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Dedication Page

To my wife Marta and my daughters Sara and Chiara withlove.

v


vi Introduction to Quantum Field Theory


Preface

This book originates from an introductory course in quantum field theorythat I have given for many years at the University of Florence. At thattime my lecture notes were written in Italian. Then, in the academic year1997/98, I was on leave at the University of Geneva where I gave the samecourse. In that occasion I decided to translate the material of my notesin English. These notes in English form the core of the present book.The course was intended to be an introduction to quantum field theoryhaving as a pre-requisite a standard course in non-relativistic quantummechanics. The language that I have chosen for this course is the one ofquantum operators avoiding any recourse to the path integral formalism.My personal feeling is that the path integral approach is a fundamentalone, both in quantum mechanics and even more in quantum field theory.However, I also think that this approach is better appreciated after beingexposed to the operator formalism.

Today quantum field theory is one of the most important tools inphysics. It represents a unified way of approaching many different problemsin different areas of physics, as elementary particles, statistical mechanics,critical phenomena, cosmology, etc. For this reason I have tried to keep thebook at the level of a non-sophisticated approach pointing more to clarifythe relevant aspects.

Entering in some detail, the books begins with a discussion of the quan-tization of the one-dimensional string. By considering the string as thecontinuum limit of a discrete chain of harmonic oscillators, it is possibleto use ordinary quantum mechanics and show the peculiarities of takingthe continuum limit. Then, the book enters in the hearth of quantum fieldtheory discussing the free scalar field together with its locality properties.After the scalar field there is a long introduction to the Dirac equation and

vii


viii Introduction to Quantum Field Theory

its properties ending with the quantization of the Dirac field in terms ofanticommutators. The part of the book dedicated to the free fields endswith the quantization of the electromagnetic field. For completeness a briefdiscussion of the massive vector fields is presented. With respect to myoriginal lecture notes I have inserted at the beginning of the book a presen-tation of the Lorentz group and of his spinor representations, together witha very brief discussion of the Poincare group. Although the derivation ofthe Dirac equation is done following the traditional way, the introductionof the spinor representations of the Lorentz group offers the possibility ofobtaining the same results in a much simpler way, both for the massive andthe massless case. In Chapter 6 the reader will find an extensive discussionof the symmetry properties in field theory, ranging from exact to sponta-neously broken symmetries. The Goldstone theorem is discussed and, afterintroducing the local (or gauge) symmetry for the electromagnetic field, Ipresent the generalization to non-abelian symmetries. The Chapter endswith the Higgs mechanism, the key to the present understanding of weakinteractions. This Chapter should be useful also to students interested incritical phenomena. The next two Chapters are dedicated to perturbationtheory and its applications to the scattering of elementary particles and tothe mechanism of emission and absorption of the electromagnetic radiationby an atomic system. In the last Chapter the reader can find a discussionof the infinities in quantum field theory and how to take care of them. Bytaking spinor electrodynamics as an example I give a detailed discussion ofregularization and renormalization processes at one-loop level.

I would like to thank particularly, Daniele Dominici, Giulio Pettini andMatthias Steinhauser for a careful reading of the manuscript and for severalsuggestions.

A particular thank goes to my wife Marta, for her patience and love inallowing me to spend many weekends to write this book.


Contents

Preface vii

1. Introduction 1

1.1 Notation and units . . . . . . . . . . . . . . . . . . . . . . 11.2 Major steps in quantum field theory . . . . . . . . . . . . 41.3 The Lorentz group . . . . . . . . . . . . . . . . . . . . . . 61.4 Spinor representations of the Lorentz group . . . . . . . . 121.5 The Poincare group . . . . . . . . . . . . . . . . . . . . . 151.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

2. Lagrangian formalism for continuum systems and quantization 17

2.1 Many degrees of freedom . . . . . . . . . . . . . . . . . . . 172.2 Linear atomic string . . . . . . . . . . . . . . . . . . . . . 182.3 String quantization . . . . . . . . . . . . . . . . . . . . . . 242.4 The canonical formalism for continuum systems . . . . . . 282.5 The canonical quantization of a continuum system . . . . 322.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

3. The Klein-Gordon field 39

3.1 The problems in relativistic quantum mechanics . . . . . 393.2 Quantization of the Klein-Gordon field . . . . . . . . . . . 423.3 Noether’s theorem for relativistic fields . . . . . . . . . . . 483.4 Energy and momentum of the Klein-Gordon field . . . . . 523.5 Locality and causality in field theory . . . . . . . . . . . . 563.6 The charged scalar field . . . . . . . . . . . . . . . . . . . 613.7 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 66

ix


x Introduction to Quantum Field Theory

4. The Dirac field 69

4.1 The Dirac equation . . . . . . . . . . . . . . . . . . . . . . 694.2 Covariance properties of the Dirac equation . . . . . . . . 724.3 The Dirac equation and the Lorentz group . . . . . . . . . 774.4 Free particle solutions of the Dirac equation . . . . . . . . 794.5 Wave packets and negative energy solutions . . . . . . . . 864.6 Electromagnetic interaction of a relativistic point-like par-

ticle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 884.7 Nonrelativistic limit of the Dirac equation . . . . . . . . . 944.8 Charge conjugation, time reversal and PCT transformations 974.9 Dirac field quantization . . . . . . . . . . . . . . . . . . . 1044.10 Massless spin 1/2 particles . . . . . . . . . . . . . . . . . . 1124.11 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 114

5. Vector fields 117

5.1 The electromagnetic field . . . . . . . . . . . . . . . . . . 1175.2 Quantization of the electromagnetic field . . . . . . . . . . 1215.3 Massive vector fields . . . . . . . . . . . . . . . . . . . . . 1295.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 131

6. Symmetries in field theories 133

6.1 The linear σ-model . . . . . . . . . . . . . . . . . . . . . . 1336.2 Spontaneous symmetry breaking . . . . . . . . . . . . . . 1406.3 The Goldstone theorem . . . . . . . . . . . . . . . . . . . 1446.4 QED as a gauge theory . . . . . . . . . . . . . . . . . . . 1476.5 Non-abelian gauge theories . . . . . . . . . . . . . . . . . 1496.6 The Higgs mechanism . . . . . . . . . . . . . . . . . . . . 1546.7 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 160

7. Time ordered products 163

7.1 Time ordered products and propagators . . . . . . . . . . 1637.2 A physical application of the propagators . . . . . . . . . 1727.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 175

8. Perturbation theory 177

8.1 The electromagnetic interaction . . . . . . . . . . . . . . . 1778.2 The scattering matrix . . . . . . . . . . . . . . . . . . . . 1808.3 Wick’s theorem . . . . . . . . . . . . . . . . . . . . . . . . 188


Contents xi

8.4 Evaluation of the S matrix at second order in QED . . . . 1918.5 Feynman diagrams in momentum space . . . . . . . . . . 2008.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 207

9. Applications 209

9.1 The cross-section . . . . . . . . . . . . . . . . . . . . . . . 2099.2 The scattering e+e− → µ+µ− . . . . . . . . . . . . . . . . 2129.3 Coulomb scattering . . . . . . . . . . . . . . . . . . . . . . 2189.4 Application to atomic systems . . . . . . . . . . . . . . . 2219.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 224

10. One-loop renormalization 227

10.1 Divergences of the Feynman integrals . . . . . . . . . . . . 22710.2 Dimensional regularization of the Feynman integrals . . . 23610.3 Integration in arbitrary dimensions . . . . . . . . . . . . . 23810.4 One loop regularization of QED . . . . . . . . . . . . . . . 24110.5 One loop renormalization . . . . . . . . . . . . . . . . . . 25010.6 Lamb shift and g − 2 . . . . . . . . . . . . . . . . . . . . . 25610.7 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 261

Bibliography 263

Index 267


Chapter 1

Introduction

1.1 Notation and units

In relativistic quantum theories the two fundamental constants, the velocityof light, c, and the Planck constant, ~, appear almost everywhere. There-fore it is convenient to choose a unit system (natural units) where theirnumerical value is one

c = ~ = 1. (1.1)

Our conventions are as follows: the Minkowski space-time metric tensorgµν , µ, ν = 0, 1, 2, 3, is diagonal with eigenvalues (+1,−1,−1,−1). In gen-eral we say that a metric in m + n dimensions has a signature (m,n) if ithas m negative and n positive eigenvalues. Therefore gµν has a signature(3, 1). The position and momentum four-vectors are given by

xµ = (t, ~x), pµ = (E, ~p), µ = 0, 1, 2, 3, (1.2)

where ~x and ~p are the three-dimensional position and momentum. Wewill adopt the convention of summing over repeated indices unless statedotherwise. Then the scalar product between two four-vectors can be writtenas

a · b = aµbνgµν = aµbµ = aµbµ = a0b0 − ~a ·~b, (1.3)

where the indices have been lowered using the metric tensor

aµ = gµνaν (1.4)

and can be raised by using the inverse metric tensor gµν ≡ gµν . The four-gradient is defined as

∂µ =∂

∂xµ=(∂

∂t,∂

∂~x

)=(∂t, ~∇

). (1.5)

1


2 Introduction to Quantum Field Theory

The four-momentum operator in position space is

pµ → i∂

∂xµ= (i∂t,−i~∇). (1.6)

The following relations will be useful

p2 = pµpµ → − ∂

∂xµ∂

∂xµ≡ −, (1.7)

x · p = Et− ~p · ~x. (1.8)

The Ricci tensor in 4-dimensions, εµνρσ is completely antisymmetric andsuch that

ε0123 = +1. (1.9)

For the electromagnetism we will use the Heaviside-Lorentz system,where the vacuum dielectric constant is fixed to one

ε0 = 1. (1.10)

From the relation ε0µ0 = 1/c2 it follows

µ0 = 1. (1.11)

In these units the Coulomb force is given by

|~F | = e1e2

4π1

|~x1 − ~x2|2(1.12)

and there are no visible constants in the Maxwell equations. For instance,the Gauss law is

~∇ · ~E = ρ, (1.13)

where ρ is the charge density. The dimensionless fine structure constant

α =e2

4πε0~c(1.14)

is given by

α =e2

4π. (1.15)

Any physical quantity can be expressed by using as fundamental unitenergy, mass, length or time in an equivalent fashion. In fact, from ourchoice the following equivalence relations follow

ct ≈ ` =⇒ time ≈ length,

E ≈ mc2 =⇒ energy ≈ mass,

E ≈ pv =⇒ energy ≈ momentum,

Et ≈ /h =⇒ energy ≈ (time)−1 ≈ (length)−1. (1.16)


Introduction 3

In practice, it is enough to notice that the product c~ has dimensions [E ·`].Therefore

c~ = 3 · 108 mt · sec−1 · 1.05 · 10−34 J · sec = 3.15 · 10−26 J ·mt. (1.17)

Recalling that

1 eV = e · 1 = 1.602 · 10−19 J, (1.18)

where e is the electric charge of the proton expressed in Coulomb, it follows

c~ =3.15 · 10−26

1.6 · 10−13MeV ·mt = 197 MeV · fm, (1.19)

where 1 fm = 10−13 cm. From which

1 MeV−1 = 197 fm. (1.20)

Using this relation we can easily convert a quantity given in MeV (thetypical unit used in elementary particle physics) in fermi. For instance,since the elementary particle masses are usually given in MeV, the Comptonwave length of an electron is simply given by

λeCompton =1me≈ 1

0.5 MeV≈ 200 MeV · fermi

0.5 MeV≈ 400 fermi. (1.21)

Therefore the approximate relation to keep in mind is 1 ≈ 200 MeV · fermi.Furthermore, using

c = 3 · 1023 fermi · sec−1, (1.22)

we get

1 fermi = 3.3 · 10−24 sec (1.23)

and

1 MeV−1 = 6.58 · 10−22 sec. (1.24)

Also, using

1 barn = 10−24 cm2, (1.25)

it follows from (1.20)

1 GeV−2 = 0.389 mbarn. (1.26)



1.2 Major steps in quantum field theory

For a more complete history of quantum field theory see, for example[Pais (1986)] and [Weinberg (1995a)].

1924

• Bose and Einstein introduce a new statistics for light-quanta(photons) [Bose (1924); Einstein (1924)].

1925

• January - Pauli formulates the exclusion principle [Pauli(1925)].

• July - Heisenberg’s first paper on quantum mechanics (matrixmechanics) [Heisenberg (1925)].

• September - Born and Jordan extend Heisenberg’s formulationof quantum mechanics to electrodynamics [Born and Jordan(1925)]. A more complete treatment was given in the famousthree men’s paper by Born, Heisenberg and Jordan [Born et al.(1925)].

1926

• January - Schrodinger’s formulation of the wave equation[Schroedinger (1926a)].

• February - Fermi introduces a new statistics, known today asthe Fermi-Dirac statistics [Fermi (1926)].

• August - Dirac relates statistics and symmetry properties ofthe wave function, and shows that the quantized electromag-netic field is equivalent to a set of harmonic oscillators satis-fying the Bose-Einstein statistics [Dirac (1926)].

1927

• March - Davisson and Germer detect the electron diffractionby a crystal [Davisson and Germer (1927)].

• October - Jordan and Klein show that the electromagneticfield satisfies commutation rules [Jordan and Klein (1927)].

1928

• January - Dirac’s generalization of quantum mechanics to therelativistic case (Dirac equation) [Dirac (1928a,b)].


Introduction 5

• January - Jordan and Wigner introduce anticommuting fieldsfor describing particles satisfying Fermi-Dirac statistics [Jor-dan and Wigner (1928)].

1929

• January - Pauli and Heisenberg develop the analog for fields ofLagrangian and Hamiltonian methods of analytical mechanics[Heisenberg and Pauli (1929, 1930)].

• March - Weyl formulates gauge invariance and its relation tocharge conservation [Weyl (1929a)].

• Klein and Nishina complete the theory of the scatteringCompton based on the Dirac equation [Klein and Nishina(1929)].

1931

• Dirac introduces the idea of anti-electron (positron) to in-terpret the negative energy solutions of his equation [Dirac(1931)].

1932

• Anderson detects the positron [Anderson (1932)].

1934

• Dirac and Heisenberg evaluate the vacuum polarization of thephoton.

• First battle with infinities in quantum field theory [Dirac(1934); Heisenberg (1934)].

1936

• Serber introduces the concept of renormalized charge [Serber(1936)].

1946

• Tomonaga begins the program of renormalization but, dueto the war, his work is completely ignored by the westernphysicists [Tomonaga (1946)] (see also the next papers [Kobaet al. (1947)]).

1947

• Lamb and Retherford discover the ”Lamb shift” [Lamb andRetherford (1947)]



• Bethe performs the first calculation of the Lamb shift [Bethe(1947)].

1948

• Schwinger evaluates the corrections to the g−2 of the electron[Schwinger (1948a)].

• The renormalization program starts in the west [Schwinger(1948b, 1949a,b, 1951a,b,c,d)], [Kanesawa and Tomonaga(1948a,b); Tomonaga (1948); Ito et al. (1948); Koba andTomonaga (1948)], [Feynman (1948a,b,c, 1949a,b, 1950)],[Dyson (1949a,b)].

1.3 The Lorentz group

Since the main subject of this book is quantum relativistic field theory wewill provide a brief introduction to some of the properties of the Lorentzgroup.

The Lorentz transformations are defined as the ones leaving invariantthe distance between two points infinitesimally close in Minkowski spacedxµ = (dx0, d~x) = (dt, d~x)

ds2 = dxµdxνgµν = dxµdxµ. (1.27)

Then, a Lorentz transformation

dx′µ = Λµνdx

ν (1.28)

is such that

dx′µdx′µ = dxµdxµ, (1.29)

that is

ΛµρgµνΛνσ = gρσ. (1.30)

By lowering the upper index of Λµν

Λµν = gµρΛρν , (1.31)

we can write (1.30) as

ΛµρgµνΛνσ = gρσ, (1.32)

or, in matrix terms

ΛT gΛ = g, (1.33)


Introduction 7

where we have introduced the matrices Λ and g with matrix elements Λµνand gµν = gµν respectively. These matrices are orthogonal with respect toa metric with signature (3, 1). They form a group called O(3, 1). From theprevious relation it follows

det|Λ| = ±1. (1.34)

Lorentz transformation with determinant equal to +1 are said to beproper1, whereas the ones with determinant equal to −1 are said improper.From eq. (1.32) we have

1 = g00 = Λ200 −

3∑1

Λ2i0, (1.35)

implying

Λ00 ≥ 1, or Λ00 ≤ −1. (1.36)

From this it follows that the Lorentz group branches in four parts, withonly one, L↑+, connected to the identity.

Table 1.1 The four different parts of the Lorentz groupSymbol Λ00 det|Λ| Type of transformation

L↑+ ≥ +1 +1 proper orthochronous

L↓+ ≤ −1 +1 proper nonorthochronous

L↑− ≥ +1 −1 improper orthochronous

L↓− ≤ −1 −1 improper nonorthochronous

The other parts can be reached, starting from the identity, throughdiscrete transformations, namely parity, P and time-reversal, T (see Table1.1) as shown in Fig. 1.1

P =

1−1−1−1

, T =

−1

11

1

. (1.37)

We will denote by g(Λ) the abstract group element associated to thetransformation Λ. It follows from eq. (1.28) that the group compositionlaw for two transformations Λ1 and Λ2 is

g(Λ1)g(Λ2) = g(Λ1Λ2). (1.38)1These matrices form a subgroup of O(3, 1) called SO(3, 1) because it consists of special

matrices with unit determinant.



T

T

-L

L+

L-

L+

P P

PT

PT

Fig. 1.1 The different parts of the Lorentz group as connected via parity, P , and time-reversal, T .

It is worth to notice that due to the condition (1.33) a generic Lorentztransformation depends on six parameters, corresponding to three rotationsand three special Lorentz transformations along the three axes.

Let us now study infinitesimal Lorentz transformations. Since they aredefined as the transformations leaving invariant the norm of a four-vector

x2 = x′2, (1.39)

for an infinitesimal transformation

x′ = x+ δx, (1.40)

we get

x2 ≈ x2 + 2x · δx =⇒ x · δx = 0. (1.41)

Since Lorentz transformations are linear

x′µ = Λµνxν ≈ xµ + εµνxν , (1.42)

we get

x · δx = 0 =⇒ xµεµνxν = 0. (1.43)

The most general solution for the parameters εµν of the transformation isthat they form a second order antisymmetric tensor

εµν = −ενµ. (1.44)

We recover, in this way, the result that the number of independent param-eters characterizing a Lorentz transformation is six.


Introduction 9

From these considerations it follows that the transformations connectedto the identity can be written as

g(Λ) = e−i2ωµνJ

µν

, (1.45)

with ωµν = −ωνµ the parameters of the transformation (Λ ≈ 1 + ω) andJµν = −Jνµ the infinitesimal generators. The latter satisfy the followingcommutation relations

[Jµν , Jρσ] = i[Jνρgµσ − Jµρgνσ + Jµσgνρ − Jνσgµρ]. (1.46)

The generators of rotations and boosts can be expressed in terms of Jµν as

Ji =12εijkJjk, Ki = Ji0, (1.47)

with commutation relations

[Ji, Jj ] = −[Ki,Kj ] = iεijkJk,

[Ji,Kj ] = iεijkKk. (1.48)

A crucial observation in the study of the Lorentz group is that it is homo-morphic to the group of complex 2× 2 matrices with determinant equal toone, the group SL(2, C)2. To prove this statement let us introduce

σµ = (1, ~τ), σµ = (1,−~τ), (1.49)

with ~τ the Pauli matrices and 1 the identity matrix in the 2-dimensionalcomplex space. Then we define

x = xµσµ = x0 + ~x · ~τ =(x0 + x3 x1 − ix2

x1 + ix2 x0 − x3

), (1.50)

with xµ a four-vector. Let us notice that

Tr(x) = 2x0, det|x| = xµxµ. (1.51)

On this space we can define the following linear transformations

x′ = UxU†, (1.52)

with det|U | = 1. Clearly this leaves invariant the determinant of x andtherefore the norm of the four-vector xµ. Furthermore x′ is hermitian as x.That is to say, we are doing a linear transformation on the real four-vectorxµ preserving its norm and its reality properties. In other words the lineartransformation induced by U is nothing but a Lorentz transformation

x′µ = Λµνx. (1.53)

2Here S stay for special, that is determinant equal to one, L for linear, 2 for 2 × 2

matrices and C for complex valued.



From this it follows

x′ = x′µσµ = Λµνx

ν σµ = Uxν σνU†. (1.54)

Using

Tr[σµσν ] = 2gµν , (1.55)

we find

Λµν =12

Tr[σµUσνU†]. (1.56)

This relation shows that there is a 2 to 1 correspondence between SL(2, C)and O(3, 1), since both U and −U correspond to the same Lorentz trans-formation, Λ. It is not difficult to evaluate the matrices U giving rise torotations around the axis i, and boosts (special Lorentz transformations)along the direction i

Ui(α) = eiαiJi = cos(αi/2) + iτi sin(αi/2),

Ui(χ) = eiχiKi = cosh(χi/2)− τi sinh(χi/2). (1.57)

In particular, we see that the generators of rotations and boosts, in this2× 2 representation, are respectively

Ji =τi2, Ki = i

τi2. (1.58)

Notice that the Ji’s are hermitian, whereas the Ki’s are antihermitian.Therefore this representation is not unitary. The same happens3 for allthe finite-dimensional representations of SL(2, C). In fact, due to the noncompact nature of the Lorentz group, it is possible to show that the uni-tary representations are infinite-dimensional. Let us also observe that therelation between SL(2, C) and O(3, 1) is the same existing between SU(2)and O(3). In fact, if we restrict the SL(2, C) transformations to be unitary,we see from the first of eqs. (1.51) that they do not change the fourth com-ponent of the four-vector and therefore they are pure rotations. It followsfrom these considerations that instead of studying the representations ofO(3, 1) we can study the representations of SL(2, C). However, the repre-sentations of O(3, 1), or tensor representations, will be characterized by thecondition T (U) = T (−U), where T is the representation and U ∈ SL(2, C).

In the case of finite-dimensional representations one can define two her-mitian combinations of the generators, namely

~J (1) =12

( ~J − i ~K), ~J (2) =12

( ~J + i ~K), (1.59)

3In particular the hermiticity of ~J and the anti-hermiticity of ~K.


Introduction 11

with commutation rules[J (1)i , J

(1)j ] = iεijkJ

(1)k , [J (2)

i , J(2)j ] = iεijkJ

(2)k

[J (1)i , J

(2)j ] = 0. (1.60)

This is the algebra of SU(2)⊗SU(2). Therefore a generic finite-dimensionalirreducible representation of the Lorentz group (or SL(2, C)) is character-ized by assigning the pair (s1, s2) with s1 and s2 the spin corresponding tothe two groups SU(2). The lowest dimensional representations are

ζα ∈(

12, 0), ζα ∈

(0,

12

). (1.61)

The elements of these representations are called undotted and dottedspinors respectively. The following relation is also important

(s1, s2) = (s1, 0)⊗ (0, s2). (1.62)It should be noticed that the physical angular momentum ~J is the sum ofthe two SU(2) generators. This means that the physical spin, j, containedin a given representation (s1, s2) is obtained by summing the two angularmomenta, and it will get one of the following values

|s1 − s2| ≤ j ≤ (s1 + s2). (1.63)For instance, the representation (1/2, 1/2) has a spin content 0 and 1 andit corresponds to a four-vector. Another relevant observation is that theinfinitesimal generators ~J and ~K transform under parity as a pseudo-vectorand a proper vector respectively

P ~JP−1 = ~J, P ~KP−1 = − ~K, (1.64)showing that parity interchanges the two SU(2). Said in a different way,parity transforms undotted spinors in dotted ones and vice versa. Fromthis it follows that a representation invariant under parity should containpairs of representations connected by parity. The simplest example is therepresentation (

12, 0)⊕(

0,12

). (1.65)

This is a four-dimensional representation of the Lorentz group and we shallsee that it corresponds to the Dirac representation.

Analogously the representation(1, 0)⊕ (0, 1) (1.66)

describes two spin 1 and corresponds to an antisymmetric tensor of rank 2,as the electromagnetic field strength tensor.

In the next section we will show how to construct all the finite-dimensional representations of SL(2, C).



1.4 Spinor representations of the Lorentz group

We can regard SL(2, C) as a representation of itself. In this case we in-troduce a two-dimensional complex vector space (spinor space) and definetransformations according to

ζ ′α = aαβζ

β , a ∈ SL(2, C), α, β = 1, 2. (1.67)

We see from eqs. (1.58) and (1.59) that

~J (1) =~τ

2, ~J (2) = 0. (1.68)

Therefore the spinors transform according to

ζα ∈(

12, 0). (1.69)

It is convenient to introduce the following special 2× 2 matrix

εαβ = εαβ = i(τ2)αβ =(

0 1−1 0

), (1.70)

with the following properties

ε−1 = εT = ε† = −ε. (1.71)

Furthermore, for any matrix a ∈ SL(2, C) we have

a = ε−1aT−1ε. (1.72)

Together with the representation specified by a we can introduce the rep-resentations defined by aT−1, a∗ and a†−1. In fact we can easily verify thatthey satisfy the same product rule as the one defined by a. For instance,

aT−11 aT−1

2 = (a1a2)T−1 (1.73)

and analogously for the other cases. It follows from (1.72) that the repre-sentation specified by aT−1 is equivalent to the one specified by a. In thesame way a∗ and a†−1 are equivalent. On the other hand, a and a∗ are notequivalent4. Then, it is convenient to introduce explicitly a second spinorspace transforming according to a∗. We will use a special notation for thecomponents of the elements of this space

ζ ′α = (aαβ)∗ζ β ≡ aα

βζ β . (1.74)

4Notice that if we restrict the matrices to SU(2) then, due to the unitarity condition,

it follows that a∗ = aT−1 and correspondingly a∗ is equivalent to a.


Introduction 13

The first spinor space is said to be the space of the undotted spinors,whereas the second one the space of the dotted spinors. By using thehermiticity of ~J and the anti-hermiticity of ~K we have(

ei~α·~J)†−1

= ei~α·~J ,

(ei~α·

~K)†−1

= e−i~α·~K . (1.75)

Since a∗ and a†−1 are equivalent representations, the previous relationsshow that going from the a representation to the a∗ is equivalent to send~J → ~J and ~K → − ~K, that is to say to exchange ~J (1) and ~J (2). But thismeans that the complex conjugate of an undotted spinor transforms as adotted spinor and vice versa. Therefore

ζα ∈(

0,12

). (1.76)

We can make use of the matrix ε to lower and raise the spinor indices

ζα = ζβεβα, ζα = ζ βεβα, (1.77)

with

εαβ = εαβ = εαβ = εαβ . (1.78)

The dotted and undotted spinors with lower indices transform respectivelyas aT−1 and a†−1. Inverting eqs. (1.77) we find

ζα = εαβζβ , ζα = εαβζβ . (1.79)

Since spinors with upper and lower indices transform according to a andaT−1 we see that the expression

ζαηα = ζαε

αβηβ = −ζαηα (1.80)

is invariant under SL(2, C) transformations. Analogous result is found fordotted spinors.

In agreement with eq. (1.56) we define spinor indices for the matricesσµ and σµ as

(σµ)αβ = (1, ~τ), (σµ)αβ = (1,−~τ). (1.81)

We can define higher order spinors via tensor products. For instance, thefollowing rank two spinors

ζαβ , ζβα , ζαβ , ζαβ (1.82)

transform respectively as

aT−1 ⊗ aT−1, a⊗ aT−1, a⊗ a, aT−1 ⊗ a†−1. (1.83)



These representations can be generally expressed as direct sum of irre-ducible representations. The latter are given by spinors of the type

ζα1α2···αm;β1β2···βn (1.84)

transforming as

ζ ′α1α2···αm;β1β2···βn = aα1

γ1 aα2γ2 · · · a

αmγm a

β1γ1aβ2γ2· · · aβmγm ζ

α1α2···αm;β1β2···βn

(1.85)symmetrized in both the m undotted and the n dotted indices. Here

aβα = (aβα)∗. (1.86)

The number of undotted indices, m, and the number of dotted indices, n,are respectively twice the spin s1 and s2 characterizing the representationof SU(2)⊗ SU(2). We say also that these spinors are of type [m,n].

As an example, let us consider the representation (1/2, 1/2). This hasangular momentum 0 and 1 and corresponds to a four-vector. This followsfrom the observation that combining one undotted and one dotted spinorsvia the σµ matrices it is possible construct the combination

ηα(σµ)αβζβ (1.87)

transforming as a four-vector. The generators of SL(2, C) for the two in-equivalent spinor representations are given by Jµν = iσµν/2 and Jµν =iσµν/2, where

(σµν)αβ =12

(σµ)αγ(σν)γβ − (σν)αγ(σµ)γβ (1.88)

and

(σµν)βα =12

(σµ)αγ(σν)γβ − (σν)αγ(σµ)γβ . (1.89)

One can show that lowering one spinor index to σµν

(σµν)αβ = (σµν)γαεγβ , (1.90)

the resulting matrices are symmetric in the spinor indices. Now, let usconsider the tensor product of two undotted spinors. According to the rulefor adding two angular momenta one finds(

12, 0)⊗(

12, 0)

= (0, 0)⊕ (1, 0) . (1.91)

This corresponds to take the antisymmetric and the symmetric part of theproduct of two undotted spinors with

ζαεαβηβ ∈ (0, 0), ζα(σµν)αβηβ ∈ (1, 0). (1.92)


Introduction 15

By doing the same operation with two dotted spinors one gets the represen-tations (0, 0) and (0, 1). The combination of (1, 0) and (0, 1) gives rise to ageneric antisymmetric tensor containing two spin 1. As an example thinkto the electromagnetic field strength consisting of ~E and ~B. Finally noticethat, in order to have tensor representations we need s1 + s2 to be integeror, m + n even. Since in this case the transformation a appears an evennumber of times, the condition T (a) = T (−a) is automatically satisfied.

1.5 The Poincare group

The Poincare group is defined as the set of transformations leaving invariantthe distance between two points in Minkowski space

(y − x)2 = (y − x)µ(y − x)νgµν . (1.93)

Therefore it consists of Lorentz transformations and translations

x′µ = Λµνxν + aµ. (1.94)

As such it is characterized by the pair (Λ, a). Denoting the correspondinggroup element by g(Λ, a) and applying two Poincare transformations onecan easily derive the multiplication rule for two group elements

g(Λ1, a1)g(Λ2, a2) = g(Λ1Λ2,Λ1a2 + a1). (1.95)

The number of parameters associated to a generic Poincare transformationis ten. Six of them correspond to Lorentz transformations and four to trans-lations. By taking again the component of the Poincare group connectedto the identity we will write

g(Λ, a) = e−i2ωµνJ

µν+iεµPµ

. (1.96)

The commutation rules involving the new 4 generators Pµ are

[Jρσ, Pµ] = i(gσµPρ − gρµPσ) (1.97)

and

[Pµ, Pν ] = 0. (1.98)

We can define the Pauli-Lubanski four-vector as

Wµ =12εµνρσJ

νρPσ. (1.99)

One can easily show that Wµ commutes with the momentum operator

[Wµ, Pν ] = 0 (1.100)



and that it has commutation relations of the same type of the momentumwith the Lorentz generators

[Jρσ,Wµ] = i(gσµWρ − gρµWσ). (1.101)It is possible to show that P 2 and W 2 are Casimir operators of the Poincaregroup, that is they commute with all Poincare generators and, in particular,with the Lorentz ones (see Exercises)

[Jµν , P 2] = [Jµν ,W 2] = 0. (1.102)On a state vector for a particle of mass m, momentum ~p = ~0, spin s andz-component sz, we haveW 0|~p = ~0,m, s, sz〉 = 0, W i|~p = ~0,m, s, sz〉 = −mJi|~p = ~0,m, s, sz〉

(1.103)and

W 2|~p = ~0,m, s, sz〉 = −m2s(s+ 1)|~p = ~0,m, s, sz〉. (1.104)

1.6 Exercises

(1) Convert 1/gram in units of length (cm.) and time (sec.).(2) Show that in the units used for writing eq. (1.12) the electric charge is

dimensionless. Hint: use eq. (1.14).(3) Evaluate the masses of the electron and the proton in GeV.(4) Using the commutation relations for the generators of the Poincare

generators show that P 2 and W 2 are Casimir operators for this algebra,that is they commute with both Jµν and Pµ. Show that JµνJµν is aCasimir of the Lorentz group, but it is not a Casimir of the Poincaregroup and give an argument to explain the last sentence on physicalgrounds.

(5) Using the 2×2 representation for a boost along the third axis, showthat this corresponds to the special Lorentz transformation

x′0 = x0 coshχ+ x3 sinhχ, x′3 = x0 sinhχ+ x3 coshχ. (1.105)(6) Associate to a four-vector in Minkowski space, xµ, the 5-dimensional

column vector (x

1

). (1.106)

Show that the 5× 5 matrix (Λ a

0 1

)(1.107)

gives a representation of the Poincare group.


Chapter 2

Lagrangian formalism for continuumsystems and quantization

2.1 Many degrees of freedom

Aim of this book is to extend ordinary quantum mechanics, describing nonrelativistic particles in interaction with a given force, to the relativistic casewhere forces are described by fields, as in the case of the electromagnetism.The most relevant differences between the two cases are that the forcesbecome dynamical degrees of freedom, and that one needs a relativistictreatment of the problem. In order to get a consistent description we willneed to quantize the field degrees of freedom.

The concept of field is a very general one. A field represents a physicalquantity depending on the space-time point. Examples are the distribu-tion of the temperature in a room, the distribution of the pressure in theatmosphere, the particle velocities inside a fluid, the electric and magneticfields, ~E and ~B, in a given region of space. The common physical feature ofthese systems is the existence of a fundamental or ground state. For someof the examples listed above, the ground states are given in Table 2.1:

Table 2.1 Examples of fields

and ground states.Field Ground state

Temperature T = constant

Pressure P = constant~E, ~B Vacuum

Then, the field is defined in terms of fluctuations around the groundstate. If these deviations are small one gets linear equations for the fields.Of course the approximation can be improved by using a perturbative ap-proach. If the linear approximation is considered one gets, in general, dy-

17



namical equations which are very similar for many different physical sys-tems. Often the fluctuations satisfy a second order differential equationdescribing wave propagation. The quantization of such systems leads to adescription in terms of particles corresponding to various classical excita-tions.

It happens that many physical systems, both with finite or infinite num-ber of degrees of freedom, can be put in Hamiltonian form. This is truealso for the case of systems described by a wave equation. The Hamiltoniandescription is what we need to proceed to the canonical quantization of thedynamical system.

We will start the discussion of field quantization with a simple one-dimensional system, a vibrating string. We will consider N linear oscillators(for instance a one-dimensional string of atoms). In this way we get avibrating string in the continuum limit. The advantage is to start witha finite number of degrees of freedom, a problem well known in ordinaryquantum mechanics. After that we will take the continuum limit by sendingN → ∞ with a separation among the atoms going to zero, obtaining avibrating string and its quantized properties.

2.2 Linear atomic string

Let us consider a string of N + 1 harmonic oscillators, or N + 1 atoms (ofunit mass) interacting through a harmonic force, as in Fig. 2.1. The length

a aaa

qn-1 qn q n+1

Fig. 2.1 In the upper line the atoms are in their equilibrium position, whereas in the

lower line they are displaced by the quantities qn.


Lagrangian formalism for continuum systems and quantization 19

of the string is L and the interatomic distance is a. Therefore L = Na.The equations of motion for the displacements of the atoms with respectto the equilibrium configurations are

qn = ω2 [(qn+1 − qn) + (qn−1 − qn)] = ω2 [qn+1 + qn−1 − 2qn] , (2.1)

Here ω is the common frequency of the oscillation of the atoms. The po-tential energy of the system is

U =12ω2

N∑n=1

(qn − qn+1)2. (2.2)

In order to define the problem one has to specify the boundary conditions.Usually one considers two possibilities:

• Periodic boundary conditions, that is qN+1 = q1.• Fixed boundary conditions, that is qN+1 = q1 = 0.

To quantize the problem is convenient to go to the Hamiltonian formulation.The Hamiltonian is given by (pn = qn)

H = T + U =12

N∑n=1

(p2n + ω2 (qn − qn+1)2

), pn = qn (2.3)

The equations of motion can be diagonalized by looking for the eigenmodes.Let us put

q(j)n = Aje

ikjane−iωjt, (2.4)

where the index j enumerates the possible eigenvalues. Notice that in thisequation the dependence on the original equilibrium position has been madeexplicit through

q(j)n = q(j)(xn) = q(j)(na) ≈ eikjxn . (2.5)

Here xn = na is the equilibrium position of the nth atom. By substitutingeq. (2.4) into the equations of motion we get

−ω2j q

(j)n = −4ω2q(j)

n sin2

(kja

2

), (2.6)

from which

ω2j = 4ω2 sin2

(kja

2

). (2.7)

The relation between ωj and kj (frequency of oscillation), shown graphicallyin Fig. 2.2, is called a dispersion relation. Although kj is a scalar, it will be



- 3 - 2 - 1 1 2 3akj

1

2

3

4 j2

Fig. 2.2 The first Brillouin zone in arbitrary units.

called wave vector, since in more than one dimension it is indeed a vectorialquantity. We may notice that wave vectors differing by multiples integer of2π/a, that is, such that

k′j = kj + 2mπ

a, m = ±1,±2, ..., (2.8)

correspond to the same ωj . This allows us to restrict kj to be in the socalled first Brillouin zone, defined by |kj | ≤ π/a. Let us now take intoaccount the boundary conditions. Here we will choose periodic boundaryconditions, that is qN+1 = q1, or, more generally, qn+N = qn. This gives us

q(j)n+N = Aje

ikja(n+N)e−iωjt = q(j)n = Aje

ikjane−iωjt, (2.9)

from which

kjaN = 2πj, (j = integer). (2.10)

Since aN = L (L is the length of the string)

kj =2πaN

j =2πLj, j = 0,±1,±2, · · · ± N

2, (2.11)

where we have assumed N to be even. The restriction on j follows fromconsidering the first Brillouin zone (|kj | ≤ π/a). Notice that the possiblevalues of kj are 2(N/2)+1 = N+1, and that j = 0 corresponds to a uniformtranslation of the string (with zero frequency). Since we are interested onlyin the oscillatory motions, we will omit this solution in the following. Itfollows that we have N non trivial independent solutions

q(j)n = Aje

−iωjteiakjn. (2.12)



The most general solution is obtained by a linear superposition (we includethe time dependence in Qj and Pj)

qn =∑j

ei2πNjn Qj√

N, (2.13)

pn =∑j

e−i2π

Njn Pj√

N. (2.14)

From the reality of qn and pn we get

Q?j = Q−j , P ?j = P−j . (2.15)

In the following we will make use of the following relation (see Exercise 1in this Section)

N∑n=1

ei2πN

(j′ − j)n= Nδjj′ . (2.16)

By using this equation we can invert the previous expansions

N∑n=1

qne−i2π

Nj′n

=∑j

∑n

ei2πN

(j − j′)n Qj√N

=√NQj′ , (2.17)

obtaining

Qj =1√N

N∑n=1

qne−i2π

Njn, (2.18)

Pj =1√N

N∑n=1

pnei2πNjn. (2.19)

Notice that (see eq. (2.3))

Pj = Q−j = Q†j . (2.20)

Substituting inside the Hamiltonian we find

H =N/2∑j=1

(|Pj |2 + ω2

j |Qj |2). (2.21)

This is nothing but the Hamiltonian of N decoupled harmonic oscillatorseach having a frequency ωj , as it can be seen by putting

Pj =1√2

(Xj + iYj), Qj =1√2

(Zj + iTj). (2.22)



The result we have obtained so far shows that the string of N atoms isequivalent to N decoupled harmonic oscillators. The oscillator modes havebeen obtained through the expansion of the displacements from the equilib-rium in normal modes. We are now in the position to introduce the conceptof displacement field. Let us define a function of the equilibrium positionof the atoms

xn = na, L = Na, (2.23)

as the displacement of the nth atom from its equilibrium position

u(xn, t) = qn(t). (2.24)

The field u(xn, t) satisfies the following equation of motion

u(xn, t) = ω2 [u(xn+1, t) + u(xn−1, t)− 2u(xn, t)]

= ω2 [(u(xn+1, t)− u(xn, t))− (u(xn, t)− u(xn−1, t))] .(2.25)

Let us now consider the continuum limit of this system. Physically this isequivalent to say that we are looking at the system at a scale much biggerthan the interatomic distance. We will define the limit by taking a → 0and keeping fixed the length of the string, that is to say

a→ 0, N →∞, aN = L fixed. (2.26)

In the limit, u(xn, t) defines a function of the variable x varying in theinterval (0, L). Furthermore

u(xn, t)− u(xn−1, t)a

→ u′(x, t) (2.27)

and

(u(xn+1, t)− u(xn, t))− (u(xn, t)− u(xn−1, t))→ a(u′(xn+1, t)− u(xn, t))

→ a2u′′(x, t). (2.28)

The equation of motion becomes

u(x, t) = a2ω2u′′(x, t). (2.29)

In order to give a sense to this equation we let ω to diverge in such a waythat

lima→0

aω = v, v finite, (2.30)

where v has the dimension of a velocity. What we get is the equation forthe string as the propagation of waves with velocity v

u(x, t) = v2u′′(x, t). (2.31)



In the limit we have alsoN+1∑n=1

a→∫ L

0

dx, (2.32)

from which

H =12a

∫ L

0

dx[(u(x, t))2 + v2 (u′(x, t))2

]. (2.33)

To get finite energy we need also a redefinition of the field variable

u(x, t) =√aφ(x, t), (2.34)

from which

H =12

∫ L

0

dx

[(φ(x, t)

)2

+ v2 (φ′(x, t))2]. (2.35)

The normal modes decomposition for the field φ(x, t) becomes

φ(x, t) =u(x, t)√

a≈ qn(t)√

a≈∑j

eikjanQj√aN

, (2.36)

kj =2πLj, −∞ < j < +∞, (2.37)

that is

φ(x, t) =1√L

+∞∑j=−∞

ei2πLjxQj(t). (2.38)

The eigenfrequencies are given by

ω2j = 4ω2 sin2

(πaLj)→ 4ω2

(πj

L

)2

a2 = (aωkj)2 → v2k2

j . (2.39)

In the continuum limit the frequency is proportional to the wave vector kj .The relation between the normal modes Qj(t) and the field φ(x, t) can beinverted by using the following relation∫ L

0

dx eix(k − k′) = Lδk,k′ , (2.40)

which holds for k and k′ of the form (2.37). The Hamiltonian is easilyobtained as

H =∞∑j=1

(|Qj |2 + v2k2

j |Qj |2). (2.41)

The main result here is that in the continuum limit the Hamiltonian of thesystem describes an infinite set of decoupled harmonic oscillators. In thefollowing we will show that the quantization of field theories of the typedescribed in this Section gives rise, naturally, to a description in terms ofparticles.



2.3 String quantization

We have shown that a string of N atoms can be described in terms of a setof decoupled harmonic oscillators, and this property holds true also in thecontinuum limit (N → ∞). In the discrete case we have shown that theHamiltonian of the system can be written as

H =N/2∑j=1

(|Pj |2 + ω2

j |Qj |2), (2.42)

where

Pj = Q†j , Q†j = Q−j , P †j = P−j , (2.43)

ω2j = 4ω2 sin2 kja

2, kj =

2πLj, |j| = 1, 2, . . . ,

N

2, (2.44)

whereas in the continuum limit

H =∞∑j=1

(|Pj |2 + ω2

j |Qj |2), (2.45)

with

ωj = v|kj |, |j| = 1, 2, . . . ,∞ (2.46)

and kj given by eq. (2.44). In both cases the quantization is trivially donethrough the canonical commutation relations

[Qj , Pk] = iδjk, [Q†j , P†k ] = iδjk. (2.47)

Having to do with harmonic oscillators it is convenient to introduce anni-hilation and creation operators

aj =√ωj2Qj + i

1√2ωj

P †j , a†j =√ωj2Q†j − i

1√2ωj

Pj , (2.48)

with j assuming a finite or an infinite number of values according to thesystem being discrete or continuum. In both cases we have

[aj , a†k] = δjk (2.49)

and

[aj , ak] = [a†j , a†k] = 0. (2.50)

Notice that

a−j =√ωj2Q†j + i

1√2ωj

Pj 6= a†j (2.51)



and

a†−j =√ωj2Qj − i

1√2ωj

P †j 6= aj . (2.52)

Therefore aj e a†j are 2N (in the discrete case) independent operators asQj and Pj . The previous relations can be inverted to give

Qj =1√2ωj

(aj + a†−j), Pj = −i√ωj2

(a−j − a†j). (2.53)

The Hamiltonian expressed in terms of aj and a†j is written as

H =N/2∑j=1

ωj [a†jaj + a†−ja−j + 1] =

N/2∑j=−N/2

ωj

[a†jaj +

12

]. (2.54)

The ground state is characterized by the equation

aj |0〉 = 0, (2.55)

with a corresponding energy

E0 =∑j

ωj2. (2.56)

In the continuum limit the energy of the fundamental state is infinite (wewill come back later on this point). The generic energy eigenstate is ob-tained by applying creation operators to the ground state

|n−N/2, · · · , nN/2〉 =1

(n−N/2! · · ·nN/2!)1/2(a†−N/2)n−N/2 · · · (a†N/2)nN/2 |0〉.

(2.57)The Hilbert space spanned by these vectors is called a Fock space. Thestate given above can be thought of being formed by n−N/2 quanta withenergy ω−N/2, up to nN/2 quanta with energy ωN/2. In this interpretationthe nj quanta (or particles) of energy ωj are indistinguishable one fromthe other. Furthermore, in a given state, we can put as many particleswe want. Therefore the particles we are describing here satisfy the Bose-Einstein statistics. Formally this follows from the commutation relation

[a†i , a†j ] = 0, (2.58)

implying the symmetry of the states with respect to the exchange of twoquanta. For instance a two-particle state is given by

|i, j〉 = a†ia†j |0〉 = a†ja

†i |0〉 = |j, i〉. (2.59)



As we have already noticed the energy of the fundamental state becomesinfinite in the continuum limit. This is perhaps the most simple of theinfinities that we will encounter in our study of field quantization. Wewill learn much later in this course about the possibility of keeping themunder control. For the moment being let us notice that normally onlyrelative energies are important and the value of E0 (see eq. (2.56)) isnot physically relevant. However, there are situations, as in the Casimireffect (see later), where it is indeed relevant. Forgetting momentarily aboutthese special situations we can define a new Hamiltonian by subtracting E0.This can be done in a rather formal way by defining the concept of normalordering. Given an operator which is a monomial in the creation andannihilation operators, we define its normal ordered form by pushing allthe annihilation operators to the right of the creation operators. We thenextend the definition to polynomials by linearity. For instance, in the caseof the Hamiltonian (2.45) we have

: H : ≡ N(H) = N

∑j

ωj2

(a†jaj + aja

†j

) =∑j

ωja†jaj . (2.60)

Coming back to the discrete case, recalling eqs. (2.13) and (2.14) and usingthe canonical commutators (2.47), we get

[qn, pm] =∑jk

ei2πN

(jn− km) 1N

[Qj , Pk] =i

N

∑j

ei2πNj(n−m)

= iδnm.

(2.61)In the continuum case we use the expansion (2.38).

