Introduction to quantum field theory, based on the book by Peskin and Schroeder

5/20/2018 Introduction to quantum field theory, based on the book by Peskin and Schroeder

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October 12, 2011

Quantum Field Theory I

Ulrich Haisch

Rudolf Peierls Centre for Theoretical Physics, University of Oxford

OX1 3PN Oxford, United Kingdom


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Abstract

This course deals with modern applications of quantum field theory with emphasize onthe quantization of theories involving scalar and spinor fields.


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Recommended Books and Resources

There is a vast array of quantum field theory texts, many of them with redeeming features.Here I mention a few of them, mostly the ones that I used or looked at when preparing thiscourse. To a large extent, I will follow the first section of

M. Peskin and D. Schroeder, An Introduction to Quantum Field TheoryThis is a very clear and comprehensive book, covering essentially everything in this course

as well as many advanced aspects of quantum field theory that go (far) beyond the scope ofthis lecture.

S. Weinberg, The Quantum Theory of Fields: Volume 1, Foundations

This is the first in a three volume series by one of the masters of quantum field theory.It takes a unique route through the subject, focussing initially on particles rather than fields.Since it has a very particular viewpoint, it is difficult to digest, but certainly worth reading.

L. Ryder, Quantum Field Theory

This elementary text has a nice discussion of much of the material in this course. It isgood for a first reading.

A. Zee, Quantum Field Theory in a Nutshell

This is a charming book, where emphasis is placed on physical understanding and theauthor isnt afraid to hide the ugly truth when necessary. It contains many gems.

By browsing the web, I also found interesting material. Nice introductions to quantumfield theory (of different length and viewpoint) have been written by C. Anastasiou and D.Tong. The corresponding scripts can be found at:

http://www.phys.ethz.ch/babis/Teaching/QFTI/qft1.pdfhttp://www.damtp.cam.ac.uk/user/tong/qft/qft.pdf

Other links to useful resources can be found on the web page of D. Tong:

http://www.damtp.cam.ac.uk/user/tong/qft.html

For completeness, I will also give relevant references at the end of each section of thisscript. The interested reader can consult them for further details on the discussed topics.

1
http://www.phys.ethz.ch/~babis/Teaching/QFTI/qft1.pdfhttp://www.phys.ethz.ch/~babis/Teaching/QFTI/qft1.pdfhttp://www.phys.ethz.ch/~babis/Teaching/QFTI/qft1.pdfhttp://www.damtp.cam.ac.uk/user/tong/qft/qft.pdfhttp://www.damtp.cam.ac.uk/user/tong/qft.htmlhttp://www.damtp.cam.ac.uk/user/tong/qft.htmlhttp://www.damtp.cam.ac.uk/user/tong/qft/qft.pdfhttp://www.phys.ethz.ch/~babis/Teaching/QFTI/qft1.pdf


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Contents

1 Introduction 31.1 Why QFT? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.2 Scales and Units . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

2 Elements of Classical Field Theory 82.1 Dynamics of Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82.2 Noethers Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92.3 Example: Electrodynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122.4 Space-Time Symmetries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142.5 Problems. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

3 Klein-Gordon Theory 223.1 Klein-Gordon Field as Harmonic Oscillators . . . . . . . . . . . . . . . . . . . 233.2 Structure of Vacuum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

3.3 Particle States. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 303.4 Two Real Klein-Gordon Fields. . . . . . . . . . . . . . . . . . . . . . . . . . . 343.5 Complex Klein-Gordon Field. . . . . . . . . . . . . . . . . . . . . . . . . . . . 373.6 Heisenberg Picture . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 383.7 Klein-Gordon Correlators. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 423.8 Non-Relativistic Limit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 483.9 Problems. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52

4 Interacting Fields 564.1 Classification of Interactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 564.2 Interaction Picture . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59

4.3 First Look at Scattering Processes. . . . . . . . . . . . . . . . . . . . . . . . . 614.4 Wicks Theorem. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 644.5 Second Look at Scattering Processes . . . . . . . . . . . . . . . . . . . . . . . 674.6 Feynman Diagrams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 684.7 Third Look at Scattering Processes . . . . . . . . . . . . . . . . . . . . . . . . 734.8 Yukawa Potential . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 764.9 Connected and Amputated Feynman Diagrams . . . . . . . . . . . . . . . . . 784.10 From Correlation Functions to Scattering Matrix Elements . . . . . . . . . . . 854.11 Decay Widths and Cross Sections . . . . . . . . . . . . . . . . . . . . . . . . . 964.12 Problems. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101

5 Dirac Theory 1075.1 Spinor Representation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1085.2 Discrete Symmetries of Dirac Theory . . . . . . . . . . . . . . . . . . . . . . . 1175.3 Continuous Symmetries of Dirac Theory . . . . . . . . . . . . . . . . . . . . . 1235.4 Solutions to Dirac Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1255.5 Quantization of Dirac Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . 1305.6 Problems. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143

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1 Introduction

As the term quantum field theory (QFT) suggests, QFT is the application of quantum me-chanics (QM) to dynamical systems offields, in the same sense that QM is concerned mainlywith the quantization of dynamical systems of particles. QFT is not only a subject that is

absolutely essential to understand the current state of elementary particle physics as well asmodern aspects of cosmology, but also plays a crucial role in many active areas of research,ranging from atomic over nuclear and condensed-matter physics to pure mathematics. Sincethe ultimate goal of this course is to gain a basic understanding of the fundamental laws ofnature, we will in the following focus mainly on the physics of elementary particles and hencedeal mostly with relativisticfields.

1.1 Why QFT?

The primary reason for introducing the concept of fields in classical physics is to construct lawsof nature that are local. The old laws of Newton (Coulomb) involve action at a distance.

This means that the force felt by a planet (an electron) changes immediately if a distantstar (proton) moves. The laws of Newton and Coulomb thus feature non-localinteractions.The field theories of Einstein (general relativity) and Maxwell (electrodynamics) remedied thesituation, with all interactions mediated in a local fashion by fields. The requirement of localityremains a strong motivation for studying QFTs. However, there are further good reasons totreat the quantum field (and not the particle) as fundamental(or as Steven Weinberg puts itin [1]: Quantum fields are the basic ingredients of the universe, and particles are just bundlesof energy and momentum made out of them.).

QM and Special Relativity

A first reason is that the combination of QM and special relativity implies that particle numberis not conserved. Consider a particle of mass m trapped in a box of size L. Heisenbergsuncertainty principle tells us that the uncertainty in the momentum of our particle is p/L. In the relativistic limit, momentum and energy can be treated on equivalent footing,and one has an uncertainty in the energy of order E c/L. Yet, if E= 2mc2, there isenough energy available to create a virtualparticle-antiparticle pair from the vacuum (Diracsea). This little exercise shows that when a particle with massm is localized within a distanceCompton = /(mc), talking about a single particle loses its sense. For distances smaller thanthis Compton wavelengththere is a high probability that we will detect particle-antiparticlepairs swarming around the single particle that we initially put into the box. Notice that

Compton is always smaller than the de Broglie wavelength given by de Broglie = /|p|.1

Ifyou like, the de Broglie wavelength is the distance at which the wavelike nature of particlesbecomes apparent, while the Compton wavelength is the distance at which the concept ofa single pointlike particle breaks down and one has to start thinking about how to describemultiparticle states.

1Throughout this course we will use boldface type (ordinary italic type) to denote 3-vectors (4-vectors).

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The presence of a multitude of particles and antiparticles at short distances (or highenergies) tells us that any attempt to write down a relativistic version of the one-particleSchrodinger equation is doomed to fail. There is no mechanism in standard non-relativisticQM to deal with changes in the particle number. Indeed, any attempt to naively write down arelativistic version of the one-particle Schrodinger equation meets serious problems: negative

probabilities, infinite towers of negative energy states, or a breakdown of causality are thecommon issues that arise.

QM and Causality

Let us have a closer look at the issue ofbreakdown of causality. Consider the amplitude

A(t) =yeiEt/x , (1.1)

that describes the propagation of a free particle from the point x to y. In non-relativistic QMone has E=p2/(2m) and hence2

A(t) = y exp i p2/(2m) t/ x=

d3p

(2)3y exp i p2/(2m) t/ ppx

=

d3p

(2)3 exp

i p2/(2m) t/ exp ip (y x)/= m

2it

3/2exp

im (y x)2/(22 t) .(1.2)

Here we have made use of the completeness d3p/(2)3 |pp| = 1 of|p and a little bitof algebra. The expression (1.2) is non-zero for all y and t, indicating that a particle canpropagate between any two points in an arbitrarily short time. In a relativistic theory, thisconclusion would signal a violation of causality. One might hope that using the relativisticexpressionE=

p2c2 + m2c4 for the energy would cure the problem, but it does not. In fact,

in the relativistic case one has

A(t) =y exp it/p2c2 + m2c4 x

=

d3p

(2)3 exp

it/p2c2 + m2c4 exp ip (y x)/=

1

222 |y x| 0 dp p sin (p |y x| /) exp it/p2c2 + m2c4 .(1.3)

This integral can be evaluated explicitly in terms of Bessel functions, but for our purposes itis sufficient to consider its asymptotic behavior for L2 = |yx|2 c2t2,i.e., separations welloutside the light-cone. We use the method of stationary phase. The relevant phase function

2The symbolp denotes the momentum operator, which in many QM books is indicated by a . To avoidclutter, I will not use the latter notation, but simply write p.

