider one way to write the program:

Program Problem 12.3–1real x1, x2, xm, xmoldreal toler, f1, f2, fm, finteger iter

C** Initialize iterations number.iter=0


C** Initial guesses.x1=0.x2=1.xm=(x1+x2)/2.xmold=x1

C** Maximal error in the approximation.toler=0.00001

C** If the new iteration does not give the same result of the previous iterationC** within toler do the following:

do while (abs (xm–xmold).gt.toler)iter=iter+1

C** Evaluate the f(x) at the different points.f1=f(x1)f2=f(x2)fm=f(xm)if ((fm*f1).lt.0) then

C** If the sign of f(xm) is similar to that of f(x2) then:x2=xmelse

C** If the sign of f(xm) is similar to that of f(x1) then:xl=xmendif

C** Remember the result of the previous iteration.xmold=xm

C** new iteration:xm=(x1+x2)/2.end do

C** Print result.write (*,*) ’The zero of f(x) is:’ ,xm

10 format (’obtained after’, i3, ’iterations.’)write (*,10) iterstopend

x

y

y = x

y = cos(x)

Fig. 12.3


C** Function for which we want to find the zero.real function f(x)real xf=cos(x)–xreturnend

This gives the result:

The zero of f(x) is 0.7390823364 obtained after 16 iterations.

Similarly, for the Newton–Raphson method we need only one starting guess x1, and we will use the same criterion for stopping the iterations. Recall that x i + 1 = xi − f (xi) / f ′(xi). If f (xi) / f ′(xi) is less than the toler-ance, the iterations will stop.

Program Problem 12.3–2real x1real f1, df1, tolerinteger iter

C** Initialize iterations.iter=0


C** Initial guess.x1=1.

C** Evaluate the function f(x)=cos(x)-x and its derivative at x=x1:f1=cos(x1)–x1df1=–1.*sin(x1)–1.


do while (abs (f1/df1).gt.toler)

C** New iteration.iter=iter+1

C** Evaluate the function f(x)=cos(x)-x and its derivative at x=x1:x1=x1–f1/df1f1=cos(x1)–x1df1=–1.*sin(x1)–1end do

C** Print result:write (*,*) ’The zero of f(x) is:’ ,x1

10 format (’obtained after’, i3, ’iterations.’)write (*,10) iterstopend

This gives the result:


We see that both methods give the same result. The bisection method converged after 16 iterations. The Newton–Raphson method gave the result after only three iterations, confirming that this method converges faster than the bisection method. This is usually the case, though sometimes the Newton–Raphson method diverges, while the bisection method converges.


12.4. Find the lowest bound state energy for an electron moving under the potential

V x

V z a

z( ) =− ≤ ≤∞ <

⎧⎨⎪

⎩⎪

0 0

00 otherwise

(12.4.1)

where a = 2 Å and V0 = 10 eV (see Fig. 12.4).

SOLUTION

The Schrödinger equation for bound states, −V0 < E < 0, is (see Chap. 3)

ψ = <0 0for z (12.4.2)

− − = ≤ ≤2

�2

2 02md

dzV E z a

ψ ψ ψ for 0 (12.4.3)

− = <�2 2

22md

dzE a z

ψ ψ for (12.4.4)

Equation (12.4.3) yields d

dz

mE V

2

2 2 0

2ψ ψ= − +�

( ) with E + V0 > 0. Thus,

ψ( ) sin ( ) cos ( )z A k z B k z z a= + < <1 1 1 1 for 0 (12.4.5)

where k m E V1 022= +( ) ./� Similarly, Eq. (12.4.4) yields

d

dz

mE2

2 22ψ ψ= −�

. The solution is

ψ = + <−A e B e a z

k z k z2 2

2 2 for (12.4.6)

where k mE222= − /� . The wavefunction should satisfy the boundary conditions ψ( )z → −∞ = 0 and

ψ( ) .z → ∞ = 0 The boundary condition z → − ∞ is already satisfied, while the second boundary condition z → ∞ is imposed by A2 = 0. Now ψ must be continuous, so we must satisfy the conditions B1 = 0 at z = 0 and

A k a B k a B e z ak a

1 1 1 1 22sin ( ) cos ( ) ( )+ = =−

at (12.4.7)

This yields A k a B ek a

1 1 22sin ( ) .= −

Similarly, ψ ′ must be continuous; hence,

A k k a B k e z ak a

1 1 1 2 22cos ( ) ( )= − =−

at (12.4.8)

az

V(z)

–V0

0

Fig. 12.4


So, together we have

k k a k1 1 2cot ( ) = − (12.4.9)

Solving Eq. (12.4.9) gives the eigenenergy states for the electron. Note that minimal energy corresponds to minimal k1 and k2; thus, we should solve this equation numerically to find the minimal values of k1 and k2. To do this we write k2 in terms of k1:

km

E V kmE

12

2 0 22

22 2= + + = −� �

( ) (12.4.10)

so, 2 02

12

22mV k k/� = + or, k

mVk2

02 1

22= −

�. Thus, we obtain − = −cot ( ) .k a mV k k1 0

212

12 / /� Replacing k1a by x we arrive at

− =−

cot xmV a x

x

2 02 2 2/�

(12.4.11)

To find the minimal energy we draw a graph of y1 = − cot x and ymV a x

x20

2 2 22=

−/� and compute the

value of x in the first intersection point between y1 and y2; see Fig. 12.5.

The value of 2 02 2mV a /� is 10.498 597. We can use the program written in Problem 12.3 with the following

function:

C** Function for which we want to find the zero.real function f(x)real xf=tan(x)+x/sqrt(10.49859654100631–x*x)returnend

From Fig. 12.5 we see that the x value lies in the interval [2, 3]; so these will be our initial guesses in the program. Running the program with the appropriate change gives the following result:


This means that k a ma E V12

022 2 336 28= + =( ) . ./� Thus, the minimal energy eigenvalue is

E V+ =0 5 2. eV (12.4.12)

x

y1y2

y

Fig. 12.5


12.5. Using the Numerov algorithm, write a program to solve the Schrödinger equation for an electron in a potential well:

V xx a

( ) =≤ ≤

∞{0 0otherwise

(12.5.1)

It is given that a = 1 Å. The program takes as input an initial guess for the energy value and gives as output the closest higher-energy eigenvalue. Compare your results to the analytical ones.

SOLUTION

The Schrödinger equation for this case is

d

dx

mE2

2 22

0ψ ψ+ =

� (12.5.2)

Introducing the nondimensional variable x = x/a, we arrive at

d

d

ma E2

2

2

22

0ψ ψ

ζ+ =

� (12.5.3)

This equation is of the form

d y

dxk x y S x

2

22+ =( ) ( ) (12.5.4)

where S(x) = 0 and k x mEa2 2 22( ) . .= =const /� In our program, we put in as an input the value of k and compute the initial value of ψ( ) ( )ζ = −1 psip . Then, using the Numerov method, we integrate the equation for ki. We start from ψ( ) ( )0 0= − psim and add to k an amount dk and integrate again. In each iteration we add dk until we get a value of psip that has the opposite sign of the initial value of psip. At this point, we back up the value of k and jump in smaller steps dk than the value of 1 × 10−5. We do this since we know that we passed over the value of k that we are trying to converge to. Running the following program with ki = 0 gives the results shown at the end of the program. Note that we expect the convergence to correspond to the ground state, since it is the highest eigenvalue that is close to E = 0.

Program Problem 12.5real kreal toler,psip,psiold

C** Get initial value of the wave numberwrite (*,*) ’Enter the starting value of the wave number k:+ (k<0 to stop)’read (*,*) kif (k.lt.0.) goto 20

C** Initial value of the step.dk=1.toler=1.E–05

C** Integrate the equation with initial value of k.call intgrt (k,psip)psiold=psip

C** Change the value of k.10 k=k+dk

C** integrate again with different values of k.call intgrt (k,psip)

C** If psip changes value backup (the secant method).if ((psip*psiold).lt.0) thenk=k-dk


dk=dk/2endif

C** If convergence is not achieved try again.if (abs (dk).gt.toler) goto 10write (*,*) ’’write (*,*) ’ The result is:’write (*,*) k

20 stopend

C** Subroutine to integrate the Schrödinger equation using the Numerov method

Subroutine intgrt (k,psip)real k,psia,psiz,h,constinteger nstep

C** Number of stepsnstep=100

C** Step value of normalized x.h=1.nstep

C** Left boundary condition.psim=0.psize=.01const=(k*h)**2/12.do 10 ix=1,nstep–1

C** Numerov method equation:psip=2* (1.–5. *const) *psiz –(1.+const)*psimpsip=psip/(1+const)psim=psizpsiz=psip10 continue

C** The result achieved.write (*,*) ’The wave number:’ ,kreturnend

Running the program with a starting value k = 0.0 yields the following:

The wave number:1.000000000 3.250000000 3.142578125 3.1416015622.000000000 3.187500000 3.141601562 3.1415405273.000000000 3.156250000 3.141113281 3.1416015624.000000000 3.140625000 3.141601562 3.1415710453.500000000 3.156250000 3.141357422 3.1416015623.250000000 3.148437500 3.141601562 3.1415863043.125000000 3.144531250 3.141479492 3.141601562

The result is 3.141586304.

Consider now the analytical solution. The eigenenergies are given by

Ema

nn = π 2 2

22

2

� (12.5.5)

The ground state is Ema1

2 2

22= π �

, which corresponds to kma

E1

2

2 1

23 141 592 6= = =

�π . . . . .


12.6. Consider the Schrödinger equation for potentials with radial symmetry V(x, y, z) = V(r) and cylindrical symmetry V(x, y) = V( r), and demonstrate how to solve these equations.

SOLUTION

For a problem with central potentials V(r), the solution of the Schrödinger equation can be written as

ψ( )( )

( , )r = R rr

Yl m θ ϕ (12.6.1)

where Ylm(q, j) is the spherical harmonic, and R(r) is a function of r satisfying the radial equation

d R

dr

ME

l l

MrV r R r

2

2 2

2

22 1

2+ − + −

⎡

⎣⎢⎢

⎤

⎦⎥⎥

=�

�( )( ) ( ) 00 (12.6.2)

where E, l, and M are the eigenenergy, angular momentum, and mass, respectively. One can see that Eq. (12.6.2) is of the form of Eq. (12.23) with

S r k rM

El l

MrV r( ) ( )

( )( )= = − + −

⎡

⎣⎢⎢

⎤0

2 1

22

2

2

2and�

�

⎦⎦⎥⎥

(12.6.3)

This equation can be solved numerically using the Numerov alogrithm (see Problem 12.5). Similarly, for problems with a potential that has a cylindrical symmetry V( r), the solution is of the form

ψ( , , )( )ρ ϕ ρρ

ϕzR

e ei m inz= (12.6.4)

where m is the angular momentum in the z direction, n is an integer, and R( r) is the solution of

d R

d

M

M

m nM

E V2

2 2

2

2

2 2 22

8 2ρ ρρ+ − + + −

⎡

⎣⎢⎢

⎤

⎦⎥

�

� � ( )( )

⎥⎥=R( )ρ 0 (12.6.5)

where E and M are the energy eigenvalue and the mass, respectively. Also in this case it can be seen that this equation is of the form of Eq. (12.23) with

S kM

M

m nM

E V r( ) ( )( )

(ρ ρρ

= = − + + −02

8 22

2

2

2

2 2 2

and�

� �))

⎡

⎣⎢⎢

⎤

⎦⎥⎥

(12.6.6)

Hence, this equation can also be solved with the Numerov method.


12.7. Write a FORTRAN subroutine:

Subroutine Trapez (FUNC, N, A, B, S)INTEGER NREAL FUNC (0 : 1000), A, B, S

which computes the value of the integral of a function whose values are N + 1 points, given in the array FUNC in the interval [A, B]. The points are separated by h = (B − A)/N.

Ans.

C** Subroutine to compute the value of a definite integral.Subroutine Trapez (func,n,a,b,s)integer nreal func(0:1000),a,b,ss=0.s=(func(0)+func(n))/2.do 1 i=1,n–1

C** (1–mod(i,2))equals 0 if i even and equals 1 if i odd s=s+func(i)


1 continues=s*(b–a)/nreturnend

12.8. Solve numerically the integral obtained in Problem 5.3:

P e d00

1

12 2

= − −∫πηη (12.8.1)

Use your preferred method.

Ans. Using the program from Problem 12.2, and changing the line func(i) = exp (x*x) into

func(i) = 2/sqrt(a cos(–1.))∗exp(–x∗x) (12.8.2)

we obtain P0 = 0.1578. After calling the subroutine Simpson, we add the line

s = 1 − s (12.8.3)

12.9. Solve the equation x2 − 5 = 0 with the initial guesses x1 = 2 and x2 = 3. Use the secant method

x x f xx x

f x f xi i i

i i

i i+

−

−= − −−

11

1( )( ) ( )

(12.9.1)

Obtain an accuracy of 1 × 10−5.

Ans.

Program Problem 12.9real x1, x2, xtmpreal f1, f2, tolerinteger iter

C** Initialize iterations’ number.iter=0


C ** Initial guesses.x1=2.x2=3.

C** Evaluate the function f(x)=x**2–5 at x=xl and at x=x2:f1=x1*x1–5.f2=x2*x2–5.


do while (abs x2–x1).gt.toler)C** New iteration.

iter=iter+1xtmp=x2–f2*(x2–x1)/(f2–f1)x1=x2x2=xtmp

C** Evaluate the function f(x)= x**2–5 at x=x1 and at x=x2:f1=x1*x1–5.f2=x2*x2–5.end do

C** Print result:write (*,*) ’The zero of f(x) is:’,x1.

10 format (’obtained after ’,i3,’ iterations.’)write (*,10) iterstopend

264

CHAPTER 13

Identical Particles

13.1 IntroductionSuppose you have a basketball and your friend has a soccer ball with the same mass; you kick them toward each other, simultaneously, with the same velocity. Two things can happen: (a) The balls collide and each ball goes back to its owner. (b) The balls travel along parallel paths without touching and exchange hands. Since the balls have different shapes and colors you can tell which possibility occurred, (a) or (b). But if the balls were identical, you would not be able to tell what happened! When we consider identical quantum particles the situation gets even worse as we cannot even trace the exact trajectories of colliding particles. In this chapter we examine the special properties of a system composed of identical particles.

13.2 Permutations and Symmetries of WavefunctionsDefinition: We say that the particles of a system are identical (or indistinguishable) if no observer can detect any permutation of these particles.

The property of indistinguishability gives rise to symmetries in the system. Consider a system of n identi-cal particles with the eigenvector |φi ⟩ for the particle i (i = 1, . . . , n). We denote the state of the system by a vector of eigenvectors |φ1⟩, |φ2⟩, . . . , |φn ⟩, keeping in mind that different ordering of the |φi ⟩’s in two vectors corresponds to different vectors, e.g., if n = 2, ( , ) ( , ).| | | |φ φ φ φ1 2 2 1⟩ ⟩ ≠ ⟩ ⟩ If s is a permutation on the letters 1, . . . , n, then it can be written as

σ σ σ σ= ⎛⎝⎜

⎞⎠⎟

⋅ ⋅ ⋅⋅ ⋅ ⋅

1 2 31 2 3

nn( ) ( ) ( ) ( )σ (13.1)

meaning that the vector 1, 2, . . . , n becomes ( , , . . . , )| | |σ σ σ1 2⟩ ⟩ ⟩n after the action of s. Thus, s permutes the eigenvectors:

σ σ σ( , . . . , ) ( , . . . , )( ) ( )| | | |φ φ φ φ1 1⟩ ⟩ = ⟩ ⟩n n (13.2)

One can see that s acts as a linear operator. A permutation σ may be written as a product of transpositions, i.e., permutations that swap two letters. If the decomposition of σ consists of an even number of transposi-tions, then σ is called an even permutation, and we write sgn( ˆ ) ;σ = 1 if this number is odd, then σ is called an odd permutation denoted sgn( ˆ ) .σ = −1 The vector | | |u n⟩ = ⟩ ⟩, . . . ,φ φ1 is said to be symmetrical if σ | |u u⟩ = ⟩ for an arbitrary permutation σ . The same vector is said to be antisymmetric if ˆ sgn ( ˆ )σ σ| |u u⟩ = ⟩ for an arbi-trary permutation σ . We define two operators:

ˆ!

ˆSn

= ∑1 σσ permutation

(13.3)


and

ˆ!

(sgn ˆ) Ân

= ∑1 σ σσ permutation

(13.4)

S and A project the entire space of wavefunctions H on two subspaces: the space of symmetric wavefunctions aS, and the space of antisymmetric wavefunctions aA:

H SH H AHs A= =ˆ ˆ (13.5)

and in addition H H HA S= ⊕ ; that is, every vector is a unique sum of a completely symmetrical vector and a completely antisymmetric vector. The verification is given in Problem 13.2.

An arbitrary antisymmetric wavefunction can be written | |u A uA ⟩ = ⟩ˆ for a wavefunction | | |u n⟩ = ⟩ ⟩( , . . . , ).φ φ1 Hence, if { }( )|φ j ⟩ is a basis of the single-particle space of states, then a basis of the antisymmetric space of all n particles is given by applying A on a basis of the entire space, spanned by | |φ φj j

n1 ⟩ ⟩, . . . , ; thus,

| | | | |α j jj j

nA

n1

1 2

1 2

1, ,

ˆ ( , , . . . , )… ⟩ = ⟩ ⟩ ⟩ =φ φ φnjn

!!(sgn ˆ)( , . . . , )( )σ σ σ

σ

| |φ φ1

j j(n)1 ⟩ ⟩∑

or

|

| | |

| |α j j

n n1

1, , !… ⟩ =

⟩ ⟩ ⟩

⟩

⋅ ⋅ ⋅φ φ φ

φ φ1 1 1

2

1 2

1

j j j

j

n

22 22

1 2

j j

nj

nj

nj

n

n

⟩ ⟩

⟩ ⟩ ⟩

⋅ ⋅ ⋅⋅⋅⋅

⋅⋅⋅⋅ ⋅ ⋅

|

| | |

φ

φ φ φ

(13.6)

is a basis of HA. The last equality comes from the properties of the determinant. (Note that this is sometimes given as the definition of a determinant.) This determinant is known as the Slater determinant and is the solution for the Schrödinger equation for noninteracting fermions.

13.3 Bosons and FermionsFrom experimental observations it seems there are two kinds of particles. The first kind consists of particles that have completely symmetrical wavefunctions; they are called bosons. The second kind consists of par-ticles with completely antisymmetric wavefunctions; they are called fermions. There are no particles with mixed symmetry. The Pauli exclusion principle is a basic principle that is valid only for identical particles that are fermions. This principle states that two identical fermions cannot be in the same quantum state. An alternative formulation of this principle asserts that the probability of finding two identical fermions with the same quantum numbers is zero.

SOLVED PROBLEMS

13.1. (a) Compute the number of permutations on n letters. (b) Show that a product of two permutations is also a permutation.

SOLUTION

(a) The number of permutations equals the number of different orderings of n distinguished letters; the first letter has n places, the second letter has n − 1 places, etc., and the nth letter has only one place. Hence, there are n n n n( )( ) !− − =1 2 1� permutations.

(b) A permutation is a function σ from the set {1, . . . , n} to itself that is bijective, i.e., ˆ ( ) ˆ ( )σ σi j≠ if i ≠ j and every i equals ˆ ( )σ j for some j. A composition of two such functions is also a bijective function from {1, . . . , n} to itself, and hence a permutation.

CHAPTER 13 Identical Particles266

13.2. Show that S and A are Hermitian operators.

SOLUTION

Let σ be any permutation, and denote | | |u n⟩ = ⟩ ⟩( , . . . , )φ φ1 and | | |v ⟩ = ⟩ ⟩( , . . . , )θ θ1 n ; then

⟨ ⟩ = ⟨ ⟨ ⟩ ⟩v | | | | |ˆ ( , . . . , )( , . . . ,( ) ( )σ σ σu n nθ θ φ φ1 1 )) ( ) ( ) ( )

(

= ⟨ ⟩ ⟨ ⟩ ⟨ ⟩

= ⟨

⋅ ⋅ ⋅

−

θ φ θ φ θ φ

θ

1 1 2 2

1

| | |σ σ σ

σ

n n

11 1 2 21

1 1) ( ) ( )ˆ| | | |φ θ φ θ φ⟩ ⟨ ⟩ ⟨ ⟩ = ⟨− −⋅ ⋅ ⋅ −

σ σσ

n n uv ⟩⟩

(13.2.1)

Hence, ˆ ˆ†σ σ= −1 and therefore

ˆ!

ˆ!

ˆ!

† †Sn n n

= = =∑ −1 1 11σ σσ σpermutation permutattion permutation

∑ ∑ =ˆ ˆσσ

S (13.2.2)

Also,

ˆ!

(sgn ˆ) ˆ!

(sgn ˆ) ˆ!

(sg† †An n n

= = =∑ ∑ −1 1 11σ σ σ σσ σ

nn ˆ ) ˆ!

(sgn ˆ) ˆ ˆσ σ σ σσ σ

− − = =∑ ∑1 1 1n

A (13.2.3)

13.3. Prove that S u| ⟩ is a symmetric vector and A u| ⟩ is an antisymmetric vector for an arbitrary |u⟩.

SOLUTION

We prove that S u| ⟩ is a symmetrical vector by showing that for an arbitrary permutation ˆ, ˆ( ˆ ) ˆτ τ S u S u| |⟩ = ⟩, so

ˆ ˆ ˆ!

ˆ!

ˆ ˆ!

ˆτ τ σ τσ σσ σ

S un

un

un

u| | | |⟩ = ⟩ = ⟩ = ′ ⟩∑ ∑1 1 1 == ⟩′

∑ S u|σ

(13.3.1)

Similarly,

ˆ ˆ ˆ!

(sgn ˆ) ˆ!

(sgn ˆ) ˆ ˆτ τ σ σ σσ σ

A un

un

| |⟩ = ⟩ =∑ ∑1 1 σ ττ

τ σ τ σ τσ

|

|

u

nu

⟩

= ⟩ =− ∑(sgn ˆ)!

(sgn ˆ)(sgn ˆ) ˆ ˆ (1 1ssgn ˆ)

!(sgn ˆ ˆ) ˆ ˆ

(sgn ˆ)!

(sgn

τ σ τ σ τ

τ

σ

1

1

nu

n

| ⟩

=

∑′′ ′ ⟩ = ⟩

′∑ ˆ ) ˆ (sgn ˆ) ˆσ σ τ

σ

| |u A u

(13.3.2)

Note that the permutations form a group; thus every element has an inverse, and therefore

ˆ ˆ ˆ ˆσ σ τ σ∑ ∑ ∑= =−1 (13.3.3)

We used the fact that sgn (st) = (sgn s)(sgn t), which can be verified.

13.4. Show that: (a) S2 = S ; (b) A2 = A; (c) AS = S A = 0.

SOLUTION

(a) Using the results of Problem 13.3, we can write

ˆ!

ˆ ˆ!

ˆ ˆ!

ˆ ˆSn

Sn

Sn

S2 1 1 1 1=⎛

⎝⎜⎜

⎞

⎠⎟⎟

= = =∑ ∑ ∑σ σ σσ σ σ

nnS

nn

S S!

ˆ !!

ˆ ˆ= =∑σ

(13.4.1)


(b) As in part (a) we have

ˆ!

(sgn ˆ) ˆ ˆ!

(sgn ˆ)(sgn ˆ) Ân

An

A2 1 1 1= = =′

∑ σ σ σ σσ

nnA A

!ˆ ˆ

σ σ∑ ∑ ⋅ =1 (13.4.2)

(c) By definition,

ˆ ˆ!

(sgn ˆ) ˆ ˆ!

ˆ sgn ÂSn

Sn

S= = =′

∑ ∑1 10σ σ σ

σ σ

(13.4.3)

and

ˆ ˆ!

ˆ ˆ!

ˆ sgn ˆSAn

An

A= = =∑ ∑1 10σ σ

σ σ

(13.4.4)

13.5. Use the symmetrization postulate for fermions to derive the Pauli exclusion principle.

SOLUTION

The symmetrization postulate for fermions states that the wavefunction of a system of n identical fermi-ons is completely antisymmetric. Thus, it is a linear combination of vectors of the form |α j jn1 … ⟩. These normalized vectors can be written as

|

| | |

| |α j j

j j j

j j

n

n

n1

1 2

1

1

1 1 1

2 2

… ⟩ =

⟩ ⟩ ⟩

⟩

⋅ ⋅ ⋅

!

φ φ φ

φ φ 22

1 2

2⟩ ⟩

⟩ ⟩ ⟩

⋅ ⋅ ⋅⋅⋅⋅

⋅⋅⋅⋅ ⋅ ⋅

|

| | |

φ

φ φ φ

j

nj

nj

nj

n

n

(13.5.1)

Hence, if two particles are in the same quantum state, two columns are the same, forcing the determinant to vanish; consequently, no nontrivial wavefunction exists in this case. This result proves the Pauli exclu-sion principle.

13.6. Show explicitly that Slater’s determinant for two particles (fermions) is antisymmetric.

SOLUTION

The Slater determinant for two fermions is given by

|| |

| ||u

j j

j j( , )

!(1 2

12

12

1 1

2 21

1 2

1 2

⟩ =⟩ ⟩

⟩ ⟩=

φ φ

φ φφ jj j j j1 2 2 1

2 1 2⟩ ⟩ − ⟩ ⟩| | |φ φ φ ) (13.6.1)

and

| | | | |uj j j j

( , ) ( )2 112 2 1 2 1

1 2 2 1⟩ = ⟩ ⟩ − ⟩ ⟩φ φ φ φ (13.6.2)

thus, | |u u( , ) ( , )2 1 1 2⟩ = − ⟩.

13.7. Show that the Slater determinant is a zero-order approximation to the Schrödinger equation of a system of n identical fermions.

SOLUTION

Consider the Schrödinger equation ˆ ( , , . . . , ) .H n E1 2 | |ψ ψ⟩ = ⟩ Neglecting the interactions between the particles, we write H0(1, 2, . . . , n) for the zero-order approximation of H:

ˆ ( , , . . . , ) ˆ ( ) ˆ ( )H n H Hs s0 0 01 2 1 2= ⊕ ⊕⋅ ⋅ ⋅ (13.7.1)


For every ˆ ( )H is0 we have ˆ ( ) ,H i Es

ij

ij

ij

0 | |φ φ⟩ = ⟩ where i stands for particle number and j counts the different

eigenvectors and eigenfunctions. Since the Slater determinant is a combination of different eigenfunctions

such as | |φ φ11j

njn⟩ ⟩⋅ ⋅ ⋅ and since the particles do not interact, the function

|

| | |

| |u

nA

j j j

j j

n

1

1 1 1

2 21

1 2

1 2

⟩ =

⟩ ⟩ ⟩

⟩ ⟩

⋅ ⋅ ⋅⋅

!

φ φ φ

φ φ ⋅⋅ ⋅⋅⋅⋅

⋅⋅⋅⋅ ⋅ ⋅

⟩

⟩ ⟩ ⟩

|

| | |

φ

φ φ φ

2

1 2

j

nj

nj

nj

n

n

(13.7.2)

is a solution to the equation H E0 | |ψ ψ⟩ = ⟩.

13.8. Three imaginary “spinless” fermions are confined to a one-dimensional box of length L. The confinement potential is

Vx L

=≤ ≤

∞{0 0otherwise

(13.8.1)

We assume that there is no interaction between the fermions. (a) What is the ground state of the system? (b) Find the state of the system.

SOLUTION

(a) As shown in Chap. 3, the eigenstates of this system are

ψn nLnxL

En

mL= ⎛

⎝⎜⎞⎠⎟ =2

2

2 2 2

2sinπ π �

(13.8.2)

Since two fermions cannot occupy the same state, the three fermions are in distinct states, and since the

system is in the ground state, the states will be ψ1, ψ2, and ψ3 with a total energy π 2 2

22 2 2

21 2 3

�

mL( ).+ +

Schematically, the structure of the system is depicted in Fig. 13.1.

n = 4

n = 3

n = 2

n = 1

Energy

∗

∗

∗

Fig. 13.1

(b) The antisymmetric state is given by

ψ = ×(normalizing factor) (Slater determinant) ==⟩ ⟩ ⟩⟩1

3

1 1 2 1 3 1

1 2 2 2!

( ) ( ) ( )

( ) (

| | || |ψ ψ ψψ ψ

x x x

x x )) ( )

( ) ( ) ( )

⟩ ⟩⟩ ⟩ ⟩

|| | |

ψψ ψ ψ

3 2

1 3 2 3 3 3

x

x x x

(13.8.3)


13.9. Repeat Problem 13.8 for three electrons. Ignore the Coulomb interaction between the electrons.

SOLUTION

(a) An electron has spin 1/2; thus, the eigenstates and eigenvalues are

ψ ψn nLn x

L Ln x

L+ −= ⎛

⎝⎜⎞⎠⎟

⎛⎝⎜

⎞⎠⎟ = ⎛

⎝⎜2 1

02

sin sinπ π ⎞⎞

⎠⎟⎛⎝⎜

⎞⎠⎟

01

(13.9.1)

where En = p2�

2n2/2mL2.The additional degree of freedom, namely, the spin, allows us to put two electrons in the first energy level, since this energy level corresponds now to two different eigenstates: spin up and spin down. Thus, there are two possible configurations for the ground state; they are depicted in Fig. 13.2.

n = 4

n = 3

n = 2 or,

n = 1

Energy

n

n

n

n

Energy

Fig. 13.2

(b) There are three basic functions for each diagram in Fig. 13.2. For the left diagram we have ψ ψ ψ1 1 2+ − +, , and

for the right diagram we have ψ ψ ψ1 1 2+ − −, , . Using the Slater determinant, we get

Ψ left =

⟩ ⟩ ⟩+ − +

+16

1 1 1 1 2 1

1 2

| | |

|

ψ ψ ψ

ψ

( ) ( ) ( )

( )

x x x

x ⟩⟩ ⟩ ⟩

⟩ ⟩

− +

+ − +

| |

| | |

ψ ψ

ψ ψ ψ1 2 2 2

1 3 1 3 2

( ) ( )

( ) ( ) (

x x

x x xx3)⟩

(13.9.2)

and

Ψ right =

⟩ ⟩ ⟩+ − −

+16

1 1 1 1 2 1

1 2

| | |

|

ψ ψ ψ

ψ

( ) ( ) ( )

(

x x x

x )) ( ) ( )

( ) ( )

⟩ ⟩ ⟩

⟩ ⟩

− −

+ − −

| |

| | |

ψ ψ

ψ ψ ψ1 2 2 2

1 3 1 3 2

x x

x x (( )x3 ⟩

(13.9.3)

13.10. A system is composed of two fermions with spin 1/2. Find the “two-particle density function” and the “one-particle density function” if both electrons are in different normalized orthogonal states.

SOLUTION

Suppose that each of the electrons has a different spin |φ1( );r +⟩ and |φ2( );r −⟩, respectively. In this case the common wavefunction has the form

| | | |ψ φ φ φ( , , , ) ( ); ( ); (1 2 1 2 1 1 2 2 1 2r r r r r⟩ = +⟩ −⟩ − )); ( );+⟩ −⟩|φ2 1r (13.10.1)

So,

ρtwo par.( , ) ( ) ( )r r r r1 2 1 12

2 22= ⟨ ⟩ = +ψ ψ φ φ| | | | | ( ) ( )| | | |φ φ1 2

22 1

2r r (13.10.2)


and

ρ ρone par. two par.( ) ( , ) (r r r r1 1 23

2 1 1= =∫ d r |φ )) ( )| | |22 1

2+ φ r (13.10.3)

If both electrons have spin |+⟩, we obtain

| | | |ψ φ φ φ( , , , ) ( ); ( );1 2 1 2 1 1 2 2 1r r r r⟩ = +⟩ +⟩ − (( ); ( );r r2 2 1+⟩ +⟩|φ (13.10.4)

and

ρtwo par.( , ) ( ) ( ) (r r r r r1 2 1 12

2 22

12= −| | | |φ φ φ 11 2 1 1 2 2 2 1 22

2 12) ( ) ( ) ( ) ( ) ( )φ φ φ φ φr r r r r+ | | | | (13.10.5)

13.11. A system contains two identical spinless particles. The one-particle states are spanned by an orthonormal system { }|φk ⟩ . Suppose that the particles’ states are |φi ⟩ and |φj i j⟩ ≠( ). (a) Find the probability of finding the particles in the states |ξ⟩ and |η⟩ (not necessarily eigenstates). (b) What is the probability that one of them is in the state |ξ⟩? (c) Suppose now that the particles are not identical and they are measured with an instrument that cannot distinguish between them. Answer parts (a) and (b) for this case.

SOLUTION

(a) The symmetric state of the system is given by

Φ11 2 1 21

2= ⟩ ⟩ + ⟩ ⟩( )| | | |φ φ φ φi j j i

( ) ( ) ( ) ( ) (13.11.1)

The new state is also symmetric; it is given by

Φ21 2 2 11

2= ⟩ ⟩ + ⟩ ⟩( )| | | |ξ η ξ η( ) ( ) ( ) ( ) (13.11.2)

Thus, the probability is

P i j i j1 1 22= ⟨ ⟩ = ⟨ ⟩ ⟨ ⟩ + ⟨ ⟩ ⟨ ⟩| | | | | | | | |Φ Φ ξ φ η φ η φ ξ φ 22 (13.11.3)

(b) Consider the symmetric state that corresponds to |ξ⟩ and to the eigenstate |φk ⟩:

| | | | |Φξ φ φ φ ξ,k

k k⟩ = ⟩ ⟩ + ⟩ ⟩( )( ) ( ) ( )12

1 2 1ξ (13.11.4)

In order to find the probability of one particle being in the state |ξ⟩, we multiply |Φξ φ,k

⟩ on the left-hand side with ⟨Φ1 | (the original state) and sum over all k:

Pk

k

i j k j i2 12= ⟨ ⟩ = ⟨ ⟩ ⟨ ⟩ + ⟨ ⟩ ⟨∑ | | | | | | |Φ Φξ φ ξ φ φ φ ξ φ, φ || |

| | | | | |

φ

φ ξ φ ξ φ ξ

k

k

i jk i ik

k

i

⟩

= ⟨ ⟩ + ⟨ ⟩ = ⟨ ⟩

∑∑

2

2δ δ || | | |2 2+ ⟨ ⟩φ ξj

(13.11.5)

(c) The state of the system is now |Φ1⟩. Hence, by multiplying with the final state ⟨ ⟨ + ⟨ ⟨ξ η ξ η( ) ( ) ( ) ( )1 2 2 1| | | | we obtain

P31 2

12 2 1

12= ⟨ ⟩ + ⟨ ⟩ =( ) ( ) ( ) ( )| | | | | | | |ξ η ξ ηΦ Φ || | | | | |⟨ ⟩ ⟨ ⟩ + ⟨ ⟩ ⟨ ⟩ =ξ φ η φ ξ φ η φi j j i P2

1 (13.11.6)


and

P k

k

k31 2

12 1 2

1= ⟨ ⟩ = ⟨ ⟩∑ ′| | | | | | | |ξ φ φ ξ( ) ( ) ( ) ( )Φ Φ 2

′′∑ =

k

P2 (13.11.7)

13.12. Suppose that a domain D contains n identical particles, and outside D there are additional identical particles such that the interaction between particles not in the same domain is negligible. Show that in region D it is enough to do antisymmetrization of the n particles in D without considering the rest of the identical particles. In your answer, refer only to the case of n = 2 fermions. (The result for bosons is the same.)

SOLUTION

Let |χ⟩ and | φ⟩ be physical antisymmetric states of the D-particles. Those functions vanish outside D. Neglecting the other identical particles, the probability of getting an eigenvalue of |χ⟩ when the system is at state | φ⟩ is ω χ φ= ⟨ ⟩ .| | | 2 We now show that the same result is obtained when we do not neglect the other N − 2 fermions. Let { }|θi ⟩ be a complete set of orthonormal physical (antisymmetric) vectors of the N − 2 particles outside of D; that is, |θi ⟩ vanishes in D. Define F as a permutation between two particles in D or between two particles not in D. Also, define G as a permutation between particles from D and particles not in D. There are 2!(N − 2)! permutations of the F-kind, and N! − 2!(N − 2)! permutations of the G-kind. The total physical state of the system must be antisymmetric for all N particles. In the basis |χθi ⟩,

| |X CA i⟩ = ⟩phyˆ χθ (13.12.1)

A is the antisymmetrization operator where C is a normalization constant, which we now compute:

⟨ ′ ′ ⟩ = ⟨ ′ ′ ⟩

=

∑

∑

χ θ χθ χ θ χθ| | | |ˆ!

sgn ˆ

!sgn

AN

A

NF

1

1

σσ

FF F G G

G

⟨ ′ ′ ⟩ + ⟨ ′ ′ ⟩⎛

⎝⎜⎜

⎞

⎠⎟⎟∑χ θ χθ χ θ χθ| | | |sgn (13.12.2)

By the definitions of |χ⟩ and |θ⟩, the second term vanishes; so,

⟨ ′ ′ ⟩ = ⟨ ′ ′ ⟩ = −∑χ θ χθ χ θ χθ| | |ˆ!