φ(x, t) =1√L

+∞∑j=−∞

ei2πLjxQj , φ(x, t) =

1√L

+∞∑j=−∞

e−i2π

LjxPj , (2.62)

from which

[φ(x, t), φ(y, t)] =1L

∑jk

ei2πL

(jx− ky)iδjk

=i

L

+∞∑j=−∞

ei2πLj(x− y)

= iδ(x− y). (2.63)

This relation could have been obtained starting from the discrete case andgoing to the continuum limit by recalling that

φ(x, t) ≈ u(x, t)√a≈ qn√

a, (2.64)



implying

[φ(xn, y), φ(xm, t)] = iδnma. (2.65)

In the continuum limit

lima→0

δnma

= δ(x− y), (2.66)

if for a→ 0, xn → x, and xm → y. In fact,

1 =∑n

a

(δnma

)→∫dx

(lima→0

δnma

), (2.67)

showing that the properties of a delta-approximation are indeed satisfied.It follows that we have the following correspondence between the discreteand the continuum case

qn → φ(x, t), pn → φ(x, t). (2.68)

Said in different words, φ(x, t) e φ(x, t) appear to be the analog of a pairof canonical variables. This remark suggests a way to approach the quan-tization of a field theory different from the one followed so far. The waywe have used is based upon the construction of the normal modes of theoscillators, but in the discrete case this is not necessary at all. In fact, thequantization can be made starting from the commutation relations amongthe canonical variables, [q, p] = i, without having an a priori knowledgeof the dynamics of the system. This suggests that one could start directlyby the field operators φ(x, t), φ(x, t), and quantize the theory by requiring[φ(x, t), φ(y, t)] = iδ(x − y). To make this approach a consistent one, weneed to extend the Hamiltonian and Lagrangian descriptions to a contin-uum system. Let us recall how we proceed in the discrete case. We start bygiving a Lagrangian function L(qn, qn, t). Then we define the conjugatedmomenta by the equation

pn =∂L

∂qn. (2.69)

We then go to the Hamiltonian formalism by taking the conjugated mo-menta, pn, as independent variables. The previous equation is used in orderto solve the velocities in terms of qn and pn. Next we define the Hamiltonianas

H(qn, pn) =∑n

pnqn − L. (2.70)



At the classical level the time evolution of the observables is obtainedthrough the equation

A = A,H, (2.71)

where the Poisson brackets can be defined starting form the brackets be-tween the canonical variables qn, pm = δnm. The theory is then quantizedthrough the ”correspondence” rule

., . → −i[., .]. (2.72)

with ., . the classical Poisson bracket and [., .] the commutator. In thenext Section we will learn how to extend the Lagrangian and Hamiltonianformalisms to the continuum case.

2.4 The canonical formalism for continuum systems

We can evaluate the Lagrangian for the string using the kinetic and thepotential energy. Let us start with the discrete case. The kinetic energy isgiven by

T =12

N∑n=1

p2n =

12

N∑n=1

u2(xn, t) =12

N∑n=1

aφ2(xn, t)→12

∫ L

0

φ2(x, t)dx.

(2.73)and the potential energy by

U =12

N∑n=1

ω2(qn − qn+1)2 =12

N∑n=1

ω2a(φ(xn, t)− φ(xn+1, t))2 (2.74)

and using that for a→ 0, v = aω, is finite, it follows

U =12

N∑n=1

av2

(φ(xn, t)− φ(xn+1, t)

a

)2

→ v2

2

∫ L

0

φ′2(x, t)dx. (2.75)

Therefore the total energy and the Lagrangian are

E = T + U =12

∫ L

0

dx[φ2(x, t) + v2φ′

2(x, t)]

(2.76)

and

L = T − U =12

∫ L

0

dx[φ2(x, t)− v2φ′

2(x, t)], (2.77)



respectively. The important result is that in the continuum limit, the La-grangian can be written as a spatial integral of a function of the field φ andof its first derivative, φ, called Lagrangian density

L =12

(φ2 − v2φ′

2). (2.78)

with the total Lagrangian given by

L =∫ L

0

L dx. (2.79)

Of course, this is not the most general situation that can be imagined, butwe will consider only the case in which the Lagrangian density dependsonly on the field and on its first derivatives

L =∫L(φ, φ, φ′, x, t)dx. (2.80)

We will restrict ourselves to theories in which the Lagrangian contains atmost first order time derivatives of the fields, since theories with higherorder derivatives have potential problems with probability conservation.

Given the Lagrangian, the next step is to build up the action functional.The extrema of the action give rise to the equations of motion. The actionis given by

S =∫ t2

t1

Ldt =∫ t2

t1

dt

∫dxL(φ, φ, φ′, x, t). (2.81)

We require S to be stationary with respect to those variations that areconsistent with the boundary conditions satisfied by the fields. If Σ is thespatial surface delimiting the region of spatial integration (for the string itreduces to the end points), we will require

δφ(x, t) = 0 on Σ. (2.82)

Furthermore we will require the variations at the times t1 and t2 to be zerofor any space point x

δφ(x, t1) = δφ(x, t2) = 0, for any x. (2.83)

In the discrete case we have only boundary conditions of the second type,but here the first ones are necessary in order to be consistent with theboundary conditions chosen for the field. Let us now require S to be sta-tionary against variations satisfying the boundary conditions (2.82) and(2.83)

0 = δS =∫ t2

t1

dt

∫dxδL =

∫ t2

t1

dt

∫dx

(∂L∂φ

δφ+∂L∂φ

δφ+∂L∂φ′

δφ′).

(2.84)



Integrating by parts

0 =∫

dx

[∂L∂φ

δφ

]t2t1

+∫ t2

t1

dt

[∂L∂φ′

δφ

]L0

+∫ t2

t1

dt

∫dx

[∂L∂φ− ∂

∂t

∂L∂φ− ∂

∂x

∂L∂φ′

]δφ. (2.85)

The boundary terms are zero due to eqs. (2.82) and (2.83). Then fromthe arbitrariness of δφ within the region of integration, we get the Euler-Lagrange equations

∂L∂φ− ∂

∂t

∂L∂φ− ∂

∂x

∂L∂φ′

= 0. (2.86)

In fact δφ can be chosen to be zero everywhere except for a small regionaround any given point x (see Fig. 2.3).

x

Fig. 2.3 Here the arbitrary variation δφ(x) is chosen to be zero all along the string,

except for a small region around the point x.

This discussion can be easily extended to the case of N fields φi, i =1, . . . , N (think, as an example, to the electromagnetic field), and to thecase of n spatial dimensions with points labelled by xα, α = 1, . . . , n. Inthis case the structure of the action will be

S =∫ t2

t1

dt

∫V

dnxL(φi, φi,

∂φi∂xα

). (2.87)

Here V is the integration space volume. We will require again the sta-tionarity of the action with respect to variations of the fields satisfying theboundary conditions

δφi(xα, t) = 0, on Σ, for any t, t1 ≤ t ≤ t2, (2.88)

where Σ is the boundary of V , and

δφi(xα, t1) = δφi(xα, t2) = 0, for any xα ∈ V. (2.89)

As before, the first boundary conditions are required in order to be consis-tent with the boundary conditions for the fields as, for instance, in the case



of spatial regions extended up to infinity, where the fields are required togo to zero. The Euler-Lagrange equations one gets in this case are

∂L∂φi− ∂

∂t

∂L∂φi− ∂

∂xα

∂L∂φ,αi

= 0, i = 1, . . . , N, α = 1, . . . , n, (2.90)

where

φ,αi ≡∂φi∂xα

. (2.91)

To go to the Hamiltonian description one introduces the momentum den-sities conjugated to the fields φi:

Πi =∂L∂φi

(2.92)

and the Hamiltonian density

H =∑i

Πiφi − L. (2.93)

In the case of the string one gets from eq. (2.77)

∂L∂φ

= φ,∂L∂φ′

= −v2φ′,∂L∂φ

= 0. (2.94)

From which one recovers the equations of motion for the field φ. Further-more

Π = φ, (2.95)

implying

H = Πφ− L = Π2 −(

12

Π2 − 12v2φ′

2)

=12

(Π2 + v2φ′

2), (2.96)

which coincides with the energy density given in eq. (2.76), after using eq.(2.95).

The greatest advantage of using the Lagrangian formalism instead ofthe Hamiltonian one is the possibility to formulate in a simple way thesymmetry properties of the theory. We shall see that this is due to thefirst theorem of Emmy Noether [Noether (1918)] which makes possible tocorrelate in a simple way the symmetry properties of the Lagrangian withthe existence of conservation laws1. Due to this correspondence it is possibleto use the Noether theorem in a constructive way. That is, to restrict theform of the Lagrangian requiring a particular type of symmetry. We will1The first Noether’s theorem deals with global symmetries, whereas the second one has

to do with local symmetries which will be discussed in Chapter 6.



discuss the theorem later on. For the moment being we will show how theequations of motion of the string generate the conservation laws.

The energy contained in the segment [a, b] of the string, with 0 ≤ a ≤b ≤ L is given by

E(a, b) =12

∫ b

a

dx[φ2 + v2φ′

2]. (2.97)

We can evaluate its time variationdE(a, b)

dt=∫ b

a

dx[φφ+ v2φ′φ′

]= v2

∫ b

a

dx[φφ′′ + φ′φ′

]= v2

∫ b

a

dx∂

∂x

[φφ′]

= v2[φφ′]ba, (2.98)

where we have made use of the equations of motion in the second step.Defining the local quantity

P(x, t) = −v2φφ′, (2.99)

which is the analogous of the Poynting’s vector in electrodynamics, we get

−dE(a, b)dt

= [P(b, t)− P(a, t)]. (2.100)

This is the classical energy conservation law, expressing the fact that if theenergy decreases in the segment [a, b], then there must be a flux of energy atthe end points a and b. The total energy is conserved due to the boundaryconditions, P (0, t) = P (L, t). But the previous law says something more,because it gives us a local conservation law, as it follows by taking the limitb→ a. In fact, in this limit

E(a, b)→ (b− a)H, (2.101)

with H given by (2.96), and∂H∂t

+∂P∂x

= 0. (2.102)

This conservation law can be checked by using the explicit expressions ofH and P , and the equations of motion.

2.5 The canonical quantization of a continuum system

As we have seen in Section 2.4, in field theory the momentum densities aredefined as

Πi =∂L∂φi

. (2.103)



It is then natural to assume the following commutation relations

[φi(xα, t),Πj(yα, t)] = iδijδn(xα − yα) α = 1, . . . , n, i, j = 1, . . . , N

(2.104)and

[φi(xα, t), φj(yα, t)] = 0, [Πi(xα, t),Πj(yα, t)] = 0. (2.105)

In the string case we have Π = φ and we reproduce eq. (2.63). Startingfrom the previous commutation relations and expanding the fields in termsof normal modes one gets back the commutation relations for the creationand annihilation operators. Therefore we reconstruct the particle interpre-tation. Using the Heisenberg representation (but omitting from now on thecorresponding index for the operators), the expansion of the string field interms of creation and annihilation operators is obtained using eqs. (2.62)and (2.53).

φ(x, t) =1√L

∑j

ei2πLjxQj

=1√L

∑j

1√2ωj

ei2πL jxaj(t) + e

−i2πLjxa†j(t)

. (2.106)

From the equations of motion of the string

φ− v2φ′′ = 0, (2.107)

we find the equations of motion for QjQj + ω2

jQj = 0. (2.108)

These equations are consistent with the Hamilton equations:

Qj = P †j , Pj = −ω2Q†j . (2.109)

Then, using the decomposition (2.48) of aj in terms of Qj and Pj = Q†j ,we get easily

aj + iωjaj = 0, (2.110)

from which

aj(t) = aj(0)e−iωjt ≡ aje−iωjt, a†j(t) = a†j(0)eiωjt ≡ a†jeiωjt (2.111)

and

φ(x, t) =1√L

∑j

1√2ωj

ei(

2πLjx− ωjt

)aj + e

−i(

2πLjx− ωjt

)a†j

.(2.112)



Let us investigate the structure of this expansion. It can be written in thefollowing way

φ(x, t) =∑j

[fj(x, t)aj + f∗j (x, t)a†j

], (2.113)

with

fj(x, t) =1√

2ωjLei

(2πLjx− iωjt

)=

1√2ωjL

ei(kjx− iωjt), (2.114)

where we have made use of the definition of kj (see eq.(2.37)). The functionsfj(x, t) and their complex conjugated satisfy the wave equation

∂2fj(x, t)∂t2

− v2 ∂2fj(x, t)∂x2

= 0 (2.115)

with periodic boundary conditions

fj(0, t) = fj(L, t). (2.116)

It is immediate to verify that they are a complete set of orthonormal func-tions ∑

j

f∗j (x, t)i∂(−)t fj(y, t) = δ(x− y), (2.117)

∫ L

0

dxf∗j (x, t)i∂(−)t fl(x, t) = δjl, (2.118)

where

A∂(−)t B = A(∂tB)− (∂tA)B. (2.119)

Let us consider the first relation. We have∑j

f∗j (x, t)i∂(−)t fj(y, t) =

∑j

2ωjf∗j (x, t)fj(y, t)

=∑j

1Le−ikj(x− y) = δ(x− y). (2.120)

Evaluating this expression with two fj(x, t)’s or two f∗j (x, t)’s one gets zero.As far as the second relation is concerned we get∫ L

0

dxf∗j (x, t)i∂(−)t fl(x, t) =

∫ L

0

dx[ωl + ωj ]f∗j (x, t)fl(x, t)

]=

1L

ωl + ωj2√ωjωl

∫ L

0

dxeix(kl − kj)ei(ωj − ωl)t

=1L

ωl + ωj2√ωjωl

ei(ωj − ωl)tLδjl = δjl. (2.121)



Also in this case, by taking two fj(x, t)’s or two f∗j (x, t)’s, the result is zerodue to the appearance of the factor (ωl − ωj).

An interesting question is why the operator ∂(−)t appears in these rela-

tions. The reason is that the scalar product should be time independent,otherwise the statement of orthonormality of the solutions would be timedependent. For instance, in the case of the Schrodinger equation, we definethe scalar product as ∫

d3xψ∗(~x, t)ψ(~x, t), (2.122)

because for hermitian Hamiltonians this is indeed time independent, as itcan be checked by differentiating the scalar product with respect to time.In the present case, we can define a time independent scalar product, byconsidering two solutions f and f of the wave equation, and evaluating thefollowing two expressions∫ L

0

dxf

[∂2f

∂t2− v2 ∂

2f

∂x2

]= 0, (2.123)

∫ L

0

dx

[∂2f

∂t2− v2 ∂

2f

∂x2

]f = 0. (2.124)

Subtracting these two expressions one from the other we get∫ L

0

dx

[∂

∂t

(f∂f

∂t− ∂f

∂tf

)− v2 ∂

∂x

(f∂f

∂x− ∂f

∂xf

)]= 0. (2.125)

If both f and f satisfy periodic boundary conditions, the second term iszero, and it follows that the quantity∫ L

0

dxf∂(−)t f (2.126)

is a constant of motion. Using eq. (2.118), we can invert the relationbetween the field and the creation and annihilation operators. We get

aj =∫ L

0

dxf∗j (x, t)i∂(−)t φ(x, t) (2.127)

and

a†j =∫ L

0

dxφ(x, t)i∂(−)t fj(x, t). (2.128)

From the field commutation relations we find

[aj , a†k] = δjk. (2.129)



In analogous way we get

[aj , ak] = [a†j , a†k] = 0. (2.130)

We have seen that the total energy of the string is a constant of motion.There is another constant of motion corresponding to the total momentumof the string

P =∫ L

0

dxP = −∫ L

0

dx φφ′. (2.131)

where P has been defined in eq. (2.99). We will show in the followingthat the conservation of the total momentum of the string derives from theinvariance of the theory under spatial translations. For the moment beinglet us check that this is in fact a conserved quantity:

dP

dt= −

∫ L

0

dx(φφ′ + φφ′) = −∫ L

0

dx12∂

∂x(φ2 + v2φ′

2) = 0, (2.132)

where we have used the equations of motion of the string and the boundaryconditions. By using the field expansion, we find

P = −∑l

12kl[a−lale−2iωlt + a†−la

†l e

2iωlt − ala†l − a†l al]. (2.133)

The first two terms give zero contribution because they are antisymmetricin the summation index (kl ≈ l). Therefore

P =12

∑j

kj [aja†j + a†jaj ] =

∑j

kja†jaj , (2.134)

where we have used ∑j

kj = 0, (2.135)

since kj → −kj when j → −j. We see that P has an expression similar tothat of H (see eq. (2.60). Therefore the states

(a†−N/2)n−N/2 · · · (a†j)nj · · · |0〉 = |ψ〉 (2.136)

have energy

H|ψ〉 = (n−N/2ω−N/2 + · · ·+ njωj + · · · )|ψ〉 (2.137)

and a momentum

P |ψ〉 = (n−N/2k−N/2 + · · ·+ njkj + · · · )|ψ〉, (2.138)

as it follows from

[H, a†j ] = ωja†j , [P, a†j ] = kja

†j . (2.139)



2.6 Exercises

(1) Prove Eq. (2.16). (Hint: add and subtract 1 to the equation.)(2) Evaluate the expressions (2.127), (2.128), for the annihilation and cre-

ation operators and their commutation relations given in eqs. (2.129)and (2.130).

(3) Derive eq. (2.133) for the momentum operator.(4) Show that the addition of a divergence to the Lagrangian density does

not change the equations of motion.(5) Given an operator A of the form

A =∑j

αja†jaj , (2.140)

with

[aj , ak] = [a†j , a†k] = 0, [aj , a

†k] = δik, (2.141)

and an eigenstate of A, |ρ〉, corresponding to the eigenvalue ρ. Showthat a†j |ρ〉 is an eigenstate of A with eigenvalue ρ + αj and that aj |ρ〉corresponds to the eigenvalue ρ− αj .


Chapter 3

The Klein-Gordon field

3.1 The problems in relativistic quantum mechanics

The extension of quantum mechanics to the relativistic case gives rise toseveral problems. The difficulties originate from the relativistic dispersionrelation

E2 = |~p|2 +m2. (3.1)

This relation gives rise to two solutions

E = ±√|~p|2 +m2. (3.2)

It is not difficult to convince himself that the solutions with negative en-ergy have unphysical behavior. For instance, increasing the momentum,the energy decreases! But these solutions do not create a real problem atthe classical level. In fact, we see from eq. (3.2) that there is a gap of 2mbetween the energies corresponding to the two types of solutions. At theclassical level, the energy is always transferred in a continuous fashion. Sothere is no way to start with a positive energy particle and end up with oneof negative energy. On the contrary, in quantum mechanics, through theemission of a quantum of energy E > 2m, a positive energy state may looseenergy up to the point of becoming a state with negative energy. Since aphysical system has the tendency to decrease its energy, the positive energystates would migrate to the states of negative energy, causing a collapse ofthe usual matter. However, we shall see that although it is not possibleto ignore these solutions, they can be interpreted in terms of antiparticles.In this way one can get rid of the problems connected with the negativeenergy solutions. From this new interpretation, new phenomena emerge asannihilation of particles and antiparticles and creation of particle antipar-ticle pairs. These new effects have deep consequences on the interpretation

39



of the theory. Suppose that we wish to localize a particle on a distancesmaller than its Compton wave-length, that is smaller than 1/m. Corre-spondingly, due to the uncertainty principle, we will have an uncertaintyon the momentum greater than m. This means that the momentum (andthe energy) of the particle could well reach values of order 2m, enough tocreate a particle-antiparticle pair. Of course, this is possible only violat-ing the conservation of energy and momentum. But this is indeed possiblesince the uncertainty on both energy and momentum will be greater than1/∆x ≈ m. Therefore the attempt of localization at the Compton scale hasthe result of pair creating particles and antiparticles, meaning that we willbe unable to define the concept of a localized single particle. At the Comp-ton scale there is no such a thing as a particle, but the picture we get fromthe previous considerations is the one of a cloud of particles and antiparti-cles surrounding our initial particle, and there is no way to distinguish theoriginal particle from the many around it.

These considerations imply that the relativistic theories cannot be seenas theories with a fixed number of particles, as in the case of ordinaryquantum mechanics. In this sense a field theory looks as the most naturalway to describe such systems. In fact, since the Fock space has not a definitenumber of particles, it offers the possibility of describing the physics of avariable number of particles.

There are other justifications for using field theories. For instance, bylooking at the quantization of the electromagnetic field, physicists realizedthat one obtains a natural explanation of the particle-wave duality, and thatin the particle description one has to do with a variable number of photons.On the contrary, physical entities as the electrons, were always describedin particle terms till 1924 when De Broglie, in his PhD thesis [De Broglie(1924)], made the hypothesis that particles would exhibit a wave-like char-acter, correlating the particle momentum with the wave-length of the waveassociated to it. This hypothesis was verified experimentally by [Davissonand Germer (1927)]. This suggested that the particle-wave duality would bea feature valid for any type of waves and/or particles. Therefore, based onthe analogy with the electromagnetic field, it became natural to introducea field for any kind of particle.

Historically, the attempt of making quantum mechanics a relativis-tic theory was pursued by looking for relativistic generalizations of theSchrodinger equation. Later it was realized that these equations should berather used as equations for the fields describing the corresponding parti-cles. As we shall see, these equations describe correctly the energy disper-


The Klein-Gordon field 41

sion relation and the spin of the various particles. Therefore their solutionscan be used as a basis for the expansion of the fields in terms of creationand annihilation operators. In order to illustrate this procedure, let us startconsidering the Schrodinger equation for a free particle

i∂ψ

∂t= Hψ, (3.3)

where H is the Hamiltonian operator

H =|~p|2

2m= − 1

2m|~∇|2. (3.4)

If we take a wave function describing an eigenstate of energy and momentum

ψ ≈ e−iEt+ i~p · ~x, (3.5)

all the informations in the equation amount to reproduce correctly theenergy-momentum relation

E =|~p|2

2m. (3.6)

In the relativistic case one could try to reproduce the positive energy branchof the dispersion relation (3.1). In that case one could start from theHamiltonian

H =√|~p|2 +m2, (3.7)

giving rise to the wave equation

i∂ψ

∂t=(√−|~∇|2 +m2

)ψ. (3.8)

There are two obvious problems with this equation:

• spatial and time derivatives appear in a non symmetric way;• the equation is non-local in space since it depends on an infinite number

of spatial derivatives(√−|~∇|2 +m2

)ψ = m

√1− |~∇|2m2

ψ = m

∞∑k=0

ck

(|~∇|2

)kψ.

(3.9)

Both these difficulties are eliminated by iterating eq. (3.8)

−∂2ψ

∂t2=(−|~∇|2 +m2

)ψ. (3.10)



This equation is both local and invariant under Lorentz transformations,in fact we can write it in the following form(

+m2)ψ = 0, (3.11)

where

=∂2

∂t2− |~∇|2 (3.12)

is the d’Alembert operator in (3 + 1) dimensions defined in eq. (1.7).Notice that in order to solve the difficulties we have listed above wehave been obliged to consider both types of solutions: positive energy,E =

√|~p|2 +m2, and negative energy E = −

√|~p|2 +m2. The equation

we have obtained in this way is known as the Klein-Gordon equation [Klein(1926); Gordon (1926); Schroedinger (1926b)]. As a relativistic extensionof the Schrodinger theory it was initially discarded because it gives rise toa non definite positive probability. In fact, if ψ and ψ? are two solutions ofsuch an equation, we can write the following identity

0 = ψ?( +m2)

)ψ − ψ

( +m2)

)ψ? = ∂µ [ψ?∂µψ − (∂µψ?)ψ] . (3.13)

Therefore the current

Jµ = ψ?∂µψ − (∂µψ?)ψ (3.14)

has zero four-divergence and the quantity∫d3x J0 =

∫d3x(ψ?ψ − ψ?ψ) (3.15)

is a constant of motion. But we cannot interpret the time-component of thecurrent as a probability density, as we do in the Schrodinger case, becauseit is not positive definite.

3.2 Quantization of the Klein-Gordon field

In this Section we will discuss the quantization of the Klein-Gordon field,that is a field satisfying the equation (3.11). The quantization will beperformed by following the steps we have previously outlined, that is

• construction of the Lagrangian density and determination of the canon-ical momentum density Π(x);

• quantization through the requirement of canonical commutation rela-tions

[φ(x, t),Π(y, t)] = iδ3(x− y),

[φ(x, t), φ(y, t)] = [Π(x, t),Π(y, t)] = 0 , (3.16)



• expansion of φ(x, t) in terms of a complete set of solutions of the Klein-Gordon equation. This in order to define the appropriate creation andannihilation operators;

• construction of the Fock space.

We start by looking for the Lagrangian density, requiring that the re-lated Euler-Lagrangian equation gives rise to the Klein-Gordon equation.To this end let us recall how one proceeds in the discrete case. Suppose tohave a system of N degrees of freedom satisfying the following equations ofmotion

miqi = −∂V∂qi

. (3.17)

Multiplying these equations by some arbitrary variations δqi, satisfying thefollowing boundary conditions

δqi(t1) = δqi(t2) = 0, (3.18)

summing over i, and integrating in time between t1 and t2, we get∫ t2

t1

dt

[N∑i=1

miqiδqi

]= −

∫ t2

t1

dt

N∑i=1

δqi∂V

∂qi. (3.19)

Integration by parts leads to

δ

∫ t2

t1

dt

[12

N∑i=1

miq2i − V

]−

[N∑i=1

miqiδqi

]t2t1

= 0. (3.20)

Using the boundary conditions we see that, if the equations of motion aresatisfied, then the action

S =∫ t2

t1

[12

N∑i=1

miq2i − V

]dt (3.21)

is stationary. Conversely from the requirement that the action is stationaryunder variations satisfying eq. (3.18), the equations of motion follow. Anal-ogously, in the Klein-Gordon case, we multiply the equation by arbitrarylocal variations of the field δφ(x) = φ(x)− φ(x), with boundary conditionsfor the fields

δφ(x, t1) = δφ(x, t2) = 0, lim~x→∞

δφ(x, t) = 0. (3.22)

Then we integrate over space and time. After integrating by parts we find

0 =∫ t2

t1

dt

∫d3x

[∂

∂t

(φδφ

)− φδφ− ~∇ ·

(~∇φδφ

)+ ~∇φ · ~∇δφ+m2φδφ

].

(3.23)



Using the boundary conditions we get

0 = δ

∫ t2

t1

dt

∫d3x

[12φ2 − 1

2~∇φ · ~∇φ− 1

2m2φ2

]. (3.24)

Therefore the Lagrangian will be given by

L =∫

d3xL, (3.25)

with

L =12[∂µφ∂

µφ−m2φ2]. (3.26)

We have just shown that the quantity (the action)

S =∫ t2

t1

dtL (3.27)

is stationary along the path where the equations of motion are satisfied.We can now write down the canonical momentum density

Π =∂L∂φ

= φ (3.28)

and the canonical commutation relations

[φ(x, t), φ(y, t)] = iδ3(x− y), [φ(x, t), φ(y, t)] = [φ(x, t), φ(y, t)] = 0.(3.29)

Let us now construct a complete set of solutions of the Klein-Gordon equa-tion. First of all we need a scalar product. But we have already shown inthe previous Section that the Klein-Gordon equation admits a conservedquantity (see eq. (3.15)).Therefore, if f and g are two solutions, the scalarproduct can be defined as

〈f |g〉 = i

∫d3xf∗∂

(−)t g, (3.30)

where the operator ∂(−)t has been defined in eq. (2.119). Let us now look

for plane-wave solutions

f(x) = A(k)e−ikx = A(k)e−i(k0x0 − ~k · ~x). (3.31)

From the wave equation we get

( +m2)f = (−k2 +m2)f = 0, (3.32)

implying

k2 = m2 =⇒ k20 = |~k|2 +m2. (3.33)



To fix the normalization, we proceed as in the one-dimensional case bytaking a finite volume and requiring periodic boundary conditions (normal-ization in the box). By taking a cube of side L we require

φ(x+ L, y, z, t) = φ(x, y + L, z, t) = φ(x, y, z + L, t) = φ(x, y, z, t), (3.34)

implying

~k =2πL~n, (3.35)

where

~n = n1~i1 + n2

~i2 + n3~i3, (3.36)

is a vector with integer components (n1, n2, n3). The normalization condi-tion is

〈f~k|f~k′〉 = i

∫V

d3xf∗~k∂(−)t f~k′ = δ~k,~k′ , (3.37)

where the delta is a Kronecker symbol defined as

δ~k,~k′ =3∏i=1

δni,n′i , (3.38)

with ~n e ~n′ two vectors with integer components, related to ~k and ~k′, bythe relation (3.35). It follows∫

V

d3xA∗~kA~k′ei(k0 − k0

′)x0 − i(~k − ~k′) · ~x(k′0 + k0) = δ~k,~k′ . (3.39)

Using ∫L

dxei2πL

(n1 − n′1)x= Lδn1,n1′ , (3.40)

we get

i

∫V

d3xf∗~k∂(−)t f~k′ = |A~k|

22k0L3δ~k,~k′ , (3.41)

where

k20 =

(2πL

)2

|~n|2 +m2. (3.42)

By considering the positive solutions of this equation, we obtain

A~k =1

L3/2

1√2ωk

, ωk =

√(2πL

)2

|~n|2 +m2 =√|~k|2 +m2 (3.43)



and

f~k(x) =1

L3/2

1√2ωk

e−ikx. (3.44)

Often we will make use of the so called normalization in the continuum.Thespace integration is then extended to all of R3 and we require

〈f~k|f~k′〉 = i

∫d3xf∗~k∂

(−)t f~k′ = δ3(~k − ~k′). (3.45)

In this case the spatial momentum can assume all the possible values inR3. It follows∫

d3xA∗~kA~k′eikx− ik′x(k0 + k′0) = (2π)3δ3(~k − ~k′)|A~k|

22k0 (3.46)

and the corresponding normalization is

A~k =1√

(2π)3

1√2ωk

, (3.47)

where

ωk =√|~k|2 +m2. (3.48)

We see that one goes from the normalization in the box to the normalizationin the continuum through the formal substitution

1√V→ 1√

(2π)3. (3.49)

The wave function in the continuum is

f~k(x) =1√

(2π)3

1√2ωk

e−ikx. (3.50)

In both cases the dispersion relation

k20 = |~k|2 +m2 (3.51)

is obviously satisfied, but we have to remember that there are two indepen-dent solutions

k0 = ±√|~k|2 +m2 = ±ωk. (3.52)

As a consequence we get two different types of wave functions with positiveand negative energy behaving respectively as e−iωkx0 and eiωkx0 , ωk > 0.The second kind of solutions has negative norm in the scalar product wehave defined. This would be a big problem if this equation had the sameinterpretation as the Schrodinger equation. However in field theory thisproblem does not arise. In fact, the physical Hilbert space is the Fock



space, where the scalar product is between the states built up in termsof creation and annihilation operators. Correspondingly, for the negativeenergy solutions we will make use of the same normalization coefficientfound for the positive one.

Having two types of solutions the most general expansion for the fieldoperator (in the Heisenberg representation) is

φ(x) =1√

(2π)3

∫d3k

1√2ωk

[a(~k)e−iωkx0 + i~k · ~x + a(~k)eiωkx0 + i~k · ~x

].

(3.53)Exchanging ~k → −~k in the second term we get

φ(x) =1√

(2π)3

∫d3k

1√2ωk

[a(~k)e−ikx + a(−~k)eikx

]=∫

d3k[f~k a(~k) + f∗~k a(−~k)]. (3.54)

Notice that the energy positive and negative solutions are orthogonal (re-member the one-dimensional case discussed in Section 2.5). We can theninvert the previous expansion with the result

a(~k) = i

∫d3xf∗~k (x)∂(−)

t φ(x), a(−~k) = i

∫d3xφ(x)∂(−)

t f~k(x). (3.55)

If φ(x) is hermitian, we have

a(−~k) = a†(~k) (3.56)

and the expansion becomes

φ(x) =∫

d3k[f~k(x)a(~k) + f∗~k (x)a†(~k)]. (3.57)

From the equations (3.55) one can easily evaluate the commutators amongthe operators a(~k) e a†(~k), obtaining

[a(~k), a†(~k′)] = δ3(~k − ~k′), (3.58)

[a(~k), a(~k′)] = [a†(~k), a†(~k′)] = 0. (3.59)

These commutation relations depend on the normalization defined for thef~k’s. For instance, if we change this normalization by a factor N~k

〈f~k|f~k′〉 = i

∫d3xf∗~k∂

(−)t f~k′ = N~k δ

3(~k − ~k′), (3.60)

leaving unchanged the expansion for the field

φ =∫

d3k[f~k a(~k) + f∗~k a†(~k)], (3.61)



we get

a(~k) =i

N~k

∫d3xf∗~k∂

(−)t φ, a†(~k) =

i

N~k

∫d3xφ∂

(−)t f~k (3.62)

and therefore

[a(~k), a†(~k′)] =i

N~kN~k′

∫d3xf∗~k∂

(−)t f~k′ =

1N~k

δ3(~k − ~k′). (3.63)

For instance, a normalization which is used very often is the covariant one

φ(x) =1

(2π)3

∫d3k

12ωk

[A(~k)e−ikx +A†(~k)eikx]. (3.64)

The name of this normalization is because the factor 1/2ωk makes theintegration over the three-momentum Lorentz invariant. In fact, one has

1(2π)3

∫d3k

12ωk

=1

(2π)4

∫d4k(2π)δ(k2 −m2)θ(k0), (3.65)

as it follows by noticing that for k0 ≈ ωk

k2 −m2 ≈ 2ωk

(k0 −

√|~k|2 +m2

). (3.66)

In this case the basis functions for the expansion are

f~k(x) =1

(2π)3

12ωk

e−ikx, (3.67)

with normalization

i

∫d3xf∗~k∂

(−)t f~k′ =

1(2π)3

12ωk

δ3(~k − ~k′) (3.68)

and therefore

[A(~k), A†(~k′)] = (2π)32ωkδ3(~k − ~k′). (3.69)

3.3 Noether’s theorem for relativistic fields

We will now review Noether’s theorem [Noether (1918)]. This theorem re-lates symmetries of the action with conserved quantities. More precisely,for any continuous transformation of fields and/or coordinates leaving in-variant the action, there exists a conserved quantity. Transformations, thatdo not involve the coordinates, are called internal transformations. Now



consider the case of a generic transformation. The variation of a local quan-tity F (x) (that is a function of the space-time point) is given, at first order,by

∆F (x) = F (x′)− F (x) = F (x+ δx)− F (x)

∼= F (x)− F (x) + δxµ∂F (x)∂xµ

. (3.70)

The total variation ∆ takes into account both the variation of the coordi-nates and the form variation of F (F → F ). It is then convenient to definea local variation δF , depending only on the variation in form of F (x)

δF (x) = F (x)− F (x). (3.71)Therefore the total variation can be written as the sum of two transforma-tions

∆F (x) = δF (x) + δxµ∂F (x)∂xµ

. (3.72)

Let us now start form a generic four-dimensional action

S =∫V

d4x L(φi, x), i = 1, . . . , N (3.73)

and let us consider an arbitrary variation of fields and coordinates,x′µ = xµ + δxµ, (3.74)

∆φi(x) = φi(x′)− φi(x) ≈ δφi(x) + δxµ∂φi

∂xµ. (3.75)

If the action is invariant under the transformation, thenSV ′ = SV . (3.76)

The variation of SV under (3.74) and (3.75) is given by (here φi,µ =∂φi/∂xµ),

δSV =∫V ′

d4x′L(φi, x′)−∫V

d4xL(φi, x)

=∫V

d4xL(φi, x+ δx)∂(x′)∂(x)

−∫V

d4xL(φi, x), (3.77)

where we have performed a change of variables x′ = x+ δx. Then

δSV ≈∫V

d4xL(φi, x+ δx)(1 + ∂µδxµ)−

∫V

d4xL(φi, x)

=∫V

d4x[L(φi, x+ δx)− L(φi, x)] +∫V

d4xL(φi, x)∂µδxµ

≈∫V

d4x

[∂L∂φi

δφi +∂L∂φi,µ

δφi,µ +∂L∂xµ

δxµ]

+∫V

d4xL∂µδxµ

=∫V

d4x

[∂L∂φi− ∂µ

∂L∂φi,µ

]δφi

+∫V

d4x∂µ

[Lδxµ +

∂L∂φi,µ

δφi]. (3.78)



When we choose variations such that the surface term is vanishing (see eqs.(2.82 and 2.83)) we get the Euler-Lagrange equations of motion

∂L∂φi− ∂µ

∂L∂φi,µ

= 0. (3.79)

Considering transformations of fields satisfying the equations of motion andleaving invariant SV , we find, using eq. (3.75)]∫

V

d4x ∂µ

[Lδxµ +

∂L∂φi,µ

∆φi − δxν ∂L∂φi,µ

φi,ν

]= 0. (3.80)

This proves Noether’s theorem, asserting that for any infinitesimal transfor-mation of fields and coordinates leaving invariant the action, there exists acurrent with zero four-divergence and a corresponding conserved quantity,the space integral of the zero component of the current. According to thechoice one does for the variations δxµ and ∆φi, and for the correspondingsymmetries of the action, one gets different conserved quantities.

Let us start with an action invariant under space and time translations.In this case we have δxµ = aµ with aµ independent on x and ∆φi = 01.From the general result in eq. (3.80) we get the following local conservationlaw

Tµν =∂L∂φi,µ

φi,ν − Lgµν , ∂µTµν = 0. (3.81)

Tµν is called the energy-momentum tensor of the system. From its localconservation we get four constants of motion

Pν =∫

d3xT 0ν . (3.82)

Pµ is the four-momentum of the system. In the case of internal symmetrieswe take δxµ = 0. The conserved current is

Jµ =∂L∂φi,µ

∆φi =∂L∂φi,µ

δφi, ∂µJµ = 0, (3.83)

with an associated constant of motion given by

Q =∫

d3xJ0. (3.84)

In general, if the system has more that one internal symmetry, we may havemore than one conserved charge Q, that is we have a conserved charge foreach independent infinitesimal transformation ∆φi.1There is no variation in form for a field under a space-time translation.



The last case we will consider is the invariance with respect to Lorentztransformations. For an infinitesimal transformation (see eq. (1.42))

x′µ = xµ + εµνxν . (3.85)

In general, the relativistic fields are chosen to belong to a representationof the Lorentz group (for instance the Klein-Gordon field belongs to thescalar representation). This means that under a Lorentz transformationthe components of the field mix together as, for instance, the componentsof a vector field under rotations. Therefore, the transformation law of thefields φi under an infinitesimal Lorentz transformation can be written ingeneral as

∆φi = −12

Σijµνεµνφj , (3.86)

where we have required that the transformation of the fields is of first orderin the Lorentz parameters εµν . The coefficients Σµν (antisymmetric in theindices (µ, ν)) define a matrix in the indices (i, j) which can be shown tobe the representative of the infinitesimal generators of the Lorentz groupin the field representation. Using this equation and the expression for δxµwe get the local conservation law

0 = ∂µ

[(∂L∂φi,µ

φi,ν − Lgµν)ενρxρ +

12∂L∂φi,µ

Σijνρενρφj

]=

12ενρ∂µ

[(Tµν xρ − Tµρ xν

)+

∂L∂φi,µ

Σijνρφj

](3.87)

and defining

Mµρν = xρT

µν − xνTµρ −

∂L∂φi,µ

Σijρνφj , (3.88)

we obtain six locally conserved currents (one for each Lorentz transforma-tion)

∂µMµνρ = 0 (3.89)

and consequently six constants of motion (notice that the lower indices areantisymmetric)

Mνρ =∫

d3xM0νρ. (3.90)

Three of these constants (the ones with ν and ρ assuming spatial values)are nothing but the components of the angular momentum of the field.



In the Klein-Gordon case

Tµν = ∂µφ∂νφ−12(∂ρφ∂

ρφ−m2φ2)gµν , (3.91)

from which

T 00 =

12φ2 +

12

(|~∇φ|2 +m2φ2

). (3.92)

This current corresponds to the invariance under time translations, and itmust be identified with the energy density of the field (compare with theequation (2.96) for the one-dimensional case). In analogous way

T 0i = φ

∂φ

∂xi(3.93)

is the momentum density of the field. Using φ = Π, the energy and mo-mentum of the Klein-Gordon field can be written in the form

P 0 = H =∫

d3xT 00 =12

∫d3x

(Π2 + |~∇φ|2 +m2φ2

), (3.94)

P i =∫

d3xT 0i = −∫

d3xΠ∂φ

∂xi, (~P = −

∫d3xΠ~∇φ). (3.95)

3.4 Energy and momentum of the Klein-Gordon field

It is very easy to verify that the energy density found previously coincideswith the Hamiltonian density evaluated in the canonical way through theLegendre transformation of the Lagrangian density

H = Πφ− L. (3.96)

Then, we will show that the four-momentum Pµ is the generator, as itshould be, of space-time translations. This amounts to say that it satisfiesthe following commutation relation with the field

[φ(x), Pµ] = i∂φ

∂xµ. (3.97)

In fact,

[φ(~y, t), H] =12

∫d3x[φ(~y, t),Π2(~x, t)] = iΠ(~y, t) = iφ(~y, t). (3.98)

Analogously

[φ(~y, t), P i] = −∫

d3x[φ(~y, t),Π(~x, t)∂φ(~x, t)∂xi

] = −i∂φ(~y, t)∂yi

= i∂φ(~y, t)∂yi

.

(3.99)



Therefore the operator

U = eiaµPµ (3.100)

generates translations in x. In fact, by looking at the first order in aµ, itfollows

eia · P φ(x)e−ia · P ≈ φ(x) + iaµ[Pµ, φ(x)] = φ(x) + aµ∂φ(x)∂xµ

≈ φ(x+ a).

(3.101)With a calculation completely analogous to the one done in Section 2.5

we can evaluate the Hamiltonian and the momentum in terms of creationand annihilation operators

H =12

∫d3k ωk[a†(~k)a(~k) + a(~k)a†(~k)], (3.102)

~P =∫

d3k~k a†(~k)a(~k). (3.103)

They satisfy the following commutation relations with a†(~k)

[H, a†(~k)] = ωka†(~k), [~P , a†(~k)] = ~ka†(~k). (3.104)

This shows that the operators a†(~k), acting on the vacuum, create statesof momentum ~k and energy ωk = (|~k|2 +m2)1/2, whereas the annihilationoperators a(~k) annihilate the vacuum states corresponding to the sameenergy and momentum. In the case of the box normalization, for any~k = (2π/L)~n (that is for any choice of the three integer components of thevector ~n), one can build up a state |n~k〉

|n~k〉 =1√n~k!

(a†(~k)

)n~k |0〉, (3.105)

containing n~k particles of momentum ~k and energy ωk. The most generalstate is obtained by tensor product of states similar to the previous one.Any of these states is characterized by a triple of integers defining themomentum ~k, that is

|n~k1 . . . n~kα〉 =∏⊗|n~ki〉 =

1√n~k1 ! · · ·n~kα !

(a†(~k1))n~k1 · · · (a†(~kα))n~kα |0〉.

(3.106)The fundamental state is the one with zero particles in any cell of themomentum space (vacuum state)

|0〉 =∏⊗|0〉i, (3.107)



where |0〉i is the fundamental state for the momentum in the cell i. Thatis

a~ki |0〉i = 0. (3.108)

In this normalization the Hamiltonian is given by

H =12

∑~k

ωk[a†(~k)a(~k) + a(~k)a†(~k)] (3.109)

and therefore

H|0〉 =12

∑~k

ωk|0〉. (3.110)

This sum is infinite. Recalling that ~k = (2π/L)~n, it follows that the cell inthe ~k-space has a volume

∆V~k =(2π)3

L3, (3.111)

from which12

∑~k

ωk =12

∑~k

∆V~kωk∆V~k

=⇒ 12

L3

(2π)3

∫d3k

√|~k|2 +m2, (3.112)

which is divergent.Let us recall that this problem can be formally avoided through the use

of the normal product. In other words by subtracting the infinite energy ofthe vacuum from the Hamiltonian. In this case

: H :=∑~k

ωk a†(~k)a(~k), (3.113)

whereas in the continuum

: H :=∫

d3k ωk a†(~k)a(~k). (3.114)

As we see, the energy of the vacuum depends on the quantization volume.This implies that it depends on the boundary conditions of the problem.In the real vacuum this is not a difficulty, but this point should be carefullyconsidered when quantizing fields living inside a finite volume. In this casethe dependence on the boundary produces measurable effects, as it waspointed out theoretically by [Casimir (1948)], and then proved experimen-tally by [Sparnaay (1958)]. We will discuss very briefly the Casimir effectarising when we have an electromagnetic field confined between two largeperfectly conducting plates. We idealize the two plates as two large parallel



L

LR

L

Fig. 3.1 The Casimir effect.

squares of side L at a distance R L. The theory shows that there is anattractive force per unit surface, p, between the two plates given by

p = − π2

240~cR4

(3.115)

We can understand the origin of this force in a very qualitative way, inparticular its dependence on the plate distance, by quantizing the electro-magnetic field (that we will take here as a Klein-Gordon field with zeromass, m = 0) in a box of side L. In this case the vacuum energy is

E0 ≈ L3

∫ kmax

1/L

k d3k, (3.116)

with the integration between a lower momentum of order 1/L and an ar-bitrary upper momentum which is necessary in order to make the integralfinite. The energy contained in the region bounded by the two surfaces ofside L (at the left in Fig. 3.1) is given by

E ≈ L2R

∫ kmax

1/L

k d3k. (3.117)

If we now insert two plates of side L, as shown in Fig. 3.1, at a distanceR, we get an analogous result, but with a lower momentum of order 1/R.Therefore the variation of energy results to be

∆E ≈ L2R

∫ 1/L

1/R

k d3k = L2R

∫ 1/L

1/R

k3 dk =L2R

4

[(1L

)4

−(

1R

)4].

(3.118)



Then, for R L, we get

∆E ≈ −L2

R3. (3.119)

The energy per unit surface behaves as 1/R3, and the pressure is given by

p ≈ −∂∆E/L2

∂R≈ − 1

R4. (3.120)

3.5 Locality and causality in field theory

For a free particle there are generally three conserved quantum numbers as,for instance, the spatial momentum or, in alternative, energy, the square ofthe angular momentum and its third component. All these quantities canbe expressed as space integrals of local functions of the fields. The localityproperty is crucial and it is strictly connected to causality. To understandthis point, let us consider the example of a free Klein-Gordon field. In thiscase there is a further constant of motion, the number of particles

N =∫

d3ka†(~k)a(~k). (3.121)

We will show that this cannot be written as the spatial integral of a localquantity, and that this implies the non observability of the number of par-ticles. We know that for the free Klein-Gordon theory there is the followingconserved current

Jµ = φ†(∂µφ)− (∂µφ†)φ. (3.122)

This expression vanishes for a hermitian field but, nevertheless, the numberof particles, N , can be expressed in terms of the positive and negative energycomponents of the field

φ(+)(x) =∫

d3k1√

2ωk(2π)3e−i(ωkt− ~k · ~x)a(~k), φ(−)(x) = φ(+)†(x).