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pL t

p2c2 + m2c4 has a stationary point p = imcL/

L2 c2t2. Plugging this value into(1.3), we find that (up to a rational function ofL and t),

A(t) expm/

L2 c2t2

. (1.4)

This expression is small but non-zero outside the light-cone and causality is again violated.In both cases, the observed failure is telling us that we need a new formalism to preserve

causality. This formalism is QFT. It solves the causality problem in a miraculous way. Wewill see later that in QFT the propagation of a particle across a space-like interval is indis-tinguishable from the propagation of an antiparticle in the opposite direction. When we askwhether an observation made at point x can affect an observation made at point y, we willfind that the amplitudes for particle and antiparticle propagation cancel in such a way thatcausality is preserved.

What else is QFT good for?

Besides solving the causality problem, QFT also provides an elegant framework to describetransitions between states of different particle number and type. An example physical pro-cesses, exhaustivelly studied (from 1989 until 2000) at the Large Electron Positron (LEP)collider in Geneva, is the production of a muon () and its antiparticle (+) out of theannihilation of an electron (e) and its antiparticle (the positron e+):

e + e+ + + . (1.5)

The experimental confirmation of the QFT predictions for processes such as (1.5), often to anunprecedented level of accuracy, is our real reason for studying QFT. But the power of QFTdoes not end here. In traditional QM the relation between spin and statistics has to be putin by hand. To agree with experiment, one should choose Bose statistics (no minus sign ifone exchanges two identical particles) for integer spin particles, and Fermi statistics (minussign if one exchanges two identical particles) for half-integer spin particles. On the otherhand, in QFT the relationship between spin and statistics is a consequence of the framework,following from the commutation quantization conditions for boson fields and anticommutationquantization conditions for fermion fields.

1.2 Scales and Units

There are three fundamental dimensionful constants in nature: the speed of light c, Plancks

constant

(divided by 2), and Newtons constant GN. Their dimensions are

[c] = length time1 ,

[] = length2 mass time1 ,

[GN] = length3 mass1 time2 .

(1.6)

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In order to avoid unnecessary clutter, we will work throughout this course in natural units,defined by c = = 1.3 This allows us to express all dimensionful quantities in terms of asingle scale which we choose to be mass or, equivalently, energy (since E=mc2 has becomeE=m). Energies will be given in units of eV (the electron volt) or more often GeV = 109 eVor TeV = 1012 eV, since we are typically dealing with high energies. To convert the unit of

energy back to units of length or time, we have to insert the relevant powers ofc and . E.g.,the length scale associated to a mass m is = h/(mc). Remembering that

hc 1.24 106 eV m , (1.7)

one finds that the length scale corresponding to the electron with mass me 511 keV ise 2 1012 m.

Throughout this course we will refer to the dimension of a quantity, meaning the massdimension. Newtons constant, e.g., has [GN] = 2 and defines a mass scale

GN=M2P , (1.8)

where MP 1019 GeV is the Planck scale. This energy corresponds to a length scale LP1035 m the Planck length. The Planck length is believed to be the smallest length scale thatmakes sense: beyond this scale quantum gravity effects are likely to become important andits no longer clear that the concept of space-time can be applied. The largest length scale wecan talk of is the size of the cosmological horizon, roughly 1060 LP.

A number for particle physics and cosmology relevant masses and the corresponding lengthscales are shown in Table 1. Let me go through the list and spend some words on the mostimportant quantities. After the size of the observable universe, the first scale we encounter isthecosmological constant() measured to be around 103 eV. Since nobody can really explainwhy the cosmological constant has this particular value, lets forget about it real quick and

turn our attention to the masses of the known elementary particles. These range from lessthan 1 eV for the neutrinos (s) to around 175 GeV for the top quark (t). The (in)famousHiggs boson (h), which is the only not yet observed ingredient of the standard model (SM) ofelementary particle physics, is believed to weigh in at about 100 to 200 GeV. For scales around1 TeV, i.e., the terascale, the predictive power of the SM is expected to break down. This isprecisely the energy regime that the Large Hadron Collider (LHC) at CERN in Geneva hasstarted to explore, having a design center-of-mass energy of 14 TeV. Beyond the electroweakscale (v) of around 250 GeV, again nobody knows with certainty what is going on. One couldfind a plethora of new (elementary) particles or a great desert. There are experimentalhints that the coupling constants of electromagnetism, and the weak and strong forces unify

at aroundMGUT= 1016

GeV,i.e., thegrand unification scale(GUT). Everything is topped offat the Planck scale where a QFT description might no longer be possible and a quantum theoryincluding the effects of gravity is needed to describe the physics of fundamental interactions.The most likely possibility for such a theory seems to be some kind of string theory. Butalso many other ideas such as loop quantum gravity, Horava-Lifshitz gravity, etc. exist. Infact, the theory of everything (TOE) could also be a QFT, but one in which the finite or

3The whole point of units is that you can choose whatever units are most convenient!

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Quantity Mass Length

Observable universe 1033 eV 1027 m 2 1010 lyCosmological constant () 103 eV 103 m

Neutrinos (s) 1 eV 106 m

Electron (e) 511 keV 2 1012 mMuon () 106 MeV 1014 m

Charm quark (c) 1.3 GeV 1015 m

Tau () 1.78 GeV 7 1016 mBottom quark (b) 4.6 GeV 3 1016 m

Top quark (t) 175 GeV 7 1018 mHiggs boson (h) [100, 200] GeV [6, 12] 1018 m

Electroweak scale (v) 250 GeV 5 1018 mLHC energy 14 TeV 9

1020 m

GUT scale (MGUT) 1016 GeV 1031 m

Planck scale (MP) 1019 GeV 1035 m

Table 1: An assortment of masses and corresponding lengths scales that appear in thecontext of particle physics and cosmology.

infinite number of renormalized couplings do not run off to infinity with increasing energy,but hit a fixed point of the renormalization group equation. This possibility goes by the

name of asymptotic safety. Dont worry if havent understood a single word of what I havemumbled about possible TOEs. All this is way too advanced to be covered in this course. Ionly mentioned it, to make propaganda for the research of Joe Conlon (string theory), AndreLukas (string theory), and John Wheater (quantum gravity), which work on such theorieshere in Oxford. Ask them if you want to know more about it.

References

[1] S. Weinberg, What is quantum field theory, and what did we think it was?, arXiv:hep-th/9702027.

[2] S. Weinberg, The Search for Unity: Notes for a History of Quantum Field Theory,Daedalus, Vol. 106, No. 4, Discoveries and Interpretations: Studies in ContemporaryScholarship, Volume II (1977), 17 p.

[3] Chapter 1 of S. Weinberg, The Quantum theory of fields. Vol. 1: Foundations, Cam-bridge, UK, Univ. Pr. (1995), 609 p.

[4] F. Wilczek, Rev. Mod. Phys. 71, S85 (1999) [arXiv:hep-th/9803075].

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2 Elements of Classical Field Theory

In this second section we will discuss various aspects of classical fields. We will cover onlythe bare minimum ground necessary before turning to the quantum theory, and will returnto classical field theory at several later stages in this course when we need to introduce new

concepts or ideas.

2.1 Dynamics of Fields

A field is a quantity defined at every space-time point x = (t,x). While classical particlemechanics deals with a finite number ofgeneralized coordinatesqa(t), indexed by a label a, infield theory we are interested in the dynamics of fields

a(t,x) , (2.1)

where botha and x are considered as labels. We are hence dealing with an infinite number of

degrees of freedom (dofs), at least one for each point x in space. Notice that the concept ofposition has been relegated from a dynamical variable in particle mechanics to a mere labelin field theory.

Lagrangian and Action

The dynamics of the fields is governed by the Lagrangian. In all the systems we will studyin this course, the Lagrangian is a function of the fields a and their derivatives a,

4 andgiven by

L(t) =

d3x L(a, a) , (2.2)

where the official name for L isLagrangian density. Like everybody else we will, however, sim-ply call it Lagrangian from now on. For any time interval t [t1, t2], theactioncorrespondingto (2.2) reads

S=

t2t1

dt

d3x L =

d4x L . (2.3)

Recall that in classical mechanics L depends only on qa and qa, but not on the second timederivatives of the generalized coordinates. In field theory we similarly restrict to LagrangiansL depending on a and a. Furthermore, with an eye on Lorentz invariance, we will onlyconsider Lagrangians depending ona and not higher derivatives.

Notice that since we have set = 1, using the convention described in Section 1.2, the

dimension of the action is [S] = 0. With (2.3) and [d4x] = 4, it follows that the Lagrangianmust necessarily have [L] = 4. Other objects that we will use frequently to construct La-grangians are derivatives, masses, couplings, and most importantly fields. The dimensions ofthe former two objects are [] = 1 and [m] = 1, while the dimensions of the latter two quan-tities depend on the specific type of coupling and field one considers. We therefore postpone

4If there is no (or only little) room for confusion, we will often drop the arguments of functions and writea= a(x) etc. to keep the notation short.

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the discussion of the mass dimension of couplings and fields to the point when we meet therelevant building blocks.