(sgn )!( )!

AN

FN

F

1 2 22

NN !⟨ ′ ′ ⟩χ θ χθ| (13.12.3)

Thus CN

N= −

!!( )!

.2 2

The probability of getting an eigenvalue of |χ⟩ for the two fermions when the

(N − 2)-state is |ψ⟩ and the D-state is |φ⟩ will be

P X X A CCA Ci

i

i

i

i= ⟨ ⟩ = ⟨ ⟩ = ⟨∑ ∑phy phy| | | |χθ φψ χθˆ† 2 4 ˆ Â C A

C C

i

i

i

i

2 | | |

|

φψ χθ φψ

χθ φψ

⟩ = ⟨ ⟩

= ⟨ ⟩

∑ ∑−

2 4 2

4 2 2

ii

i

i

i

i∑ ∑ ∑= ⟨ ⟩ = ⟨ ⟩ ⟨ ⟩ = ⟨ ⟩χθ φψ χ φ θ ψ χ φ| | | |

2 2 2 2

(13.12.4)


13.13. Prove that the Pauli exclusion principle does not hold for bosons.

13.14. Show explicitly that the Slater determinant for three fermions is antisymmetric.


13.15. Show that any function on the real line is a sum of symmetric and antisymmetric functions.

Ans. f xf x f x f x f x

( )( ) ( ) ( ) ( )= + − + − −

2 2

13.16. What happens to the Slater determinant if there is a linear dependency between | |φ φj jn1 ⟩ ⟩⋅ ⋅ ⋅ ?

Ans. It vanishes.

13.17. Three particles are confined within the potential

V x y

x a y b( , ) =

≤ ≤ ≤ ≤∞{0 0 0and

otherwise

(13.17.1)

Find the ground state of the system when the particles are bosons.

Ans. | |ψ φ φ φ0 1 2 3 1 1 1 1 1 2 1 1 3( , , ) ( ) ( ) ( ), , ,r r r r r r⟩ = ⟩, where φn nx y

x yx y

abn x

a

n y

b, ( , ) sin sin= ⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

4 π π..

13.18. Refer to Problem 13.14 and find the ground state of the system when the particles are “spinless” fermions. (That is, use Pauli’s exclusion principle, but neglect the additional degree of spin.)

Ans. || | |

ψφ φ φ

0 1 2 3

11 1 12 1 21 113

( , , )!

( ) ( ) (r r r

r r r⟩ =

⟩ ⟩ ))( ) ( ) ( )( )

⟩⟩ ⟩ ⟩⟩

| | || |φ φ φφ φ

11 2 12 2 21 2

11 3 12

r r rr (( ) ( )r r3 21 3⟩ ⟩|φ

13.19. Solve Problems 13.14 and 13.15 without neglecting the spin.

Ans. |

| | |

ψ

φ φ φ

0 1 2 3

1 113

( , , )!

( ) ( )

r r r

r r

⟩ =

⟩ ⟩11+

11−

12− (( )

( ) ( ) ( )

(

r

r r r

r

1

2 2 2

⟩

⟩ ⟩ ⟩| | |

|

φ φ φ

φ11+

11−

12−

11+

33 3 3) ( ) ( )⟩ ⟩ ⟩| |φ φ11−

12−r r

and, three additional states by substituting

φ φ12+

21−, , and φ21

+ for φ12− .

13.20. Solve Problem 13.10 for two bosons.

Ans. | | | |Φ( , , , ) ( ); ( ); (1 2 1 2 1 1 1 2 2 2 1r r r r⟩ = ⟩ ⟩ +φ φ φS S rr r

r rr

2 1 2 1 2

1 21 1

); ( );

( , )( )

S S⟩ ⟩

=

|

|

φ

ρφ

two par.

|| | | | | | |

|

22 2

21 2

22 1

21 2

1 1

φ φ φ

φ

( ) ( ) ( )

( )

r r r

r

+ ≠S S

φφ φ φ

ρ

2 2 1 2 2 12

1 2( ) ( ) ( )

(

r r r

r

+ =

⎧⎨⎪

⎩⎪ | S S

one par. 11 1 23

2 1 12

2 1) ( , ) ( ) ( )= = +∫ ρ φ φtwo par. r r r rd r | | | ||2

273

CHAPTER 14

Addition of Angular Momenta

14.1 IntroductionConsider two angular momenta j1 and j2. These momenta can be angular momenta relating to two different particles or angular momenta relating to one particle (angular momentum and spin). These two momenta act in different state spaces, and all their components commute with one another. The individual eigenstates of j1 and j2 will be denoted, as usual, by | j m1 1⟩ and | j m2 2 ⟩, so that (see Chap. 6)

ˆ ( )ˆj12

1 12

1 1 1 1

1 1 1 1

1| |

| |

j m j j j m

j j m m jz

⟩ = + ⟩

⟩ =

�

� 11 1m ⟩

⎧⎨⎪

⎩⎪ (14.1)

and similarly for j2. The state space of the compound system is obtained by taking the direct product (tensor product) of the individual state space of the two angular momenta:

| | | |j m j m j j m m m m1 1 2 2 1 2 1 2 1 2⟩ ⊗ ⟩ = ⟩ ≡ ⟩;

(14.2)

For fixed j1 and j2, m1 and m2 have the values

m j j j

m j j j1 1 1 1

2 2 2 2

11

= − − += − − +

⎧⎨⎩

, , . . . ,, , . . . ,

(14.3)

where the set of numbers { j1, m1} and { j2, m2} are either integers or half-integers. The state space of the com-pound system is (2j1 + 1)(2j2 + 1)-dimensional space. The states |m m1 2 ⟩ are, according to their construction, eigenstates of the operators {ˆ , ˆ , ˆ , ˆ }.j j1

222

1 2j jz z

14.2 { j , j , , J }12

22

zˆ ˆ ˆ ˆJ2 Basis

In the absence of interaction between j1 and j2, the operators j1 and j2 commute with the Hamiltonian, and thus, | j m1 1⟩ and | j m2 2 ⟩ are also eigenstates of the system. However, if j1 and j2 interact with

ˆ ˆ ˆ ˆ ( )H H= +0 1 2α αj ji where is a coupling constant

(14.4)

then j1 and j2 are not conserved, but ˆ ˆ ˆJ j j= +1 2 is conserved. Thus, it is better to transform to an eigenstate basis of the operators {ˆ , ˆ , ˆ , ˆ }j j J J1

222 2

z . The eigenstates in this basis will be denoted by | |j j J M J M1 2 ⟩ ≡ ⟩, and satisfy

ˆ ( )ˆJ2 2 1| |

| |

JM J J JM

J JM M JMz

⟩ = + ⟩⟩ = ⟩

⎧⎨⎪

⎩⎪

�

�

(14.5)

274

In this case,

J j j j j j j= − − + +| | | |1 2 1 2 1 21, , . . . ,

(14.6)

and for each value of J,

M J J J= − − +, , . . . ,1 (14.7)

Note that

ˆ ( )j12 2

1 1 1| |JM j j JM⟩ = + ⟩� (14.8)

Therefore, using the identity

2ˆ ˆ ˆ ˆ ˆj j J j j1 22

12

22i = − − (14.9)

we have

ˆ ˆ [ ( ) ( ) ( )]j j1 2

2

1 1 2 221 1 1i | |JM J J j j j j J⟩ = + − + − +�

MM⟩ (14.10)

As a result, |JM⟩ are also eigenstates of the operators ˆ ˆj j1 i 2. In a commonly employed terminology, one refers to |JM⟩ as an eigenstate in the coupled representation and to |m m1 2 ⟩ as an eigenstate in the uncoupled representation.

14.3 Clebsch–Gordan CoefficientsThe two sets of orthonormal states |m m1 2 ⟩ and |JM⟩ are related by a unitary transformation; that is, we can write the eigenstates |JM⟩ in terms of |m m1 2 ⟩ by

| | |JM m m JM m m

m m

⟩ = ⟨ ⟩ ⟩∑ 1 2 1 2

1 2,

(14.11)

where ⟨ ⟩m m JM1 2 | are the Clebsch–Gordan coefficients. It is possible to obtain a general expression for the Clebsch–Gordan coefficients. However, it is often simpler to construct the coefficients for particular cases. They can be calculated by successive applications of the step operators ˆ ˆ ˆJ J iJx y± = ± on the vectors |JM⟩, using the following relations:

ˆ ( ) ( )ˆJ JM J J M M J M

J m m J±

±

⟩ = + − ± ± ⟩

⟩ =

| |

|

�

�

1 1 1

1 1 2 1

,

(( ) ( )J m m m m1 1 1 1 21 1 1+ − + ± ⟩

⎧⎨⎪

⎩⎪ | ,

(14.12)

together with the phase condition,

| |J J J M j j m j m j= + = ± + ⟩ = = ± = ± ⟩1 2 1 2 1 1 2 2, ( ) ,

(14.13)

Some properties of the Clebsch–Gordan coefficients are given below:

⟨ ⟩ = = +m m JM M m m1 2 1 20| unless

(14.14)

⟨ ⟩m m JM1 2 | is real

(14.15)

⟨ ⟩ ⟨ ′ ′ ⟩ = ′ ′=−=−∑ JM m m m m J M JJ MM

m j

j

m

| |1 2 1 2

2 2

2

1

δ δ,

jj

j

1

1

∑ (14.16)

CHAPTER 14 Addition of Angular Momenta


⟨ ⟩ ⟨ ′ ⟩ = ′ ′=−=

∑ m m JM JM m m m m m m

M J

J

J j

1 2 1 21 1 2 2

1

| ||

δ δ−−

+

∑j

j j

2

1 2

|

(14.17)

J J M M m m J M j j m m( ) ( ) , ( ) ( )+ − ± ⟨ + ⟩ = + −1 1 1 1 11 2 1 1 1 1| ∓ || |m m JM1 21∓ , ⟩

+ j j2 2( ++ − ⟨ − ⟩1 1 12 2 1 2) ( )m m m m JM∓ , |

(14.18)

⟨ ⟩ = − ⟨ ⟩+ −m m JM m m JM

j j J2 1 1 21 1 2| |( ) (14.19)

⟨− − − ⟩ = − ⟨ ⟩+ −m m J M m m JM

j j J1 2 1 21 1 2, , ( )| |

(14.20)

SOLVED PROBLEMS

14.1. Consider two angular momenta of magnitudes j1 and j2. The total angular momentum of this system is then ˆ ˆ ˆJ J J= +1 2, where J1 and J2 are commuting operators. Let |m m1 2 ⟩ be the common eigenstates of the observables {ˆ , ˆ , ˆ , ˆ }J J1

222

1 2J Jz z . Let |JM⟩ be the common eigenstates of {ˆ , ˆ , ˆ , ˆ }J J J12

22 2 Jz . (a) Find

all possible values for m1 and m2. (b) Find the possible values for J and M. (c) Show that the state space of the compound system has dimensionality

( ) ( )( )2 1 2 1 2 11 2

1 2

1 2

J j j

J j j

j j

+ = + += −

+

∑| |

(14.1.1)

where j1 and j2 are fixed quantum numbers.

SOLUTION

(a) Let us denote by | j m1 1⟩ the eigenvectors common to the observables {ˆ ˆ },J12

1, J z of respective eigenvalues

�j1( j1 + 1) and �m1. Similarly, let | J m2 2⟩ be the eigenvectors common to {ˆ ˆ }.J22, J z2 The state space of

the compound system is obtained by taking the tensor product of individual state spaces of the two angular momenta. Thus,

| | | |j m j m j j m m m m1 1 2 2 1 2 1 2 1 2⟩ ⊗ ⟩ = ⟩ ≡ ⟩, ; ,

(14.1.2)

where j1 and j2 are fixed quantum numbers. The possible values of |m m1 2 ⟩ are given by

m j j j

m j j j1 1 1 1

2 2 2 2

11

= − − += − − +

⎧⎨⎩

, , . . . ,, , . . . ,

(14.1.3)

where the set of numbers { j1, m1} and {j2, m2} are either integers or half-integers. The dimension of the state space of the compound system is (2j1 + 1)(2j2 + 1) (according to the number of independent eigenstates for basis |m m1 2 ⟩).

(b) The state space of the system is a direct sum of orthogonal subspaces of definite total angular momen-tum J. Thus,

| | |JM m m JM m m

m m

⟩ = ⟨ ⟩ ⟩∑ 1 2 1 2

1 2,

(14.1.4)

CHAPTER 14 Addition of Angular Momenta276

where ⟨ ⟩ ∝ +m m JM M m m1 2 1 2| δ , are Clebsch–Gordan coefficients. Assuming that j1 ≥ j2, we have

J j j j j j j= − − + +1 2 1 2 1 21, , . . . ,

(14.1.5)

Consequently, the possible values of M for each value of J are

M J J J= − − +, , . . . ,1 (14.1.6)

Clearly, each value of J in Eq. (14.1.4) corresponds to a subspace of dimension (2J + 1) of definite total angular momentum.

(c) Consider the left side of Eq. (14.1.1). Using me results of part (b) and setting J = j1 − j2 + i, we find

( ) [ ( ) ]2 1 2 11 2

0

2 2

1 2

1 2

J j j i

i

j

J j j

j j

+ = − + +== −

+

∑∑| |

== − − + + + +

= +

12

2 2 1 2 2 1 2 1

2 1

1 2 1 2 2

1

[( ) ( )]( )

(

j j j j j

j ))( )2 12j +

(14.1.7)

14.2. Two angular momenta of respective magnitudes j1 and j2 and total angular momentum ˆ ˆ ˆJ j j= +1 2, are described by the basis | | |m m j m j m1 2 1 1 2 2⟩ ≡ ⟩ ⊗ ⟩. By construction, the states |m m1 2 ⟩ are eigenstates

of {ˆ ˆ ˆ ˆ }j j12

22

1 2, , ,J Jz z and ˆ ˆ ˆJ J Jz z z= +1 2 . (a) Find all the eigenvalues of the operator Jz and their degree of degeneracy. (b) Consider the states

| || |ψψ

+

−

⟩ = = = ⟩⟩ = = − = − ⟩

⎧⎨⎩

m j m j

m j m j1 1 2 2

2 1 2 2

,

,

(14.2.1)

for which m1 and m2 both assume either maximal or minimal values. Show that the states |ψ+ ⟩ and | |ψ− ⟩ are eigenstates of J2 (as well as of Jz) and find the corresponding eigenvalues.

SOLUTION

(a) The basis states |m m1 2 ⟩ satisfy

ˆ ( )ˆ (

j

j12

1 22

1 1 1 2

1 22

1| |

|

m m j j m m

m m j

⟩ = + ⟩

⟩ =

�

�22

2 jj m m2 1 21+ ⟩

⎧⎨⎪

⎩⎪ )| (14.2.2)

where j1 and j2 are fixed quantum numbers and

ˆ , . . . ,

ˆ

J m m m m m m j j j

J

z

z

1 1 2 1 1 2 1 1 1 1

2

1| |⟩ = ⟩ = − − +� ,

|| |m m m m m m j j j1 2 2 1 2 2 2 2 21⟩ = ⟩ = − − +

⎧⎨⎪

⎩⎪ � , , . . . , (14.2.3)

where m1 and m2 are either integers or half-integers. Using Eq. (14.2.2), we immediately find

ˆ ( ˆ ˆ ) ( )J m m J J m m m m m mz z z| | |1 2 1 2 1 2 1 2 1 2⟩ = + ⟩ = + ⟩ ≡� ��M m m| 1 2⟩ (14.2.4)

Consequently, the eigenvalues of Jz are �M, where the quantum number M = m1 + m2 takes the values

M j j j j j j= − + − + + +( ) ( ) , . . . ,1 2 1 2 1 21,

(14.2.5)

The degree of degeneracy g(M) of these values has the following properties:

1. The value M = Mmax = ( j1 + j2) is not degenerate:

g j j( )1 2 1+ =

(14.2.6)


2. The degree of degeneracy is increased by 1 as M decreases by 1 until a maximum degeneracy is reached for the value M = j1 − j2. The degeneracy remains constant as long as | M | ≤ j1 − j2 and is equal to

g M j j j M j j( ) ( )= + − − ≤ ≤ −2 12 1 2 1 2 (14.2.7)

3. For M < − ( j1 − j2), g(M) decreases by 1 as M decreases by 1. The value M = Mmin = − ( j1 + j2) is not degenerate. Generally, g(M) is an even function of M:

g M g M( ) ( )− = (14.2.8)

(b) From Eqs. (14.2.7) and (14.2.8), the states |ψ± ⟩ are eigenvectors of Jz, with respective nondegenerate eigenvalues l ± = ±�( j1 + j2). Since the operators Jz and J2 commute, we have

ˆ ˆ ˆ ( ˆ ) ( ˆ )J Jz zJ J J2 2 2| | |ψ ψ ψ± ± ± ±⟩ = ⟩ = ⟩λ

(14.2.9)

Consequently, the vectors | |ψ ψ± ±⟩ ≡ ⟩J2 are also eigenvectors of Jz with the same eigenvalues l±. How-ever, due to the nondegeneracy of l+ (or l−), the eigenvectors | �ψ+ ⟩ must be proportional to |ψ+ ⟩ (and similarly, | �ψ− ⟩ is proportional to |ψ− ⟩). Therefore, J2 | |ψ ψ± ±⟩ ∝ ⟩, so that |ψ+ ⟩ and |ψ− ⟩ are eigenvectors

of J2 as well as of Jz. Indeed, since |ψ± ⟩ both correspond to the extreme possible values of m1 and m2,

( ˆ ˆ ˆ ˆ ) ( ˆ ˆJ J J J m j m j J J1 2 1 2 1 1 2 2 1 2+ − − + + −+ = = ⟩ = +| , ˆ ˆ )J J m j m j1 2 1 1 2 2 0− + = − = − ⟩ ≡| ,

(14.2.10)

and

( ˆ ˆ )

( ˆ

J J m j m j j j m j m jz z1 2 1 1 2 2 1 2 1 1 2 2| |= = ⟩ = = = ⟩, ,

JJ J m j m j j j m j m jz z1 2 1 1 2 2 1 2 1 1 2 2ˆ ) | |= − = − ⟩ = = − = −, , ⟩⟩

⎧⎨⎪

⎩⎪ (14.2.11)

Therefore,

ˆ ( ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆJ J J212

22

1 2 1 2 12|ψ± + − −⟩ = + + + +J J J J Jz z JJ

j j j j j j

2

21 1 2 2 1 21 1 2

+ ±

±

⟩

= + + + + ⟩

=

)

[ ( ) ( ) ]

|

|

ψ

ψ�

�22

1 2 1 2 1[( ) ( )]j j j j+ + + + ⟩±|ψ (14.2.12)

Thus, |ψ+ ⟩ and |ψ− ⟩ both correspond to the same eigenvalue of J2 given by �2J(J + 1) = �2( j1 + j2) × ( j1 + j2 + 1).

14.3. Consider two angular momenta, both of magnitude J. Let ˆ ˆ ˆJ J J= +1 2 be the total angular momentum

and P the interchange operator defined by P m m m m| |1 2 2 1⟩ = ⟩. (a) Find the eigenvalues of P. (b) Show

that P commutes with J; i.e., [ ˆ ˆ]P, J = 0 (and [ ˆ ˆ ]P, J2 0= ). (c). Obtain the simultaneous eigenvalues of J2 and P.

SOLUTION

(a) Let us denote by |ψ⟩ an eigenvector of P with an eigenvalue l; namely, P | |ψ ψ⟩ = ⟩λ . Therefore,

( ˆ) ˆ ˆP PP2 2| | |ψ ψ ψ⟩ = ⟩ = ⟩λ (14.3.1)

Expanding |ψ⟩ in the (complete) |m m1 2⟩ basis, we have

| | |ψ ψ⟩ = ⟨ ⟩ ⟩∑ m m m m

m m

1 2 1 2

1 2,

(14.3.2)


However, by the definition of P, ( ˆ)P m m m m21 2 1 2| |⟩ = ⟩, and then

( ˆ) (( ˆ) )

,

P m m P m m

m m

21 2

21 2

1 2

| | | |ψ ψ ψ⟩ = ⟨ ⟩ ⟩ = ⟩∑ (14.3.3)

Comparing Eqs. (14.3.1) and (14.3.3) we find that l2 = 1 and, as a result, the eigenvalues of P must be l = ±1.

(b) The action of the operator ˆ ˆ ˆJ J J= +1 2 on the basis states

| | | |m m j m j m j m j m1 2 1 1 2 2 1 1 2 2⟩ = ⟩ ≡ ⟩ ⊗ ⟩, (14.3.4)

can be written as

( ˆ ˆ ) ( ˆ ) (J J J1 2 1 2 1 1 1 2 2 1 1+ ⟩ = ⟩ ⊗ ⟩ + ⟩ ⊗| | | |m m j m j m j m ˆ )J2 2 2| j m (14.3.5)

Therefore,

ˆ ˆ ( ˆ ) ( ˆ )P m m j m j m j mJ J J| | | | |1 2 2 2 2 1 1 1 2 2⟩ = ⟩ ⊗ ⟩ + ⟩ ⊗ jj m1 1⟩ (14.3.6)

Similarly [using Eq. (14.3.5) with the interchange m1 ↔ m2],

ˆ ˆ ˆ ( ˆ )J J JP m m m m j m j m j m| | | | |1 2 2 1 1 2 2 1 1 2⟩ = ⟩ = ⟩ ⊗ ⟩ + 22 2 1 1⟩ ⊗ ⟩( ˆ )J | j m (14.3.7)

Clearly, the last two expressions, Eqs. (14.3.7) and (14.3.6), coincide. Hence,

[ ˆ ˆ] [ ˆ ˆ ]P P, ,J J= → =0 02 (14.3.8)

(c) The results of part (b) imply that the | JM⟩ basis vector can be taken as simultaneous eigenvectors of the

set {ˆ ˆ ˆ ˆ ˆ}J J J12

22 2, , , ;Jz P . This means that | JM⟩ states have definite parity ±1 under the operator interchange

P. Indeed, using Eq. (14.3.2) together with the symmetry property of the Clebsch–Gordan coefficients,

⟨ ⟩ = − ⟨ ⟩+ −m m JM m m JM

j j J2 1 2 11 1 2| |( ) (14.3.9)

we find

ˆ ( ˆ )P JM m m JM P m m m m JM m m

m

| | | | |⟩ = ⟨ ⟩ ⟩ = ⟨ ⟩ ⟩1 2 1 2 1 2 2 1

11 21 2

1 2

2 1 1 2

,,

,

mm m

m m

m m JM m m

∑∑

∑= ⟨ ⟩ ⟩| | (14.3.10)

where in the last line we interchanged the order of the summation index. Therefore,

ˆ ( ) ( )P JM m m JM m mj j J j j| | |⟩ = − ⟨ ⟩ ⟩ = −+ − +

1 11 2 1 2

1 2 1 2−− ⟩∑ J

m m

JM|

1 2, (14.3.11)

In particular, for j1 = j2 = j, the number J assumes the values J = 0, 1, . . . , 2j, and then

ˆ ( )P JM JMj J| |⟩ = − ⟩−1 2 (14.3.12)

14.4. A system of two independent spin 1/2 particles whose orbital motion can be neglected is described by the basis | | |S m S m m m1

12 1 2

12 2 1 2= ⟩ ⊗ = ⟩ ≡ ⟩, , where |m m1 2 ⟩ are common eigenstates of

ˆ ˆ ˆ ˆS S12

22

1 2, , ,S Sz z . Consider the total spin operator ˆ ˆ ˆ ,S S S= ′ +1 2 with components ˆ ˆ ˆ ˆS = ( , , )S S Sx y z

and magnitude ˆ ˆ ˆS S S21 2

2= +| | . (a) Apply the operators ˆ ˆ ˆS S iSx y± = ± and Sz on states |m m1 2 ⟩ and


calculate the results. (b) As in part (a), apply ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆS S S212

22

1 2 1 2 1 22= + + + ++ − − +S S S S S Sz z on |m m1 2 ⟩

and calculate the results. (c) Construct the states |sms ⟩, which are eigenstates of ˆ ˆ ˆ ,S S S12

22 2, , and

Sz, as linear combinations of |m m1 2 ⟩. Find the corresponding eigenvalues and verify that ˆ ( )S2 2 1| |sm s s sms s⟩ = + ⟩� and S sm m smz s s s| |⟩ = ⟩� . (d ) Discuss the symmetry properties of the

|sms ⟩ under the interchange of the particles P m m m m| |1 2 2 1⟩ = ⟩.

SOLUTION

(a) To calculate the action of ˆ ˆ ˆ ,S S S= +1 2 on the states |m m1 2⟩, we introduce the following notations:

| | | | |m m112 2

12= ± = ± ⟩ = + +⟩ + −⟩ − +⟩ − −⟩, { , , , }

So,

ˆ ( ˆ ˆ ) ( )

ˆ

S

S

1 1 2 1 2 1 2 1 22 2| | | |m m I m m m m⟩ = ⊗ ⟩ = ⟩ ⟩� �σ σ

22 1 2 1 2 1 2 1 22 2| | | |m m I m m m m⟩ = ⊗ ⟩ = ⟩ ⟩

⎧

⎨⎪

� �( ˆ ˆ ) ( )σ σ⎩⎩

⎪ (14.4.1)

Here, I1,2

and ˆ ( ˆ ˆ ˆ ), ,σ σ σ σ1 2 1 2= x y z, , denote single-spin operators, which are represented by the 2 × 2 unit matrix and the three Pauli matrices (respectively) and satisfy

ˆ ˆ ˆ ˆσ σ σ σz z| | | | | | | |+⟩ = +⟩ −⟩ = − −⟩ −⟩ = +⟩ +⟩ = −⟩+ −2 2

ˆ ˆ ( ˆ ˆ ˆ ˆ ˆ)σ σ σ σ σ σ+ − ±+⟩ = −⟩ = ≡ ± =| |0 0 32x yi I

(14.4.2)

The total spin operator, ˆ ˆ ˆ ,S S S= +1 2 takes the form ˆ ( ˆ ˆ ˆ ˆ )S = ⊗ + ⊗�2 1 2 1 2σ σI I and consequently,

ˆ ( ˆ ˆ ˆ ˆ )

ˆ ( ˆ ˆ ˆ

S

S

z z zI I

I I

= ⊗ + ⊗

= ⊗ +± ±

�

�

2

2

1 2 1 2

1 2

σ σ

σ 11 2⊗

⎧

⎨⎪⎪

⎩⎪⎪ ±

ˆ )σ

(14.4.3)

(14.4.4)

Therefore, using Eqs. (14.4.3) and (14.4.4),

ˆ

ˆ ˆ

ˆ

S

S S

S

z

z z

z

| |

| |

| |

+ +⟩ = + +⟩

+ −⟩ = − +⟩ =

− −⟩ = − − −

�

�

0

⟩⟩

⎧

⎨⎪⎪

⎩⎪⎪

(14.4.5)

Similarly, using Eqs. (14.4.4) and (14.4.2),

ˆ ˆ ( )

ˆ ˆ

S S

S S

+ −

+ +

− −⟩ = + +⟩ = + −⟩ + − +⟩

+ −⟩ = − +⟩

| | | |

| |

�

== + +⟩

+ −⟩ = − +⟩ = − −⟩

+ +⟩ = − −⟩− −

+ −

�

�

|

| | |

| |

ˆ ˆ

ˆ ˆS S

S S ==

⎧

⎨

⎪⎪

⎩

⎪⎪⎪ 0

(14.4.6)

(b) In the notations of part (a) the operator S S S2 2= +| |ˆ ˆ1 2 equals

ˆ ( ˆ ˆ ˆ ˆ ˆ ˆ )S2

2

1 2 1 2 1 246 2= + ⊗ + ⊗ + ⊗+ − − +

� σ σ σ σ σ σz z (14.4.7)


where the identities ˆ ˆ ˆ ˆ ( ˆ ˆ ˆ ˆ )S S S S S S S Sx x y y1 2 1 2 1 2 1 2

12

+ = ++ − − + and ˆ ( ) ˆσ 2 3= I have been used. Therefore, from

Eqs. (14.4.2) and (14.4.7) we get

ˆ ( )

ˆ [(

S

S

22

2

22

46 2 2

46

| | |

|

+ +⟩ = + + +⟩ = + +⟩

+ −⟩ = −

��

�22 4 2

2 2

) ] ( )

ˆ

ˆ

| | | |

| |

+ −⟩ + − +⟩ = + −⟩ + − +⟩

− +⟩ = + −⟩

�

S S

SS22

2

46 2 2| | |− −⟩ = − − −⟩ = − −⟩

⎧

⎨

⎪⎪⎪⎪

⎩

⎪⎪⎪⎪

��( )

(14.4.8)

(c) By direct inspection of Eqs. (14.4.5) and (14.4.8) and in accordance with the results of Problem 14.2, we find

ˆ ˆ

ˆ

S

S

2 2

2 2

2

2

| | | |

| |

+ +⟩ = + +⟩ + +⟩ = + +⟩

− −⟩ = − −⟩

� �

�

Sz

ˆSz | |− −⟩ = − − −⟩�

(14.4.9)

Moreover,

ˆ ( ) ( )

ˆ ( )

S

S

2 2

2

2| | | |

| |

+ −⟩ + − +⟩ = + −⟩ + − +⟩

+ −⟩ − − +⟩

�

==

⎧⎨⎪

⎩⎪ 0

(14.4.10)

and

ˆ ( ) ˆ ( )S Sz z| | | |+ −⟩ + − +⟩ = + −⟩ − − +⟩ = 0

(14.4.11)

Therefore, up to the unimportant global phase we obtain

| |

| | |

|

s m

s m

s

s

s

= = ⟩ = + +⟩

= = ⟩ = + −⟩ + − +⟩

=

1 1

1 012

1

,

, ( )

,,triplet

,

m

s

s m

s

s

= − ⟩ = − −⟩

⎫

⎬⎪⎪

⎭⎪⎪

=

= = ⟩ =

1

1

0 012

|

| (( )| |+ −⟩ − − +⟩ ⎫⎬⎭

=s 0singlet

(14.4.12)

where the states { }|sms ⟩ are orthonormal and they all satisfy

ˆ ( )

ˆ

S2 2 1 1 0| |

| |

sm s s sm s

sm m sm

s s

z s s s

⟩ = + ⟩ =

⟩ =

�

�

,

S ⟩⟩ = −ms 1 0 1, ,

(14.4.13)

(d ) The symmetry properties of applying the interchange m1 ↔ m2 to the |sms ⟩ states follow from the expres-sions in Eq. (14.4.1). By direct observation of these equations we can see that the s = 1 (triplet) states are not affected by the interchange operation, whereas the S = 0 (singlet) state changes its sign. That is,

ˆ ( )

ˆ ( )

P

P

triplet triplet

singlet singlet

=

= −

⎧⎨⎪

⎩⎪⎪ (14.4.14)

where P m m m m| |1 2 2 1⟩ ≡ ⟩ is the interchange operator.

Note: The expressions of Eq. (14.4.14) are in accordance with Eq. (14.3.13), where one only needs to replace 2j → 1 and J → s = 0, 1.


14.5. Let ˆ ˆ ˆS S S= +1 2 be the total angular momentum of two spin 1/2 particles ( )s s1 2 1 2= = / . Calculate the Clebsch–Gordan coefficients ⟨ ⟩m m sms1 2 | by successive applications of ˆ ˆ ˆS S iSx y± = ± on the vectors |sms ⟩. Work separately in the two subspaces s = 1 and s = 0.

SOLUTION

In order to find the Clebsch–Gordan coefficients for the addition of spin s1 = s2 = 1/2, we shall use the following relations [see Eqs. (14.12) in Sec. 14.3]:

,ˆ ( ) ( )

ˆ

S sm s s m m s m

S m m

s s s s±

±

⟩ = + − ± ± ⟩| |

|

� 1 1 1

1 1 22 1 1 1 1 1 2

2 1 2

1 1 1⟩ = + − ± ± ⟩

⟩ =±

�

�

s s m m m m

S m m

( ) ( )

ˆ

|

|

,

ss s m m m m2 2 2 2 1 21 1 1( ) ( )+ − ± ±

⎧⎨⎪

⎩⎪ | ,

(14.5.1)

We shall also use the phase condition

| |s s s m s s m s m ss= + = ± + ⟩ = = ± = ± ⟩1 2 1 2 1 1 2 2, ,( )

(14.5.2)

Note: The states |s s s m s ss= + = ± + ⟩1 2 1 2, ( ) are eigenstates of S2 and Sz, with nondegenerate eigenvalues l± = ±� (s1 + s2), respectively (see Prob. 14.2). Therefore,

| |s s s m s s e m s m ssi= + = ± + ⟩ = = ± = ± ⟩1 2 1 2 1 1 2 2, ,( ) φ

and the phase φ may be chosen as φ = 0. For subspace s = 1: From Eq. (14.5.2) we immediately have

| |1 1 12

12, ,⟩ = = + +⟩

(14.5.3)

Then, operating with ˆ ˆ ˆS S S− − −= +1 2 on both sides of Eq. (14.5.3) and using Eq. (14.5.1), we obtain

ˆ , ( ) ( ) , ,

ˆ ,

S

S

−

−

⟩ = + − − ⟩ = ⟩| | |

|

1 1 1 1 1 1 1 1 1 0 2 1 0

1 1

� �

⟩⟩ = + = + −

⎧

⎨⎪⎪

⎩⎪ − −( ˆ ˆ )S S1 2

12

112

112

,12

,12

,12

� �−⎪⎪

(14.5.4)

Thus,

| | |1 012

12

12

12

12

12

, , ,⟩ = − + −⎛⎝⎜

⎞⎠⎟

= + −⟩ + − +⟩( )

(14.5.5)

Similarly, operating with S− once again on the state |1 0, ⟩, we find

ˆ ( ) ( )

ˆ

S

S

−

−

⟩ = + − − − ⟩ = − ⟩| | |

|

1 0 1 1 1 0 0 1 1 1 2 1 1

1

, , ,� �

,, , , ,01

2

12

12

12

12

1

2

12

121 2⟩ = − + −⎛

⎝⎜⎞⎠⎟

+ −− −ˆ ˆS S ++ −⎛

⎝⎜⎞⎠⎟

= − − + − −⎛⎝⎜

⎞⎠⎟

= −

12

12

2

12

12

12

12

2

2

1

,

, ,�

2212

, −

⎧

⎨

⎪⎪⎪⎪

⎩

⎪⎪⎪⎪

(14.5.6)

Therefore, in accordance with the condition in Eq. (14.5.2),

| |1 112

12

, ,− ⟩ = − − = − −⟩

(14.5.7)


For subspace s = 0: Since ms = m1 + m2 (in this case ms = 0), we have

|0 012

12

12

12

, , ,⟩ = − + −α β

(14.5.8)

Next, due to the orthonormality of |sms ⟩ basis we get

⟨ ⟩ = → + = → = −

⟨ ⟩ = → +

1 0 0 0 012

0

0 0 0 0 1 2

, ,

, ,

|

| | |

( )α β β α

α || | | |β α α2 21 2 1 1 2= → = → = /

Therefore,

|0 012

12

12

12

12

, , ,⟩ = − − −⎛⎝⎜

⎞⎠⎟

(14.5.9)

14.6. Let ˆ ˆ ˆS S S= +1 2 be the total angular momentum of two spin 1 particles. (a) Represent the vectors |sms ⟩ as linear combinations of | | | |s m s m m m1 1 2 2 1 2⟩ ⊗ ⟩ ≡ ⟩ ⟩ in the subspace s = 2. (b) Repeat part (a), working in the subspace s = 1. (c) Repeat part (a), working in the subspace s = 0.