(3.123)In fact, it is not difficult to show that

N =∫

d3xφ(−) i∂(−)t φ(+). (3.124)

This is a constant of motion, because φ(+) and φ(−) are both solutions ofthe equation of motion, and therefore

jµ = iφ(−)(∂µφ(+))− (∂µφ(−))φ(+) (3.125)



is a conserved current. However, φ(+) and φ(−) are not local functions ofφ and therefore also the conserved current is not local. In fact, to relatethe negative and positive frequency components to the field we need a timeintegration. Let us define

φ(x) =∫d4kφ(k)e−ikx, (3.126)

with

φ(k) =

√2ωk

(2π)3δ(k2 −m2) (a(k)θ(k0) + a(k)θ(−k0)) , (3.127)

one has

φ(+)(x) =∫d4kφ(+)(k)e−ikx, (3.128)

with

φ(+)(k) = θ(k0)φ(k). (3.129)

Using the convolution theorem for the Fourier transform we get

φ(+)(x) =∫d4x′θ(x− x′)φ(x′). (3.130)

But

θ(x− x′) =1

(2π)4

∫d4keik(x− x′)θ(k0)

= δ3(~x− ~x′)∫dk0

2πeik0(x0 − x′0)θ(k0)

= δ3(~x− ~x′)θ(x0 − x′0). (3.131)

Therefore

φ(+)(~x, x0) =∫dx′0θ(x0 − x′0)φ(~x, x′0). (3.132)

To show the implications of having to do with a non local current, let usdefine a particle density operator

N (x) = iφ(−)∂(−)t φ(+). (3.133)

This operator does not commute with itself at equal times and differentspace points

[N (~x, t),N (~y, t)] 6= 0, ~x 6= ~y. (3.134)



However, for local operators2, O(φ), this commutator is automatically zero,due to the canonical commutation relations

[O(φ(~x, t)),O(φ(~y, t))] = 0, ∀ ~x 6= ~y. (3.135)

We want to argue that the vanishing of this commutator is just the necessarycondition in order that O represents an observable quantity. In fact, if thecommutator of a local operator with itself is not zero at space-like distances,the measurement of the observable at some point, x, would influence themeasurements done at points with space-like separation from x, because wecannot diagonalize the operator simultaneously at two such points. But thiswould imply the propagation of a signal at a velocity greater than the lightvelocity, in contrast with the causality principle. We see that the vanishingof the commutator of a local observable with itself at space-like distancesis a necessary condition in order to satisfy the causality principle. We shownow that this is automatically satisfied if the operator under considerationis a local function of the fields. We will start showing that the commutatorof the field with itself is a Lorentz invariant function. Therefore, from thevanishing of the commutator for separations between points of the typexµ = (t, ~x), and yµ = (t, ~y), it follows the vanishing for arbitrary space-likeseparations. Let us evaluate the commutator

[φ(x), φ(y)] =∫

d3k1d3k2

(2π)3√

2ωk12ωk2

×[[a(~k1), a†(~k2)]e−ik1x+ ik2y

+[a†(~k1), a(~k2)]eik1x− ik2y]

=∫

d3k

(2π)32ωk

[e−ik(x− y) − eik(x− y)

]= −2i

∫d3k

(2π)32ωksin(ωk(x0 − y0))ei

~k(~x− ~y). (3.136)

Using eq. (3.65), this expression can be written in invariant form

[φ(x), φ(y)] =∫

d4k

(2π)3θ(k0)δ(k2 −m2)

[e−ik(x− y) − eik(x− y)

]=∫

d4k

(2π)3ε(k0)δ(k2 −m2)e−ik(x− y). (3.137)

2A local operator is a function of the fields and of a finite number of derivatives of the

fields.



Since the sign of the fourth component of a time-like four-vector is invariantunder proper Lorentz transformations, we see that defining

[φ(x), φ(y)] = i∆(x− y), (3.138)

the function

∆(x− y) = −i∫

d4k

(2π)3ε(k0)δ(k2 −m2)e−ik(x− y) (3.139)

is Lorentz invariant and, as such, it depends only on (x− y)2. Since ∆(x−y) vanishes at equal times, it follows that it is zero for arbitrary space-like separations. Therefore the canonical commutation relations ensure theobservability of the Klein-Gordon field. For the negative and positive energycomponents we get

∆(+)(x− y) ≡ [φ(+)(x), φ(−)(y)] =∫

d3k

(2π)32ωke−ik(x− y)

=∫

d4k

(2π)3θ(k0)δ(k2 −m2)e−ik(x− y). (3.140)

Also in this case we have a Lorentz invariant function, and therefore it isenough to study its equal times behavior:

∆(+)(0, ~x) =∫

d3k

(2π)32ωkei~k · ~x =

∫k2dkd(cos θ)dϕ

(2π)32ωkeikr cos θ

= −i 14π2r

∫ ∞0

kdk

2ωk

[eikr − e−ikr

]= − 1

8π2r

d

dr

∫ +∞

−∞dk

eikr√|~k|2 +m2

. (3.141)

By putting k = m sinh θ, we get

∆(+)(0, ~x) = − 18π2r

d

dr

∫ +∞

−∞dθeimr sinh θ. (3.142)

The integral defines a Hankel function H(1)0∫ +∞

−∞dθeimr sinh θ = iπH

(1)0 (imr). (3.143)

Using the following relation between H(1)0 and H

(1)1

d

drH

(1)0 (imr) = −imH(1)

1 (imr), (3.144)

we obtain

∆(+)(0, ~x) = − m

8πrH

(1)1 (imr). (3.145)



The asymptotic behavior of the Hankel function H(1)1 (imr) for large and

small values of r is given by

limr→∞

H(1)1 (imr) ≈ −

√2

πmre−mr, lim

r→0H

(1)1 (imr) ≈ − 2

mr, (3.146)

from which

limr→∞

∆(+)(0, ~x) ≈ m

8πr

√2

πmre−mr, lim

r→0∆(+)(0, ~x) ≈ 1

4πr2. (3.147)

We see that for space-like separations this commutator does not vanish.But for space separations larger than the Compton wavelength 1/m, ∆(+)

is practically zero. Remember that for an electron the Compton wavelengthis about 3.9 · 10−11 cm. Clearly, an analogous result is obtained for thecommutator of the particle density operator. From this we can derive theimpossibility of localize a Klein-Gordon particle (but the result can beextended to any relativistic particle) over distances of the order of 1/m. Toprove this let us introduce the following operators

N(V ) =∫V

d3xN (x), (3.148)

where V is a sphere. Suppose we want to localize the particle around a

R

riV0

Vi

x0

Fig. 3.2 In order to localize a particle inside V0, there should be no other particles

within a distance ri ≈ R.

point x0 with an uncertainty R. We consider a sphere V0 centered at x0

with radius R. We then take other spheres Vi not connected to V0, that is



with centers separated by x0 by a distance ri > 2R, as shown in Fig. 3.2.The requirement to localize the particle within V0, with a radius R < 1/m,is equivalent to ask for the existence of a state with eigenvalue 1 for theoperator N(V0) and eigenvalues 0 for all the N(Vi) with ri ≈ 1/m. Butsuch an eigenstate does not exist because, as we have shown previously[N(V0), N(Vi)] 6= 0. On the contrary, if we take volumes Vi at distances rimuch bigger than 1/m, the corresponding operators N(Vi) commute, andwe can construct the desired state. Therefore it is possible to localize theparticle only over distances much bigger than the Compton wavelength.The physical explanation is that to realize the localization over distancesmuch smaller than 1/m, we need energies much bigger than m. But in thiscase there is a non zero probability to create particle antiparticle pairs.

To summarize, a local quantity can be an observable if and only if thecommutator with itself vanishes for space-like distances, otherwise therewould be a violation of the causality principle. If the quantity is a localfunction of the fields, the previous condition is automatically satisfied dueto the canonical commutation relations. As a consequence, a relativisticparticle described by a hermitian Klein-Gordon field, cannot be localizedover distances of the order of 1/m, because the particle density is not alocal function of the fields. From these considerations we see also that thenegative energy components of the fields are essential for the internal con-sistency of the theory. Otherwise the particle interpretation of the field(that is the commutation relations among creation and annihilation opera-tors) and the locality properties (vanishing of the commutators at space-likedistances) would not be compatible.

3.6 The charged scalar field

We have shown that a hermitian Klein-Gordon field describes a set of iden-tical scalar particles. If we want to describe different kind of particles weneed to introduce different kind of fields. As an example, consider two dif-ferent hermitian scalar fields. The free Lagrangian is simply the sum of thetwo free Lagrangian densities describing separately the fields

L =12

2∑i=1

[(∂µφi)(∂µφi)−m2

iφ2i

](3.149)



and we can write immediately the canonical commutation relations, sincethe two fields refer to different degrees of freedom,

[φi(~x, t), φj(~y, t)] = iδijδ3(~x− ~y), (3.150)

[φi(~x, t), φj(~y, t)] = [φi(~x, t), φj(~y, t)] = 0. (3.151)

All the considerations made for a single field can be trivially extended tothe case of two fields. In particular there will be two kind of creation andannihilation operators, a†i(~k), i = 1, 2, which we will assume mutuallycommuting. The two particle state made with different creation operatorsis not any more symmetric

a†1(~k1)a†2(~k2)|0〉 6= a†2(~k1)a†1(~k2)|0〉. (3.152)

This means that the two fields correspond to distinguishable particles.Something new happens when the mass term for the two fields is the

same (m1 = m2 = m)

L =12

[(∂µφ1)(∂µφ1) + (∂µφ2)(∂µφ2)]− 12m2[φ2

1 + φ22

]. (3.153)

Then, the theory acquires a symmetry under rotations in the plane of thetwo fields φ1 and φ2

φ′1 = φ1 cos θ + φ2 sin θ,

φ′2 = −φ1 sin θ + φ2 cos θ. (3.154)

In fact, the Lagrangian is a function of the norm of the following two-dimensional vectors

~φ = (φ1, φ2), ∂µ~φ = (∂µφ1, ∂µφ2) (3.155)

and their norm is invariant under rotations in the plane. For infinitesimaltransformations we have

δφ1 = φ2θ, δφ2 = −φ1θ, (3.156)

or, in a more compact form

δφi = εijφjθ, (3.157)

where εij is the two-dimensional antisymmetric Ricci tensor, defined by

ε12 = −ε21 = 1. (3.158)

From the Noether theorem, we have a conserved current, associated to thissymmetry, given by (see eq. (3.83))

Jµ =∂L∂φi,µ

∆φi = φ,µi εijφjθ. (3.159)



It is convenient to factorize out the angle of the infinitesimal rotation anddefine a new current

jµ =1θJµ = φ,µi εijφj = φ,µ1 φ2 − φ,µ2 φ1. (3.160)

The conservation of the current follows from the equality of the masses ofthe two fields, as it can be verified directly from the equations of motionfollowing from eq. (3.149):

∂µjµ = (φ1)φ2 − (φ2)φ1 = −(m2

1 −m22)φ1φ2. (3.161)

The conserved charge associated to the current is

Q =∫

d3x j0 =∫

d3x(φ1φ2 − φ2φ1) (3.162)

and it is the generator of the infinitesimal transformations of the fields

[Q,φ1] = −iφ2, [Q,φ2] = iφ1. (3.163)

Correspondingly a finite transformation is given by

U = eiQθ. (3.164)

In fact,

eiQθφ1e−iQθ = φ1 + iθ[Q,φ1] +

i2

2!θ2[Q, [Q,φ1]] + . . .

= φ1 + θφ2 −12φ1θ

2 + . . .

= φ1 cos θ + φ2 sin θ. (3.165)

In analogous way one can derive the transformation properties of φ2. Theinvariance of L under rotations in the plane (φ1, φ2) is referred to as theinvariance under the group O(2)3. The real basis for the fields used so faris not the most convenient one. In fact, the charge Q mixes the two fields.One can understand better the properties of the charge operator in a basisin which the fields are not mixed. This basis is a complex one and it isgiven by the combinations

φ =1√2

(φ1 + iφ2), φ† =1√2

(φ1 − iφ2), (3.166)

(the factor 1/√

2 has been inserted for a correct normalization of the fields).It follows

[Q,φ] =1√2

[Q,φ1 + iφ2] =1√2

(−iφ2 − φ1) = −φ (3.167)

3The notation O(N) is used to characterize the orthogonal transformations in N di-

mensions.



and analogously

[Q,φ†] = φ†. (3.168)

Therefore the field φ lowers the charge of an eigenstate of Q by one unit,whereas φ† increases the charge by the same amount. In fact, if Q|q〉 = q|q〉

Q(φ|q〉) = ([Q,φ] + φQ)|q〉 = (−1 + q)φ|q〉 (3.169)

and

φ|q〉 ∝ |q − 1〉. (3.170)

In analogous way

φ†|q〉 ∝ |q + 1〉. (3.171)

Inverting the relations (3.166) we get

φ1 =1√2

(φ+ φ†), φ2 = − i√2

(φ− φ†), (3.172)

from which

L = ∂µφ†∂µφ−m2φ†φ (3.173)

and

jµ = i[(∂µφ)φ† − (∂µφ†)φ

]. (3.174)

The charge operator results to be

Q = i

∫d3x φ†∂

(−)t φ. (3.175)

The commutation relations in the new basis are given by

[φ(~x, t), φ†(~y, t)] = iδ3(~x− ~y) (3.176)

and

[φ(~x, t), φ(~y, t)] = [φ†(~x, t), φ†(~y, t)] = 0,

[φ(~x, t), φ(~y, t)] = [φ†(~x, t), φ†(~y, t)] = 0. (3.177)

Let us notice that these commutation relations could have also been ob-tained directly by canonical quantization of the Lagrangian (3.173), since

Πφ =∂L∂φ

= φ†, Πφ† =∂L∂φ†

= φ. (3.178)

The original O(2) symmetry becomes now an invariance of the La-grangian (3.173) under a phase transformation of the fields. This followsfrom (3.173) but it can be seen also directly

φ→ 1√2

(φ′1 + iφ′2) = e−iθφ (3.179)



and

φ† → eiθφ†. (3.180)

In this basis we speak of invariance under the group U(1)4.Using the expansion for the real fields

φi(x) =∫

d3k[f~k(x)ai(~k) + f∗~k (x)a†i(~k)

], (3.181)

we get

φ(x) =∫

d3k

[f~k(x)

1√2

(a1(~k) + ia2(~k)) + f∗~k (x)1√2

(a†1(~k) + ia†2(~k))]

(3.182)and introducing the combinations

a(~k) =1√2

(a1(~k) + ia2(~k)), b(~k) =1√2

(a1(~k)− ia2(~k)), (3.183)

it follows

φ(x) =∫d3k

[f~k(x)a(~k) + f∗~k (x)b†(~k)

]φ†(x) =

∫d3k

[f~k(x)b(~k) + f∗~k (x)a†(~k)

]. (3.184)

Using these relations we can evaluate the commutation relations for thecreation and annihilation operators in the complex basis

[a(~k), a†(~k′)] = [b(~k), b†(~k′)] = δ3(~k − ~k′), (3.185)

[a(~k), b(~k′)] = [a(~k), b†(~k′)] = 0. (3.186)

We get also

: Pµ :=∫

d3kkµ

2∑i=1

a†i(~k)ai(~k) =∫

d3kkµ

[a†(~k)a(~k) + b†(~k)b(~k)

].

(3.187)Therefore both the operators a†(~k) e b†(~k) create particles states with mo-mentum kµ, as the original operators a†i. The charge Q is given by

Q = i

∫d3xφ†∂

(−)t φ

=∫d3k1d

3k2δ3(~k1 − ~k2)

[a†(~k1)a(~k1)− b(~k1)b†(~k1)

], (3.188)

4U(N) is the group of unitary transformations acting on complex vectors of dimension

N .



where we have used the orthogonality relations (3.45). For the normalordered charge operator we get

: Q :=∫

d3k[a†(~k)a(~k)− b†(~k)b(~k)

], (3.189)

showing explicitly that a† and b† create particles of charge +1 and −1respectively. The two types of particles, with same mass and oppositecharges are called the one the antiparticle of the other. It is important tonotice that the current density j0 is local in the fields, and therefore

[j0(x), j0(y)] = 0, ∀ (x− y)2 < 0. (3.190)

By following the discussion done in Section 3.5, we construct the operators

Q(Vi) =∫Vi

d3x j0(x), (3.191)

which, for any Vi and Vj , commute at equal times. Therefore it is possibleto localize a state of definite charge in an arbitrary spatial region. Thisagrees with the argument we have developed in the case of the number ofparticles, because the pair creation process does not change the charge ofthe state.

Finally we notice that the field operator is a linear combination of anni-hilation, a(~k), and creation, b†(~k), operators, meaning that a local theorydeals in a symmetric way with the annihilation of a particle and the creationof an antiparticle. For instance, to annihilate a charge +1 is equivalent tothe creation of a charge −1.

3.7 Exercises

(1) Show that the two different definitions of the number particle operator,N , given in eqs. (3.121) and (3.124) coincide.

(2) Show that the number of particles N and the Hamiltonian for the Klein-Gordon case are commuting operators.

(3) Derive the commutators in eqs. (3.185) and (3.186) using the expres-sions (3.184) for the field and its hermitian conjugate.

(4) Derive the expressions (3.187) and (3.189) for the momentum and thecharge operators for a charged field.

(5) The Hamiltonian of the Klein-Gordon field is given by eq. (3.94). Usingthe quantum equations of motion for the operators in the Heisenbergrepresentation (Heisenberg equations),

A = i[H,A], (3.192)



for the field and its conjugate momentum, show that the Klein-Gordonequation follows.

(6) Show that for a scalar field the angular momentum density has zerodivergence

∂µMµνρ = 0, (3.193)

if the energy-momentum tensor is symmetric

Tµν = Tνµ. (3.194)

(7) Given a classical Klein-Gordon field satisfying the boundary conditions

φ(t = 0, ~x) = 0, φ(t = 0, ~x) = c, (3.195)

where c is a constant, determine the form of φ at a generic time. (Hint:Use eq. (3.55) in order to find out the coefficients of φ in a plane waveexpansion).

(8) Prove the following equation for a Klein-Gordon field

φ(x) =∫d3y

[φ(y)∂x0∆(x− y)− φ(y)∆(y − x)

](3.196)

Here i∆(x − y) = [φ(x), φ(y)]. Give an argument to justify the factthat this equation holds for any value of y0. The consequence is thatthrough this formula we can solve the Cauchy problem for the field φ

given the boundary conditions on φ and φ

(9) The Lagrangian density for a charged scalar field is invariant under thesubstitution

φ→ ηCφ†, (3.197)

where ηC is a phase factor. It is possible to define a unitary operatorC such that

CφC−1 = ηCφ†. (3.198)

Show that

C2 = 1, Ca(k)C−1 = ηCb(k), Cb(k)C−1 = η∗Ca(k). (3.199)

Furthermore prove that the charge operator Q, changes sign under C

CQC−1 = −Q (3.200)



(10) Given a Klein-Gordon field φ, introduce the 2-component vector

ψ =(θ

χ

)(3.201)

where

θ =12

(φ+

i

mφ

), χ =

12

(φ− i

mφ

)(3.202)

Show that the Klein-Gordon equation can be re-written in the form

i∂ψ

∂t= Hψ (3.203)

with H a 2 × 2 matrix. Find the form of H and show that it is nothermitian in the positive definite metric defined as

〈ψ|ψ〉 =∫d3~xψ†ψ (3.204)

On the other hand it is hermitian in the non positive definite metricdefined as

〈ψ||ψ〉 =∫d3~xψ†σ3ψ (3.205)

where σ3 is a Pauli matrix.


Chapter 4

The Dirac field

4.1 The Dirac equation

In 1928 Dirac tried to solve the problem of a non positive probability den-sity, present in the Klein-Gordon case, formulating a new wave equation.Dirac thought that in order to get a positive probability it was necessaryto have a wave equation of first order in the time derivative (as it happensfor the Schrodinger equation). Therefore Dirac looked for a way to reducethe Klein-Gordon equation (of second order in the time derivative) to afirst-order differential equation. The Pauli formulation of the electron spinput Dirac on the right track. In fact, Pauli showed that in order to describethe spin, it was necessary to generalize the Schrodinger wave function (acomplex number) to a two-component object

ψ → ψα =(ψ1

ψ2

), (4.1)

modifying also the wave equation to a matrix equation

i∂ψα∂t

=2∑

β=1

Hαβψβ , (4.2)

where the Hamiltonian H is, in general, a 2× 2 matrix. The electron spinis then described by a special set of 2× 2 matrices, the Pauli matrices, ~σ

~S =12~σ. (4.3)

Dirac realized that it was possible to write the squared norm of a spatialvector as

|~k|2 = (~σ · ~k)2, (4.4)

69



as it follows from

[σi, σj ]+ = 2δij , (4.5)

where [A,B]+ = AB + BA is the anticommutator of the operators A andB.

Following this suggestion Dirac tried to write down a first order differ-ential equation for a many-component wave function

i∂ψ

∂t= −i~α · ~∇ψ + βmψ ≡ Hψ, (4.6)

where ~α and β are matrices. The requirements that this equation shouldsatisfy are

• the wave function ψ, solution of the Dirac equation, should satisfyalso the Klein-Gordon equation in order to get the correct dispersionrelation between energy and momentum;

• the equation should admit a conserved current with the fourth compo-nent being positive definite;

• the equation should be covariant with respect to Lorentz transforma-tions (see later).

In order to satisfy the first requirement, we iterate the Dirac equationand ask that the resulting second order differential equation coincides withthe Klein-Gordon equation

−∂2ψ

∂t2= (−i~α · ~∇+ βm)2ψ

=(−αiαj ∂2

∂xi∂xj+ β2m2 − i(β~α+ ~αβ) · ~∇

)ψ

=(−1

2[αi, αj

]+

∂2

∂xi∂xj+ β2m2 − i(β~α+ ~αβ) · ~∇

)ψ. (4.7)

We see that it is necessary to require the following matrix relations[αi, αj

]+

= 2δij ,[αi, β

]+

= 0, β2 = 1. (4.8)

We would like the Hamiltonian, H, to be hermitian. To this end we require~α and β to be hermitian matrices. Since for any choice of the index i,(αi)2 = 1, it follows that the eigenvalues of ~α and β must be ±1. We canalso prove the following relations

Tr(β) = Tr(αi) = 0. (4.9)

For instance, from αiβ = −βαi, we get αi = −βαiβ, and therefore

Tr(αi) = −Tr(βαiβ) = −Tr(αi) = 0, (4.10)


The Dirac field 71

where we have made use of the cyclic property of the trace. The conse-quence is that the matrices αi and β can be realized only in a space ofeven dimensions. This is perhaps the biggest difficulty that Dirac had tocope with. In fact, the αi’s enjoy the same properties of the Pauli matri-ces. However, in a 2 × 2 matrix space, a further anticommuting matrix β

does not exist. It took some time before Dirac realized that the previousrelations could have been satisfied by 4× 4 matrices.

An explicit realization of the Dirac matrices is the following

αi =(

0 σiσi 0

), β =

(1 00 −1

), (4.11)

as it can be easily checked.Let us now show that also the second of our requirements is satisfied.

We multiply the Dirac equation by ψ† from the left, and then we considerthe equation for ψ†

−i∂ψ†

∂t= i(~∇ψ†) · ~α+mψ†β, (4.12)

multiplied from the right by ψ. Subtracting the resulting equations we get

iψ†∂ψ

∂t+i

∂ψ†

∂tψ = ψ†(−i~α· ~∇+βm)ψ−(i~∇ψ† ·~α+ψ†βm)ψ = −i~∇·(ψ†~αψ),

(4.13)that is

i∂

∂t(ψ†ψ) + i

∂

∂xj(ψ†αjψ) = 0. (4.14)

We see that the current

jµ = (ψ†ψ,ψ†αiψ) (4.15)

is a conserved one:

∂jµ

∂xµ= 0. (4.16)

Furthermore its fourth component j0 = ψ†ψ is positive definite. Of coursewe have still to prove that jµ is a four-vector, implying that∫

d3x ψ†ψ (4.17)

is invariant under Lorentz transformations.



4.2 Covariance properties of the Dirac equation

To discuss the transformation properties of the Dirac equation underLorentz, it is convenient to write the equation in a slightly different form.Let us multiply the equation by β

iβ∂ψ

∂t= −iβ~α · ~∇ψ +mψ (4.18)

and define the following matrices

γ0 = β =(

1 00 −1

), γi = βαi =

(0 σi−σi 0

). (4.19)

Then the equation becomes(iγ0 ∂

∂x0+ iγi

∂

∂xi−m

)ψ = 0, (4.20)

or, in a more compact way

(i∂ −m)ψ = 0, (4.21)

where

∂ = γµ∂

∂xµ= γµ∂µ. (4.22)

The matrices γµ satisfy the following anticommutation relations[γi, γj

]+

= βαiβαj + βαjβαi = −[αi, αj

]+

= −2δij , (4.23)[γ0, γi

]+

=[β, βαi

]+

= αi + βαiβ = 0, (4.24)

or

[γµ, γν ]+ = 2gµν . (4.25)

Notice that

(γi)† = (βαi)† = αiβ = −γi (4.26)

and

(γi)2 = −1. (4.27)

In order the Dirac equation to be covariant, the following two conditionshave to be satisfied:

• given the Dirac wave function ψ(x) in the Lorentz frame, S, an observerin a different frame, S′ should be able to evaluate, in terms of ψ(x), thewave function ψ′(x′) describing the same physical state as ψ(x) does inS;


The Dirac field 73

• according to the relativity principle, ψ′(x′) must be a solution of anequation that in S′ has the same form as the Dirac equation in S.That is to say (

iγµ∂

∂x′µ−m

)ψ′(x′) = 0. (4.28)

The matrices γµ should satisfy the same algebra as the matrices γµ, becausein both cases the wave functions should satisfy the Klein-Gordon equation(which is invariant in form). Therefore, neglecting a possible unitary trans-formation, the two sets of matrices can be identified. As a consequence,the Dirac equation in S′ will be(

iγµ∂

∂x′µ−m

)ψ′(x′) = 0. (4.29)

Since both the Dirac equation and the Lorentz transformations are linear,we will require the wave functions in two different Lorentz frames to belinearly correlated

ψ′(x′) = ψ′(Λx) = S(Λ)ψ(x), (4.30)

where S(Λ) is a 4×4 matrix operating on the complex vector ψ(x) and Λ isthe Lorentz transformation. On physical grounds, the matrix S(Λ) shouldbe invertible

ψ(x) = S−1(Λ)ψ′(x′), (4.31)

but using the relativity principle, since one goes from the frame S′ to theframe S through the transformation Λ−1, we must have

ψ(x) = S(Λ−1)ψ′(x′), (4.32)

from which

S−1(Λ) = S(Λ−1). (4.33)

Considering the Dirac equation in the frame S(iγµ

∂

∂xµ−m

)ψ(x) = 0, (4.34)

we can write (iγµ

∂

∂xµ−m

)S−1(Λ)ψ′(x′) = 0. (4.35)

Multiplying from the left by S(Λ) and using

∂

∂xµ=∂x′ν∂xµ

∂

∂x′ν= Λνµ

∂

∂x′ν, x′ν = Λνµxµ, (4.36)



it follows (iS(Λ)γµS−1(Λ)Λνµ

∂

∂x′ν−m

)ψ′(x′) = 0. (4.37)

Comparing with eq. (4.28), we get

S(Λ)γµS−1(Λ)Λνµ = γν , (4.38)

or

S−1(Λ)γνS(Λ) = Λνµγµ. (4.39)

For an infinitesimal transformation we write

Λµν = gµν + εµν , (4.40)

with εµν = −ενµ (see eq. (1.44)). By expanding S(Λ) to the first order inεµν

S(Λ) = 1− i

4σµνε

µν (4.41)

and using (4.39), we find the following condition on σµν(1 +

i

4σρλε

ρλ

)γν

(1− i

4σαβε

αβ

)= (gµν + ενµ)γµ, (4.42)

from whichi

4ερλ[σρλ, γν ] = ενµγ

µ =12ερλ(gρνγλ − gλνγρ) (4.43)

and finally

[σρλ, γν ] = −2i(gρνγλ − gλνγρ). (4.44)

It is not difficult to show that the solution of this equation is given by

σρλ =i

2[γρ, γλ]. (4.45)

In fact,

[σρλ, γν ] =i

2[γργλ − γλγρ, γν ] =

i

2[γργλγν − γνγργλ − γλγργν + γνγλγρ]

=i

2[(2gρλ − γλγρ)γν − γν(2gρλ − γλγρ)− γλγργν + γνγλγρ]

= −i[γλγργν − γνγλγρ]= −i[γλ(2gρν − γνγρ)− (2gνλ − γλγν)γρ]

= −2i[gρνγλ − gνλγρ]. (4.46)


The Dirac field 75

A finite Lorentz transformation is obtained by exponentiation

S(Λ) = e− i

4σµνε

µν

, (4.47)

with

σµν =i

2[γµ, γν ]. (4.48)

We can now verify that the current jµ, defined in eq. (4.15), transforms asa four-vector. To this end we introduce the following notation

ψ(x) = ψ†(x)β = ψ†(x)γ0. (4.49)

It follows

j0 = ψ†ψ = ψγ0ψ, ji = ψ†αiψ = ψβαiψ = ψγiψ, (4.50)

or

jµ = ψγµψ. (4.51)

The transformation properties of ψ under Lorentz transformations are par-ticularly simple. By noticing that

γ0γµ†γ0 = γµ (4.52)

and

σµν† = − i

2[γµ, γν ]† =

i

2[γµ†, γν†], (4.53)

it follows

γ0σµν†γ0 = σµν (4.54)

and therefore

γ0S†(Λ)γ0 = S−1(Λ), (4.55)

from which

ψ′(x′) = ψ(x)S−1(Λ). (4.56)

Finally we get

j′µ(x′) = ψ′(x′)γµψ′(x′) = ψ(x)S−1(Λ)γµS(Λ)ψ(x) = Λµν j

ν(x). (4.57)

Therefore jµ has the desired transformation properties. The representationfor the Lorentz generators, in the same basis used previously for the γµmatrices, is

σ0i =i

2[γ0, γi] =

i

2(β2αi − βαiβ) = −iαi = −i

(0 σiσi 0

), (4.58)



σij =i

2[γi, γj ] = − i

2[αi, αj ] = εijk

(σk 00 σk

). (4.59)

The generators of the spatial rotations are nothing but the Pauli matrices,as one should expect for spin 1/2 particles.

The behavior of the Dirac wave function under parity ~x → −~x can beobtained in analogous way. In this case

ΛP νµ =

1−1−1−1

(4.60)

and therefore

S−1(ΛP )γµS(ΛP ) = γµ. (4.61)

This relation is satisfied by the choice

S(ΛP ) = ηP γ0, (4.62)

where ηP is a non observable arbitrary phase. Then

ψ(x)→ ψ′(x′) = ηP γ0ψ(x), x′µ = (x0,−~x). (4.63)

It is useful to classify the bilinear expressions in the Dirac wave functionunder Lorentz transformations. Let us consider expressions of the typeψAψ, where A is an arbitrary 4 × 4 matrix. As a basis for the 4 × 4matrices we can take the following set of 16 linearly independent matrices

ΓS = 1,

ΓVµ = γµ,

ΓAµ = γ5γµ,

ΓTµν = σµν ,

ΓP = γ5, (4.64)

where the matrix γ5 is defined as

γ5 = γ5 = iγ0γ1γ2γ3. (4.65)

This matrix has the following properties

γ5† = iγ3γ2γ1γ0 = iγ0γ1γ2γ3 = γ5, (4.66)

γ52 = 1, [γ5, γµ]+ = 0, (4.67)


The Dirac field 77

and, in the previous representation of the γ-matrices:

γ5 =(

0 11 0

). (4.68)

One can easily verify that the bilinear expressions have the following be-havior under Lorentz transformations

ψψ ≈ scalar,

ψγµψ ≈ four− vector,

ψγ5γµψ ≈ axial four− vector,

ψσµνψ ≈ antisymmetric 2nd rank tensor,

ψγ5ψ ≈ pseudoscalar. (4.69)

As an example, let us verify the last of these statements through a paritytransformation:

ψ(x)γ5ψ(x)→ ψ′(x′)γ5ψ′(x′) = η?P ηP ψ(x)γ0γ5γ0ψ(x) = −ψ(x)γ5ψ(x).

(4.70)

4.3 The Dirac equation and the Lorentz group

In this Section we will show how the Dirac equation follows naturally fromthe theory of the representations of the Lorentz group.

We have seen in Section 1.3 that parity interchanges the spinor rep-resentation (s1, s2) with (s2, s1). If one is interested in representations ofthe complete Lorentz group, that is the Lorentz group extended by parity(L↑+ ⊕ L

↑−), we have the following two possibilities

(s, s), (s1, s2)⊕ (s2, s1). (4.71)

Representations corresponding to the first possibilities are, for instance,(0, 0), the scalar representation, or (1/2, 1/2), the four-vector representa-tion. The simplest example of the second possibility is(

12, 0)⊕(

0,12

). (4.72)

This case corresponds to the Dirac representation. To understand betterthis point, let us first consider the four-vector representation. We have saidthat this is characterized by the bispinor (1/2, 1/2). More explicitly, if weare given a four-vector V µ, we can construct the bispinor



V ≡ Vαβ = (σµ)αβV µ =(

V 0 − V 3 −V 1 + iV 2

−V 1 − iV 2 V 0 + V 3

). (4.73)

This equation can be simply inverted

V µ =12

Tr (σµV ) . (4.74)

Now, suppose we would like to construct a first-order wave equation foran undotted spinor. We can act upon the undotted spinor by the gradientoperator in its bispinorial representation

i∂µ(σµ)αβζβ . (4.75)

By covariance, this expression can be zero, or else, it must be a dottedspinor. That is

i∂µ(σµ)αβζβ = 0, (4.76)

or

i∂µ(σµ)αβζβ = mηα. (4.77)

The first case corresponds to a massless fermion (see later) and was con-sidered for the first time in [Weyl (1929b)] . However Pauli criticized thisequation on the basis that it was not parity invariant. Weyl was vindicatedonly in 1956 after the discovery of the parity nonconservation in weak in-teractions. In the second case one is forced to introduce the dotted spinorηα which, in turn, must also satisfy a first order wave equation that can bewritten, without loss of generality, as1

i∂µ(σµ)αβηβ = mζα. (4.78)

Let us now introduce the four-component spinor

ψ =(ζα

ηα

). (4.79)

The two spinor equations can then be written as a single equation

(iγµ∂µ −m)ψ = 0, (4.80)

where

γµ =

(0 (σµ)αβ

(σµ)αβ 0

). (4.81)

1In principle two different constants m1 and m2 could appear in the two spinor equa-tions. However, by a convenient rescaling of the spinors, the two equations can be written

in terms of a single constant.


The Dirac field 79

One can easily verify that these matrices satisfy the algebra (4.25), andfurthermore that

γ5 = iγ0γ1γ2γ3 =(

1 00 −1

). (4.82)

In the basis that we have considered in the previous Section, γ0 was diago-nal, whereas in the present basis γ5 is the diagonal one. The two basis arecalled respectively the energy and the chiral basis. The first name corre-sponds to the fact that for a particle at rest, the Dirac operator is diagonal.The second denomination comes from the name chirality operator for γ5

(see later). If we write the four-component field in the energy basis as(φ

χ

), (4.83)

the relation with the chiral basis is

φ =1√2

(ζ + η), χ =1√2

(ζ − η). (4.84)

4.4 Free particle solutions of the Dirac equation

In this Section we will study the wave plane solutions of the Dirac equation.In the rest frame of the particle we look for solutions of the type

ψ(t) = ue−imt, (4.85)

where u is a four-component spinor. This solution has positive energy.Substituting inside the Dirac equation we get

(i∂ −m)ψ(t) = (mγ0 −m)ue−imt = 0, (4.86)

that is

(γ0 − 1)u = 0. (4.87)

Since γ0 has eigenvalues ±1 it follows that there also solutions of the typeeimt, corresponding to negative energy states. More generally we can lookfor solutions of the form

ψ(+)(x) = e−ikxu(k), positive energy,

ψ(−)(x) = eikxv(k), negative energy. (4.88)

Inserting into the Dirac equation

(k −m)u(k) = 0,

(k +m)v(k) = 0. (4.89)



In the rest frame we get

(γ0 − 1)u(m,~0) = 0,

(γ0 + 1)v(m,~0) = 0. (4.90)

There are two independent spinors of type u and two of type v satisfyingthese equations. In the basis where γ0 is a diagonal matrix we can choosethe following solutions

u(1)(m,~0) =

1000

, u(2)(m,~0) =

0100

,

v(1)(m,~0) =

00010

, v(2)(m,~0) =

0001

. (4.91)

In a general Lorentz frame the solutions can be obtained by boosting thesolutions in the rest frame. That is, applying a Lorentz transformationfrom the rest frame to a generic one. Alternatively, we can notice that thefollowing expression

(k −m)(k +m) = k2 −m2 (4.92)

vanishes for k2 = m2. Therefore we can solve our problem (except for anormalization constant), by putting

u(α)(k) = cα(k +m)u(α)(m,~0),

v(α)(k) = dα(−k +m)v(α)(m,~0), (4.93)

with k2 = m2. In order to determine the normalization constants cα anddα we make use of the orthogonality conditions satisfied by the rest framesolutions (see eq. (4.91))

u(α)(m,~0)u(β)(m,~0) = δαβ ,

v(α)(m,~0)v(β)(m,~0) = −δαβ ,u(α)(m,~0)v(β)(m,~0) = 0. (4.94)

Since these relations involve Lorentz scalars, ψψ, we can ask that they aresatisfied also for u(α)(k) e v(α)(k). Let us start with the u spinors:

u(α)(k)u(β)(k) = c∗αcβ u(α)(m,~0)(k +m)2u(β)(m,~0)

= c∗αcβ u(α)(m,~0)(2m2 + 2mk)u(β)(m,~0). (4.95)


The Dirac field 81

By taking into account that u(α)(m,~0) and u†(α)(m,~0) are eigenstates of

γ0 with eigenvalue +1, we get

u(α)(m,~0)γµu(β)(m,~0) = u†(α)

(m,~0)γ0γµγ0u

(β)(m,~0)

= u†(α)

(m,~0)γµu(β)(m,~0), (4.96)

from which

u(α)(m,~0)γµu(β)(m,~0) = gµ0u†(α)

(m,~0)γ0u(β)(m,~0) = gµ0δαβ , (4.97)

that is

u(α)(k)u(β)(k) = |cα|2(2m2 + 2mE)δαβ . (4.98)

Then we choose

cα =1√

2m(m+ E), E =

√|~k|2 +m2. (4.99)

In analogous way we have

v(α)(k)v(β)(k) = d∗αdβ v(α)(m,~0)(−k +m)2v(β)(m,~0)

= d∗αdβ v(α)(m,~0)(2m2 − 2mk)v(β)(m,~0) (4.100)

and using the fact that v(α)(m,~0) and v†(α)(m,~0) are eigenstates of γ0 witheigenvalue −1, we get

v(α)(m,~0)γµv(β)(m,~0) = −v†(α)(m,~0)γ0γ

µγ0v(β)(m,~0)

= v(α)(m,~0)γµv(β)(m,~0), (4.101)

implying

v(α)(m,~0)γµv(β)(m,~0) = gµ0δαβ . (4.102)

Therefore we obtain

v(α)(k)v(β)(k) = −|dα|2(2m2 + 2mE)δαβ (4.103)

and, finally

dα = cα =1√

2m(m+ E). (4.104)

The normalized solutions we have obtained are

u(α)(k) =k +m√

2m(m+ E)u(α)(m,~0), v(α)(k) =

−k +m√2m(m+ E)

v(α)(m,~0).

(4.105)



Notice that positive and negative energy spinors are orthogonal. In thefollowing it will be useful to express our solutions in terms of two componentspinors, φ(α)(m,~0) and χ(α)(m,~0)

u(α)(m,~0) =(φ(α)(m,~0)

0

), v(α)(m,~0) =

(0

χ(α)(m,~0)

). (4.106)

From the explicit representation (4.19) of the γµ matrices we get

k =

(E −~k · ~σ~k · ~σ −E

), (4.107)

from which

u(α)(k) =

√m+ E

2mφ(α)(m,~0)

~k · ~σ√2m(m+ E)

φ(α)(m,~0)

, (4.108)

v(α)(k) =

~k · σ√

2m(m+ E)χ(α)(m,~0)√

m+ E

2mχ(α)(m,~0)

. (4.109)

In the following we will need the explicit expression for the projectorsof the positive and negative energy solutions. To this end, let us observethat

2∑α=1

u(α)(m,~0)u(α)(m,~0) =

1000

( 1 0 0 0)

+

0100

(0 1 0 0)

=1 + γ0

2

(4.110)and analogously

2∑α=1

v(α)(m,~0)v(α)(m,~0) = −1− γ0

2. (4.111)

Using γ0γµ = 2gµ0 − γµγ0 e k2 = m2, we get

(k +m)γ0(k +m) = (k +m)(2E − kγ0 +mγ0)

= 2E(k +m) + (k +m)(−k +m)γ0

= 2E(k +m). (4.112)


The Dirac field 83

Therefore the positive energy projector is given by

Λ+(k) =2∑

α=1

u(α)(k)u(α)(k) =k +m√

2m(m+ E)1 + γ0

2k +m√

2m(m+ E)

=1

2m(m+ E)(k +m)2 + 2E(k +m)

2=k +m

2m. (4.113)

In similar way we get the expression for the negative energy projector

Λ−(k) = −2∑

α=1

v(α)(k)v(α)(k) =−k +m

2m. (4.114)

It is easy to verify that the matrices Λ±(k) verify all the properties of acomplete set of projection operators

Λ2± = Λ±, Λ+Λ− = 0, Λ+ + Λ− = 1. (4.115)

In this normalization the density ψ†ψ has the correct Lorentz transforma-tion properties

ψ(+)(α)

†(x)ψ(+)

(β) (x) = u(α)(k)γ0u(β)(k)

=1

2m(m+ E)u(α)(m,~0)(k +m)γ0(k +m)u(β)(m,~0)

=1

2m(m+ E)2Eu(α)(m,~0)(k +m)u(β)(m,~0)

=2E(m+ E)2m(m+ E)

δαβ =E

mδαβ , (4.116)

where we have used eq. (4.112). Therefore the density for positive energysolutions transforms as the fourth component of a four-vector. The sameis true for the negative energy solutions

ψ(−)(α)

†(x)ψ(−)

(β) (x) =E

mδαβ . (4.117)

We find also

u†(α)

(k)v(β)(k) = 0, kµ = (E,~k), kµ = (E,−~k). (4.118)

In fact,

u†(α)

(k)v(β)(k) = u(α)(m,~0)(k +m)γ0(−ˆ

k +m)2m(m+ E)

v(β)(m,~0)

= u(α)(m,~0)(k +m)(−k +m)γ0

2m(m+ E)v(β)(m,~0)

= 0. (4.119)



It follows that solutions with opposite energy and same three momentumare orthogonal

ψ(+) = e−i(Ex0 − ~k · ~x)u(k), kµ = (E,~k),

ψ(−) = e+i(Ex0 + ~k · ~x)v(k), kµ = (E,−~k). (4.120)

The positive and negative energy solutions are doubly degenerate. It ispossible to remove the degeneration through the construction of projectorsfor states with definite polarization. Let us consider again the solutions inthe rest frame. The generator of the rotations along the z-axis is given by

σ12 =(σ3 00 σ3

). (4.121)

Clearly u(1)(m,~0) and v(1)(m,~0) are eigenstates of this operator (and there-fore of the third component of the spin operator) with eigenvalues +1,whereas u(2)(m,~0) and v(2)(m,~0) belong to the eigenvalue −1. The projec-tor for the eigenstates with eigenvalues +1 can be written as

1 + σ12

2=

1 + σ12n3R

2, (4.122)

where nµR = (0, 0, 0, 1) is a unit space-like four-vector. Also we have

σ12 =i

2[γ1, γ2] = iγ1γ2 = −γ0γ5γ

3 = γ5γ3γ0 (4.123)

and

σ12n3R = γ5nRγ0. (4.124)

The presence of γ0 forbids a simple extension of this expression to a genericLorentz frame. We can avoid this, by changing the definition of the projec-tion in the rest frame system. Let us put

Σ(±nR) =1± σ12n

3Rγ0

2=

12

(1± σ3 0

0 1∓ σ3

). (4.125)

In this case Σ(nR) and Σ(−nR) project out u(1)(m,~0), v(2)(m,~0) andu(2)(m,~0), v(1)(m,~0), respectively. That is, Σ(±nR) projects out the posi-tive energy solutions with spin ±1/2 and the negative energy solutions withspin ∓1/2. Then, we have

Σ(±nR) =1± γ5nR

2. (4.126)


The Dirac field 85

In the rest frame n2R = −1, nR · k = 0. We can go to a generic frame

preserving these conditions

Σ(±n) =1± γ5n

2, n2 = −1, n · k = 0. (4.127)

The projector Σ(±n) projects out energy positive states that in the restframe have a polarization given by ~S · ~n = ±1/2, and the negative energystates with polarization ~S · ~n = ∓1/2.

In the following we will use the following notation

u(kR, nR) = u(1)(m,~0),

u(kR,−nR) = u(2)(m,~0),

v(kR,−nR) = v(1)(m,~0),

v(kR, nR) = v(2)(m,~0). (4.128)

These spinors satisfy the following relations

Σ(±nR)u(kR,±nR) = u(kR,±nR), Σ(±nR)v(kR,±nR) = v(kR,±nR)(4.129)

and

Σ(±nR)u(kR,∓nR) = Σ(±nR)v(kR,∓nR) = 0. (4.130)

All these relations generalize immediately to an arbitrary reference frame(always requiring n2 = −1 e n · k = 0)

Σ(±n)u(k,±n) = u(k,±n), Σ(±n)v(k,±n) = v(k,±n), (4.131)

Σ(±n)u(k,∓n) = Σ(±n)v(k,∓n) = 0. (4.132)

The properties of the spin projectors are

Σ(n) + Σ(−n) = 1, Σ(±n)2 = Σ(±n), Σ(n)Σ(−n) = 0. (4.133)

Let us just verify the second equation(1 + γ5n

2

)2

=1 + (γ5n)2 + 2γ5n

4=

2 + 2γ5n

4= Σ(n), (4.134)

where we have made use of n2 = −1. In analogous way

Σ(n)Σ(−n) =1 + γ5n

21− γ5n

2=

1− (γ5n)2

4= 0. (4.135)



4.5 Wave packets and negative energy solutions

As we have shown the Dirac equation leads to a positive probability density.This solves the problem present in the Klein-Gordon case. On the otherhand the Dirac equation does not solve the problem of the negative energysolutions (and it should not, as we have seen their relevance for locality inthe Klein-Gordon case). In fact, the completeness of the spinors involvesall the solutions

2∑α=1

[u(α)(k)u(α)(k)− v(α)(k)v(α)(k)

]= Λ+(k) + Λ−(k) = 1. (4.136)

In the case of a non interacting theory there are no possibilities of transi-tions among positive and negative energy states but, when an interactionis turned on, such a possibility cannot be excluded. In fact, if we try tolocalize a Dirac particle within distances of order 1/m the negative energysolutions cannot be ignored. To clarify this point let us consider the timeevolution of a gaussian wave packet, assigned at time t = 0,

ψ(~x, 0) =1

(πd2)3/4e−|~x|

2

2d2 w, (4.137)

where w is a fixed spinor, w = (φ, 0), with w†w = 1. As one can check, thewave packet is correctly normalized to one∫

d3x ψ†ψ =1

(πd2)3/2

∫d3x e

−|~x|2

d2 = 1. (4.138)

The solution of the Dirac equation with this boundary condition is obtainedby expanding over all the plane wave solutions

ψ(~x, t) =∫

d3k1√

(2π)3

√m

E

2∑α=1

[b(k, α)u(α)(k)e−ikx

+ d?(k, α)v(α)(k)eikx]

(4.139)

and evaluating the expansion coefficients b(k, α) and d?(k, α), by requiringthat the solution coincides with eq. (4.137) at time t = 0, that is

ψ(~x, 0) =∫

d3k1√

(2π)3

√m

E

2∑α=1

[b(k, α)u(α)(k)

+ d?(k, α)v(α)(k)]ei~k · ~x =

1(πd2)3/4

e−|~x|

2

2d2 w. (4.140)


The Dirac field 87

where kµ = (E,−~k). Fourier transforming both sides of this equation∫ψ(~x, 0)e−i~k·~xd3~x =

√(2π)3

√m

E

2∑α=1

[b(k, α)u(α)(k) + d?(k, α)v(α)(k)

]

=1

(πd2)3/4

∫d3x e

−|~x|2

2d2 e−i~k · ~xw

=1

(πd2)3/4(2πd2)3/2e

−|~k|2d2

2 w. (4.141)

From which√m

E

2∑α=1

[b(k, α)u(α)(k) + d?(k, α)v(α)(k)

]=(d2

π

)3/4

e−|~k|2d2

2 w.