Principle of Least Action

The dynamical behavior of fields can be determined by the principle of least action. Thisprinciple states that when a system evolves from one given configuration to another betweentimes t1 and t2 it does so along the path in cofiguration space for which the action is anextremum (usually a minimum) and hence satisfies S= 0. This condition can be rewritten,using partial integration, as follows

S=

d4x

La

a+ L

(a)(a)

=

d4x

La

L(a)

a+

L

(a)a

= 0 .

(2.4)

The last term is a total derivative and vanishes for any a that decays at spatial infinity andobeysa(t1, x) =a(t2, x) = 0. For all such paths, we obtain the Euler-Lagrange equationsof motion(EOMs) for the fieldsa, namely

L

(a)

L

a= 0 . (2.5)

Hamiltonian Formalism

The link between the Lagrangian formalism and the quantum theory goes via the path integral.While this is a powerful formalism, we will for the time being usecanonical quantization, sinceit makes the transition to QM easier. For this we need the Hamiltonian formalism of fieldtheory. We start by defining the momentum densitya(x) conjugate to a(x),

a = La

. (2.6)

In terms ofa, a, andLthe Hamiltonian densityis given byH =aa L , (2.7)

where, as in classical mechanics, we have eliminated a in favor ofa everywhere inH. The

Hamiltonian then simply takes the form

H= d3x H . (2.8)2.2 Noethers Theorem

The role of symmetries in field theory is possibly even more important than in particle mechan-ics. There are Lorentz symmetry, internal symmetries, gauge symmetries, supersymmetries,etc.We start here by recasting Noethers theorem in a field theoretic framework.

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Currents and Charges

Noethers theorem states that every continuous symmetryof the Lagrangian gives rise to aconserved currentJ(x), so that the EOMs (2.5) imply

J = 0 , (2.9)

or in componentsdJ0/dt + J= 0. To every conserved current there exists also a conserved(global) chargeQ,i.e., a physical quantity which stays the same value at all times, defined as

Q=

R3

d3x J0 . (2.10)

The latter statement is readily shown by taking the time derivative ofQ,

dQ

dt =

R3

d3xdJ0

dt =

R3

d3x J, (2.11)

which is zero, if one assumes that Jfalls off sufficiently fast as |x| . Notice, however, thatthe existence of the conserved current J is much stronger than the existence of the (global)charge Q, because it implies that charge is in fact conserved locally. To see this, we define thecharge in a finite volume V by

QV =

V

d3x J0 . (2.12)

Repeating the above analysis, we find

dQVdt

= V

d3x J= S

dS J, (2.13)

whereSdenotes the area bounding V, dSis a shorthand for n dSwith n being the outwardpointing unit normal vector of the boundary S, and we have used Gauss theorem. In physicalterms the result means that any charge leaving V must be accounted for by a flow of thecurrent 3-vector Jout of the volume. This kind of local conservation law of charge holds inany local field theory.

Proof of Theorem

In order to prove Noethers theorem, well consider infinitesimal transformations. This isalways possible in the case of a continuous symmetry. We say thata is a symmetry of thetheory, if the Lagrangian changes by a total derivative

L(a) =J(a) , (2.14)for a set of functions J. We then consider the transformation ofL under anarbitrarychangeof fielda. Glancing at (2.4) tells us that in this case

L =

La

L(a)

a+

L

(a)a

. (2.15)

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When the EOMs are satisfied than the term in square bracket vanishes so that we are simplyleft with the total derivative term. For a symmetry transformation satisfying (2.13) and (2.14),the relation (2.15) hence takes the form

J =

L=

L(a)

a , (2.16)or simplyJ

= 0 with

J = L

(a)a J , (2.17)

which completes the proof. Notice that if the Lagrangian is invariantunder the infinitesimaltransformation a, i.e., L = 0, thenJ = 0 and J contains only the first term on theright-hand side of (2.17).

We stress that that our proof only goes through for continuous transformations for whichthere exists a choice of the transformation parameters resulting in a unit transformation, i.e.,no transformation. An example is a Lorentz boost with some velocity v, where for v = 0

the coordinatesx remain unchanged. There are examples of symmetry transformations wherethis does not occur. E.g., a parity transformation does not have this property, and Noetherstheorem is not applicable then.

Energy-Momentum Tensor

Recall that in classic particle mechanics, spatial translationinvariance gives rise to the con-servation of momentum, while invariance under time translations is responsible for the con-servation of energy. What happens in classical field theory? To figure it out, lets have a lookat infinitesimal translations

x x = a(x) a(x + ) =a(x) + a(x) , (2.18)where the sign in the field transformation is plus, instead of minus, because we are doing anactive, as opposed to passive, transformation. If the Lagrangian does not explicitly dependon x but only through a(x) (which will always be the case in the Lagrangians discussed inthis course), the Lagrangian transforms under the infinitesimal translation as

L L + L . (2.19)

Since the change inL is a total derivative, we can invoke Noethers theorem which givesus four conserved currents T = (J

)one for each of the translations ( = 0, 1, 2, 3).

From (2.18) we readily read off the explicit expressions forT,

T= L

(a)a L . (2.20)

This quantity is called the energy-momentum (or stress-energy) tensor. It has dimension[T] = 4 and satisfies

T= 0 . (2.21)

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The four conserved charges are (= 0, 1, 2, 3)

P =

d3x T0 , (2.22)

Specifically, the time component ofP is

P0 =

d3x T00 =

d3x

aa L

, (2.23)

which (looking at (2.7) and (2.8)) is nothing but the Hamiltonian H. We thus conclude thatthe chargeP0 is the total energy of the field configuration, and it is conserved. In fields theory,energy conservation is thus a pure consequence of time translation symmetry, like it was inparticle mechanics. Similarly, we can identify the chargesPi (i= 1, 2, 3),

Pi =

d3x T0i =

d3x aia , (2.24)

as the momentum components of the field configuration in the three space directions, and theyare of course also conserved.

2.3 Example: Electrodynamics

As a simple application of the formalism we have developed so far in this section, let us tryto derive Maxwells equations using the field theory formulation. In terms of the electric andmagnetic fields E and B and the charge density and 3-vector current j, these equationstake the well-known form

B= 0 , (2.25)

E+ Bt

= 0 , (2.26)

E= , (2.27)

B Et

=j . (2.28)

TheEandB fields are spatial 3-vectors and can be expressed in terms of the componentsof the 4-vector fieldA = (,A) by

E=

A

t , B=

A . (2.29)

This definition ensures that the first two homogeneous Maxwell equations (2.25) and (2.26)are automatically satisfied,

(A) =ijk ijAk = 0 , (2.30)

At

+

t(A) = () = ijk jk= 0 . (2.31)

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Hereijk is the fully antisymmetric Levi-Civita tensor with 123 = 123 = +1 and the indicesi,j,k= 1, 2, 3 are summed over if they appear twice.

The remaining two inhomogeneous Maxwell equations (2.27) and (2.28) follow from theLagrangian

L=

1

2(A) (

A) +1

2(A

)2

AJ

, (2.32)

withJ = (,j). From the rules presented in Section2.1,we gather that the dimension of thevector field and current is [A] = 1 and [J

] = 3, respectively. The funny minus sign of the firstterm on the right-hand side is required to ensure that the kinetic term 1 /2 A2i is positive usingthe Minkowski metric. Notice also that the Lagrangian (2.32) has no kinetic term 1/2 A20 andhence A0 is not dynamical. Why this is and necessarily has to be the case will only becomefully clear if you attend the advanced QFT course. Yet, we can already get an idea whatis going on by remembering that the photon (the quantum of electrodynamics) has only twopolarization states,i.e., two physical dofs, while the massless vector field Ahas obviously fourdofs. The fact that time component A0 is not dynamical reduces the number of independentdofs in A

from four to three. But this is still one too many. The last unwanted dof can be

gauged away using the gauge symmetryof the quantum version of electromagnetism akaquantum electrodynamics (QED).

Enough said, lets do serious business and compute something. To see that the statementmade before (2.32) is indeed correct, we first evaluate

LA

= J , L(A)

= A + (A) , (2.33)

from which we derive the EOMs,

0 = L(A) LA = 2A + (A) + J = (A A) + J . (2.34)Introducing now the field-strength tensor

F=A A , (2.35)

we can write (2.32) and (2.34) quite compact,

L = 14

FF JA . (2.36)

F =J , (2.37)

Does this look familiar? I hope so. Notice that [F] = 2. In order to see that (2.37) indeedcaptures the physics of (2.27) and (2.28), we compute the components ofF. We find

F0i = Fi0 =0Ai iA0 = +

A

t

i= Ei ,

Fij = Fji =iAj jAi = ijkBk ,(2.38)

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while all other components are zero. With this in hand, we then obtain from F0 = and

F1 =j 1,

F0 =0F

00 + iFi0 = E= ,

F1 =0F01 + iFi1 = E1

t + B3

x2 B2

x3 = B Et 1

=j 1 .

(2.39)

Similar relations hold for the remaining components i= 2, 3. Taken together this proves thesecond inhomogeneous Maxwell equation (2.28).