SOLUTION

(a) For s = 2, ms = 2 (ms = m1 + m2), we immediately have

| | |2 2 1 1, ⟩ = ⟩ ⟩ (14.6.1)

Applying ˆ ˆ ˆS S S− − −= +1 2 to both sides of Eq. (14.6.1) we find

ˆ ( ) ( )

ˆ (

S

S

−

−

⟩ = + − − ⟩

⟩ ⟩ = +

| |

| |

2 2 2 2 1 2 2 1 2 1

1 1 1 1

, ,�

� 11 0 1 1 0)( )| | | |⟩ ⟩ + ⟩ ⟩

⎧⎨⎪

⎩⎪

(14.6.2)

Thus,

| | | | |2 112

0 1 1 0, ⟩ = ⟩ ⟩ + ⟩ ⟩( )

(14.6.3)

Applying ˆ ˆ ˆS S S− − −= +1 2 once more to both sides of Eq. (14.6.3), we obtain

ˆ ( ) ,

ˆ ( )

S

S

−

−

⟩ = + ⟩

⟩ ⟩ + ⟩ ⟩ =

| |

| | | |

2 1 2 2 1 2 0

12

0 1 1 0

, �

�

222 1 1 0 0 0 0 1 1[ ( )]| | | | | | | |− ⟩ ⟩ + ⟩ ⟩ + ⟩ ⟩ + ⟩ − ⟩

⎧

⎨⎪

⎩⎪⎪

(14.6.4)

Hence,

| | | | | | |2 016

1 1 2 0 0 1 1, ⟩ = − ⟩ ⟩ + ⟩ ⟩ + ⟩ − ⟩[ ]

(14.6.5)

Similarly, we obtain

| | | | |2 112

0 1 1 0, − ⟩ = ⟩ − ⟩ + − ⟩ ⟩( )

(14.6.6)

and finally,

| | |2 2 1 1, − ⟩ = − ⟩ − ⟩ (14.6.7)

Note: One can, obviously, take Eq. (14.6.7) as a starting point and calculate the state |2 1, − ⟩ “up the ladder”

with the help of the operator ˆ ˆ ˆS S S+ + += +1 2.


(b) For S = 1, the state |1 1, ⟩ can be written as

| | | | |1 1 1 0 0 1, ⟩ = ⟩ ⟩ + ⟩ ⟩α β (14.6.8)

where the constants a and b are determined by orthonormality. Thus,

⟨ ⟩ = → + = → = −

⟨ ⟩ = → +

2 1 1 1 012

0

1 1 1 1 1 2

, ,

, ,

|

| | |

( )α β β α

α || | | |β α2 21 2 1= → = (14.6.9)

which leads to

| | | | |1 112

1 0 0 1, ⟩ = ⟩ ⟩ − ⟩ ⟩( )

(14.6.10)

Now,

ˆ ( )

ˆ ( )

S

S

−

−

⟩ = + ⟩

⟩ ⟩ + ⟩ ⟩ =

| |

| | | |

1 1 1 1 1 1 0

12

1 0 0 1

, ,�

�

222 0 0 1 1 0 0 1 1( )| | | | | | | |⟩ ⟩ + ⟩ − ⟩ − ⟩ ⟩ − − ⟩ ⟩

⎧⎨⎪

⎩⎪

(14.6.11)

so that

| | | | |1 012

1 1 1 1, ⟩ = ⟩ − ⟩ − − ⟩ ⟩( )

(14.6.12)

Repeating the process once again, we obtain

| | | | |1 112

1 0 0 1, − ⟩ = − ⟩ ⟩ − ⟩ − ⟩( )

(14.6.13)

(c) The subspace s = 0 contains only one state that can be written as

| | | | | | | |0 0 1 1 1 1 0 0⟩ ⟩ = ⟩ − ⟩ + − ⟩ ⟩ + ⟩ ⟩γ δ ρ (14.6.14)

where g, d, and r are arrived at from orthonormality conditions:

⟨ ⟩ =⟨ ⟩ =⟨ ⟩ =

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪

2 0 0 0 01 0 0 0 00 0 0 0 1

, ,, ,, ,

|||

⇒⇒ = = − =γ δ ρ1

3

1

3

1

3, ,

(14.6.15)

Therefore,

| | | | | | |0 013

1 1 0 0 1 1, ⟩ = ⟩ − ⟩ − ⟩ ⟩ + − ⟩ ⟩( )

(14.6.16)

Note: The states s = 0, 2 are symmetric under the exchange of particles, whereas in s = 1 they are antisymmetric.

14.7. A system of two angular momenta, of respective magnitudes j1 = 1 and j2 = 2, is described by the basis | |j m j m1 1 2

12 21= ⟩ ⊗ = ⟩, , . The system is in a state |JM⟩, where J is the total angular momentum and

M is the z component of J. Consider, in particular, the states (a) |J M= = ⟩32

32, and (b) | J M= = ⟩1

212, .

For each state calculate the probability of measuring each pair of possible values (m1, m2), and find the expectation values of J z1 and ˆ .J z2 (c) Calculate the expectation value of Jy in the state |J M= = ⟩1

212, .

SOLUTION

(a) The possible values for (m1, m2) are

j m

j m

1 1

2 2

1 1 0 112

12

12

= → = −

= → = −

⎧⎨⎪

⎩⎪

, ,

,

(14.7.1)


The possible values of | JM⟩ are then

J M

J M

= → = − −

= → = −

⎧

⎨⎪

⎩⎪

32

32

12

12

32

12

12

12

, , ,

,

(14.7.2)

In particular, for | J M= = ⟩32

32, we have

J M j m j m= = = = = ⟩ = =32

32 1 1 2

12 2

121 1, , ,|

(14.7.3)

Therefore,

prob , , / ,m m m m J M1 2 1 2

32

32

21

12

1 1 2= =⎛⎝⎜

⎞⎠⎟ = = = = = = 11

112

01 2prob orm m≠ ≠⎛⎝⎜

⎞⎠⎟ =

⎧

⎨⎪⎪

⎩⎪⎪

(14.7.4)

The expectation values of J z1 and J z2 are given by

⟨ ⟩ =

⟨ ⟩ =

JM J JM m

JM J JM m

z

z

| |

| |

ˆ

ˆ

1 1

2 2

�

�

(14.7.5)

Hence, for m1 = 1, m2

12

= , we find

⟨ ⟩ = ⟨ ⟩ =ˆ , ˆJ Jz z1 2 2� �/ (14.7.6)

(b) First, let us write the state | J M= = ⟩12

12, as a linear combination of |m m1 2⟩ states. Starting from

Eq. (14.7.3) we find

J J M J− = = = +⎛⎝⎜

⎞⎠⎟ − −⎛

⎝⎜⎞⎠⎟ =3

232

32

32

32

132

32

1, � ,,

, ,

M

J m m m m

=

= = = + = = +−

12

1 212 1 2

121 1 1 1 0

12

12

ˆ ( )� � ++⎛⎝⎜

⎞⎠⎟ + = = −

⎧

⎨⎪⎪

⎩⎪⎪ ⋅1

12

12

11 212m m,

(14.7.7)

Thus,

J M m m m m= = = = = + = = −32

12 1 2

12 1 2

12

23

013

1, , , (14.7.8)

Consequently, due to orthogonality,

J M m m m m= = = = = − = = −12

12 1 2

12 1 2

12

13

023

1, , , (14.7.9)

Hence,

prob , , ,

pr

m m m m J M1 212 1 2

12

12

12

2130 0= =( ) = = = = = =

oob , , ,m m m m J M1 212 1 2

12

12

12

2231 1= = −( ) = = = − = = =

⎧

⎨⎨⎪

⎩⎪

(14.7.10)


where for all the other pairs P(m1, m2) = 0. The expectation values of J z1 and J z2 in the state | J M= = ⟩12

12,

are given by Eqs. (14.7.5) and (14.7.10). Indeed, substituting Eq. (14.7.9) into Eq. (14.7.5), we obtain

⟨ ⟩ = = + = =

⟨ ⟩ = =

ˆ ( ) ( )

ˆ (

J m m

J m

z

z

1 1 1

2 2

13

023

123

13

� � �

112

23

12

162) ( )� � �+ = − = −

⎧

⎨⎪

⎩⎪ m

(14.7.11)

(c) The operator Jy can be written as

ˆ ( ˆ ˆ )Ji

J Jy = −+ −12

(14.7.12)

Therefore,

⟨ ⟩ = ⟨ + − + + ⟩ +JM J JM

iJM J J M M J M JJy| | | |ˆ ( ( ) ( ) ,

�2

1 1 1 ++ − − − ⟩ =1 1 1 0MM J M| ,

(14.7.13)

Alternatively, for | J M= = ⟩12

12, we can choose the well-known spin 1/2 representation (see Chap. 7):

J M J M Jy= = ⎛⎝⎜

⎞⎠⎟ = = − = ⎛

⎝⎜⎞⎠⎟ =

−12

12

12

12

10

01 2

0, ,

� iii 0

⎛⎝⎜

⎞⎠⎟

(14.7.14)

This leads to the following expression:

⟨ ⟩ =−⎛

⎝⎜⎞⎠⎟

⎛⎝⎜

⎞⎠⎟ =ˆ ( )J

iiy 1 0

20

010

0�

(14.7.15)

which coincides with Eq. (14.7.13).

14.8. Consider a system of two spin 1/2 particles whose orbital variables are ignored. The Hamiltonian of the system is ˆ ˆ ˆH z z= +ε ε1 1 2 2σ σ , where e1 and e 2 are real constants, and ˆ , ˆσ σ1 2z z are the projections of

the spins ˆ ˆS1 12= � σ and ˆ ˆS2 22

= � σ of the two particles onto the z axis. (a) The initial state of the system,

at t = 0, is | | |ψ( ) ( )01

2⟩ = + −⟩ + − +⟩ . ˆ ( ˆ ˆ )S S S2

1 22= + is measured at time t. What are the values that can

be arrived at and what are their probabilities? (b) If the initial state of the system is arbitrary, what Bohr frequencies might appear in the evolution of ⟨ ⟩S2 ? (c) Answer parts (a) and (b) for ˆ ˆ ˆS S Sx x x= +1 2 .

SOLUTION

(a) The eigenstates of ˆ ( ˆ ˆ )S S S21 2

2= + (and Sz) are the |sms ⟩ states, where s = 1, 0 correspond to the triplet and singlet states, respectively. The results of the measurement of S2 are, therefore,

s

s

= →

= →

⎧⎨⎪

⎩⎪

1 2

0 0

2�

(14.8.1)

However, the states |sms ⟩ are not eigenstates of the Hamiltonian and consequently the probabilities, prob (s = 1) and prob (s = 0), are changed as a function of time. The stationary states of the system are

m m1 2

12

12

= ± = ± = + +⟩ + −⟩ − +⟩ − −⟩, , , ,{ }| | | | (14.8.2)

and its energy levels are given by

ˆ ( ˆ ˆ ) ( )H z z| | |−⟩ = + −⟩ = ± ± −⟩ε ε ε ε1 1 2 2 1 2σ σ

(14.8.3)

Therefore, taking into account the initial state of the system,

| | | |ψ( ) ( )012

1 0⟩ = + −⟩ + − +⟩ = = = ⟩s ms, (14.8.4)


we find

| | |ψ( ) exp expt i t i t⟩ = − − + −⟩ + − − +⟩12 1 2 1 2ε ε ε ε/ /� �{{ } (14.8.5)

Now, writing the states of Eq. (14.8.5) in the form

| | |

| | |

+ −⟩ = ⟩ + ⟩

− +⟩ = ⟩ − ⟩

⎧

⎨⎪

12

1 0 0 0

12

1 0 0 0

( )

( )

, ,

, ,

⎪⎪

⎩⎪⎪

(14.8.6)

and substituting Eq. (14.8.6) into Eq. (14.8.5) yields

| | |ψ( ) cos( )

sin( )

tt

it

⟩ =−

⟩ −−

⟩ε ε ε ε1 2 1 21 0 0 0

� �, , (14.8.7)

Hence, the probabilities prob (s = 1) and prob (s = 0) are

prob ,( ) ( ) cos(

, ,

s s m t

m

s

s

= = ⟨ = ⟩ == −∑1 1

1 0 1

2

2 1|ψε −−

= = ⟨ = = ⟩ =

ε

ε

2

2 2 10 0 0

)

( ) ( ) sin(

t

s s m ts

�

prob , |ψ−−

⎧

⎨

⎪⎪⎪

⎩

⎪⎪⎪

ε2) t�

(14.8.8)

(14.8.9)

Moreover, the expectation value of S2 is

⟨ ⟩ = = + =⋅ ⋅ψ ψ( ) ˆ ( ) ( ) )t t s s| |S2 22 1 0 0� prob prob (

== − = +2 12 21 2

2� � �cos [( ) ] [ cos )]ε ε ωt tB/ (

(14.8.10)

where wB = 2 (e1 − e2) t/� is the Bohr frequency. Note that Eqs. (14.8.6) and (14.8.7) contain linear combination of the |sms ⟩ states with ms = 0. Indeed, from Eqs. (14.8.3) and (14.8.4) we find that the

operator ˆ ˆ ˆS S Sz z z= +1 2 commutes with the Hamiltonian and ˆ ( ) .Sz |ψ 0 0⟩ = Thus,

| | |ψ ( ) ( ) ( )t C t C t⟩ = ⟩ + ⟩1 21 0 0 0, , (14.8.11)

where C1 (t) and C2 (t) are time-dependent (complex) coefficients. As a result, we also have

prob ,

prob

( ) ( ) ( )

(

s s m t C ts= = ⟨ = = ⟩ =1 1 02

12| | |ψ

ss s m t C ts= = ⟨ = = ⟩ =

⎧⎨⎪

⎩⎪ 0 0 02

22) ( ) ( ), | | |ψ

(14.8.12)

where the sum in Eq. (14.8.8) is reduced to a single term (ms = 0).

(b) We consider an arbitrary initial state of the form

| | | | |ψ( )0 ⟩ = + +⟩ + + −⟩ + − +⟩ + − −⟩α β γ δ (14.8.13)

where a, b, g, and d are complex constants. In this case, the evolution of |ψ( )t ⟩ is given by

| | |ψ( )( ) ( )

t e ei t i t⟩ = + +⟩ + + −− + − −α βε ε ε ε1 2 1 2/ /� � ⟩⟩ + − +⟩ + − −⟩− +γ δε ε ε ε

e ei t i t( ) ( )1 2 1 2/ /� �| | (14.8.14)

Using Eq. (14.8.6), we then find

ψ( ) , ,( ) ( )

t e ei it t

= ⟩ + −− + +α δε ε ε ε1 2 1 21 1 1 1/ /� �

| | ⟩⟩

+ + ⟩

+

− − −12

1 0

12

1 2 1 2( )( ) ( )β γε ε ε ε

e ei t i t/ /

,� � |

(( )( ) ( )β γε ε ε ε

e ei t i t− − −− ⟩1 2 1 2 0 0

/ /,

� � | (14.8.15)


Therefore, the expectation value of S2 is

⟨ ⟩ = + + + +ψ ψ( ) ˆ ( ) ( )t t| | | | | | | | | |S2 212

2 2 2 2 2� α δ β γ RRe [ * ]

( )β γ ε εe

i t2 1 2−{ }/�

(14.8.16)

Clearly, ⟨ ⟩S2 is characterized by a single Bohr frequency that is identical to the one from part (a) and equals wB = 2 (e1 − e2)/�.

Note: For a = d = 0 and β γ= = 2 , Eq. (14.8.16) is reduced to

⟨ ⟩ = +−⎛

⎝⎜⎞⎠⎟

⎡

⎣⎢

⎤ψ ψ( ) ˆ ( ) cos

( )t t t| |S2 2 1 21

2�

�

ε ε

⎦⎦⎥

(14.8.17)

which coincides with Eq. (14.8.10).

(c) To find the expectation value Sx, we return to Eq. (14.8.14) and calculate the ket ( ˆ ˆ ) ( ) .S S tx x1 2+ ⟩|ψ This gives

ˆ ( ) {[ ]

( ) ( )S t e ex

i t i t|ψ ⟩ = +− + +� � �

21 2 1 2α δε ε ε ε/ /

(( )

[ ]( ) ( )

| |+ −⟩ + − +⟩

+ +− − −β γε ε ε εe e

i t i t1 2 1 2/ /� �(( )}| |+ +⟩ + − −⟩ (14.8.18)

Therefore,

⟨ ⟩ = + +ψ ψ( ) ˆ ( ) { * *t S t e exi t i t| |

� � �

22 22 1α β α γε ε/ / δδ β δ γ

βα β δ

ε ε

ε

* *

* *

e e

e

i t i t

i t

− −

−

+

+ +

2 2

2

1 2

2

/ /

/

� �

�ee e e

i t i t i t2 2 21 1 2ε ε εγ α γ δ

α

/ / /� � �

�

+ +

=

−* * }

Re{( * ββ γ δ α γ β δε ε+ + +* ) ( * * ) }e ei t i t2 22 2/ /� � (14.8.19)

In this case, the Bohr frequencies that appear in the evolution of ⟨ ⟩Sx are wB1 = 2e1/� and wB2 = 2e2/�.

14.9. The total angular momentum of a spin 1/2 particle is J = L + S, where L is the orbital momentum and S is the spin (l is an integer, s = 1/2). Let | | |l m s m m ml s l s, / , ,⟩ ⊗ = ⟩ ≡ ⟩1 2 be the eigenstates

of {ˆ ˆ ˆ ˆ }L S2 2, , ,L Sz z and |JM⟩ the eigenstates of { J2ˆ , ˆ }.J z Find the Clebsch–Gordan coefficients

⟨ ⟩m m JMl s, | by successive applications of ˆ ˆ ˆJ L S± ± ±= + to the vectors |JM⟩. Work separately in the two subspaces J = l + 1/2 and J = l − 1/2.

SOLUTION

First we notice that if l = 0 the vectors |m mt s= = ± ⟩0 1 2, / are eigenstates of {ˆ , ˆ }.J2 Jz In this case J = s = 1/2 and M = ms = ±1/2. Therefore,

| | |J M m m ml s= = ⟩ = = = ± ⟩ ≡ = ±⟩1 2 1 2 0 1 2 0/ , / , / ,

(14.9.1)

The only nonzero Clebsch–Gordan coefficients are then ⟨ + ⟩ = ⟨ − − ⟩ =012

12

012

12

1, , ,| |, . On the other hand, for l ≠ 0, there are two possibilities:

J l M l l l

J l M l

= + = + − − +⎛⎝⎜

⎞⎠⎟

= − = −

12

12

12

12

12

, , . . . ,

112

32

12

, l l− − −⎛⎝⎜

⎞⎠⎟

⎧

⎨⎪⎪

⎩⎪⎪

, . . . ,

(14.9.2)

In this case (l ≠ 0), we will consider the subspaces J l= + 12

and J l= − 12

separately, and show that

| | |J l Ml

l M m M l M m M= + ⟩ =+

+ + = − +⟩ + − + =12

12 1

12

12

12

, , ++ −⟩⎡

⎣⎢

⎤

⎦⎥

= − ⟩ =+

+ + = +

12

12

12 1

12

12

,

, ,| |J l Ml

l M m M −−⟩ + − + = − +⟩⎡

⎣⎢

⎤

⎦⎥

⎧

⎨

⎪⎪

⎩

⎪⎪ l M m M

12

12

| ,

(14.9.3)

(14.9.4)


where | |m m ml s, ,⟩ ≡ ±⟩ and M m m ml s= + = ± 12

.

For J l= + 12 : The subspace J l= + 1

2 contains a multiple of 2l + 1 independent eigenfunctions. As

usual (see Problem 13.2), the maximal eigenvalue, M lmax = + 12

is nondegenerate. Therefore,

| |J l M l m l= + = + ⟩ = = +⟩12

12

, ,

(14.9.5)

Now, applying the operator ˆ ˆ ˆJ L S− − −= + to Eq. (14.9.5), we get

ˆ ( )( )J l l J M J M l l l− + + ⟩ = + − + + − ⟩ =| |

12

12

112

12

2, ,� � ++ + − ⟩

+⟩ = + +⟩ =− − −

112

12

|

| | |

l l

J l L S l l

,

, , ( 2ˆ ( ˆ ˆ ) � ll l− +⟩ + −⟩

⎧

⎨⎪

⎩⎪ 1, ,| )

(14.9.6)

Thus, in accordance with Eq. (14.9.3)

| | |J l M ll

l l l= + = − ⟩ =+

− +⟩ + −⟩12

12

12 1

2 1, , ,[ ] (14.9.7)

Expression (14.9.7) can be generalized by recurrence. In general,

J l M l M l M l− + ⟩ = + +⎛⎝⎜

⎞⎠⎟ − +⎛

⎝⎜⎞⎠⎟ +| |

12

12

32

12

, ,� MM

J M l M l M

− ⟩

− +⟩ = + +⎛⎝⎜

⎞⎠⎟ − +⎛

⎝⎜⎞⎠⎟−

1

12

12

32

ˆ | |, � MM M

J M l M l

− +⟩ + − −

+ −⟩ = + +⎛⎝⎜

⎞⎠⎟−

32

12

12

12

, ,

,

�

�

|

|ˆ −− +⎛⎝⎜

⎞⎠⎟ − −⟩

⎧

⎨

⎪⎪

⎩

⎪⎪ M M

32

12

| ,

(14.9.8)

Therefore, the application of ˆ ˆ ˆJ L S− − −= + to both sides of Eq. (14.9.3) leads to

|l Ml

l M l M l M

+ − ⟩ =+

+ +⎛⎝⎜

⎞⎠⎟ + −⎛

⎝⎜⎞⎠⎟ −

12

11

2 1

12

12

,++⎛

⎝⎜⎞⎠⎟

+ +⎛⎝⎜

⎞⎠⎟ − +⎛

⎝⎜⎞⎠⎟

− +⟩

⎧ 32

12

32

32

l M l M

M| ,⎨⎨⎪⎪

⎩⎪⎪

++ + + −⎛

⎝⎜⎞⎠⎟

+ +⎛⎝⎜

⎞⎠⎟ −

l M l M

l M l M

12

112

12

++⎛⎝⎜

⎞⎠⎟

− −⟩

⎫

⎬⎪⎪

⎭⎪⎪

=+

+ − −

32

12

12 1

12

32

|

|

M

ll M M

,

, ++⟩ + − + − −⟩⎡

⎣⎢

⎤

⎦⎥l M M

32

12

| , (14.9.9)

Indeed, the resulting expression in Eq. (14.9.9) is identical to Eq. (14.9.3), where M is changed to M − 1.

For J l= − 12

: The state | J l M l= − = − ⟩12

12

, is a linear combination of |l, −⟩ and |l − +⟩1, . Note that

this is the only possible way to obtain m m ll s+ = − 12

, as in Eq. (14.9.7). Thus,

| | |J l M l l l= − = − ⟩ = − +⟩ + −⟩12

12

1, , ,α β (14.9.10)


This state must be orthogonal to Eq. (14.9.7). Therefore,

α β=+

= −+

22 1

12 1

l

l l

Namely,

| | |J l M ll

l l l= − = − ⟩ =+

−⟩ − − +⟩12

12

12 1

2 1, [ , , ] (14.9.11)

Now, we can apply ˆ ˆ ˆJ L S− − −= + to Eq. (14.9.11) and find all the other coefficients. A calculation similar to that of J l= + 1 2/ yields

| | |l Ml

l M M l M M− − ⟩ =+

+ − − −⟩ + − − −12

11

2 112

12

12

32

, , , ++⟩⎡

⎣⎢

⎤

⎦⎥ (14.9.12)

The last expression is identical to Eq. (14.9.4), where M is changed into M − 1.

14.10. Two spin 1/2 particles (who orbital variables are ignored) are described by an unperturbed Hamiltonian ˆ ( ˆ ˆ ).H A z z0 1 2= − +σ σ We add the perturbation ˆ ( ˆ ˆ ˆ ˆ ),H x x y y1 1 2 1 2= +ε σ σ σ σ where ˆ ( ˆ ˆ ˆ )σ σ σ σ= x y z, , are Pauli matrices, and ε << A are positive constants. (a) Find the eigenvalues and eigenfunctions of ˆ .H0 (b) Calculate (exactly) the energy levels and the corresponding eigenfunctions of the total Hamiltonian ˆ ˆ .H H0 1+ (c) Using the perturbation theory, calculate the first-order corrections to the energy levels of ˆ .H0 Compare them to the exact results of part (b).

SOLUTION

(a) The two spin 1/2 particles are described by the standard basis | | | |m m1 2

12

12

= ± = ± ⟩ = + +⟩ + −⟩ − +⟩, , ,{ ,

|− −⟩}. Since ˆ ( ˆ ˆ )H A S Sz z0 1 2

2= − +�

, we find

ˆ

ˆ ˆ

ˆ

H A

H H

H A

0

0 0

0

2

0

2

| |

| |

| |

+ +⟩ = − + +⟩

+ −⟩ = − +⟩ =

− −⟩ = −− −⟩

⎧

⎨

⎪⎪

⎩

⎪⎪

(14.10.1)

Thus, the eigenvalues of H0 are − 2A, 0, and 2A.

(b) The total Hamiltonian ˆ ˆH H0 1+ can be written as

ˆ ( ˆ ˆ ) [( ˆ ˆ ) ˆ ˆ ˆH A z z= − + + + − − −s s s s s s1 2 1 22

12

22

22

ε ss s1 2z zˆ ] (14.10.2)

Clearly, H commutes with the operators { ˆ ˆ ˆ ˆ }S S Sz12

22 2, , ,S , where ˆ ˆ ˆS S S= +1 2 is the total spin of the par-

ticles. Therefore, the eigenstates of the total Hamiltonian are the following triplet and singlet states:

| |

| | |

| |

1 1

1 012

1 1

,

,

,

⟩ = + +⟩

⟩ = + −⟩ + − +⟩

− ⟩ = − −⟩

⎧

⎨

⎪

( )⎪⎪

⎩

⎪⎪

⟩ = + −⟩ − − +⟩,| | |0 012

( )

(14.10.3)


The corresponding energy levels are, then,

ˆ ( )

ˆ

H A A

H

| | |

|

1 1 2 4 3 1 1 2 1 1

1 0

, , ,

,

⟩ = − + − − ⟩ = − ⟩

⟩ =

ε ε ε

(( )

ˆ (

0 4 3 1 0 2 1 0

1 1 2 4 3

+ − + ⟩ = ⟩

− ⟩ = + −

ε ε ε ε

ε ε

| |

|

, ,

,H A −− − ⟩ = − ⟩

⎧

⎨

⎪⎪

⎩

⎪⎪ ε) | |1 1 2 1 1, ,A

(14.10.4)

and

ˆ ( )H | | |0 0 0 0 3 0 0 2 0 0, , ,⟩ = + − + ⟩ = − ⟩ε ε ε (14.10.5)

(c) The matrix elements of ˆ ( ˆ ˆ ˆ ˆ )H1 1 2 1 22= ++ − − +

ε s s s s in the unperturbed basis { }| | | |+ +⟩ + −⟩ − +⟩ − −⟩, , ,

are given by the following matrix:

| | | |

||||

+ +⟩ + −⟩ − +⟩ − −⟩

+ +⟩+ −⟩− +⟩− −⟩

0 0 0 00 0 2 00 2 0 00 0 0 0

εε

⎛

⎝

⎜⎜⎜

⎞

⎠

⎟⎟⎟

(14.10.6)

The unperturbed energy levels are e1 = −2A, e2 = e3 = 0, and e4 = 2A. Therefore,

Δ

Δ

ε

ε

1 1

4 1

0

0

= ⟨+ + + +⟩ =

= ⟨− − − −⟩ =

⎧⎨⎪

⎩⎪

| |

| |

ˆ

ˆ

H

H

(14.10.7)

where e1 and e4 are the nondegenerate energy levels. As for the degenerate zero eigenvalue, we consider the determinant

det ( ˆ ˆ)H I12 22

24− = −

− = −λ λ εε λ λ ε

(14.10.8)

Thus,

Δε ε2 3 2, = ±

(14.10.9)

Equations (14.10.7) and (14.10.9) lead to

E A

E

E

1 1 1

2 3 2 3 2 3

4 4 4

2

2

2

= + = −

= + = ±

= + =

ε ε

ε ε ε

ε ε

Δ

Δ

Δ

, , ,

AA

⎧

⎨⎪⎪

⎩⎪⎪

(14.10.10)

which agrees with the exact result of part (b).

14.11. The motion of an electron in a central field of force is described by a Hamiltonian of the form

ˆ ˆ ˆ ,H H H= +0 so where ˆ ˆ ˆ( )Hm

V r0

2

2= +p

and ˆ ( ) ˆ ˆ.H rso = ζ L Si The spin-orbit coupling leads to energy

differences between levels with the same values of L2 and S2 but different values of J2, where J = L + S. (a) Show that [ ˆ ˆ ] [ ˆ ˆ ]H H, ,L S2 0= = but [ ˆ ˆ ]H Lz, ≠ 0 and [ ˆ ˆ ]H Sz, ≠ 0. (b) Show that [ ˆ ˆ ] [ ˆ ˆ ] .H H, ,J2 0= =Jz (c) Consider the stationary states of H that are also eigenstates of the observables { , , ,2ˆ ˆ ˆ ˆ }L S J2 2 Jz . Express the angular part of these eigenfunctions in terms of spherical harmonics and two-component spinors. (d ) Let the eigenfunctions of part (c) be characterized by the quantum numbers l, J, and M (which are related to the eigenvalues of L2, J2, and Jz

, respectively). Determine the possible values of Lz and Sz and find their probabilities and average values.


SOLUTION

(a) The Hamiltonian H0 commutes with all the components of L and S , and the operator L acts only on the angular variables (q, j) (see Chap. 6). Therefore,

[ ˆ ˆ ] [ ˆ ( ) ˆ ˆ ˆ ] ( )[ ˆ ˆ ˆH H r r, , ,L L S L L S L20

2 2= + =ζ ζi i ]]

(14.11.1)

In addition, [ ˆ ˆ ]S Li j, = 0 and [ ˆ ˆ ]Li , L2 0= for all the components i, j = x, y, z = 1, 2, 3. Thus,

[ ˆ ˆ ] ( ) ([ ˆ ˆ ] ˆ ˆ [ ˆ ˆ ])H r L S L Si i i i

i

, , ,L L L2 2 2 0= + =ζ==

∑1

3

(14.11.2)

Similarly, by changing the roles of L and S in Eq. (14.11.2), we obtain

[ ˆ ˆ ] ( ) ([ ˆ ˆ ] ˆ ˆ [ ˆ ˆ ])H r L S L Si i i i

L

, , ,S S S2 2 2 0= + =ζ==

∑1

3

(14.11.3)

Furthermore, using the relations [ ˆ , ˆ ]L L i Li j ij k k= �ε , we obtain

[ ˆ ˆ ] [ ˆ ( ) ˆ ˆ ˆ ] ( )[ ˆ ˆ ˆH L H r L r L S Lz z x x, ,= + = +0 ζ ζL Si yy y z z z

x z x y

S L S L

r L L S r L

ˆ ˆ ˆ ˆ ]

( )[ ˆ ˆ ] ˆ ( )[ ˆ

+

= +

,

,ζ ζ ˆ ] ˆ ( ) ( ˆ ˆ ˆ ˆ )L S i r L S L Sz y y x x y= − + ≠�ζ 0 (14.11.4)

and finally,

[ ˆ ˆ ] ( ) ˆ [ ˆ ˆ ] ( ) ˆ [ ˆ ˆ ]H S r L S S r L S Sz x x z y y z, , ,= + =ζ ζ ii r L S L Sx y y x�ζ ( ) ( ˆ ˆ ˆ ˆ )− + ≠ 0

(14.11.5)

(b) From Eqs. (14.11.4) and (14.11.5), we immediately find

[ ˆ ˆ ] [ ˆ ˆ ˆ ]H J H L Sz z z, ,= + = 0 (14.11.6)

Moreover,

[ ˆ ˆ ] [ ˆ ( ˆ ˆ ) ] [ ˆ ˆ ˆ ˆ ˆ ]H H H, , ,J L S L S L S2 2 2 2 2= + = + + =i 22

2 0 0

[ ˆ ˆ ˆ ]

[ ˆ ( ) ˆ ˆ ˆ ˆ ] [ ˆ ˆ

H

H r H

,

, ,

L S

L S L S L

i

i i= + =ζ ii i iˆ ] ( )[ ˆ ˆ ˆ ˆ ]S L S L S+ =2 0ζ r , (14.11.7)

(c) The results of parts (a) and (b) imply that one can find the basis of states | |nl s J M R r JMnl, , , ⟩ = ⟩( ) , which is

made up of the simultaneous eigenfunctions of the mutually commuting observables { ˆ , ˆ , ˆ , ˆ , ˆ }.H JzL S J2 2 2 The angular part of these eigenfunctions, | JM⟩, had already been worked out in Problem 14.10, where we found the following expressions:

| | |J l Ml

l M M l M M= + ⟩ =+

+ + − +⟩ − − + +12

12 1

12

12

12

12

, , , −−⟩⎡

⎣⎢

⎤

⎦⎥

= − ⟩ =+

+ + + −⟩ − −| |J l Ml

l M M l M12

12 1

12

12

, , ++ − +⟩⎡

⎣⎢

⎤

⎦⎥

⎧

⎨

⎪⎪⎪

⎩

⎪⎪⎪

12

12

| M ,

(14.11.8)

The states | |M M± ±⟩ ≡ ±⟩12

, , on the right side of Eq. (14.11.8) denote the product-basis eigenstates

| |lm s⟩ ⊗ = ±⟩1 2/ , for an electron of an orbital angular momentum l and spin s = 1/2. In the coordinate rep-

resentation, ⟨ ⟩ =r, , , ,( ) ( )r lm Ylmθ ϕ θ ϕ| , where Yl

m( , )θ ϕ are the spherical harmonic functions (see Chap. 6). Therefore,

| |J l Ml M

lY

l MlM= + ⟩ =

+ ++ +⟩ +

− +−12

12

2 1

121 2, ,/ ( )θ ϕ

22 1

12

12

2 1

1 2

lY

J l Ml M

lY

lM

l

+ −⟩

= + ⟩ =+ +

+

+ / ( )θ ϕ,

,

|

| MMlM

l M

lY+ −−⟩ +

− ++ +⟩

⎧

⎨

⎪

1 2 1 2

12

2 1/ /( ) ( )θ ϕ θ ϕ, ,| |

⎪⎪⎪

⎩

⎪⎪⎪

(14.11.9)


where − J ≤ M ≤ J. By construction, the angular wavefunctions in Eq. (14.11.8) or Eq. (14.11.9) satisfy

ˆ ( )

ˆ

ˆ

J

L

2 2

2

1| |

| |

|

JM J J JM

J JM M JMz

⟩ = + ⟩

⟩ = ⟩

⎧⎨⎪

⎩⎪

�

�

JJM l l JM

JM JM

⟩ = + ⟩

⟩ = ⟩

⎧⎨⎪

⎩⎪

�

�

2

2

1

3 4

( )

ˆ ( )

|

| |S2 /

(14.11.10)

Consequently,

ˆ ˆ ( ˆ ˆ ˆ ) [ ( ) (L S J L Si | |JM JM j j l⟩ = − − ⟩ = + −12 2

12 2 22

�ll JM+ − ⟩1 3 4) ]/ |

(14.11.11)

(d) The operator Lz can assume the values �m, where m is an integer and − l ≤ m ≤ l. The operator Sz can assume the values ± �/2, where ± corresponds to up/down spin states. The probabilities of these values are

determined by the Clebsch–Gordan coefficients of Eq. (14.11.8), and depend on the state of the system.