(4.142)Using the orthogonality relations for the spinors we find the amplitudes

b(k, α) =√m

E

(d2

π

)3/4

e−|~k|2d2

2 u†(α)

(k)w, (4.143)

d?(k, α) =√m

E

(d2

π

)3/4

e−|~k|2d2

2 v†(α)

(k)w. (4.144)

Expressing u and v in terms of two-component spinors (see eqs. (4.108)and (4.109)) we get

b(k, α) =√m

E

(d2

π

)3/4

e−|~k|2d2

2√m+ E

2mφ(α)†(m,~0)φ, (4.145)

d?(k, α) =√m

E

(d2

π

)3/4

e−|~k|2d2

2 1√2m(m+ E)

χ(α)†(m,~0)~k · ~σφ,

(4.146)from which we can evaluate the ratio of the negative energy amplitudes tothe positive energy ones

d?(k, α)b(k, α)

≈ |~k|m+ E

. (4.147)

The amplitudes (for both signs of the energy) contribute only if |~k| 1/d(due to the gaussian exponential). Suppose that we want to localize the



particle over distances larger than 1/m, that is we require d 1/m. Sincethe negative energy state amplitudes are important only for |~k| > m 1/d,their contribution is depressed by the gaussian exponential. On the otherhand, if we try to localize the particle over distances d ≈ 1/m, the negativeenergy states contribution becomes important for values of |~k| of order m,or of order 1/d, that is in the momentum region in which the correspondingamplitudes are not negligible. We see that the negative energy solutionsare necessary in order to be consistent with the uncertainty principle.

4.6 Electromagnetic interaction of a relativistic point-likeparticle

In order to understand better the physics behind the Dirac equation wewill now introduce the interaction with the electromagnetic field. For themoment we will do it at the level of classical field theory. This will allowus to discuss the spin of the Dirac particle and also the properties of theantiparticles. To this end we will start considering the interaction of a point-like particle with the electromagnetic field in the relativistic formalism.From this discussion we will be able to derive a recipe for introducing thecoupling of any charged particle to the electromagnetic field. Later on wewill see that this prescription is equivalent to the requirement of gaugeinvariance of the theory (see Section 6.4).

Let us recall that the classical expression for the electromagnetic four-current is given by

jµ = (ρ, ρ~v), (4.148)

where ρ is the charge density, and ~v the velocity field. A point-like particle,following a given world line, can be described in a parametric form by fourfunctions xµ(τ), with τ an arbitrary line parameter. The correspondingcharge density at the time t is localized at the position ~x(τ), evaluated atthe parameter value τ such that t = x0(τ) (see Fig. 4.1). Therefore

ρ(~y, t) = eδ3(~y − ~x(τ))|t=x0(τ). (4.149)

It follows

jµ(y) = edxµ

dx0δ3(~y − ~x(τ))|y0=x0(τ). (4.150)

This expression can be put in a covariant form by using the following rela-


The Dirac field 89

t = x ( )0

t

x ( )µ

Fig. 4.1 The space-time trajectory of a point-like particle.

tion2 ∫dτf(τ)δ(y0 − x0(τ)) =

(dx0

dτ

)−1

f(τ)∣∣∣x0(τ)=y0

. (4.151)

We get

jµ(y) = e

∫ +∞

−∞dτdxµ

dτδ4(y − x(τ)). (4.152)

This four-current is conserved:

∂µjµ(y) = −e

∫ +∞

−∞dτdxµ

dτ

∂

∂xµδ4(y − x(τ)) = 0 (4.153)

The only possible contributions could come from the end points x(±∞).We recall also that the equations of motion for a free relativistic scalar

particle can be derived by the following action

S = −m∫ τf

τi

dτ√x2, xµ =

dxµ

dτ. (4.154)

We will be interested in deriving the Lagrangian describing the interac-tion between a charged particle and the electromagnetic field (we assumethat the particle has charge e). We should be able to derive the followingequations of motion

d

dt

m~v√1− |~v|2

= e( ~E + ~v ∧ ~B). (4.155)

2We assume dx0/dτ > 0 in order to have a correct parameterization of the trajectory.



We will show that the Lagrangian depends on the four-vector potential Aµand not on the fields ~E and ~B. In fact we will verify that the followingaction reproduces the previous equations of motion

S = −m∫ τf

τi

dτ√x2 −

∫d4yAµ(y)jµ(y)

= −m∫ τf

τi

dτ√x2 − e

∫ τf

τi

dτAµ(x(τ))xµ(τ). (4.156)

Using

∂L

∂xµ= −e∂Aν

∂xµxν ,

∂L

∂xµ= −m xµ√

x2− eAµ (4.157)

and the Euler-Lagrangian equations

∂L

∂xµ− d

dτ

∂L

∂xµ= 0, (4.158)

we get

−e∂Aν∂xµ

xν +md

dτ

xµ√x2

+ e∂Aµ∂xν

xν = 0. (4.159)

Therefore

md

dτ

xµ√x2

= e(∂µAν − ∂νAµ)xν . (4.160)

Since ds = dτ√x2, where ds is the line element measured along the trajec-

tory, we see that the four-velocity of the particle is

Uµ =xµ√x2, (4.161)

from which we get the equations of motion in a covariant form

md

dsUµ = eFµνU

ν . (4.162)

Here we have introduced the electromagnetic strength tensor

Fµν = ∂µAν − ∂νAµ. (4.163)

Using the relations between the potentials and the electric and magneticfields

~E = −~∇A0 − ∂ ~A

∂t, ~B = ~∇∧ ~A, (4.164)

we get

~E = (F 10, F 20, F 30), ~B = (−F 23,−F 31,−F 12), (4.165)


The Dirac field 91

(we can also write F ij = −εijkBk). By choosing τ = x0 in the eq. (4.160)we find

md

dt

−vk√1− |~v|2

= eFk0 + eFkidxi

dt= −eEk − εkijBjvi, (4.166)

reproducing eq. (4.155).There are various ways to convince oneself about the necessity of the ap-

pearance of the four-potential inside the Lagrangian. For instance, considerthe Maxwell equations

∂µFµν = jν , ∂µFµν = 0, (4.167)

where Fµν = 12εµνρσF

ρσ is the dual tensor. In Section 3.2 we have shownhow to deduce the expression for the Lagrangian multiplying the field equa-tions by an infinitesimal variation of the fields. In the actual case, toconsider Fµν as the fields to be varied, would create a problem becausemultiplying both sides of the Maxwell equations by δFµν we would not geta Lorentz scalar. This difficulty is avoided by taking Aµ as the indepen-dent degrees of freedom of the theory (we will show in the following thatalso this point of view has its own difficulties). In this case, due to thedefinition (4.163) of the electromagnetic tensor, the homogeneous Maxwellequations become identities (in fact, it is just solving these equations thatone originally introduces the vector and the scalar potentials)

Fµν = εµνρσ∂ρAσ =⇒ ∂µFµν = 0, (4.168)

whereas the inhomogeneous ones give rise to

∂µ(∂µAν − ∂νAµ) = Aν − ∂ν∂µAµ = jν . (4.169)

The previous difficulty disappears because both sides of this equation mult-plied by Aµ are Lorentz scalars. By regarding jµ as a given external current,independent on Aµ, we can now get easily the expression for the Lagrangian.



By multiplying eq. (4.169) by δAν and integrating in d4x we get

0 =∫V

d4xδAν(Aν − ∂ν∂µAµ − jν)

=∫V

d4x[− δ(∂µAν)∂µAν + ∂µ(δAν∂µAν)

+δ(∂µAν)∂νAµ − ∂µ(δAν∂νAµ)− δ(Aµjµ)]

=∫V

d4x[− 1

2δ(∂µAν)∂µAν − 1

2δ(∂νAµ)∂νAµ

+12δ(∂µAν)∂νAµ +

12δ(∂νAµ)∂µAν − δ(Aµjµ)

]+ surface terms

= −12

∫V

d4x(δFµν)Fµν −∫V

d4xδ(Aµjµ) + surface terms

= δ

[∫V

d4x (−14FµνF

µν −Aµjµ)]

+ surface terms. (4.170)

We see that the action for an electromagnetic field interacting with anexternal current jµ is given by (here Fµν must be though as a function ofAµ)

S = −14

∫V

d4x FµνFµν −

∫V

d4x jµAµ. (4.171)

Notice that the interacting term has the same structure we found for thepoint-like particle.

We stress again that the Aµ’s are the canonical variables of the elec-trodynamics. In principle, one could reintroduce the fields by inverting therelations between fields and potentials. However, in this way, one wouldend up with a non-local action. From these considerations one can arguethat the potentials play an important role in quantum mechanics, muchmore than in the classical case, where they are essentially a convenienttrick. Recall also that the canonical variables satisfy local commutationrelations (the commutator vanishes at space-like distances), implying thatlocal observables should be local functions of the potentials. This is go-ing to create us some problem because the theory is invariant under gaugetransformations, whereas the potentials are not

Aµ(x)→ Aµ(x) + ∂µΛ(x), (4.172)

(where Λ(x) is an arbitrary function). Therefore, the observables of thetheory should be gauge invariant implying that the potentials cannot beobserved. An example of an observable which is both local in the potentialsand gauge invariant is the electromagnetic tensor strength Fµν .


The Dirac field 93

The pure electromagnetic part of the action (4.171) is naively gauge in-variant, being a function of Fµν . As far as the interaction term is concerned,we have (assuming that the current is gauge invariant)

jµAµ → jµA

µ + jµ∂µΛ = jµA

µ + ∂µ(jµΛ)− (∂µjµ)Λ. (4.173)

Adding a four-divergence to the Lagrangian density does not change theequations of motion∫ t2

t1

dt

∫d3x ∂µχ

µ =∫ t2

t1

dt∂

∂t

∫d3 xχ0. (4.174)

Therefore the invariance of the Lagrangian under gauge transformations(neglecting a four-divergence) is guaranteed, if the potentials are coupledto a gauge invariant and conserved current

∂µjµ = 0. (4.175)

We have shown that this condition is indeed satisfied for the point-likeparticle.

In order to derive the general prescription to couple the electromagneticpotentials to a charged particle, let us go back to the action for the point-like particle. This prescription is known as the minimal substitution.By choosing x0 = τ in the Lagrangian of eq. (4.156), we get

L = −m√

1− |~v|2 − e(A0 − ~v · ~A), (4.176)

from which

~p =∂L

∂~v=

m~v√1− |~v|2

+ e ~A. (4.177)

The Hamiltonian is obtained through the usual Legendre transform

H = ~p · ~v − L = m|~v|2√

1− |~v|2+ e~v · ~A+m

√1− |~v|2 + e(A0 − ~v · ~A)

=m√

1− |~v|2+ eA0. (4.178)

Therefore, we get the following relations

~p− e ~A = m~v√

1− |~v|2, H − eA0 =

m√1− |~v|2

, (4.179)

where the quantities in the right hand sides of these two equations are thesame as in the free case. It follows that we can go from the free case to theinteracting one, through the simple substitution (minimal substitution)

pµ → pµ − eAµ. (4.180)



In the free case, inverting the relations between momenta and velocities

|~v|2 =|~p|2

m2 + |~p|2, 1− |~v|2 =

m2

m2 + |~p|2, (4.181)

we get the Hamiltonian as a function of the canonical momenta

Hfree =√m2 + |~p|2. (4.182)

By performing the minimal substitution we get

H − eA0 =√m2 + (~p− e ~A)2, (4.183)

from which

H = eA0 +√m2 + (~p− e ~A)2, (4.184)

which is nothing but eq. (4.178), after using eq. (4.177). From the pointof view of canonical quantization, the minimal substitution corresponds tothe following substitution in the space-time derivatives

∂µ → ∂µ + ieAµ. (4.185)

4.7 Nonrelativistic limit of the Dirac equation

In order to understand better the role of the spin in the Dirac equation wewill now study the nonrelativistic limit in presence of an electromagneticfield.

(i∂ −m)ψ(x) = 0 =⇒ (i∂ − eA−m)ψ(x) = 0. (4.186)

Notice that the Dirac equation is invariant under the transformation (4.172)

Aµ(x)→ Aµ(x) + ∂µα(x), (4.187)

if, at the same time, we perform the following local phase transformationon the wave function

ψ(x)→ e−ieα(x)ψ(x). (4.188)

The eq. (4.186) is also invariant under Lorentz transformations if, in goingfrom the frame S to the frame S′ (x → x′ = Λx), the field Aµ transformsas

Aµ(x)→ A′µ(x′) = (Λ−1)νµAν(x). (4.189)

This shows that Aµ(x), under Lorentz transformations behaves as ∂µ:

∂

∂xµ→ ∂

∂x′µ=

∂xν

∂x′µ∂

∂xν= (Λ−1)νµ∂ν . (4.190)


The Dirac field 95

Eq. (4.189) says simply that Aµ transforms as a four-vector under Lorentztransformations.

In order to study the non relativistic limit is better to write ψ(x) in thefollowing form

ψ(x) =(φ(x)χ(x)

), (4.191)

where φ(x) and χ(x) are two-component spinors. By defining

~π = ~p− e ~A (4.192)

and using the representation in 2× 2 blocks of the Dirac matrices given ineq. (4.11), we get, after multiplication by γ0,

i∂

∂t

(φ(x)χ(x)

)=(

0 ~σ · ~π~σ · ~π 0

)(φ(x)χ(x)

)+(m 00 −m

)(φ(x)χ(x)

)+eA0

(φ(x)χ(x)

).

(4.193)This gives rise to two coupled differential equations

i∂φ

∂t= ~σ · ~πχ+ (m+ eA0)φ,

i∂χ

∂t= ~σ · ~πφ− (m− eA0)χ. (4.194)

In the non relativistic limit and, for weak fields, the mass term is the dom-inant one, and the energy positive solution will behave roughly as e−imt.With this consideration in mind we put(

φ(x)χ(x)

)= e−imt

(φ(x)χ(x)

), (4.195)

and assume that φ and χ are functions slowly varying with time. In thisway we obtain

i∂φ

∂t= ~σ · ~πχ+ eA0φ,

i∂χ

∂t= ~σ · ~πφ− (2m− eA0)χ. (4.196)

Assuming eA0 2m, and ∂χ/∂t ≈ 0 we have

χ ≈ ~σ · ~π2m

φ, (4.197)

from which

i∂φ

∂t=[

(~σ · ~π)2

2m+ eA0

]φ. (4.198)



One must be careful in evaluating (~σ · ~π)2, because the components of thevector ~π do not commute among themselves. In fact

[πi, πj ] = [pi − eAi, pj − eAj ] = ie∂Aj

∂xi− ie∂A

i

∂xj, (4.199)

where we have made use of

[pi, f(~x)] = −i∂f(~x)∂xi

. (4.200)

From ~B = ~∇∧ ~A it follows

[πi, πj ] = ieεijkBk (4.201)

and

(~σ · ~π)2 = σiσjπiπj =

(12

[σi, σj ] +12

[σi, σj ]+

)πiπj

= |~π|2 +14

[σi, σj ][πi, πj ] = |~π|2 +i

2εijkσk(ie)εijlBl, (4.202)

that is

(~σ · ~π)2 = |~π|2 − e~σ · ~B. (4.203)

The equation for φ becomes

i∂φ

∂t=

[(~p− e ~A)2

2m− e

2m~σ · ~B + eA0

]φ. (4.204)

This is nothing but the Pauli equation for an electron interacting with anelectromagnetic field. In particular, the term proportional to the magneticfield represents the interaction with a magnetic dipole given by

~µ =e

2m~σ =

e

m~S, (4.205)

where we have introduced the spin matrices ~S = ~σ/2. Recalling that thegyromagnetic ratio, g, for a particle of spin ~S and magnetic moment ~µ isdefined by the equation

~µ =e

2mg~S, (4.206)

we see that the Dirac equation predicts a gyromagnetic ratio equal to two.We may see this also in a slightly different way, by considering the interac-tion with a weak uniform magnetic field. In this case the vector potentialis given by

~A =12~B ∧ ~x. (4.207)


The Dirac field 97

Neglecting the quadratic term in the fields we have

(~p− e ~A)2 ≈ |~p|2 − e(~p · ~A+ ~A · ~p). (4.208)

Using ∑i

[pi, Ai] = −i~∇ · ~A = 0, (4.209)

it follows

(~p− e ~A)2 ≈ |~p|2 − 2e~p · ~A = |~p|2 − e~p · ( ~B ∧ ~x)

= |~p|2 − epiεijkBjxk = |~p|2 − eεkijxkpiBj

= |~p|2 − e(~x ∧ ~p) · ~B = |~p|2 − e~L · ~B (4.210)

and finally

i∂φ

∂t=

[|~p|2

2m− e

2m(~L+ 2~S) · ~B + eA0

]φ, (4.211)

which exhibits explicitly the value of the gyromagnetic ratio. Experimen-tally this is very close to two, and we shall see, in the following, that thedifference is explained by quantum electrodynamics (QED). This is in fact,one of the most important results of this theory. However, let us noticethat, from the point of view of the Dirac equation, to find a value of the gy-romagnetic ratio so close to the experimental value is not a real prediction.In fact, one could think to add to the theory a further interaction termproportional to Fµνψσµνψ. This term is both Lorentz and gauge invariant.Such a term would give a further contribution to the magnetic moment ofthe electron, and therefore it would change the gyromagnetic ratio. Weshall see that the requirement that QED is a renormalizable3 forbids, infact, the appearance of such a term.

4.8 Charge conjugation, time reversal and PCT transfor-mations

Dirac equation had a great success in explaining the fine structure of thehydrogen atom, but the problem of negative energy solutions that, in prin-ciple, makes the theory unstable, was still there. Dirac looked for a solutionto this problem by taking advantage of the Pauli exclusion principle, whichapplies to half-integer spin particles. Dirac made the hypothesis that all3Renormalizable means that there exists a consistent procedure to eliminate the infini-

ties arising when calculating higher order terms in perturbation theory (see later).



the negative energy states were occupied by electrons. In such a situation,the Pauli principle forbids to any electron with positive energy to make atransition to a negative energy state. This solves the stability problem, butat the same time new phenomena are implied. For instance, an electron ina negative energy state could get enough energy (bigger than 2m which isthe minimal energy gap between the negative and positive energy states)to make a transition to a state of positive energy. If we imagine that inthe state of energy −E are present N electrons (we are simplifying things,because due to the momentum degeneracy there is actually an infinite num-ber of electrons), and that one of these electrons undergoes the transition,the energy of the state changes as follows

E −NE → E − (N − 1)E = E −NE + E, (4.212)where E is the energy of all the other electrons (with energy different from−E) in the fundamental state. Notice that in the Dirac theory the fun-damental state is the one with all the negative energy states occupied andzero electrons in the energy positive states. In a sense this is the physi-cal explanation of the infinite energy of the vacuum that we found in thecase of the Klein-Gordon field, and that we will find also in the Dirac case(see later). In a complete analogous way, also the charge of the vacuum isinfinite and its variation in the previous transition is given by

Q+Ne→ Q+ (N − 1)e = Q+Ne− e, (4.213)where e is the charge of the electron (e < 0). We see that the vacuumenergy and the electric charge charge increase respectively by E and −ein the transition. We can interpret this by saying that the hole left in thevacuum by the electron has charge −e and energy E. In other words, wecan think to the hole being a particle of positive energy and positive charge.This is the way in which the idea of antiparticles came around: the hole isthought as the antiparticle of the electron. The transition of an electronof negative energy to a state of positive energy is then seen as the creationof a particle antiparticle (the hole) pair. Of course this may happen onlyif a convenient amount of energy (for instance electromagnetic energy) isabsorbed by the energy negative state. In the same way, once we have ahole in the vacuum, it may happen that a positive energy electron makes atransition to that particular hole state. In this case both the electron andthe hole disappear. This is the pair annihilation phenomenon happeningwith a release of energy.

The hole theory is nowadays reinterpreted in terms of antiparticles, butthis way of thinking has been extremely fruitful in many fields, as in thestudy of electrons in metals, in nuclear physics and so on.


The Dirac field 99

Assuming seriously the hole theory means that the Dirac equationshould admit, beyond the positive energy solutions corresponding to anelectron, other positive energy solutions with the same mass of the elec-tron, but with opposite charge. To see this point in a formal way, we lookfor a transformation of the electron wave function, ψ(x), to the antielectron(positron) wave function ψC(x), such that, if ψ satisfies

(i∂ − eA−m)ψ(x) = 0, (4.214)

then ψC satisfies

(i∂ + eA−m)ψC(x) = 0. (4.215)

ψC is said the charge conjugated of ψ and the operation relating the twowave functions is called charge conjugation. We will require the transfor-mation to be local and such that the transformed of the antiparticle wavefunction gives back, except for a possible phase factor, the electron wavefunction. To build up ψC we will start by taking the complex conjugateof ψ. This is clearly the only way of changing a negative energy solution,described by eiEt, in a positive energy solution, described by e−iEt. Bytaking the hermitian conjugate, multiplying by γ0 (from the right) andtransposing, we get

(i∂ − eA−m)ψ(x) = 0→ −i∂µψγµ − eψA−mψ = 0

→ [γµT (−i∂µ − eAµ)−m]ψT = 0, (4.216)

where

ψT = γ0Tψ?. (4.217)

If there is a matrix, C, such that

CγµTC−1 = −γµ, (4.218)

multiplying eq. (4.216) by C, we get the wanted result

(i∂ + eA−m)CψT = 0. (4.219)

This describes a particle with charge −e. Therefore, apart a phase factorηC , we can identify ψC with CψT :

ψC = ηCCψT . (4.220)

In the representation where γ0 is diagonal we have

γ0T = γ0, γ1

T = −γ1, γ2T = γ2, γ3

T = −γ3. (4.221)



It is enough to choose C commuting with γ1 and γ3 and anticommutingwith γ0 and γ2. It follows that C must be proportional to γ2γ0. Let uschoose

C = iγ2γ0 =(

0 −iσ2

−iσ2 0

). (4.222)

In this way, C satisfies

−C = C−1 = CT = C†. (4.223)

To understand how the transformation works, let us consider, in the restframe, a negative energy solution with spin down

ψ(−)down = eimt

0001

, (4.224)

ψ(−)Cdown = ηCCψ

Tdown = ηCCγ0ψ

(−)down

?= ηCiγ

2ψ(−)down

?

= ηCe−imt

0 0 0 10 0 −1 00 −1 0 01 0 0 0

0001

(4.225)

and

ψ(−)Cdown = ηCe

−imt

1000

= ηCψ(+)up . (4.226)

That is, given a negative energy wave function describing an electron withspin down, its charge conjugated is a positive energy wave function describ-ing a positron with spin up. For an arbitrary solution with definite energyand spin, by using the projectors of Section 4.4, we write

ψ =εp+m

2m1 + γ5n

2ψ, (4.227)

where p0 > 0 and ε = ±1 selects the energy sign. Since C commutes withγ5, γ5

? = γ5, and

γ0γµ?γ0 = γµT , (4.228)

as it follows from

γ0γµ†γ0 = γµ, γT0 = γ0, (4.229)


The Dirac field 101

we obtain

ψC = ηCCγ0εp? +m

2m1 + γ5n

?

2ψ? = ηCC

εpT +m

2m1− γ5n

T

2γ0ψ?

=−εp+m

2m1 + γ5n

2ψC . (4.230)

We see that ψC is described by the same four-vectors pµ and nµ appearingin ψ, but with opposite sign of the energy. Then

u(p, n) = ηCvC(p, n), v(p, n) = ηCu

C(p, n). (4.231)

Since the spin projector selects the states of spin ±1/2 along ~n accordingto the sign of the energy, it follows that the charge conjugation inverts thespin projection of the particle. Notice also that, being ψC a solution of theDirac equation with e→ −e, the transformation

ψ → ψC , Aµ → −Aµ (4.232)

is a symmetry of the Dirac equation. Since in this transformation we changesign to the four-potential, we say that the photon has charge conjugation−1.

Another discrete transformation we will consider here is time reversal.The physical meaning of this transformation can be illustrated in terms ofa movie were we record all the observations made on the state described bythe wave function ψ(x). If we run the movie backward and we make a seriesof observations which are physically consistent, we say that the theory isinvariant under time reversal. From a mathematical point of view this isa symmetry if, sending t → t′ = −t, it is possible to transform the wavefunction in such a way that it satisfies the original Dirac equation. If thishappens, the transformed wave function describes an electron propagatingbackward in time. To build up explicitly the time reversal transformation,let us consider the electron in interaction with the electromagnetic field.It is convenient to write the Dirac equation in Hamiltonian form (see eq.(4.6))

i∂ψ(~x, t)∂t

= Hψ(~x, t), (4.233)

with

H = eA0 + γ0~γ · (−i~∇− e ~A) + γ0m. (4.234)

Let us define our transformation through the following equation

ψ′(~x, t′) = ηTKψ(~x, t), t′ = −t, (4.235)



where ηT is a phase factor, From eq. (4.233), omitting the spatial argument

i∂

∂tK−1ψ′(t′) = HK−1ψ′(t′). (4.236)

Multiplying this equation by K we get

∂

∂t′K(−i)K−1ψ′(t′) = KHK−1ψ′(t′). (4.237)

The invariance can be realized in two ways

K(−i)K−1 = i; KHK−1 = H, (4.238)

or

K(−i)K−1 = −i; KHK−1 = −H. (4.239)

The second possibility can be excluded immediately, since under time re-versal we have

~∇ → ~∇, ~A→ − ~A, A0 → A0. (4.240)

As it follows recalling that the vector potential is generated by a distributionof currents (changing sign under time reversal), whereas the scalar potentialis generated by a distribution of charges. Then, let us define K as a 4× 4matrix T times the operation of complex conjugation

K = T × (complex conjugation). (4.241)

From eq. (4.237) we get

i∂

∂t′ψ′(t′) = TH?T−1ψ′(t′). (4.242)

The theory is invariant if

TH?T−1 = H. (4.243)

By taking into account the transformation properties of the potentials weget

TH?T−1 = T (eA′0 + (γ0~γ)? · (i~∇+ e ~A′) + γ0?m)T−1. (4.244)

Since we want to reproduce H, we need a matrix T such that

Tγ0(~γ)?T−1 = −γ0~γ, Tγ0T−1 = γ0, (4.245)

where we have used the reality properties of γ0. In conclusion, T mustcommute with γ0 and satisfy

T~γ?T−1 = −~γ. (4.246)


The Dirac field 103

In our representation, the matrices γ1 and γ3 are real, whereas γ2 is pureimaginary. Therefore

Tγ0T−1 = γ0, Tγ1T−1 = −γ1, Tγ2T−1 = γ2, Tγ3T−1 = −γ3.

(4.247)By choosing the phase, we put

T = iγ1γ3. (4.248)

With this choice T satisfies

T † = T, T 2 = 1. (4.249)

To understand the correspondence with the classical results, where momen-tum and angular momentum change sign under time reversal, let us studyhow a positive energy solution transforms:

K

[p+m

2m1 + γ5n

2ψ(t)

]= T

[p? +m

2m1 + γ5n

?

2

]ψ?(t)

= T

[p? +m

2m

]T−1T

[1 + γ5n

?

2

]T−1Tψ?(t)

=ˆp+m

2m1 + γ5

ˆn2

ψ′(t′), (4.250)

where, again t′ = −t, and

p = (p0,−~p), n = (n0,−~n). (4.251)

The three discrete symmetry operations described so far, parity, P ,charge conjugation, C, and time reversal, (T ), can be combined togetherinto a symmetry transformation called PCT . Omitting all the phases, andrecalling that the matrices P , C and T are defined respectively in eqs.(4.61), (4.222) and (4.241)

ψPCT (−x) = PC[Kψ(x)]T

= PCγ0(Kψ(x))? = iγ0γ2(−iγ1?γ3?)ψ(x)

= iγ5ψ(x) (4.252)

and it suggests a simple correspondence between the wave function of apositron moving backward in time (ψPCT (−x)), and the electron wave func-tion. For a free particle of negative energy we have

ψPCT (−x) = iγ5−p+m

2m1 + γ5n

2ψ(x)

=p+m

2m1− γ5n

2(iγ5ψ(x))

=p+m

2m1− γ5n

2ψPCT (−x). (4.253)



Comparison with eq. (4.230), giving the charge conjugated of an energynegative state

ψC =p+m

2m1 + γ5n

2ψC , (4.254)

we see that the two expressions differ only in the spin direction. Similarconclusion can be reached by starting from the Dirac equation multipliedby iγ5. We get (x′ = −x, and A′(x′) = A(x))

iγ5(i∂x − eA(x)−m)ψ(x) = (−i∂x + eA(x)−m)ψPCT (x′)

= (i∂x′ + eA(x′)−m)ψPCT (x′), (4.255)showing that a positron moving backward in time satisfies the same equa-tion as an electron moving forward. Eq. (4.255) tells us that the PCTtransformation on ψ, combined with the PCT transformation on the four-vector potential, that is Aµ(x)→ −Aµ(−x), is a symmetry of the theory.

The interpretation of the positrons as negative energy electrons movingbackward in time is the basis of the positron theory by [Stueckelberg (1942);Feynman (1948c, 1949b,c)]. In this approach it is possible to formulate thescattering theory without using field theory. In fact, the pair creationand pair annihilation processes can be reinterpreted in terms of scatteringprocesses among electrons moving forward and backward in time.

4.9 Dirac field quantization

In this Section we abandon the study of the Dirac wave equation thoughtas a generalization of the Schrodinger equation, due to its difficulties tocope with many particle states. We will adopt here the point of view ofquantum field theory. That is the relativistic equation will be reinterpretedas an equation for an operator-valued field. In the Klein-Gordon case wehave shown that after quantization, we get a many particle system satisfyingBose-Einstein statistics. On the other hand we have also seen that the Diracequation describes spin 1/2 particles, which should satisfy the Fermi-Diracstatistics. As a consequence we expect to run into troubles if we wouldinsist in quantizing the Dirac field as we did for the Klein-Gordon case.To show how these troubles come about we will follow the canonical wayof quantization, showing that this leads to problems with the positivity ofthe energy. Looking for a solution we will find also the way of solving thewrong statistics problem.

We will begin our study by looking for the action giving rise to theDirac equation. We will take the quantities ψ and ψ as independent ones.


The Dirac field 105

Following the usual procedure, we multiply the Dirac equation by δψ (insuch a way to form a Lorentz scalar) and integrate over the space-timevolume V

0 =∫V

d4x δψ(i∂ −m)ψ = δ

∫V

d4x ψ(i∂ −m)ψ, (4.256)

where we have assumed that ψ and ψ are independent variables. We willthen assume the following action

S =∫V

d4x ψ(i∂ −m)ψ. (4.257)

It is simply verified that this action gives rise to the correct equation ofmotion for ψ. In fact,

∂L∂ψ

= −mψ, ∂L∂ψ,µ

= ψiγµ, (4.258)

from which

−mψ − i∂µψγµ = 0. (4.259)

The canonical momenta result to be

Πψ =∂L∂ψ

= iψ†, Πψ† =∂L∂ψ†

= 0. (4.260)

The canonical momenta do not depend on the velocities. In principle,this creates a problem for the Hamiltonian formalism. In fact a rigoroustreatment requires an extension of the classical Hamiltonian treatment,which was performed by Dirac himself [Dirac (2001)]. In this particularcase, the result one gets is the same as proceeding in a naive way. For thisreason we will avoid to describe this extension, and we will proceed as inthe standard case. Then the Hamiltonian density turns out to be

H = Πψψ − L = iψ†ψ − ψ(iγ0∂0 + iγk∂k −m)ψ = ψ†(−i~α · ~∇+ βm)ψ.(4.261)

If one makes use of the Dirac equation, it is possible to write the Hamilto-nian density as

H = ψ†i∂ψ

∂t. (4.262)

Contrarily to the Klein-Gordon case (see eq. (3.92)), the Hamiltonian den-sity is not positive definite. Let us recall the general expression for theenergy momentum tensor (see eq. (3.81))

Tµν =∂L∂φi,µ

φi,ν − gµνL. (4.263)



In our case we get

Tµν = iψγµψ,ν − gµν (ψ(i∂ −m)ψ) (4.264)

and using the Dirac equation

Tµν = iψγµψ,ν , (4.265)

we can immediately verify that this expression has vanishing four-divergence. Also

T 0k = iψ†∂kψ, (4.266)

from which we get the momentum of the field

P k =∫

d3x T 0k =⇒ ~P = −i∫

d3x ψ†~∇ψ. (4.267)

Now, let us consider the angular momentum density (see eq. (3.88))

Mµρν = xρT

µν − xνTµρ −

∂L∂φi,µ

Σijρνφj . (4.268)

The matrices Σijµν are defined in terms of the transformation properties ofthe field (see eq. (3.86))

∆φi = −12

Σijµνεµνφj . (4.269)

From eqs. (4.30) and eq. (4.41) for an infinitesimal Lorentz transformation,we get

∆ψ(x) = ψ′(x′)− ψ(x) = [S(Λ)− 1]ψ(x) ≈ − i4σµνε

µνψ(x), (4.270)

giving

Σµν =i

2σµν = −1

4[γµ, γν ]. (4.271)

Our result is then

Mµρν = iψγµ

(xρ∂ν − xν∂ρ −

i

2σρν

)ψ

= iψγµ(xρ∂ν − xν∂ρ +

14

[γρ, γν ])ψ. (4.272)

By taking the spatial components we obtain

~J = (M23,M31,M12) =∫

d3x ψ†(−i~x ∧ ~∇+

12~σ ⊗ 12

)ψ, (4.273)


The Dirac field 107

where 12 is the identity matrix in 2 dimensions, and using eq. (4.59) wehave defined

~σ ⊗ 12 =(~σ 00 ~σ

). (4.274)

The expression of ~J shows the decomposition of the total angular momen-tum in the orbital and in the spin part. The theory has a further conservedquantity, the current ψγµψ.

We will need the decomposition of the Dirac field in plane waves. Tothis end we will make use of the spinors u(p,±n) e v(p,±n) that we havedefined at the end of the Section 4.4. The expansion is similar to the oneused in eq. (4.139), but with operator valued coefficients, b and d.

ψ(x) =∑±n

∫d3 p√(2π)3

√m

Ep

[b(p, n)u(p, n)e−ipx + d†(p, n)v(p, n)eipx

],

(4.275)

ψ†(x) =∑±n

∫d3 p√(2π)3

√m

Ep[d(p, n)v(p, n)e−ipx

+ b†(p, n)u(p, n)eipx]γ0, (4.276)

where Ep =√|~p|2 +m2. For convenience we will collect here the main

properties of the spinors:

• Dirac equation

(p−m)u(p, n) = u(p, n)(p−m) = 0,

(p+m)v(p, n) = v(p, n)(p+m) = 0. (4.277)

• Orthogonality

u(p, n)u(p, n′) = −v(p, n)v(p, n′) = δnn′ ,

u†(p, n)u(p, n′) = v†(p, n)v(p, n′) =Epmδnn′ ,

v(p, n)u(p, n′) = v†(p, n)u(p, n′) = 0, (4.278)

where, if pµ = (Ep, ~p), then pµ = (Ep,−~p).• Completeness ∑

±nu(p, n)u(p, n) =

p+m

2m,

∑±n

v(p, n)v(p, n) =p−m

2m. (4.279)



We are now in the position to express the Hamiltonian in terms of theoperators b(p, n) and d(p, n). Using eq. (4.262) and integrating over thespace coordinates, we find

H =∑±n,±n′

∫d3p d3p′

Ep′m√EpEp′

×[d(p, n)b(p, n′)e−i(Ep + Ep′)tv†(p, n)u(p, n′)δ3(~p+ ~p′)

+ b†(p, n)b(p, n′)e+i(Ep − Ep′)tu†(p, n)u(p, n′)δ3(~p− ~p′)

− d(p, n)d†(p, n′)e−i(Ep − Ep′)tv†(p, n)v(p, n′)δ3(~p− ~p′)

− b†(p, n)d(p, n′)e+i(Ep + Ep′)t

×u†(p, n)v(p, n′)δ3(~p+ ~p′)]. (4.280)

Performing one of the momentum integrations and using the orthogonalityrelations, we get

H =∑±n

∫d3p Ep[b†(p, n)b(p, n)− d(p, n)d†(p, n)]. (4.281)

In analogous way we find

~P =∑±n

∫d3p ~p[b†(p, n)b(p, n)− d(p, n)d†(p, n)]. (4.282)

If we try to interpret these expressions as we did in the Klein-Gordoncase, we would assume that the operator d(p, n) creates from the vacuum astate of energy −Ep and momentum −~p. Dirac tried to solve the problemassuming that the vacuum was filled up by the negative energy solutions.Due to the Pauli principle this would make impossible for any other negativeenergy state to be created. Let us called the vacuum filled up by thenegative energy solutions the Dirac vacuum. Then the operator d(p, n)should give zero when acting upon this state. We then define as the truevacuum of the theory the Dirac vacuum and require

d(p, n)|0〉Dirac = 0. (4.283)

That is, in the Dirac vacuum the operator d(p, n) behaves as an annihilationoperator (as we have anticipated in writing). Since the Dirac vacuum isobtained by applying to the original vacuum a bunch of operators d(p, n),with all possible pµ and nµ, the previous equation can be satisfied assumingthe algebraic identity

(d(p, n))2 = 0. (4.284)


The Dirac field 109

We can satisfy this relation in a uniform algebraic way by requiring thatthe operators d(p, n) anticommute among themselves

[d(p, n), d(p′, n′)]+ = 0. (4.285)

This led [Jordan and Wigner (1928)] to the idea of quantizing the Diracfield in terms of anticommutators[

b(p, n), b†(p′, n′)]+

=[d(p, n), d†(p′, n′)

]+

= δnn′δ3(~p− ~p′). (4.286)

The problem of positivity is then solved automatically, since the four-momentum operator can be written as

Pµ =∑±n

∫d3p pµ

[b†(p, n)b(p, n) + d†(p, n)d(p, n)−

[d(p, n), d†(p, n)

]+

].

(4.287)Due to the anticommutation relations, the last term is an infinite negativeconstant which, physically, can be associated to the energy of the infiniteelectrons filling up the Dirac vacuum (called also the Dirac sea). If weignore this constant (as we did in the Klein-Gordon case, and with thesame warning), the energy operator is positive definite. The use of theanticommutators solves also the problem of the wrong statistics. In fact,the wave functions are now antisymmetric for the exchange of two Diracparticles (from now on we will put |0〉Dirac = |0〉):

b†(p1, n1)b†(p2, n2)|0〉 = −b†(p2, n2)b†(p1, n1)|0〉. (4.288)

Therefore, the quanta of the Dirac field satisfy the Fermi-Dirac statistics.Once we have realized all that, we can safely forget about the hole theoryand related stuff. In fact, looking at the four-momentum operator, we cansimply say that d†(p, n) creates and d(p, n) annihilates a positron state.Then we think to the vacuum as a state with no electrons and/or positrons(that is without electrons and holes).

In the Klein-Gordon case we interpreted the conserved current as theelectromagnetic current. We shall show now that in the Dirac case theexpression ψγµψ, has the same interpretation. Let us start evaluating thespatial integral of the density ∫

d3x ψ†ψ, (4.289)

in terms of the creation and annihilation operators∫d3x ψ†ψ =

∑±n

∫d3p

[b†(p, n)b(p, n) + d(p, n)d†(p, n)

]. (4.290)



As we know, this expression is formally positive definite. However, if wecouple the Dirac field to the electromagnetism through the minimal substi-tution we find that the free action (4.257) becomes

S =∫V

d4xψ(i∂ − eA−m)ψ. (4.291)

Therefore the electromagnetic field is coupled to the conserved current

jµ = eψγµψ. (4.292)

This forces us to say that the integral of the fourth component of the currentshould be the charge operator, and as such it should not be positive definite.In fact, we find

Q = e

∫d3x ψ†ψ =

∑±n

∫d3p e[b†(p, n)b(p, n)− d†(p, n)d(p, n)

+[d(p, n), d†(p, n)

]+

]. (4.293)

The subtraction of the infinite charge associated to the Dirac sea leaves uswith an operator which is not anymore positive definite. We see also thatthe operators b† create particles of charge e (electrons) whereas d† createparticles of charge −e (positrons). Notice that the interpretation of Q asthe charge operator would not have worked by using commutation relations.

A further potential problem is connected with causality. In the Klein-Gordon case we have shown that the causality properties are guaranteed,for local observables, by the canonical commutation relations for the fields.But this is just the property we have given up in the Dirac case. In orderto discuss this point, let us start evaluating the equal time anticommutatorfor the Dirac field[

ψ(~x, t), ψ†(~y, t)]+

=∫

d3p

(2π)3

m

Ep

×[(

p+m

2m

)ei~p · (~x− ~y) +

(p−m

2m

)e−i~p · (~x− ~y)

]γ0

=∫

d3p

(2π)3

m

Ep

[(p+m

2m

)+

(ˆp−m2m

)]γ0e

i~p · (~x− ~y)

=∫

d3p

(2π)3

m

Ep

2Ep2m

e−i~p · (~x− ~y) = δ3(~x− ~y), (4.294)

where the four-vector pµ is defined as in eq. (4.278), that is pµ = (p0,−~p).In the same fashion we get

[ψ(~x, t), ψ(~y, t)]+ =[ψ†(~x, t), ψ†(~y, t)

]+

= 0. (4.295)


The Dirac field 111

By using eq. (4.260), the anticommutator between ψ and ψ† can be writtenas

[Πψ(~x, t), ψ(~y, t)]+ = iδ3(~x− ~y). (4.296)

This shows that also in the Dirac case one can use the canonical formalismof quantization, but replacing commutators with anticommutators. Evalu-ating the anticommutator for arbitrary space-time separations we obtain[

ψ(x), ψ†(y)]+

=∫

d3p

(2π)3

m

Ep

×[(

p+m

2m

)e−ip(x− y) +

(p−m

2m

)eip(x− y)

]γ0

=[(i∂ +m

)xγ0

] ∫ d3p

(2π)3

12Ep

[e−ip(x− y) − eip(x− y)

]=[(i∂ +m

)xγ0

]i∆(x− y), (4.297)

where ∆(x) is the invariant function obtained in eq. (3.139) in the evalu-ation of the commutator for the Klein-Gordon field. From the propertiesof ∆(x), it follows that the anticommutator of the Dirac fields vanishesat space-like distances. Furthermore by evaluating the commutator of theDirac field one can show that this does not vanish for a space-like separa-tion. It follows that the Dirac field cannot be an observable quantity. Thisobservation by itself would put in a serious trouble the idea of quantizingthe Dirac field via anticommutation relations. So, we have to solve thecausality problem. The crucial observation lies in the following identity

[AB,C] = A[B,C] + [A,C]B = A[B,C]+ − [A,C]+B, (4.298)

which holds for arbitrary operators. The identity shows that AB commuteswith C if A and B both commute or anticommute with C. An immediateconsequence is that a local quantity containing an even number of Diracfields commutes with itself at space-like distances. So, in order to recon-cile the causality with the quantization of the Dirac field we have to giveup the possibility that this field is an observable. However, all the impor-tant physical quantities, as energy-momentum tensor and electromagneticcurrent are bilinear in the Fermi fields, and therefore they are observablequantities.

What we have shown here is that, in order to give a sense to the quanti-zation of the Dirac field, we have to use anticommutation relations which, inturn, imply that the corresponding quanta obey the Fermi-Dirac statistics.This is nothing but an example of the celebrated spin statistics theorem



that was proved by [Pauli (1940)]. This theorem says that in a Lorentzinvariant local field theory, integer and half-integer particles must satisfyBose-Einstein and Fermi-Dirac statistics respectively.

4.10 Massless spin 1/2 particles

Let us consider the Dirac equation for a massless particle, m = 0,

i∂ψ(x) = 0. (4.299)

By taking a positive energy solution

ψ(x) = e−ikxψ(k), (4.300)

it follows from (4.299), k2 = 0. Therefore, for positive energy solutions,E = k0 = |~k|, we can write

Eγ0 ψ(k) = ~k · ~γ ψ(k)→ E ψ(k) = ~k · ~αψ(k). (4.301)

Recalling from Section 4.2 that

αi =(

0 σiσi 0

), β =

(1 00 −1

), γ5 =

(0 11 0

), (4.302)

we see that multiplying by γ5 the second equation in (4.301) we get

γ5ψ(k) =~k · ~Σ|~k|

ψ(k), (4.303)

where

~Σ = γ5~α =(~σ 00 ~σ

). (4.304)

The operator ~Σ · ~k/|~k| is called ”helicity operator” since it tells us thedirection of the spin with respect to the momentum. In fact, its eigenvaluesare ±1, as it follows from (

~Σ · ~k|~k|

)2

= 1. (4.305)

The operator γ5 is also called chirality. Therefore eq. (4.303) says that,in the massless case, for positive energy solutions chirality and helicitycoincide. For the negative energy solutions they differ in sign since E =−|~k|. Notice that in the massless case there is no point in using 4 × 4matrices. In fact the Dirac equation can be written as

i∂ψ(x)∂t

= −i~α · ~∇ψ(x). (4.306)


The Dirac field 113

Since the only matrices appearing here are the αi’s, with the algebra

[αi, αj ]+ = 2δij , (4.307)

it is clear that they can be represented in terms of 2 × 2 Pauli matrices,±σi’s. Here the choice of the sign is related to positive or negative chirality.This can be seen also starting by the 4 × 4 representation but choosing adifferent basis for the algebra of the γ matrices, the chiral basis, where γ5

is taken diagonal. Then the γ-matrices are given by (see Section 4.3)

αi =(σi 00 −σi

), γ0 =

(0 −1−1 0

), γ5 =

(1 00 −1

). (4.308)

In this representation the fields with positive and negative chiralities aregiven respectively by

ψ+ =(φ

0

), ψ− =

(0χ

), (4.309)

with the two component fields satisfying

(−k0 + ~k · ~σ)φ = 0, (k0 + ~k · ~σ)χ = 0. (4.310)

The relativistic description of a massless particle in terms of two-componentspinors was first given by [Weyl (1929b)]. However this equation was re-jected since it does not preserve parity. In fact in this representation theparity operator γ0 is not diagonal and exchanges the two chirality solutions.Also the charge conjugation operator exchanges the two solutions. In factthe charge conjugation matrix C (see eq. (4.222)) is given by

C = −iα2 =(σ2 00 −σ2

). (4.311)

This matrix is diagonal but in the definition of the charge conjugationthere is also a conjugation and a γ0 matrix (see eq. (4.220)). As a resultthe charge conjugation exchanges the chiralities and send positive energysolutions into negative energy ones. However, the combined operation CP

leaves the Weyl equation invariant. Today we know that parity is violatedand in fact, the fermions in the standard models are described by masslessDirac equations and only after symmetry breaking they become massive.As a consequence the fermions can be described by Weyl equations thatafter the breaking pair together in such a way to generate massive Diracfields.



The Weyl equations can be easily obtained from the considerations madein Section 4.3. In the massless case (and forgetting about a representa-tion with definite parity) we do not need to introduce both representations(1/2, 0) and (0, 1/2). In fact, the equation (4.76)

i∂µ(σµ)αβζβ = 0, (4.312)for spinors belonging to the (1/2, 0) representation is consistent by itselfand we do not need to introduced a dotted spinor. By comparison with theresults found in Section 4.3, we see that the two component spinors definedin eq. (4.309), φ and χ coincide respectively with ζα and ηα.

We will not study in detail the solutions of the massless Dirac equation(4.299). However the normalization condition should be modified in thiscase. A convenient choice is

u†(p, n)uβ(p, n′) = v†(p, n)vβ(pn′) = 2Eδnn′ (4.313)In alternative, when evaluating physical observable as a cross-section, wecan assume m 6= 0 at the beginning, and take the limit m → 0 at the endof the calculations. We will se an example of this in Section 9.2.

4.11 Exercises

(1) Verify the relation

γ5σµν =

i

2εµνρλσρλ. (4.314)

(2) Verify the transformation properties of the Dirac bilinears in eq.(4.69).