Let me also derive the energy-momentum tensor T of electrodynamics, ignoring for themoment the source term AJ

. Using (2.33) one finds

T = (A)(A) (A)(A) +1

4FF

. (2.40)

Notice that the first term in (2.40) is not symmetric, which implies that T =T . In fact,

this is not really surprising since the definition of the energy-momentum tensor (2.20) doesnot exhibit an explicit symmetry in the indices and . Nevertheless, there is typically a wayto massage the energy-momentum tensor of any theory into a symmetric form.5 To learn howthis can be done in the case under consideration is the objective of a homework problem.

2.4 Space-Time Symmetries

One of the main motivations to develop QFT is to reconcile QM with special relativity. Wethus want to construct field theories in which space and time are placed on an equal footingand the theory is invariant under Lorentz transformations,

x

(x

)

=

x

, (2.41)with

=, (2.42)

so that the distance ds2 = dxdx is preserved. Here = = diag (1, 1, 1, 1)

denotes the Minkowski metric. E.g., a rotation by the angle about the z-axis, and a boostbyv


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by Andre Lukas. Alternatively, you can study the group theory crash course written byMartin Bauer (a PhD student at Mainz University). It can be found on my Oxford homepage.

The various fields belong to different representations of the Lorentz group. The simplestexample is the scalar field, which under the Lorentz transformation x x,6 transforms as

(x)

(x) =(

1

x) . (2.44)The inverse 1 appears in the argument because we are dealing with an active transformation,in which the field is truly shifted. To see why this means that the inverse appears, it will sufficeto consider a non-relativistic example such as a temperature field. Suppose we start with aninitial field(x) which has a hotspot at, say,x = (1, 0, 0). Lets now make a rotationx Rxabout the z-axis so that the hotspot ends up at x= (0, 1, 0). If we want to express the newfield (x) in terms of the old field (x), we have to place ourselves at x = (0, 1, 0) and askwhat the old field looked like at the point R1x = (1, 0, 0) we came from. This R1 is theorigin of the 1 factor in the argument of the transformed field in (2.44).

The Lagrangian formulation of field theory makes it especially easy to discuss Lorentz

invariance, since an EOM is automatically Lorentz invariant if it follows from a Lagrangianthat is aLorentz scalar. This is an immediate consequence of the principle of least action. If aLorentz transformation leaves the Lagrangian unchanged, the transformation of an extremumin the action will be another extremum. To give an example, lets look at the followingLagrangian

L =12

()2 1

2m22 . (2.45)

where is a real scalar and, as we will see later, m is the mass of (for now on just thinkaboutm as a parameter). Obviously, the dimension of the field is [] = 1. You will show in ahomework assignment that the EOM corresponding to (2.45) takes the form

+ m2= 0 . (2.46)This equation is the famous Klein-Gordon equation. The Laplacian in Minkowski space issometimes denoted by . In this notation, the Klein-Gordon equation reads ( + m2)= 0.

Let us first check that a Lorentz transformation leaves the Lagrangian ( 2.45) and its ac-tion invariant. According to (2.44), the mass term transforms as 1/2 m22(x) 1/2 m22(x)withx = 1x. The transformation ofis

(x) ((x)) = (1)()(x) . (2.47)Using (2.43) we thus find that the derivative term in the Klein-Gordon Lagrangian behaves as

12((x))2 12(1)()(x)(1)()(x)

=1

2()(x

)()(x)

=1

2((x

))2

,

(2.48)

6To shorten the notation we will often use matrix notation and drop the indices , etc.

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under the Lorentz transformation . Putting things together, we find that the action of theKlein-Gordon theory is indeed Lorentz invariant,

S=

d4x L(x)

d4x L(x) =

d4x L(x) =S . (2.49)

Notice that changing the integration variables from d4xto d4x, in principle introduces an Ja-cobian factor det (). This factor is, however, equal to 1 for Lorentz transformation connectedto the identity, that we are dealing with.

A similar calculation also shows that, as promised, also the EOM of the Klein-Gordon field is invariant,

2 + m2

(x) 2 + m2(x)=

(1)(1)+ m

2

(x)

= + m

2

(x) = 0 .

(2.50)

In the case of the Klein-Gordon theory, we hence conclude that the statements made before(2.45) are indeed correct.

Representations of Lorentz Group

The transformation law (2.44) is the simplest possible transformation law for a field. In fact,it is the only possibility for a one-component fieldakaa real scalar. Yet, it is also clear that inorder to describe nature (think only about electromagnetism) we need multicomponent fields,which have more complicated transformation properties. The most familiar case is that of avector field, such as the vector potential A, which we have already met in Section 2.3. In

this case the quantity that is distributed in space-time also carries an orientation which mustbe rotated and/or boosted.

In fact, we will learn in this course that the Lorentz group has a variety of representa-tions, corresponding to particles with integer (bosons) and half-integer spins (fermions) inQFT. These representations are normally constructed out ofspinors. To start this general(and somewhat formal) discussion, let me examine the allowed possibilities for linear fieldtransformations

a(x) a(x) =M()abb(x) , (2.51)under (2.41). The first important point to notice is that the Lorentz transformations form agroup. This means that two successive Lorentz transformations,

x x = x , x x = x , (2.52)can also be described in terms of a single one

x x = x , (2.53)with

= . (2.54)

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What happens to (2.51) under this set of Lorentz transformations? For x x = x, wehave (in matrix notation)

(x) (x) =M()(x) . (2.55)On the other hand, for x x = x x = x, we get

(x) (x) =M()(x) =M()M()(x) . (2.56)In order for the last two equations to be consistent with each other, the field transformationsMmust obviously fulfill

M() =M()M() . (2.57)

In group theory terminology, this means that the matrices Mfurnish a representation of theLorentz group. Field Lorentz transformations are therefore not random, but they can be foundif we find all (finite dimensional) representations of the Lorentz group.

So how do the common representation of the Lorentz group look like and how do we getall of them? While both questions will be answered in this lecture, I believe it is best to do

it case-by-case whenever we will meet a new type of (quantum) field. Since we already talkedabout the real scalar (Klein-Gordon field) and the vector A (potential in electrodynamics),it makes nevertheless sense to give the representations for these two types of fields already atthis point.

Since a scalar field by definition does not change under Lorentz transformations, (x)(x) =(x), the scalar representation of the Lorentz group is simply

M() = 1 . (2.58)

This was easy! The representation of the vector A is also not difficult to figure out. Let mefor the time being only state the result. One finds

M() = , (2.59)

which means that a vector field A transforms under a Lorentz transformation as (restoringindices)

A(x) (A)(x) = A(x) . (2.60)It is important to notice that the latter transformation property implies that any term buildout ofA and , where all Lorentz indices are contracted is invariant under Lorentz trans-formations. As an exercise you are supposed to show this explicitly for terms like A

,etc.

Angular Momentum

In classical particle mechanics, rotational invariance gives rise to conservation of angularmomentum. What is the analogy in field theory? Moreover, we now have further Lorentztransformations, namelyboosts. What conserved quantity do they correspond to? In order toaddress these questions, we first need the infinitesimal form of the Lorentz transformations

=+

, (2.61)

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whereis infinitesimal. The condition (2.42) for to be a Lorentz transformation becomesin infinitesimal form

= (+ ) (

+

) =

+ + + O(2) , (2.62)which implies that must be an antisymmetric matrix,

= . (2.63)Notice that an antisymmetric 4 4 matrix has six independent parameters, which agrees withthe number of different Lorentz transformations, i.e., three rotations and three boosts.

Applying the infinitesimal Lorentz transformation to our real scalar field , we have

(x) (x x) =(x) x(x) , (2.64)where the minus sign arises from the factor 1 in (2.43). The variation of the field underan infinitesimal Lorentz transformation is hence given by

= x . (2.65)By the same line of reasoning, one shows that the variation of the Lagrangian is

L = xL = (xL), (2.66)where in the last step we used the fact that = 0 due to its antisymmetry. The Lagrangianchanges by a total derivative, so we can apply Noethers theorem (2.17) withJ = xLto find the conserved current,

J =

L

()

x +

x

L=

L

() L

x = Tx .

(2.67)

Stripping off, we obtain six different currents, which we write as

(J) =xT xT . (2.68)These currents satisfy

(J) = 0 , (2.69)and give (as usual) rise to six conserved charges. For ,

= 0, the Lorentz transformation

is a rotation and the three conserved charges give the total angular momentumof the field(i, j= 1, 2, 3):

Qij =

d3x

xiT0j xjT0i . (2.70)

Whats about the boosts? In this case, the conserved charges are

Q0i =

d3x

x0T0i xiT00 . (2.71)18


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The fact that these are conserved tells us that

0 =dQ0i

dt =

d3x T0i + t

d3x

dT0i

dt d

dt

d3x xiT00

=Pi

+ t

dPi

dt d

dt d3x xiT00 .(2.72)

Yet, also the momentum Pi is conserved, i.e., dPi/dt= 0, and we conclude that

d

dt

d3x xiT00 = const. . (2.73)

This is the statement that the center of energy of the field travels with a constant velocity. Ina sense its a field theoretical version of Newtons first law but, rather surprisingly, appearinghere as a conservation law. Notice that after restoring the label a our results for (J) etc.also apply in the case of multicomponent fields.