In the state | J l M= + ⟩12

, , we find

prob ,/

prob ,

m Ml M

l

m M

= − +⎛⎝⎜

⎞⎠⎟ = + +

+

= +

12

1 22 1

12

−−⎛⎝⎜

⎞⎠⎟ = − +

+

⎧

⎨⎪⎪

⎩⎪⎪

l Ml

1 22 1

/ (14.11.12)

Where |M| ≤ l + 1/2. If M = l + 1/2, then prob (m = l, + ) = 1, and all the other combinations have zero probability. The expectation value of Lz is

| | |ˆ ˆL l M L l M m Mz z⟩ = ⟨ + + ⟩ = = −⎛⎝⎜

⎞⎠⎟

12

12

12

, , prob � MM

m M M

l

−⎛⎝⎜

⎞⎠⎟

+ = +⎛⎝⎜

⎞⎠⎟ +⎛

⎝⎜⎞⎠⎟

=

12

12

12

2

prob �

�++ + +⎛

⎝⎜⎞⎠⎟ −⎛

⎝⎜⎞⎠⎟ + − +⎛

⎝⎜⎞⎠⎟ +⎛

112

12

12

12

l M M l M M⎝⎝⎜⎞⎠⎟

⎡⎣⎢

⎤⎦⎥

= +�lM

l2

2 1 (14.11.13)

Similarly, the average value of Sz is

⟨ ⟩ = ⟨ + + ⟩ = + + +⎛⎝⎜

⎞⎠⎟

ˆ ˆS l M S l Ml

l Mz z

12

12 2 1

12

, ,| |� ⋅⋅ + − +⎛

⎝⎜⎞⎠⎟ −⎛

⎝⎜⎞⎠⎟

⎡⎣⎢

⎤⎦⎥

= +

12

12

12

22

2 1

l M

Ml

� (14.11.14)

In the state | J l M= − ⟩12

, , we find

prob

/

prob

m Ml M

l

m M

= − −⎛⎝⎜

⎞⎠⎟ = + +

+

= − +⎛

12

1 22 1

12

,

,⎝⎝⎜⎞⎠⎟ = − +

+

⎧

⎨⎪⎪

⎩⎪⎪

l Ml

1 22 1

/ (14.11.15)


where |m| ≤ l − (1/2). Therefore, the expectation values of Lz and Sz in this state are,

⟨ ⟩ = ⟨ − − ⟩ =

⟨ ⟩ = ⟨ −

ˆ ˆ

ˆ ˆ

L l M L l M M

S l M

z z

z

12

12

12

, ,

,

| |

|

�

SS l MM

lz | − ⟩ = − +

⎧

⎨⎪⎪

⎩⎪⎪

12 2

22 1

,�

(14.11.16)

14.12. The spin-orbit interaction for the electron in a hydrogen-like atom is given by ˆ ( ) ˆ ˆH rso = ζ L Si , where

ζ ( )( )

rm c r

dV rdr

e

= ⎛⎝⎜

⎞⎠⎟

1

2

12 2 and V(r) = − Ze2/r. (a) Derive an equation for the energy levels of such atoms in

terms of the quantum numbers l and J. (b) Show that the spin-orbit correction to the unperturbed energy levels is proportional to Z 4.

SOLUTION

(a) The complete Hamiltonian of our problem is ˆ ˆ ˆH H H= +0 so, where

ˆ ˆ( ) ˆ ( ,H

mV r

m r rr

m re e e0

2 2 2

2 22

2 21 1

2= + = ∂

∂+P

L� θ ϕ)) ˆ( )+ V r (14.12.1)

and Hso is treated as a small perturbation. For convenience, we take the unperturbed wavefunctions of H0

to be the simultaneous eigenfunctions of {ˆ ˆ ˆ ˆ }L S J2 2 2, , , Jz , where ˆ ˆ ˆJ L S≡ + . Thus,

ˆ ( ) ( )H R r JM E R r JMnl nl nl00 0 0| |⟩ = ⟩

(14.12.2)

where R rnl0 ( ) are the radial wavefunctions of H0, and Enl

0 are the corresponding energy levels (see Chap. 8).

The kets | JM⟩ in Eq. (14.12.2) represent the angular part of the wavefunctions of H0, including the spin

states. In this representation [see Eqs. (14.11.7), (14.11.8), and (14.11.10)], and for l ≠ 0, we have

ˆ ( )( )

( )[ ( ) ( )H R r JMr

R r J J l lnl nl00

20

21 1| ⟩ = + − +� ζ −− ⟩3 4/ ] | JM

(14.12.3)

where J = l ± 1/2 and |M| ≤ J. Expression (14.12.3) shows that the perturbation Hso is already diagonal in

the subspace {n, l = J ± 1/2}, which corresponds to a degenerate energy level Enl0 . Using the first-order

perturbation theory we therefore find

E n l J E nl JM H nl JMnl( ) ˆ, , , ,so= + ⟨ ⟩0 | | (14.12.4)

where ⟨ ⟩ ≡r nl R rnl| 0 ( ). Defining the integral over the radial functions to be

ζ ζ ζnl nl nlnl r nl r R r r R r dr≡ ⟨ ⟩ = ∫| |( ) ( ) ( ) ( )*2 0 0 (14.12.5)

and using Eq. (14.12.3) we obtain

E n l J E J J l lnl nl( ) [ ( ) ( ) ], , /= + + − + −02

21 1 3 4

� ζ (14.12.6)

Since J = l ± 1/2, we can distinguish between two cases:

E n l JE l J l

E lnl nl

nl nl

( )(

, ,/ /

=+ = +

+ +

0 2

0 2

2 1 2

1

ζ

ζ

�

� ))/ /2 1 2J l= −

⎧⎨⎪

⎩⎪ (14.12.7)

Each of these energy levels is (2J + 1) degenerate. The degeneracy can be removed by a magnetic field (see Problem 14.13).


(b) The first-order energy correction due to spin-orbit interaction is proportional to the radial integral xnl. For l ≠ 0, we have

ξnl

e e

nlm c r

dVdr

nlm c

nl= ⟨ ⎛⎝⎜

⎞⎠⎟ ⟩ = ⟨| | |

1

2

1 1

22 2 2 211

2

2

2

2 23

rdVdr

Zer

nl

Ze

m cnl r nl

e

−⎛

⎝⎜⎞

⎠⎟⟩

= ⟨ ⟩−

|

| | == ⟨ ⟩−Z

m enl r nl

e

α 2 2

2 23

2

�| | (14.12.8)

A detailed calculation of ⟨ ⟩−r nl3 yields (see Chap. 8)

⟨ ⟩ = ⎛⎝⎜

⎞⎠⎟ + +

−nl r nla n l l lB

| |33

1 11 2 1( ) ( )/ (14.12.9)

where aB = �2/Ze2me is the Bohr radius. Therefore,

ζ αnl

eZe m

n l l l= + +

( )( ) ( )

4

4

2

32

11 2 1� /

(14.12.10)

14.13. A hydrogen-like atom is placed in a weak magnetic field B = B k, where the interaction is described by the Zeeman Hamiltonian, ˆ ( ˆ ˆ )′ = +H L SB z zμ B 2 /�. (a) Assume that in the absence of B, the

wavefunctions of the atom are eigenfunctions of ˆ ˆ ˆL S J2 2 2, , , and Jz, where ˆ ˆ ˆJ L S= + . Use the first-

order perturbation theory to calculate the energy splittings due to the magnetic field. (b) The electron of such an atom is excited to a p-state. How many components does each of the levels split into when a weak magnetic field is applied?

SOLUTION

(a) The perturbing Zeeman Hamiltonian can be written in the following form:

ˆ ( ˆ ˆ ) ( ˆ ˆ )′ =

+=

+H

L S J SB z z B z zμ μB B2� �

(14.13.1)

where μB is the Bohr magneton. The energy levels, E = E (n, l, J ) + DE, of the complete Hamiltonian ˆ ˆ ˆ ˆH H H Hso= + + ′0 are then given by

Δ = ⟨ = ± + = ± ⟩ = + ⟨E J l M J S J l M M JB z z Bμ μB B

12

12

, ( ˆ ˆ ) ,| | == ± = ± ⟩⎡⎣⎢

⎤⎦⎥

l M S J l Mz

12

12

, ˆ ,| |

(14.13.2)

The matrix element of Sz was already calculated in Prob. 14.11. Combining the appropriate results from Eqs. (14.11.12) and (14.11.14), we find

⟨ = ± = ± ⟩ = ± +J l M S J l MM

lz

12

12 2 1

, ˆ ,| |�

(14.13.3)

Hence,

Δ = ± +⎡⎣⎢

⎤⎦⎥

ElBμ B 11

2 1 (14.13.4)

(b) In the absence of a magnetic field there are two degenerate energy levels, which are specified by the quantum numbers (l = 1, J = 1/2), respectively [see Eq. (14.12.7)]. When the magnetic field B is applied, the degeneracy is removed. The J = 3/2 level is split into four components since M = − 3/2, −1/2,


+1/2, +3/2. Similarly, the J = 1/2 level is split into two components corresponding to M = −1/2, +1/2. The energy changes are given by Eq. (14.13.4).Thus,

Δ =E l J g l J MB( , ) ( , )μ B (14.13.5)

g l Jl

( , )( )

= ± +⎡⎣⎢

⎤⎦⎥

11

2 1 (14.13.6)

where g is the Lande factor. In particular, g (1, 3/2) = 4/3, and g (1, 1/2) = 2/3.


14.14. Show that the Clebsch–Gordan coefficients satisfy the following recurrence relations:

J J M M m m J M j j m m( ) ( ) , ( ) ( )+ − ± ⟨ + ⟩ = + −1 1 1 1 11 2 1 1 1 1| ∓ ⟨⟨ ⟩

+ + − ⟨ − ⟩

m m JM

j j m m m m JM

1 2

2 2 2 2 1 2

1

1 1 1

∓

∓

,

( ) ( ) ,

|

| (14.14.1)

14.15. Consider a deuterium atom composed of a nucleus of spin I = 1 and an electron. The electronic angular momentum is J = L + S, where L is the orbital angular momentum of the electron and S is its spin. The total

angular momentum of the atom is ˆ ˆ ˆF J I= + , where Î is the nuclear spin. The eigenvalues of J2 and F2 are J

(J + 1) �2 and F (F + 1), respectively. (a) What are the possible values of the quantum numbers J and F for a deuterium atom in the 1s ground state? (b) Answer the same question in the 2p excited state. (c) What are the possible values of the quantum numbers J and F for a hydrogen atom in the 2p level? The hydrogen atom’s nucleus is a pr oton of spin I = 1/2.

Ans. (a) J = 1/2, F = 1/2, 3/2

(b) if J = 1/2, F = 1/2, 3/2; if J = 3/2, F = 1/2, 3/2, 5/2

(c) if J = 1/2, F = 0, 1; if J = 3/2, F = 1, 2

14.16. Let ˆ ˆ ˆ ˆS S S S= + +1 2 3 be the total spin of three independent spin 1/2 particles, and let |m m m1 2 3⟩ be the common

eigenstates of ˆ , ˆ ,S Sz z1 2 and S z3 (there are 23 = 8 states), (a) What are the possible values of the total spin?

(b) Find a basis of eigenstates common to S2 and Sz, in terms of the |m m m1 2 3⟩. Hint: First consider the addition

of two spins, then add the results to the third spin. (c) Do the operators S2 and Sz form a complete basis?

Ans.

( ) ,

( )

a

b

1 2 3 2

12

23

12

12

/ /

, = + − +⟩ − − + +( ) = + + −⟩| | | −− + − +⟩ + − + +⟩( )

− = + − −⟩ − − + −⟩( ) =

16

12

16

12

12

| |

| |

,

, || | | |+ − −⟩ + − + −⟩( ) − − − +⟩ = + + +⟩

=

23

13

32

32

32

12

, ,,

, || | | | | |+ + −⟩ + + − +⟩ + − + +⟩( ) − = + − −⟩ + − + −⟩ +, 32

12

13

, −− − +⟩( ) − = − − −⟩, 32

32, |

(c) No, since the states 12

12, ± do not have a unique decomposition in |m m m1 2 3⟩ basis.

296

Scattering Theory

15.1 Cross SectionConsider the typical scattering problem depicted in Fig. 15.1.

θ0

dA = r2dW

r

UnscatteredPotential V(r)

Incident beam

dW

Scat

tered

ϕ

Fig. 15.1

A beam of particles scatters from the potential V(r) with coordinate origin at point O. We define the dif-ferential cross section ds /dW as the ratio of the number of scattered particles dn(q, j) per unit time within the solid angle dW divided by the incident particle flux F:

dd

dnFd

σ θ ϕΩ Ω= ( , ) (15.1)

CHAPTER 15


where ds /dW has the dimensions of a surface. We assume:

1. Any interaction between the scattered particles themselves is neglected.

2. Multiple scattering processes are neglected. A multiple scattering process is a process in which a scat-tered particle can be scattered multiple times in the same target range.

3. The incident beam width is much larger than a typical range of the scattering potential, so that the particle will have a well-defined momentum.

The total cross section is obtained by integrating over dW :

σ σT

dd

d= ∫ Ω Ω (15.2)

When the scattering is from a potential, say, V(r), the differential cross section is the same in the Lab and center-of-mass (CM) frames:

dd

dd

σ σΩ Ω

⎛⎝⎜

⎞⎠⎟ = ⎛

⎝⎜⎞⎠⎟

Lab CM

(15.3)

However, if we consider electric scattering of particle 1 from particle 2, then the differential cross section in the two frames will be different, and is given by

dd

dd

σ γ γ θγ θ

σΩ Ω

⎛⎝⎜

⎞⎠⎟ = + +

+

Lab ( cos )cos

/1 21

2 3 2

| |⎛⎛⎝⎜

⎞⎠⎟

CM

(15.4)

where q is the scattering angle in the CM frame and g = m1/m2.

15.2 Stationary Scattering StatesConsider a scattering problem relating to particles with mass m (in this section we use the reduced mass m and not the standard mass m) and well-defined momentum p = �k, which scatters from a time-independent potential V = V(r). The Hamiltonian of the system is

ˆ ˆ ˆ( )H H V= +0 r (15.5)

where H0 is the free Hamiltonian, H0 = �2k 2/2m. The wavefunction for a scattered particle with energy E > 0 is obtained by solving the stationary Schrödinger equation:

[ ˆ ( )] ( )∇ + − =2 2 0k U r rφ (15.6)

where

kE

U V= =2 22 2

μ μ� �

ˆ ( ) ˆ( )r r (15.7)

For a collision between two particles, ˆ ( )V r is the interaction potential between them (r = r1 − r2), and E is the kinetic energy associated with the particle of reduced mass m in the CM frame.

For a potential V(r) of shorter range than the Coulomb potential [rV(r) → 0 where r → ∞], the solution of the Schrödinger equation can be written as a composition of an incident plane wave and a spherical wave of amplitude f (q, j):

φ( ) ( , )r e fe

rri k z

ikr

→∞ → + θ ϕ (15.8)

CHAPTER 15 Scattering Theory298

The scattering amplitude is given by

f e U d r krk

i

ff( , ) ( ) ( )θ ϕ π φ= − ′ ′ ′ =⎛

⎝− ′1

43k r

r rk ri i

⎜⎜⎞⎠⎟∫ (15.9)

The amplitude fk(q, j) depends on the potential and the scattering angles q and j. This quantity is directly related to the differential cross section

d

dfk

σ θ ϕ θ ϕ( , )( , )Ω = | |2 (15.10)

15.3 Born ApproximationThe Born approximation is obtained by treating the potential U(r) as a small perturbation. Equation (15.9) then gives

f e U d rkB i( , ) ( ) ( )θ ϕ φ= − ′ ′ ′− ′∫1

43

πq i r r r (15.11)

where q = kf − ki and kf, i are the final and initial momentum, respectively. Note that in the Born approxima-tions the scattering amplitude fk

B is proportional to the Fourier transform of the potential U(r) with respect to q. If the potential has spherical symmetry, U(r) = U(r); Eq. (15.11) is simplified by taking q as the polar axis and integrating over dW ′. For this case, we obtain

fq

qr rV r drkB ( ) sin( ) ( )θ μ= −

∞

∫22

0� (15.12)

where q = 2k sin (q / 2) is the momentum transfer and k = | kf | = | ki

|; see Fig. 15.2

Fig. 15.2

q

ki

kfq

The Born approximation is valid when either of the two following conditions holds:

1.

2.

Va

ka

Va

ka ka

<< ≤

<<⎡

⎣⎢

⎤

⎦⎥ >>

�

�

2

2

2

2

1

1

μ

μ

(15.13)

where a is the range of the potential and V is the “averaged” potential defined by

Va

V rr

d r= ∫1

4 23

π( )

(15.14)

The second condition shows that the Born approximation is always applicable for sufficiently fast (high-energy) particles. This condition is weaker than the first one; hence, if the potential can be regarded as a perturbation at low energies, it can always be so regarded at high energies, whereas the converse is not necessarily true.


15.4 Partial Wave ExpansionsConsider a potential with spherical symmetry, V(r) = V(r). In this case, the stationary wavefunction φk r( , )θ and the scattering amplitude fk(q) can be expanded in terms of Legendre polynomials Pl(cosq):

φχ

k ll l

l

r Ar P

r( , )

( ) (cos )θ

θ=

=

∞

∑0

(15.15)

and

f l f Pk l l

l

( ) ( ) (cos )θ θ= +=

∞

∑ 2 1

0

(15.16)

where the coefficients Al, fl, and the functions χl (r) are to be determined. χl (r) satisfies the radial Schrödinger equation,

d

drk U r

l l

rrl

2

22

2

10+ − − +⎡

⎣⎢

⎤

⎦⎥ =( )

( )( )χ (15.17)

where the boundary conditions are χl(0) = 0. In the asymptotic region r → ∞,

χl r l l l l lr A j kr B n kr rk

C krl

( ) [ ( ) ( )] sin→∞ + = −∼1

2π ++⎛

⎝⎜⎞⎠⎟δl (15.18)

where jl and nl are the spherical Bessel and Neumann functions, respectively. The parameter dl is called the phase shift, since it determines the difference in phase between this solution and the solution of the free radial Schrödinger equation:

χl r lrk

C krl0 1

2( ) sin→∞ −⎛

⎝⎜⎞⎠⎟∼

π (15.19)

Similarly, we can expand the plane waves in terms of the Legendre polynomials:

e e i l j kr Pikz ikr ll l

l

= = +=

∞

∑cos ( ) ( ) (cos )θ θ2 1

0

(15.20)

Now, substituting the expansions of eikz, fk(q), and φk r( , )θ in Eq. (15.15) we obtain Al = (2l + 1) i lei

lδ

, and

fik

l e Pki

l

l

l( ) ( )( ) (cos )θ θδ= + −=

∞

∑12

2 1 12

0

(15.21)

Thus, the differential cross section is given by

dd k

l e Pi

l l

l

lσ δ θδΩ = +

=

∞

∑12 12

0

2

( ) sin (cos ) (15.22)

and total cross section is

σ π θ θ θ π δπ

T l

l

f dk

l= = +∫ ∑=

∞

24

2 12

02

2

0

| |( ) sin ( ) sin (15.23)

From Eqs. (15.21) and (15.23) we verify directly that

σ πT kk

f= 40Im ( ( )) (15.24)


The last result is called the optical theorem. The phase shifts dl are completely determined by the asymptotic form of the radial function χl (r). Expansion of Eq. (15.23) is particularly useful for a short-range potential that vanishes outside the region r < a. In this case, the partial waves that satisfy the condition l (l + 1) > ka may be neglected. Moreover, since the radial wavefunction Rl(r) = χl (r)/r and its derivative are continuous at the boundary r = a, we have

tan( ) ( )

( ) ( )δ

γ

γl

l l l

l l l

kj ka j ka

kn ka n ka=

′ −

′ − (15.25)

where g l is the logarithmic derivative, defined as

γ ll

l

r al lR

dRdr

R rr

r= =⎡⎣⎢

⎤⎦⎥= −

1 1( ) ( )χ (15.26)

For sufficiently weak potential for which the Born approximation holds, all the phase shifts are small and are given by

sin ( ) ( )δ δ μl l lV r j kr dr≈ = − ∫2

22

� (15.27)

15.5 Scattering of Identical ParticlesThe case where two identical particles collide requires special consideration. If the total spin of the system is even, the differential cross section is

dd

f fσ θ π θΩ = + −| |( ) ( ) 2 (15.28)

while if the total spin is odd, the differential cross section is

dd

f fσ θ π θΩ = − −| |( ) ( ) 2 (15.29)

For example, if s = 1/2, the spin wavefunction can be in singlet (total spin is zero) or triplet (total spin is one) states.

For an unpolarized beam of particles with spin s, the system can be in (2s + 1)2 spin states that are distrib-uted with equal probabilities. From the total number of possibilities, (2s + 1) spin states are antisymmetric. Therefore, the differential cross section is

dd

f fs

f fsσ θ π θ θΩ = + − + −

+| | | |( ) ( )( )

[ ( ) *2 221

2 12Re (( )]π θ− (15.30)

SOLVED PROBLEMS

15.1. A particle of mass m and momentum p = �k is scattered by the potential Ve

rV a

r a

( )/

r =−

0 , where V0

and a > 0 are real constants (Yukawa potential). (a) Using the Born approximation, calculate the

differential cross section. (b) Obtain the total cross section.

SOLUTION

(a) The range of the Yukawa potential is characterized by the distance a. We assume that V a022 << � /μ , so

that the Born approximation is valid for all the values of ka [see Eq. (15.13)]. The scattering amplitude is then given by

fV a

ee

rd ri

r r

( , )/

θ ϕμ

= − −−

∫14

2 02

30

π �

q ri (15.1.1)


Since the potential has spherical symmetry V (r) = V (r), we can carry out the integration using the relation

r e V r dr dq

qr V r rdri2

0

4−∞

∫ ∫=q ri ( ) sin ( ) ( )Ω π (15.1.2)

where r = ⏐ r ⏐ and dW = sinq dq dj. Therefore,

fV a

kr kr e dr

V ar a( ) sin( ) /θμ μ

= − = −−∞

∫2 2 102

0

03

2� � 11

2 1

1 2 22 20

3

2 2+= −

+q a

V a

ka

μθ� [ sin ( )]/

(15.1.3)

Finally, the differential cross section dd

fσ θΩ = | |( ) 2 is

d

dV a

k a

σ θ μθ

( )

[ sin ( )]Ω =+

4 1

1 4 2

202 6

4 2 2 2 2� /

(15.1.4)

Note that due to the spherical asymmetry the cross section does not depend on the azimuthal angle.(b) The total scattering cross section is obtained by integration:

σ σ θ μ π= =+∫ d

dd

V a

k a

( )Ω Ω

4 4

1 4

202 6

4 2 2�

(15.1.5)

Note that the infinite range limit (a → ∞, V0 → 0, and V0 a = Z1Z2e2 = constant) of the Yukawa potential

corresponds to the Coulomb interaction between two ions of charges Z1e and Z2e. At this point, Eq. (15.14) is reduced to the well-known Rutherford formula,

dd

Z Z e

k

Z Z e

E

σ μθΩ = =4

16 2 16

2

412

22 4

4 412

22 4

� sin ( )/ 22 4 2sin ( )θ / (15.1.6)

where E = (�2k2) /2m is the kinetic energy of the particles in the CM frame and m is their reduced mass.

15.2. Using the Born approximation, calculate the differential cross section ds /d W for a central Gaussian

potential of the form VV

e r a( ) /r = −0 4

4

2 2

π. Compare your result with the differential cross section for

the Yukawa potential VV a

re r a( ) /r = −0 .

SOLUTION

For the Gaussian potential, in the Born approximation we have

fq

qr V r r drV

q q( ) sin ( ) ( ) coθ μ μ

π= − = ∂

∂

∞

∫2 2

420

02

� �ss ( ) /qr e dr

V

q qae

r a

q a

−∞

−

∫= ∂

∂ =

2 2

2 2

4

0

02

2

4

42

μπ

π�

−− −2 03

2

2 2μV a q

qe q a

� (15.2.1)

where q = 2k sin (q/2). Therefore,

dd

V ae q aσ μ

Ω⎛⎝⎜

⎞⎠⎟ = −

Gaussian

4 202 6

42 2 2

� (15.2.2)

For the Yukawa potential, we found in Problem 15.1 that

dd

V a

q a

σ μΩ

⎛⎝⎜

⎞⎠⎟ =

+⎡⎣ ⎤⎦Yukawa

4 1

1

202 6

4 2 2 2�

(15.2.3)

The differential cross section of Eqs. (15.2.2) and (15.2.3) are plotted in Fig. 15.3 (in a2 units). Note that for qa << 1, both cross sections coincide and are given by

dd

V aq a

σ μΩ ≈ −

41 2

202 6

42 2

�( ) (15.2.4)


Thus, for a small momentum transfer (i.e., large distance) the specific form of the short-range scattering potential is not important. On the other hand, for a large momentum transfer, the Gaussian cross section decreases more rapidly as compared to the Yukawa cross section. This is expected since for short distances the Gaussian potential is much weaker than the Yukawa potential.

15.3. Show that if the scattering potential has a translation invariance property, V(r + R) = V(r), where R is a constant vector, then the Born approximation scattering vanishes unless q i R = 2pn, where n is an integer.

SOLUTION

The translation symmetry of the potential, V(r) = V(r + R), implies

e V d r e V d ri i− −∫ ∫= +q r q rr r Ri i( ) ( )3 3 (15.3.1)

By changing variables r → r + R on the right-hand side of Eq. (15.3.1) we obtain

e V d r e V d ri i i− − ′+∫ ∫= ′ ′q r q r q Rr ri i i( ) ( )3 3 (15.3.2)

Therefore,

e V e d ri i−∫ − =q R q Rri i( ) [ ]1 03 (15.3.3)

Equation (15.3.3) holds when either of the following two conditions is satisfied

e V d ri− =q r r qi ( ) ( )3 0 arbitrary∫∫

= → =e n niq R q Ri i1 2π ( )is an integer

(15.3.4)

(15.3.5)

The Born scattering amplitude f B(q) is proportional to the Fourier transform of the potential V(r). We there-fore conclude that f B(q) vanishes identically, unless the condition of Eq. (15.3.5) is satisfied. Normally,

f f nB( ) ( )q k Rq q, k

k

= =∑ δ πi 2 (15.3.6)

Note that the translation symmetry of the scattering potential corresponds to the scattering form of a lattice. For any vector R of the lattice, the set of vectors k that satisfy k i R = 2 p n constitutes the reciprocal lattice.

Fig. 15.3

qa1 2 3 4 50

0.2

0.4

0.6

0.8

1.0

Gaussian

Yukawa

ds/dΩ


Therefore, as a result of the conditions of Eq. (15.3.5), the scattering amplitude vanishes unless the momentum transfer q is equal to some vector of the reciprocal lattice. This is precisely the Bragg–von Laue scattering condition.

15.4. Using the Born approximation, express the differential cross section for nonrelativistic scattering of an electron from a spherical symmetric charge distribution r (r) as the product of the cross section for a point charge q (Rutherford scattering) and the square of a form factor F (k), where k is the momentum transfer. Evaluate F (k) explicitly for uniform charge distribution of radius R and for a Gaussian charge distribution.

SOLUTION

In the Born approximation, the differential cross section is given by

d

de V d riσ θ μ

π( )

( )Ω =⎡

⎣⎢

⎤

⎦⎥

−∫2

2 43

2

4 �

k r ri (15.4.1)

where m denotes the nonrelativistic electron mass and k is the transferred momentum. By definition, the potential of the electron due to a symmetric charge distribution V(r) = V(r) can be written as the convolu-tion integral:

Ve

d reqr

rq

( )( ) ( )

rr

r r= − ′

− ′ ′ = −⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞∫ ρ ρ| |

3

⎠⎠⎟ (15.4.2)

The first term of the right-hand side of Eq. (15.4.2) corresponds to the Coulomb interaction, and leads, after regularization (see Problem 15.1), to the Rutherford cross section. Therefore,

dd

dd q

e r diσ θ σ ρ( )( )Ω Ω= ⎛

⎝⎜⎞⎠⎟

−∫Rutherford

1 3k ri rr

F k

⎡

⎣⎢

⎤

⎦⎥

( )� ��

2

(15.4.3)

For a uniform charge distribution of radius R we have

ρ π( )rq

Rr R

r= <

>

⎧⎨⎪

⎩⎪

3

40 0

3 (15.4.4)

Thus, we obtain

F kk R

r kr drR

( ) sin ( )= ⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢⎢

⎤

⎦⎥⎥

=∫4 34

13

0

2π

π333 2

2

R k

krk

r krsin ( )

cos ( )−⎛⎝⎜

⎞⎠⎟

⎡⎣⎢

⎤⎦⎥

(15.4.5)

Similarly, for a Gaussian distribution ρπ

( ) :/rq

Re r R= −

3 2 3

2 2/

F kk R

r kr e drr R( ) sin ( )//=

⎡

⎣⎢⎢

⎤

⎦⎥⎥

∞−∫4 1

3 2 30

2 2ππ

22

4

2

14

2 2

=⎡

⎣⎢⎢

⎤

⎦⎥⎥

−

π er k

(15.4.6)

15.5. Consider scattering from a spherical symmetric potential. The solution of the Schrödinger equation

is given by the expansion φ( , ) ( ) (cos )r R r Pl l

l

θ θ==

∞

∑0

, where R(r) is the solution of the radial wave

equation and Pl (cosq) is the Legendre polynomial of order l. In the limit r → ∞ the asymptotic form of the wavefunction is

φ ( , ) ~ ( )r er

f eri k z i krθ θ→∞ + 1

(15.5.1)


where f (q) is the scattering amplitude. Similarly, the asymptotic form of R(r) is

R r Akr l

krr l

l( ) ~

sin

→∞

− +⎛⎝⎜

⎞⎠⎟

π δ2

(15.5.2)

where dl are phase shifts. (a) Use Eqs. (15.5.1) and (15.5.2) to obtain the Legendre expansion of f (q). (b) Show that the total cross section is given by

σ π δT l

lk

l= +=

∞

∑42 12

2

0

( ) sin (15.5.3)

SOLUTION

(a) The asymptotic form of the wavefunction is given in Eqs. (15.5.1) and (15.5.2):

φ ( , ) ~sin

(cosr Akr l

krPr l

l

l

lθ

π δ→∞

=

∞ − +⎛⎝⎜

⎞⎠⎟∑ 2

0

θθ θ) ( )= +er

f eikz ikr1 (15.5.4)

Using the Legendre expansion of eikz,

e e l i j kr Pikz ikr ll l

l

= = +=

∞

∑cos ( ) ( ) (cos )θ θ2 1

0

(15.5.5)

we find that

Akr l

krP l il

l

l

l

sin(cos ) ( )

− +⎛⎝⎜

⎞⎠⎟

= +=

∞

∑π δ

θ2

2 1

0

lll

likr

kr l

kr rf e

sin( )

− +⎛⎝⎜

⎞⎠⎟

+

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥π δ

θ2 1 ⎥⎥

⎥⎥=

∞

∑l

lP

0

(cos )θ

(15.5.6)

where f f Pl l

l

( ) (cos )θ θ==

∞

∑0

. Now we write sin xe e

i

ix ix

= − −

2, and obtain

A e l i el

i krl

l i krl

l− +⎛

⎝⎜⎞⎠⎟ −⎛

⎝⎜⎞⎠⎟− + =

π δ π2 22 1( ) 22ik f el

ikr (15.5.7)

A e l i el

ikrl

l ikrl

l− − + − −

− + =π δ π

2 22 1 0 (15.5.8)

Therefore, from Eq. (15.5.8) we obtain A l i ell i

l= +( )2 1δ

and then, by substituting back into Eq. (15.5.8),

f ik l e Pi

l

l

l( ) ( ) ( ) ( ) (cos )θ θδ= + −−

=

∞

∑2 2 1 11 2

0

(15.5.9)

(b) The total cross section is

σ θ π θ δ= = + −−∫| |f d

d

kl e

il( )

(cos )( )( )2

21

12

24

2 1 1Ω PP

kd l l e

l

l

(cos )

(cos ) ( )( )(

θ

π θ

=

∞

∑∫= ′ + +

0

2

222 1 2 1

22 2

0

1 1i i

l l

l l

l le P Pδ δ θ θ− − ′

′=

∞

−∑ )( ) (cos ) (cos )

,11

1

∫ (15.5.10)


Now, d P Pll l l l(cos ) (cos ) (cos )

( )θ θ θ δ′ ′

−= +∫ 2

2 11

1

. Therefore,

σ π πδ δT

i i

kl e e

kll l= + − − = +−

22 2

222 1 2

42 1( )( ) ( ) sin δδl

ll =

∞

=

∞

∑∑00

(15.5.11)

15.6. Consider the hard sphere potential of the form

V rr r

r r( ) =

>∞ <

⎧⎨⎩

0 0

0

where k r0 0 1<< . (a) Assume only s-wave scattering and calculate d0(k), f (k), ds /d W, and s T. (b) Write the radial Schrödinger equation for l = 1, and show that the solution for the p-wave scattering is of the form

χk r Akr

krkr a

krkr

kr1( )sin ( )

cos ( )cos ( )

sin (= − + + ))⎛⎝⎜

⎞⎠⎟

⎡⎣⎢

⎤⎦⎥

where A and a are constants. (c) Determine d1(k) from the condition imposed on χk1(r0). (d) Show that in the limit k → 0, d1(k) ~ (k0r0)

3 and δ δ1 0( ) ( )k k<< .

SOLUTION

(a) The radial Schrödinger equation for r > r0 is

d

drk

l l

rr rkl

2

22

21

0 0+ − +⎡

⎣⎢⎢

⎤

⎦⎥⎥

= >( )( ) ( )χ for (15.6.1)

which due to the infinitely repelling potential must be constrained by the condition χkl (r0) = 0. Therefore, the s-wave general solution is,

χk rC k r r r r

r r0

0 0 0

00( )

sin ( )=

− >

<

⎧⎨⎪

⎩⎪ (15.6.2)

The phase shift d0 (k) is, by definition, given by the asymptotic form of the equation; namely, d0(k) = −kr0. Thus, in the s-wave approximation,

f k e kr r e

dd

k

ki kr i k r

( ) sin ( )

sin

θ

σ

= ≈

=

− − −

−

10 0

2

0 0

Ω22

0 02

22

0 024

4

( )

sin ( )

kr r

kkr rT

≈

= ≈

⎧

⎨

⎪⎪⎪

⎩

⎪⎪⎪σ π π

(15.6.3)

(b) From Eq. (15.6.1), the p-wave radial equation is

d

drk

rrk

2

22

2 1

20+ −

⎡

⎣⎢⎢

⎤

⎦⎥⎥

=χ ( ) (15.6.4)

The general solution is given by [see Eq. (15.18) in Sec. 15.4]

χk r c r j kr d rn kr1 1 1 1 1( ) ( ) ( )= + (15.6.5)

where j1(kr) and n1(kr) are the spherical Bessel functions:

j xx

x

xx

n xx

x

xx1 2 1 2( )

sin cos( )

cos sin= − = − − (15.6.6)


For A = c1/k, Aa = −d1/k we obtain

χk r Akr

krkr a

krkr

kr1( )sin ( )

cos ( )cos ( )

sin (= − + + ))⎛⎝⎜

⎞⎠⎟

⎡⎣⎢

⎤⎦⎥

(15.6.7)

where A and a are both r-independent constants.(c) From Eq. (15.6.7) and the condition χk1(r0) = 0, we find

akr

krkr

krkr

kr=

−

+

cos ( )sin ( )

cos ( )sin ( )

00

0

0

00

(15.6.8)

Furthermore, the asymptotic form of χk1(r) is

χk rr A kr a kr C kr1 1 2( ) ~ [ cos ( ) sin ( )] sin→∞ − + = − +π δ11

⎛⎝⎜

⎞⎠⎟

= −⎛⎝⎜

⎞⎠⎟ +C kr1 12

sin cos cosπ δ kkr −⎛

⎝⎜⎞⎠⎟

⎡⎣⎢

⎤⎦⎥

π δ2 1sin

= −C1 1cos [ cδ oos ( ) sin ( ) tan ]kr kr+ δ1 (15.6.9)

Identifying in Eq. (15.6.9), A = c1cos d1 and a = tan d1, and using Eq. (15.6.8) we obtain

tan ( )cos ( )

sin ( )

cos ( )si

δ1

00

0

0

0

kkr

krkr

krkr

=−

+ nn ( )kr0

(15.6.10)

(d) In the limit kr0 1<< we have

sin ( )( )

cos ( )( )

kr krkr

krkr

0 00

3

00

2

61

2≈ + ≈ − (15.6.11)

and Eq. (15.6.10) takes the form

tan ( ) ( ) ( ) ( )δ δ1 03

1 031

313

k kr k kr≈ ⇒ = − (15.6.12)

15.7. A (point) particle is scattered by a second particle with a rigid core; that is, the scattering potential is V(r) = 0 for r > a and V(r) = ∞ for r < a. The energy of the scattered particle satisfies ka = 1. (a) Find the expression for dl . Complete Table 15.1 (express dl in radians).

(b) Calculate the differential cross section ds /dW for angles 0 and p, taking into account only the waves l = 0 and l = 1. (c) Calculate the total cross section s T taking into account only the waves l = 0 and l = 1. (d) What is the accuracy of part (c)?