(3) The 16 matrices, Γα = (1, γµ, γµγ5, σµν), form a complete basis in the

space of the 4×4 matrices. Show that any 4×4 matrix can be writtenas

A =∑α

bαΓα, (4.315)

with

bα =14

Tr[AΓα]. (4.316)

(4) Show that the complex conjugate of the matrix element of a 4×4matrix between Dirac spinors can be written as

(ufOui)∗ = uiOuf (4.317)where

O = γ0O†γ0 (4.318)

Then, evaluate the expression Γα for the 16 matrices Γα.


The Dirac field 115

(5) Verify the following identity

u(p′)σµν(p+ p′)νu(p) = iu(p′)(p′ − p)µu(p). (4.319)

(6) The Foldy-Wouthuysen transformation [Foldy and Wouthuysen(1950)]

H ′ = UHU†, (4.320)

where H is the Dirac Hamiltonian and

U = ei~α·~pθ(p) (4.321)

is such to make H ′ proportional to the matrix β for a convenient choiceof the function θ(p). Determine this function.

(7) Derive eqs. (4.282) and (4.293) for the three momentum ~P and forthe charge operators, using the expressions (4.275) and (4.276) for thefields ψ and ψ†.

(8) In the form given by eq. (4.257), the Lagrangian density of the Diracfield is not hermitian. Show that it is hermitian up to a four-divergenceand devise a way to write it in an explicit hermitian way.

(9) Evaluate the following equal time commutators

[Mµρν(x), ψ(y)]x0=y0 , [Tµν(x), ψ(y)]x0=y0 , [jµ(x), ψ(y)]x0=y0 .

(4.322)(10) Given an operator A of the form

A =∑j

αja†jaj , (4.323)

with

[aj , ak]+ = [a†j , a†k]+ = 0, [aj , a

†k]+ = δik, (4.324)

and an eigenstate of A, |ρ〉, corresponding to the eigenvalue ρ, showthat a†j |ρ〉 is an eigenstate of A with eigenvalue ρ+ αj and that aj |ρ〉corresponds to the eigenvalue ρ− αj .


Chapter 5

Vector fields

5.1 The electromagnetic field

In Section 4.6 we have shown that the action for the electromagnetic fieldmust be expressed in terms of the four-vector potential Aµ. We recall alsothat the Lagrangian density for the free case is given by (see eq. (4.171))

L = −14FµνF

µν , (5.1)

where

Fµν = ∂µAν − ∂νAµ. (5.2)

The resulting equations of motion are:

Aµ − ∂µ(∂νAν) = 0. (5.3)

We recall also that the potentials are defined up to a gauge transformation

Aµ(x)→ A′µ(x) = Aµ(x) + ∂µΛ(x). (5.4)

In fact, Aµ and A′µ satisfy the same equations of motion (and give rise tothe same electromagnetic field). In other words, the action for the elec-tromagnetic field is gauge invariant. Up to now, we have considered onlysymmetries depending on a finite number of parameters. For instance, inthe case of the O(2) symmetry for the charged scalar field, the transforma-tion depends on a single parameter, the rotation angle. In the case of thegauge symmetry, one deals with a continuous number of parameters, givenby the function Λ(x). In fact, in each space-time point, we can changethe definition of Aµ by adding the gradient of Λ evaluated at that point.The main consequence of this type of invariance is to reduce the effectivedegrees of freedom of the theory from 4 to 2. In order to prove this state-ment, let us start from the classical theory. We recall that it is possible to

117



use the gauge invariance to require some particular condition on the fieldAµ (gauge fixing). For instance, we can perform a gauge transformation insuch a way that the transformed field satisfies

∂µAµ = 0. (5.5)

In fact, given an arbitrary Aµ, we can gauge transform it by choosing Λ(x)such that

Λ + ∂µAµ = 0. (5.6)

Then, the transformed field A′µ = Aµ + ∂µΛ has vanishing four-divergence.When Aµ satisfies the condition ∂µA

µ = 0, we say that the potential is inthe Lorenz gauge1. However Aµ is not yet completely determined. Infact, we might perform a further gauge transformation

Aµ → A′µ = Aµ + ∂µΛ′, (5.7)

with

Λ′ = 0. (5.8)

Then the new field A′µ satisfies ∂µA′µ = 0 and we are still in the Lorenzgauge. In this gauge the equations of motion simplify and reduce to thewave equation, which is nothing but the Klein-Gordon equation withm = 0.Since the Lorenz gauge condition is relativistically invariant, the theoryformulated in this gauge is explicitly covariant, making this choice of gaugeparticularly convenient. On the other hand, the counting of the effectivedegrees of freedom is not so evident. From this point of view a betterchoice is the Coulomb gauge, which is defined (in the non interactingcase) as the gauge where the scalar potential and the spatial divergence ofthe vector potential vanish. To see that such a gauge exists, let us performthe following gauge transformation

A′µ(x) = Aµ(x)− ∂µ∫ t

0

A0(~x, t′)dt′. (5.9)

Clearly

A′0 = 0. (5.10)

Then we perform a second gauge transformation

A′′µ = A′µ − ∂µΛ, (5.11)1This gauge was introduced by Ludvig Valentin Lorenz (1829-1891), a Danish mathe-

matician and physicist, to not be confused with Hendrich Antoon Lorentz (1853-1928),

the author of the Lorentz transformations.


Vector fields 119

in such a way that ~∇ · ~A′′ = 0. To this end we choose Λ(x) such that

~∇ · ~A′′ = ~∇ · ~A′ + ~∇2Λ = 0. (5.12)

This equation can be solved by recalling that

~∇2 1|~x|

= −4πδ3(~x). (5.13)

Then

Λ(~x, t) =1

4π

∫d3x′

|~x− ~x′|~∇ · ~A′(~x′, t) (5.14)

and

∂Λ(~x, t)∂t

=1

4π

∫d3x′

|~x− ~x′|~∇ · ∂

∂t~A′( ~x ′, t). (5.15)

From the Gauss equation for the electric field it follows

~∇ · ~E = −~∇2A0 − ~∇ · ∂∂t~A = 0 (5.16)

and, in terms of A′µ

~∇ · ~E = −~∇ · ∂∂t~A′ = 0. (5.17)

Therefore

∂Λ(~x, t)∂t

= 0, (5.18)

and A′′0 = A′0 = 0. This shows that the second gauge transformation doesnot destroy the vanishing of the scalar potential. In conclusion, we haveshown that it is possible to choose a gauge such that

A0 = ~∇ · ~A = 0. (5.19)

It follows that the electromagnetic field has only two independent degreesof freedom. Another way of showing that Aµ has two degrees of freedomis through the equations of motion. Let us consider the four-dimensionalFourier transform of Aµ(x)

Aµ(x) =∫

d4k eikxAµ(k). (5.20)

Substituting this expression inside the equations of motion we get

−k2Aµ(k) + kµ(kνAν(k)) = 0. (5.21)



Let us now decompose Aµ(k) in terms of four independent four-vectors,which can be chosen as kµ = (E,~k), kµ = (E,−~k), and two further four-vectors eλµ(k), λ = 1, 2, orthogonal to kµ

kµeλµ = 0, λ = 1, 2. (5.22)

The decomposition of Aµ(k) reads

Aµ(k) = aλ(k)eλµ + b(k)kµ + c(k)kµ. (5.23)

From the equations of motion we get

−k2(aλeλµ + bkµ + ckµ) + kµ(bk2 + c(k · k)) = 0. (5.24)

The term in b(k) cancels, therefore it is left undetermined by the equationsof motion. For the other quantities we have

k2aλ(k) = c(k) = 0. (5.25)

The arbitrariness of b(k) is a consequence of the gauge invariance. In factif we gauge transform Aµ(x)

Aµ(x)→ Aµ(x) + ∂µΛ(x), (5.26)

then

Aµ(k)→ Aµ(k) + ikµΛ(k), (5.27)

where

Λ(x) =∫

d4k eikxΛ(k). (5.28)

Since the gauge transformation amounts to the translation b(k) → b(k) +iΛ(k), we can always choose b(k) = 0. Therefore we are left with the twodegrees of freedom described by the amplitudes aλ(k), λ = 1, 2. Further-more these amplitudes are different from zero only if the dispersion relationk2 = 0 is satisfied. This shows that the corresponding quanta have zeromass. With the choice b(k) = 0, the field Aµ(k) becomes

Aµ(k) = aλ(k)eλµ(k), (5.29)

proving that kµAµ(k) = 0. Therefore the choice b(k) = 0 is equivalent tofix the Lorenz gauge.


Vector fields 121

5.2 Quantization of the electromagnetic field

Let us now consider the quantization of the electromagnetic field. If wecould ignore the constraints on Aµ, which imply that the gauge field hasonly two independent components, we could require non trivial commuta-tion relations for all the components of the field. That is

[Aµ(~x, t),Πν(~y, t)] = igνµδ3(~x− ~y), (5.30)

[Aµ(~x, t), Aν(~y, t)] = [Πµ(~x, t),Πν(~y, t)] = 0, (5.31)

with

Πµ =∂L∂Aµ

. (5.32)

To evaluate the conjugated momenta, it is better to write the Lagrangiandensity (see eq. (5.1)) in the following form

L = −14

[Aµ,ν −Aν,µ][Aµ,ν −Aν,µ] = −12Aµ,νA

µ,ν +12Aµ,νA

ν,µ. (5.33)

Therefore∂L∂Aµ,ν

= −Aµ,ν +Aν,µ = Fµν , (5.34)

implying

Πµ =∂L∂Aµ

= Fµ0. (5.35)

It follows

Π0 =∂L∂A0

= 0. (5.36)

We see that it is impossible to satisfy the condition

[A0(~x, t),Π0(~y, t)] = iδ3(~x− ~y). (5.37)

We can try to find a solution to this problem modifying the Lagrangiandensity in such a way that Π0 6= 0. But doing so we will not recoverthe Maxwell equations. However, we can take advantage of the gauge in-variance, modifying the Lagrangian density in such a way to recover theequations of motion in a particular gauge. For instance, in the Lorenz gaugewe have the equations of motion

Aµ(x) = 0, (5.38)



which can be obtained from the Lagrangian density

L = −12Aµ,νA

µ,ν , (5.39)

(just remember the Klein-Gordon Lagrangian). We will see that the mi-nus sign is necessary to recover a positive Hamiltonian density. We nowexpress this Lagrangian density in terms of the gauge invariant one, givenin eq. (5.1). To this end we observe that the difference between the twoLagrangian densities is nothing but the second term of eq. (5.33)

12Aµ,νA

ν,µ = ∂µ[

12Aµ,νA

ν

]− 1

2(∂µAµ,ν)Aν

= ∂µ[

12Aµ,νA

ν

]− ∂ν

[12

(∂µAµ)Aν

]+

12

(∂µAµ)2. (5.40)

Then, up to a four-divergence, we can write the new Lagrangian density inthe form

L = −14FµνF

µν − 12

(∂µAµ)2. (5.41)

One can check that this form gives the correct equations of motion. In factfrom

∂L∂Aµ,ν

= −Aµ,ν +Aν,µ − gµν(∂λAλ),∂L∂Aµ

= 0, (5.42)

we get

0 = −Aµ + ∂µ(∂νAν)− ∂µ(∂λAλ) = −Aµ. (5.43)

The term

−12

(∂µAµ)2, (5.44)

which is not gauge invariant, is called the gauge fixing term. Moregenerally, we could add to the original Lagrangian density a term of theform

−λ2

(∂µAµ)2. (5.45)

The corresponding equations of motion would be

Aµ − (1− λ)∂µ(∂λAλ) = 0. (5.46)

For λ = 1, these equations coincide with the Maxwell equations in theLorenz gauge. Therefore, in the following we will use λ = 1. From eq.(5.42) we see that

Π0 =∂L∂A0

= −∂µAµ. (5.47)


Vector fields 123

In the Lorenz gauge we find again Π0 = 0. To avoid the correspondingproblem we can require ∂µAµ = 0 not to hold as an operator equation, butrather as a condition upon the physical states

〈phys|∂µAµ|phys〉 = 0. (5.48)

Correspondingly the price to pay in order to quantize the theory in a co-variant way is to work in a Hilbert space bigger than the physical one. Infact, the Hilbert space will be given by the subspace of physical states, theones satisfying eq. (5.48), plus the ones that do not satisfy the constraint(5.48). A bonus of this procedure is that it allows us of making use of localcommutation relations. On the contrary, in the Coulomb gauge, one needsto introduce non local commutation relations for the canonical variables.We will come back later to the condition (5.48).

Since we don’t have to worry any more about the equation Π0 = 0, wecan proceed with our program of canonical quantization. The canonicalmomentum densities are

Πµ =∂L∂Aµ

= Fµ0 − gµ0(∂λAλ), (5.49)

or, explicitly

Π0 = −∂λAλ = −A0 − ~∇ · ~A,Πi = ∂iA0 − ∂0Ai = −Ai + ∂iA0. (5.50)

Since the spatial gradient of the field commutes with the field itself at equaltime, the canonical commutator (5.30) gives rise to

[Aµ(~x, t), Aν(~y, t)] = −igµνδ3(~x− ~y). (5.51)

To get the quanta of the field, we look for plane wave solutions of the waveequation. We need four independent four-vectors in order to expand thesolutions in the momentum space. In a given frame, let us consider the unitfour-vector defining the time axis. This must be a time-like vector, n2 = 1,and we will choose n0 > 0. For instance, nµ = (1, 0, 0, 0). Then we taketwo four-vectors ε(λ)

µ , λ = 1, 2, in the plane orthogonal to nµ and kµ:

kµε(λ)µ = nµε(λ)

µ = 0, λ = 1, 2. (5.52)

Notice that now k2 = 0, since we are considering solutions of the waveequation. The four-vectors ε(λ)

µ , being orthogonal to nµ are space-like, thenthey will be chosen orthogonal and normalized in the following way

ε(λ)µ ε(λ

′)µ = −δλλ′ . (5.53)



Next, we define a unit space-like four-vector, orthogonal to nµ and lying inthe plane (k, n)

nµε(3)µ = 0, (5.54)

with

ε(3)µ ε(3)µ = −1. (5.55)

By construction ε(3)µ is orthogonal to ε

(λ)µ . This four-vector is completely

fixed by the previous conditions, and we get

ε(3)µ =

kµ − (n · k)nµ(n · k)

. (5.56)

As a last unit four-vector we choose nµ

ε(0)µ = nµ. (5.57)

These four-vectors are orthonormal and we can write

ε(λ)µ ε(λ

′)µ = gλλ′. (5.58)

and being linearly independent, they satisfy the completeness relation

ε(λ)µ ε(λ

′)ν gλλ′ = gµν . (5.59)

Using the explicit expressions for ε(3)µ and ε(0)

µ , together with the complete-ness we find the following result for the transverse physical polarizationvectors ∑

λ=1,2

ε(λ)µ ε(λ)

ν = −gµν −kµkν − (n · k)(nµkν + nνkµ)

(n · k)2(5.60)

In the frame where nµ = (1,~0) and kµ = (k, 0, 0, k), we have

ε(1)µ = (0, 1, 0, 0), ε(2)µ = (0, 0, 1, 0), ε(3)µ = (0, 0, 0, 1). (5.61)

The plane wave expansion of Aµ is

Aµ(x) =∫

d3k√2ωk(2π)3

3∑λ=0

ε(λ)µ (k)

[aλ(k)e−ikx + a†λ(k)eikx

], (5.62)

where we have imposed the hermiticity condition for Aµ(x) (classically it isa real field). For any fixed µ, this expansion is the same as the one that wewrote for the Klein-Gordon field, with the substitution ε

(λ)µ aλ(k) → a(k).

Then, from eq. (3.55)

ε(λ)µ (k)aλ(k) = i

∫d3x f∗~k (x)∂(−)

t Aµ(x), (5.63)


Vector fields 125

with the functions f~k(x) defined as in Section 3.2. Using the orthogonalityof the ε(λ)µ’s we find

aλ(k) = igλλ′

∫d3x ε(λ

′)µ(k)f∗~k (x)∂(−)t Aµ(x) (5.64)

and analogously

a†λ(k) = igλλ′

∫d3x ε(λ

′)µ(k)Aµ(x)∂(−)t f~k(x). (5.65)

Comparison with the calculation done in eq. (2.129) shows that

[aλ(k), a†λ′(k′)] = −

∫d3x f∗~k (x)i∂(−)

t f~k′(x)gλλ′ (5.66)

and using the orthogonality relations given in eq. (3.45)

[aλ(k), a†λ′(k′)] = −gλλ′δ3(~k − ~k′). (5.67)

Analogously

[aλ(k), aλ′(k′)] = [a†λ(k), a†λ′(k′)] = 0. (5.68)

Again, comparison with the Klein-Gordon commutators implies

[Aµ(x), Aν(y)] = −igµν∆(x− y), (5.69)

with the invariant function ∆(x) defined as in eq. (3.139), but with m = 0.The commutation rules we have derived for the operators aλ(k) create someproblem. Let us consider a one-particle state

|1, λ〉 =∫

d3k f(k)a†λ(k)|0〉, (5.70)

its norm is given by

〈1, λ|1, λ〉 =∫

d3k d3k′ f∗(k)f(k′)〈0|aλ(k)a†λ(k′)|0〉

=∫

d3k d3k′ f∗(k)f(k′)〈0|[aλ(k), a†λ(k′)]|0〉

= −gλλ∫

d3k |f(k)|2. (5.71)

Therefore the states with λ = 0 have negative norm. This problem is notcompletely unexpected. In fact, physically we know that the only physicalstates are the transverse ones (λ = 1, 2).

To clarify the situation let us go back to the gauge fixing condition〈phys|∂µAµ|phys〉 = 0. Remember that its aim is to select the physicalpart of the total Hilbert space. Therefore the relevant question is if thephysical states satisfying this condition have positive norm. To discuss



this point notice that the gauge fixing condition is bilinear in the states,therefore it could destroy the linearity of the Hilbert space. So we will tryto formulate this condition in a linear way. A possibility would be to write

∂µAµ|phys〉 = 0. (5.72)

But this would be a too strong requirement. Not even the vacuum statesatisfies it. However, if we consider the positive and negative frequencyparts of the field

A(+)µ (x) =

∫d3k√

2ωk(2π)3

3∑λ=0

ε(λ)µ (k)aλ(k)e−ikx, A(−)

µ (x) = (A(+)(x))†,

(5.73)it is possible to weaken the condition requiring

∂µA(+)µ (x)|phys〉 = 0. (5.74)

In this way the original requirement is automatically satisfied

〈phys.|(∂µA(+)µ + ∂µA(−)

µ )|phys〉 = 0. (5.75)

To make this condition more explicit let us evaluate the four-divergence ofA

(+)µ

i∂µA(+)µ (x) =

∫d3k√

2ωk(2π)3e−ikx

∑λ=0,3

kµε(λ)µ (k)aλ(k). (5.76)

Using eq. (5.56), we get

kµε(3)µ = −(n · k), kµε(0)

µ = (n · k), (5.77)

from which

[a0(k)− a3(k)]|phys〉 = 0. (5.78)

Notice that

[a0(k)− a3(k), a†0(k′)− a†3(k′)] = −δ3(~k − ~k′) + δ3(~k − ~k′) = 0. (5.79)

Let us denote by |Φ~k(n0, n3)〉 the state with n0 scalar photons (that iswith polarization ε

(0)µ (k)), and with n3 longitudinal photons (that is with

polarization ε(3)µ (k)). Then the following states satisfy the condition (5.74)

|Φ(m)~k〉 =

1m!

(a†0(k)− a†3(k))m|Φ~k(0, 0)〉. (5.80)

These states have vanishing norm

|||Φ(m)~k〉||2 = 0. (5.81)


Vector fields 127

More generally we can make the following observation. Let us consider thenumber operator for scalar and longitudinal photons

N =∫

d3k (a†3(k)a3(k)− a†0(k)a0(k)). (5.82)

Notice the minus sign arising from the commutation relations, and whichensures that N has positive eigenvalues. For instance

Na†0(k)|0〉 = −∫

d3k′ a†0(k′)[a0(k′), a†0(k)]|0〉 = a†0(k)|0〉. (5.83)

Let us consider a physical state |ϕn〉 with a total number, n, of scalar andlongitudinal photons. Then

〈ϕn|N |ϕn〉 = 0, (5.84)

as it follows from eq. (5.78)). Therefore

n〈ϕn|ϕn〉 = 0. (5.85)

We see that all the physical states with a definite number of scalar andlongitudinal photons have zero norm, except for the vacuum state (n = 0),that is

〈ϕn|ϕn〉 = δn,0. (5.86)

A generic physical state with zero transverse photons is a linear superposi-tion of the states |ϕn〉

|Ψ〉 = c0|ϕ0〉+∑n 6=0

cn|ϕn〉. (5.87)

This state has a positive definite norm

〈Ψ|Ψ〉 = |c0|2 ≥ 0. (5.88)

The proof that a physical state has a positive norm can be extended tostates with transverse photons. Of course, the coefficients cn, appearing inthe expression of a physical state, are completely arbitrary, but this is notgoing to modify the values of the observables. For instance, consider theHamiltonian. We have

H =∫

d3x : [ΠµAµ − L] :

=∫

d3x :[Fµ0Aµ − (∂λAλ)A0

+14FµνF

µν +12

(∂λAλ)2]

: . (5.89)



One can easily show that the Hamiltonian is given by the sum of all thedegrees of freedom appearing in Aµ (see the Klein-Gordon case, eq. (3.94))

H =12

∫d3x :

[3∑i=1

(A2i + (~∇Ai)2

)− A2

0 − ~∇A2

0

]:

=∫

d3k ωk :

[3∑

λ=1

a†λ(k)aλ(k)− a†0(k)a0(k)

]: . (5.90)

Since a0 and a3 act in the same way on the physical states, we get

〈phys|H|phys〉 = 〈phys|∫

d3k ωk

2∑λ=1

a†λ(k)aλ(k)|phys〉. (5.91)

The generic physical state is a linear combination of terms of the form|ϕT 〉⊗|Ψ〉, with |Ψ〉 defined as in eq. (5.87). Since only |ϕT 〉 contributes tothe evaluation of an observable quantity, we can always choose |Ψ〉 propor-tional to |ϕ0〉. However, this does not mean that we are always working inthe restricted physical space, because in a sum over the intermediate stateswe need to include all the degrees of freedom. This is crucial for the explicitcovariance and locality of the theory.

The arbitrariness in defining the state |Ψ〉 has a very simple interpre-tation. It corresponds to add to Aµ a four-gradient, that is to perform agauge transformation. Consider the following matrix element

〈Ψ|Aµ(x)|Ψ〉 =∑n,m

c?ncm〈ϕn|Aµ(x)|ϕm〉. (5.92)

Since Aµ change the occupation number by one unit and all the states |ϕn〉have zero norm (except for the state with n = 0), the only non vanishingcontributions come from n = 0, m = 1 and n = 1, m = 0

〈Ψ|Aµ(x)|Ψ〉 = c?0c1〈0|∫

d3k√2ωk(2π)3

e−ikx[ε(3)µ (k)a3(k)

+ ε(0)µ (k)a0(k)]|ϕ1〉+ c.c.. (5.93)

In order to satisfy the gauge condition the state |ϕ1〉 must be of the form

|ϕ1〉 =∫

d3q f(~q)[a†3(q)− a†0(q)]|0〉 (5.94)

and therefore

〈Ψ|Aµ(x)|Ψ〉 =∫

d3k√2ωk(2π)3

[ε(3)µ (k) + ε(λ)

µ (k)][c?0c1e−ikxf(~k) + c.c.].

(5.95)


Vector fields 129

From eqs. (5.56) and (5.57) we have

ε(3)µ + ε(0)

µ =kµ

(k · n), (5.96)

implying

〈Ψ|Aµ(x)|Ψ〉 = ∂µΛ(x), (5.97)

with

Λ(x) =∫

d3k√2ωk(2π)3

1n · k

(ic?0c1e−ikxf(~k) + c.c.). (5.98)

It is important to notice that this gauge transformation leaves Aµ in theLorenz gauge. In fact

Λ = 0, (5.99)

since the momentum k inside the integral satisfies the mass-shell conditionk2 = 0.

5.3 Massive vector fields

For completeness we will comment briefly about the massive vector fields,also called Proca fields. We will not discuss them any further along thisbook, but it will be important to understand some of their properties as,for instance, the fact that they have 3 independent degrees of freedom2.In particular, this point will be relevant when we will discuss the Higgsmechanism (see next Chapter).

Massive vector bosons are described by a vector field Vµ satisfying anequation of motion which generalizes the Maxwell equation

( +m2)Vµ − ∂µ(∂νVν) = 0. (5.100)

The Lagrangian density giving rise to these equation is

L = −14FµνF

µν +12m2V 2, (5.101)

where

Fµν = ∂µVν − ∂νVµ, (5.102)

Notice the change of sign with respect to the Klein-Gordon Lagrangiandensity due to the choice of the metric gµν , implying a minus sign for the2A more detailed treatment can be found in [Itzykson and Zuber (1980)].



spatial components. If we take the four-divergence of the equation (5.100)we get

m2∂µVµ = 0. (5.103)

Therefore Vµ has zero four-divergence as a consequence of the equations ofmotion. Let us now count the number of independent degrees of freedomof Vµ. To this end let us consider the Fourier transform of Vµ(x)

Vµ(x) =∫

d4keikxVµ(k). (5.104)

Let us introduce four independent four-vectors3,

kµ = (E,~k), εµ(i) = (0, ~ni), i = 1, 2, εµ(3) =1m

(|~k|, E

~k

|~k|

), (5.105)

with ~k · ~ni = 0, ~n1 · ~n2 = 0 and |~ni|2 = 1. Then we can write Vµ(k) as

Vµ(k) = ε(λ)µ aλ(k) + kµb(k). (5.106)

By inserting this expression inside eqs. (5.100)

(k2 −m2)(ε(λ)µ aλ(k) + kµb(k))− kµk2b(k) = 0, (5.107)

that is

(k2 −m2)aλ = 0, m2b = 0. (5.108)

Remember that in the electromagnetic case the field b was not determineddue to the gauge invariance, whereas in the massive case it is fixed to be zeroby the equations of motion. Therefore the field has three degrees of freedomcorresponding to two transverse and to one longitudinal polarizations. Allthese degrees of freedom satisfy a Klein-Gordon equation with mass m. Itis important to stress that the massive vector field has one extra-degree offreedom (the longitudinal), with respect to the massless case. We see alsothat the three independent fields described by aλ(k) satisfy a Klein-Gordonequation with mass m. Correspondingly the field Aµ can be expressed as4

Aµ(x) =∑

λ=1,2,3

∫d3k[f~k aλ(~k)ε(λ)

µ + f∗~k a†λ(~k)ε(λ)∗

µ ], (5.109)

with the non vanishing commutators between creation and annihilationoperators given by

[aλ(~k), a†λ′(~k)] = δλλ′δ3(~k − ~k′). (5.110)

3The following discussion is very similar to the one done for the electromagnetic field.

However, notice the differences arising from the fact that in the present case m 6= 0.4Compare with eq. (3.57).


Vector fields 131

As a final remark, let us notice that introducing a fourth normalizedfour-vector

ε(0)µ =

kµm, (5.111)

we get a basis in the four-dimensional Minkowski space, satisfying the com-pleteness relation

3∑λ,λ′=0

ε(λ)µ ε(λ

′)ν gλλ′ = gµν . (5.112)

Therefore the sum over the physical polarizations is given by3∑

λ=1

ε(λ)µ ε(λ)

ν = −gµν +kµkνm2

. (5.113)

5.4 Exercises

(1) Derive the relation (5.56).(2) Derive the expressions (5.64) and (5.65) for the annihilation and cre-

ation operators of the electromagnetic field.(3) Derive the the expression (5.90) for the Hamiltonian operator of the

electromagnetic field.(4) Show that, for k2 = 0, the following quantities

Pµν = gµν −kµkν + kν kµ

k · k, P⊥µν =

kµkν + kν kµk · k

, (5.114)

where kµ = (k0,−~k), are orthogonal projectors that is

PµνPνρ = δρµ, P⊥µνP

⊥νρ = δρµ, P⊥µνPνρ = 0,

P⊥µν + Pµν = gµν . (5.115)


Chapter 6

Symmetries in field theories

6.1 The linear σ-model

In this Section we will study, from a classical point of view, some fieldtheory with particular properties of symmetry. We will start consideringthe linear σ-model. This is a model for N scalar fields, with a symmetryO(N). The Lagrangian density is given by

L =12

N∑i=1

∂µφi∂µφi −

12µ2

N∑i=1

φiφi −λ

4

(N∑i=1

φiφi

)2

. (6.1)

L is invariant under linear transformations acting upon the vector ~φ =(φ1, · · · , φN ) and leaving invariant its norm

|~φ|2 =N∑i=1

φiφi. (6.2)

Consider an infinitesimal transformation (from now on we will omit theindex of sum over repeated indices)

δφi = εijφj . (6.3)

The condition for the norm being invariant is

|~φ+ δ~φ|2 = |~φ|2, (6.4)

from which

~φ · δ~φ = 0, (6.5)

or, in components,

φiεijφj = 0. (6.6)

133



This is satisfied by

εij = −εji, (6.7)

showing that the rotations in N dimensions depend on N(N −1)/2 param-eters. For a finite transformation we have

|~φ ′|2 = |~φ |2, (6.8)

with

φ ′i = Sijφj , (6.9)

implying

SST = 1. (6.10)

In fact, by exponentiating the infinitesimal transformation one gets

S = eε, (6.11)

with

εT = −ε. (6.12)

From this expression it follows that S is an orthogonal transformation. Thematrices S form the rotation group in N dimensions, O(N).

Noether’s theorem says that there is a conserved current for any contin-uous symmetry of the theory. In this case we will get N(N−1)/2 conservedquantities. It is useful, for further generalizations, to write the infinitesimaltransformations in the form

δφi = εijφj = − i2εABT

ABij φj , i, j = 1, · · · , N, A,B = 1, · · · , N,

(6.13)which is similar to what we did in Section 3.3 when we discussed the Lorentztransformations. By comparison we see that the matrices TAB are givenby

TABij = i(δAi δBj − δAj δBi ). (6.14)

It is not difficult to show that these matrices satisfy the algebra

[TAB , TCD] = −iδACTBD + iδADTBC − iδBDTAC + iδBCTAD. (6.15)

This is nothing but the Lie algebra of the group O(N), and the TAB are theinfinitesimal generators of the group. Applying now the Noether theoremwe find the conserved current

jµ =∂L

∂(∂µφi)δφi = − i

2φi,µεABT

ABij φj , (6.16)


Symmetries in field theories 135

since the N(N −1)/2 parameters εAB are linearly independent, we find theN(N − 1)/2 conserved currents

JABµ = −iφi,µTABij φj . (6.17)

In the case of N = 2 the symmetry is the same we studied in Section 3.6,and the only conserved current is given by

J12µ = −J21

µ = φ1,µφ2 − φ2,µφ1, (6.18)

in agreement with (3.160). One can easily check that the charges associatedto the conserved currents close the same Lie algebra as the generators TAB .More generally, if we have conserved currents given by

jAµ = −iφi,µTAijφj , (6.19)

with

[TA, TB ] = ifABCTC , (6.20)

using the canonical commutation relations, we get the commutators for theconserved charges

QA =∫

d3x jA0 (x) = −i∫

d3x φiTAijφj , (6.21)

that is

[QA, QB ] = ifABCQC . (6.22)

An important example is the σ-model for N = 4. In this case weparameterize our fields in the form

~φ = (π1, π2, π3, σ) = (~π, σ). (6.23)

These fields can be arranged into a 2× 2 matrix

M = σ + i~τ · ~π, (6.24)

where ~τ are the Pauli matrices. Recalling that τ2 is pure imaginary, τ1 andτ3 real, and that τ2 anticommutes with τ1 and τ3, we get

M = τ2M∗τ2. (6.25)

Furthermore the following relation is satisfied

|~φ|2 = σ2 + |~π|2 =12Tr(M†M). (6.26)

Then we can write the Lagrangian (6.1) in the form

L =14Tr(∂µM†∂µM)− 1

4µ2Tr(M†M)− 1

16λ(Tr(M†M)

)2. (6.27)



This Lagrangian is invariant under the following transformation of the ma-trix M

M → LMR†, (6.28)

where L and R are two special (with determinant equal to 1) unitary ma-trices. That is L,R ∈ SU(2). The reason to restrict these matrices to bespecial is that only in this way the transformed matrix satisfies the condi-tion (6.25). In fact, if A is a 2× 2 matrix with detA = 1, then

τ2AT τ2 = A−1. (6.29)

Therefore, for M ′ = LMR† we get

τ2M′∗τ2 = τ2L

∗M∗RT τ2 = τ2L∗τ2(τ2M∗τ2)τ2RT τ2 (6.30)

and from (6.29)

τ2L∗τ2 = τ2L

†T τ2 = L†−1

= L,

τ2RT τ2 = R−1 = R†, (6.31)

since L and R are independent transformations, the invariance group inthis basis is SU(2)L ⊗ SU(2)R. In fact, this group and O(4) are relatedby the following observation: the transformation M → LMR† is a lineartransformation on the matrix elements of M , but from the relation (6.26)we see that M → LMR† leaves the norm of the vector ~φ = (~π, σ), Sincethe same conditions are satisfied by the orthogonal transformation actingupon the field ~φ = (σ, ~π), the transformation M → LMR† must belongto O(4). This shows that the two groups SU(2) ⊗ SU(2) and O(4) arehomomorphic1(actually there is a 2 to 1 relationship, since −L and −Rdefine the same S as L and R).

We can evaluate the effect of an infinitesimal transformation. To thisend we will consider separately left and right transformations. We param-eterize the transformations as follows

L ≈ 1− i

2~θL · ~τ , R ≈ 1− i

2~θR · ~τ , (6.32)

then we get

δLM = (− i2~θL ·~τ)M = (− i

2~θL ·~τ)(σ+ i~π ·~π) =

12~θL ·~π+

i

2(~θL∧~π−~θLσ) ·~τ ,

(6.33)where we have used

τiτj = δij + iεijkτk. (6.34)1The relation between O(4) and SU(2)L⊗SU(2)R is similar to the one existing between

the Lorentz group O(3, 1) and SL(2, C) (see Section 1.3).



Since

δLM = δLσ + iδL~π · ~τ , (6.35)

we get

δLσ =12~θL · ~π, δL~π =

12

(~θL ∧ ~π − ~θLσ). (6.36)

Analogously we obtain

δRσ = −12~θR · ~π, δR~π =

12

(~θR ∧ ~π + ~θRσ). (6.37)

The combined transformation is given by

δσ =12

(~θL − ~θR) · ~π, δ~π =12

[(~θL + ~θR) ∧ ~π − (~θL − ~θR)σ] (6.38)

and we can check immediately that

σδσ + ~π · δ~π = 0, (6.39)

as it must be for a transformation leaving the form σ2 + |~π|2 invariant. Ofparticular interest are the transformations with ~θL = ~θR ≡ θ. In this casewe have L = R and

M → LML†. (6.40)

These transformations span a subgroup SU(2) of SU(2)L ⊗ SU(2)R calledthe diagonal subgroup, and the corresponding infinitesimal transformationsare

δσ = 0, δ~π = ~θ ∧ ~π. (6.41)

We see that the transformations corresponding to the diagonal SU(2) arerotations in the 3-dimensional space spanned by ~π. These rotations definea subgroup O(3) of the original symmetry group O(4). From Noether’stheorem we get the conserved currents

jLµ =12σ,µ~θL · ~π +

12~π,µ · (~θL ∧ ~π − ~θLσ) (6.42)

and extracting the coefficients of ~θL

~JLµ =12

(σ,µ~π − ~π,µσ − ~π,µ ∧ ~π). (6.43)

Analogously

~JRµ =12

(−σ,µ~π + ~π,µσ − ~π,µ ∧ ~π). (6.44)



Using the canonical commutation relations one can verify that the corre-sponding charges satisfy the Lie algebra of SU(2)L ⊗ SU(2)R

[QLi , QLj ] = iεijkQ

Lk , [QRi , Q

Rj ] = iεijkQ

Rk , [QLi , Q

Rj ] = 0. (6.45)

By taking the following combinations of currents

~JVµ = ~JLµ + ~JRµ , ~JAµ = ~JLµ − ~JRµ , (6.46)

one has

~JVµ = ~π ∧ ~π,µ (6.47)

and

~JAµ = σ,µ~π − ~π,µσ. (6.48)

The corresponding algebra of conserved charges is

[QVi , QVj ] = iεijkQ

Vk , [QVi , Q

Aj ] = iεijkQ

Ak , [QAi , Q

Aj ] = iεijkQ

Vk . (6.49)

These equations show that QVi are the infinitesimal generators of the diag-onal subgroup SU(2) of SU(2)L ⊗ SU(2)R. This can be seen also from

[QVi , πj ] = iεijkπk, [QVi , σ] = 0. (6.50)

In the following we will be interested in treating the interacting fieldtheories by using perturbation theory. As in the quantum mechanical case,this is a well defined procedure only when we are considering the theoryclose to a minimum of the energy of the system. In fact, if we expandaround a maximum the oscillations of the system can become very largeleading outside of the domain of perturbation theory. In the case of thelinear σ-model the energy is given by

H =∫

d3xH =∫

d3x

[N∑i=1

∂L∂φi

φi − L

]

=∫

d3x

[12

N∑i=1

(φ2i + |~∇φi|2) + V (|~φ|2)

]. (6.51)

Since in the last equation the first two terms are positive definite, it followsthat the absolute minimum is obtained for constant field configurations,that is for

∂V (|~φ|2)∂φi

= 0. (6.52)

Let us call by vi the generic solution of this equation (in general it couldhappen that the absolute minimum is degenerate). Then the condition



for getting a minimum is that the eigenvalues of the matrix of the secondderivatives of the potential at the stationary point are definite positive. Inthis case we define new fields by shifting the original ones by vi

φi → φ′i = φi − vi. (6.53)

The Lagrangian density becomes

L =12∂µφ

′i∂µφ′i − V (|~φ′ + ~v|2). (6.54)

Expanding V in series of φ′i we get

V = V (|~v|2) +12

∂2V

∂φi∂φj

∣∣∣~φ=~v

φ′iφ′j + · · · . (6.55)

This equation shows that the particle masses are given by the eigenvaluesof the second derivative of the potential at the minimum. In the case ofthe linear σ-model we have

V =12µ2|~φ|2 +

λ

4(|~φ|2)2. (6.56)

Therefore

∂V

∂φi= µ2φi + λφi|~φ|2. (6.57)

In order to have a solution to the stationary condition we must have φi = 0,or

|~φ|2 = −µ2

λ. (6.58)

This equation has real solutions only if µ2/λ < 0. However, in order tohave a potential bounded from below one has to require λ > 0, thereforewe may have non zero solutions to the minimum condition only if µ2 < 0.But notice that in this case µ2 cannot be identified with a physical masssince the masses are given by the eigenvalues of the matrix of the secondderivatives of the potential at the minimum and they are positive definiteby definition. We will study this case in the following Sections. Whenµ2 > 0 the minimum is at φi = 0 and one can study the theory by takingthe term λ(|~φ|2)2 as a small perturbation (that is requiring both λ andthe values of φi, the fluctuations around the minimum to be small). Thefree theory is given by the quadratic terms in the Lagrangian density, anddescribes N particles of common mass m. Furthermore, both the free andthe interacting theories are O(N) symmetric.



6.2 Spontaneous symmetry breaking

In this Section we will show that the linear σ-model with µ2 < 0 is an exam-ple of a general phenomenon named spontaneous symmetry breaking.This phenomenon lies at the basis of the modern description of phase tran-sitions and it has acquired a capital relevance in the last years in all fieldsof physics. In particular, in the case of the theory of elementary particles,it is the basis for the description of the weak interactions.

The idea is very simple and consists in the observation that a theory withHamiltonian invariant under a symmetry group may not show explicitly thesymmetry at the level of the solutions. As we shall see this may happenwhen the following conditions are realized:

• The theory is invariant under a symmetry group G.• The fundamental state of the theory is degenerate and transforms in

a non trivial way under the symmetry group.

Just as an example consider a scalar field described by a Lagrangian invari-ant under parity

P : φ→ −φ. (6.59)

The Lagrangian density will be of the type

L =12∂µφ∂

µφ− V (φ2). (6.60)

If the vacuum state is non degenerate, barring a phase factor, we must have

P |0〉 = |0〉, (6.61)

since P commutes with the Hamiltonian. It follows

〈0|φ|0〉 = 〈0|P−1PφP−1P |0〉 = 〈0|PφP−1|0〉 = −〈0|φ|0〉, (6.62)

from which

〈0|φ|0〉 = 0. (6.63)

The case is different if the fundamental state is degenerate. This would bethe case in the example (6.60), if

V (φ2) =µ2

2φ2 +

λ

4φ4, (6.64)

with µ2 < 0. In fact, this potential has two minima located at

φ = ±v, v =

√−µ

2

λ. (6.65)



By denoting with |R〉 and |L〉 the two states corresponding to the classicalconfigurations φ = ±v, we have

P |R〉 = |L〉 6= |R〉. (6.66)

Therefore

〈R|φ|R〉 = 〈R|P−1PφP−1P |R〉 = −〈L|φ|L〉, (6.67)

which, contrarily to the previous case, does not imply that the expectationvalue of the field vanishes. In the following we will be rather interested inthe case of continuous symmetries. So let us consider two scalar fields, anda Lagrangian density with symmetry O(2)

L =12∂µ~φ · ∂µ~φ−

12µ2~φ · ~φ− λ

4(~φ · ~φ)2, (6.68)

where

~φ · ~φ = φ21 + φ2

2. (6.69)

For µ2 > 0 there is a unique ground state (minimum of the potential) ~φ = 0,whereas for µ2 < 0 there are infinite degenerate states corresponding to thepoints belonging to the manifold

|~φ|2 = φ21 + φ2

2 = v2, (6.70)

with v defined as in (6.65). By denoting with R(θ) the operator rotatingthe fields in the plane (φ1, φ2) we have, for the non-degenerate case,

R(θ)|0〉 = |0〉 (6.71)

and

〈0|φ|0〉 = 〈0|R−1RφR−1R|0〉 = 〈0|φθ|0〉 = 0, (6.72)

since φθ 6= φ. In the case µ2 < 0 (degenerate case), we have

R(θ)|0〉 = |θ〉, (6.73)

where |θ〉 is one of the infinitely many degenerate fundamental states lyingon the circle |~φ|2 = v2. Then

〈0|φi|0〉 = 〈0|R−1(θ)R(θ)φiR−1(θ)R(θ)|0〉 = 〈θ|φθi |θ〉, (6.74)

with

φθi = R(θ)φiR−1(θ) 6= φi. (6.75)

Again, the expectation value of the field (contrarily to the non-degeneratecase) does not need to vanish. The situation can be described qualitatively



saying that the existence of a degenerate fundamental state forces the sys-tem to choose one of the equivalent states, and consequently to break thesymmetry. But the breaking is only at the level of the solutions, the La-grangian and the equations of motion preserve the symmetry. One caneasily construct classical systems exhibiting spontaneous symmetry break-ing. For instance, a classical particle in a double-well potential. This systemis parity invariant, x → −x, where x is the particle position. The equilib-rium positions are at the points, ±x0, corresponding to the minima of thepotential. If we put a particle close to x0, it will perform oscillations aroundthat point and the original symmetry is lost. A further example is givenby a ferromagnet extending in all 3-dimensions. The corresponding Hamil-tonian is invariant under rotations in the 3-dimensional space. However,below the Curie temperature, the ferromagnet exhibits spontaneous mag-netization, breaking the rotational symmetry in space to the symmetry ofrotations around the direction of the magnetization. These situations aretypical for the so-called second order phase transitions. One can describethem through the Landau free-energy, which depends on two different kindof parameters:

• Control parameters, as µ2 for the scalar field, or as the temperaturefor the ferromagnet.

• Order parameters, as the expectation value of the scalar field or asthe magnetization.

The system goes from one phase to another varying the control parameters,and the phase transition is characterized by the order parameter whichassumes different values in different phases. In the previous examples, theorder parameter was zero in the symmetric phase and different from zeroin the broken phase.

The situation looks more involved at the quantum level, since sponta-neous symmetry breaking cannot happen in finite systems. This followsfrom the existence of the tunnel effect. Let us consider again a particle ina double-well potential, and recall that we have defined the ground statesthrough the correspondence with the classical minima

x = x0 → |R〉,x = −x0 → |L〉. (6.76)

But the tunnel effect gives rise to a transition between these two states andas a consequence it removes the degeneracy. In fact, due to the nonvanishingtransition probability between the two degenerate states, the Hamiltonian



acquires a non zero matrix element between the states |R〉 and |L〉. Bydenoting with H the matrix of the Hamiltonian between these two states,we get

H =(ε0 ε1ε1 ε0

). (6.77)

The eigenvalues of H are

(ε0 + ε1, ε0 − ε1). (6.78)

We have no more degeneracy and the eigenstates are

|S〉 =1√2

(|R〉+ |L〉), (6.79)

with eigenvalue ES = ε0 + ε1, and

|A〉 =1√2

(|R〉 − |L〉), (6.80)

with eigenvalue EA = ε0 − ε1. One can show that ε1 < 0 and therefore theground state is the symmetric one, |S〉. This situation gives rise to the socalled quantum oscillations. We can express the states |R〉 and |L〉 in termsof the energy eigenstates

|R〉 =1√2

(|S〉+ |A〉),

|L〉 =1√2

(|S〉 − |A〉). (6.81)

Then, let us prepare a state, at t = 0, by putting the particle in the rightminimum. This is not an energy eigenstate and its time evolution is givenby

|R, t〉 =1√2

(e−iESt|S〉+ e−iEAt|A〉

)=

1√2e−iESt

(|S〉+ e−it∆E |A〉

), (6.82)

with ∆E = EA − ES . Therefore, for t = π/∆E the state |R〉 transformsinto the state |L〉 and comes back to |R〉 for t = 2π/∆E. As a consequencethe state oscillates between states described by the kets |R〉 and |L〉, witha period

T =2π∆E

. (6.83)

In nature there are finite systems as sugar molecules, which seem to exhibitspontaneous symmetry breaking. In fact, we observe both right-handed



and left-handed sugar molecules. The explanation is simply that the energydifference ∆E is so small that the oscillation period is of the order of 104−106 years.

The splitting of the fundamental states decreases with the height of thepotential between two minima. Therefore, for infinite systems, the previ-ous mechanism does not work, and we may have spontaneous symmetrybreaking. In fact, coming back to the scalar field example, its expectationvalue on the vacuum must be a constant, as it follows from the translationalinvariance of the vacuum

〈0|φ(x)|0〉 = 〈0|eiPxφ(0)e−iPx|0〉 = 〈0|φ(0)|0〉 = v (6.84)

and the energy difference between the maximum at φ = 0, and the minimumat φ = v, becomes infinite in the limit of infinite volume

H(φ = 0)−H(φ = v) = −∫V

d3x

[µ2

2v2 +

λ

4v4

]=µ4

4λ

∫V

d3x =µ4

4λV.

(6.85)

6.3 The Goldstone theorem

One of the most interesting consequences of spontaneous symmetry break-ing is the Goldstone theorem. This theorem says that for any spontaneouslybroken continuous symmetry, there exists a massless particle2 (the Gold-stone boson). The theorem holds rigorously in a relativistic local fieldtheory, under the following hypotheses:

• The spontaneous broken symmetry must be a continuous one.• The theory must be manifestly covariant.• The Hilbert space of the theory must have a definite positive norm.