Poincare Invariance

We now require that a physical system possesses both space-time translation (2.18) and Lorentztransformation symmetry (2.41). The symmetry group that includes both transformations iscalled the Poincare group. Notice that for any Poincare-invariant theory the two chargeconservation equations (2.21) and (2.69) should hold. This is only possible if the energy-momentum tensorT is symmetric. Indeed,

0 =(J) =

xT xT=xT

+ Tx

xT

Tx

=T T =T T .

(2.74)

Since Maxwells theory is Poincare invariant, this general result tells us that the expressionof the energy-momentum tensor in (2.40) can be made symmetric without changing physics.The key to actually do it, lies in making use of the conservation law (2.21) in an appropriateway.

2.5 Problems

i) Suppose that a no further specified Lagrangian

L depends not only on and but

also on the second derivatives of the fields:7

L = L(, , ) . (2.75)

For the case that the variations vanish at the endpoints and that (1. . . N) =1. . . N() holds, derive the Euler-Lagrange EOMs for such a theory.

7For the sake of brevity, we have omitted the subscript a labelling the different fields.

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Apply your result to obtain the EOMs for the field with Lagrangian

L =12

(t) (x) +

6(x)

3 2

()2 . (2.76)

ii) Let us study the dynamics of acoustic waves in an elastic medium (e.g. air), as describedby the Lagrangian

L =12

y

t

2 1

2v2sound(y)

2 , (2.77)

with the density of the medium and vsound the speed of sound.

Find the Euler-Lagrange EOMs for the system and their solutions. What do they de-scribe? Calculate the HamiltonianH.

iii) Consider the Klein-Gordon Lagrangian (2.45). Derive the kinetic and potential energy(T and V with L = T V) as well as the Euler-Lagrange EOMs for the field . Writedown the energy-momentum tensor T and show that it indeed satisfies

T = 0.

Give the expressions for the conserved energyEand momentum Pi.

iv) Using (2.60) show that the terms A, (A

)2, and (A)(A) are Lorentz invariant.

What are the dimensions of these terms?

v) We saw that in the case of electrodynamics in vacuum using (2.20) leads to an energy-momentum tensor T that is not symmetric. To remedy that, one can add to T

a term of the form , where is antisymmetric in its first two indices, i.e.,

= .Show that such an object is automatically divergenceless, i.e., it obeys

= 0.

This feature implies that instead ofT

one can also use

=T + , (2.78)

without changing the physics, since has the same globally conserved energy andmomentum asT.

Show that this construction, with

=FA , (2.79)

leads to an energy-momentum tensor that is symmetric and yields the standard

formulas for the electromagnetic energy and momentum densities:

E=12

E2 + B2

, S=EB . (2.80)

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References

[1] Chapter 4 of L. D. Landau and E. M. Lifshitz, The Classical Theory of Fields, FourthEdition: Vol. 2 (Course of Theoretical Physics Series), Butterworth-Heinemann (1975),481 p.

[2] Chapter 5 of B. Thide, Electromagnetic Field Theory, revised and extended 2nd edi-tion, http://www.plasma.uu.se/CED/Book/index.html

21
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3 Klein-Gordon Theory

In QM, canonical quantization is a recipe that takes us from the Hamiltonian formalismof classical dynamics to the quantum theory. The recipe tells us to take the generalizedcoordinates qa and their conjugate momenta p

a = L/qa and promote them to operators.The Poisson bracket structure of classical mechanics descends to the structure of commutationrelations between operators, namely

[qa, qb] = [pa, pb] = 0 ,

[qa, pb] =ia

b ,(3.1)

where [a, b] =ab bais the usual commutator.If one wants to construct a QFT, one can proceed in a similar fashion. The idea is to

start with the classical field theory and then to quantize it, i.e., reinterpret the dynamicalvariables asoperators that obey canonical commutation relations,8

[a(x

), b(y

)] = [

a

(x

),

b

(y

)] = 0 ,[a(x),

b(y)] =i(3)(x y)ab .(3.2)

Herea(x) are field operators and the Kronecker delta in (3.1) has been replaced by a deltafunction since the momentum conjugates a(x) are densities. Notice that for now, we areworking in the Schrodinger picturewhich means that the operators a(x) and

a(x) do onlydepend on the spatial coordinates but not on time. The time dependence sits in the states| which obey the usual Schodinger equation

id

dt| =H| . (3.3)

While all this looks pretty much the same as good old QM there is an important difference.The wavefunction| in QFT, is a functional, i.e., a function of every possible configurationof the field a, and not a simple function.

9 So things are more complicated in QFT than inQM after all.

The HamiltonianH, being a function ofa and a, also becomes an operator in QFT. In

order to solve the theory, one task is to find the spectrum, i.e., the eigenvalues and eigenstatesofH. This is usually very difficult, since there is an infinite number of dofs within QFT, atleast one for each point x in space. However, for certain theories, called free theories, one canfind a way to write the dynamics such that each dof evolves independently from all the others.Free field theories typically have Lagrangians which are quadratic in the fields, so that theEOMs are linear.

8This procedure is sometimes referred to as second quantization. We will not use this terminology here.9In functional analysis, a functional is a map from a vector space to the field underlying the vector space,

which is usually the real numbers. In other words, it is a function that takes a vector as its argument or inputand returns a scalar. Commonly, the vector space is a space of functions, so the functional takes a functionas its argument, and so it is sometimes referred to as a function of a function. The use of functionals goesback to the calculus of variations where one searches for a function which minimizes a certain functional. Aparticularly important application in physics is to search for a state of a system which minimizes the energyfunctional.

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3.1 Klein-Gordon Field as Harmonic Oscillators

So far the discussion in this section was rather general. Let us be more specific and considerthe simplest relativistic free theory as a practical example. It is provided by the classicalKlein-Gordon equation (2.45). To exhibit the coordinates in which the dofs decouple from

each other, we only have to Fourier transform the field ,

(t,x) =

d3p

(2)3eipx (t,p) . (3.4)

In momentum space (2.45) simply reads2

t2+p2 + m2

(t,p) = 0 , (3.5)

which tells us that for each value ofp, the Fourier transform (t,p) solves the equation of aharmonic oscillatorwith frequency

p= |p|2 + m2 . (3.6)We see that the most general solution of the classical Klein-Gordon equation is a linear super-position of simple harmonic oscillators, each vibrating at a different frequency with a differentamplitude. In order to quantize the field , we must hence only quantize this infinite numberof harmonic oscillators (as Sidney Coleman once said [1]: The career of a young theoreticalphysicist consists of treating the harmonic oscillator in ever-increasing levels of abstraction.).Lets recall how to do it in QM.

Harmonic Oscillator in QM

Consider the QM Hamiltonian

H=1

2p2 +

1

22q2 , (3.7)

with the canonical commutation relations [q, p] = i. In order to find the spectrum of thesystem, we define annihilation and creation operators(also known as lowering and raising orladder operators)

a=

2q+

i2

p , a =

2q i

2p . (3.8)

Expressingqand p through a and a gives

q= 12

(a + a) , p= i2

(a a) . (3.9)

The commutator of the operators introduced in (3.8) is readily computed. One finds [a, a] = 1.Expressing the Hamiltonian (3.7) throughaanda gives

H=

2

aa + aa

=

aa +

1

2

. (3.10)

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It is also easy to show that the commutator ofH with a and a takes the form

[H, a] = a , [H, a] =a . (3.11)These relations imply that if | is an eigenstate ofHwith energyE,i.e.,H| =E|, thenwe can construct other eigenstates by acting with the operators a and a on|:

Ha| = (E ) a| , Ha| = (E+ ) a| , (3.12)This feature explains why a (a) is called annihilation (creation) operator. From the latterequation it is also clear that the spectrum of (3.7) has a ladder structure, . . . , E 2, E,E,E+, E+2 , . . .. If the energy is bounded from below, there must be aground state|0,which satisfiesa|0 = 0. This state has the ground state orzero-pointenergy H|0 =/2|0.Excited states|nare then created by the repeated action ofa,

(a)n |0 =

n (n 1) . . . 1 |n =

n! |n , (3.13)and satisfy

H|n

= N+ 12 |n = n +12 |n , (3.14)where N = aa is the number operatorwith N|n = n|n. The prefactor on the right-handside of (3.13) is needed to guarantee that the states|nare normalized to 1, i.e.,n|n = 1.

Quantization of Real Klein-Gordon Field

If we treat each Fourier mode of the fieldas an independent harmonic oscillator, we can applycanonical quantization to the real Klein-Gordon theory, and in this way find the spectrum ofthe corresponding Hamiltonian. In analogy to (3.9), we write and as a linear sum of aninfinite number of operators ap anda

p , labelled by the 3-momentum p,

(x) = d3p(2)3

12p

ap eipx + ap eipx ,(x) =

d3p

(2)3(i)

p2

ap e

ipx ap eipx

.