SOLUTION

(a) The phase shifts for a rigid sphere are given by the equation

tan( )( )

δll

l

j kan ka

= (15.7.1)

Table 15.1

tandl dl sindl

l = 0

l = 1

l = 2


Using the known expressions of the spherical Bessel functions (see the Mathematical Appendix)

j xx

xn x

xx

j xx

x

xx

n

0 0

1 2 1

( )sin

( )cos

( )sin cos

= = −

= − (( )cos sin

( ) sinc

xx

x

xx

j xx x

x

= − −

= −⎛⎝⎜

⎞⎠⎟

−

2

2 23 1 3 oos

( ) cossinx

xn x

x xx

x

x2 2 2 23 1 3= − −⎛

⎝⎜⎞⎠⎟

−

(15.7.2)

and substituting x = ka = 1, we find tan d0 = −1.56, tan d1 = − 0.22, and tan d2 = − 0.02. Therefore,

(b) The differential cross section is given by

dd k

l e Pi

l l

l

lσ δ θδΩ = +

=

∞

∑12 12

0

2

( ) sin (cos ) (15.7.3)

For l = 0, l and k = a−1,

dd

a e e

a

i iσ δ δ θ

δ

δ δΩ = +

=

20 1

2

2 2

0 13| |sin sin cos

[sin 00 0 1 0 12

126 9+ − +sin sin cos ( ) cos sin cos ]δ δ δ δ θ δ θ (15.7.4)

Substituting q = 0, p, we obtain

dd

aσ δ δ δ δ δπΩ 0

2 20 0 1 0 16 9

,sin sin sin cos( ) si= ± − +| nn2

1δ | (15.7.5)

with d1 and d0 given in Table 15.2.(c) The total cross section is given by

σ π δT l

lk

l= +=

∞

∑42 12

2

0

( ) sin (15.7.6)

For l = 0, 1 and k = a−1,

σ π δ δ πT a a= + ≈4 3 0 8542 20

21

2[sin sin ] . (15.7.7)

(d) A rough estimate on the accuracy of the result in part (c) is given by calculating the additional term l = 2:

σ πT a≈ +( . . )0 85 0 002 4 2 (15.7.8)

15.8. Consider the potential of a square well of depth V0:

V rV r a

r a( ) =

− <

>

⎧⎨⎪

⎩⎪

0

0 (15.8.1)

Table 15.2

tan dl dl sin dl

l = 0 −1.56 −1.00 − 0.84

l = 1 − 0.22 − 0.22 − 0.22

l = 2 − 0.02 − 0.02 − 0.02


Set k mE= 2 2/� , k mV0 022= /� , and K k k2 2

02= + . (a) Calculate the phase shifts d0 and d1 for

low-energy scattering ( ).ka << 1 (b) Find the condition for resonance scattering of the s-waves and p-waves. (c) Calculate the total cross section for “off-resonance” scattering at low energies (ka << 1 and δ δ1 0 1<< << ).

SOLUTION

(a) We begin from the radial part of the Schrödinger equation:

d

drK

l l

rr r al

2

22

211

0+ − +⎡

⎣⎢⎢

⎤

⎦⎥⎥

= >( )( ) ( )( )χ for

dd

drk

l l

rr r al

2

22

221

0+ − +⎡

⎣⎢⎢

⎤

⎦⎥⎥

= >( )( ) (( )χ for ))

⎧

⎨

⎪⎪⎪

⎩

⎪⎪⎪

(15.8.2)

where χl(1) and χl

(2) denote the solutions at r < a and r > a, respectively. The general form of χl(1) and χl

(2) is given by

χ

χ

l l l l l

l l l l

A rj Kr B rn Kr

C rj kr D

( )

( )

( ) ( )

( )

1

2

= +

= + rrn krl ( )

(15.8.3)

where jl and nl are the spherical Bessel functions, and Al, Bl, Cl, and Dl are constants. In particular,

j x

xx

n xx

x

j xx

x

xx

n

0 0

1 2 1

( )sin

( )cos

( )sin cos

= = −

= − (( )cos sin

xx

x

xx

= − −

⎧

⎨⎪⎪

⎩⎪⎪ 2

(15.8.4)

Actually, since the radial wavefunction Rl (r) = χl (r)/r must be regular at the origin (r = 0), we can set Bl = 0 in Eq. (15.8.3). Hence, in the interior region (r < a), we have

R r A j Krl l l( ) ( )= (15.8.5)

The phase shifts dl can now be determined by calculating the logarithmic derivative of Rl(r) [see Sec. 15.4 Eqs. (15.25) and (15.26)]:

tan( ) ( )( ) ( )

δγγl

l l l

l l l

kj ka j kakn ka n ka

=′ −′ − (15.8.6)

where γ ll

l

r aR

dRdr

=⎛⎝⎜

⎞⎠⎟

= −

1. Substituting Eq. (15.8.5) for Rl(r) into Eq. (15.8.6) and using Eq. (15.8.4),

we find

γ

γ

0

1

2

10

12

1

= − =

= − − =

K Kaa

l

K aKa Ka a

l

cot ( ) ( )

cot ( )( ))

⎧

⎨⎪⎪

⎩⎪⎪

(15.8.7)

so that, for l = 0,

tan

cos ( ) sin ( )

( )

sin (

δγ

0

2 0

=

−⎛⎝⎜

⎞⎠⎟

−kka ka ka

ka

kaaka

kka ka ka

ka

ka

)

sin ( ) cos ( )

( )

cos (+⎛⎝⎜

⎞⎠⎟

+2 0γ ))ka

(15.8.8)

In the limit ka → 0, Eq. (15.8.8) reduces to

tanδγ

γ00

2

011≈ − + <<

kaa

ka (15.8.9)


and similarly,

tan( )δ

γγ1

31

1312

1≈ −−+ >>ka a

aka (15.8.10)

where g0 and g1 are given by Eq. (15.8.7). Note that unless g0a = −1 or g1a = −2, both d0 and d1 vanish as k → 0 and δ δ0 1 1<< << .

(b) Resonance scattering occurs when a particular phase shift becomes exponentially large. Reso-nance scattering of s-waves ( ,ka << 1 l = 0) is found by using Eqs. (15.8.7) and (15.8.9). Thus, 1 0 00+ = ⇒ =γ a Ka kacot ( ) and the resonance condition is

Ka n n= + =( ) ( , , . . .)2 12

1 2π

(15.8.11)

Similarly, resonance scattering of p-waves ( , )ka l<< =1 1 is 2 01+ = ⇒ = ±∞γ a Ka kacot ( ) and the resonance condition is

Ka n n= =π ( , , . . .)1 2 (15.8.12)

(c) The total cross section is given by

σ π δT l

lk

l= +=

∞

∑42 12

2

0

( ) sin (15.8.13)

Recall that unless g0a = −1 or g1a = −2 [see Eqs. (15.8.9) and (15.8.10)] both d0 and d1 vanish as k → 0. However, as a consequence of the 1/k2 factor in Eq. (15.8.3), only the l = 0 partial wave gives a finite contribution to the cross section. Thus, for off-resonance scattering ( , )Ka << << <<1 10 1δ δ

σ π δπγ

γπT k

a

aa

Ka≈ =+

= −4 4

14 12

20

02 4

02

2sin( )

tan ( )KKa

⎛⎝⎜

⎞⎠⎟

2

(15.8.14)

where K k k na

= + ≠ +02 2 2 1

2( ) .

π

15.9. Refer to the potential in Problem 15.8. (a) Find the differential cross section ds/dW for s-wave resonance scattering ( , ).ka l<< =1 0 (b) Find ds/dW for p-wave resonance scattering ( , ).ka l<< =1 1

SOLUTION

(a) From Problem 15.8, we have

tan

cot( )

δγ

γγ

00

2

0

0

1

1

= − += −

⎧⎨⎪

⎩⎪

kaa

a Ka Ka

(15.9.1)

where K k k kk

kk

k

k= + = +

⎛

⎝⎜

⎞

⎠⎟ ≈ +

⎛

⎝⎜

⎞

⎠02 2

0

2

02

1 2

0

2

021 1

2

/

⎟⎟ . Near the resonance, 1 − r0a ≈ 0 and d0 is not neces-

sarily small. However, using Eq. (15.9.1) and the identity sind0 = 1/(1 + cot2d0),

sin( ) ( )

( ) ( ) ( )2

0

20

2

20

20

21δ

γγ γ

=+ +

ka a

ka a a (15.9.2)

Furthermore, if ka << 1, we can expand g0(K) in a Taylor series about K = k0:

γ γ γ0 0 00

0

( ) ( )KK

K kK kK k

= + ∂∂ − +=

=| � (15.9.3)


where K kkk

− =⎛⎝⎜

⎞⎠⎟0

0

212

. Using Eq. (15.9.1) we then find that in the leading terms of Ka, g0 = a0 + b0k2, with

α

β

0 0 0

00

00

20

1

12

= −

= −

k cot ( )

cot ( )sin (

k aa

kk a

k a

k a))

⎡

⎣⎢⎢

⎤

⎦⎥⎥

≈ −

⎧

⎨⎪⎪

⎩⎪⎪

a2

(15.9.4)

Hence,

1 120 0 0

20 0

2

+ ≈ + + = −γ α βa a ak k a k aka

cot ( )( )

(15.9.5)

Finally, substituting Eq. (15.9.5) back into Eq. (15.9.2) we obtain

dd

a

k aka

lσ

ξΩ =

+ −⎡

⎣⎢⎢

⎤

⎦⎥⎥

=2

2 20

2 2

2

0( )

( ) (15.9.6)

where ka << 1 and x0 = k0a cot (k0a). Recall that near the resonance |x0| ≤ 1, ds /dW for x0 = 1 and l = 0 as is shown in Fig. 15.4 (in a2 units).

(b) From Problem 15.8 we have

tan

( )

cot ( )

δγγ

γ

1

31

1

1

2 2

312

12

=−+

= − −

ka aa

aK a

Ka Ka

⎧⎧

⎨⎪⎪

⎩⎪⎪

(15.9.7)

In this case we find

sin

( )

21

1

1

2

6

1

121

9δ

γγ

=+

+−

⎡⎣⎢

⎤⎦⎥

aa ka

(15.9.8)

where 1 − g1a ≈ 3. Repeating the calculations in part (b), we have

K kkk

K k= +⎛⎝⎜

⎞⎠⎟

≈ +00

2

1 1 121

2γ α β( ) (15.9.9)

where near the resonance, the coefficients a1 and b1 are

α

β

1 0 0

1

2

2

= − −

= −

⎧

⎨⎪

⎩⎪

k k aa

a

tan ( ) (15.9.10)

ka2 4 6 8 100

0.2

0.4

0.6

0.8

1.0

ds /dΩ

Fig. 15.4


Hence,

2 221 1 1

20

2

1+ = + + = − − ≡ −γ α β ξa a ak ka k aka ka

tan ( )( ) ( ))2

2 (15.9.11)

where ka << 1 and |x1| ≤ 1. The contribution of p-wave resonance scattering is

dd k

σ θ δΩ = 92

2 21cos sin (15.9.12)

Therefore, substituting Eq. (15.9.8) into Eq. (15.9.12) with the help of Eq. (15.9.11),

dd

k a

ka kal

σ θ

ξΩ =

+ −⎡⎣⎢

⎤⎦⎥

=9

2

14 6 2

6

1

2 2cos

( ) ( )( ) (15.9.13)

15.10. Using the Born approximation, calculate the phase shifts d1 for scattering in a centrally symmetric field.

SOLUTION

The scattering amplitude for a central field in the Born approximation is given by

fk

r kr U r drB( ) sin( ) ( )θ = −∞

∫1

0

(15.10.1)

This approximation is valid for a sufficiently weak potential when all the phase shifts δl << 1. The expres-sion of f (q) in terms of dl is

fk

l e PB il l

l

l( ) ( ) sin (cos )θ δ θδ= +=

∞

∑12 1

0

(15.10.2)

which reduces to

fk

l PBl l

l

( ) ( ) (cos )θ δ θ≈ +=

∞

∑12 1

0

(15.10.3)

Multiplying Eq. (15.10.3) by Pl (cos q) and integrating (using the orthogonality relations of Legendre

polynomials) we find f P dk

Bl

l( ) (cos ) (cos )θ θ θδ

≈ +−∫

2

1

1

. Therefore, by comparing with Eq. (15.10.1)

+ = −∞

−∫2 1

0 1

δθ θl

lk kr kr U r dr P dsin ( ) ( ) (cos ) (cos )

112

0 1

1

∫ ∫= −∞

−r U r dr

krkr

P dl( )sin ( )

(cos ) (cos )θ θ∫∫∫= −

∞

2 2 2

0

j kr r U r drl ( ) ( )

(15.10.4)

Namely, if all the phase shifts are small, one has

δlB

lk r U r j kr dr= −∞

∫ 2 2

0

( ) ( ) (15.10.5)

Using the relations U = 2mV/�2 and j rr

J rl l( ) ( )/

/= ⎛⎝⎜

⎞⎠⎟ +

π2

1 2

1 2 leads finally to

δ π μlB

l

kr V r J kr dr= − +

∞

∫�2 1 2

2

0

( )[ ( )]/ (15.10.6)

15.11. For the potential V(r) = V0Rd (r − R): (a) Calculate using the Born approximation, the quantities f (q) and d dσ / Ω. Specify the limits of validity of your calculation for both high- and low-energy scattering, respectively. (b) Calculate the phase shifts dl for all the partial waves in the approximation that correspond to the Born approximation. (c) Find the condition for which s-wave scattering is dominant. Obtain the differential cross section for this case and compare with the result from part (a).


SOLUTION

(a) The scattering amplitude in the Born approximation is given by

f e V d rB i( ) ( )θ πμ= − ∫1

42

23

�

q ri r (15.11.1)

where m is the mass of the particle and q = kf + ki is its momentum transfer. For a spherical symmetric potential, the angular integration can always be performed and Eq. (15.11.1) reduces to

fq

r qr V r drB( ) sin ( ) ( )θ μ= −∞

∫22

0� (15.11.2)

Substituting in Eq. (15.11.2) the potential V(r) = V0Rd (r − R)

fV R

qqRB( ) sin( )θ

μ= −

2 02

2�

(15.11.3)

where q = 2k sin (q /2) and k E= 2 2μ /� . Therefore,

dd

fV R

qqRBσ θ

μΩ = =| |( ) sin ( )2

202 4

4 224

� (15.11.4)

The Born approximation is applicable in cases where the scattering potential can be considered as a perturbation, namely, under the condition

( ) ( )e V r drkikr2

0

1− <<∞

∫ �μ (15.11.5)

In our problem, this condition of validity can be written as

e kr V r dr V R kRkikr sin ( ) ( ) sin ( )

00

2

2

∞

∫ = << �μ (15.11.6)

We can now distinguish between two limiting cases that depend on the value of kR:

2 210

20μV R

k

V R

� �= <<v ( )high energies (15.11.7)

2

1 102

μV RkR

�<< <<( , )low energies (15.11.8)

We note that the first condition (kR arbitrary) is less restrictive than the second one ( )kr << 1 . Equation (15.11.7) indicates that the Born approximation is applicable for scattering at sufficiently high energies. Equation (15.11.8) shows, on the other hand, that if kR << 1, then the Born approximation is valid for all velocities v = �k /m (in both cases one must, of course, consider scattering from a relatively weak and short-ranged potential). We can also verify, from Eqs. (15.11.4) and (15.11.5), that in the low-energy limit (qR → 0), the Born scattering cross section is completely isotropic.

(b) Recall (Problem 15.10) that the Born approximation corresponds to the case where all the phase shifts are relatively small (dl ≈ sin dl ≤ 1). Thus, we obtain

δ π μ π μl l lrV r J kr dr

R VJ= − = −+ +� �

2 1 22

20

2 1( )[ ( )] [/ // ( )]22

0

kR∞

∫ (15.11.9)

Note that using the asymptotic Bessel function expressions

J xx

x ll x+ →∞ → −1 2

22/ ( ) sin( )π π / (15.11.10)

J xxll x

l

+ →

+

→ +1 2 0

1 222 1/

/

( )( )!!π (15.11.11)


we can recover the conditions shown in Eqs. (15.11.7) and (15.11.8) of part (a). Substituting expression (15.11.11) into Eq. (15.11.9), we find that for arbitrary value of kR,

| |δπ μ

πl

R VkR

< <<2

02

21

� (15.11.12)

Thus, the condition | |δl << 1 for all coincides with Eq. (15.11.8). Similarly, substituting expression (15.11.11) into Eq. (15.11.9) for small value kR << 1,

δπ μ

πl

lR V kR

l≈

+

+20

2

2 1

22

2 1�

( )

[( )!!] (15.11.13)

Hence, the condition δ δ0 >> l coincides with Eq. (15.11.8).

(c) From Eq. (15.11.13) we find that if kR << 1 then δ δ0 >> l. This result is in agreement with the general analysis of partial wave expansion, which states that for a finite-range potential the main contribution to the scattering amplitude comes from values of l < kR, where R is the range of the potential. Using Eq. (15.11.13) for s-wave scattering (l = 0) we obtain

δ δμ

0 0

30

2

2≈ ≈ −sin

R V k

� (15.11.14)

Thus, the leading term of the differential cross section is

dd k

eR Viσ δ

μδΩ = =1 4

2 0

2 2 602

20 sin

� (15.11.15)

As expected from the previous discussion, Eqs. (15.11.4) and (15.11.14) coincide in the limit qR → 0. In this case, both the s-wave approximation and the Born approximation lead to a differential cross section that depends neither on the angle of scattering nor on the energy of the incident particle.

15.12. A particle of mass m is scattered from a spherical repelling potential of radius R:

V rV r R

r R( ) =

≤≥

⎧⎨⎩

0

0

(a) Using the Born approximation, calculate the total cross section in the limit of low energies. (b) Repeat the calculation of s T by using the partial wave expansion, and considering only the s-wave contribution.

SOLUTION

(a) The scattering amplitude in the Born approximation is

fV

qqr r dr

V

qkB( ) sin( )

sin (θ πμ π μ

= − = −14

2 4 202

02

� �

qqR

q

R qRq

) cos ( )2

0

−⎡

⎣⎢

⎤

⎦⎥

∞

∫ (15.12.1)

This leads in the limit qR → 0 to the isotropic cross section:

dd

fV RB

kBσ θ

μΩ = =| |( ) 2

202 6

4

4

9� (15.12.2)

so that the total cross section is given by

σ σ π μTB

Bdd

dV R

= =∫ Ω Ω16

9

202 6

4�

(15.12.3)

(b) In the limit E → 0 it is sufficient to consider only s-wave scattering. In order to determine the phase shift d0 we examine the radial Schrödinger equation for χ0(r):

ddr

Er r A kr

2

2 2 01

012

0+⎡⎣⎢

⎤⎦⎥

= ⇒ = +μ δ�

χ χ( ) ( ) sin ( 00

2

2 2 0 02

022

0

)

( ) ( ) ( )

r R

ddr

E V r r

>

+ −⎡⎣⎢

⎤⎦⎥

= ⇒μ�

χ χ == − ≤

⎧

⎨⎪⎪

⎩⎪⎪

B kr V E r Rsinh ( ( ) )0 1/

(15.12.4)


where k = (2mE/�2)1/2 and V E0 >> . These solutions satisfy the boundary conditions

χ χ χ χ01

02

01

02( ) ( ) ( ) ( )( ) ( )R R R R= ′ = ′ (15.12.5)

Namely,

A kR B KR

A kR B

sinh ( ) sinh ( )

cos ( ) cosh

+ =+ =

δδ

0

0 (( )KR⎧⎨⎩

(15.12.6)

where K = k ( [V0 /E) − 1]1/2 ≈ k (V0 /E)1/2), and thus

tan ( ) ( ) tanh ( )/KR E V KR+ ≈δ0 01 2/ (15.12.7)

and since kR << 1,

δ

θ δδ0 0

1 2

0 0

10

= −

=

⎧⎨⎪ ( ) tanh ( )

( ) sin

/E V KR kR

fk

ei

/

⎩⎩⎪ (15.12.8)

Finally, the total cross section is given by

σ π δ π δ πT k kR

KRKR

02

20 2 0

2 24 44 1≈ ≈ = −⎡

⎣⎢⎤sin

tanh ( )

⎦⎦⎥2

(15.12.9)

Note that Eqs. (15.12.3) and (15.12.9) coincide only in the limit of a very short-range potential ( ).kR << 1 Note also that although both methods lead to isotropic differential cross sections, the Born approximation involves a violation of the optical theorem.

15.13. Particles are scattered from the potential V(r) = g/r2, where g is a positive constant. (a) Write the radial wave equations and give their regular solutions. (b) Prove that the phase shifts are given by

δ π μl l l

g= + − +⎛⎝⎜

⎞⎠⎟ +

⎡

⎣⎢⎢

⎤

⎦⎥⎥2

12

12

22

2�

(15.13.1)

(c) Find the energy dependence of the cross section for a fixed scattering angle, (d) Find dl for 2 12μg/� << and show that the differential cross section is

dd

gE

σθ

π μ θ= ⎛⎝⎜

⎞⎠⎟

3

2

2

2 2�cot (15.13.2)

where E is the energy of the scattered particle. (e) For the same potential, calculate the differential cross section in the Born approximation and compare it with the above result.

SOLUTION

(a) The radial wave equation is

d

drk U r

l l

rrkl

2

22

21

0+ − − +⎡

⎣⎢⎢

⎤

⎦⎥⎥

=( )( )

( )χ (15.13.3)

with k E= 2 2μ /� , U = 2mg/�2r2, and φ χl l

mkl l

mR r Y r Y= =( ) ( ) . Substituting the given potential we get

d

drk

r

gl l rkl

2

22

2 21 2

1+ − + +⎡⎣⎢

⎤⎦⎥

⎛

⎝⎜⎞

⎠⎟=μ

�( ) ( )χ 00 (15.13.4)

For g = 0, the solution of Eq. (15.13.4) is given by a free spherical wave:

χkl l rrk r

j krkr l

kr0

2 22 2( ) ( )

sin ( )= −→∞π

π∼

/ (15.13.5)


Therefore, the asymptotic solution for R(r) = χkl/r is

R rkr l

krkr l

krkll( ) ~

sin ( ) sin ( )+ += −π δ π/ /2 2�

(15.13.6)

where �l is given by the relation

� �l l l lg

( ) ( )+ = + +1 12

2

μ�

(15.13.7)

(b) By comparing the two sides of Eq. (15.13.6), we obtain

δ πl l l= −( )�

2 (15.13.8)

where �l is found by solving the quadratic equation (15.13.7):

�l l lg

l l= − ± + + +⎡⎣⎢

⎤⎦⎥

= − ± + +12

12

1 4 12 1

214

12( ) (μ�

))

( )

+

= − ± +⎛⎝⎜

⎞⎠⎟ + + → >

2

12

12

20

2

2

2

μ

μ

g

lg

l

�

�sign � (15.13.9)

Finally, substituting Eq. (15.3.8) into Eq. (15.13.7) leads to Eq. (15.13.1).

(c) The cross section is given by

dd k

l e Pi

l l

l

lσ δ θδΩ = +

=

∞

∑12 12

0

2

( ) sin (cos ) (15.13.10)

Since the dl are not dependent on k (in our particular case), we have

dd k E

σθΩ ∝ ∝

=

1 12 const.

(15.13.11)

(d) For 2 12μg/� << , using Eq. (15.13.1) we obtain

δ π μl l l

g

l

= + − +⎛⎝⎜

⎞⎠⎟ +

+⎛⎝⎜

⎞⎠⎟

⎡

⎣

⎢⎢⎢2

12

12

12

12

22

�⎢⎢

⎤

⎦

⎥⎥⎥⎥

≈ + − +⎛⎝⎜

⎞⎠⎟ +

+⎛⎝⎜

⎞

1 2

22

12

12

112

/

π μl l

g

l� ⎠⎠⎟

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

= −+⎛

⎝⎜⎞⎠⎟

<<222 1

2

1π μg

l� (15.13.12)

Thus, substituting into Eq. (15.13.10),

dd k

lg

lPl

l

σ π μ θΩ = ++

=

=

∞

∑12 1

2 1 2

1

2 2

0

2

( )( )

(cos )� /

kk

gP

g

kPl

l

l

l

2 2

0

22 2 2

4 2π μ θ π μ θ� �

(cos ) (cos )

=

∞

=∑ =

00

2∞

∑ (15.13.13)

In order to sum the series, we will use the generating function of Pl(x):

P x ttx t

Pll

l

ll

( ) (cos )sin( )

=− +

⇒ ==

∞

∑1

1 2

12 22

0

θ θ /==

∞

∑0

(15.13.14)


Therefore,

dd

g

k

g

E

σ π μθ

π μθΩ = =

2 2 2

4 2 2

2 2

2 24 2 8

1

� �sin ( ) sin (/ /22) (15.13.15)

Finally, using dW = sinq dq dj, we get

dd

g

E

g

E

σθ

π μ θθ

π μ θ= = −

3 2

2 2

3 2

24 2 2 1� �

sin

sin ( )

sin

/ ccosθ (15.13.16)

This result coincides with Eq. (15.13.2).

(e) In the Born approximation, the scattering amplitude is

f e U d rg

rek

i i( , ) ( )θ ϕ πμ

π= − ′ ′ = −− ′ −1

42

4

132 2

q r ri

�

qq ri ′∞

′ = −

= −

∫∫∫ d rg qr

qrdr

g

qq

32

2 2

2μ

π μ

θ�

�

sin ( )

[ == 2 2k sin( )]θ / (15.13.17)

Therefore,

dd

fg

kk

σ θ ϕ π μθΩ = =| |( , )

sin ( )2

2 2 2

4 2 24

1

2� / (15.13.18)

which coincides with Eq. (15.13.15). This result is expected since all the phase shifts are small [see Sec. 15.4, Eq. (15.27)].

15.14. Calculate the total cross section for scattering from a completely absorbing sphere of radius a ka( ).>> 1

SOLUTION

The problem of scattering in the presence of absorbing can be treated phenomenologically by introducing the complex scattering potential: V(r) → VR − iVl (Vl ≥ 0). Then, one ends up with complex phase shifts dl = xl + ihl, such that | sl | ≡ | ei 2d l | ≤ 1, where | sl

| = 0 correspond to the case of complete absorption. (Note that the equality | sl

| = 1 is satisfied for real phase shifts, i.e., nonabsorbing media.) By introduc-ing complex phase shifts one then finds that the total cross section consists of two parts, s T = s el + s abs, which are given as

σ πel = + −

=

∞

∑kl l sl

l

22

0

2 1( ) | | (15.14.1)

and

σ πabs = + −

=

∞

∑kl l sl

l

22 2

0

2 1( )( )| | (15.14.2)

Recall that for a potential of finite range a, if

l l ka sl l( ) ( )+ > ⇒ = → =1 0 12 δ (15.14.3)

Hence, in our problem we have

s l l ka

s l l kal

l

= + <

= + >

⎧⎨⎪

⎩⎪

0 1

1 1

2

2

( ) ( )

( ) ( ) (15.14.4)

Setting L (L + 1) = (ka)2 and substituting sl of Eq. (15.14.3) back into Eq. (15.14.1) yields

σ σ π π π πel abs= = + = + = ==

kl

kL L

kka a

l

2 2 22 22 1 1( ) ( ) ( )

00

L

∑ (15.14.5)


Therefore,

σ πT a= 2 2 (15.14.6)

Note that this result is two times larger than the classical result. However, it is half the result of scattering from a hard sphere.

15.15. Two ions of He+ are scattered from each other. The nuclear spin of the ions is zero. The interaction between the ions is Coulombic. (a) Write the scattering amplitude in the frame of the center of mass. (b) Find the differential cross section if the total spin is zero (singlet). (c) Repeat part (b), with total spin of one (triplet). (d) What is the differential cross section for a system of unpolarized ions?

SOLUTION

(a) The scattering amplitude for Coulombic interactions (see Problem 15.19) is

fn

ke

in i i( )

sin ( )ln(sin ( / ))θ

θθ π η= − + +

2 222 22

0

/ (15.15.1)

where η0 1≡ +arg ( )Γ in and n Z Ze

k≡ ′μ

2

2�

. Here, m = mHe/2 is the reduced mass, k E= 2 2μ /� , and Z = Z′ = 1.

(b) The nuclear spin is zero. Thus, the ions are identical fermions (each has spin 1/2 contributed by its electron). If the total spin is zero, the system is in an antisymmetric spin state, and hence the orbital wavefunction must be symmetric

dd

f fn

k

es inσ θ π θΩ⎛⎝⎜

⎞⎠⎟ = + − =

=2

−0 2

2

2

4| |( ) ( )

ln (siin ( / )) ln (cos ( / ))

sin ( ) cos (

θ θ

θ θ

2

2

2 2

22/ /+

−e in

22

4

1

2

1

2

2

2

2

2 4 4

)

sin ( ) cos ( )

cos[ ln tan= + +n

k

n

θ θ/ /

22

2 22

2 2

( )]

sin ( ) cos ( )

θθ θ

/

/ /

⎡

⎣⎢⎢

⎤

⎦⎥⎥

(15.15.2)

(c) This is the same as in part (b), but now the system is in a symmetric spin state (triplet). Hence, the orbital wavefunction must be antisymmetric:

dd

f fn

k

sσ θ π θθΩ

⎛⎝⎜

⎞⎠⎟ = − − =

=12

2

2 44

1( ) ( )

sin (| |

/22

1

2

2 2

24

2

2) cos ( )

cos [ ln tan ( )]

sin ( )+ −

θθ

θ/

/

/

n

ccos ( )2 2θ /

⎡

⎣⎢⎢

⎤

⎦⎥⎥

(15.15.3)

(d) For an unpolarized ion beam, the probability of having total spin s = 0 is 1/4, and the probability of total spin s = 1 is 3/4. Therefore,

dd

dd

dd

s sσ σ σΩ Ω Ω= ⎛

⎝⎜⎞⎠⎟ + ⎛

⎝⎜⎞⎠⎟

= =14

34

0 1

(15.15.4)

Substituting the results from Eqs. (15.15.2) and (15.15.3) leads to

dd

n

k

nσθ θΩ = + −

2

2 4 44

1

2

1

2sin ( ) cos ( )

cos [ ln tan

/ /

22

2 22

2 2

( )]

sin ( ) cos ( )

θθ θ

/

/ /

⎡

⎣⎢⎢

⎤

⎦⎥⎥

(15.15.5)

Note: In the limit of low energies n >> 1, Eq. (15.15.5) differs from the classical result. However, the term oscillates rapidly so averaging over very small angles destroys the interference.

15.16. The interaction potential of two identical particles of spin 1 2/ is

ˆ ( ) ( )[( ) ˆ ˆ ˆ ]V r V r I= + ⋅3 1 2σ σ

where σ i are Pauli matrices and I is the unit operator (see Chap. 7) in the spin space. V(r) is given by

V r rr R

r R

m( ) = − <

>=⎛

⎝⎜⎞⎠⎟

⎧⎨⎪

⎩⎪

�2

240

2μ μ


(a) What is the result of applying the spin operator [ ˆ ˆ ˆ ]3 1 2I + ⋅σ σ on the singlet state and on the triplet

state? (b) Two such particles are scattered on each other at low energies, kR k E>> =1 2 2( ( )).μ /� What is the dominant phase shift that contributes to the scattering amplitude (and the cross section) if the total spin of the system is s = 0? (c) For the conditions in part (b), what is the dominant phase shift if s = 1? (d) Calculate the phase shift of part (b) in the limit kR << 1. Find the cross section. (e) Calculate the phase shift of part (c) in the limit kR << 1. Find the cross section. ( f ) Find the cross section for an unpolarized beam.

SOLUTION

(a) The total spin of the system is ˆ ˆS = �σ /2, where ˆ ˆ ˆσ σ σ= +1 2. Hence,

ˆ ˆ ˆ ˆ ˆσ σ σ σ σ212

22

1 22= + + ⋅ (15.16.1)

From the properties of Pauli matrices,

ˆ ˆ ˆ ˆ ˆ ˆ ˆσ σ σ σ σ σ12

1 1 12

12

12 3= = + + =⋅ x y z I (15.16.2)

and similarly, ˆ ˆ ˆ ˆσ σ σ22

2 2 3= =⋅ I . In the singlet state σ 2 0|singlet⟩ = ; therefore, using Eq. (15.16.1) we find

( ˆ ˆ ) ( ) ˆ ˆσ σ1 2

12

0 3 3 3⋅ ⟩ = − − ⟩ = −| |sin singlet I glet II glet|sin ⟩ (15.16.3)

In the triplet state σ 2 8| |triplet triplet⟩ = ⟩; hence, similar to the previous calculation,

( ˆ ˆ ) ( ) ˆ ˆσ σ1 2

12

8 3 3⋅ ⟩ = − − ⟩ =| | |triplet I triplet I ttriplet⟩ (15.16.4)

Finally, for the operator [ ˆ ˆ ˆ3 1 2I + ⋅σ σ ], we obtain

[ ˆ ˆ ˆ ] ( )

[ ˆ

3 3 3 0

3

1 2I glet glet

I

+ ⟩ = − ⟩ =⋅σ σ | |sin sin

++ ⟩ = + ⟩ =⋅ˆ ˆ ] ( )σ σ1 2 3 1 4| | |triplet triplet triplett⟩ (15.16.5)

(b) For total spin s = 0 the system is in the antisymmetric singlet state. Since the total wavefunction must be antisymmetric (fermions), the orbital wavefunction is symmetric. In general, only even partial waves contribute to a symmetric orbital wavefunction. In our case V = 0 so all the phase shifts vanish.

(c) For total spin s = 1, the system is in one of the three triplet states. The spin wavefunction is symmetric and the orbital wavefunction must be antisymmetric. Thus, only odd partial waves contribute, and for kR << 1 the dominant phase shift is d1.

(d) Let us consider explicitly the phase shift d0 for the s = 0 state. The radial wavefunction Rkl (r) = χkl (r) /r is found by solving

d

dr

l l

rk rkl

2

2 2 221 2

0− + − +⎡

⎣⎢⎢

⎤

⎦⎥⎥

=( ) ˆ ( )μ

�V χ (15.16.6)

The solution Rk0 (r) is found by setting V = 0 and l = 0 in Eq. (15.16.8). Therefore,

R rkr

krj krk0 0( )

sin ( )( )= = (15.16.7)

The logarithmic derivative g0 is then

γ 00

0 1

0

1 1= = − = −=

RdRdr

k kRR

kj kRj kR

r R

cot ( )( )( )

(15.16.8)

where ′ = −j x j x0 1( ) ( ) has been used. Since V = 0 we expect that all phase shifts vanish. In particular,

tan( ) ( )( ) ( )

δγγ0

0 0 0

0 0 0=

′ −′ − =

−kj kR j kRkn kR n kR

kjj kR kj kRkn kR n kR

1 1

0 0 00

( ) ( )( ) ( )

−′ − =γ (15.16.9)

and ( ˆ ) .σ s= =0 0


(e) The dominant contribution for states s = 1 comes from d1. Substituting l = 1 and ˆ ( )V = =4 2 2V r r� /μ [see Eqs. (15.16.2) and (15.16.5)] back into Eq. (15.16.6), we have

d

dr r rk rk

2

2 2 2

2

22

1

2 20− + +

⎡

⎣⎢⎢

⎤

⎦⎥⎥

=μμ�

� χ ( ) (15.16.10)

Therefore, Rk1(r) = Rk0(r) = j0(kr) and g1 = g0. The phase shift d1 is now given by

tan( ) ( )( ) ( )

δγγ1

1 1 1

1 1 1=

′ −′ −

kj kR j kRkn kR n kR

(15.16.11)

In the limit kR → 0, we have

j xx

j xx

n xx

xj x0

2

1 1

2

2 113 3

1 2 1( )

!( ) ( ) ( )≈ − = ≈ − +

′ =/33

21 3′ ≈ +n x

x( ) (15.16.12)

Substituting g1R ≈ − (kR)2/3 in Eq. (15.16.11), we find tan ( )δ131

61≈ <<kR . The scattering amplitude

for d1 is

fk

e Pi

( ) sin (cos )θ δ θδ= 31

1 1 (15.16.13)

After antisymmetrization, we obtain

dd

f f fk

sσ θ π θ θΩ⎛⎝⎜

⎞⎠⎟ = + − = =

=12 22

36( ) ( ) ( )| | | | 22

21

2sin cosδ θ (15.16.14)

Finally, substituting sind1 = tand1 gives

dd

R k Rsσ

Ω⎛⎝⎜

⎞⎠⎟ =

=12 4( ) (15.16.15)

( f ) The cross section for an unpolarized beam is

dd

dd

dd

R ks sσ σ σ

Ω Ω Ω= ⎛⎝⎜

⎞⎠⎟ + ⎛

⎝⎜⎞⎠⎟ =

= =14

34

34

0 12( RR)4 (15.16.16)

15.17. Consider the scattering of two identical spinless particles of mass m. The interaction potential depends on the distance r between the particles and is given by

V rhm R a

r R

r R

( ) = − +⎛⎝⎜

⎞⎠⎟ <

>

⎧⎨⎪

⎩⎪16

1 1

0

2

(15.17.1)

where R a<< are constants. (a) Find the phase shift d0 (k) in the low-energy limit,

kRRa

kE<< << =1

22

μ�

(15.17.2)

where E is the energy in the center-of-mass frame, and μ is the reduced mass. (b) Calculate the total cross section. (c) Repeat your calculation for scattering of two identical spin 1/2 fermions that are polarized in the singlet state. (d) Calculate [in the approximation of parts (b) and (c)] the total cross section for unpolarized spin 1/2 fermions.