We will limit ourselves to analyze the theorem in the case of a classicalscalar field theory. Let us start considering the Lagrangian for the linearσ-model with invariance O(N)

L =12∂µ~φ · ∂µ~φ−

µ2

2~φ · ~φ− λ

4(~φ · ~φ)2, (6.86)

where, according to our previous discussione, we assume λ > 0. The con-ditions for V to be stationary is

∂V

∂φl= µ2φl + λφl|~φ|2 = 0, (6.87)

2In the non relativistic case the counting of the massless states is different and it depends

on the dispersion relation for the Goldstone particles [Nielsen and Chada (1976)].



with solutions

φl = 0, |~φ|2 = v2, v =

√−µ2

λ. (6.88)

The character of the stationary points can be studied by evaluating thesecond derivatives

∂2V

∂φl∂φm= δlm(µ2 + λ|~φ|2) + 2λφlφm. (6.89)

We have two possibilities:

• µ2 > 0, we have only one real solution given by ~φ = 0, which is aminimum, since

∂2V

∂φl∂φm= δlmµ

2 > 0. (6.90)

• µ2 < 0, there are infinite solutions, among which ~φ = 0 is a maximum.The points of the sphere |~φ|2 = v2 are degenerate minima. In fact, bychoosing φl = vδlN as a representative point, we get

∂2V

∂φl∂φm= 2λv2δlNδmN > 0. (6.91)

Expanding the potential around this minimum we get

V (~φ) ≈ V∣∣∣min

+12

∂2V

∂φl∂φm

∣∣∣min

(φl − vδlN )(φm − vδmN ). (6.92)

If we are interested in the perturbative expansion, this must be made arounda minimum and the right fields to be used are φl − vδlN . The mass matrixis given by the coefficients of the quadratic term

M2lm =

∂2V

∂φl∂φm

∣∣∣min

= −2µ2δlNδmN =

0 0 · 00 0 · 0· · · ·0 0 · −2µ2

. (6.93)

Therefore the masses of the fields φa, a = 1, · · · , N − 1 , and χ = φN − v,are given by

m2φa = 0, m2

χ = −2µ2. (6.94)

By defining

m2 = −2µ2, (6.95)



we can write the potential as a function of the new fields

V =m4

16λ+

12m2χ2 +

√m2λ

2χ

(N−1∑a=1

φ2a + χ2

)+λ

4

(N−1∑a=1

φ2a + χ2

)2

.

(6.96)In this form the original symmetry O(N) is broken. However a residualsymmetry O(N − 1) is left. In fact, V depends only on the combination∑N−1a=1 φ2

a, and this is invariant under rotations around the axis we havechosen as representative for the fundamental state, (0, · · · , v). However,it must be stressed that this is not the most general potential invariantunder O(N − 1). In fact the most general potential (up to the fourth orderin the fields) describing N scalar fields with a symmetry O(N − 1) is apolynomial in

∑N−1a=1 φ2

a and χ depending on 7 coupling constants, whereasthe one we have obtained depends only on the two original parameters mand λ. Therefore spontaneous symmetry breaking puts heavy constraintson the dynamics of the system. We have also seen that there are N − 1massless scalars. Clearly the rotations along the first N − 1 directionsleave the potential invariant, whereas the N − 1 rotations on the planes(a,N), a = 1, · · · , N − 1 move away from the surface of the minima. Thiscan be seen also in terms of generators. Since the field we have chosenas representative of the ground state is φi|min = vδiN , we have, using eq.(6.14),

T abij φj |min = i(δai δbj − δbi δaj )vδjN = 0, (6.97)

since a, b = 1, · · · , N − 1, and

T aNij φj |min = i(δai δNj − δNi δaj )vδjN = ivδai 6= 0. (6.98)

Therefore we have N − 1 broken symmetries and N − 1 massless scalars.The generators of O(N) divide up naturally in the generators of the vac-uum symmetry (here O(N − 1)), and in the so called broken generators,each of them corresponding to a massless Goldstone boson. In general, ifthe original symmetry group G of the theory is spontaneously broken downto a subgroup H (which is the symmetry of the vacuum), the Goldstonebosons correspond to the generators of G which are left after subtractingthe generators of H. Intuitively one can understand the origin of the mass-less particles noticing that the broken generators allow transitions from apossible vacuum to another. Since these states are degenerate the opera-tion does not cost any energy. From the relativistic dispersion relation thisimplies that we must have massless particles. One can say that Goldstonebosons correspond to flat directions in the potential.



6.4 QED as a gauge theory

Many field theories possess global symmetries. These are transformationsleaving invariant the action of the system and are characterized by a certainnumber of parameters which are independent on the space-time point. Asa prototype we can consider the free Dirac Lagrangian density

L0 = ψ(x)[i∂/ −m]ψ(x), (6.99)

which is invariant under the global phase transformation

ψ(x)→ ψ′(x) = e−iQαψ(x), (6.100)

where Q is the electric charge of the field in units of e which, only for thisChapter, will be assumed as the charge of the proton. If there are manycharged fields, Q is a diagonal matrix with eigenvalues equal to the chargesof the different fields (always measured in units of e). For instance, a termas ψ2ψ1φ, with φ a scalar field, is invariant by choosing Q(ψ1) = Q(φ) = 1,and Q(ψ2) = 2. This is said to be an abelian symmetry because differenttransformations commute among themselves

e−iαQe−iβQ = e−i(α+ β)Q = e−iβQe−iαQ. (6.101)

It is also referred to as a U(1) symmetry. The physical meaning of thisinvariance lies in the possibility of assigning the phases of fields in arbitraryway, without changing the observable quantities. This way of thinking isin some sort of contradiction with causality, since it requires to assign thephase of the fields simultaneously in all space-time points. It looks morephysical to be able to assign the phase arbitrarily in each space-time point.If the theory does not depend by this arbitrary choice of the phases we saythat it is gauge invariant3. The free Lagrangian (6.99) cannot be invariantunder the transformation (6.100), with a parameter α depending on thespace-time point, due to the derivative present in the kinetic term. Then,the idea is simply to generalize the derivative ∂µ to a so called covariantderivative Dµ having the property that Dµψ transforms as ψ, that is

Dµψ(x)→ [Dµψ(x)]′ = e−iQα(x)Dµψ(x). (6.102)

In this case the term

ψDµψ (6.103)3This was the main idea underlying the work of [Yang and Mills (1954)] that lead to

the formulation of non-abelian gauge symmetries, see later.



will be invariant as the mass term under the local phase transformation.To construct the covariant derivative, we need to enlarge the field contentof the theory, by introducing a vector field, the gauge field Aµ, in thefollowing way

Dµ = ∂µ + ieQAµ. (6.104)

The transformation law of Aµ is obtained from eq. (6.102)

[(∂µ + ieQAµ)ψ]′ = (∂µ + ieQA′µ)ψ′(x)

= (∂µ + ieQA′µ)e−iQα(x)ψ

= e−iQα(x)[∂µ + ieQ(A′µ −

1e∂µα)

]ψ, (6.105)

from which

A′µ = Aµ +1e∂µα. (6.106)

The Lagrangian density

Lψ = ψ[iD/ −m]ψ = ψ[iγµ(∂µ+ieQAµ)−m)ψ = L0−eψQγµψAµ (6.107)

is then invariant under gauge transformations or under the local groupU(1). We see that the requirement of local invariance gives rise to theelectromagnetic interaction as obtained through the minimal substitutionwe discussed before.

In order to determine the kinetic term for the vector field Aµ we noticethat eq. (6.102) implies that under a gauge transformation, the covariantderivative transforms according to the unitary transformation

Dµ → Dµ′ = e−iQα(x)Dµe

iQα(x). (6.108)

Then, also the commutator of two covariant derivatives

[Dµ, Dν ] = [∂µ + ieQAµ, ∂ν + ieQAν ] = ieQFµν , (6.109)

with

Fµν = ∂µAν − ∂νAµ, (6.110)

transforms exactly in the same way

Fµν → e−iQα(x)FµνeiQα(x) = Fµν . (6.111)

The last equality follows from the commutativity of Fµν with the phasefactor. The complete invariant Lagrangian density can be written as

L = Lψ + LA = ψ[iγµ(∂µ + ieQAµ)−m]ψ − 14FµνF

µν . (6.112)



The gauge principle has automatically generated an interaction betweenthe gauge field and the charged field. We notice also that gauge invarianceprevents any mass term for the vector field Aµ. As a consequence thephoton field is necessarily massless. Also, since the local invariance impliesthe global one, by using the Noether theorem we find the conserved current

jµ =∂L∂ψ,µ

δψ = ψγµ(Qα)ψ, (6.113)

from which, eliminating the infinitesimal parameter α,

Jµ = ψγµQψ. (6.114)

6.5 Non-abelian gauge theories

The approach of the previous Section can be easily extended to non-abelianlocal symmetries. We will consider the case of N Dirac fields. The freeLagrangian density

L0 =N∑a=1

ψa(i∂/ −m)ψa (6.115)

is invariant under the global transformation

Ψ(x)→ Ψ′(x) = AΨ(x), (6.116)

where A is a unitary N ×N matrix, and we have denoted by Ψ the columnvector with components ψa. Notice that U(N) is the maximal symmetrygroup of eq. (6.115). More generally the actual symmetry could be asubgroup of U(N). For instance, when the masses are not all equal. So wewill consider here the gauging of a subgroup G of U(N). The fields ψa(x)will belong, in general, to some reducible representation of G. Denotingby U the generic element of G, we will write the corresponding matrix Uabacting upon the fields ψa as

U = e−iαATA, U ∈ G. (6.117)

where TA denote the generators of the Lie algebra associated to G, Lie(G),(that is the vector space spanned by the infinitesimal generators of thegroup) in the fermion representation. The generators TA satisfy the algebra

[TA, TB ] = ifABC TC , (6.118)



where fABC are the structure constants of Lie(G). For instance, if G =SU(2), and we take the fermions in the fundamental representation,

Ψ =(ψ1

ψ2

). (6.119)

we have

TA =σA

2, A = 1, 2, 3, (6.120)

where σA are the Pauli matrices. In the general case the TA’s are N ×Nhermitian matrices that we will choose normalized in such a way that

Tr(TATB) =12δAB . (6.121)

To make local the transformation (6.117), means to promote the parametersαA to space-time functions

αA → αA(x). (6.122)

Notice, that the group does not need to be abelian, and therefore, in general

e−iαATAe−iβAT

A6= e−iβAT

Ae−iαAT

A. (6.123)

In the case of a local symmetry we will define a covariant derivative in strictanalogy to what we have done in the abelian case. That is we will requirethe following transformation properties

DµΨ(x)→ [DµΨ(x)]′ = U(x)[Dµψ(x)]. (6.124)

We will put again

Dµ = ∂µ + igBµ, (6.125)

with Bµ is a N ×N matrix acting upon Ψ(x). In components

Dµab = δab∂

µ + ig(Bµ)ab. (6.126)

The eq. (6.124) implies

DµΨ → (∂µ + igB′µ)U(x)Ψ

= U(x)∂µΨ + U(x)[U−1(x)igB′µU(x)]Ψ + (∂µU(x))Ψ

= U(x)[∂µ + U−1(x)igB′µU(x) + U−1(x)∂µU(x)]Ψ, (6.127)

from which

U−1(x)igB′µU(x) + U−1(x)∂µU(x) = igBµ (6.128)



and

B′µ(x) = U(x)Bµ(x)U−1(x) +i

g(∂µU(x))U−1(x). (6.129)

For an infinitesimal transformation

U(x) ≈ 1− iαA(x)TA, (6.130)

we get

δBµ(x) = −iαA(x)[TA, Bµ(x)] +1g

(∂µαA(x))TA. (6.131)

Since Bµ(x) acquires a term proportional to TA, the transformation law isconsistent with Bµ being linear in the generators of the Lie algebra, that is

(Bµ)ab ≡ AµA(TA)ab. (6.132)

The fields AµA are called non abelian gauge fields and their transformationlaw is

δAµC = fABC αAAµB +

1g∂µαC . (6.133)

The difference with respect to the abelian case is that the transformationacts on the fields not only with the inhomogeneous piece, but also witha homogeneous one depending on the non trivial commutation relationsamong the generators of the Lie algebra of the non abelian group. In fact,for an abelian group the structure constants are zero and we reproduce theabelian transformation.

The kinetic term for the gauge fields is constructed as in the abeliancase. In fact the quantity

[Dµ, Dν ] (6.134)

transforms as

[Dµ, Dν ]→ U(x)[Dµ, Dν ]U−1(x), (6.135)

in virtue of the eq. (6.124), that implies

Dµ → U(x)DµU−1(x). (6.136)

Therefore, defining as in the abelian case

[Dµ, Dν ] ≡ igFµν , (6.137)

we see that

F ′µν = U(x)FµνU−1(x). (6.138)



This time the tensor Fµν is not invariant but transforms homogeneously,since it does not commute with the gauge transformation as in the abeliancase.

To construct invariant quantities from objects transforming homoge-neously, it is enough to take the trace. The invariant kinetic term is then

LA = −12Tr[FµνFµν ]. (6.139)

Let us now evaluate Fµν

igFµν = [Dµ, Dν ] = [∂µ + igBµ, ∂ν + igBν ]

= ig(∂µBν − ∂νBµ)− g2[Bµ, Bν ], (6.140)

or

Fµν = (∂µBν − ∂νBµ) + ig[Bµ, Bν ]. (6.141)

In components

Fµν = FµνC TC , (6.142)

with

FµνC = ∂µAνC − ∂νAµC − gf

ABC AµAA

νB . (6.143)

The main feature of the non-abelian gauge theories is the bilinear term inthe previous expression. Such a term originates from the fact that G is notabelian, corresponding to fABC 6= 0. The kinetic term for the gauge fields,expressed in components, is given by

LA = −14

∑A

FµνAFµνA . (6.144)

Therefore, whereas in the abelian case LA is a free Lagrangian (it containsonly quadratic terms), in the case of non abelian symmetries it containsinteraction terms both cubic and quartic in the fields. The physical motiva-tion lies in the fact that the gauge fields couple to all the fields transformingin a non trivial way under the gauge group. Therefore they couple also tothemselves (remember the homogeneous piece of transformation).

To derive the equations of motion for the gauge fields, let us considerthe total action ∫

V

d4x[Ψ(i∂/ −m)Ψ− gΨγµBµΨ

]+ SA, (6.145)

where

SA = −12

∫V

d4xTr(FµνFµν). (6.146)



The variation of SA is

δSA = −∫V

d4xTr(FµνδFµν). (6.147)

Using the definition (6.141) for the field strength we get

δFµν = ∂µδBν + ig(δBµ)Bν + igBµδ(Bν)− (µ↔ ν), (6.148)

from which

δSA = −2∫V

d4xTr[Fµν(∂µδBν + ig(δBµ)Bν + igBµ(δBν))], (6.149)

where we have taken into account the antisymmetry properties of Fµν .Integrating by parts we obtain

δSA = −2∫V

d4xTr[−(∂µFµν)δBν − igBµFµνδBν + igFµνBµδBν ]

= 2∫V

d4xTr [(∂µFµν + ig[Bµ, Fµν ]) δBν ]

=∫V

d4x (∂µFµν + ig[Bµ, Fµν ])A δAνA. (6.150)

Here we have made use of the cyclic property of the trace. By taking intoaccount also the free term for the Dirac fields and the interaction we findthe equations of motion

∂µFµνA + ig[Bµ, Fµν ]A = gΨγνTAΨ,

(i∂/ −m)Ψ = gγµBµΨ. (6.151)

From the first equation we see that the currents ΨγµTAΨ are not conserved.In fact, the conserved currents turn out to be

JAν = ΨγνTAΨ− i[Bµ, Fµν ]A, (6.152)

as it follows from the first equation in (6.151). The reason is that undera global transformation of the symmetry group, the gauge fields are notinvariant, said in different words they are charged fields with respect to thegauge fields. In fact, we can verify immediately that the previous currentsare precisely the Noether currents. Under a global variation we have

δAµC = fABC αAAµB , δΨ = −iαATAΨ (6.153)

and we get

jµ =∂L∂Ψ,µ

δΨ +∂L

∂Aν,µCδAνC , (6.154)



from which

jµ = ΨγµαATAΨ− FµνC fABC αAAνB . (6.155)

In the case of simple compact Lie groups one can define fABC = fABC withthe property fABC = fBCA. It follows

FµνC fABCAνBTA = i[Bν , Fνµ] ≡ i[Bν , Fνµ]ATA. (6.156)

Therefore

jµ = ΨγµαATAΨ− i[Bν , F νµ]AαA. (6.157)

After division by αA we get the Noether currents (6.152). The contributionof the gauge fields to the currents is also crucial for their conservation. Infact, the divergence of the fermionic contribution is given by

∂µ(ΨγµTAΨ) = −igΨγµTABµΨ + igΨγµBµTAΨ = −igΨγµ[TA, Bµ]Ψ,(6.158)

which vanishes for abelian gauge fields, whereas it is compensated by thegauge fields contribution in the non abelian case.

6.6 The Higgs mechanism

We have seen that in the case of continuous symmetries, the spontaneoussymmetry breaking mechanism leads to massless scalar particles, the Gold-stone bosons. Also gauge theories lead to massless vector bosons. In fact,as in the electromagnetic case, gauge invariance forbids the presence in theLagrangian of terms quadratic in the gauge fields. Unfortunately in naturethe only massless particles we know are the photons4. But once sponta-neous symmetry breaking and gauge symmetry are present at the sametime, things change and one gets completely different results. In fact, if welook back at the hypotheses underlying a gauge theory, it turns out thatthe Goldstone theorem does not hold in this context. The reason is that itis impossible to quantize a gauge theory in a way which is at the same timemanifestly covariant and with a Hilbert space with positive definite metric.In fact, remember that for the electromagnetic field one has to choose thegauge before quantization. What happens is that, if one chooses a physicalgauge, as the Coulomb gauge, in order to have a Hilbert space spanned byonly the physical states, than the theory looses the manifest covariance. If4In 1998 the SuperKamiokande collaboration has observed neutrino oscillations, proving

that also the neutrinos, that for a very long time were supposed to be massless, have

indeed a mass.



one goes to a covariant gauge, as the Lorenz one, the theory is covariant butone has to work with a bigger Hilbert space with non-definite positive met-ric, and where the physical states are obtained through a supplementarycondition. This situation is common to all the gauge theories.

The way in which the Goldstone theorem is evaded is that the Goldstonebosons disappear and, at the same time, the gauge bosons correspondingto the broken symmetries acquire mass. This is the famous Higgs mech-anism.

Let us start with a scalar theory invariant under O(2)

L =12∂µ~φ · ∂µ~φ−

µ2

2~φ · ~φ− λ

4(~φ · ~φ)2 (6.159)

where ~φ = (φ1, φ2). Let us now analyze the spontaneous symmetry breakingmechanism. In the case µ2 < 0 the symmetry is broken and we can choosethe vacuum as the state

~φ0 = (v, 0), v =

√−µ2

λ. (6.160)

After the translation φ1 = χ + v, with 〈0|χ|0〉 = 0, we get the potential(m2 = −2µ2)

V =m4

16λ+

12m2χ2 +

√m2λ

2χ(φ2

2 + χ2) +λ

4(φ2

2 + χ2)2. (6.161)

Correspondingly the group O(2) is completely broken (except for the dis-crete symmetry φ2 → −φ2). The Goldstone particle is described by φ2.This field has a peculiar way of transforming under O(2). In fact, theoriginal fields transform as

δφ1 = −αφ2, δφ2 = αφ1, (6.162)

from which

δχ = −αφ2, δφ2 = αχ+ αv. (6.163)

We see that the Goldstone field, φ2, under the transformation, performsboth a rotation and a translation, αv. This is the main reason why theGoldstone particle is massless. In fact, one can have invariance under trans-lations of the field, only if the potential is flat in the corresponding direction.This is what happens when one moves in a way which is tangent to the sur-face of the degenerate vacua (in this case a circle). How do things changeif our theory is gauge invariant? In that case the translation becomes an



arbitrary function of the space-time point and it should be possible to elim-inate the Goldstone field from the theory. This is better seen by using polarcoordinates for the fields, that is

ρ =√φ2

1 + φ22, sin θ =

φ2√φ2

1 + φ22

. (6.164)

Under a finite rotation, the new fields transform as

ρ→ ρ, θ → θ + α. (6.165)

Notice that the two coordinate systems coincide when we are close to thevacuum. In fact, in that case we can perform the following expansion

ρ =√φ2

2 + χ2 + 2χv + v2 ≈ v + χ, θ ≈ φ2

v + χ≈ φ2

v. (6.166)

Therefore, if we make the theory invariant under a local transformation,we will have invariance under

θ(x)→ θ(x) + α(x). (6.167)

By choosing α(x) = −θ(x) we can eliminate the θ field from the theory.The only remaining degree of freedom in the scalar sector is ρ(x).

Let us study the gauging of this model. It is convenient to introducecomplex variables

φ =1√2

(φ1 + iφ2), φ† =1√2

(φ1 − iφ2). (6.168)

The O(2) transformation becomes a phase transformation on φ

φ→ eiαφ (6.169)

and the Lagrangian density (6.159) can be written as

L = ∂µφ†∂µφ− µ2φ†φ− λ(φ†φ)2. (6.170)

We know that it is possible to promote a global symmetry to a local oneby introducing the covariant derivative

∂µφ→ (∂µ + igAµ)φ, (6.171)

from which

L = (∂µ − igAµ)φ†(∂µ + igAµ)φ− µ2φ†φ− λ(φ†φ)2 − 14FµνF

µν . (6.172)

In terms of the polar coordinates (ρ, θ) we have

φ =1√2ρeiθ, φ† =

1√2ρe−iθ. (6.173)



By performing the following gauge transformation on the scalars

φ→ φ′ = φe−iθ (6.174)

and the corresponding transformation on the gauge fields

Aµ → A′µ = Aµ +1g∂µθ, (6.175)

the Lagrangian will depend only on the fields ρ and A′µ (we will put A′µ =Aµ in the following)

L =12

(∂µ − igAµ)ρ(∂µ + igAµ)ρ− µ2

2ρ2 − λ

4ρ4 − 1

4FµνF

µν . (6.176)

In this way the Goldstone boson disappears. We have now to translate thefield ρ

ρ = χ+ v, 〈0|χ|0〉 = 0 (6.177)

and we see that this generates a bilinear term in Aµ, coming from thecovariant derivative, given by

12g2v2AµA

µ. (6.178)

Therefore the gauge field acquires a mass

m2A = g2v2. (6.179)

It is instructive to count the degrees of freedom in the symmetric and in thebroken case. In the symmetric case we had 4 degrees of freedom, two fromthe scalar fields and two from the gauge field. In the broken case, after thegauge transformation, we have only one degree of freedom from the scalarsector and three degrees of freedom from the vector field which is nowmassive. This is as it should be, since the number of degrees of freedomcannot change simply changing the phase. Notice that the reason whywe may read clearly the number of degrees of freedom only after the gaugetransformation is that before performing the transformation the Lagrangiancontains a mixing term

Aµ∂µθ, (6.180)

between the Goldstone field and the gauge vector which forbids to readdirectly the mass of the states from the Lagrangian. The previous gaugetransformation realizes the purpose of making that term vanish and, at thesame time, makes diagonal the mass matrix of the fields (ρ,Aµ). The gaugein which such a thing happens is called the unitary gauge.



We will consider now the further example of a symmetry O(N). TheLagrangian density, invariant under local transformations, is

L =12

(Dµ)ijφj(Dµ)ikφk −µ2

2φiφi −

λ

4(φiφi)2, (6.181)

where

(Dµ)ij = δij∂µ + ig

2(TAB)ijWAB

µ , (6.182)

where (TAB)lm = i(δAl δBm − δAmδBl ). In the instance of broken symmetry

(µ2 < 0), we choose again the vacuum along the direction N , with v definedas in (6.160)

φi|min = vδiN . (6.183)

Recalling that

T abij φj

∣∣∣min

= 0, T aNij φj

∣∣∣min

= ivδai , a, b = 1, · · · , N − 1, (6.184)

the mass term for the gauge field is given by

−18g2TABij φj

∣∣∣min

(TCD)ikφk∣∣∣min

WABµ WµCD

= −12g2T aNij φj

∣∣∣min

(T bN )ikφk∣∣∣min

W aNµ WµbN

=12g2v2δai δ

biW

aNµ WµbN =

12g2v2W aN

µ WµaN . (6.185)

Therefore, the fields W aNµ associated to the broken directions T aN acquire

a mass g2v2, whereas the fields W abµ , associated to the unbroken symmetry

O(N − 1), remain massless.In general, if G is the global symmetry group of the Lagrangian, H the

subgroup of G leaving invariant the vacuum, and GW the group of local(gauge) symmetries, GW ∈ G, one can divide up the broken generators intwo categories. In the first category we put the broken generators lying inGW ; they have associated massive vector bosons. In the second category weput the other broken generators; they have associated massless Goldstonebosons. Finally the gauge fields associated to generators of GW lying inH remain massless. From the previous derivation this follows noticingthat the generators of H annihilate the minimum of the fields, leaving thecorresponding gauge bosons massless, whereas the non zero action of thebroken generators generate a mass term for the corresponding gauge fields.The situation is represented in Fig. 6.1. Now let us show how to eliminate



G H GW

Fig. 6.1 This figure shows the various groups, G, the global symmetry of the La-

grangian, H ∈ G, the symmetry of the vacuum, and GW , the group of local symmetries.

The broken generators in GW correspond to massive vector bosons. The broken gen-erators not belonging to GW correspond to massless Goldstone bosons. The unbroken

generators in GW correspond to massless vector bosons.

the Goldstone bosons corresponding to the broken generators in the gaugegroup. In fact, we can define new fields ξa and χ as

φi =(e−iT

aNξa)iN

(χ+ v), (6.186)

where a = 1, · · · , N − 1, that is with the sum restricted to the brokendirections. The other degree of freedom is in the second factor. The corre-spondence among the fields ~φ and (ξa, χ) can be seen easily by expandingaround the vacuum(

e−iTaNξa

)iN

≈ δiN − i(T aN )iNξa = δiN + δai ξa, (6.187)

from which

φi ≈ (vξa, χ+ v) (6.188)

showing that the ξa’s are really the Goldstone fields. The unitary gauge isdefined through the transformation

φi →(eiT

aNξa)ij

φj = δiN (χ+ v), (6.189)

Wµ → eiTaNξaWµe

−iT aNξa − i

g

(∂µe

iT aNξa)e−iT

aNξa . (6.190)

This transformation eliminates the Goldstone degrees of freedom and theresulting Lagrangian depends on the field χ, on the massive vector fields



W aNµ and on the massless fields W ab

µ . Notice again the counting of thedegrees of freedom: N scalar fields + 2N(N − 1)/2 massless vector fields= N2 fields in the symmetric case, and 1 massive scalar field + 3(N − 1)massive vector fields + 2(N−1)(N−2)/2 massless vector fields = N2 fieldsafter the breaking of the symmetry.

6.7 Exercises

(1) Consider the Lagrangian density

∂µφ†∂µφ−m2φ†φ (6.191)

with (φ1

φ2

)(6.192)

a doublet of SU(2). Show the invariance under SU(2) and evaluate theconserved currents.

(2) Verify the commutation relations given in eq. (6.15).(3) Derive the commutation relations among the non-abelian charges of

eq. (6.22), using the definition (6.21) and the canonical commutationrelations for the fields.

(4) Derive the potential given in eq. (6.96).(5) Using the equations of motion for the fields verify explicitly that the

currents in eq. (6.152) have zero divergence.(6) Consider the electromagnetic tensor Fµν . Show that the gauge invariant

quantity

A = εµνρλFµνF ρλ (6.193)

is the divergence of a four-vector Kµ. Find the expression of Kµ.(7) Show that the Dirac equation is invariant under the chiral transforma-

tion

ψ → eiθγ5ψ, (6.194)

if and only if the mass of the Dirac field vanishes. For m 6= 0 evaluatethe Noether current and its divergence.

(8) Consider the O(4) σ-model interacting with a doublet of Dirac fieldsΨ:

L =12

[∂µσ∂µσ + ∂µ~π∂µ~π] + iΨ∂Ψ + Ψ(σ + i~τ · ~π)Ψ

−m2

2(σ2 + ~π2) +

λ

4(σ2 + ~π2)2. (6.195)



Check the invariance of L under the following SU(2)transformations(see 6.41)

δσ = 0, δ~π = ~θ ∧ ~π, (6.196)

δΨ = −i~θ · ~τ

2Ψ. (6.197)

Then, evaluate the Noether currents.(9) Consider the following dilation transformation

xµ → x′µ = eλxµ, φ(x)→ φ′(x′) = eλφ(x) (6.198)

Show that the classical Lagrangian for a massless scalar field is invariantunder this transformation. Evaluate the infinitesimal transformationand derive the Noether current.


Chapter 7

Time ordered products

7.1 Time ordered products and propagators

In the next Section we will discuss perturbation theory in the context ofquantum fields. We will see that one of the most relevant quantities willbe the propagator, that is the vacuum expectation value of a time orderedproduct of two fields. To introduce this quantity from a physical point ofview we will begin considering a charged Klein-Gordon field. As we knowfrom Section 3.6, the field φ destroys a particle of charge +1 and creates aparticle of charge −1. In any case the net variation of the charge is −1. Inanalogous way the field φ† gives rise to a net variation of the charge equalto +1. Let us now construct a state with charge +1 applying φ† to thevacuum

|ψ(~y, t)〉 = φ†(y)|0〉 =∫

d3k√2ωk(2π)3

eiky |~k,m, 1〉, (7.1)

where |~k,m, 1〉 is the single particle state with charge +1, momentum ~k

and mass m. We want to evaluate the probability amplitude for the state,|ψ(~y, t)〉, to propagate to the same state, |ψ(~x, t′)〉, at a later time t′ > t.This is given by the matrix element

θ(t′ − t)〈ψ(~x, t′)|ψ(~y, t)〉 = θ(t′ − t)〈0|φ(~x, t′)φ†(~y, t)|0〉. (7.2)

It is convenient to think to this matrix element as the one corresponding tothe creation of a charge +1 at the point ~y and time t, and to its annihilationat the point ~x and time t′. This interpretation is correct since the state|ψ(~y, t)〉 is an eigenstate of the normal ordered charge density operator,j0(~x, t), given in eq. (3.174)

ρ(~x, t)|ψ(~y, t)〉 = [ρ(~x, t), φ†(~y, t)]|0〉= +δ3(~x− ~y)φ†(~y, t)|0〉 = δ3(~x− ~y)|ψ(~y, t)〉, (7.3)

163



where we have used eq. (3.168) and the fact that the vacuum has zerocharge.

However for t′ < t we could reach the same result by creating a particleof charge -1 at (~x, t′), and annihilating it at (~y, t). The correspondingamplitude is

θ(t− t′)〈0|φ†(~y, t)φ(~x, t′)|0〉. (7.4)

The situation is represented in Fig. 7.1, where we have considered thecase of a charged particle exchanged between a proton (charge +1) and aneutron (charge 0). The total amplitude is obtained by adding the twocontributions. We define the vacuum expectation value of the time orderedproduct (T product) of two fields as

−i∆F (x− y) ≡ 〈0|T (φ(x)φ†(y)|0〉 = 〈0|T (φ†(y)φ(x))|0〉= θ(x0 − y0)〈0|φ(x)φ†(y)|0〉

+θ(y0 − x0)〈0|φ(y)†φ(x)|0〉. (7.5)

Fig. 7.1 The two probability amplitudes contributing to the process np→ np.

The function ∆F (x − y) is called the Feynman propagator, and wewill show immediately that it depends indeed on the difference of the two


Time ordered products 165

coordinates x and y. Using the expressions (3.184) for the fields, we find

−i∆F (x− y) =∫

d3k d3k′[θ(x0 − y0)f∗~k (y)f~k′(x)〈0|a(~k′)a†(~k)|0〉

+θ(y0 − x0)f~k(y)f∗~k′(x)〈0|b(~k′)b†(~k)|0〉]

=∫

d3k

(2π)3

12ωk

[θ(x0 − y0)e−ik(x− y)

+θ(y0 − x0)eik(x− y)], (7.6)

where, we recall that k0 = ωk. This expression can be written in a moreconvenient way by using the following integral representation of the stepfunction

θ(t) = limη→0+

i

2π

∫ +∞

−∞dω

e−iωt

ω + iη. (7.7)

This representation can be verified immediately by noticing that for t < 0the integral is convergent in the upper complex half-plane of ω. Since thereare no singularities in this region (the integral has a pole at ω = −iη)), wesee that the integral vanishes. In the case t > 0 the integral is convergentin the lower half-plane. Then we pick up the contribution of the pole(in a clockwise direction) and we find 1 as a result. By substituting thisrepresentation in eq. (7.6) we get (omitting the limit that, however, shouldbe taken at the end of the integration)

−i∆F (x− y) = i

∫d3k

(2π)4

∫dω

12ωk

[e−iω(x0 − y0)

ω + iηe−ik(x− y)

+eiω(x0 − y0)

ω + iηeik(x− y)

]. (7.8)

By performing the change of variable k0 = ω + ωk, we find

−i∆F (x− y) = i

∫d4k

(2π)4

12ωk

[e−ik(x− y)

k0 − ωk + iη+

eik(x− y)

k0 − ωk + iη

]

= i

∫d4k

(2π)4

12ωk

e−ik(x− y)[

1k0 − ωk + iη

− 1k0 + ωk − iη

]= i

∫d4k

(2π)4

e−ik(x− y)

k2 −m2 + iε, (7.9)

where we have defined ε = 2ηωk. Notice that ε is a positive quantity and,again, we omit to write explicitly the limit. Then

∆F (x− y) = −∫

d4k

(2π)4

e−ik(x− y)

k2 −m2 + iε. (7.10)



From this representation it follows that ∆F (x) is a Green function for theKlein-Gordon operator

( +m2)∆F (x) = δ4(x). (7.11)

This property of the T -product is a simple consequence of its very definition.In fact, by using the canonical commutators

( +m2)x〈0|T (φ(x)φ†(y))|0〉 = ∂20〈0|T (φ(x)φ†(y))|0〉

+〈0|T ((−~∇2x +m2)φ(x)φ†(y))|0〉

= ∂0〈0|δ(x0 − y0)[φ(x), φ†(y)]|0〉+ ∂0〈0|T (φ(x)φ†(y))|0〉+〈0|T ((−~∇2

x +m2)φ(x)φ†(y))|0〉= 〈0|δ(x0 − y0)[φ(x), φ†(y)]|0〉+ 〈0|T ((x +m2)φ(x)φ†(y))|0〉= −iδ4(x− y). (7.12)

It is easily seen that an analogous result holds for the hermitian Klein-Gordon field, φ† = φ, that is

∆F (x− y) = i〈0|T (φ(x)φ(y))|0〉. (7.13)

It is possible to specify different Green’s functions by choosing in aconvenient way the integration path of the following integral

∆C(x) =∫C

d4k

(2π)4

e−ikx

k2 −m2. (7.14)

Im( )

Re( )

kk-

C+

Fig. 7.2 The integration path C+.

In fact, the integrand has two poles in k0, which are located at k0 = ±ωk,with ωk = (|~k|2 +m2)1/2. To define the integral we need to specify how theintegration path goes around the poles. The result is different by taking



paths closed around the poles, rather than paths extending from −∞ to+∞. In fact, one gets solutions for the homogeneous Klein-Gordon equationin the first case and Green’s functions in the second one. Let us considerthe first case and let us define the two integration paths C± given in Figs.7.2 and 7.3. Correspondingly we define the integrals

∆(±)(x) = −i∫C±

d4k

(2π)4

e−ikx

k2 −m2. (7.15)

We find

Im( )

Re( )

kk-

C_

Fig. 7.3 The integration path C−.

∆(+)(x) = −i∫C+

d4k

(2π)4

e−ikx

(k0 − ωk)(k0 + ωk)=∫

d3k

(2π)3

e−ikx

2ωk

= [φ(+)(x), φ(−)(y)], (7.16)

where we have used eq. (3.140). Also

∆(−)(x) = −∫

d3k

(2π)3

eikx

2ωk= −∆(+)†(x) = [φ(−)(x), φ(+)(y)]. (7.17)

It follows

∆(+)(x) + ∆(−)(x) = i∆(x) (7.18)

and using eq. (3.138), i∆(x− y) = [φ(x), φ(y)]. Therefore the commutatorcan be represented as

∆(x) = −∫C

d4k

(2π)4

e−ikx

k2 −m2, (7.19)

with C given in Fig. 7.4. Quite clearly ∆(x) satisfies the homogeneousKlein-Gordon equation.



Im( )

Re( )

kk-

C

Fig. 7.4 The integration path C.

In the case of ∆F , the position of the poles is the one in Fig. 7.5.Then, the integral can be defined by shifting the poles to the real axis andchoosing an integration path CF , as specified in Fig. 7.6. That is

∆F (x) = −∫CF

d4k

(2π)4

e−ikx

k2 −m2. (7.20)

Using eqs. (7.6), (7.16) and (7.17), we see that

∆F (x) = iθ(x0)∆(+)(x)− iθ(−x0)∆(−)(x). (7.21)

The reason why the ∆C invariant functions, defined on a closed path C,satisfy the homogeneous Klein-Gordon equation, is a simple consequenceof the action of the Klein-Gordon operator. This action removes the singu-larities from the integrand, leaving the integral of an analytic function ona closed path, which vanishes due to the Cauchy theorem.

For a free Dirac field, the Feynman propagator has a similar definition

SF (x− y)αβ = −i〈0|T (ψα(x)ψβ(y))|0〉, (7.22)

but with the T -product defined as follows

T (ψα(x)ψβ(y)) = θ(x0 − y0)ψα(x)ψβ(y)− θ(y0 − x0)ψβ(y)ψα(x). (7.23)

Notice that

T (ψα(x)ψβ(y)) = −T (ψβ(y)ψα(x)), (7.24)



Im( )

Re( )k

k-

Fig. 7.5 The position of the poles in the definition of ∆F (x).

Im( )

Re( )

kk-

CF

Fig. 7.6 The integration path CF .

(the analogous property holds for the Klein-Gordon case, but with a plussign, see eq. (7.5)). The minus sign, introduced in the definition of theT -product for the Dirac field, is needed because only in this way it mayrepresent the Green function for the Dirac operator. In fact,

(i∂x −m)αβT (ψβ(x)ψγ(y)) = i(γ0)αβδ(x0 − y0)[ψβ(x), ψγ(y)

]+

= iδ(x0 − y0)(γ0)αβ(γ0)βγδ3(~x− ~y)

= iδ4(x− y)δαγ , (7.25)

that is

(i∂ −m)SF (x) = δ4(x). (7.26)

It should be clear that the choice of sign in the definition of the T -productis related to the way in which we perform the canonical quantization. From



the property of SF of being the Green function of the Dirac operator, wecan see that

SF (x) = −(i∂ +m)∆F (x) (7.27)

and

SF (x) =∫

d4k

(2π)4e−ikx k +m

k2 −m2 + iε=∫CF

d4k

(2π)4e−ikx k +m

k2 −m2. (7.28)

Let us consider the photon propagator. The only difference with theKlein-Gordon case is that the polarization vector gives an extra factor−gµν ,and therefore

〈0|T (Aµ(x)Aν(y)|0〉 = −igµν∫

d4k

(2π)4

e−ikx

k2 + iε. (7.29)

Defining

D(x) = −∫

d4k

(2π)4

e−ikx

k2 + iε, (7.30)

we get

〈0|T (Aµ(x)Aν(y)|0〉 = +igµνD(x− y). (7.31)

The choice of the integration paths in eq. (7.14) allows us to definevarious types of Green’s functions according to the boundary conditionswe require. For instance, consider the Klein-Gordon equation in a givenexternal source

( +m2)φ(x) = j(x). (7.32)

The solution can be given in terms of the Green’s function defined by

( +m2)G(x) = δ4(x). (7.33)

In fact, the solution can be written as

φ(x) = φ(0)(x) +∫

d4y G(x− y)j(y), (7.34)

where φ(0)(x) satisfies the homogeneous Klein-Gordon equation and it ischosen in such a way that φ(x) satisfies the boundary conditions of theproblem. For instance, if we give the function φ(x) at t = −∞ we require

limt→−∞

φ(x) = limt→−∞

φ(0)(x). (7.35)

Then, in order to satisfy the boundary conditions it is enough to choose forG(x) the retarded solution defined by

Gret(~x, x0 < 0) = 0. (7.36)



Im( )

Re( )

kk-

C

C

ret

adv

Fig. 7.7 The integration paths for Gret and Gadv.

Such a solution can be found easily by applying the method we have illus-trated in the previous Section. By choosing the integration path as in Fig.7.7, that is, leaving both poles below the path we get

Gret(x) = −∫Cret

d4k

(2π)4

e−ikx

k2 −m2. (7.37)

Clearly Gret(x) vanishes for x0 < 0. In fact, in this case, we can closethe path on the half-plane, Im ω > 0, without surrounding any singularity.Therefore the function vanishes. In analogous way we can define a functionGadv(x) vanishing for x0 > 0, by choosing a path below the poles (seeFig. 7.7). By integrating explicitly over ω one sees easily that the retardedsolution propagates forward in time both the positive and negative energysolutions, whereas the advanced one propagates both solutions backwardin time. Both Gret(x) and Gadv(x) can be connected to the invariant ∆functions defined previously. In particular, using Fig. 7.6, we can establishthe following relations between paths in the frequency complex plane

Cret = CF − C−. (7.38)

Then, from eqs (7.37) and (7.20)

Gret(x) = ∆F (x) +∫C−

d4

(2π)4

e−ikx

k2 −m2= ∆F (x) + i∆(−)(x), (7.39)

where we have used eq. (7.15). Then, from (7.21)

Gret(x) = iθ(x0)∆(+)(x)− iθ(−x0)∆(−)(x) + i∆(−)(x)

= iθ(x0)(

∆(+)(x) + ∆(−)(x))

= iθ(x0)∫

d3k

(2π)3

12ωk

(e−ikx − eikx

). (7.40)



In analogous way we find

Gadv = −iθ(−x0)∫

d3k

(2π)3

12ωk

(e−ikx − eikx

). (7.41)

By using the expression (7.6) for the Feynman propagator we see thatit propagates the positive energy solutions forward in time and the nega-tive energy ones backward in time. In fact, [Stueckelberg (1942); Feynman(1948c, 1949b,c)] showed that the backward propagation of the negative en-ergy solutions is equivalent to the forward propagation of the anti-particles.

7.2 A physical application of the propagators

The Feynman propagator acquires its full meaning only in quantum theorywhere, as we shall prove, it represents the central element of the perturba-tion theory.

In order to stress the relevance of the Feynman propagator we willconsider now a simple application of what discussed so far. Let us considertwo static point-like electric charges placed at ~x1 and ~x2. They can bedescribed by the charge density

j0(~x) =∑m=1,2

emδ3(~x− ~xm), (7.42)

where the em’s are the values of the electric charges. The current densityvanishes because we have supposed the charges to be static. Therefore theinteraction Hamiltonian is (see eq. (4.171))

Hint = −Lint =∫

d3x jµ(x)Aµ(x) =∑m=1,2

emA0(~xm, 0). (7.43)

In this equation we have taken the electromagnetic field operator A0 at thetime t = 0, since we will make use of the Schrodinger representation. So far,we have worked in the Heisenberg representation, since it is more convenientfrom the point of view of the relativistic covariance of the formalism, but theproblem we are interested here is the evaluation of the interaction energybetween two static electric charges. We recall that the relation between anoperator, AH(t) in the Heisenberg representation and the same operator inthe Schrodinger representation, AS , is

AH(t) = eiHtASe−iHt, (7.44)

Therefore the two operators are the same at t = 0. To evaluate the in-teraction energy we can do a perturbative calculation by determining the



energy shift induced by the interaction Hamiltonian. We will think to thestatic electric charges as classical objects. Therefore we will quantize onlythe photon field A0(~x, 0). Furthermore we will consider the effect of theperturbation on the vacuum state (the state without photons). Since

〈0|Hint|0〉 = 0, (7.45)

we must evaluate the energy shift at the second order in perturbation the-ory. We have

∆E =∑n

〈0|Hint|n〉〈n|Hint|0〉E0 − En

=∫

d3k

−ωk〈0|Hint|~k〉〈~k|Hint|0〉. (7.46)

The only state which contributes in the sum is the state with a singlephoton of energy ωk. By using the expression for Hint and the followingrepresentation for 1/ωk

1ωk

= limε→0+

i

∫ ∞0

dt e−i(ωk − iε)t, (7.47)

we can write ∆E in the form (notice that we have suppressed the limitbecause the resulting expression is regular for ε→ 0+):

∆E = −i∑m,n

emen

∫d3k

∫ ∞0

dt

×e−iωkt〈0|A0(~xn, 0)|~k〉〈~k|A0(~xm, 0)|0〉

= −i∑m,n

emen

∫d3k

∫ +∞

−∞dt θ(t)

×〈0|eiH0tA0(~xn, 0)e−iH0t|~k〉〈~k|A0(~xm, 0)|0〉

= −i∑m,n

emen

∫ +∞

−∞dt θ(t)〈0|A0(~xn, t)A0(~xm, 0)|0〉, (7.48)

where H0 is the free Hamiltonian for the electromagnetic field. To getthis expression we have used the completeness (recalling again that A0

couples only states that differ by a photon). Then, notice that the operatorsappearing in the last line can be interpreted as operators in the Heisenbergrepresentation. By writing explicitly the various terms in the sum we get

∆E = −i∫ +∞

−∞dtθ(t)

[e2

1〈0|A0(~x1, t)A0(~x1, 0)|0〉

+e1e2〈0|A0(~x1, t)A0(~x2, 0)|0〉+e1e2〈0|A0(~x2, t)A0(~x1, 0)|0〉

+e22〈0|A0(~x2, t)A0(~x2, 0)|0〉

]. (7.49)



The intermediate states description of these four contributions is given inFig. 7.8. Notice that the first and the fourth diagram describe a correctionto the intrinsic properties of the charges, and since we are interested inthe evaluation of the interaction energy we can omit these terms from ourcalculation. Then, the energy interaction ∆E12 is given by

Fig. 7.8 The graphical description of eq. (7.49).

∆E12 = −ie1e2

∫ +∞

−∞dt θ(t)

[〈0|A0(~x1, t)A0(~x2, 0)|0〉

+〈0|A0(~x2, t)A0(~x1, 0)|0〉]. (7.50)

Sending t→ −t in the second term, we get

∆E12 = −ie1e2

∫ +∞

−∞dt[θ(t)〈0|A0(~x1, t)A0(~x2, 0)|0〉

+θ(−t)〈0|A0(~x2,−t)A0(~x1, 0)|0〉]

(7.51)

and using H0|0〉 = 0,

〈0|A0(~x2,−t)A0(~x1, 0)|0〉 = 〈0|e−iH0tA0(~x2, 0)eiH0tA0(~x1, 0)e−iH0t|0〉= 〈0|A0(~x2, 0)A0(~x1, t)|0〉. (7.52)

Therefore our final result is

∆E12 = −ie1e2

∫ +∞

−∞dt 〈0|T (A0(~x1, t)A0(~x2, 0))|0〉. (7.53)

We see that the energy interaction is expressed in terms of the Feynmanpropagator. Recalling the eqs.(7.30) and (7.31) we get (here we put x0

1 = t

and x02 = 0)

∆E12 = −e1e2

∫ +∞

−∞dt

∫d4k

(2π)4

e−ik(x1 − x2)

k2 + iε

= e1e2

∫d3k

(2π)3

ei~k(~x1 − ~x2)

~k2=e1e2

4π1

|~x1 − ~x2|. (7.54)



For the final equality, see Exercise (4) in this Chapter. This result tellsus that the interaction energy between two classical charges evaluated atthe lowest order in quantum theory coincides with the Coulomb energy.Therefore we can expect quantum corrections from higher orders. We shallsee in the following that, in fact, this is the case.