(3.15)

The commutation relations (3.2) become

[ap, aq] = [ap, a

q] = 0 ,

[ap, aq] = (2)

3 (3)(p q) .(3.16)

Let us assume that the latter equations hold, it then follows that

[(x), (y)] = d3p d3q(2)6

i2q

p

[ap, aq] eipxiqy + [ap, aq] eipx+iqy=

d3p d3q

(2)6i2

qp

(2)3 (3)(p q)

eipxiqy eipx+iqy

=

d3p

(2)3i2

eip(xy) eip(xy)

= i(3)(x y) ,

(3.17)

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where we have dropped terms [ap, aq] = [ap, a

q] = 0 from the very beginning. To show that

[(x), (y)] = [(x), (y)] = 0 is left as an exercise.In terms of the ladder operatorsapanda

pthe Hamiltonian of the real Klein-Gordon theory

takes the form

H=1

2 d3x 2 + ()2 + m22=

1

2

d3x d3p d3q

(2)6

pq

2

ap e

ipx ap eipx

aq eiqx aq eiqx

+

1

2

pq

ip ap e

ipx ip ap eipx iq aq eiqx iq aq eiqx

+ m2

2

pq

ap e

ipx + ap eipx

aq e

iqx + aq eiqx

=

1

4 d3p(2)3 1p(2p+p2 + m2)(apap+ apap) + (2p+p2 + m2)(apap+ apap) ,

(3.18)

where we have first used the expressions for and given in (3.15) and then integrated overd3x to get delta functions (3)(p q), which, in turn, allows us to perform the d3qintegral.Inserting finally the expression (3.6) for the frequency, the first term in (3.18) vanishes andwe are left with

H=1

2

d3p

(2)3 p

apa

p+ a

pap

=

d3p

(2)3 p

apap+

1

2[ap, a

p]

= d3p

(2)3

papap+

1

2

(2)3 (3)(0) .(3.19)

We see that the result contains a delta function, evaluated at zero where it has an infinitespike. This contribution arises from the infinite sum over all modes vibrating with the zero-point energy p/2. Moreover, the integral over p diverges at large momenta|p|. To betterunderstand what is going on let us have a look at the ground state |0 where the former infinityfirst becomes apparent.

3.2 Structure of Vacuum

As in the case of the harmonic oscillator in QM, we define the vacuum |0 through the condition

that it is annihilated by the action ofal lap ,ap|0 = 0 , p . (3.20)

With this definition the energy E0 of the vacuum comes entirely from the second term in thelast line of (3.19),

H|0 =E0 |0 =

d3p

(2)3p2

(2)3 (3)(0)

|0 = |0 . (3.21)

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In fact, the latter expression contains not only one but two infinities. The first arisesbecause space is infinitely large. Infinities of this kind are often referred to as infrared (IR)divergences. To isolate this infinity, we put the theory into a box with sides of length L andimpose periodic boundary conditions (BCs) on the field. Then, taking the limit L , weget

(2)3 (3)(0) = limL

L/2L/2

d3x eipxp=0

= limL

L/2L/2

d3x= V , (3.22)

whereV denotes the volume of the box. This result tells us that the delta function singularityarises because we try to compute the total energy E0 of the system rather than its energydensityE0. The energy density is simply calculated fromE0 by dividing through the volumeV. One finds

E0= E0V

=

d3p

(2)3p2

, (3.23)

which is still divergent and resembles the sum of zero-point energies for each harmonic oscil-lator. SinceE0 in the limit|p| , i.e., high frequencies (or short distances), thissingularity is anultraviolet (UV) divergence. This divergence arises because we want too much.We have assumed that our theory is valid to arbitrarily short distance scales, correspondingto arbitrarily high energies. Recalling the discussion of energy scales in Section1.2, this as-sumption is clearly absurd. The integral should be cut off at high momentum, reflecting thefact that our theory presumably breaks down at some point (most likely far below the GUTor Planck scale).

Fortunately, the infinite energy shift in (3.19) is harmless if we want to measure the energydifference of the energy eigenstates from the vacuum. We can therefore recalibrate ourenergy levels (by an infinite constant) removing from the Hamiltonian operator the energy ofthe vacuum,

: H: =H

E0 = H

0

|H

|0

. (3.24)

With this definition one has : H: |0 = 0. In fact, the difference between the latter Hamiltonianand the previous one is merely an ordering ambiguity in moving from the classical to thequantum theory. E.g., if we would have defined our Hamiltonian to take the form

H=1

2(q ip) (q+ ip), (3.25)

which is classically the same as our original definition (3.7), then after quantization instead of(3.10), we would have gotten

H= aa . (3.26)

This type of ordering ambiguity arises often in field theories. The method that we have

used above to deal with it is called normal ordering. In practice, normal ordering works byplacing all annihilation operatorsapin products of field operators to the right. Applied to theHamiltonian of the real Klein-Gordon theory this prescription leads to

: H: =

d3p

(2)3 p a

pap . (3.27)

In the remainder of this section, we will normal order all operators in this manner (droppingthe : : for simplicity).

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Cosmological Constant

Above we concluded that as long as we are interested in the differences between energy levelsthe infinite total energy E0 of the vacuum does not matter (which effectively means that E0has no effect on particle physics phenomenology). So is the value ofE0 unobservable then?

No, in fact, not at all, since gravity is supposed to see all energy densities. In particular, thesum of all the zero-point energies should contribute to Einsteins equations,

R R2

g+ g= 8GNT, (3.28)

in the form of a cosmological constant = E0/V. Here R is the Ricci curvature tensor, Rthe scalar curvature (for their definitions please consult a text on general relativity), g isthe metric tensor (not to be mixed up with the Minkowski metric ),GNdenotes Newtonsconstant, which we have already met in (1.8), and Tis the energy-momentum tensor in itssymmetric form. Unfortunately, I do not have time to explain (3.28) in detail. If you want tolearn more about Einsteins equation, I suggest that you attend Andrew Steanes course on

general relativity. In order to be able to follow this lecture, it is sufficient to know that theseequations contain a term proportional to E0/V.

An assortment of observations (cosmic microwave background, type-Ia supernovae, baryonacoustic oscillations,etc.) tells us that 74% of the energy density in the universe has the prop-erties of a cosmological constant. This constant energy density filling space homogeneouslyis one form of dark energy. Another possibility of dark energy would be a scalar field suchas quintessence, a dynamic quantity whose energy density can vary in space. The rest of thecomposition of todays cosmos is made up by dark matter, amounting to 22%, and visiblematter (atoms, etc.), giving the missing 4%. Dark matter is dark in the sense that it is in-ferred to exist from gravitational effects on visible matter and background radiation, but is

undetectable by emitted or scattered electromagnetic radiation. So in conclusion, fully 96%of the universe seems to be composed of stuff weve never seen directly on earth.But our lack of understanding does not end there. In the last subsection, we have argued

that integrating in (3.23) up to infinity is not the right thing to do, but that one shouldonly consider modes up to a certain UV cut-off UV, where one stops trusting the underlyingtheory. The resulting energy density E0then scales like 4UV. While it is not clear which precisevalue we should take for UV, let us be not very ambitious and take a value for this scale, upto which we truly believe that we understand the physics of fundamental interactions. Theelectron mass me = 511 keV could be such a choice. In consequence,

Epredicted0

(511 keV)4

6

1022 eV , (3.29)

where the superscript predicted should probably better read guessed. Glancing at Table1,we see that the observed value ofE0 is

Eobserved0 (103 eV)4 1012 eV , (3.30)

so it is clearly non-zero but unfortunately also roughly 34 orders of magnitude smaller thanour prediction. Notice that the choice UV=methat lead to (3.29) was, in fact, a conservative

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one, because other educated guesses such as UV=v, MGUT,etc. would have lead to a muchbigger disagreement of up to 120 orders of magnitude for the choice UV= MP.

From the point of view of QFT, the net cosmological constant, is the sum of a number ofapparently disparate contributions, including zero-point fluctuations of each field theory dofand potential energies from scalar fields, as well as a bare cosmological constant. There is no

obstacle to imagining that all of the large and apparently unrelated contributions add together,with different signs, to produce a net cosmological constant consistent with the limit (3.30),other than the fact that it seems ridiculous. We know of no special symmetry which couldenforce a vanishing vacuum energy while remaining consistent with the known laws of physics.This conundrum is the cosmological constant problem. While no widely accepted solution tothis problem exists, there are many proposed ones ranging from the anthropic principleto thestring-theory landscape. Dont bother if you have never even heard of any of them, it is notimportant at all for what follows.

Casimir Effect

Using the normal ordering prescription we can happily setE0 = 0, while chanting the mantrathat only energy differences can be measured. However, it should be possible to see that thevacuum energy is different if, for a reason, the fields vanish in some region of the space-volumeor if some frequencies p do not contribute to the vacuum energy. Such a set-up can berealized, by forcing the real Klein-Gordon field to satisfy appropriate BCs. Let us assume,that vanishes on the planes with x = 0 andx = L,

(0, y , z ) =(L,y,z) = 0 , (3.31)

The presence of these BCs affects the Fourier decomposition of the field and, in particular,leads to a quantization of the momentum of the field inside the planes (k

Z+),

p=

k

L, py, pz

. (3.32)

For simplicity let us consider a masslessreal scalar field. In this case the ground-state energyper unit area Sbetween the planes is given by the following expression

E0(L)

S =

k=1

dp2(2)2

1

2

k

L

2+p2 . (3.33)

Notice that we only integrate over the perpendicular directionsp

= (py, pz), since the mo-mentum px is discretized. Consequently, the volume integral has to be replaced by a surfaceintegral of the planes. In analogy to (3.22), this gives a factor S/(2)2 instead ofV /(2)3.