SOLUTION

(a) The reduced mass of the two identical particles of mass m is m = m/2. Setting

V r R aV r a

r R

( ) = − +⎛⎝⎜

⎞⎠⎟ ≡ − <

>

⎧⎨⎪

⎩⎪

πμ

2 2 2

081 1

0

� (15.17.3)


and defining the constants

KE V

kR a

kR a

=+

= + +⎛⎝⎜

⎞⎠⎟ ≡ +

2 14

1 12

1 102

2 22

0

μπ π( )

�

⎛⎛⎝⎜

⎞⎠⎟ (15.17.4)

we have (see Problem 15.8)

tan ( )δγ

γ00

2

011= − + <<

kRR

kR (15.17.5)

where g0 is given by γ 0

1= −K KRR

cot ( ) . Thus, substituting g0 into Eq. (15.17.5) we find

tan ( ) tan ( )δ0 k KRkK

KR= − + (15.17.6)

In the limit of Eq. (15.17.2), K → k0. Hence,

KR k RRa

kK

kR Ra

KR→ ≈ +⎛⎝⎜

⎞⎠⎟ ≈ −⎛

⎝⎜⎞⎠⎟0 2

12

1π

π tan ( ) ≈≈ +⎛⎝⎜

⎞⎠⎟

⎡⎣⎢

⎤⎦⎥

≈ −tanπ2

1Ra

aR

(15.17.7)

Therefore, keeping the leading terms of orders kR and R/a in Eq. (15.17.6), we find

tanδ π π0 12 2≈ − + −⎡

⎣⎢⎤⎦⎥

kRaR

(15.17.8)

which in the limit R a/ << 1 leads to tan .δ π0

21= − <<ka

(b) The scattering amplitude in the s-wave approximation is

fk

ei

0 0

10( ) sinθ δδ= (15.17.9)

For identical spinless particles, the amplitude must be symmetrical and therefore the differential cross section is

dd

f fk

σθ π θ δ0

0 02

22

0

4Ω = + − =( ) ( ) sin| | (15.17.10)

Thus, substituting d0 in tan d0 = − (2ka)/p and integrating over dW we obtain

σ π0264= a (15.17.11)

(c) For identical fermions in the singlet spin state the orbital wavefunction must be symmetric. Therefore, the result of part (b) remains unchanged:

( )σ π00 264s a= = (15.17.12)

(d) For unpolarized spin 1/2 particles, the probability of total spin 0 is 1/4 and the probability of spin 1 is 3/4. Therefore,

σ σ σ0 00

011

434

= += =( ) ( )s s (15.17.13)

However, due to antisymmetrization of the orbital wavefunction in the triplet state, d0 does not contrib-ute and (s 0)s = 1 = 0; hence,

σ σ π0 00 21

416= ==( )s a (15.17.14)

15.18. A particle of mass m1 and speed v1 is scattered inelastically by a particle of mass m2 at rest in the Lab frame, where m2 > m1 (Fig. 15.5), resulting in two particles of mass m3 and m4, m1 + m2 = m3 + m4. Assume that in this process, an amount of energy Q is converted from the rest energy of m1 + m2 into kinetic energy of m3 + m4. (a) Find the relations between the scattering angles of m3 in CM (q) and Lab (q0). (b) Find the relation between the differential cross section ds (q)/dW and ds (q0)/dW in the CM and Lab frames, respectively.


Fig. 15.5

m1

Before impact

m2

v

u

w

m3

m4

After impact

Lab frame

CM frame

Before impact

m1m2

v1 v2

v1′

v2′

m3

After impact

m4

q0

q

q

SOLUTION

(a) From conservation of momentum and the definition of CM we obtain

m m

m m

1 1 2 2

3 1 4 2

v v

v v

=

′ = ′

⎧⎨⎪

⎩⎪ (15.18.1)

Similarly, from conservation of energy in the CM frame,

E m m

E Q m m

= +

+ = ′ + ′

⎧

⎨⎪⎪

⎩

12

12

12

12

1 12

2 22

3 12

4 22

v v

v v⎪⎪⎪

(15.18.2)


where E is the initial kinetic energy in the CM frame, and Q is the kinetic energy gain from the collision. Taking v1 = v − v and v2 = −v, where v is the velocity of the CM frame relative to the Lab, we find that

v v v v= + = +m

m mm

m m1

1 21

2

1 2

(15.18.3)

Therefore, in the center-of-mass frame where the total momentum is zero, u v vcos cos ,θ θ0 1= ′ + u vsin sin .θ θ0 1= ′ Hence

tansin

cos /sin

cosθ θ

θθ

θ γ01

= + ′ ≡ +v v (15.18.4)

where γ = ′v v/ 1 .

(b) From Eqs. (15.18.1) and (15.18.2), we have

v v v v21

21 2

3

41= ′ ′ = ′

mm

mm

(15.18.5)

E

E Qm mm m

m mm m

m m

m+ =++

⎛⎝⎜

⎞⎠⎟ ′

=1 4

3 2

1 2

3 4

12

12

1 4v

v ( 11 22

3

2

12+ m m)

v

v (15.18.6)

Therefore,

γ 22

12

1 3

2 4=

′= +

v

v

m mm m

EE Q

(15.18.7)

The relation between the cross section is found from the condition

dd

d ddd

dσ θ θ ϕ σ θ θΩ Ω

⎛⎝⎜

⎞⎠⎟ = ⎛

⎝⎜⎞⎠⎟

Lab CM

sin sin0 0 0 ddϕ (15.18.8)

However, ϕ ϕCM Lab= and from Eq. (15.18.4) we have

cos

(cos )

cos

sin ( cos

20

2

2

0 0

1 2

1 2

θ θ γγ θ γ

θ θ γ

= ++ +

+d θθ γ γ θ θ θ+ = +

⎧

⎨⎪⎪

⎩⎪⎪

2 3 2 1) cos sin/ | | d

(15.18.9)

Hence,

dd

dd

σ γ θ γγ θ

σ02 3 21 2

1Ω⎛⎝⎜

⎞⎠⎟

= + ++

Lab( cos )

cos

/

| | ΩΩ⎛⎝⎜

⎞⎠⎟

CM

(15.18.10)

15.19. (a) Write the Schrödinger equation for a system of two charged particles with charges Ze and Z ′e that interact by Coulombic interaction (use parabolic coordinates). (b) Write the solution in the form

u v( , ) ( , )ξ η ξ η= eikz (15.19.1)

Show that the asymptotic solution of v (x, h) for an outgoing wave in the limit r → ∞ does not depend on the coordinate h. (c) Express v (x) in terms of the confluent hypergeometric function and find the asymptotic solution that is regular at the origin. (d) Find the differential cross sec-tion and show that it coincides with the Rutherford formula.

SOLUTION

(a) The Schrödinger equation in the CM frame is

− ∇ + ′⎡

⎣⎢⎢

⎤

⎦⎥⎥

=�2

22

2μZ Z e

rr E ru u( ) ( ) (15.19.2)


where E = �2k2/2m and m is the reduced mass. In parabolic coordinates,

ξ θ

η θ

ϕ ϕ

ξ= − = −

= + = +

=

⎧

⎨⎪⎪

⎩⎪⎪

=r r z

r r z

x( cos )

( cos )

1

1

ηη ϕ

ξη ϕ

η ξ

cos

sin

( )

y

z

=

= −

⎧

⎨

⎪⎪⎪

⎩

⎪⎪⎪

12

(15.19.3)

The Laplacian is given by

∇ = +∂

∂∂

∂⎛⎝⎜

⎞⎠⎟

+ ∂∂

∂∂

⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢

⎤

⎦⎥ + ∂2

24ξ η ξ ξ ξ η η η ∂∂ϕ2 (15.19.4)

and Eq. (15.19.2) is written in the following form:

4 2

2ξ η ξ ξ ξ η η ημ

+∂

∂∂

∂⎛⎝⎜

⎞⎠⎟

+ ∂∂

∂∂

⎛⎝⎜

⎞⎠⎟

− ′⎡

⎣⎢

Z Z e

h⎢⎢

⎤

⎦⎥⎥

+ ∂∂

+⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪=

2

22 0

ϕk ru( ) (15.19.5)

(b) From the azimuthal symmetry of the solution we have u = u(x, h). In the limit r → ∞ the outgoing wave is of the form r−1eikr. Therefore,

u v v v( , ) ( , ) ( ) (ξ η ξ= − + → − =→∞e r z r z e r z eikzr

ikz ikz )) (15.19.6)

Equation (15.19.6) leads to a separation of variables in the form

u v v( , ) ( ) ( )/ /ξ η ξ ξη ξ= = −e e eikz ik ik2 2 (15.19.7)

(c) Substituting Eq. (15.19.7) into Eq. (15.19.5), the Schrödinger equation is reduced to an equation for v (x ):

ξξ

ξ ξμd

dik

dd

nk nZ Z e

k

2

2

2

21 0v v

v+ − − = ≡ ′⎛

⎝⎜⎞

⎠⎟( )

� (15.19.8)

This equation is of the form

zd F

dzb z

dFdz

aF2

2 0+ − − =( ) (15.19.9)

and its solution is the confluent hypergeometric function F(a, b, z), where z = ikx, a = −in, and b = 1:

v ( ) ( , , )ξ ξ= −AF in ik1 (15.19.10)

The asymptotic solution of F(a, b, z), which is regular at the origin, behaves like

F bi z

b aa a b

zie z

z

a z a

→∞

−

→ −− + −⎛

⎝⎜⎞⎠⎟ −Γ Γ( )

( )( )

( )1

−−

+ − −⎛⎝⎜

⎞⎠⎟ +

⎡

⎣⎢⎢

⎤

⎦⎥⎥

⋅ ⋅ ⋅b

aa b a

zΓ ( )( )( )

11

(15.19.11)

Substituting z = ikx, a = −in, and b = 1, we find

v ( )( )

( )/

ln ( lξ θξ

πξ ξ

→∞−→ + +A

ein

e fr

en

n kc

i k n2

11

Γnn )2kr⎡

⎣⎢⎤⎦⎥

(15.19.12)

where

fin

i ine

kc

in

( )( )

( ) sin (

ln sin ( / )

θθ

θ= +

−Γ

Γ1

2

2 2

2 /22) (15.19.13)


Therefore,

ur

ni kz n k r z

c

Aein

e fr

e→∞+ −→ + +

πθ

/[ ln ( )]

( )( )

2

11

Γii kr n kr[ ln ]−⎡

⎣⎢⎤⎦⎥

2 (15.19.14)

(d) Note that u is not of the form

uri kz

i kr

e fe

r→∞ → + ( )θ (15.19.15)

The reason is that the Coulombic potential does not decrease rapidly enough. However, we can general-ize the result and find the cross section by

dd

fn

k

σ θθΩ ( )

sin ( )= =| |2

2 24 2/ (15.19.16)

This is the famous Rutherford formula.

15.20. The scattering amplitude for neutron-proton scattering is given by

f A BfP N

i( ) ( ˆ ˆ )θ ξ σ σ ξ= ⟨ + ⟩⋅| | (15.20.1)

where A and B are “constants,” σ are Pauli matrices, and { , }| |ξ ξi f⟩ ⟩ are the initial and final spin states of the system | | | | |ξ⟩ = + + ⟩ + − ⟩ − + ⟩ − + ⟩i f p N p N p N p N, { , , , }. (a) Calculate the scattering amplitude for

each of the 16 possibilities. (b) Find the differential cross section for scattering of | |+⟩ → +⟩N N and | |+⟩ → −⟩N N when the spin of the emergent proton is not measured by the detector. (c) Find the cross section for scattering in the states | | | |singlet singlet triplet triplet⟩ → ⟩ ⟩ → ⟩, , .| |singlet triplet⟩ → ⟩

SOLUTION

(a) The operator ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆσ σ σ σ σ σ σ σp Nxp

xN

yp

yN

zp

zN⋅ = + + operates separately on the proton and neutron states.

For example:

⟨+ ⟨+ + +⟩ +⟩ = + ⟨+ +⟩⋅| | | | | |p Np N

p N p xA B A B( ˆ ˆ ) ( ˆσ σ σ pp N x N

p y p N y N N

⟨+ +⟩

+ ⟨+ +⟩ ⟨+ +⟩ + ⟨+

| |

| | | | |

ˆ

ˆ ˆ

σ

σ σ ˆ ˆ )σ σz N p z p| | |+⟩ ⟨+ +⟩ (15.20.2)

Using the results σ x | |+⟩ = −⟩, σ y i| |+⟩ = −⟩, ˆ ,σ z | |+⟩ = +⟩ and the orthogonality of the spin states ⟨+ −⟩ =| 0, we obtain

⟨+ ⟨+ + +⟩ +⟩ = +⋅| | | |p Np N

p NA B A B( ˆ ˆ )σ σ (15.20.3)

Similarly,

⟨+ ⟨+ + +⟩ −⟩ = ⟨+ +⟩ ⟨+⋅| | | | | | |p Np N

p N xA B B( ˆ ˆ ) [ ˆσ σ σ ˆ ˆ ˆ ˆ ˆσ σ σ σ σx y y z z| | | | | | | | |−⟩ + ⟨+ +⟩ ⟨+ −⟩ + ⟨+ +⟩ ⟨+ −−⟩ =] 0

(15.20.4)

All the 16 possible scattering processes are then summarized in Table 15.3.

(b) We consider the case where the incident proton is in a general spin state α β α β| | | | | |+⟩ + −⟩ + =p p( ).2 2 1 Since the state of the emergent proton is not measured (and different components do not interfere),

dd

f fN N

σΩ

⎛⎝⎜

⎞⎠⎟ = +

+⟩ → +⟩+ −

| |

| | | |2 2 (15.20.5)


where f± is the amplitude for having a proton in | ± ⟩p final states, respectively. Using Table 15.3 for the entries that correspond to the process | |+⟩ → −⟩N N we find

f

fA B

A BM+

−

⎛⎝⎜

⎞⎠⎟

=+

−⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟ ≡ ⎛

⎝⎜⎞⎠

00

αβ

αβ

ˆ ⎟⎟ (15.20.6)

Therefore,

dd

M MN N

σ α β αβ αΩ

⎛⎝⎜

⎞⎠⎟ = ⎛

⎝⎜⎞⎠⎟ =

+⟩ → +⟩| |( ) ˆ ˆ (* * † *ββ α

β

α

*)| |

| |

| | | |

A B

A B

A B

+−

⎛

⎝⎜

⎞

⎠⎟

⎛⎝⎜

⎞⎠⎟

= +

2

2

2

0

0

22 2 2+ −| | | |β A B

We assume now that the incident protons are not polarized. This means that we have equal probabilities of 1/2 of finding a proton in |+⟩p or |−⟩p initial states. Therefore, substituting a = 1, b = 0, and a = 0, b = 1 into Eq. (15.20.7) we obtain

dd

A B A B AN N

σΩ

⎛⎝⎜

⎞⎠⎟ = + + − =

+⟩ → +⟩| |

( )12

2 2 2| | | | | | ++ | |B 2 (15.20.7)

Note that this result can be written as

dd

Tr M MN N

σΩ

⎛⎝⎜

⎞⎠⎟ =

+⟩ → +⟩| |

12

( ˆ ˆ )† (15.20.8)

which is valid for the case of unpolarized particles of spin 1/2. In order to use Eq. (15.20.9) for the process | |+⟩ → −⟩N N

, we replace the matrix M of Eq. (15.20.6) by

f

fB

M+

−

⎛⎝⎜

⎞⎠⎟

= ⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟ ≡ ⎛

⎝⎜⎞⎠⎟

0 20 0

αβ

αβ

ˆ (15.20.9)

This immediately gives us

dd

B BN N

σΩ

⎛⎝⎜

⎞⎠⎟ = + =

+⟩ → −⟩| |

( )12

0 4 22 2| | | | (15.20.10)

(c) Consider the system in a singlet state where the total spin is zero:

| | | | |singlet⟩ = +⟩ −⟩ − −⟩ +⟩12

( )p N p N (15.20.11)

Table 15.3

⟨+ ⟨+| |p N ⟨− ⟨+| |p N ⟨+ ⟨−| |p N ⟨− ⟨−| |p N

⟨+ ⟨+| |p NA + B 0 0 0

⟨− ⟨+| |p N0 A − B 2B 0

⟨+ ⟨−| |p N0 2B A − B 0

⟨− ⟨−| |p N0 0 0 A + B


Therefore, the scattering amplitude is

f f f| | | | | |singlet singlet⟩→ ⟩ + −⟩→ + −⟩ + −⟩→ −= −12

( ++⟩ − +⟩→ + −⟩ − +⟩→ − +⟩− +

= − − − + −

f f

A B B B A

| | | | )

[ (12

2 2 BB A B)] = − 3 (15.20.12)

Indeed, a great deal of algebra can be saved by noting the singlet state and the triplet states are eigenstates of the operator ˆ ˆσ σN p⋅ with eigenvalues −3 and 1, respectively:

( ˆ ˆ )

( ˆ ˆ )

σ σ

σ σ

N p

N p

⋅⋅

⟩ = − ⟩| |

|

singlet singlet

tri

3

pplet triplet⟩ = ⟩| (15.20.13)

Thus, the scattering amplitudes are

f A B N p( ) ( ˆ ˆ )singlet singlet singlet si→ = ⟨ + ⋅| |σ σ nnglet

triplet triplet triplet

⟩ = −

→ = ⟨ +

A B

f A

3

( ) (| BB A B

f

N pˆ ˆ )

( )

σ σ⋅ ⟩ = +

→ = ⟨

| triplet

singlet triplet ttriplet singlet| |( ˆ ˆ )A B N p+ ⟩ =⋅σ σ 0

(15.20.14)

The cross sections are therefore

dd

A B

dd

σ

σ

Ω

Ω

⎛⎝⎜

⎞⎠⎟ = −

⎛⎝⎜

⎞⎠

→singlet singlet

| |3 2

⎟⎟ = +→triplet triplet

| |A B 2

(15.20.15)

15.21. Consider the time-independent Schrödinger equation in one dimension with V(x) = 0 only in the finite region | |x x< 0. (a) Show that for any solution of the Schrödinger equation φ( )x , the probability current is a constant that does not depend on the position x. (b) The asymptotic form of the wavefunction φ( )x can be written as

φ

φL

ikx ikx

Rikx ikx

x Ae Be x x

x Ce De x

( )

( )

= + << −

= +

−

−0

>>> −

⎧⎨⎪

⎩⎪ x0

(15.21.1)

where A, B, C, and D are complex constants that are related by the scattering matrix S:

BC

SAD

⎛⎝⎜

⎞⎠⎟ = ⎛

⎝⎜⎞⎠⎟

ˆ

Show that S is a unitary matrix.

SOLUTION

(a) The time-independent Schrödinger equation is

− ∇ + − =�2

2

20

mV Eφ φ( ) (15.21.2)

Since V and E are real, any solution φ satisfies the conjugate equation as well:

− ∇ + − =�2

2

20

mV Eφ φ∗ ∗( )

Multiplying these equations by φ∗ and φ, respectively, and then subtracting the resulting expressions, leads to

φ φ φ φ φ∗ ∗ ∗− ∇ + −⎛

⎝⎜⎞

⎠⎟− − ∇ + −

⎛

⎝� �

22

22

2 2mV E

mV E( ) ( )⎜⎜

⎞

⎠⎟= − ∇ − ∇ =φ φ φ φ φ∗ ∗�

22 2

20

m( ( ) ) (15.21.3)


Thus,

− ∇ ∇ − = − ∇ =⋅im

V i�

�2

20[ ( ) ] ˆφ φ φ φ∗ ∗ i J (15.21.4)

Namely,

dJ x

dxJ x

imddx

ddx

( )( )= = −

⎡

⎣⎢⎢

⎤

⎦⎥0

2where

� φ φ φ φ∗∗

⎥⎥ (15.21.5)

(b) In the previous section we showed that the probability current J(x) is conserved and does not depend on x for any solution φ(x) of the Schrödinger equation. For large negative values of x we have

2imJ A e B e ikAe ikBeL

ikx ikx ikx ikx

�= + − −− −( ) ( ) (∗ ∗ AAe Be ikA e ikB e

ik A

ikx ikx ikx ikx+ − +

=

− −) ( )

(

∗ ∗

2 | |22 2− | |B ) (15.21.6)

Therefore,

Jk

mA BL = −�

( )| | | |2 2 (15.21.7)

Similarly, for large positive values of x,

Jk

mC DR = −�

( )| | | |2 2 (15.21.8)

Now, from current conservation, JL = JR, and

| | | | | | | | | | | | | | | |A B C D B C A D2 2 2 2 2 2 2 2− = − ⇒ + = + (15.21.9)

from which it follows that the scattering matrix is unitary, that is,

ˆ ˆ ˆ†S S I= (15.21.10)


15.22. A particle with mass m1 is scattered elastically by a particle of mass m2 at rest in the Lab frame. (a) Find the relation between the scattering angle of m2 in the Lab frame and the scattering angle in the CM frame. Show that in the Lab frame the particle m2 will always recoil in the front half of the sphere. (b) Find the relation between the scattering angle of m1 in the Lab and CM frames. (c) What is the range of possible angles for scattering of particle m1 in the Lab frame for the following conditions: i. m m1 2 1/ < ; ii. m m1 2 1/ = ; iii. ( ) .m m1 2 1/ >

Ans. (a) tansin

costanθ θ

θ θ π θ θ π2 2 21

0 0 02

= − ≤ < → ≥ → ≤ ≤⎛⎝⎜

⎞⎠⎟

(b) tansin

cosθ θ

θ γ γ11

2= + =⎛

⎝⎜⎞⎠⎟

mm

(c) i. γ θ π< → ≤ ≤1 0 1 ; ii. γ θ π θ π= → = → ≤ ≤12

021 1

iii. /γ θ θ γ> → ≤ ≤ = −1 0 11 11max sin ( )

15.23. A particle of mass m1 is elastically scattered by a particle of a mass m2 at rest in the Lab frame. (a) Consider particle m1. Find the relation between the differential cross section in the center of the mass frame, d dσ θ ϕ( , ) ,/ Ω and the differential cross section in the Lab frame, d dσ θ ϕ0 0 0( , )/ .Ω (b) Assume m1 = m2. Find the differential cross section for scattering m1 in the Lab frame, if it is given that the cross section in the center-of-mass frame is symmetrical. (c) Calculate the total cross section for part (b). Show explicitly that the total cross section is not dependent on the related frame.

Ans. (a) dd

dd

mm

σ γ γ θγ θ

σ γ02 3 2

1

2

1 21Ω Ω= + +

+ =⎛( cos )cos

;/

| | ⎝⎝⎜⎞⎠⎟


(b) dd

dd

σθ σ θ π ϕ π0

0 0 04 02

0 2Ω Ω= < ≤ < ≤⎛⎝⎜

⎞⎠⎟cos ,

(c) σσ σ

π θ θ θ ϕ σππ

00

00 0 0 0 0

0

2

0

244

= = =∫dd

d d dΩ Ω cos sin/

∫∫∫15.24. Assuming azimuthal symmetry, find the relation between the scattering amplitude and the differential cross

section.

Ans. d d fkσ θ/ Ω = ( ) .| |2

15.25. Using Fermi’s golden rule calculate the probability density (per unit time) of the transition of a particle (mass m and energy E ) from initial state | pi ⟩ to final state | pf ⟩. Show that the cross section d d W p p Ji f iσ / /Ω = ( , ) , where W is

the transition probability density and Ji, the probability current density JEmi

i= ⎛⎝⎜

⎞⎠⎟

12

33

π�, coincides with the

Born approximation.

15.26. Consider the potential V V= − 0 for r a≤ and V = 0 for r a> . Show that the s-bound states ( , )l E= <0 0 satisfy the quantization condition: tan ( ) ,Ka K k= − / where

K k k2

02 2= +( )

k

V0

02

1 22= ⎛

⎝⎜⎞⎠⎟

μ�

/

k

E= ⎛⎝⎜

⎞⎠⎟

22

1 2μ�

/

15.27. Building on Problem 15.26, show that the phase shift for s-wave scattering states ( , )l E= =0 0 is δ ξ0 0= −( ) ,k kr where tan ( ) tan [ ( )].Ka K k k= + / ξ

15.28. Find the condition for resonance scattering at low energies (ka << 1), and demonstrate that for near resonance dd k k k a

σΩ ≈

+1

202 2

0cot ( ).

15.29. Given the potential well V r V( ) = − 0 for r R< and V r( ) = 0 for r R> , (a) show that for q k= 2 2sin ( )θ / the scattering amplitude f ( )θ is

fV

qRq

q

R Rqq

( )sin ( ) cos ( )θ

μ= −

⎛⎝⎜

⎞⎠⎟

02 2

2

� (15.29.1)

(b) Show that the differential cross section in the limit Rq << 1 is a constant that is not dependent on k or q. (c) Consider the potential V B( ) ( ).r r= δ Use the Born approximation to find the differential cross section. Calculate the constant B so that the results from parts (b) and (c) coincide.

Ans. (c) dd

Bσ μπΩ =

2 2

2 44 �; B V R= 4

3 03π

15.30. For the potential V = − V0 for r ≤ a and V = 0 for r > a, find the conditions for d0 = p.

Ans. ka << 1 and tan (ka) = ka; k mV≡ 2 02/�

15.31. Derive the optical theorem s total = 4p Im fk (0) / k. Hint: Use the partial wave expansion of f (q) and

σ π σ πδel abs= + − = +

=

∞

∑kl e

kl

i

l

l

22 2

0

22 1 1 2 1( ) ( )(| | 112 2

0

−=

∞

∑ | |ei

l

lδ

)

15.32. Consider scattering of two identical spinless particles of mass m. The energy of the scattered particle is E0 = �2k2/2m, whereas the target particle is at rest. The interaction potential between the particles is V(r) = V0/r2, where r is the relative coordinate. (a) Find the phase shifts dl. (b) Write the differential cross section ds (q)/dW in the center-of-mass frame. (c) Obtain ds0(q0)/dW in the Lab frame, where the target particle is initially at rest.


Ans. (a) δ π μl l l l

V= − − + + + +⎡

⎣⎢⎤⎦⎥

⎛

⎝⎜

⎞

⎠⎟

⎡

⎣⎢

212

1 1 4 12 0

2( )�⎢⎢

⎤

⎦⎥⎥

(b) dd k

L e Pi

L L

L

Lσ δ θδΩ = +

⎡

⎣

⎢

=

∞

∑164 12 2 2

0

2( ) sin (cos )⎢⎢

⎤

⎦

⎥⎥

2

; (c) d

dd

dσ θ

θ σ θ( )cos

( )004

2Ω Ω=

15.33. (a) Calculate, in Born approximation, the scattering amplitude of a particle with mass m from a spherical well potential with radius a and depth V0; calculate the boundary of a spherical well “point” when a → 0 and V0 → ∞ with V0a

3 = C for a given constant C. (b) A neutron is scattered by a neutron. The neutron mass is m, and we assume that the potential between the two neutrons satisfies the conditions given in part (a). Calculate the scattering amplitude and the differential cross section (in the center-of-mass frame) when the neutron pair is in singlet state and in triplet state.

Ans. (a) fC qa qa qa

qaq k( )

sin( ) cos ( )

( );θ μ= −⎡

⎣⎢

⎤

⎦⎥ =3

22 3�

ssin( ); ; ( )θ μ θ μ/2

22

30 2= →→m

fC

qa �

(b) sin :glet f f fmC d

ds( ) ( ) ( ) ;θ θ π θ σ= + − = ⎛⎝⎜

⎞⎠⎟

23 2

� Ω

ss m C=

=0 2 2

449 �

triplet f f fddA

s

: ( ) ( ) ( ) ;θ θ π θ σ= − − = ⎛⎝⎜

⎞⎠⎟ =

=

0 01

Ω

15.34. Consider elastic scattering of two helium atoms in their ground state. Assume that we can describe them as impenetrable spheres, each of radius a. Designate by s 43, s 33, and s 44 the total cross section of (He4, He3), (He3, He3), and (He4, He4), respectively. (a) Using partial waves expansions, derive the three differential cross sections. (b) Prove that for ka << 1, the relations s 43 : s 33 : s 44 = 1 : 1 : 4 hold.

Ans. (a) tan( )( )

( )

( )[δl

l

l

lj kan ka

kaka

l= << → −

+

+

for 12 1

2 1

(( )!!]; sin

(co

2 1

2 12

43

lf

lk

e

dd

f P

li

l

l l

l

−≡ +

=

δ δ

σΩ ss ) ; (cos )

,

θσ

θl

l l

l

dd

f P

=

∞

=∑

⎡

⎣

⎢⎢

⎤

⎦

⎥⎥

=0

2

33

0Ω

evenn odd

∞

=

∞

∑ ∑⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

+⎡

⎣

⎢⎢⎢

⎤

⎦

2

1

3 f Pl l

l

(cos )

,

θ ⎥⎥⎥⎥

=⎡

⎣

⎢⎢

⎤

⎦

⎥⎥

=

∞

∑

2

44

0

2

4

;

(cos )dd

f Pl l

l

σθΩ

(b) s 43 = 4pa2; s 33 = 2pa2 [2 + 3 (ka)4]; s 44 = 16pa2

CHAPTER 16

330

Semiclassical Treatment of Radiation

16.1 The Interaction of Radiation with Atomic SystemsThe classical Hamiltonian of a particle with mass m, charge e, and spin S in an external electromagnetic field is given by

Hm

ec

V ee

mc= −⎛

⎝⎜⎞⎠⎟ + + −1

2

2

p A r S( ) φ iB (16.1)

where A is the vector potential, φ is the scalar potential, and B = ∇ × A is the magnetic field. It is possible to choose a gauge for which H will be simpler. The gauge generally employed in problems dealing with radiation is the Coulomb gauge. This gauge is also called the radiation gauge or the transversal gauge. In this gauge one chooses ∇ = =i A 0 0φ (16.2)

Thus, the Hamiltonian obtained in this gauge is

ˆ ˆ ˆ( ) ˆ ˆHm

Ve

mce

mc

emc

= +⎡

⎣⎢

⎤

⎦⎥ + − + −p

r A p A2

2 2

2

22i SS iB

⎡

⎣⎢

⎤

⎦⎥ (16.3)

with H0 as the unperturbed Hamiltonian (in the absence of an external field) and ˆ ′H (t) as the perturbation

Hamiltonian

ˆ ˆ ˆH H H= + ′0

For a semiclassical treatment of the radiation we assume that the term A2 is very small and negligible (see Problem 16.2). In this case,

ˆ ( ) ˆ ˆ′ = − −H te

mce

mcA p Si iB (16.4)

This limit is called the low intensity limit.

16.2 Time-Dependent Perturbation TheoryIn the low-intensity limit, ˆ ′H (t) can be treated as a small time-dependent perturbation. If the system is ini-tially in the state |i⟩ and the perturbation is turned on at t = 0, the first-order amplitude for finding the system in the state | f ⟩ at time t is given by

a ti

e f H t i dtfi

i tt

f i( ) ( ) ˆ ( )1

0

1= ⟨ ′ ′ ⟩ ′′∫�

ω| | (16.5)


where �ω f i f iE E= − . In a semiclassical treatment one usually assumes that the electromagnetic field A is described by a plane wave:

A r k r k r( , ) cos ( ) exp[ ( )t A t A i t= − + = −2 0 0| |d di iω θ ω ]] exp[ ( )]*+ − −A i t0d k ri ω (16.6)

where A0 = | A0 | eiq is a complex number, d is a unit vector in the direction of polarization, k is the wave or

propagation vector, and d i k = 0 (transversal gauge). Therefore,

a te T e

fi

i t

f i

f ii

f i f i( )

( ) ( )

( )1 1= − −− −

− + +ω ω ω ω

ω ω �

tt

f i

f iGT−

+1

ω ω � (16.7)

where

T

emc

f e l A i i

T

fii

f i

+ ≡ − ⟨ + × ⟩[ ˆ ˆ ( )]| |k r p S ki i i0 d d

GG iemc

f e A i i≡ − ⟨ − × ⟩

⎧

⎨⎪⎪

− [ ˆ ˆ ( )]*| |k r p S ki i i0 d d⎩⎩⎪⎪

(16.8)

See Problem 16.4.

16.3 Transition RateConsider the transition amplitude a tfi

( ) ( )1 . A resonant transition is obtained when the frequency of the external radiation field is close to one of the characteristic frequencies of the unperturbed system, i.e., ω ≈ ± f

i. In

this case, one can neglect the interference term in Eq. (16.7) and distinguish between resonant absorption (w f i > 0) and resonant emission (w f i < 0). The transition probability is then given by

PT t

fif i f i

f i≅

−−

⎛

⎝⎜⎞⎠⎟

+| |

�2

22

2

sin[( ) ]

( )

ω ωω ω

/

/ωω f i > 0 (16.9)

(see Problem 16.4) and for induced emission:

PT t

fif i f i

f i≅

−+

⎛

⎝⎜⎞⎠⎟

−| |

�2

22

2

sin[( ) ]

( )

ω ωω ω

/

/ωω f i < 0 (16.10)

For a strictly monochromatic field, these transition probabilities depend strongly on the difference w − | w f i |, and lead to a nonstationary transition rate. A transition probability that is linear in time (constant transition rate) is obtained if one considers the transition from an initial state |i⟩ to a continuum of final states | f ⟩. In this case, the transition rate is obtained by using a Fermi golden rule:

WdP t

dtf T i E Efi f i

±±

±= = ⟨ ⟩ = ±( ) ˆ ( )2 2π ρ ω�

�| | | | (16.11)

where r (Ef ) is the density of the final states. Similarly, when the radiation field is not monochromatic, but rather contains a spectrum of frequencies u(w), the transition rate is

We

m

uf e ifi

f i

f i

= ⟨ ±±4 2 2

2 2 2

π ω

ω�

( )[ ˆ ˆ (/| k r p Si i id kk × ⟩d)] | |i 2 (16.12)

where |i⟩ and | f ⟩ are the initial and final (discrete) states, and the plus/minus signs correspond to absorption and induced emission, respectively.

CHAPTER 16 Semiclassical Treatment of Radiation332

16.4 Multipole TransitionsIn the long wavelength approximation, e ii± ≈ + ⋅ ⋅ ⋅k r k ri i1 so Tfi

± is given by the following multipole expansion:

T im f ii

f if i f i± ≈ ⟨ ⟩ + ⟨ + × ⟩ω | | | |d di iˆ ( ˆ ˆ ) ( )r L S k

22 −− ⟨ ⟩

mf ifiω

2| |( ˆ)( ˆ)k r ri id (16.13)

The first term in Eq. (16.13) corresponds to an electric-dipole transition. The second term corresponds to a magnetic-dipole transition, and the third term corresponds to an electric-quadrupole transition. Usually, the transition rate is dominated by the electric–dipole term; in this case the transition rate is

We

u f ifi f i= ⟨ ⟩4 2 2

22π ω

�( ) ˆ| | | |d i r (16.14)

However, for particular states |i⟩ and | f ⟩, ⟨ ⟩f i| |d i r may vanish. This state is called the forbidden transition. Note that for an isotropic external radiation field, the polarization vector ε is randomly oriented. Averaging the components of the unit vector d over all angles gives

We

u f i B u ufi f i f i f i= ⟨ ⟩ ≡4

3

2 2

22π ω

�( ) ˆ ( )| | | |r (16.15)

Bf i are known as the Einstein coefficients for absorption and induced emission.