7.3 Exercises

(1) Consider a Klein-Gordon field interacting with a classical source j(x)described by the following Lagrangian density

L =12∂µφ(x)∂µφ(x)− 1

2m2φ2(x) + j(x)φ(x). (7.55)

a) Find the solutions to the equations of motion in the limit x0 → ∞and show that in this limit the solution can be written in the form

φ(x) =∫

d3k√2π)3

1√2ωk

[(a(k) +

i√2ωk

j(k))e−ikx

+(a†(k)− i√

2ωkj(−k)

)eikx

], (7.56)

where

j(±k) =∫

d4y√(2π)3

e∓ikxj(x), k2 = m2. (7.57)

(Hint: Write φ(x) = φ(0) +∫Gret(x− y)j(y)d4y with φ(0) a free field)

b) Evaluate the v.e.v.’s of the Hamiltonian and of the number of par-ticles operator.

(2) Prove that the Feynman propagator of a massless scalar field is givenby

∆F (x) =1

4π2δ(x2)− i

4π2P 1x2, (7.58)

where P denotes the principal part.(3) Evaluate the Feynman propagator for a massive vector field and show

that

〈0|T (Vµ(x)Vν(y)|0〉 = −i∫

d4k

(2π)4e−ik(x−y)

(gµν − kµkν/m2

k2 −m2 + iε

).

(7.59)(4) Prove the equation (7.54). In order to perform the integration over ~k

the following integral is useful∫ ∞0

sin yy

dy =π

2. (7.60)



(5) Repeat the calculation made in Section 7.2 in the case of a fixed scalarcharge interacting with a massive scalar field, φ according to the Hamil-tonian

Hint =∫d3~xj(x)φ(x) =

∑m=1,2

gφ(~xm, 0). (7.61)

Show that the interaction energy between the two sources is given bythe Yukawa potential

∆E12 =g2

4π1

|~x1 − ~x2|e−m|~x|, (7.62)

with m the mass of the field.(6) Consider the Lagrangian density for the electromagnetic field with the

gauge fixing of eq. (5.45):

L = −14FµνF

µν − λ

2(∂µAµ)2. (7.63)

Evaluate the Feynman propagator.


Chapter 8

Perturbation theory

Relativistic quantum field theory does not offer simple calculable examplesand therefore one has to develop perturbative methods in order to dis-cuss interacting fields. In the following we will be interested in describingscattering processes, since they are the typical processes appearing in theexperiments in particle physics. The methods we will use for introducingthe perturbation theory are quite general but, for sake of simplicity and forthe particular physical interest, we will discuss in detail only the case ofthe electromagnetic interactions of charged spin 1/2 particles described bya Dirac field. The results we will obtain can be easily generalized to otherinteracting theories.

8.1 The electromagnetic interaction

As we have already discussed, the electromagnetic interaction of an ar-bitrary charged particle is obtained through the minimal substitution or,equivalently, by invoking the gauge principle

∂µ → ∂µ + ieAµ. (8.1)

For a charged Klein-Gordon particle, following this prescription, we get thefollowing Lagrangian density

Lfree = ∂µφ†∂µφ−m2φ†φ− 1

4FµνF

µν

→ [(∂µ + ieAµ)φ)]† [(∂µ + ieAµ)φ]−m2φ†φ− 14FµνF

µν . (8.2)

The interacting part is given by the following two terms

Lint. = −ie[φ†∂µφ− (∂µφ†)φ

]Aµ + e2A2φ†φ. (8.3)

177



In the first term the gauge field is coupled to the current

jµ = ie[φ†∂µφ− (∂µφ†)φ

], (8.4)

but another interacting term appears. This term is a straight consequenceof the gauge invariance. In fact, the current jµ, which is conserved in theabsence of the interaction, is neither conserved, nor gauge invariant, whenthe electromagnetic field is turned on. In fact, consider the infinitesimalgauge transformation

δφ(x) = −ieΛ(x)φ(x), δAµ(x) = ∂µΛ(x), (8.5)

then

δLfree = ieΛ,µφ†∂µφ− ie∂µφ†Λ,µφ = jµ∂µΛ (8.6)

and writing Lint in the form

Lint = −jµAµ + e2A2φ†φ, (8.7)

we find

δLint = −jµ∂µΛ− (δjµ)Aµ + 2e2Λ,µAµφ†φ. (8.8)

The first term cancels with the variation of Lfree, whereas the other twoterms cancel among themselves. In fact,

δjµ = ie[φ†(−ie)Λ,µφ− ieΛ,µφ†φ

]= 2e2Λ,µφ†φ. (8.9)

This shows that the A2 term is necessary to compensate the non invarianceof the current under gauge transformations. In fact, the conserved andgauge invariant current can be obtained through the Noether theorem

Jµ = ie[φ†(∂µ + ieAµ)φ− (∂µ − ieAµ)φ†)φ

]= jµ − 2e2Aµφ

†φ. (8.10)

The situation is much simpler in the case of the Dirac equation where

Lfree = ψ(i∂ −m)ψ − 14FµνF

µν → ψ(i∂ − eA−m)ψ − 14FµνF

µν , (8.11)

giving the interaction term

Lint = −eψγµψAµ. (8.12)

Here the gauge field is coupled to a conserved and gauge invariant current.As a consequence −jµAµ is the only interaction term. The gauge variationsof the free and of the interacting Lagrangian densities are respectively

δLfree = eψγµψΛ,µ = jµΛ,µ (8.13)


Perturbation theory 179

and

δLint = −jµδAµ = −jµΛ,µ. (8.14)

The canonical quantization for an interacting system follows the same pro-cedure as in the non interacting case. We require canonical commutationand/or anticommutation relations at equal times for the various fields. Fordifferent fields we require equal time vanishing commutation (anticommu-tation) relations for spin integer (half-integer) fields, whereas we requirezero commutation relations among fields of integer spin and fields of half-integer spin. Usually the canonical commutation relations among the fieldsare not changed by the interactions with respect to the free case. Howeverthis is not the case if the interaction term involves derivatives of the fields.This follows from the definition of the canonical momentum densities

Πi =∂L∂φi

=∂Lfree

∂φi+∂Lint

∂φi. (8.15)

For instance, for the charged scalar field we get

Π ≡ Πφ =∂L∂φ

= φ† − ieφ†A0, Π† = Πφ† = φ+ ieφA0. (8.16)

Since the canonical momenta contain the time component of the gaugefield, one can verify that the canonical commutators among the fields andtheir derivatives are changed by the interaction. Also the propagators aremodified. However we will not insist on this point, because in practice it hasno consequences on the perturbation theory (see later). When derivativeinteractions are not present, the canonical momentum densities coincidewith the free ones, and we get

H = Πφ− L = Πφ− Lfree − Lint = Hfree − Lint (8.17)

and therefore

Hint = −Lint. (8.18)

This is what happens for the electromagnetic interaction between a Diracfield and the electromagnetic one. The corresponding theory is called QED(Quantum Electro Dynamics). We recall also that in general the Hamil-tonian and the electromagnetic current are normal ordered in such a waythat the vacuum is an eigenstate of the energy and of the charge operatorswith vanishing eigenvalues. Therefore the interaction term is written as

Lint = −e : ψγµψ : Aµ. (8.19)



We can verify that this is equivalent to write

Lint = −e2

[ψ, γµψ]Aµ. (8.20)

For instance, if we consider the electric charge as defined indirectly in eq.(8.20), we get

Q =e

2

∫d3x(ψ†ψ − γ0ψψ

†γ0)

=e

2

∑±n

∫d3p[b†(p, n)b(p, n) + d(p, n)d†(p, n)

−b(p, n)b†(p, n)− d†(p, n)d(p, n)]

=e

2

∑±n

∫d3p[2b†(p, n)b(p, n)− 2d†(p, n)d(p, n)

−[b†(p, n), b(p, n)

]+

+[d(p, n), d†(p, n)

]+

]=: Q :, (8.21)

since the two anticommutators cancel among themselves.

8.2 The scattering matrix

The scattering processes are a central element in the study of the elemen-tary particles. In the typical scattering process the incoming particles areprepared in a state of definite momentum. These particles are sent eitheron a fixed target, or into another beam moving in the opposite direction.After the collision has taken place, one looks at the final states produced bythe scattering process. In ordinary quantum mechanics this situation is welldescribed by using free wave functions for the initial and final states. Thisdescription is certainly correct if one has to do with short-range potentials.However, in field theory, this representation is not really correct since, alsoin absence of reciprocal interactions, the particles have self interactions aswe have already observed in Section 7.2. For instance, a real electron can bethought of as being surrounded by a cloud of photons which can be emittedand reabsorbed. The point is that, in general, the number of particles isnot a conserved quantity. Particles can be created and annihilated and thiscan happen also outside the strict region of interaction. A rigorous treat-ment of these problems is non trivial and it goes beyond the scope of thiscourse. Therefore we will confine ourselves to a rather intuitive treatmentof the problem. On the other hand the limitations of the method will berather obvious so it may well give the basis for a more refined treatment.



To simplify the matter we will make use of the adiabatic hypothesis.This consists in looking at a scattering process in the following way. Attime t = −∞ we will suppose that our system can be described in terms offree particles, that is with the interaction turned off. Between t = −∞ anda time t = −T , much before the scattering process takes place, we let thecoupling describing the interaction to grow from zero to its actual value.In the interval −T < t < +T , the coupling stays constant, and then fromt = +T and t = +∞ the coupling goes again to zero. In practice, onedefines the interacting part of the Hamiltonian as

Hint(t, ε) = e−ε|t|Hint, (8.22)

performing all the calculations and taking the limit ε → 0+ at the end.The consistency of this procedure has been shown by various authors anda detailed discussion can be found, for instance, in the book by [Jauch andRohrlich (1980)].

By using the adiabatic hypothesis we can now discuss the perturbativecalculation of the scattering amplitudes. The perturbative expansion willbe possible only if the interaction term is small. For instance in QED theperturbative expansion gives rise to a series of powers in the fine structureconstant e2/4π ≈ 1/137. Therefore, if the coefficients of the expansiondo not grow too much, the expansion is justified. Let us start with theequation of motion for the states in the Schrodinger representation

i∂|ΦS(t)〉

∂t= HS |ΦS(t)〉. (8.23)

Suppose also that we have two interacting fields A and B. Then we canwrite

HS = H0S +HI

S , (8.24)

with

H0S = H0

S(A) +H0S(B) (8.25)

and

HIS ≡ HI

S(A,B), (8.26)

where H0S(A) and H0

S(B) are the free Hamiltonians for the fields A and B,and HI

S is the interaction Hamiltonian. It turns out convenient to introducea new representation for the vectors of state, the interaction representation.This is defined by the following unitary transformation upon the states andthe operators in the Schrodinger representation

|Φ(t)〉 = eiH0St|ΦS(t)〉, O(t) = eiH

0StOSe

−iH0St. (8.27)



Of course the matrix elements of any operator in the interaction represen-tation are the same as in the Schrodinger representation

〈Φ′(t)|O(t)|Φ(t)〉 = 〈Φ′S(t)|OS |ΦS(t)〉. (8.28)

We have also H0S = H0, where H0 is the free Hamiltonian in the interaction

representation. Notice also that the interaction representation coincideswith the Heisenberg representation when we switch off the interaction. Inthe interaction representation the time evolution of the states is dictatedby the interaction Hamiltonian

i∂|Φ(t)〉∂t

= −H0SeiH0

St|ΦS(t)〉+ eiH0St(H0

S +HIS)|ΦS(t)〉

= eiH0StHI

Se−iH0

St|Φ(t)〉, (8.29)

from which

i∂|Φ(t)〉∂t

= HI |Φ(t)〉, (8.30)

where HI is the interaction Hamiltonian in the interaction representation.On the other hand the operators evolve with the free Hamiltonian. There-fore, in the interaction representation they coincide with the Heisenbergoperators of the non-interacting case.

In order to describe a scattering process we will assign to the vector ofstate a condition at t = −∞

|Φ(−∞)〉 ≡ |Φi〉, (8.31)

where the state Φi will be specified by assigning the set of incoming freeparticles in terms of eigenstates of momentum, spin and other possiblequantum numbers. For instance, in QED we will have to specify how manyelectrons, positrons and photons are in the initial state and we will have tospecify their momenta, the spin projection of the fermions and the polar-ization of the photons. The equations of motion will tell us how this stateevolves in time and it will be possible to evaluate the state at t = +∞,where, ideally, we will detect the final states. In practice the preparationand the detection processes are made at some finite times. It follows thatour ideal description will be correct only if these times are much biggerthan the typical interaction time of the scattering process. Once we know|Φ(+∞)〉, we are interested to evaluate the probability amplitude of detect-ing at t = +∞ a given set of free particles (see the adiabatic hypothesis)specified by a vector of state |Φf 〉. Therefore the probability amplitude (orscattering amplitude) we are interested in is

Sfi = 〈Φf |Φ(+∞)〉. (8.32)



We will define the S matrix as the operator giving |Φ(+∞)〉 when appliedto |Φ(−∞)〉

|Φ(+∞)〉 = S|Φ(−∞)〉. (8.33)

The amplitude Sfi is then

Sfi = 〈Φf |S|Φi〉. (8.34)

We see that Sfi is the S matrix element between free states. To evaluatethe S matrix we first transform the Schrodinger equation in the interactionrepresentation into an integral equation

|Φ(t)〉 = |Φ(−∞)〉 − i∫ t

−∞dt1 H

I(t1)|Φ(t1)〉. (8.35)

The right hand side of this equation satisfies both the Schrodinger equa-tion and the boundary condition at t = −∞. The perturbative expansionconsists in evaluating |Φ(t)〉 by iterating this integral equation

|Φ(t)〉 = |Φ(−∞)〉

−i∫ t

−∞dt1 H

I(t1)[|Φ(−∞)〉

−i∫ t1

−∞dt2 H

I(t2)|Φ(t2)〉]. (8.36)

Continuing the iteration we get

|Φ(t)〉 =[1− i

∫ t

−∞dt1 H

I(t1) + (−i)2

∫ t

−∞dt1

∫ t1

−∞dt2 H

I(t1)HI(t2)

+ · · ·]|Φ(−∞)〉. (8.37)

Of course this is meaningful only if the expansion is convergent. By takingthe limit for t = +∞ we get the perturbative expansion of the S matrix

S = 1 +∞∑n=1

(−i)n∫ +∞

−∞dt1

∫ t1

−∞dt2 · · ·

∫ tn−1

−∞dtn

×[HI(t1)HI(t2) · · ·HI(tn)

]. (8.38)

We can rewrite this expression in terms of T -products

S = 1 +∞∑n=1

(−i)n

n!

∫ +∞

−∞dt1 · · ·

∫ +∞

−∞dtn T

(HI(t1) · · ·HI(tn)

). (8.39)

The T -product of n terms means that the factors have to be written from theleft to the right with decreasing times. For instance, if t1 ≥ t2 ≥ · · · ≥ tn,then

T(HI(t1) · · ·HI(tn)

)= HI(t1) · · ·HI(tn). (8.40)



The equality of the two expressions (8.38) and (8.39) holds term by term.As an example, consider n = 2 in eq. (8.39) for finite times t1 and t2. Thisterm can be written as

A =∫ t2

t1

∫ t2

t1

dt ds T(HI(t)HI(s)

)=∫ t2

t1

dt HI(t)(∫ t

t1

ds HI(s))

+∫ t2

t1

dt

(∫ t2

t

ds HI(s))HI(t). (8.41)

By looking at Figs. 8.1 and 8.2 one sees easily that exchanging the inte-grations on s and t one gets∫ t2

t1

dt

∫ t2

t

ds =∫ t2

t1

ds

∫ s

t1

dt. (8.42)

t1 t2 t

t

t2

Fig. 8.1 The figure represents schematically the integralR t2t1

dtR t2t ds.

Therefore

A =∫ t2

t1

dt

∫ t

t1

ds HI(t)HI(s) +∫ t2

t1

ds

∫ s

t1

dt HI(s)HI(t) (8.43)

and exchanging s↔ t in the second integral,

A = 2∫ t2

t1

dt

∫ t

t1

ds HI(t)HI(s). (8.44)

By letting t2 → +∞ and t1 → −∞, we prove the equality of the n = 2terms in (8.38) and (8.39).



t1 t2 t

t1 s

Fig. 8.2 The figure represents schematically the integralR t2t1

dsR st1

dt.

The result for the nth term in the series can be obtained in a completelyanalogous way.

Since the S matrix connects the set of free states at t = −∞ with aset of free states at t = +∞, it should represent simply a change of basisand as such it should be unitary. From this point of view the unitarityproperty of the S matrix is very important because it has to do with thefundamental properties of quantum mechanics. So we have to check thatat least formally1 the expression (8.39) represents a unitary operator. Inorder to prove this point we rewrite S in the form

S = T(e−i∫ +∞

−∞dt HI(t))

. (8.45)

This expression is a symbolic one and it is really defined by its series ex-pansion

S =∞∑n=0

(−i)n

n!T

(∫ +∞

−∞dt HI(t)

)n=∞∑n=0

(−i)n

n!

∫ +∞

−∞dt1 · · · dtnT

(HI(t1) · · ·HI(tn)

). (8.46)

The motivation for introducing the T -ordered exponential is that it satisfiesthe following factorization property

T(e

∫ t3

t1

O(t)dt)= T

(e

∫ t3

t2

O(t)dt)T(e

∫ t2

t1

O(t)dt). (8.47)

1This observation refers to the fact that we don’t really know if the series which we

have found for the S matrix is convergent or not.



To prove this relation we first consider the following expression (t1 ≤ t2 ≤t3)

T(∫ t3

t1

O(t)dt)n≡∫ t3

t1

· · ·∫ t3

t1

ds1 · · · dsn T (O(s1) · · ·O(sn))

=(∫ t3

t2

+∫ t2

t1

)(∫ t3

t2

+∫ t2

t1

)· · ·(∫ t3

t2

+∫ t2

t1

)ds1 · · · dsnT (O(s1) · · ·O(sn))

=n∑k=0

n!(n− k)!k!

∫ t3

t2

· · ·∫ t3

t2

ds1 · · · dsn−k

×∫ t2

t1

· · ·∫ t2

t1

dz1 · · · dzk T (O(s1) · · ·O(sn−k)O(z1) · · ·O(zk))

=n∑k=0

n!(n− k)!k!

∫ t3

t2

· · ·∫ t3

t2

ds1 · · · dsn−kT (O(s1) · · ·O(sn−k))

×∫ t2

t1

· · ·∫ t2

t1

dz1 · · · dzk T (O(z1) · · ·O(zk)) . (8.48)

In the last term we have used the fact that all the times zi are smaller thanthe times si. Therefore we have shown the relation

T

(∫ t3

t1

dt O(t))n

=n∑k=0

n!(n− k)!k!

×T(∫ t3

t2

dt O(t))n−k

T

(∫ t2

t1

dt O(t))k

.(8.49)

The factorization property (8.47) follows immediately if we remember thatthe analogous property for the ordinary exponential,

ea+ b = eaeb, (8.50)

follows from the binomial expansion

ea+ b =∞∑n=0

1n!

(a+ b)n =∞∑n=0

1n!

n∑k=0

n!(n− k)!k!

an−kbk, (8.51)

using∞∑n=0

n∑k=0

=∞∑k=0

∞∑n=k

(8.52)

and replacing k with h = n−k in eq (8.51). Since eq. (8.49) generalizes thebinomial formula to T -products of powers of time integrals of operators, by



the same token we get the formula (8.47). With this property we can nowprove the unitarity of any operator of the form

U = T(e−i∫ tf

ti

dt O(t)), (8.53)

with O(t)a hermitian operator. To this end let us divide the time interval(ti, tf ) in N infinitesimal intervals ∆t with

ti ≡ t1 ≤ t2 ≤ · · · ≤ tN = tf , (8.54)

then we can write

U = limN→∞

e−i∆tO(tN )e−i∆tO(tN−1) · · · e−i∆tO(t1), (8.55)

from which

U† = limN→∞

e+i∆tO(t1)e+i∆tO(t2) · · · e+i∆tO(tN ) (8.56)

and the unitarity follows immediately.Let us notice that if there are no derivative interactions we have

S = T(e−i∫ +∞

−∞dt HI(t))

= T(e+i∫

d4x Lint). (8.57)

From this expression we can prove that the Lorentz invariance of the theoryimplies the invariance of the S matrix. Since in a relativistic theory thelagrangian is Lorentz invariant, the previous statement would be trivial butfor the presence of the T -product in the expression of the S matrix. Forsimplicity, let us consider the second order contribution (but the argumentis valid at any order)∫

d4x1 d4x2

[θ(x0

1 − x02)Lint(x1)Lint(x2) + θ(x0

2 − x01)Lint(x2)Lint(x1)

].

(8.58)The T -product is Lorentz invariant for time-like separations of the twooperators (remember that a proper Lorentz transformation cannot changethe sign of the time component of a four-vector). On the other hand weintegrate over x1 and x2 and the separation between the two points canbe either space-like or time-like. In the case of space-like separations, andassuming that the Lagrangian density is a local function of the fields, itfollows that

[Lint(x1),Lint(x2)] = 0 for (x1 − x2)2 < 0 (8.59)



and therefore for (x1 − x2)2 < 0 we get

T (Lint(x1)Lint(x2)) = Lint(x1)Lint(x2). (8.60)

This shows that the T -product of local invariant Lorentz operators isLorentz invariant. We saw an example of this property when we haveevaluated the propagator for a scalar field.

One could think that for theories with derivative interactions theLorentz invariance is lost. However, it is possible to show that also inthese theories the S matrix is given by the expression on the right handside of eq. (8.57) (see, for instance, [Itzykson and Zuber (1980)]), that is itdepends on Lint.

8.3 Wick’s theorem

In the previous Section we have shown that the S matrix can be evalu-ated in terms of matrix elements of T -products. As we shall see in theapplications, the matrix elements of the S matrix between free particlestates can in turn be expressed as vacuum expectation values (VEV’s) ofT -products. These VEV’s satisfy an important theorem due to Wick thatstates that the VEV of a T -product of an arbitrary number of free fields(the ones that appear in the interaction representation) can be expressedas combinations of VEV’s of T -products among two fields, that is in termsof Feynman propagators. In order to prove the theorem we will use thetechnique of generating functionals. That is we will start by proving thefollowing identity

T(

exp(−i∫

d4x j(x)φ(x)))

=: exp (−i∫

d4x j(x)φ(x)) :

× exp (−12

∫d4x d4y j(x)j(y)〈0|T (φ(x)φ(y))|0〉), (8.61)

where φ(x) is a free real scalar field and j(x) an ordinary real function.This formula can be easily extended to charged scalar, fermion and photonfields. The Wick theorem is then obtained by expanding both sides of thisequation in powers of j(x) and taking the VEV of both sides. Let us startby expanding the left hand side of the (8.61), by using the factorizationproperty in eq. (8.47). Let us also define

O(t) =∫

d3x j(x)φ(x) (8.62)



and notice that for free fields [O(t), O(t′)] is just an ordinary number (calleda c-number, to be contrasted with operators which are called q-numbers).Dividing again the interval (ti, tf ) in N pieces of amplitude ∆t as in theprevious Section, we get

T(

exp (−i∫ tf

ti

dt O(t)) = limN→∞

exp (−i∆tO(tN )) exp (−i∆tO(tN−1))

· · · × exp (−i∆tO(t1)) (8.63)and using

exp (A) exp (B) = exp (A+B) exp (+12

[A,B]), (8.64)

valid if [A,B] commutes with A and B, we get

T(

exp (−i∫ tf

ti

dt O(t)))

= limN→∞

N∏i=3

exp (−i∆tO(ti))

× exp (−i∆t(O(t2) +O(t1))) exp (−12

∆t2[O(t2), O(t1)])

= limN→∞

N∏i=4

exp (−i∆tO(ti)) exp (−i∆t(O(t3) +O(t2) +O(t1)))

× exp (−12

∆t2[O(t3), O(t2) +O(t1)]− 12

∆t2[O(t2), O(t1)])

= limN→∞

exp (−i∆tN∑i=1

O(ti)) exp (−12

∆t2∑

1≤i≤j≤N

[O(tj), O(ti)])

= exp(−i∫ tf

ti

dt O(t))

× exp (−12

∫ tf

ti

dt1 dt2 θ(t1 − t2)[O(t1), O(t2)]), (8.65)

from which

T(

exp (−i∫

d4x j(x)φ(x)))

= exp (−i∫

d4x j(x)φ(x))

× exp (−12

∫d4x d4y j(x)j(y)θ(x0 − y0)[φ(x), φ(y)]). (8.66)

The next step is to expand the first exponential on the right hand side ofthis equation in normal products. We have

exp (−i∫

d4x j(x)φ(x)) = exp (−i∫

d4x j(x)(φ(+)(x) + φ(−)(x)))

= exp (−i∫

d4x j(x)φ(−)(x)) exp (−i∫

d4x j(x)φ(+)(x))

× exp (+12

∫d4x d4y j(x)j(y)[φ(−)(x), φ(+)(y)]), (8.67)



where we have used again the equation (8.64). Therefore

exp (−i∫

d4x j(x)φ(x)) =: exp (−i∫

d4x j(x)φ(x)) :

× exp (+12

∫d4x d4y j(x)j(y)[(φ(−)(x), φ(+)(y)]). (8.68)

Substituting in eq. (8.66)

T(

exp (−i∫

d4x j(x)φ(x)))

=: exp (−i∫

d4x j(x)φ(x)) :

× exp (+12

∫d4x d4y j(x)j(y)A(x, y)), (8.69)

where

A(x, y) = [φ(−)(x), φ(+)(y)]− θ(x0 − y0)[φ(x), φ(y)] (8.70)

This is a c-number because it contains commutators of free fields, andtherefore it can be evaluated by taking its VEV

A(x, y) = 〈0|[φ(−)(x), φ(+)(y)]− θ(x0 − y0)[φ(x), φ(y)]|0〉. (8.71)

Since φ(+)|0〉 = 0, we can write

A(x, y) = −〈0| [φ(y)φ(x) + θ(x0 − y0)φ(x)φ(y)− θ(x0 − y0)φ(y)φ(x)] |0〉= −〈0| [θ(y0 − x0)φ(y)φ(x) + θ(x0 − y0)φ(x)φ(y)| 0〉= −〈0|T (φ(x)φ(y))|0〉. (8.72)

This proves our identity (8.61). Let us now expand both sides of thisidentity in a series of j(x) and compare term by term. We will use thesimplified notation φi ≡ φ(xi). We get

T (φ) = : φ :, (8.73)

T (φ1φ2) = : φ1φ2 : +〈0|T (φ1φ2)|0〉, (8.74)

T (φ1φ2φ3) = : φ1φ2φ3 : +3∑

i6=j 6=k=1

: φi : 〈0|T (φjφk)|0〉, (8.75)

T (φ1φ2φ3φ4) = : φ1φ2φ3φ4 : +4∑

i 6=j 6=k 6=l=1

[: φiφj : 〈0|T (φkφl)|0〉

+〈0|T (φiφj)|0〉〈0|T (φkφl)|0〉]

(8.76)



and analogous expressions for higher order monomials in the scalar field.By taking the VEV of these expressions, and recalling that the VEV of anormal product is zero, we get the Wick theorem. The T -product of twofield operators is sometimes called the contraction of the two operators.Therefore to evaluate the VEV of a T -product of an arbitrary number of freefields, it is sufficient to consider all the possible contractions of the fieldsappearing in the T -product. For instance, from the last of the previousrelations we get

〈0|T (φ1φ2φ3φ4)|0〉 =4∑

i 6=j 6=k 6=l=1

〈0|T (φiφj)|0〉〈0|T (φkφl)|0〉. (8.77)

An analogous theorem holds for the photon field. For the fermions one hasto remember that the T -product is defined in a slightly different way. Thisgives a minus sign any time we have an odd permutation of the fermionicfields with respect to the original ordering. As an illustration the previousformula, in the case of fermions, becomes

〈0|T (ψ1ψ2ψ3ψ4)|0〉 =4∑

i 6=j 6=k 6=l=1

σP 〈0|T (ψiψj)|0〉〈0|T (ψkψl)|0〉, (8.78)

where σP = ±1 is the sign of the permutation (i, j, k, l) with respect to thefundamental one (1, 2, 3, 4) appearing on the left hand side. More explicitly

〈0|T (ψ1ψ2ψ3ψ4)|0〉 = +〈0|T (ψ1ψ2)|0〉〈0|T (ψ3ψ4)|0〉−〈0|T (ψ1ψ3)|0〉〈0|T (ψ2ψ4)|0〉+〈0|T (ψ1ψ4)|0〉〈0|T (ψ2ψ3)|0〉. (8.79)

8.4 Evaluation of the S matrix at second order in QED

In the case of QED the S matrix is given by

S = 1 +∞∑n=1

(+i)n

n!

∫· · ·∫

d4x1 · · · d4xnT (LI(x1) · · · LI(xn)) , (8.80)

with (see eq. (8.19))

LI = −e : ψAψ : . (8.81)

We have now to understand how to use the Wick theorem in the actualsituation with normal ordered operators inside the T -product. Consider,



for simplicity, two scalar fields. From eq. (8.74) we get (at equal times theT -product and the usual product coincides)

φa(x)φb(x) =: φa(x)φb(x) : +〈0|T (φa(x)φb(x))|0〉, (8.82)

from which

: φa(x)φb(x) := φa(x)φb(x)− 〈0|T (φa(x)φb(x))|0〉. (8.83)

Therefore

T (: φa(x)φb(x) : φ1(x1) · · ·φn(xn)) = T (φa(x)φb(x)φ1(x1) · · ·φn(xn))

−〈0|T (φa(x)φb(x))|0〉T (φ1(x1) · · ·φn(xn)). (8.84)

Since the second term subtracts the contraction between the two operatorstaken at the same point, we can generalize the Wick expansion by sayingthat, when normal products are contained inside a T -product, the Wickexpansion applies with the further rule that the contractions of operatorsat the same point, inside the normal product, vanish. With this conventionwe can write

T (: A(x1)B(x1) · · · : · · · : A(xn)B(xn) · · · :)= T (A(x1)B(x1) · · ·A(xn)B(xn) · · · ) . (8.85)

In the case of QED one can get convinced more easily by recalling that (seeeqs. (8.19) and (8.20))

: ψγµψ :=12

[ψ, γµψ] (8.86)

and noticing that, inside a T -product, the fields can be freely commutedexcept for taking into account their statistics. Therefore

T (: ψγµψ :) =12T ([ψ, γµψ]) = T (ψγµψ). (8.87)

For the following analysis it is useful to remember how the various fieldoperators act on the kets

ψ+ annihilates e−, ψ− creates e+,

ψ+ annihilates e+, ψ− creates e−,

A+ annihilates γ, A− creates γ. (8.88)

Decomposing LI in positive and negative frequency components2

LI = −e : (ψ+ + ψ−)(A+µ +A−µ )γµ(ψ+ + ψ−) :, (8.89)

2Remember that e is the electric charge of the particle with its own sign. Therefore,

for the electron, e < 0.



we get 8 terms with non vanishing matrix elements. For instance,

: ψ+α A−αβψ

−β := −ψ−β A

−αβψ

+α (8.90)

has the following non vanishing matrix element

〈e+γ|ψ−A−ψ+|e+〉. (8.91)

This process corresponds to a positron emitting a photon. This and theother seven processes, described by the S matrix at first order in perturba-tion theory

S(1) = −ie∫

d4x : ψ(x)A(x)ψ(x) : . (8.92)

are represented by the diagrams of Fig. 8.3.

e+

e-e+ e+

e+

e-e+ e+

e+

e-e- e-

e+

e-e- e-

Fig. 8.3 Diagrams for the processes described by the S matrix at first order.

However none of these contributions corresponds to a physically possibleprocess since the four-momentum is not conserved. We will show later thatthe conservation of the four-momentum is a consequence of the theory. Forthe moment we will assume it and we will show that for real particles (thatis for particles on mass shell, p2 = m2) these processes cannot happen. Forinstance, consider

e−(p)→ e−(p′) + γ(k), (8.93)

If the four-momentum is conserved

p′ = p− k, (8.94)



from which

m2 = m2 − 2p · k, (8.95)

where we have used k2 = 0 for the photon. In the rest frame of the initialelectron we get mk0 = 0. Therefore the process is possible only for aphoton with vanishing four-momentum. Let us now consider the 2nd ordercontribution

S(2) =(−ie)2

2!

∫d4x1 d

4x2 T(ψ(x1)A(x1)ψ(x1)ψ(x2)A(x2)ψ(x2)

).

(8.96)We can expand S(2) using Wick’s theorem, and classifying the various con-tributions according to the number of contractions. If we associate to acontraction of two fields at the points x1 and x2 a line, we see that theterms originating from S(2) can be obtained by connecting among themthe diagrams depicted in Fig. 8.3 in all possible ways. In this case the onlynon vanishing contractions are the ones between ψ and ψ, and between Aµand Aν . Recalling from Section 7.1

〈0|T (ψα(x)ψβ(y))|0〉 = iSF (x− y)αβ〈0|T (Aµ(x)Aν(y))|0〉 = igµνD(x− y), (8.97)

we get

S(2) =6∑i=1

S(2)i , (8.98)

where

S(2)1 =

(−ie)2

2!

∫d4x1 d

4x2 : ψ(x1)A(x1)ψ(x1)ψ(x2)A(x2)ψ(x2) :, (8.99)

S(2)2 =

(−ie)2

2!

∫d4x1 d

4x2 : ψ(x1)A(x1)iSF (x1 − x2)A(x2)ψ(x2) :

+(−ie)2

2!

∫d4x1 d

4x2 : ψ(x2)A(x2)iSF (x2 − x1)A(x1)ψ(x1) :

= (−ie)2

∫d4x1 d

4x2 :

× : ψ(x1)A(x1)iSF (x1 − x2)A(x2)ψ(x2) :, (8.100)

S(2)3 =

(−ie)2

2!

∫d4x1 d

4x2 : ψ(x1)γµψ(x1)igµνD(x1−x2)ψ(x2)γνψ(x2) :,

(8.101)



S(2)4 =

(−ie)2

2!

∫d4x1 d

4x2

× : ψ(x1)γµiSF (x1 − x2)igµνD(x1 − x2)γνψ(x2) :

+(−ie)2

2!

∫d4x1 d

4x2 : ψ(x2)γµiSF (x2 − x1)igµνD(x2 − x1)γνψ(x1) :

= (−ie)2

∫d4x1 d

4x2

× : ψ(x1)γµiSF (x1 − x2)igµνD(x1 − x2)γνψ(x2) :, (8.102)

S(2)5 =

(−ie)2

2!

∫d4x1 d

4x2(−1)

× : Tr[iSF (x1 − x2)A(x2)iSF (x2 − x1)A(x1)] :, (8.103)

S(2)6 =

(−ie)2

2!

∫d4x1 d

4x2(−1)

×Tr[iSF (x1 − x2)γµiSF (x2 − x1)γνigµνD(x2 − x1)]. (8.104)

The term S(2)1 is nothing but the product of two processes of type S(1)

and it does not give rise to real processes. The term S(2)2 is obtained by

contracting two fermionic fields, meaning that we are connecting a fermionline with two of the vertices of Fig. 8.3. The possible external particles aretwo γ, two e−, two e+, or a pair e+e−. Selecting the external states wecan get different physical processes. One of these processes is the Comptonscattering γ + e− → γ + e−. In this case we must select in S

(2)2 , ψ+(x2)

to destroy the initial electron and ψ−(x1) to create the final electron. Asfor the photons are concerned, since Aµ is a real field, we can destroy theinitial photon both in x2 and x1 and create the final photon in the otherpoint. Therefore we get two contributions

S(2)2 (γe− → γe−) = Sa + Sb, (8.105)

with

Sa = (−ie)2

∫d4x1 d

4x2 ψ−(x1)γµiSF (x1 − x2)γνA−µ (x1)A+

ν (x2)ψ+(x2)

(8.106)and

Sb = (−ie)2

∫d4x1 d

4x2 ψ−(x1)γµiSF (x1 − x2)γνA−ν (x2)A+

µ (x1)ψ+(x2).

(8.107)



e-x2 x1

e-

Sa

e-x2 x1

e

Sb

Fig. 8.4 Diagrams for the Compton scattering.

The corresponding diagrams are given in Fig. 8.4.The terms corresponding to the Compton scattering for a positron are

obtained from the previous ones by substituting ψ+ (annihilates an elec-tron) with ψ− (creates a positron) and ψ− (creates an electron) with ψ+

(annihilates a positron). The other two processes coming from S(2)2 are

2γ → e+e− (pair creation) and e+e− → 2γ (pair annihilation). The S

matrix element for the pair creation is given by

S(2)2 (2γ → e+e−) = (−ie)2

∫d4x1 d

4x2

×ψ−(x1)γµiSF (x1 − x2)γνA+µ (x1)A+

ν (x2)ψ−(x2). (8.108)

Notice that in evaluating

A+µ (x1)A+

ν (x2)|γ(k1)γ(k2)〉, (8.109)

we get two contributions since one of the fields A+µ can annihilate any one

of the two external photons. The diagrams for these two contributions aregiven in Fig. 8.5.

e-

e+ e+

e-x1

x2

x1

x2

Fig. 8.5 Diagrams for the pair creation.



We have analogous contributions and diagrams for the pair annihilationprocess.

The next processes we consider are the ones generated by S(2)3 in which

we have contracted the photon fields. They are: electron scattering e−e− →e−e−, positron scattering e+e+ → e+e+ and electron-positron scatteringe+e− → e+e−. For the electron scattering we have

S(2)3 (2e− → 2e−) =

(−ie)2

2!

∫d4x1 d

4x2

× : ψ−(x1)γµψ+(x1)igµνD(x1 − x2)ψ−(x2)γνψ+(x2) : . (8.110)

The term

ψ+(x1)ψ+(x2)|e−(p1)e−(p2)〉 (8.111)

gives rise to two contributions, and other two come from the final state.The corresponding diagrams are given in Fig. 8.6.

e- e-

e- e-

a)

e-

e-e-

e-

b)

e-

e-e-

e-

c)

e- e-

e-e-

d)

Fig. 8.6 Diagrams for electron scattering.

The terms a) and d) differ only for the exchange x1 ↔ x2 and thereforethey are equal after having exchanged the integration variables. The same



is true for the terms b) and c). In this way we get a factor 2 which cancelsthe 2! in the denominator. This is the same kind of cancellation we haveseen for the term S

(2)2 . At the order n on has n! equivalent diagrams which

cancel the factor n! coming from the expansion of the S matrix. This meansthat it is enough to draw all the inequivalent diagrams. For the electronscattering we have two such diagrams differing for a minus sign due to theexchange of the fermion lines. This happens because field theory takesautomatically into account the statistics of the particles.

For the process e+e− → e+e− we get the diagrams of Fig. 8.7. Theinequivalent diagrams are those of Fig. 8.8 and correspond to the followingcontributions

S(2)3 (e+e− → e+e−) = Sa(e+e− → e+e−) + Sb(e+e− → e+e−), (8.112)

with

Sa(e+e− → e+e−) = (−ie)2

∫d4x1 d

4x2

× : ψ−(x1)γµψ+(x1)igµνD(x1 − x2)ψ+(x2)γνψ−(x2) : (8.113)

and

Sb(e+e− → e+e−) = (−ie)2

∫d4x1 d

4x2

× : ψ−(x1)γµψ−(x1)igµνD(x1 − x2)ψ+(x2)γνψ+(x2) : . (8.114)

The term S(2)4 gives rise to the two possibilities e− → e− and e+ →

e+. These are not scattering processes but contributions to the intrinsicproperties of the electrons and positrons, in particular to their mass. Theseare called self-energy contributions. For the electron we have the diagramof Fig. 8.9 with a contribution given by

S(2)4 = (−ie)2

∫d4x1 d

4x2 ψ−(x1)γµiSF (x1−x2)igµνD(x1−x2)γνψ+(x2).

(8.115)In analogous way the term S

(5)2 contributes to the self-energy of the photon.

However, as we shall show in the following, the vanishing of the mass of thephoton is not changed by this correction due to the gauge invariance of thetheory. We get the two equivalent diagrams of Fig. 8.10 corresponding to

S(2)5 = (−ie)2

∫d4x1 d

4x2(−1)Tr[iSF (x1 − x2)γµiSF (x2 − x1)γν ]

×A−µ (x1)A+ν (x2). (8.116)

The minus sign arises because in a fermionic loop we have to exchange twofermion fields inside the T -product. The last term is the one where all the



e- e-

e+ e+

a)

e+ e+

e-e-

b)

e+

e- e-

e+

c)

e+

e- e-

e+

d)

Fig. 8.7 Diagrams for the process e+e− → e+e−.

fields are contracted. There are no external particles and the correspondingdiagram of Fig. 8.11 is called a vacuum diagram. We can ignore it becauseit is possible to show that it contributes to a phase factor for the vacuumstate.

e- e-

e+ e+

Sa

x2

x1

e+

e- e-

e+

Sb

x2 x1

Fig. 8.8 Inequivalent diagrams for the process e+e− → e+e−.



e- e-

x2 x1

Fig. 8.9 Electron self-energy.

x2 x1 x2 x1

Fig. 8.10 Photon self-energy.

x2 x1

Fig. 8.11 Vacuum diagram.

8.5 Feynman diagrams in momentum space

As we have already discussed, in a typical experiment in particle physicsone or two beams of particles with definite momenta are prepared. Fur-thermore, the momenta of the final states are measured. For this reason itis convenient to work in momentum space. Let us recall the expressions forthe fermion and the photon propagators (see eqs. (7.28) and (7.30))

SF (x) =1

(2π)4

∫d4p e−ipxSF (p) (8.117)

and

D(x) =1

(2π)4

∫d4p e−ipxD(p), (8.118)



where

SF (p) =1

p−m+ iε(8.119)

and

D(p) = − 1p2 + iε

. (8.120)

From the expansion of the fields in terms of creation and annihilation op-erators, one can evaluate the action of the positive frequency part of ψ, ψand Aµ on the one particle states. We get

ψ+(x)|e−(p, r)〉 =∑±s

∫d3k

√m

(2π)3Eke−ikxu(k, s)b(k, s)b†(p, r)|0〉,

(8.121)where the indices r, s stands for the polarization of the fermions. Then,using [

b(k, s), b†(p, r)]+

= δrsδ3(~p− ~k), (8.122)

it follows

ψ+(x)|e−(p, r)〉 =√

m

(2π)3Eku(p, r)e−ipx|0〉. (8.123)

For later applications it is more convenient to use the normalization in abox than the continuous one. This amounts to the substitution

1√(2π)3

→ 1√V

(8.124)

and to the use of discrete momenta ~p = (2π/L)3~n. We get

ψ+(x)|e−(p, r)〉 =√

m

V Epu(p, r)e−ipx|0〉, (8.125)

ψ+(x)|e+(p, r)〉 =√

m

V Epv(p, r)e−ipx|0〉 (8.126)

and for the photon

A+µ (x)|γ(k, λ)〉 =

√1

2V Ekε(λ)µ (k)e−ikx|0〉. (8.127)

By conjugating these expressions we obtain the action of the negative fre-quency operators on the bra vectors.

As an example let us consider a process associated to S(1)

|i〉 = |e−(p)〉 → |f〉 = |e−(p′), γ(k)〉. (8.128)



From (8.92) we get3

〈f |S(1)|i〉 = −ie∫

d4x 〈e−(p′), γ(k)|ψ−(x)A−µ (x)γµψ+(x)|e−(p)〉

= −ie∫

d4x

(√m

V Ep′u(p′)eip

′x)

×(√

12V Ek

ε∗(k)eikx)(√

m

V Epu(p)e−ipx

)=

−iem√V 3EpEp′2Ek

(2π)4δ4(p′ + k − p)u(p′)ε∗(k)u(p). (8.129)

This expression can be written as

〈f |S(1)|i〉 = (2π)4δ4(p′ + k − p)√

m

V Ep

√m

V Ep′

√1

2V EkM, (8.130)

where the quantity

M = −ieu(p′)ε∗(k)u(p) (8.131)

is called the Feynman, or the invariant amplitude for the process. Noticethat M is a Lorentz invariant quantity. The term (2π)4δ4(p′ + k − p)gives the conservation of the four-momentum in the process, whereas theother factors are associated to the various external particles (incoming andoutgoing). As we said previously this process is not physically possiblesince it does not respect the four-momentum conservation.

This structure is quite general and for any process a delta-function ex-pressing the four-momentum conservation will appear. Furthermore therewill be factors as

√m/V Ep for each external fermion and

√1/2V Ek for

each boson. Now we have to investigate the rules for evaluating the in-variant amplitude M. To this end let us start considering the Comptonscattering

|i〉 = |e−(p), γ(k)〉 → |f〉 = |e−(p′), γ(k′)〉. (8.132)

The S matrix element for the Compton scattering is given in eqs. (8.105),(8.106) and (8.107)

S(2)2 (γe− → γe−) = Sa + Sb (8.133)

3Notice that the polarization vectors of the photons can be real, as for linear polariza-

tion, or complex, as for circular polarization.



and

〈f |Sa|i〉 = (−ie)2

∫d4x1 d

4x2

√m

V Ep

√m

V Ep′

√1

2V Ek

√1

2V Ek′

×u(p′)eip′x1 ε∗(k′)eik

′x1i

(2π)4

∫d4qe−iq(x1 − x2)SF (q)

×ε(k)e−ikx2u(p)e−ipx2

= (−ie)2

∫d4x1d

4x2

∫d4q

(2π)4

√m

V Ep

√m

V Ep′

√1

2V Ek

√1

2V Ek′

×ei(p′ + k′ − q)x1e−i(p+ k − q)x2ε∗µ(k′)εν(k)

×u(p′)γµi

q −m+ iεγνu(p)

= (2π)4δ4(p′ + k′ − p− k)√

m

V Ep

√m

V Ep′

√1

2V Ek

√1

2V Ek′

×ε∗µ(k′)εν(k)u(p′)(−ieγµ)i

p+ k −m+ iε(−ieγν)u(p). (8.134)

Also in this case we can write

〈f |Sa|i〉 = (2π)4δ4(p′ + k′ − p− k)√

m

V Ep

√m

V Ep′

√1

2V Ek

√1

2V Ek′Ma,

(8.135)with

Ma = ε∗µ(k′)εν(k)u(p′)(−ieγµ)i

q −m+ iε(−ieγν)u(p), q = p+ k.

(8.136)We will associate to this expression the diagram in Fig. 8.12.

p

k

q=p+k

k’

p’

Fig. 8.12 Contribution to the Compton scattering.

The four-momentum q is determined by the conservation of the four-momentum at the vertices: q = p + k = p′ + k′. However, notice that in



general q2 = m2 + 2p · k 6= m2, that is the exchanged particle (describedby the propagator) is not a real particle but a virtual one. Looking at theprevious expression one understands immediately how the various piecesare connected to the graphical elements in the diagram. In fact, we havethe following rules for any given diagram (Feynman diagram):

• write the amplitude starting from the outgoing particles,• for each vertex there is a factor −ieγµ,• for each internal fermion line there is a factor iSF (p) (fermion propa-

gator),• for each ingoing and/or outgoing fermion line there is a factor u(p)

and/or u(p),• for each ingoing and/or outgoing photon line there is a factor ελµ (or its

complex conjugate for outgoing photons if we consider complex polar-ization as the circular one).

Notice also that the spinorial factors start from the final states andend up with the initial ones. The further contribution to the Comptonscattering (given by Sb) corresponds to the diagram in Fig. 8.13 and, usingthe previous rules, we get

Mb = u(p′)(−ieγµ)i

q −m+ iε(−ieγν)u(p)εµ(k)εν(k′), q = p− k′.

(8.137)

p

k’

q=p k’

k

p’

Fig. 8.13 The crossed contribution to the Compton scattering.

If we write down the Compton amplitude for the positrons we see thatwe must associate v(p), v(p) to the initial and final states respectively.This is seen from eq. (8.126) which shows that the annihilation operatorfor the positrons is associated to v(p). For this reason, when drawing thediagrams in momentum space, is often convenient to invert the direction of



the positron lines, in such a way to follow the rule of associating unbarredspinors to incoming fermion lines and barred spinors to outgoing lines (seeFig. 8.14).

e

u(p)

e+

v(q’)

e+

v(q)

e

u(p’)

initial state final state

Fig. 8.14 The spinor conventions for antifermions.