Let us see if we are able to calculate (3.33). We first switch to polar coordinates,

E0(L)

S =

k=1

0

dpp2

1

2

k

L

2+p2 . (3.34)

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As it stands this integral is divergent in the limit p . We can regulate this singularityin a number of different ways. One way to do it, is to introduce a UV cut-offa L, so thatmodes of momentum much bigger than a1 are removed. E.g., multiplying the integrand in(3.34) by the factor exp [a ((k/L)2 +p2)1/2] would do the job, since the resulting expressionhas the property that as a

0, one regains the full, infinite result (3.33). The drawback of

this method is that the new integral is quite difficult to perform (though doable), so lets seeif we find an easier way.

The trick is to consider (3.33) not in d = 4 dimensions, but to work in less dimensions,say, d = 4 2 with > 0. While this looks very weird at first sight, let me mention thatin general there exists a value of for which the integral is well-defined. We shall performour calculation for such a value, and then try to analytically continue the result to = 0. Ind= 4 2dimensions the integral (3.34) takes the form

E0(L)

S =

k=1

0

dpp12

2

1

2

k

L

2+p2 . (3.35)

To evaluate this expression, we first change variables p k/Ll. We then obtainE0(L)

S =

1

4

L

32 k=1

k32

0

dl l12

1 + l2

= 1

8

L

32(2 3)

0

dl2(l2)

1 + l2 ,

(3.36)

where we have identified the infinite sum with a Riemann zeta function, employingk=1

1

ka =(a) . (3.37)

Performing now the change of variables l2 x/(1 x), we arrive at 0

dl2(l2)

1 + l2=

10

dx x (1 x)5/2 =B(1 , 3/2) , (3.38)where in the last step we have used the definition of the Euler beta function,

B(a, b) =(a)(b)

(a + b) =

10

dx xa1 (1 x)b1 . (3.39)Putting everything together, the final result in d = 4 2dimensions reads

E0(L)

S =

1

8

L32

(2 3) B(1 , 3/2) . (3.40)

Amazingly,10 we can even take the limit 0. Using (a+ 1) = a (a) with (1) = 1 and(1/2) =

, and recalling that (3) = 1/120, we arrive at the finite expression

E0(L)

S =

2

1440L3. (3.41)

10Many subtleties have been swept under the carpet in this calculation. E.g., the dimensions of the ex-pressions in (3.35) to (3.40) are wrong by2. All cheats will become clear when the method of dimensionalregularization is properly introduced.

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This result implies that the vacuum energy depends on the distance between the two planes,on which vanishes. Can we realize this in an experiment?

Remember that the electromagnetic field is zero inside a conductor. If we place two un-charged conducting plates parallel to each other at a distance L, then we can reproduce theBCs of the set-up that we have just studied. While the quantization of the electromagnetic

field is more complicated than the real Klein-Gordon field, which we have used to model theeffect, this difference becomes (almost) immaterial as far as the vacuum energy is concerned.Our analysis, leads to an amazing prediction. Two electrically neutral metal plates attracteach other. This is known as the Casimir-Polder force, first predicted in 1948 [4]. Notice,that the energy of the vacuum gets smaller when the conducting plates are closer, as indicatedby the minus sign in (3.41). Therefore, there is an attractive force between them. This is aneffect that has by now been verified experimentally with great precision.11 In our example,the force per unit area (pressure or rather anti-pressure) between the two conductor plates isgiven by

F= 1S

E0(L)

L =

2

480L4. (3.42)

In fact, the true Casimir-Polder force is twice as large as the latter result, due to the twopolarization states of the photon.

3.3 Particle States

After the discussion of the properties of the vacuum, we can now turn to the excitations of. Its easy to verify (and therefore left as an exercise) that, in full analogy to (3.11), theHamiltonian and the ladder operators of the real Klein-Gordon theory obey the followingcommutation relations

[H, ap] = pap , [H, ap] =pap . (3.43)These relations imply that we can construct energy eigenstates by acting on the vacuum state|0 withap(remember that they also imply that ap |0 = 0,p). We define

|p =ap |0 . (3.44)This state has energy

H|p =Ep |p =p |p , (3.45)with p given in (3.6), which is nothing but the relativistic energy of a particle with 3-momentum p and mass m. We thus interpret the state|p as the momentum eigenstateof a single scalar particleof mass m.

Let us check this interpretation by studying the other quantum numbers of|p. We beginwith the total momentumP introduced in (2.24). Turning this expression into an operator,we arrive, after normal ordering, at

P =

d3x =

d3p

(2)3p apap . (3.46)

11The first experimental test of the Casimir-Polder force was conducted by Marcus Sparnaay in 1958, in adelicate and difficult experiment with parallel plates. Due to the large experimental errors, his results couldneither prove the theoretical prediction right nor wrong.

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Acting with Pon our state|p gives

P |p =

d3q

(2)3 q aqaqa

p|0 =

d3q

(2)3 q aq

(2)3 (3)(p q) + apaq

|0 =p |p , (3.47)

where we have employed the second line in (3.16) and used the fact that an annihilationoperator acting on the vacuum is zero. The latter result tells us that the state|p hasmomentum p. Another property of|p that we can study is its angular momentum. Againwe take the classical expression for the total angular momentum (2.67) and turn it into anoperator,

Ji =ijk

d3x (J0)jk , (3.48)

It is a good exercise to show that by acting with Ji on the one-particle state with zeromomentum one gets

Ji |p= 0 = 0 . (3.49)

This result tells us that the particle carries no internal angular momentum. In other words,quantizing the real Klein-Gordon field gives rise to a spin-zero particle akaa scalar.

Multiparticle States

Acting multiple times with the creation operators on the vacuum we can create multiparticlestates. We interpret the state

|p1,...,pn =ap1 . . . apn |0 , (3.50)

as ann-particle state. Since one has [api , apj

] = 0, the state (3.50) is symmetric under exchange

of any two particles. E.g.,

|p, q =apaq |0 =aqaq |0 = |q,p . (3.51)

This means that the particles corresponding to the real Klein-Gordon theory are bosons. Wesee that, as promised already in Section 1.1, the relationship between spin and statisticsis, in fact, a consequence of the QFT framework, following, in the case at hand, from thecommutation quantization conditions for boson fields (3.2).

The full Hilbert space of our theory is spanned by acting on the vacuum with all possiblecombinations of creation operators,

|0 , a

p |0 , a

pa

q |0 , a

pa

qa

r |0 , . . . . (3.52)This space is known as theFock spaceand is simply the sum of the n-particle Hilbert spaces,for alln 0. Like in QM, there is also an operator which counts the number of particles in agiven state in the Fock space. It is the number operator

N=

d3p

(2)3apap , (3.53)

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which satisfies N|p1, . . . ,pn = n |p1, . . . ,pn. Notice that the number operator commuteswith the Hamiltonian, i.e., [N, H] = 0, ensuring that particle number is conserved. Thismeans that we can place ourselves in the n-particle sector, and will remain there. This is aproperty of free theories, but will no longer be true when we consider interactions. Interactionscreate and destroy particles, taking us between the different sectors in the Fock space.

Operator-Valued Distributions

We have referred to the states|p as particles. Yet, this name is somewhat misleading,since these states are momentum eigenstates and therefore not localized in space. Recall thatin QM both the position and momentum eigenstates are not good elements of the Hilbertspace since they are not normalizable (they normalize to delta functions). Similarly, in QFTneither the operators (x), norapand a

pare good operators acting on the Fock space. This

is because these operators all produce states that are not normalizable:

0 |(x)(x)| 0 =(3)(0) ,

0

apa

p

0

= (2)3 (3)(0) . (3.54)

This feature implies that they are operator-valued distributionsand not functions. In the caseof(x) one has that although the field operator has a well-defined vacuum expectation value(VEV),0|(x)|0 = 0, the fluctuations0|(x)(x)|0 of the operator at a fixed point areinfinite. We can construct well-defined operators by smearing these distributions over space.E.g., we can create a wavepacket

| =

d3p

(2)3eipx (p) |p , (3.55)

which is partially localized in both position and momentum space. A typical state might bedescribed by the Gaussian (p) = exp [p2/(2m2)].

Relativistic Normalization

The vacuum|0 is normalized as0|0 = 1. The one-particle states|p =ap |0 then satisfyp|q = 0 apaq 0= 0 (2)3 (3)(p q) + aqap 0= (2)3 (3)(p q) , (3.56)

where we have made use of (3.16) and (3.20) to arrive at the final answer. Since the latterexpression depends on 3-momenta, an immediate question that arises is whether it is Lorentzinvariant. What could go wrong? Suppose we perform a Lorentz transformation

p (p) = p , (3.57)such that p p. In our QFT it would be preferable, if the statep changes under this Lorentztransformation as

|p |p =U() |p , (3.58)withU() being unitary,i.e.,U()U() =U()U() = 1. In such a case the normalizationof|p would remain unchanged

p|p p|p = p U()U()p= p|p . (3.59)32


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In order to find out whether or not the original and the Lorentz-transformed state,|p and|p, are related by an unitary transformation, we should look at an object which we knowis Lorentz invariant. One such object is theidentity operator(which is really the projectionoperator onto one-particle states). With the normalization (3.56) we know that it is given by

1 = d3p(2)3

|pp| . (3.60)

This operator is Lorentz invariant, but it consists of two terms: the measure

d3p and theprojector |pp|. Are these two objects Lorentz invariant by themselves? In fact, they are not.