16.5 Spontaneous EmissionAn excited atomic system can also emit radiation in the absence of an external radiation field. The transition rate for a spontaneous transition, in the dipole approximation, is given by

We

cf i Afi

f if i

spon = ⟨ ⟩ ≡43

2 3

32

�

ω| | | |r (16.16)

where Af i is the Einstein coefficient for spontaneous emission.

SOLVED PROBLEMS

16.1. The motion of a charged particle in an external electromagnetic field is described by the classical Hamiltonian

Hm

ec

V e= −⎛⎝⎜

⎞⎠⎟ + +1

2

2

p A r( ) φ (16.1.1)

where A(r, t) and φ( , )r t are the electromagnetic potentials, e is the charge, and c is the speed of light.

Show that the time-dependent Schrödinger equation it

H�∂∂ =ψ ψˆ is invariant under the following

gauge transformation:

A A A r→ ′ = + ∇

→ ′ = − ∂∂

→ ′ =

χ

φ φ φ χ

ψ ψ χ

( , )

( , ) /

t

c t

eie r t

1

�ccψ

⎧

⎨

⎪⎪⎪

⎩

⎪⎪⎪

(16.1.2)


SOLUTION

Under the gauge transformation of Eq. (16.1.2), the Schrödinger equation it

H�∂ ′∂ = ′ ′ψ ψˆ , takes the form

it

Tm

V e r t�∂∂ = − ′⎛

⎝⎜⎞⎠⎟ + + ′( ˆ ) ˆ ˆ( ) ( , )ψ φ1

2

2

p A rec

⎡⎡

⎣⎢⎢

⎤

⎦⎥⎥

− − ∇⎛⎝⎜

⎞⎠⎟ + +

ˆ

ˆ ˆ( )

T

mec

ec

V e

ψ

χ φ=1

2p A r

2

−− ec t

T∂∂

⎡

⎣⎢⎢

⎤

⎦⎥⎥

χ ψˆ

(16.1.3)

However,

∂ ′∂ = ∂

∂ = ∂∂ + ∂

∂⎛⎝⎜

⎞⎠⎟

ψ ψ ψ ψ χt t

T Tt

iec t

( ˆ ) ˆ�

(16.1.4)

and

ˆ ( ˆ ) ( ) ˆ ˆp pT i T Tec

ψ ψ ψ ψ χ= − ∇ = + ∇⎛⎝⎜

⎞⎠⎟� (16.1.5)

Therefore, the right-hand side of Eq. (16.1.3) equals

12m

ec

ec

Tec

T Vˆ ˆ ˆ ˆ ˆ(p A p A r− − ∇⎛⎝⎜

⎞⎠⎟ −⎛

⎝⎜⎞⎠⎟ +⋅χ ψ ))

ˆ ˆ ˆ(

+ − ∂∂

⎡⎣⎢

⎤⎦⎥

= −⎛⎝⎜

⎞⎠⎟ +

eec t

Tm

ec

V

φ χ ψ

12

2

p A r)) + ∂∂

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪e

ec t

φ χ ψ−

(16.1.6)

and consequently, expression (16.1.3) reduces to

ˆ ˆ ˆT it

ec t

T Ht

�∂∂ − ∂

∂⎛⎝⎜

⎞⎠⎟

= − ∂∂

⎛⎝⎜

⎞⎠⎟

χ ψ χ ψ (16.1.7)

Multiplying Eq. (16.1.7) on the left by ˆ†T we obtain it

H�∂∂ =ψ ψˆ , which is the Schrödinger equation in

the original gauge.

16.2. An atomic electron with mass m, charge e, and spin S interacts with an external radiation field described by the vector potential A (r, t). The Hamiltonian of the system is

ˆ ˆ ˆ ( )H H H t= + ′0

(16.2.1)

where ˆ ˆ ˆ( )H p m V02 2= +/ r is the “atomic” Hamiltonian, and ˆ ( )′H t is the time-dependent remaining

interaction. (a) Show that the interaction Hamiltonian can be written as

ˆ ( ) ˆ ˆ′ = − + −H te

mce

mc

emc

A p A Si i2

22

2B (16.2.2)

where ∇ =i A 0, φ = 0, and B = ∇ × A. (b) Consider the low-intensity limit and estimate the relative magnitudes of the various interaction terms.

SOLUTION

(a) The Hamiltonian of the electron in an external electromagnetic field is

ˆ ˆ ( ) ˆHm

ec

V ee

mc= −⎛

⎝⎜⎞⎠⎟ + + −1

2

2

p A r Sφ iB (16.2.3)


where V(r) is the binding potential of the nucleus, ( , )φ A are the scalar and vector electromagnetic potentials, and B = ∇ × A is the magnetic field. Equation (16.2.3) leads to

ˆ ˆ

( ˆ ˆ ) ( )Hm

emc

e

mcV e

e= − + + + + −pp A A p A r

2 2

22

2 2 2i i φ

mmc

H eemc

e

mc

emc

ˆ

ˆ ( ˆ ˆ )

S

p A A p A

i

i i

B

= + − + + −0

2

22

2 2φ ˆS iB

(16.2.4)

This expression is further simplified if we choose the gauge ∇ =i A 0 and φ = 0 (the transversal gauge). Taking into account that p = − ∇i� , and operating with ( ˆ )p Ai on an arbitrary function ψ (r), we find

( ˆ ) ( ) ( ) ( )p A r r ri ψ ψ= − ∂∂

= − ∂∂

⎛

=∑i

xA i

xA� �α α

α

α α1

3

⎝⎝⎜⎞⎠⎟

+ ∂∂

⎡

⎣⎢⎢

⎤

⎦⎥⎥

= − ∇ + =

∑ ψ ψ

ψ ψ

Ax

i

α α

α

3

�( ) î iA A p AA p ri ˆ ( )ψ (16.2.5)

Therefore,

ˆ ˆ ˆ ( ) ˆ ˆ ˆH H H t He

mce

mc

emc

= + ′ = − + −0 0

2

22

2A p A Si iB (16.2.6)

where H0 is the unperturbed “atomic” Hamiltonian and ˆ ( )′H t is the time-dependent interaction.

(b) Consider, for example, a hydrogen-like atom interacting with a monochromatic radiation field of angu-lar frequency w = ck. In this case, the magnitude of each term in Eq. (16.2.6) can be approximated by its root-mean-square value in a given unperturbed stationary state of H0. Let us define the following root-mean-square values:

A

p

≡ ⟨ ⟩

≡ ⟨ ⟩

⎧⎨⎪

⎩⎪

ψ ψ

ψ ψ

| |

| |

ˆ ˆ

ˆ ˆ

A A

p p

i

i

(16.2.7)

and examine the relative magnitudes of the three interaction terms of Eq. (16.2.6):

ˆ ( , ) ˆ

ˆ ( , )

ˆ

′ = −

′ =

′ = −

He

mct

He

mct

He

s

1

2

2

22

2

A r p

A r

i

mmctˆ ( , )S riB

⎧

⎨

⎪⎪⎪⎪

⎩

⎪⎪⎪⎪

(16.2.8)

where |ψ⟩ is an eigenstate of H0 for which A ≠ 0 and p ≠ 0. Consider, first the ratio ˆ ˆ :′ ′H H2 1/

ˆ

ˆ ˆ′′

≈ ≈ ≈′H

H

e AeAcp

eAp mc

p m

H

H2

1

2 2

21

0

/

/ (16.2.9)

In the low-intensity limit ˆ ˆ′ ′H H1 0/ is a small perturbation. Consequently, the ratio ˆ ˆ′ ′H H2 1/ is also small.

Note that in the high-intensity limit, where the radiation is of the order of the atomic field, ˆ ′H2 can

become as large as ˆ ′H1. Consider now the ratio ˆ ˆ′ ′H Hs / 1. Since B = ∇ × ≈A kA, we obtain

ˆ

ˆ′

′≈ ≈ ≈

H

H

BAp

kAAp p

s

1

� � �

λ (16.2.10)

Due to the uncertainty relation, �/p is of the order of the Bohr radius (for hydrogen, a0 = 0.5 Å) and for optical sources λ ≅ 5000 Å. Therefore ˆ ˆ′ ′ ≈ <<−H Hs / 1

410 1.


Note: The results of Eqs. (16.2.9) and (16.2.10) suggest that ˆ ′H2 and ˆ ′H

s can be neglected in the low-

intensity limit. This conclusion does not hold in the following situations:

1. Forbidden transitions where the dipole matrix element of ˆ ′H1 vanishes and ˆ ′H

1 is reduced to the same

order of magnitude as ˆ ′Hs.

2. Strictly forbidden transitions where the matrix element of ˆ ′H1 vanishes identically.

16.3. A monochromatic radiation field of angular frequency w = ck is described by the following vector potential:

A r A k r

A k r

( , ) cos ( )

exp[ ( )]

t t

i t

= − +

= − +

2 0

0

| | i

i

ω θ

ω AA k r0* exp[ ( )]− −i ti ω

(16.3.1)

where A0 = |A0| e iq is the complex polarization vector, k is the wave vector, and A k0 0i = . (a) Calculate the electric field E ( , )r t and the magnetic field B( , )r t associated with the potential A(r, t). (b) Find the Poynting vector S E B= ×c

4π , and verify the relation

uIc c

= = ωπ

2

2 02

2| |A (16.3.2)

where I is the irradiance of the radiation and u is the energy density.

SOLUTION

(a) The electric field E = − ∂∂

1c t

A is given by

E ( , ) exp[ ( )] exp[ (*r A k r A k rtic

i tic

i= − − −ω ω ω0 0i i −−

= − − − −

ω

ω ω

t

ik i t ik i

)]

exp[ ( )] exp[ (*A k r A k r0 0i i tt

k t

)]

sin ( )= − − +2 0| |A k ri ω θ

(16.3.3)

The magnetic field B = ∇ × A is given by

B ( , ) exp[ ( )] exp[ (*r k A k r k A k rt i i t i i= × − − × −0 0i iω −−

= − × − +

ω

ω θ

t

t

)]

sin ( )2 0k A k r| | i

(16.3.4)

(b) The intensity of the radiation field is found by averaging the Poynting vector S E B= ×c4π over time.

Using Eqs. (16.3.3) and (16.3.4) we find

S = × × − +

=

ck tπ ω θ

ωπ

| | | |A k A k r

k r

0 02

2

( ) sin ( )

sin (

i

i −− +ω θt )| |A k02

(16.3.5)

Thus, after averaging S( , )r t over one oscillation period,

Ic

tc

= = − + =| | | | | |Sωπ ω θ ω

π

2

02 2

2

02

2A k r Asin ( )i (16.3.6)

where sin ( )2 1 2k ri − + =ω θt / . Similarly, the mean energy density averaged over time is

u k= + = + × × =14

12

2 2 20

20 0

2

π πω

( ) [ ( ) ( )]*E B | |A k A k Ai22 2 0

2

πc| |A (16.3.7)


Indeed, from Eqs. (16.3.6) and (16.3.7) we recover Eq. (16.3.2):

uIc c

= = ωπ

2

2 02

2| |A (16.3.8)

where | |A A A02

0 0= * .Note: Following from the definition of the Poynting vector as the energy density flux of the radiation, the irradiance I is the energy per unit area per unit time, which propagates along the k direction. This quantity can also be associated with the number of photons (i.e., the number of quanta or photons with energy �w), which propagate along the k direction. Using Eq. (16.3.6), we find that the flux of photons (i.e., the number of photons per unit area per unit time) is given by

FI

c≡ =

� �ωωπ2 0

2| |A (16.3.9)

16.4. An atomic electron of mass m, charge e = −| e |, and spin S interacts with a monochromatic radiation field of angular frequency w = ck. The Hamiltonian of the system is

ˆ ˆ ˆ ( ) ˆ ˆ ( ) ˆH H H t He

mce

mc= + ′ = − − ∇ ×0 0 A p A Si i (16.4.1)

where H ′(t) is a small perturbation (the low-intensity limit). The vector potential A(r, t) is given by the following plane wave:

A r k r

k r

( , ) cos ( )

exp[ ( )

t A t

A i t

= − +

= −

2 0

0

| |d

d

i

i

ω θ

ω ]] exp[ ( )]*+ − −A i t0d k ri ω (16.4.2)

where A0 = | A0 | ei q is a complex number, d is a unit vector in the direction of polarization, k is the wave

vector, and d i k = 0 (transversal gauge). Let | i⟩ and | f ⟩ be two eigenstates of the unperturbed Hamiltonian H

0, which correspond to the energy levels Ei and Ef , respectively (Ei ≠ Ef ). Assuming that the pertur-

bation H(t) is turned on at t = 0, calculate the probability Pf i (t) for resonant transition | |i f⟩ → ⟩.

SOLUTION

We consider the Hamiltonian in Eq. (16.4.1), and treat H ′(t) as a small time-dependent perturbation. If the system is initially in the state |i⟩ and the perturbation is turned on at t = 0, the first-order amplitude for finding the system in state | f ⟩ at t > 0 is given by

a ti

e f H t i dtie

mcf i

itf i( )( ) ˆ ( )1

0

1= ⟨ ′ ′ ⟩ ′ =′

� �ω

| |tt

i te f t t i dtf i∫ ′⟨ + ∇ × ⟩ ′

ω| |A r p S A r( , ) ˆ ˆ [ ( , )]i i

0

tt

∫ (16.4.3)

with �wf i ≡ Ef − Ei (see Chap. 10). Substituting A(r, t) from Eq. (16.4.2), and integrating over dt′ we obtain

a tie

mce f e if i

i t if i( ) ( )( ) [ ˆ1

0= ⟨ +− ′

�ω ω

|A pk ri id ˆ ( )]

( ) *

S k

A k r

i

i

× ⟩{+ ⟨

∫+ ′ −

d |

|

i

e f e

t

i t if i

0

0

ω ω[[ ˆ ˆ ( )]d di ip S k− × ⟩} ′i i dt| (16.4.4)

Therefore,

a te T e

f i

i t

f i

f ii

f i f i( )

( ) ( )

( )1 1= − −− −

− + +ω ω ω ω

ω ω �

tt

f i

f iT−+

−1

ω ω � (16.4.5)

where we define

Te

mcf e A i i

T

fii

f i

+

−

≡ − ⟨ + × ⟩

≡

| |k r p S ki i i0[ ˆ ˆ ( )]d d

−− ⟨ − × ⟩

⎧

⎨⎪⎪

⎩⎪⎪

−emc

f e A i ii| |k r p S ki i i0*[ ˆ ˆ ( )]d d

(16.4.6)


Equation (16.4.5) contains time-factors of the form (see Fig. 16.1)

e

ieti t

f i

i f if i

fi

( )( ) sin[( )ω ωω ω

ω ωω ω±

±−±

=±1 2/ /22

2

]

( )ω ωf i

+ / (16.4.7)

Consequently, the transition probability P t a tf i f i( ) ( )( )= | |1 2 is appreciable only if | | ω ω π− < ≡f i t4 / Δ (or if | | ω ω+ <fi Δ). In this case, and for Δ << 2| |ω f i , one can neglect the interference term of Pf i(t) and distinguish between two resonant transitions:

Absorption (wf i > 0):

P tT t

fif i f i

f i( )

sin[( ) ]

( )≅

−−

⎛

⎝⎜

+| |2

2

2

2�

ω ωω ω

/

/⎞⎞

⎠⎟

2

(16.4.8)

Induced emission (wf i < 0):

P tT t

f if i f i

f i( )

sin[( ) ]

( )≅

++

⎛

⎝⎜

−| |2

2

2

2�

ω ωω ω

/

/⎞⎞

⎠⎟

2

(16.4.9)

Note: It is worthwhile to emphasize that the resonance approximations in Eqs. (16.4.8) and (16.4.9) are only valid under two conditions:

1. P tf i ( ) << 1 (applicability of first-order perturbation theory).

2. 2π ω ωt f i<< ≈| | (no interference term in Pf i (t)).

These conditions are compatible if

| |Tf i f i± << �ω

16.5. A bounded spinless particle of mass m and charge e interacts with a nonmonochromatic field of radiation that covers a spread of incoherent frequencies in the range w ± dw /2. The particle is described by the Hamiltonian ˆ ˆ ˆ( )H p m V0

2 2= +/ r and the irradiance of the radiation is given by I = cu(w)dw, where u (w) is the energy density per unit angular frequency (see Problem 16.3). Consider the transition probability Pf i (t), where |i⟩ and | f ⟩ are two eigenstates of the unperturbed Hamiltonian H

0. (a) Show that the transition rates, for absorption and for induced emission, are given by

We

m

uf e if i

f i

f i

iabs = ⟨ ⟩+4 2 2

2 2 2

π ω

ω�

( )ˆ| | |k r pi id ||2 for E Ef i> (16.5.1)

We

m

uf e ii f

i f

f i

iind = ⟨ ⟩−4 2 2

2 2 2

π ω

ω�

( )ˆ| | |k r pi id ||2 for E Ef i> (16.5.2)

Δ

w fi

2p/t

t

0

Δ

–w w

sin [wfi ± w)t/2](wfi + w)/2

w ≥ 0Δ = 4p/t

Fig. 16.1


where �w f i ≡ Ef − Ei, and | | | |k = ω f i c/ . (b) Assume that Ef > Ei, and verify the principle of detailed balance:

W Wf i ifabs ind= (16.5.3)

SOLUTION

(a) We assume that the radiation covers a spread of frequencies with no phase relation between different fre-quency components. Treating each frequency component separately and using the results of Problem 16.4 we obtain

δP te A

m cf e if i

i± ±= ⟨ ⟩( ) ˆsin [2

02

2 2 22| |

| |�

k r pi iε(( ) ]

( )

ω ωω ω

f i

f i

t±±

⎛

⎝⎜⎞

⎠⎟/

/

2

2

2

(16.5.4)

where the plus/minus signs of δP tfi± ( ) correspond to absorption and induced emission, respectively. The

total transition P tfi± ( ) is then found by replacing | |A0

2 in Eq. (16.5.4) by the relation in Eq. (16.3.2):

| |Ac I c u

02

2

2

2

22 2= =π

ωπ ω δω

ω( )

(16.5.5)

and summing over all the incoherent frequencies in the range w ± d w. Thus,

P te

m

uf ef i

i f i± ±= ⟨∑( )( )

ˆ2 2

2 2 2π ω δω

ωδ ω

�| | k r pi iε ii

tf i

f i⟩

±±

⎛

⎝⎜⎞

⎠⎟2

22

2

sin [( ) ]

( )

ω ωω ω

/

/ (16.5.6)

where | k | = w /c. The time factor in Eq. (16.5.6) has a sharp peak at w = ±wf i . Therefore, we can replace the summation over dw by an integral and extend the limits of integration to ± ∞. This gives

P te

m

uf e if i

i±

−∞

∞±= ⟨ ⟩∫( )

( ) ˆ2 2

2 2 2π ω

ω�| | |k r pi iε ||2

22

2

sin [( ) ]

( )

ω ωω ω ωf i

f i

td

±±

⎛

⎝⎜⎞

⎠⎟/

/ (16.5.7)

The last term in Eq. (16.5.7) can now be replaced by p td [(wf i ± w)/2]. Hence,

P te t

m

uf e if i

i±

−∞

∞±= ⟨ ⟩∫( )

( ) ˆ4 2 2

2 2 22π ω

ω�

k r pi id δδ ω ω ω( )f i d± (16.5.8)

Finally, the transition rates, W t dP t dtfi f i± ±≡( ) ( )/ , are given by

WdP t

dte

m

uf ef i

f i f i

f i

abs = = ⟨+

+( ) ( )4 2 2

2 2 2π ω

ω�| | ii

f iik r pi id ˆ | |⟩ >2 0for ω (16.5.9)

WdP t

dte

m

uf ei f

f i i f

f i

ind = = ⟨−

−( ) ( )4 2 2

2 2 2π ω

ω�| | ii

f iik r pi id ˆ | |⟩ <2 0for ω (16.5.10)

where �w f i ≡ Ef − Ei and | | | |k = ω fi c/ .

(b) We consider the transitions | |i f⟩ ↔ ⟩, where Ef > Ei. Using expressions (16.5.1) and (16.5.2),

We

m

uf e if i

f i

f i

iabs = ⟨ ⟩4 2 2

2 2 2π ω

ω�

( )ˆ| | | |k r pi id 22 (16.5.11)

We

m

ui e fi f

f i

f i

iind = ⟨ ⟩−4 2 2

2 2 2π ω

ω�

( )ˆ| | |k r pi id ||2 (16.5.12)


The matrix element in Eq. (16.5.12) can now be written as

⟨ ⟩ = ⟨ ⟩ = ⟨− −i e f f e ii i| | | |k r k rp pi ii iε εˆ [( ˆ † *)( )] ff e ii| |( ) )ε i iˆ ( †p k r− ⟩ (16.5.13)

where † denotes the Hermitian conjugate operator. However, k i d = 0 and consequently [ ,ε i iˆ ] .p k re i± = 0 Hence,

⟨ ⟩ = ⟨ ⟩ = ⟨−f e i f e i fi i| | | | |( ˆ ) ˆ† * *d di ii ip pk r k r( ) ee ii k r pi id ˆ *| ⟩ (16.5.14)

where [ †e ei i± ±=k r k ri i] has been used. Finally, we obtain

⟨ ⟩ = ⟨ ⟩−i e f f e ii i| | | |k r k rp pi ii iε εˆ ˆ * (16.5.15)

Therefore, W Wfi ifabs ind= .

16.6. Consider the matrix element T f e i ifii± ≡ ⟨ + × ⟩| |k r p S ki i i[ ]d dˆ ˆ ( ) , for a one-electron system in a

linearly polarized radiation field. Prove that in the long wavelength approximation, Tfi± is given by

the following multipole expansion:

T im f ii

f if i f i± ≈ ⟨ ⟩ + ⟨ + × ⟩ −ω | | | |d di ir L S k

22( ˆ ˆ ) ( )

mmf ifiω

2⟨ ⟩| |( ˆ) ( ˆ)k r ri id (16.6.1)

where d is the unit polarization vector, k is the wave or propagation vector, and L = r × p is the orbital angular momentum. The three terms in Eq. (16.6.1) correspond to electric-dipole, magnetic-dipole, and electric-quadrupole transitions.

SOLUTION

In the long wavelength limit, exp (i k i r) = 1 + i k i r + …. Therefore,

T f i i f i i ff i± ≈ ⟨ ⟩ + ⟨ × ⟩ + ⟨| | | | |d d di i iˆ ˆ ( ( ) (p S k k r) ii ˆ )p | i⟩ (16.6.2)

However, k i d = 0, so that k i r and d i p are commuting operators that satisfy the relation

( ˆ ) ( ) ( ) ( ˆ ) ( ) ( ˆ )r p k k r p r k p× × = −i i i i id d d

Thus,

2( ) ( ˆ ) ˆ ( ) ( ) ( ˆ ) ( ) ( ˆk r p L k k r p r ki i i i i i id d d d= × + + pp) (16.6.3)

where L = r × p. Substituting Eq. (16.6.3) in Eq. (16.6.2) gives

T f ii

f ii

ff i± ≈ ⟨ ⟩ + ⟨ + × ⟩ + ⟨| | | |d di iˆ ˆ ˆ ( )p L S k

22

2( ) || |( ) ( ˆ ) ( ˆ ) ( )k r p k p ri i i id d+ ⟩i (16.6.4)

Therefore, in order to obtain Eq. (16.6.1) we only have to verify the following matrix identities:

⟨ ⟩ = ⟨ ⟩

⟨ +

f i im f i

f

f i| | | |

|

d d

d

i i

i i

ˆ ( )

( ) ( ˆ ) (

p r

k r p

ω

dd di i i ir k p k r r) ( ˆ ) ( ) ( ˆ)| | |i im f if i⟩ = ⟨ ⟩

⎧⎨⎪

⎩⎪ ω (16.6.5)

To that end, we recall that |i⟩ and | f ⟩ are eigenstates of ˆ ˆ / ˆ ( )H m V02 2= +p r and choose our coordinate

system so that the vector d is on the z axis and the vector k is on the y axis. In this case,

[ ˆ ˆ ] [ ˆ, ˆ ] [ ˆ, ˆ ]d i r p, 2H z Hm

zim

pz0 0

12

= = = � (16.6.6)


and

[ ˆ, ˆ ] [ ˆ ˆ, ˆ ] ( ˆ ˆ ˆ ˆ)k ri H yz Him

yp p zz y0 0= = +� (16.6.7)

Furthermore,

⟨ ⟩ = − ⟨ ⟩ = − ⟨f z H i E E f z i f z ii f f i| | | | | |[ ˆ, ˆ ] ( ) ˆ ˆ0 �ω ⟩⟩ (16.6.8)

⟨ ⟩ = − ⟨ ⟩ = − ⟨f yz H i E E f yz i fi f f i| | | | |[ ˆ ˆ, ˆ ] ( ) ˆ ˆ0 �ω ˆ ˆyz i| ⟩ (16.6.9)

Hence,

⟨ ⟩ = ⟨ ⟩

⟨ + ⟩ =

f p i im f z i

f yp p z i

z f i

z y

| | | |

| |

ˆ ˆ

( ˆ ˆ ˆ ˆ)

ω

iim f yz ifiω ⟨ ⟩

⎧⎨⎪

⎩⎪ | |ˆ ˆ (16.6.10)

which, for d = k and k = k j, coincide with expressions (16.6.5).

16.7. Find the selection rules for emission and absorption of (a) electric-dipole radiation, (b) magnetic-dipole radiation, and (c) electric-quadrupole radiation, by an electron in a central potential.

SOLUTION

(a) Electric-dipole transitions: To obtain the selection rules for electric-dipole transitions we consider matrix elements of the form ⟨ ⟩ ⟨ ⟩f x i f y i| | | |ˆ , ˆ , and ⟨ ⟩f z i| |ˆ where | i⟩ and | f ⟩ are eigenstates of an electron moving in a central potential. The unperturbed wavefunction is then given by

| |

| |

i n l m R Y

f

i i i n l m n l lm

i i i i i i

i⟩ ≡ ⟩ → =

⟩ ≡

, , ( , )ψ θ ϕ

nn l m R Yf f f n l m n l l

m

f f f i f f

f, , ( , )⟩ → =

⎧

⎨⎪

⎩⎪ ψ θ ϕ

(16.7.1)

where Ylm( , )θ ϕ are the spherical harmonic functions. In this representation,

x iy r e rY

z r rY

i± = =

= =

± ±sin ( , )

cos

θ π θ ϕ

θ π

ϕ ∓83

43

11

110( )θ

⎧

⎨⎪⎪

⎩⎪⎪

(16.7.2)

Therefore, the matrix element ⟨ ⟩f z i| | is proportional to the angular integral

Y Y Y dl

m

lm

f

f

i

i( )∫*

( , ) ( ) ( , )θ ϕ θ θ ϕ10 Ω (16.7.3)

which is different from zero only if Δl = lf − li = ±1 and Δm = mf − mi = 0. Similarly, the matrix ele-ments ⟨ ⟩f x i| | and ⟨ ⟩f y i| |

are proportional to linear combinations of the form

Y Y Y dl

m

lm

f

f

i

i( ) ±∫*

( , ) ( ) ( , )θ ϕ θ θ ϕ11 Ω (16.7.4)

which are different from zero only if Δl = ±1 and Δm = ±1. Grouping these results together we finally obtain

Δ = − = ±

Δ = − = ±

⎧⎨⎪

⎩⎪

l l l

m m m

f i

f i

1

0 1, (16.7.5)


(b) Magnetic-dipole transitions: The selection rules for magnetic-dipole transitions are found from matrix elements of the form ⟨ ⟩ ⟨ ⟩f L i f L ix y| | | |ˆ , ˆ , and ⟨ ⟩f L iz| |ˆ . From the general properties of angu-lar momentum (see Chap. 6) we immediately find

⟨ + ⟩ = ⟨ ⟩+f L iL i n l m L n l mx y f f f i i i l| | | |( ˆ ˆ ) , , ˆ , , ~ δff i f il m mδ , +1

(16.7.6)

⟨ − ⟩ = ⟨ ⟩−f L iL i n l m L n l mx y f f f i i i l| | | |( ˆ ˆ ) , , ˆ , , ~ δff i f il m mδ , −1 (16.7.7)

⟨ ⟩ = ⟨ ⟩f L i n l m L n l m mz f f f z i i i i l lf i

| | | |ˆ , , ˆ , , ~ � δ δmm mf i

(16.7.8)

Therefore, the magnetic-dipole matrix elements vanish identically unless

Δ = − =

Δ = − = ±

⎧⎨⎪

⎩⎪

l l l

m m m

f i

f i

0

0 1, (16.7.9)

Note: In the presence of spin, one also obtains Δm m ms s sf i

= − = ±0 1, . However, this selection rule is trivially satisfied by spin 1/2 particles.

(c) Electric-quadrupole transitions: For electric-dipole radiation, we have to calculate matrix elements ofthe products xz, yz, and zx. We note, for example, that yz can be expressed as a linear combination

of r Y221( , )θ ϕ and r Y2

21− ( , )θ ϕ . Therefore, the matrix element ⟨ ⟩f yz i| |ˆ ˆ contains angular integrations

of the form

Y Y Y dl

m

lm

f

f

i

i( ) ±∫*

( , ) ( ) ( , )θ ϕ θ θ ϕ21 Ω (16.7.10)

These integrals are different from zero only if Δ l = 0, ± 2 (excluding the case lf = li = 0) and Δm = ±1. The last condition becomes Δm = ± 2, ±1, 0 if one considers arbitrary polarizations. The electric-quadrupole selection rules are then found to be

Δ = − = ±

Δ = − = ± ±

⎧⎨⎪

⎩⎪

l l l

m m m

f i

f i

0 2

0 1 2

,

, , (16.7.11)

where the case of lf = li = 0 is excluded.

Notes: (1) The electric-dipole interaction is an odd operator, which connects only states of different parities. Since the parity of |nlm⟩ is given by (−1)l, Δ l must be odd in accordance with Eq. (16.7.5). (2) The magnetic-dipole and electric-quadrupole interactions are even operators that connect only states of the same parity. This is, again, compatible with Eq. (16.7.9) and Eq. (16.7.11). (3) The magnetic-dipole and electric-quadrupole transitions are never in competition with the electric-dipole transition. (4) For Δ l = 0 and Δm = 0, ± 1, there is a contribution from both the magnetic-dipole and the electric-quadrupole interactions. However, for Δ l = 2 we have pure quadrupole transition.

16.8. An isotropic three-dimensional harmonic oscillator of mass m, angular frequency w 0, and charge e is placed in a linearly polarized field of radiation. Calculate the probability per unit time for resonant transitions of frequency w f i = w 0 and w f i = ± 2w 0.

SOLUTION

Let us choose a coordinate system such that the unit vector d is on the z axis and the vector k is on the y axis. The transition rate, for absorption (and for induced emission), is then given by

We

m

un n n efi

f i

f ixf

yf

zf iky= ⟨ +4 2 2

2 2 2π ω

ω�

( ), , ˆ| | pp n n nz x

iyi

zi| |, , ⟩ 2 (16.8.1)


where |n n nx y z, , ⟩ and E n n nn n n x y zx y z

= + + +�ω0 3 2( / ) are the eigenstates and eigenvalues of the unper-

turbed harmonic oscillator. Using the results of Chap. 5, we find

ˆ , , , , ,p n n n i

mn n n n n nz x y z z x y z z x| | |⟩ = + + ⟩ −

�ω02

1 1 nn n

ky n n n km

n n n

y z

x y z y x y

,

, , , ,

− ⟩( )⟩ = + +

1

21 1

0| |

�ω nn n n n nz y x y z⟩ − − ⟩( )

⎧

⎨

⎪⎪

⎩

⎪⎪ | , ,1

(16.8.2)

Therefore,

⟨ ⟩ = ⟨+n n n e p n n n n nxf

yf

zf iky

z xi

yi

zi

xf

xi, , ˆ , ,| | | ⟩⟩ ⟨ + + ⟩ ⟨ ⟩n ky n n p ny

fyi

zf

z yi| | | |( ) ˆ1 � (16.8.3)

where the higher terms of order �ω02 1/mc << have been neglected. Note that these terms are important only

for high-order transitions in which wf i = 3w0, 4w0, . . . . Expression (16.8.3) is different from zero only if

Δ Δ Δn n nx y z= = ± = ±0 0 1 1, , , and (16.8.4)

In particular, there is no competition between the electric-dipole (Δny = 0) and the electric-quadrupole transitions (Δny = ±1). The energy difference for each transition is w f i = w 0 (Δnx + Δny + Δnz). Thus,

⟨ ⟩ =+n n n e p n n ni

mn

xf

yf

zf i ky

z xi

yi

zi z

, , ˆ , ,| |

�ω02

iif i

zi

yi

f iik

n n

+ =

+ + =

⎧

⎨⎪⎪

⎩⎪

1

21 1 2

0

0

for

for

ω ω

ω ω�⎪⎪

(16.8.5)

Finally, substituting Eq. (16.8.5) in Eq. (16.8.1) and using the relation k = w f i /c, we obtain

W e

um

n

u

m c

fi

zi

f i

= ×+ =

21

2 2

0

00

02

π

ωω ω ω

ω

( )( )

( )

�for

22 01 1 2( )( )n nzi

yi

f i+ + =

⎧

⎨⎪⎪

⎩⎪⎪

for ω ω (16.8.6)

Note that for the same incident intensity, W (2w 0) /W(w0) ∼ �w 0 / mc2.

16.9. A two-level system with eigenvalues E2 > E1 is in thermodynamic equilibrium with a heat reservoir at absolute temperature T. The system undergoes the following transitions: (i) absorption 1 → 2, (ii) induced emission 2 → 1, and (iii) spontaneous emission 2 → 1. The transition rates for each of these processes are given by

W P B u

W P B u

W

21 1 21 21

12 2 12 21

12

abs

ind

spon

=

=

( )

( )

ω

ω

==

⎧

⎨

⎪⎪

⎩

⎪⎪ P A2 12

(16.9.1)

where u(w21) is the energy distribution of the radiation field, Pj is the probability of finding the system in level j of degeneracy gj( j = 1, 2), and A12 and B12 are the Einstein coefficients for spontaneous and induced emission, respectively. (a) Calculate the probabilities P1 and P2 under equilibrium conditions. (b) Use Eq. (16.9.1) together with Planck’s formula for blackbody radiation to show that

g B g B

Ac

B

1 21 2 12

12213

2 3 12

=

=

⎧

⎨⎪⎪

⎩⎪⎪

�ω

π

(16.9.2)


SOLUTION

(a) Under thermal equilibrium at absolute temperature T, the probability of finding a system in one of its

stationary states |i⟩ with an eigenvalue ei is proportional to the Boltzman factor e ikT−ε /

. In this problem, ei assumes the values ei = E1, E2 with respective degeneracies gi = g1, g2 (a two-level system). Therefore,

P Cg e

P Cg e

E kT

E kT

1 1

2 2

1

2

=

=

⎧⎨⎪

⎩⎪

−

−

/

/ (16.9.3)

where C is the normalization constant. Since P1 + P2 = 1, we immediately find that

C g e g eE kT E kT− − −= +1

1 21 1/ /

(16.9.4)

Since E2 − E1 = �w 21, we have

P

P

g

ge

kT1

2

1

2

21= �ω / (16.9.5)

(b) Suppose that a larger number of systems, such as in part (a), form a closed cavity that is kept in equi-librium with its own thermal radiation at constant temperature T. In this case,

W W W21 12 12abs ind spon= + (16.9.6)

Therefore, from Eq. (16.9.1), we obtain

P B u P B u P A1 21 21 2 12 21 2 12

( ) ( )ω ω= + (16.9.7)

or equivalently, by using Eq. (16.9.5),

[ ] ( )/

g e B g B u g AkT

1 21 2 12 21 2 1221�ω ω− = (16.9.8)

The thermal radiation inside the cavity is distributed according to Planck’s formula:

uc e kT

( )( )/

ω ωπ ω=

−�

3

2 3 1� (16.9.9)

Therefore, Eq. (16.9.8) takes the form

g B eg B

g B cg A

kT

1 212 12

1 21

213

2 3 221

� �ω ω

π/ −

⎛

⎝⎜⎞

⎠⎟=

11221 1( )

/e

kT�ω − (16.9.10)

Hence,

g B g B

Ac

B

1 21 2 12

12213

2 3 12

=

=

⎧

⎨⎪⎪

⎩⎪⎪

�ω

π

(16.9.11)

16.10. Calculate the Einstein coefficients, A1s2p, B1s2p, and B2p1s, for an electron moving in a central potential.

Recall that 1s = (n = 1, l = 0) and 2p = (n = 2, l = 1).

SOLUTION

Let us first consider the probability per unit time for the transition 1s → 2p (absorption). Since the states 2p are degenerate with respect to the magnetic quantum number, we have

We

u m B up s p s2 1

2 2

2 212

2 1

4

321 100= ⟨ ′ ⟩ ≡π ω

�( ) ˆ| | | |r (( )ω21

1

1

′ = −∑m

(16.10.1)


However,

| | | | | | |⟨ ′ + ′ ⟩ = + ⟨ ′ +′

∑ n l m nlm l n l r n

m

, , ˆ ( ) , ˆ1 1 12r ll⟩ |2 (16.10.2)

and

| | | | | | |⟨ ′ − ′ ⟩ = ⟨ ′ −′

∑ ( , , ) ˆ , ˆn l m nlm l n l r n

m

1 12r ll⟩ |2 (16.10.3)

Therefore, substituting l = 0 in Eq. (16.10.3) and using Eq. (16.10.1), we obtain

Be

rp s2 1

2 2

224

321 10= ⟨ ⟩π

�| | | |ˆ (16.10.4)

where ⟨ ⟩ =∞

∫21 10 210

310| |ˆ ( ) ( )*r R r r R r dr is the radial integral. The coefficients for the transition 2p → 1s

(emission) are found by setting g1s = 1 and g2p = 3 in Eq. (16.9.11). This gives

B

er

Ae

c

s p

s p

1 2

2 2

22

1 2213 2

4

910 21

4

9

= ⟨ ⟩

=

π

ω

�

�

| | | |ˆ

33210 21| | | |⟨ ⟩

⎧

⎨

⎪⎪

⎩

⎪⎪ r

(16.10.5)

16.11. Find the probability per unit time of spontaneous transition for a hydrogen atom in the first excited state.

SOLUTION

The probability per unit time for the transition 2p → 1s (emission) is given by

W u B As p s p s p1 2 21 1 2 1 2= +( )ω (16.11.1)

Thus, using Eq. (16.10.5), we obtain

We

ucs p1 2

2 2

2 21213

2 34

3100= +

⎡

⎣⎢⎢

⎤

⎦⎥⎥

⟨π ωω

π�

�( ) | || | |r 21 2′⟩m (16.11.2)

where |21 ′m is one of the three 2p states. In particular, for the hydrogen atom in the first excited level,

| | | | | | | |⟨ ′⟩ = ⟨ ⟩ =100 2113

10 2159

2 202ˆ ˆr rm a (16.11.3)

where a0 = �2/me2 is the Bohr radius. Therefore,

Ae

ca

ca

s p1 2

2213

3 02 21

3

2 0220

272027

2027

= = =ω

αω ω

�

2213

4

2

2c m

�

α (16.11.4)

where a = e2/�c ≈ 1/137 is the fine structure constant. However,

�ω α21

2

0

2 234 2

34 2

= =ea

mc (16.11.5)

Hence,

Amc

s p1 2 35

23

218 120

8

548

6 25 10= = ≈ × − −α α ω�

. sec (16.11.6)

Expression (16.11.6) leads to a radiative lifetime of the order 1.5 × 10−9 sec.


16.12. A linear harmonic oscillator of mass m, angular frequency w 0, and charge e is excited by a nonresonant radiation field of the form

A tA ky t t

t( , )

cos ( )r =

− >

≤

⎧⎨⎪

⎩⎪

2 0

0 0

0k ω (16.12.1)

where w ≠ w0. Let | n⟩ and En = �w0(n + 1/2) be the eigenstates and the eigenvalues of the oscillator, and let |ψ( )t ⟩ be its time-dependent state vector in the presence of radiation. (a) Use first-order perturbation theory to find an expression for |ψ( )t ⟩. Assume that initially | |ψ( )t = ⟩ = ⟩0 0 . (b) Calculate the induced dipole moment that is proportional to the amplitude of the external electric field.

SOLUTION

(a) The time-dependent state vector can be written as

| | |ψ( ) ( )/ /

t e a t e niE t

n

n

iE t⟩ = +−

≠

−∑0 00

0

� � (16.12.2)

where an(t) = 0 for t < 0. Using first-order perturbation theory (see Chap. 10) for ˆ ( ) ( , ) ˆH te

mct= − A r pi ,

we find

a ti

e n H t dtie

mcni t

tn( )( ) ˆ ( )1

0

100= ⟨ ′ ′ ⟩ ′ =′∫�

ω | |��

e A ky t n p dti t

t

zn

ω ω0

002 0

′∫ − ⟨ ⟩ ′cos ( ) | | (16.12.3)

Therefore (see Problem 16.4),

a te T e

n

i t

n

ni t

n n( )

( ) ( )

( )1

0

00 01= − −

− −− + +ω ω ω ω

ω ω �−−

+

−1

0

0ω ωn

nT�

(16.12.4)

where

T

emc

n A e pie

eA e n en

ikyz

n iky0 0

000+ ≡ − ⟨ ⟩ = − ⟨| | |ˆ ˆ

ωzz

Te

mcn A e p

iee

A eniky

zn i

|

| |

0

00 00

0

⟩

≡ − ⟨ ⟩ = −− − −ˆω kky n ez⟨ ⟩

⎧

⎨⎪⎪

⎩⎪⎪ | |ˆ 0

(16.12.5)

Finally, multiplying the ket |ψ( )t ⟩ by a global phase factor eiE t

0/�

, we obtain

| | | |�ψ ψ( ) ( ) ( )/

t t e b t niE t

n

n

⟩ = ⟩ = ⟩ + ⟩≠

∑0 0

0

� (16.12.6)

where the coefficients b t a t en n

in( ) ( )≡ − ω

0 are given by Eq. (16.12.4):

b te e T e e

n

i t i t

n

ni t i t

n n

( ) = − −− − −− − + −ω ω ω ω

ω ω0 0

0

0� ωω ωn

nT

0

0+

−

� (16.12.7)

(b) The induced dipole moment is given by ⟨ ⟩ = ⟨ ⟩D t t ez t( ) ( ) ˆ ( )� �ψ ψ| | . Thus, up to the first order in A0, we find

⟨ ⟩ = ⟨ ⟩ + ⟨ ⟩ +≠ ≠

∑ ∑D t ez b t ez n bn

n

n

n

( ) ( ) ˆ *0 0 0

0 0

| | | | (( ) ˆt ez n⟨ ⟩0 | | (16.12.8)


Next, we substitute the coefficients bn (t) from Eq. (16.12.7) and neglect all the terms that oscillate at frequencies ±wn0. (These terms disappear in the limit t → ∞ due to the finite lifetimes of all the excited states.) This gives

⟨ ⟩ = ⟨ ⟩ −⟨ ⟩

−

+ −

D t ezT ez n en

i t

n( ) ˆ Re

ˆ0 0

2 00

0| |

| |�

ω

ω ωω ω ω

ω

+⟨ ⟩

+⎡

⎣⎢⎢

⎤

⎦⎥⎥

⎧⎨⎪

⎩⎪

⎫−

≠∑ T ez n en

i t

nn

0

00

0 | |ˆ⎬⎬⎪

⎭⎪ (16.12.9)

namely,

⟨ ⟩ = ⟨ ⟩ − ⟨ ⟩≠

∑D t ezAc

ez n

n

n( ) ˆ ˆ Im0 02

00 2

0

0| | | | | |�

ω ee ei ky t

n

i ky t

n

− − − −

− + +⎡

⎣⎢⎢

⎤

⎦⎥⎥

( ) ( )ω ω

ω ω ω ω0 0 (16.12.10)

Finally, since E = − ∂∂ = − −1

2 0c tA k ky t

Ak sin ( )ω , we obtain

⟨ ⟩ = ⟨ ⟩ + ⟨ ⟩−

D t ezez nz

nn( ) ˆ

ˆ0 0

2 0 2

02 2 0| |

| | | |E� ω ω

ωnn ≠∑

0

(16.12.11)

In particular, for the linear harmonic oscillator, expression (16.12.11) is reduced to the classical formula; i.e.,

⟨ ⟩ =−

D te

m z( )( )

2

02 2ω ω

E (16.12.12)


16.13. Refer to Problem 16.4 and find the transition rate Wf i (probability per unit time) for the transition from the initial

state |i⟩ to a continuum of final states of energies Ef ± dEf /2 (Fermi’s golden rule).

Ans. WdP t

dtf T i E Efi

f if i= = ⟨ ⟩ = ++( )

ˆ ( )2 2π ρ ω�

�| | | | , where r (Ef ) ≡ dNf /dEf is the density of final states

and Tfi+ is given in Eq. (16.4.6).

16.14. Find the transition rate for absorption and for induced emission of electric-dipole radiation by a one-electron system in an isotropic radiation field. Hint: The transition rate is found by averaging the electric-dipole matrix element over all possible directions of polarization.

Ans. We

u f ifi f i= ⟨ ⟩4

3

2 2

22π ω

�( ) ˆ| | | |r , where | | | | | | | |⟨ ⟩ ≡ ⟨ ⟩ ⟨ ⟩f i f i f iˆ ˆ ˆ *r r r2 i

16.15. The oscillator strength of a transition | |k n⟩ → ⟩ is defined as

f f f f m n knk nkx

nky

nkz

nk= + + ≡ ⟨ ⟩( / ) ˆ2 2ω � | | | |r (16.15.1)

where |n⟩ and |k⟩ are eigenstates of ˆ ˆ ˆ( )Hm

V r0

2

2= +p

. Show that fnk satisfies the sum rule fnk

n∑ = 3.

16.16. In the presence of a spin-orbit interaction, find the selection rules for emission and absorption of (a) electric-dipole radiation, (b) magnetic-dipole radiation, and (c) electric-quadrupole radiation. Note that the selection rules can be obtained by expanding the stationary states |ls JMJ; ⟩ in terms of | |lm sms⟩ ⊗ ⟩, where J = L + S.

Ans. (a) Electric-dipole transitions: Δ l = ±1; Δ J = 0, ±1; Δ MJ = 0, ±1

(b) Magnetic-dipole transitions: Δ l = 0; Δ J = 0, ±1; Δ MJ = 0, ±1

(c) Electric-quadrupole transitions: Δ l = 0, ±2; Δ J = 0, ±1, ±2; Δ MJ = 0, ±1, ±2

347

APPENDIX

Mathematical Appendix

A.1 Fourier Series and Fourier TransformIf f (x) is a periodic function with a fundamental period L, then it can be expanded in a Fourier series:

f x a en

ik x

n

n( ) ==−∞

∞

∑ (A.1)

where kn = 2p n/L. The coefficients an of the series are given by

aL

f x e dxn

Lik x

n= ∫ −1

0

( ) (A.2)

The Fourier transform of a function f (x) is defined as

F k F f x f x e dxikx( ) [ ( )] ( )= =−∞

∞−∫1

2π (A.3)

while the inverse Fourier transform is

f x F k e dkikx( ) ( )=−∞

∞

∫1

2π (A.4)

Notice that in quantum mechanics we define the transformations slightly differently, as follows:

Ψ( ) [ ( )] ( ) /k F x x e dxipx= =−∞

∞−∫ψ ψ1

2π�

� (A.5)

and

ψ( ) ( ) /x k e dkipx=−∞

∞

∫1

2π�

�Ψ (A.6)

Two formulas of Fourier transform theory are especially relevant.

Identity of norms: f x dx F k dk( ) ( )2 2=

−∞

∞

−∞

∞

∫∫ (A.7)

Parseval’s theorem: f x g x dx F k G k dk( ) ( ) ( ) ( )* *

−∞

∞

−∞

∞

∫ ∫= (A.8)

APPENDIX Mathematical Appendix348

A.2 The Dirac c -FunctionThe Dirac d-function is defined by the relation

f x x x dx f x−∞

∞

∫ − =( ) ( ) ( )δ 0 0 (A.9)

Some important and useful properties of the d-function are given below:

δ δ( ) ( )− =x x (A.10)

δ δ( ) ( )cxc

x c= >10for (A.11)

x x x x x xδ δ( ) ( )− = −0 0 0 (A.12)

Note that xd (x) = 0. Also,

f x x x f x x x( ) ( ) ( ) ( )δ δ− = −0 0 0 (A.13)

δ δ δ( ) [ ( ) ( )]x cc

x c x c c2 2 12

0− = − + + >for (A.14)

δ δ[ ( )]( )

( )f xf x

x xi

i

i= ′ −∑ 1 (A.15)

where xi are simple zeros of the function f (x).

δ δ δ( ) ( ) ( )x x x x dx x x− − = −−∞

∞

∫ 1 2 1 2 (A.16)

We define d ′(x) by the relation

f x x dx f−∞

∞

∫ ′ = − ′( ) ( ) ( )δ 0 (A.17)

Some properties that are connected to d ′(x) are given below:

′ − = − ′δ δ( ) ( )x x (A.18)

δ δ( ) ( )( ) ( ) ( )n n nx x− = −1 (A.19)

x x n xn nδ δ( ) ( )( ) ( )= − −1 (A.20)

f x x dx fn n n( ) ( ) ( ) ( )( )δ−∞

∞

∫ = −1 0 (A.21)

The d-function in three-dimensional space is defined by

f dx dy dz f∫ − =( ) ( ) ( )r r r rδ 0 0 (A.22)


where d (r − r0) = d (x − x0) d (y − y0) d (z − z0). In spherical coordinates (r, q, j) we have

δθ

δ δ θ θ δ ϕ ϕ

δ

( )sin

( ) ( ) ( )

(

r r− = − − −

=

0 2 0 0 0

2

1

1

rr r

rr −− − −r0 0 0) (cos cos ) ( )δ θ θ δ ϕ ϕ (A.23)

The integral representation of the d-function is obtained by using the definition of Fourier transform [see Sec. A.1], so that

δ π( )( )

x x e dxik x x− = −

−∞

∞

∫0

12

0 (A.24)

The step function q (x) (also called the Heaviside function) is defined as

θ( )xx

x=

>

<

⎧⎨⎪

⎩⎪

1 0

0 0

for

for (A.25)

The relation between d (x) and q (x) is

δ θ( )

( )x

d xdx

= (A.26)

Finally, we mention an important relation for d (r):

∇ ⎛⎝⎜

⎞⎠⎟ = −2 1

4r

πδ ( )r (A.27)

A.3 Hermite PolynomialsThe Hermite polynomials Hn(x) are defined by the relation

H x ed

dxe nn

n xn

nx( ) ( ) , , , . . .= −

⎛

⎝⎜⎞

⎠⎟=−1 0 1 2

2 2

(A.28)

The Hn(x) are the solutions to the differential equation

d H x

dxx

dH x

dxnH xn n

n

2

22 2 0

( ) ( )( )− + = (A.29)

The orthogonality relation for Hn(x) is

e H x H x dx nxm n

nmn

−

−∞

∞

∫ =2

2( ) ( ) !π δ (A.30)

Two important recurrence relations for Hn(x) are

dH x

dxnH x H x xH x nH xn

n n n n

( )( ) ( ) ( ) ( )= = −+ −2 2 2

1 1

The first few Hermite polynomials are given below:

H x H x x H x x

H x x x H x0 1 2

2

33

4

1 2 4 2

8 12

( ) ( ) ( )

( ) (

= = = −

= − )) = − +16 48 124 2x x


A.4 Legendre PolynomialsLegendre polynomials Pl (x) are given by Rodrigue’s formula,

P xn

d

dxx

l

l

n

l

ll( )

( )

!( )= − −1

212 (A.31)

The first few Legendre polynomials are given below:

P x P x x P x x

P x x

0 1 22

33

112

3 1

12

5 3

( ) ( ) ( ) ( )

( ) (

= = = −

= − xx P x x x) ( ) ( )4

4 218

35 30 3= − +

In terms of cos q the first few Legendre polynomials are

P P

P

0 1

2

1

14

1 3 2

(cos ) (cos ) cos

(cos ) ( cos

θ θ θ

θ θ

= =

= + )) (cos ) ( cos cos )P3

18

3 5 3θ θ θ= +

The orthogonality relation of the Legendre polynomials is

P x P x dxll l ll

−′ ′∫ =

+1

12

2 1( ) ( ) δ (A.32)

A.5 Associated Legendre FunctionsAssociated Legendre functions P x

lm( ) are defined as

P x xd

dxP x xl

m mm

m l( ) ( ) ( )= − − ≤ ≤1 1 12 for (A.33)

where m ≥ 0. Pl (x) are the Legendre polynomials. Note that

P x P x P x m ll l l

m0 0( ) ( ) ( )= = >for (A.34)

The differential equation that P xlm( ) satisfies is

( ) ( )1 2 11

22

2

2

2− − + + −−

⎛

⎝⎜⎞

⎠⎟⎡

⎣⎢⎢

⎤

⎦x

d

dxx

ddx

l lm

x⎥⎥⎥

=P xlm ( ) 0 (A.35)

The first few associated Legendre functions are given below:

P x x P x x x P x x

P x

11 2

21 2

22 2

31

1 3 1 3 1( ) ( ) ( ) ( )

(

= − = − = −

)) ( ) ( ) ( ) ( ) (= − − = − =32

5 1 1 15 1 15 12 232 2

33x x P x x x P x −− x2 3)

The orthogonality relation of the associated Legendre functions is

P x P x dx P Plm

lm

lm

lm

−′ ′∫ ∫=

1

1

0

( ) ( ) (cos ) (cos ) sπ

θ θ iin( )!( )!

θ θ δdl

l ml m ll= +

+− ′

22 1 (A.36)


A.6 Spherical HarmonicsThe spherical harmonics are defined as

Yl l m

l mP el

m mlm i( , ) ( )

( )!( )!

(cos )θ ϕ π θ= − + −+1

2 14

mm mϕ for ≥ 0 (A.37)

and

Y Ylm m

lm− = −( , ) ( ) [ ( , )]*θ ϕ θ ϕ1 (A.38)

The differential equation that Ylm satisfies is

1 1

12

2

2sinsin

sin( )θ θ θ θ θ ϕ

∂∂

∂∂

⎛⎝⎜

⎞⎠⎟

+ ∂∂

+ +⎡

⎣⎢ l l

⎤⎤

⎦⎥ =Yl

m ( , )θ ϕ 0 (A.39)

The Ylm

have well-defined parity given as follows:

Y Ylm l

lm( , ) ( ) ( , )π θ π ϕ θ ϕ− + = −1 (A.40)

The orthonormalization relation of Ylm is written as

d Y Y dlm

lm

l lϕ θ ϕ θ ϕ θ θ δ δπ π

0

2

0∫ ∫ ′′

′ ′=[ ( , )] ( , ) sin*mm m (A.41)

and the closure relation

Y Ylm

m

l

lm

=−∑ ′ ′ = − ′

1

( , )[ ( , )] (cos cos ) (*θ ϕ θ ϕ δ θ θ δ ϕϕ ϕ θ δ θ θ δ ϕ ϕ− ′ = − ′ − ′=

∞

∑ )sin

( ) ( )1

0l

(A.42)

Some important recurrence relations are given below:

e m Y l l m m Yilm

lmϕ

θ θ θ ϕ∂∂ −⎛

⎝⎜⎞⎠⎟

= + − +cot ( , ) ( ) ( )1 1 ++1( , )θ ϕ (A.43)

e m Y l l m m Yilm− − ∂

∂ −⎛⎝⎜

⎞⎠⎟

= + − −ϕθ θ θ ϕcot ( , ) ( ) ( )1 1 ll

m−1( , )θ ϕ (A.44)

Yl m l m

l lYl

ml( , ) cos

( )( )( )( )

θ ϕ θ = + + + −+ + +

1 12 1 2 3 1

mmlml m l m

l lY+ + −

+ − −( )( )( )( )2 1 2 1 1 (A.45)

The first few Ylm are given below:

Y

Y Y e

Y

i

00

10

11

20

1

43

43

85

163

=

= = −

=

π

π θ π θ

π

ϕcos sin

( ccos ) sin cos sin221

22 21

158

1532

θ π θ θ π θϕ− = − =Y e Yi ee

Y Y

i2

30 3

317

165 3

2164

5

ϕ

π θ θ π θ= − = −( cos cos ) sin ( ccos )

sin cos

2

32 2 2

33

1

10532

356

θ

π θ θ

ϕ

ϕ

−

= = −

e

Y e Y

i

i

443 3

π θ ϕsin e i


An important result for spherical harmonics is

Pl

Y Ylm

m

l

lm

lm(cos ) ( ) ( , ) ( ,α π θ ϕ θ= + −

=−∑4

2 11

1

1 1 2 ϕϕ2 ) (A.46)

where a is the angle between the directions (q1, j1) and (q2, j2). This result is known as the spherical harmon-ics addition theorem.

A.7 Associated Laguerre PolynomialsFirst we shall deal with the Laguerre polynomials given by Rodrigue’s formula,

L x ed

dxx e

lx

l

ll x( ) ( )= − (A.47)

The associated Laguerre polynomials are defined as

L xd

dxL x

lm

m

m l( ) ( )= (A.48)

where l and m are nonnegative integers. Note that

L x L x L x ml l l0 0 0 1( ) ( ) ( )= = >for (A.49)

The first few associated Laguerre polynomials are given below:

L x L x x L x

L x x x11

21

22

31 2

1 2 4 2

3 18

( ) ( ) ( )

( )

= − = − =

= − + −− = − + = −18 6 18 632

33L x x L x( ) ( )

The orthogonality relation of the associated Laguerre polynomials is

x e L x L x dxl

l mm x

lm

lm

ll−

∞

′ ′∫ = −0

3

( ) ( )( !)

( )!δ (A.50)

A.8 Spherical Bessel FunctionsBessel’s differential equation is given as

xd

dxx

ddx

x l J xl

22

22 2 0+ + −

⎡

⎣⎢⎢

⎤

⎦⎥⎥

=( ) ( ) (A.51)

where l ≥ 0. The solutions to this equation are called Bessel functions of order l. Jl (x) are given by the series expansion

J xx

l

xl

xl ll

l

l( )( ) ( ) ( )(

=+

− + + ⋅ + +2 11

2 2 2 2 4 2 2 2

2 4

Γ 441 2

1

2

0)

( ) ( )! ( )

⎡

⎣⎢

⎤

⎦⎥ = −

+ +

+

=

∞

∑k l k

R

xn l k

/Γ (A.52)

If l = 0, 1, 2, . . . , then J−l (x) = −1lJl (x). If l ≠ 0, 1, 2, . . . , then Jl (x) and J−l (x) are linearly independent. In this case, Jl (x) is bounded at x = 0, while J−l (x) is the unbounded Bessel function of the second kind. Nl (x) (also called Neumann functions) are defined by

N xJ x l J x

lll

l l( )( ) cos ( ) ( )

sin ( )( , , , . .=

−≠−π

π 0 1 2 ..) (A.53)


These functions are unbounded at x = 0. The general solution of Eq. (A.51) is

y x AJ x BJ x l

y x AJ x B

l l

l

( ) ( ) ( ) , , , . . .

( ) ( )

= + ≠

= +

− 0 1 2

NN x ll ( ) all

⎧⎨⎪

⎩⎪ (A.54)

where A and B are arbitrary constants. Spherical Bessel functions are related to Bessel functions according to

j xx

J xl l( ) ( )/= +π2 1 2

(A.55)

Also, the Neumann spherical functions are related to the Neumann function Nl (x) by

n xx

N xl l( ) ( )/= +π2 1 2

(A.56)

The functions jl (x) and nl (x) are given explicitly as

j x xx

ddx

xxl

ll

( ) ( )sin= −

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

1 (A.57)

n x xx

ddx

xxl

ll

( ) ( )cos= − −

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

1 (A.58)

The first few jl (x) and nl (x) are given below:

j xx

xn x

xx

j xx

x

xx

n

0 0

1 2 1

( )sin

( )cos

( )sin cos

= = −

= − (( )cos sin

( ) sin

xx

x

xx

j xx x

xx

= − −

= −⎛⎝⎜

⎞⎠⎟

−

2

2 3

3 1 322 2 3 2

3 1 3cos ( ) cos sinx n x

x xx

xx= − −⎛

⎝⎜⎞⎠⎟

−

The asymptotic behavior of the jl (x) and nl (x) as x → ∞ and x → 0 is given by

j x

xl

n xl

x

l x

l

l x l

( )( )!!

( )( )!!

→

→ +

→+

→ − −

⎧

⎨0

0 1

2 1

2 1

⎪⎪⎪

⎩⎪⎪

(A.59)

j x

xx

l

n xx

x

l x

l x

( ) sin

( ) cos

→∞

→∞

→ −⎛⎝⎜

⎞⎠⎟

→ − −

12

1

π

πll2

⎛⎝⎜

⎞⎠⎟

⎧

⎨

⎪⎪

⎩

⎪⎪

(A.60)

where ( )!! ( )( ).2 1 1 3 5 2 1 2 1l l l+ = ⋅ ⋅ − +⋅ ⋅ ⋅


Index


Index 357

AAbsorption, 332Aharonov-Bohm effect, 201Angular momentum, 84, 117, 273

addition of, 273algebra of, 119basis, 273matrix representation of, 121

Approximation, long wavelength, 339Associated Legendre functions, 350

BBarrier potential, 235Basis, 14

continuous, 65discrete, 65orthonormal, 119standard, 119

Bessel function, 299, 305, 312, 352Bisection method, 250Black body radiation, 1, 342Bohr model, 3

angular frequency, 3, 207, 286energy, 3hydrogen atom, 10magneton, 196, 294radius, 3, 166, 236, 242, 334, 344velocity, 3

Bohr, Niels, 3Bohr-Sommerfeld quantization rule, 234, 242, 245Bolzmann constant, 10Born approximation, 298, 300, 302, 311, 313, 328Bosons, 265Boundary conditions, 39Bra, 61, 65Bragg-Von Laue scattering, 303

CCauchy-Schwartz inequality, 16Characteristic polynomial, 16Clebsch-Gordan coefficients, 276, 278, 281, 287, 292Closure relation, 65Coefficient

reflection, 29, 44, 46, 48transmission, 29, 44, 46, 48

Commutation relations, 118canonical commutations relations, 68

Commutator, 63Commuting operators, 63Complex conjugate, 13Complex field, 13Complex number, 14Compton effect, 1, 6, 7

Conditions matching, 43Conjugate variables, 70Connection formulas, 234Conservative system, 69Constants

Bolzmann, 10fine structure, 344Rydbreg, 10

Continuity equation, 28, 59, 180Coulomb force, 3, 269Coulomb potential, 297, 324Cross section, 296

differential, 296Rutherford, 303total, 297

DDe Broglie, 4

relation, 4, 12wavelength, 10, 233

Degeneracyaccidental, 167essential, 167, 206

Deuteron, 247, 295Diffraction, 7, 8Dirac d-function, 17, 348Dirac notation, 61Distribution, 18Dual space, 61Duality

of light, 3of matter, 3

EEhrenfest

equations, 94theorem, 34, 40, 42

Eigenenergy, 43Eigenfunction, 28, 108Eigenspace, 63Eigenstate, 43Eigensubspace, 73Eigenvalue, 16, 21, 28, 235

common, 63Eigenvector, 16, 22, 82, 235Einstein coefficients, 332Einstein, Albert, 1Electric-dipole moment, 229Electromagnetic radiation, 4, 5Electron, 1, 3, 11, 230Emission

induced, 332spontaneous, 332

Index358

Energy levels, 11Euler method, 251Exact ground state, 229Expectation, value, 108, 155Experiments

double-slit, 7, 8Franck-Hertz, 3Stern-Gerlach, 145, 196

FFermion, 265Fermi’s golden rule, 207, 328, 331Forbidden band, 57Form factor, 303FORTRAN, 253, 255, 262Fourier

coefficients, 17inverse Fourier transform, 18, 347series, 16, 17, 24, 347transform, 16, 17, 23, 24, 108, 347

Franck-Hertz experiments, 3Free particle, 26Functional, 18, 61

GGauge, 179

Coulomb, 330freedom, 183invariance, 179Landau, 179radiation, 330symmetric, 179transversal, 330

Gaussian curve, 46, 75Group velocity, 5, 12Gyromagnetic relation, 181

HHamiltonian, 180, 215, 285, 289, 333

Hamiltonian, Zeeman, 294Hamiltonian operator, 16Hamilton-Jacobi equations, 94Harmonic oscillator, 98, 99, 104, 105, 110, 115,

139, 210, 236Heaviside function, 349Heisenberg, 4

uncertainty relation 4, 70, 95, 104Heisenberg picture, 70, 94Heisenberg’s microscope, 9Helium atom, 237, 329Hermite polynomials, 98, 99, 104, 191, 349Hermitian conjugation, 64Hermitian operator, 16, 21Hertz, Heinrich, 1

Hydrogen atom, 3, 10, 164, 166, 168, 177, 178, 199, 218, 225, 236, 295, 344

energy levels of, 3, 11, 167subshell, 167

Hydrogen-like atoms, 164, 168, 172, 238

IIdentical particles, 264Identity matrix, 16Inner product, 15

space, 15Interacting particles, 165Interference, 8Ionization energy, 166

Kket, 61, 65, 81

trial kets, 232Kronecker delta function, 16

LLaguerre polynomials, 167, 352Landau

gauge, 179, 189Landé factor, 295levels, 192

Legendre functions, 120, 133, 350Legendre polynomial, 120, 133, 238, 299, 350, 352Lennard-Jones potential, 114Linear combination, 14Linear momentum, 117Linear operator, 14Logarithmic derivative, 300Low intensity limit, 330

MMagnetic flux, 199, 201Magnetic moment, 181, 191Maxwell, James, 1Maxwell’s equations, 179Mean value, 168Measurement of physical quantities, 62Molecule, 230Moment of inertia, 132, 229Momentum, 3Momentum transfer, 298Muonic atoms, 168Muonium, 176

NNeumann function, 299Neutron, 10Newton, Isaac, 1Newton-Raphson method, 251, 257

Index 359

Norm, 15, 69Normalization, 127Normalized, 15, 25Nuclear spin, 317Null vector, 14Numerical quadrature, 249

Simpson’s method, 250trapezoidal method, 249

Numerov algorithm, 252, 260

OObservable, 61

commuting observables, 63Operator, 27

adjoint of, 64angular momentum, 117, 123annihilation, 100anti-Hermitian, 16, 23, 77conjugate, 16, 64, 77, 86, 128creation, 100evolution, 70function of, 64Hamiltonian, 16, 27Hermitian, 16, 21, 22, 61, 77, 86, 94, 159, 339identity, 65Laplacian, 26linear, 14, 27lowering, 100, 112, 118, 119, 146lowering spin, 146, 157matrix elements of, 66mean value of an, 27momentum, 27normal, 16, 22parity, 86, 177projection, 62raising, 100, 112, 118, 119, 146raising, spin, 146, 157representations of, 65root-mean-square deviation of an, 27, 63rotation, 122, 141, 142self-conjugate, 16skew-Hermitian, 77spatial, 27spin, 145step, 274unitary, 16, 22vector, 68

Optical theorem, 300Orthogonal 15Orthonormal, 64

PParity operator, 177Partial wave expansions, 299

Particle, free, 26Parseval-Plancherel formula, 17Parseval’s theorem, 347Pauli exclusion principle, 265, 267, 271Pauli matrices, 145, 147, 196, 317Permitted band, 57Permutation, 264Perturbation

adiabatic, 224of a degenerate state, 206of a nondegenerate level, 205sudden, 224time-dependent, 204, 206, 330time-independent, 204

Phase condition, 274Phase shift, 299Phase velocity, 5Photoelectric effect, 1, 5 Photon, 1, 3, 5, 6, 7, 9, 10, 11Planck, Max, 1

Planck-Einstein relations, 2Planck’s constant, 1, 2Plane wave, 26

Positronium, 168Postulates, of quantum mechanics, 62Potential

attraction, 239barrier, 28, 34, 235central, 164effective, 165Gaussian, 301, 302hard sphere, 305harmonic oscillator, 98isotropic harmonic, 106periodic, 54scalar, 179scattering, 302spherical repelling, 313step, 28stopping, 1, 2time-independent, 297vector, 179well, 34, 38Yukawa, 300, 301

Poynting vector, 335Probability, 107, 128, 152, 155, 173, 221, 248Probability current, 28, 36, 37, 44, 59, 180, 193, 326Probability density, 28, 36, 45, 180, 186Projector, 62Propagation number, 11

QQuantization rules, 68Quantum angular momentum, 117

Index360

Quantum number, of spin, 145Quantum particles, 3

RRadial equation, 165Radiation, 330

black body, 342external radiation field, 332gauge, 330monochromatic radiation field, 335nonresonant radiation field, 345pressure, 12semiclassical treatment of, 330

Reduced mass, 166Reflection coefficient, 29, 44, 46Region, classically allowed, 233Representation, 65

⏐ p⟩, 66⏐ r ⟩, 66algebraic, 120change of, 66coupled, 274differential, 119uncoupled, 274

Resonance scattering, 48, 308Ritz theorem, 232Rodrigue’s formula, 120, 352Root-mean-deviation, 63, 74Roots, 250Rotation

generator of, 122in spin space, 146

Rotator, 229Runge-Kutta method, 252Rutherford, cross section, 303Rutherford formula, 301, 322Rutherford scattering, 303Rydberg constant, 10

SScalar product, 27, 61Scattering, 296

amplitude, 298, 319angle, 6Bragg-Von Laue scattering condition, 303differential cross section, 296elastic, 329high-energy, 311of identical particles, 300low-energy, 311matrix, 326off-resonance, 308p-wave resonance scattering, 309potential, 297, 316

Scattering (Cont.):process, 297resonance, 48, 308, 311stationary scattering states, 297theory, 296total cross section, 297

Schrödingerequation, 4, 25, 98, 237picture, 70, 94stationary Schrödinger equation, 26

Screening charge, 237Secant method, 251Selection rules, 340Semiclassical approximation, 233, 247Simpson’s method, 250, 254Singlet, 280, 317Slater’s determinant, 265, 267, 272Space, dual, 61Spatial operators, 27Spectrum, continuous, 72Spherical harmonics, 120, 133, 165, 351

addition theorem, 352Spherical symmetry potential, 121Spin, 145

antisymmetric spin state, 317commutation relations, 145eigenfunction, 145eigenvalues, 145lowering, 157nuclear spin, 275operator, 145polarization, 168raising, 157space, 145standard basis, 145symmetric spin state, 317vector, 147

Spin 1/2, 145Spinless particles, 165, 268Spinor, 147Spin-orbit coupling, 290Square-integrable, 25Standard basis, of vector space, 119, 145Stark effect, 218State, vacuum, 100State space, 61

continuous, 64discrete, 64

Stationary, 26, 108states, 4, 46, 69

Step function, 349Stern-Gerlach experiment, 145, 196Stokes’s theorem, 200Stopping potential, 1

Index 361

Symmetric gauge, 188Symmetrization, 68Symmetry

cylindrical, 262radial, 262

TTaylor series, 160, 252, 309Tensor product, 273Time evolution, 68

for a conservative system, 69of the mean value, 69operator, 70

Total probability, 25Transformation, guage, 179Transition

electric-dipole, 332, 339, 340, 346electric-qudrupole, 332, 339, 341, 346forbidden, 335magnetic-dipole, 332, 346probability, 221rate, 331strictly forbidden, 335

Translation invariance, 302Transmission coefficient, 29, 44, 46, 49, 243, 247Transposition, 264Transversal guage, 330Triplet, 280, 317Tunnel effect, 50Turning points, 233

UUncertainty, of energy, 70Uncertainty relation, 4, 70, 85, 95, 104Unit matrix, 16Units

Gaussian system of, 181MKS units, 181Unperturbed states, 204

VVacuum state, 100Variational method, 232, 239Variational parameter, 237Vector space, 14

dimension, 14inner product, 15standard basis, 119state space, 61

Velocitydrift, 195group, 5phase, 5

WWavefunction, 25

antisymmetric, 265normalized, 25probability current of, 28probability density of, 28space, 66

Wave guide, 11Wave number, 5Wave outgoing, 322Wave packet, 4, 27, 45, 46Wave plane, 336Wave vector, 3, 4Wavelength, 4Waves, plane, 26WKB approximation, 233, 241, 245, 247Work function, 2, 3

YYukawa potential, 300, 301

ZZeeman effect, 198Zeeman Hamiltonian, 294

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