In order to prove this statement, we start with the measure

d4pwhich is obviously Lorentzinvariant. The relativistic dispersion relation for a massive particle,i.e., p2 =m2, and hence

p20 = E2p= p

2 +m2 is also Lorentz invariant. Solving forp0, there are two branches of solutions,namely p0 =Ep. But the choice of branch is another Lorentz-invariant concept. Puttingeverything together tells us that

d4p (p20 p2 m2) p0>0= d3p2p0 p0=Ep = d3

p2Ep

, (3.61)

is Lorentz invariant. From the latter result we can figure out everything else. E.g., theLorentz-invariant delta function for 3-momenta is

2Ep(3)(p q) , (3.62)

since d3p

2Ep2Ep

(3)(p q) = 1 . (3.63)

This finally tells us that the relativistically normalizedmomentum eigenstates are given by12

|p =

2Ep |p =

2Epap |0 , (3.64)

and satisfyp|q = (2)3 2Ep (3)(p q) . (3.65)

We can also express the identity operator in terms of the|p states. One has

1 =

d3p

(2)31

2Ep|pp| . (3.66)

We remark that some texts on QFT also define relativistically normalized annihilation (cre-ation) operators bya(p) =

2Epap

a(p) =

2Epap

. In order to avoid (further) confusion,

we wont make use of this notation here.

12Our notation is rather subtle here, since the relativistically normalized momentum states|p differ from|pjust by the fact that they are not set in boldface type.

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3.4 Two Real Klein-Gordon Fields

Our task is to describe all known particles and their interactions. It is then interesting tostudy the quantization of a system with more than one field. In order to keep things simple,let us try to describe a system of two real Klein-Gordon fields 1,2 which differ only in their

mass parameters (m1=m2),L =

i=1,2

1

2(i)

2 12

m2i2i

. (3.67)

This Lagrangian leads to two independent Klein-Gordon equations,

(2 + m2i ) i= 0 . (3.68)

The Hamiltonian, the total momentum, and the number operator of the system is given by

H=H1+ H2 , P =P1+ P2 , N=N1+ N2 , (3.69)

where

Hi =

d3p

(2)3 i,p a

i,pai,p , Pi =

d3p

(2)3p ai,pai,p , Ni =

d3p

(2)3ai,pai,p , (3.70)

withi,p= (p2 + m2i )

1/2. It should be clear, that we can construct particle states in the same

fashion as we did with the Lagrangian of just a single real Klein-Gordon field. Products ofa1,poperators acting on|0 create relativistic particles with mass m1, while a2,p operators createparticles with massm2. E.g., the states

|S1 =a

1,p |0 , |S2 =a

2,p |0 , (3.71)satisfy

H|Si =i,p|Si , P|Si =p |Si , N|Si = 1|Si . (3.72)These relations tell us that the states|S1,2 are degenerate in the sense that they are single-particle states with the same momentump. However, they can be distinguished by measuringthe energy of the particles as long as the masses m1,2 are different (which we have assumedfor the time being).

Equal-Mass Case

Admittedly the case of two real Klein-Gordon fields with different masses m1,2is pretty boring.Things get a little bit more interesting, if we consider the special case m1 =m2 =m. Why?Because in this case the system possesses an additional rotation symmetryin the space of fields1,2. According to Noethers theorem this should lead to a new conserved charge. In order tobe able to identify the additional charge, we first write the Lagrangian (3.67) in a form thatexhibits the symmetry

L =12

(T)() 1

2m2 T . (3.73)

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Here we have introduced the field vector = (1, 2)T.

Obviously, the latter Lagrangian is invariant under the orthogonal transformations (O(2)transformations or two-dimensional rotations),

=R , (3.74)

withRT =R1. To calculate the conserved current, we again consider infinitesimal symmetrytransformations (i, j = 1, 2)

Rij =ij+ ij+ O(2) . (3.75)The orthogonality of the matrix R,

ij+ ji = RTij =R

1ij =ij ij, (3.76)

tells us that the matrix is antisymmetric. The infinitesimal transformation of the field 1under (3.74) is

1 1 = R1ii = (1i+ 1i)i = 1+ 11 1+ 12 2 = 1+ 12 2 , (3.77)which tells us that the variation of1 is

1= 12 2 . (3.78)

An analog calculation gives2= 21 1 = 12 1 . (3.79)

Knowing the variations1,2 of the fields, the conserved current corresponding to (3.74) isreadily written down,

J = L(i)

i= 12 (1) 2 (2) 1 , (3.80)so the conserved charge is

Q=

d3x

01

2

02

1

. (3.81)

Substituting in the above expression the physical solutions (3.15) for the fields 1,2, andperforming the integration over the space coordinates, one obtains (the actual computation ispart of an exercise)13

Q= i d3p(2)3 a1,pa2,p a2,pa1,p , (3.82)which is anhermitianoperator, i.e., it satisfies Q =Q. There is an ambiguity worth noting,when applying Noethers theorem to find the conserved charge under the transformation (3.74).Obviously, if Q is conserved, then so is every other operator c1Q+ c2 with c1,2 constantnumbers. The expression for Q in (3.82) is therefore unique up to a multiplicative and an

13This expression has not be normal ordered.

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additive constant. The ambiguity on the additive constant is removed when we remove thecontribution of the vacuum to the charge of particle states (as we have done for the energy).The normal-ordered charge operator

: Q : = Q 0|Q|0 , (3.83)

is ambiguous only up to a multiplicative factor, which essentially denotes the units in whichwe measure the charge of a state. Notice that we have already used this ambiguity in (3.81)and simply ignored the factor12. In the following, we will use the normalization (3.83) ofQ,dropping as before the : : to avoid unnecessary clutter.

So far so good. Next we would like to determine the spectrum ofQ. This is most easilydone using the technique of ladder operators. We first define the following linear combinations,

a,p= 1

2(a1,p ia2,p), (3.84)

of annihilation operators (an analog definition holds for the hermitian conjugate operators).It is left as a homework problem to show that these new operators satisfy the following

commutation relations

[Q, a,p] = a,p , [Q, a,p] = a,p . (3.85)The latter relations imply that we can obtain states with charge q 1 from a state|S ofcharge q, i.e., Q |S =q|S, by the action ofa,p ,

Q

a,p |S

= (q 1)

a,p |S

. (3.86)

In other words the operators a,pare ladder operators with respect to Q. Sincea,pare linear

combinations ofa1,pand a2,p , which are ladder operators for the Hamiltonian Hand the total

momentum operator P, so are a

,p.

To find now all the common eigenstates of the charge operator Q, it is sufficient to startfrom a single common eigenstate and then to act with a,p on this state. It is not surprisingthat the vacuum|0 is also an eigenstate ofQ, namely the one with zero charge14

Q |0 = 0 |0 = 0 . (3.87)Repeated application of the ladder operators,

|Sn =ni=1

a,pi |0 , (3.88)

then createsn-particle states with positive (

|S+n

) and negative (

|Sn

) charge. Consequently,

one has

H|Sn =

ni=1

i,pi

|Sn , P |Sn =

ni=1

pi

|Sn ,

N|Sn =n |Sn , Q |Sn = n |Sn ,(3.89)

14Notice that the normal ordering (3.83) ofQ plays an essential role here.

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The main results of this subsection can be summarized as follows. The mass degeneracy ofthe Klein-Gordon fields1,2results in a newO(2) symmetry of the Lagrangian. This gives riseto a new conserved quantity, the charge Q. A particle state is then characterized by its mass(or equivalently its energy), its momentum, and its charge, which can be either positive ornegative. States with the same energy and momentum, but opposite charge, can be interpreted

as particles and antiparticles. Notice that for a single real Klein-Gordon field there is only asingle type of particle, since a real scalar particle is its own antiparticle.

3.5 Complex Klein-Gordon Field

We can gain further insight into the theory by rewriting the Lagrangian (3.73) a little bit,

L = ()() m2 , (3.90)where

= 1

2(1+ i2) , (3.91)

denotes the complex Klein-Gordon field. We could now compute the Hamiltonian and mo-mentum operators directly in terms of and , arriving at the same expressions as in therepresentation with two real fields (if you dont believe me you are free to check this yourself).In order to compute the charge Q, we then need to identify the internal symmetryof the newLagrangian. In fact, it is easy to see that (3.90) is invariant under a field phase-redefinitionakaaglobalU(1) transformation,

=ei , () =ei . (3.92)Notice that this transformation is the equivalent of the rotation symmetry transformation(3.74) that we have found earlier, in the real field representation. We verify this by using theexplicit form of the matrix R in terms of sine and cosine of the rotation angle ,1

2

cos sin sin cos 12 , = 1i2 cos isin isin cos 1i2 ,= 1+ i2 ei (1+ i2), = ei .

(3.93)

So why should we bother about the complex Klein-Gordon Lagrangian if (3.73) and (3.90)are equivalent? The reason is that the complex field representation is more suggestive to thefact that we have both particle and antiparticle states. To s

Documents

Introduction to quantum field theory, based on the book by Peskin and Schroeder