In this case one has to be careful with the direction of momenta whichare flowing in a direction opposite to the arrow in the case of antiparticles.As for the internal lines, there is no distinction between particles and an-tiparticles. Therefore, in general, one draws the arrows in a way consistentwith the flow of the momenta. Consider now e−e− scattering. The dia-grams for this process are in Fig. 8.15, and the matrix element is givenby

〈f |S(2)3 |i〉 = (2π)4δ4(p′1 + p′2 − p1 − p2)

4∏i=1

√m

V EiM, (8.138)

where M =Ma +Mb, with

Ma = u(p′1)(−ieγµ)u(p1)igµνD(p2 − p′2)u(p′2)(−ieγν)u(p2) (8.139)

and

Mb = −u(p′2)(−ieγµ)u(p1)igµνD(p2 − p′1)u(p′1)(−ieγν)u(p2). (8.140)

The relative minus sign comes from the exchange of the two electrons inthe final state, that is from the Fermi statistics. From this example weget the further rule for the Feynman diagrams

• - for each internal photon line there is a factor igµνD(p) (propagator).



p2 p2’

p2 p2’p1 p1’

a)

p2

p1’

p2 p1’p1

p2’

b)

Fig. 8.15 The Feynman diagrams for the scattering e−e− → e−e−.

As a last example let us consider the electron self-energy

|i〉 = |e−(p)〉 → |f〉 = |e−(p′)〉, (8.141)

the S matrix element is

〈f |S(2)4 |i〉 = −e2

∫d4x1 d

4x2

√m

V Ep′

√m

V Epu(p′)eip

′x1 ]

×γµ∫

d4q1

(2π)4e−iq1(x1 − x2)iSF (q1)

×γν∫

d4q2

(2π)4e−iq2(x1 − x2)iD(q2)u(p)e−ipx2

=√

m

V Ep′

√m

V Ep

∫d4q1

(2π)4

d4q2

(2π)4

×(2π)4δ4(p′ − q1 − q2)(2π)4δ4(p− q1 − q2)

×u(p′)(−ieγµ)iSF (q1)(−ieγν)u(p)igµνD(q2)

= (2π)4δ4(p′ − p)√

m

V Ep′

√m

V EpM, (8.142)

where in the last term we have integrated over q1. M is given by

M =∫

d4q2

(2π)4u(p′)(−ieγµ)iSF (p− q2)(−ieγν)u(p)igµνD(q2), (8.143)

corresponding to the diagram in Fig. 8.16.We see that the rule of conservation of the four-momentum is always

valid and also, that we have to integrate with measure d4q/(2π)4 over allthe momenta which are not determined by the conservation law. For thegeneral case it is more convenient to formulate this rule in the followingway:



p

q2

q1=p-q2

p’=p

Fig. 8.16 The Feynman diagram for the electron self-energy.

• for each vertex there is an explicit factor (2π)4δ4(∑

ent pent), where pinare the momenta entering the vertex,

• integrate all the internal momenta, pint, with a measure d4pint/(2π)4.

In this way the factor (2π)4δ4(∑pext) is automatically produced. We

can verify the previous rule in the self-energy case∫d4q1

(2π)4

∫d4q2

(2π)4(2π)4δ4(q1 + q2 − p)(2π)4δ4(p′ − q1 − q2)

= (2π)4δ4(p′ − p)∫

d4q2

(2π)4. (8.144)

As a last rule we recall

• a factor (-1) for each fermionic loop.

8.6 Exercises

(1) Consider a scalar field of mass M , coupled to the Dirac field with massm, with an interaction given by

LI = gψψφ (8.145)

Discuss under which conditions the S-matrix, at first order in pertur-bation theory, is non vanishing.

(2) Draw the Feynman diagrams and evaluate the corresponding Feynmanamplitudes for the Bhabha scattering (e+e− → e+e−) at the 2nd orderin perturbation theory.

(3) Repeat the previous exercise for the scattering Møller (e−e− → e−e−),the pair creation γγ → e+e− and for the pair annihilation e+e− → γγ.



(4) Write the amplitudeM for the Compton scattering at the second orderin the electric charge and show that it is invariant under the followingsubstitution on any of the photon polarization vectors

εµ(k) + λkµ, (8.146)

for any value of λ. What is the relation with gauge invariance?(5) The result of the previous exercise holds for any amplitude involv-

ing external photons. That is substituting a polarization vector withthe corresponding photon four-momentum the result vanishes. Usingthis result and equation (5.60) for the sum over the physical photonpolarizations, show the following relation (assuming real polarizationvectors) ∑

λ=1,2

ε(λ)µ Mµ(ε(λ)

ν Mν)∗ = −MµMν∗gµν (8.147)

whereMµ is a generic amplitude involving at least one external photon.This result is very useful when evaluating the probability for emissionor absorption of unpolarized photons where one has to sum or averageover the photon polarizations. Of course, this result can be used forperforming the polarization sum over any number of external photons.


Chapter 9

Applications

9.1 The cross-section

Let us consider a scattering process with a set of initial particles withfour-momenta pi = (Ei, ~pi) which collide and produce a set of final par-ticles with four-momenta pf = (Ef , ~pf ). From the rules of the previ-ous Chapter we know that each external photon line contributes with afactor (1/2V E)1/2, whereas each external fermionic line contributes with(m/V E)1/2. Furthermore, the conservation of the total four-momentumgives a term (2π)4δ4(

∑i pi −

∑f pf ). If we separate in the S matrix the

term δfi corresponding to no scattering events we can write

Sfi = δfi + (2π)4δ4(∑i

pi −∑f

pf )

×∏ferm

( m

V E

)1/2∏bos

(1

2V E

)1/2

M, (9.1)

where M is the Feynman amplitude which can be evaluated by drawingthe corresponding Feynman diagrams and using the rules of the previousChapter.

Let us consider the typical case of a two particle collision giving rise toN particles in the final state. Since we are interested to a situation with thefinal state different from the initial one, the probability for the transitionwill be the modulus square of the second term in eq. (9.1). In doing thisoperation we encounter the square of the Dirac delta which is not a definitequantity. However, we should recall that we are quantizing the theory ina box, and we are considering the system for a finite, although large, timeinterval that we parameterize as (−T/2, T/2). Therefore we have not really

209



to do with the delta function but rather with (Pi(f) =∑i(f) pi(f))

(2π)4δ4 (Pf − Pi)→∫V

d3x

∫ T/2

−T/2dt ei(Pf − Pi)x. (9.2)

Consider one of the factors appearing in this equation, for instance the timeintegral. By performing the integration we get (∆E = Ef − Ei))∫ T/2

−T/2dt ei∆Et =

2 sin(T∆E/2)∆E

. (9.3)

And evaluating the modulus square

|(2π)δ(Ef − Ei)|2 →4 sin2(T∆E/2)

(∆E)2. (9.4)

On the right hand side we have a function of ∆E, whose integral holds2πT , and has a peak at ∆E = 0. Therefore in the T →∞ limit we have adelta-convergent sequence

limT→∞

4 sin2(T∆E/2)(∆E)2

= 2πTδ(Ef − Ei). (9.5)

By doing the same operations also for the space integrals we get∣∣∣(2π)4δ4 (Pf − Pi)∣∣∣2 → (2π)4L3Tδ4(Pf − Pi), (9.6)

where L is the side of the volume V = L3. Therefore, the probability perunit time of the transition is

w = V (2π)4δ4(Pf − Pi)∏i

12V Ei

∏f

12V Ef

∏ferm

(2m)|M|2. (9.7)

Here, for reasons of convenience, we have writtenm

V E= (2m)

12V E

. (9.8)

w gives the probability per unit time of a transition toward a state withwell defined quantum numbers, but we are rather interested to final stateshaving momenta between ~pf and ~pf + d~pf . Since in the volume V themomentum is given by ~p = 2π~n/L, the number of final states is(

L

2π

)3

d3p. (9.9)

The cross-section is defined as the probability per unit time divided by theflux of the ingoing particles (number of particles per unit surface and unitof time), and has the dimensions of a squared length

[cross− section] = [Probability per unit time/Flux]

= [t−1/(`−2t−1)] = [`2]. (9.10)


Applications 211

The flux is given by the relative velocity of the two beams, or the velocity ofthe beam in the instance of scattering over a fixed target, times the densityof particles, ρ:

vrelρ = vrel/V, (9.11)since in our normalization we have one particle in the volume V , and vrel isthe relative velocity of the ingoing particles. For bosons this follows fromthe normalization condition (3.37). For fermions recall that, in the boxnormalization, the wave function is

ψ(x) =√

m

EVu(p)e−ipx, (9.12)

from which we get that the total number of particles in the volume V isone: ∫

V

d3x|ψ(x)|2 =∫V

d3xm

V Eu†(p)u(p) = 1. (9.13)

The cross-section for getting the final states with momenta between ~pf e~pf + d~pf is given by

dσ = wV

vreldNF = w

V

vrel

∏f

V d3pf(2π)3

. (9.14)

We obtain

dσ =V

vrel

∏f

V d3pf(2π)3

V (2π)4δ4(Pf − Pi)1

4V 2E1E2

×∏f

12V Ef

∏ferm

(2m)|M|2

= (2π)4δ4(Pf − Pi)1

4E1E2vrel

∏f

d3pf(2π)32Ef

∏ferm

(2m)|M|2. (9.15)

Notice that the dependence on the quantization volume V disappears, asit should be, in the final equation. Furthermore, the total cross-section,which is obtained by integrating the previous expression over all the finalmomenta, is Lorentz invariant. In fact, both the factors M and d3p/2Eare invariant. Furthermore

~vrel = ~v1 − ~v2 =~p1

E1− ~p2

E2, (9.16)

but in the frame where the particle 2 is at rest (laboratory frame) we havep2 = (m2,~0) and ~vrel = ~v1, from which

E1E2|~vrel| = E1m2|~p1|E1

= m2|~p1| = m2

√E2

1 −m21

=√m2

2E21 −m2

1m22 =

√(p1 · p2)2 −m2

1m22. (9.17)

We see that also this factor is Lorentz invariant.



9.2 The scattering e+e− → µ+µ−

In order to exemplify the previous techniques we will study the processe+e− → µ+µ−. The Lagrangian density describing the interaction is1

LI = −e[ψeγ

λψe + ψµγλψµ

]Aλ, (9.18)

where ψe and ψµ, are the electron and the muon fields2. Fig. 9.1 describesthe Feynman diagram for this process at the second order (in this diagramthe arrows are oriented according to the direction of the momenta). Noticethat in contrast to the process e+e− → e+e− the diagram in the crossedchannel, Fig. 9.2, is now missing.

e+p1

e-

p2

µ-

p4

µ+p3

Fig. 9.1 The Feynman diagram for the scattering e+e− → µ+µ−.

e+ e+

e e

Fig. 9.2 The crossed diagram for the scattering e+e− → e+e−.

1Here we will consider only the electromagnetic interaction disregarding the contribu-tion due to the exchange of the Z vector boson.2The muon is a particle with the same spin and electric charge of the electron, but its

mass is 105.66 Mev, about 200 times the mass of the electron.


Applications 213

The Feynman amplitude is

M = u(p4, r4)(−ieγµ)v(p3, r3)−igµν

(p1 + p2)2v(p1, r1)(−ieγν)u(p2, r2)

= ie2u(p4, r4)γµv(p3, r3)1

(p1 + p2)2v(p1, r1)γµu(p2, r2), (9.19)

where we have introduced the polarization of the fermions, ri (the direc-tion of the spin in the rest frame). Often one is interested in unpolarizedcross-sections. In that case one has to sum the cross-section over the fi-nal polarizations and to average over the initial ones. In other words, weare assuming that both the initial and the final states are in a statisticalmixture. Correspondingly we need to evaluate the following quantity

X =14

∑ri

|M|2. (9.20)

Using

γ0γ†µγ0 = γµ, (9.21)

we can write

M? = −ie2v(p3, r3)γµu(p4, r4)1

(p1 + p2)2u(p2, r2)γµv(p1, r1). (9.22)

The amplitude M can be expressed in the following form

M =ie2

(p1 + p2)2Amuonsµ (r3, r4)Aµelectrons(r1, r2), (9.23)

with

Amuonsµ = u(p4, r4)γµv(p3, r3),

Aelectronsµ = v(p1, r1)γµu(p2, r2). (9.24)

Therefore

X =14

e4

(p1 + p2)4

∑r1,r2

(Aelectronsµ A? electrons

ν

) ∑r3,r4

(AµmuonsA? νmuons) . (9.25)

Defining the quantity

Aelectronsµν =

∑r1,r2

(Aelectronsµ A? electrons

ν

)=∑r1,r2

v(p1, r1)γµu(p2, r2)u(p2, r2)γνv(p1, r1) (9.26)



and using eqs. (4.279) for the positive and negative energy projectors, weobtain

Aelectronsµν = Tr

[p1 −me

2meγµp2 +me

2meγν

](9.27)

and the analogous quantity for the muons

Amuonsµν = Tr

[p4 +mµ

2mµγµp3 −mµ

2mµγν

]. (9.28)

To evaluate the trace of Dirac matrices we will make use of their alge-braic properties. Let us start showing that the trace of an odd number ofgamma matrices is zero. In fact for odd n

Tr [a1 · · · an] = Tr [a1 · · · anγ5γ5] = Tr [γ5a1 · · · anγ5]

= (−1)nTr [a1 · · · an] , (9.29)

where we have used the cyclic property of the trace and the anticommuta-tivity of γ5 and γµ. Obviously we have

Tr[1] = 4. (9.30)

Furthermore

Tr[ab] =12Tr[ab+ ba] =

12aµbνTr([γµ, γν ]+) = 4aµbνgµν = 4a · b. (9.31)

Then, using ab = −ba+ 2a · b we can evaluate

Tr[a1a2a3a4] = Tr[(−a2a1 + 2a1 · a2)a3a4]

= −Tr[a2(−a3a1 + 2a1 · a3)a4] + 8(a1 · a2)(a3 · a4)

= Tr[a2a3(−a4a1 + 2a1 · a4)]− 8(a1 · a3)(a2 · a4) + 8(a1 · a2)(a3 · a4)

= −Tr[a2a3a4a1] + 8(a1 · a4)(a2 · a3)−−8(a1 · a3)(a2 · a4) + 8(a1 · a2)(a3 · a4), (9.32)

that is

Tr[a1a2a3a4] = 4[(a1 ·a2)(a3 ·a4)−(a1 ·a3)(a2 ·a4)+(a1 ·a4)(a2 ·a3)]. (9.33)

This relation can be easily extended, by induction, to the trace of 2n γ-matrices. Other useful properties are

γµγµ = 4, (9.34)

γµaγµ = (−aγµ + 2aµ)γµ = −2a (9.35)

and

γµabγµ = 4a · b. (9.36)


Applications 215

Let us go back to our process. Evaluating the trace we get

Aµνelectrons =1

4m2e

Tr[(p1γµp2γ

ν)−m2eγµγν ]

=1m2e

[pµ1pν2 + pν1p

µ2 − gµν(p1 · p2 +m2

e)]. (9.37)

Analogously we get

Aµνmuons =1m2µ

[pµ3pν4 + pν3p

µ4 − gµν(p3 · p4 +m2

µ)]. (9.38)

Substituting into X we find

X =e4

2m2em

2µ(p1 + p2)4

[(p1 · p3)(p2 · p4) + (p1 · p4)(p2 · p3)

+m2µ(p1 · p2) +m2

e(p3 · p4) + 2m2em

2µ]. (9.39)

This process is studied in machines, called colliders, where two beams, oneof electrons and the other of positrons with the same energy are made tocollide, and one looks for a final pair µ+ − µ−. Therefore it is convenientto use the frame of the center of mass for the pair e+− e−. We will choosethe momentum variables as in Fig. 9.3 with

p1 = (E, ~p), p2 = (E,−~p), p3 = (E, ~p ′), p4 = (E,−~p ′). (9.40)

p1 p2

p3

p4

Fig. 9.3 The kinematic for the scattering e+e− → µ+µ−.

In this frame the scalar products can be expressed as

p1 · p3 = p2 · p4 = E2− pp′ cos θ, p1 · p4 = p2 · p3 = E2 + pp′ cos θ, (9.41)



p1 · p2 = E2 + p2, p3 · p4 = E2 + p′2, (p1 + p2)2 = 4E2, (9.42)

where p = |~p| and p′ = |~p ′|. In order for the process to be kinematicallypossible we must have E > mµ ≈ 200 me. We can then neglect me withrespect to E and mµ, obtaining

X =e4

2m2em

2µ

116E4

[(E2 − pp′ cos θ)2 + (E2 + pp′ cos θ)2 +m2

µ(E2 + p2)]

=e4

16m2em

2µ

1E2

[E2 + p′2 cos2 θ +m2

µ], (9.43)

from which

dσ =e4

16m2em

2µ

1E2

[E2 + p′2 cos2 θ +m2

µ](2π)4δ4(p1 + p2 − p3 − p4)

× 14(E2 + p2)

(2me)2(2mµ)2 d3p3d

3p4

(2π)64E2= δ4(p1 + p2 − p3 − p4)

× e4

128π2

1E6

[E2 + p′2 cos2 θ +m2

µ]d3p3d3p4, (9.44)

where we have used E2 = p2, since we are neglecting the electron mass. Wecan integrate this expression over ~p4 and |~p3| using the conservation of thefour-momentum given by the delta function. Using d3p3 = p′

2dp′dΩ we get

dσ

dΩ=∫

p′2dp′δ(E1 + E2 − E3 − E4)

e4

128π2

1E6

(E2 + p′2 cos2 θ +m2

µ)

=[∂(E3 + E4)

∂p′

]−1e4

128π2

p′2

E6(E2 + p′

2 cos2 θ +m2µ). (9.45)

The derivative can be evaluated by noticing that E23 = E2

4 = m2µ + p′

2

∂(E3 + E4)∂p′

= 2p′

E. (9.46)

Using e2 = 4πα we get the differential cross-section

dσ

dΩ=α2

8p′

2

E6

E

2p′(E2 + p′

2 cos2 θ +m2µ) =

α2

16E4

p′

E(E2 + p′

2 cos2 θ +m2µ).

(9.47)In the extreme relativistic limit E mµ (p′ ≈ E) we get the expression

dσ

dΩ=

α2

16E2(1 + cos2 θ), (9.48)


Applications 217

from which the total cross-section

σ =α2

16E2

∫dΩ(1 + cos2 θ) =

α2

16E22π∫ 1

−1

dw(1 + w2) =α2π

3E2. (9.49)

In the general case we get

σ =α2

16E4

p′

E

∫dΩ(E2 + p′

2 cos2 θ +m2µ)

=πα2

4E4

p′

E

(E2 +

13p′

2 +m2µ

). (9.50)

In the high energy limit we can easily estimate the total cross-section.Recalling that

1 GeV−2 = 0.389 mbarn, (9.51)

we get

σ ≈ 5.6 · 10−5(mbarn)(E(GeV))2

· 0.389 ≈ 2.17 · 10−5(mbarn)(E(GeV))2

, (9.52)

or in terms of nanobarns, 1 nbarn = 10−6 mbarn,

σ ≈ 20 (nbarn)(E(GeV))2

. (9.53)

The cross-section e+e− → µ+µ− is extremely useful in evaluating the ratio

R =σ(e+e− → hadrons)σ(e+e− → µ+µ−)

(9.54)

In fact, it is possible to prove that this ratio, barring higher order correctionscan be expressed in therm of the charges of the elementary constituents ofhadrons, that is the quarks,

R =∑

quarks with m<E

Q2q + higher order corrections (9.55)

The measurement of R has been one of the milestones in showing the exis-tence of quarks and that quarks besides having a fractional electric charge,have another degree of freedom, called color. Each quark exists in threedifferent states of this new degree of freedom3.

Let us comment about the case of massless fermions. As we have seenthere is no problem in taking the limit me → 0 in the evaluation of thecross-section. The same would be for mµ. The point is that for any factor1/m arising from the projector there is a corresponding factor m due to thewave function of the external fermions (see eq. (9.15)).3For a history of the discovery of QCD and the formulation of the Quantum Chromody-

namics (QCD), the theory of the strong interactions, see [Pickering (1984)] and [Peskin

and Schroeder (1995)].



9.3 Coulomb scattering

Sometimes one can think to the electromagnetic field as an assigned quan-tity, in that case it will be described by a classical function rather than byan operator. This is the case for the scattering of electrons and positronsfrom an external field as the Coulomb field of a heavy nucleus. The fullelectromagnetic field will be, of course, the sum of the classical part and ofthe quantized part. The expansion of the S matrix is still given by

S = 1 +∞∑n=1

(i)n

n!

∫· · ·∫

d4x1 · · · d4xnT (LI(x1) · · · LI(xn)) , (9.56)

with

LI(x) = −e : ψ(x)γµψ(x)[Aµ(x) +Aextµ (x)] : . (9.57)

We will consider here the scattering of an electron off an infinitely heavy

Ze

p’ p

p’p

Fig. 9.4 The Feynman diagram for the Coulomb scattering of an electron. The external

field is represented by a cross.

nucleus producing a static Coulomb potential. Let us introduce the Fouriertransform of this field

Aextµ (~x) =

∫d3q

(2π)3ei~q · ~xAext

µ (~q). (9.58)

At first order in the external field we have

S(1) = −ie∫

d4x : ψ(x)γµψ(x)Aextµ (~x) : (9.59)

and the transition we consider is

|i〉 = |e−(p, r)〉 → |f〉 = |e−(p′, s)〉. (9.60)


Applications 219

This is described by the diagram of Fig. 9.4 with a contribution givenby

〈f |S(1)|i〉 = −ie∫

d4x 〈e−(p′, s)|ψ−(x)γµψ+(x)Aextµ (~x)|e−(p, r)〉

= −ie(

m

EpV

)1/2(m

Ep′V

)1/2 ∫d4x eip

′xu(p′, s)

×∫

d3q

(2π)3ei~q · ~xAext

µ (~q)γµu(p, s)e−ipx

= −ie(

m

EpV

)1/2(m

Ep′V

)1/2

(2π)δ(E′ − E)

×∫

d3q

(2π)3(2π)3δ3(~p+ ~q − ~p ′)u(p′, s)γµAext

µ (~q)u(p, s)

= (2π)δ(Ep′ − Ep)(

m

EpV

)M, (9.61)

with

M = u(p′, s)(−ieγµ)Aextµ (~p ′ − ~p)u(p, s). (9.62)

Notice that in this case the spatial momentum is not conserved. In fact,the presence of the external field violates the translational invariance of thetheory and, as a consequence, the nucleus absorbs the momentum ~p ′ − ~pfrom the electron. From the previous expression we see also that whenthere are external fields the Feynman rules are modified, and we have tosubstitute the wave function of a photon(

12EqV

)1/2

ε(λ)µ (q), (9.63)

with the Fourier transform of the external field

Aextµ (~q). (9.64)

The probability per unit time is given by

w =1T

∣∣∣〈f |S(1)|i〉∣∣∣2 = 2πδ(E′ − E)

( m

EV

)2

|M|2. (9.65)

The expression for the density of final states is (p′ = |~p ′|)

dNf =V d3p′

(2π)3=V p′

2dp′dΩ

(2π)3. (9.66)

Since E′2 = E2 = p′

2 + m2 = p2 + m2. we have p = |~p| = p′, andE′dE′ = p′dp′. Therefore

dNf = Vp′E′dE′

(2π)3dΩ. (9.67)



The incoming flux is v/V = p/V E, and we get

dσ =V wdNf

v=(m

2π

)2

dE′δ(E′ − E)|M|2dΩ. (9.68)

The differential cross-section is obtained by integrating over the final energyof the electron

dσ

dΩ=(m

2π

)2

|M|2 =(me

2π

)2 ∣∣∣u(p′, s)γµAextµ (~q)u(p, r)

∣∣∣2, (9.69)

where ~q = ~p ′ − ~p. Averaging over the initial polarizations and summingover the final ones we obtain

dσ

dΩ=(me

2π

)2 12Aextµ (~q)Aext

ν (~q)Tr[p′ +m

2mγµp+m

2mγν]. (9.70)

From the evaluation of the trace we get

Tr[...] =1m2

[p′µpν − gµν(p′ · p) + p′νpµ + gµνm

2]. (9.71)

Assuming now that the external field is of Coulomb type, we have

Aextµ (~x) =

(− Ze

4π|~x|,~0)

(9.72)

and

Aext0 (~q) = − Ze

|~q|2. (9.73)

We see that we need only the terms with µ = ν = 0 from the trace

Tr[...]µ=ν=0 =1m2

[E2 +m2 + ~p · ~p ′] (9.74)

and since ~p · ~p ′ = p2 cos θ

Tr[...]µ=ν=0 =1m2

[E2 +m2 + p2 cos θ]. (9.75)

Then, we getdσ

dΩ= 2

(Zα)2

|~q|4(E2 +m2 + p2 cos θ). (9.76)

Using

|~q|2 = |~p ′ − ~p|2 = 4p2 sin2 θ

2, (9.77)

m2 = E2 − |~p|2, and v = p/E, we finally obtaindσ

dΩ= 2

(Zα)2

(4p2 sin2(θ/2)2(E2 +m2 + p2 cos θ)

=(Zα)2

4E2v4 sin4(θ/2)[1− v2 sin2(θ/2)]. (9.78)

In the non relativistic limit v 1, E ≈ m, we finddσ

dΩ=

(Zα)2

4m2 sin4(θ/2)1v4

=(Zα)2

16T 2 sin4(θ/2), (9.79)

which is the classical Rutherford formula for the Coulomb scattering withT = mv2/2.


Applications 221

9.4 Application to atomic systems

We consider here the interaction of an atomic system with the electromag-netic field. As we have shown, when we have quantized the electromagneticfield, the physical states can be taken as containing only transverse pho-tons. The total system will be described by the following non-relativisticHamiltonian

H =∑i

(~pi(t)− e ~A(~xi(t), t))2

2mi+∑i

eA0(~xi(t), t)+Hem = H0+HI , (9.80)

where Hem is the the Hamiltonian for the non interacting electromagneticfield and

HI = −e∑i

~pi(t) · ~A(~xi(t), t)mi

+ e2∑i

~A(~xi(t), t)2

2mi(9.81)

is the interaction term4. H0 is the sum of the atomic Hamiltonian (describ-ing the interaction of the electrons with the nucleus) with Hem. The sumis over all the electrons in the atom. Notice that the part in A0 describesthe interaction of the electrons with the nucleus and it has been includedin H0. The possible processes generated at the first order in perturbationtheory are represented in Fig. 9.5. Here we will consider only the emissionand absorption processes.

Let us start considering the emission process. At the lowest order con-sider the transition

|A,nk〉 → |A′, (nk + 1)〉, (9.82)where A and A′ are the labels for the atomic system in the energy eigen-states corresponding to the eigenvalues EA and EA′ respectively. We as-sume also an initial radiation field with nk photons each with momentum~k. We want to evaluate the probability for the atom to emit a photon ofenergy ωk = |~k|. From our rules we can evaluate the relevant S-matrixelement∫ +∞

−∞dt ie 〈A′, (nk + 1)|

∑i

~vi(t) · ~A(−)(~xi(t), t)|A,nk〉. (9.83)

By using the fact that the operators are in the interaction picture we canwrite the previous expression as follows∫ +∞

−∞dt ie 〈A′, (nk + 1)|eiHatt

∑i

~vi(0) · ~A(−)(~xi(0), t)e−iHatt|A,nk〉,

(9.84)4We are neglecting the fact that [~p, ~A(~x, t)] 6= 0. This is because in the approximation

that we will use (dipole approximation, see later) ~A is evaluated at ~x = 0.



A A' A A'

A A'

emission absorption

elastic or anelastic scattering

Fig. 9.5 The possible first order processes for an atomic system interacting with a

radiation field, as described by the Hamiltonian interaction of (9.81).

where Hat is the part of H0 relative to the atom and acts only on theposition and the velocity operators. Inserting the expansion for the field ~A

we get ∫ +∞

−∞dt ie

∫d3~q√2ωqV

ei(EA′ − EA − ωq)t

×〈A′, (nk + 1)|∑i

~vi(0) ·∑λ=1,2

~ε ∗λ (~q)e−i~q · ~xi(0)a†λ(~q)|A,nk〉, (9.85)

from which

2πδ(EA′ − EA − ωk)(ie)

√(nk + 1)

2ωkV〈A′|

∑i

~vi(0) · ~ε ∗(~k)e−i~k · ~xi(0)|A〉.

(9.86)Here we have used the following properties of the creation and annihilationoperators

|n〉 =a†n√n!|0〉, (9.87)

a† |n〉 =√n+ 1|n+ 1〉, a |n〉 =

√n |n− 1〉. (9.88)

By taking the modulus square of the matrix element, dividing by the time T ,and multiplying by the final states density we get the differential probability


Applications 223

for the emission of one photon

dwe = 2πδ(EA′ − EA − ωk) e2(nk + 1)

×|〈A′|∑i

~vi(0) · ~ε ∗(~k)e−i~k · ~xi(0)|A〉|2 d3~k

2ωk(2π)3. (9.89)

For low frequencies, that is when |~k| 1/a, where a is the typical atomicsize, it is possible to make the so called dipole-approximation,

e−i~k · ~x ≈ 1. (9.90)

Furthermore we can use the property

~xA′A = −i[Hat, ~x]AA′ = −i(EA′ − EA)~xA′A = −iωkxA′A. (9.91)

Using also α = e2/4π we get

dwedΩ

∣∣∣A→A′

=α

2πe(nk + 1)ω3

k|〈A′|~d · ~ε ∗(~k)|A〉|2, (9.92)

where we have introduced the dipole-operator

~d = e∑i

~xi(0). (9.93)

If we do not detect the polarization of the emitted photon we have to sumover the polarizations and this can be done by using∑

λ=1,2

εiλ(~k) εj ∗λ (~k) = δij −kikj

|~k |2, (9.94)

following from ∑λ=0,1,2,3

εµλ(~k) εν ∗λ (~k)gλλ = gµν (9.95)

and recalling that (choosing nµ = (1,~0))

εi3 =ki − (n · k)ni

(n · k)=ki

|~k|, εi0 = ni = 0. (9.96)

Therefore, by introducing the angle θ between ~k and ~d we obtain

dwedΩ

∣∣∣A→A′

=α

2π(nk + 1)ω3

k|~dAA′ |2 sin2 θ. (9.97)

Integrating over the solid angle we find the emission probability

we =43αω3

k(nk + 1)|~dAA′ |2 (9.98)



In a completely analogous way we find the absorption probability

wa =43αω3

knk|~dAA′ |2 (9.99)

We can easily establish the relation between nk and the intensity per unitvolume of a radiation consisting of nk photons of energy ωk

I(ωk)dωk =nkVωk2

∫sol. ang.

dNf =nkVωkV ω2

kdωkπ2

= nkω3k

π2dωk, (9.100)

where we integrate over the solid angle. Here dNf is the number of states ofmomentum between ~k and ~k+ d~k and the factor 2 comes from the possiblepolarizations. We have also used∫

sol. ang.

dNf =∫

sol. ang.

d3~kV

(2π)3=ω2kdωkV

2π2(9.101)

We are now in the position to determine the black body radiation law. Infact, inside the block body the thermal equilibrium will be established when

wewa

= eωk/kT . (9.102)

However we havewewa

= 1 +1nk. (9.103)

It follows

nk =1

eωk/kT − 1, (9.104)

from which

I(ω) =ω3

π2

1

eωk/kT − 1. (9.105)

From the previous results we can also evaluate easily the life-time relativeto the radiation emission as

1τ

= we =43αω3(n+ 1)|~dAA′ |2. (9.106)

9.5 Exercises

(1) Consider the scattering e+e− → µ+µ− in the limit E me,mµ. Show,at the same order of perturbation theory considered in the text, andusing dimensional analysis, that the total cross-section must be of theform

σ ∝ α2

E2. (9.107)


Applications 225

(2) In a first approximation the proton can be though as a point-like par-ticle. In this case it can be described in terms of a Dirac field, ψp. Theinteraction of the proton with the electromagnetic field is given by

LI = −eψpγµψpAλ. (9.108)

Consider the process

e−p→ e−p (9.109)

in the reference frame where the initial proton is at rest. Then, considerthe following two cases:

(a) The electron energy E m, where m is the mass of the electron.(b) The electron energy E is of the same order or bigger than the mass

of the proton, M and therefore we can neglect m.

Evaluate the differential cross-section for the scattering ep showing thatin the first case it coincides with the differential cross-section obtainedfor the Coulomb scattering with a fixed potential.

(3) Consider the Compton scattering e−γ → e−γ for unpolarized initialphotons and electrons. Assume that the final polarizations are unde-tected. Evaluate the analog of the quantity X defined in eq. (9.20).The result obtained in eq. (8.147) of Section 8.6 will be useful. Ne-glecting the mass of the electron, express the result in terms of theMandelstam variables:

s = (pi + ki)2, t = (pf − pi)2, u = (pf − ki)2 (9.110)

Notice also, that these variables are not independent, since

s+ t+ u = 0 (9.111)

(4) Under the same conditions of the previous exercise evaluate the quan-tity X for the pair annihilation process e+e− → 2γ. The result isrelated to the one obtained in the exercise (3) via the so-called cross-ing symmetry, pi → −pf and ki → −kf . Explain why.


Chapter 10

One-loop renormalization

10.1 Divergences of the Feynman integrals

Let us consider again the Coulomb scattering. We will expand the S matrixup to the third order in the electric charge, assuming that the external fieldis weak enough so we can consider it at first order. Then, the relevantFeynman diagrams are the ones in Fig. 10.1. The Coulomb scattering canbe used, in principle, to define the physical electric charge of the electron.This is done assuming that the amplitude is linear in ephys, from which weget an expansion of the type

ephys = e+ a2e3 + · · · = e(1 + a2e

2 + · · · ), (10.1)

in terms of the parameter e which appears in the original Lagrangian. Thefirst problem we encounter is that we would like to express the results of ourcalculations in terms of measured quantities as, for instance, the electriccharge, ephys. This could be done by inverting the previous expansion but,and here comes the second problem, the coefficients of the expansion aredivergent quantities. To show this, consider, for instance, the self-energycontribution to one of the external electrons (as the one in Fig. 10.1a). Wehave

Ma = u(p′)(−ieγµ)Aextµ (~p ′ − ~p) i

p−m+ iε

[ie2Σ(p)

]u(p), (10.2)

where (recall that we are using the Lorenz gauge for the electromagneticfield)

ie2Σ(p) =∫

d4k

(2π)4(−ieγµ)

−igµν

k2 + iε

i

p− k −m+ iε(−ieγν)

= −e2

∫d4k

(2π)4γµ

1k2 + iε

1

p− k −m+ iεγµ, (10.3)

227



p’ p

p’p=

p’ p

p’p

p’ p

p’

p

p k

k

p’ p

p’

p’ k

k

p

p’ p

k k+q

p’p

p’ p

p’ k

kp’

p k

pa) b)

c) d)

Fig. 10.1 The Feynman diagrams for the Coulomb scattering at the third order in theelectric charge and at the first order in the external field, denoted by the cross.

that is

Σ(p) = i

∫d4k

(2π)4γµ

p− k +m

(p− k)2 −m2 + iεγµ

1k2 + iε

. (10.4)

For large momentum, k, the integrand behaves as 1/k3 and the integraldiverges linearly. Analogously, one can check that all the other third ordercontributions are divergent. Let us write explicitly the amplitudes for theother diagrams

Mb = u(p′)ie2Σ(p′)i

p ′ −m+ iε(−ieγµ)Aext

µ (~p ′ − ~p)u(p), (10.5)

Mc = u(p′)(−ieγµ)−igµνq2 + iε

ie2Πνρ(q)Aextρ (~p ′− ~p)u(p), q = p′− p, (10.6)

where

ie2Πµν(q) = (−1)∫

d4k

(2π)4

×Tr[

i

k + q −m+ iε(−ieγµ)

i

k −m+ iε(−ieγν)

],(10.7)


One-loop renormalization 229

(the minus sign originates from the fermion loop) and therefore

Πµν(q) = i

∫d4k

(2π)4Tr

[1

k + q −mγµ

1

k −m+ iεγν]. (10.8)

The last contribution is

Md = u(p′)(−ie)e2Λµ(p′, p)u(p)Aextµ (~p ′ − ~p), (10.9)

where

e2Λµ(p′, p) =∫

d4k

(2π)4(−ieγα)

i

p ′ − k −m+ iεγµ

× i

p− k −m+ iε(−ieγβ)

−igαβk2 + iε

, (10.10)

or

Λµ(p′, p) = −i∫

d4k

(2π)4γα

1

p ′ − k −m+ iεγµ

× 1

p− k −m+ iεγα

1k2 + iε

. (10.11)

The problem of the divergences is a serious one and in order to give somesense to the theory we have to find a way to define our integrals. This iswhat is called the regularization procedure of the Feynman integrals, con-sisting in formulating a prescription in order to make the integrals finite.This can be done in various ways, for instance, introducing an ultravioletcut-off or, as we shall see later, by the more convenient method of dimen-sional regularization. However, the theory should not depend on the wayin which we define the integrals, otherwise we would have to assign somephysical meaning to the regularization procedure. This brings us to theother problem, the inversion of eq. (10.1). Since, after regularization, thecoefficients are finite, we can proceed to perform the inversion, obtaininge as a function of ephys and therefore all the other observables in terms ofthe physical electric charge (that is the one measured in the Coulomb scat-tering). By doing so we introduce in the observables a dependence on therenormalization procedure. When this dependence cancels out, we say thatthe theory is renormalizable. Thinking to the regularization in terms ofa cut-off, this means that considering the observable quantities in termsof ephys and removing the cut-off, that is taking the limit for the cut-offgoing to the infinity, the result should be finite. Of course, this cancella-tion is not obvious at all, and in fact in most of the theories this does nothappen. However there is a restricted class of renormalizable theories as,



for instance, QED. We will not discuss neither the renormalization at allorder, nor we will prove which criteria a theory should satisfy in order tobe renormalizable. We will give these criteria without proof, but we willtry to justify them on a physical basis. As for as QED is concerned we willstudy in detail the renormalization at one-loop level.

The previous way of defining a renormalizable theory amounts to saythat the original parameters in the Lagrangian, as e, should be infinite andthat their divergences should compensate the divergences of the Feynmandiagrams. Then one can try to separate the infinite from the finite part ofthe parameters (however this separation is ambiguous, see later). The infi-nite contributions are called counterterms, and by definition they have thesame operator structure of the original terms in the Lagrangian. There-fore, the procedure of regularization can be performed by adding to theoriginal Lagrangian counterterms cooked in such a way that their contri-bution kills the divergent part of the Feynman integrals. This means thatthe coefficients of these counterterms have to be infinite when we removethe regularization. We see that the theory will be renormalizable if thecounterterms we add to make the theory finite have the same structure ofthe original terms in the Lagrangian. In fact, if this is the case, they can beabsorbed by the original parameters giving rise to finite values. Howeverthe final, finite parameters are arbitrary, since they have to be fixed by theexperiments (renormalization conditions).

In the case of QED all the divergences can be brought back to the threefunctions Σ(p), Πµν(q) and Λµ(p′, p) (see later in Section 10.4). This doesnot mean that an arbitrary diagram is not divergent, but it can be madefinite if the previous functions are finite. In such a case one has only toshow that eliminating these three divergences (primitive divergences) thetheory is automatically finite. In particular we will show that the divergentpart of Σ(p) can be absorbed into the definition of the mass of the electronand a redefinition of the electron field (wave function renormalization).The divergence in Πµν , the photon self-energy, can be absorbed in thewave function renormalization of the photon (the mass of the photon isnot renormalized due to the gauge invariance). And finally the divergenceof Λµ(p′, p) goes into a re-definition of the parameter e. To realize thisprogram we divide up the Lagrangian density in two parts, one written interms of the physical parameters and the other containing the counterterms.We will call the original parameters and fields of the theory, bare parametersand bare fields. We will make use of an index B in order to distinguishthe bare quantities from the physical ones. Therefore the two pieces of



the Lagrangian density should look as follows: the piece in terms of thephysical parameters (from now on we will omit the underscript, phys, fromthe physical parameters) Lp

Lp = ψ(i∂ −m)ψ − eψγµψAµ −14FµνF

µν − 12

(∂µAµ)2 (10.12)

and the counterterms part Lc.t.

Lc.t. = iBψ∂ψ −Aψψ − C

4FµνF

µν − E

2(∂µAµ)2 − eDψAψ. (10.13)

Of course, the sum of these two contributions should coincide with theoriginal Lagrangian written in terms of the bare quantities. Adding togetherLp and Lc.t. we get

L = (1 +B)iψ∂ψ − (m+A)ψψ − e(1 +D)ψγµψAµ −1 + C

4FµνF

µν

+ gauge− fixing, (10.14)

where, for sake of simplicity, we have omitted the gauge fixing term. Defin-ing the renormalization constant of the fields

Z1 = (1 +D), Z2 = (1 +B), Z3 = (1 + C), (10.15)

we write the bare fields as

ψB = Z1/22 ψ, AµB = Z

1/23 Aµ, (10.16)

obtaining

L = iψB ∂ψB −m+A

Z2ψBψB −

eZ1

Z2Z1/23

ψBγµψBAµB −

14FB,µνF

µνB

+ gauge− fixing. (10.17)

Defining

mB =m+A

Z2, eB =

eZ1

Z2Z1/23

, (10.18)

we get the initial Lagrangian expressed in terms of the bare fields and bareparameters

L = iψB ∂ψB −mBψBψB − eBψBγµψBAµB −14FB,µνF

µνB + gauge− fixing.

(10.19)Since the finite piece can be fixed in terms of the observable quantities,the division of the parameters in physical and counter term part is welldefined. This way of fixing the finite part of the Lagrangian gives rise tothe renormalization conditions. The counter terms A, B, ... are determined



recursively at each perturbative order in such a way to eliminate the di-vergent parts and to respect the renormalization conditions. We will seelater how this works in practice at one-loop level. Another observation isthat Z1 and Z2 have to do with the self-energy of the electron, and as suchthey depend on the electron mass. Therefore, if we consider the theoryfor a different particle, as the muon, which has the same interactions asthe electron and differs only for the value of the mass (mµ ≈ 200me), onewould get a different electric charge. Or, phrased in a different way, onewould have to tune the bare electric charges of e and µ at different values,in order to get the same physical charges. This looks very unnatural, butthe gauge invariance of the theory implies that at any order in perturbationtheory Z1 = Z2. As a consequence eB = e/Z

1/23 , and since Z3 comes from

the photon self-energy, the relation between the bare and the physical elec-tric charge is universal (that is, it does not depend on the kind of chargedparticle under consideration).

Summarizing, we start dividing up the bare Lagrangian in two pieces.Then we regularize the theory giving some prescription to get finite Feyn-man integrals. The part containing the counterterms is determined, orderby order, by requiring that the divergences of the Feynman integrals, whichcome about when removing the regularization, are cancelled out by thecounterterm contributions. Since the separation of an infinite quantity intoan infinite plus a finite term is not well defined, we use the renormaliza-tion conditions to fix the finite part. After evaluating a physical quantitywe remove the regularization. Notice that although the counterterms aredivergent quantities, when we remove the cut-off, we can order them ac-cording to the power of the coupling in which we are doing the perturbativecalculation. That is to say, we have to do with a double limit, one in thecoupling and the other in some parameter, regulator, which defines the reg-ularization. The order of the limits is first to work at some order in thecoupling, at fixed regulator, and then remove the regularization.

Before going into the calculations for QED we want to illustrate somegeneral result about the renormalization. If one considers a theory involvingscalar, fermion and massless spin 1 (as the photon) fields, it is not difficultto construct an algorithm which allows to evaluate the ultraviolet (that isfor large momenta) divergence of any Feynman diagram. For example, inthe case of the electron self-energy (see Figs. 10.1a and 10.1b) one hasan integration over the four-momentum p and a behavior of the integrand,coming from the propagators, as 1/p3, giving rise to a linear divergence(as we shall see later the divergence is only logarithmic). From this count-



ing one can see that the only terms in the Lagrangian density containingmonomials in the fields with mass dimension smaller or equal to the num-ber of space-time dimensions are renormalizable. Theories satisfying thiscriterium are renormalizable (barring some small technicalities). The massdimensions of the fields can be easily evaluated from the observation thatthe action is dimensionless in our units (~ = 1). Therefore, in n space-timedimensions, the Lagrangian density, defined as∫

dnx L (10.20)

has a mass dimension n. Looking at the kinetic terms of the bosonic fields(two derivatives) and of the fermionic fields (one derivative), we see that

dim[φ] = dim[Aµ] =n

2− 1, dim[ψ] =

n− 12

. (10.21)

In particular, in 4 dimensions the bosonic fields have dimension 1 and thefermionic 3/2. Then, we see that QED is renormalizable, since all the termsin the Lagrangian density have dimensions smaller or equal to 4

dim[ψψ] = 3, dim[ψγµψAµ] = 4. (10.22)

The condition on the dimensions of the operators appearing in the La-grangian can be translated into a condition on the coupling constants. Infact, each monomial Oi will appear multiplied by a coupling gi

L =∑i

giOi, (10.23)

with

dim[gi] = n− dim[Oi]. (10.24)

The renormalizability requires

dim[Oi] ≤ n, (10.25)

from which

dim[gi] ≥ 0, (10.26)

that is the couplings must have positive dimension in mass or be dimension-less. Therefore the renormalizability criterium can be also formulated in thefollowing way: only terms in the Lagrangian density containing couplingswith positive mass dimension, [gi] ≥ 0, are renormalizable.

In QED the only couplings are the mass of the electron and the electriccharge which is dimensionless. As a further example consider a single scalar



field. The most general renormalizable Lagrangian density is characterizedby three parameters

L =12∂µφ∂

µφ− 12m2φ2 − ρφ3 − λφ4. (10.27)

Here m2 has dimension 2, ρ has dimension 1 and λ is dimensionless. Inparticular, linear σ-models are renormalizable theories. We will call theorieswith all the couplings such that [gi] ≥ 0 and all the possibly divergentoperators present in the original Lagrangian, strictly renormalizable.

Giving these facts, let us try to understand what makes different renor-malizable from non renormalizable theories. In the renormalizable case theonly divergent diagrams are the ones corresponding to the processes de-scribed by the operators appearing in the Lagrangian. Therefore adding toL the counter term

Lc.t. =∑i

δgiOi, (10.28)

we can choose the δgi in such a way to cancel, order by order, the diver-gences. The theory depends on a finite number of arbitrary parametersequal to the number of parameters gi. Therefore the theory is a predictiveone. In the non renormalizable case, the number of divergent diagramsincreases with the perturbative order. At each order we have to introducenew counterterms having an operator structure different from the originalone. Therefore the theory depends on an infinite number of arbitrary pa-rameters. As an example, consider a fermionic theory with an interactionof the type (ψψ)2. Since this term has dimension 6, the relative couplinghas dimension −2:

Lint = −g2(ψψ)2. (10.29)

At one loop the theory gives rise to the divergent diagrams of Fig. 10.2.The divergence of the first diagram can be absorbed into a counter term ofthe original type

−δg2(ψψ)2. (10.30)

The other two diagrams need counter terms of the type

δg3(ψψ)3 + δg4(ψψ)4. (10.31)

These counterterms give rise to new one-loop divergent diagrams, as forinstance the ones in Fig. 10.3. The first diagram modifies the alreadyintroduced counterterm (ψψ)3, but the second one needs a new counterterm

δg5(ψψ)5. (10.32)



Fig. 10.2 Divergent diagrams coming from the interaction (ψψ)2.

a) b)

Fig. 10.3 Divergent diagrams coming from the interactions (ψψ)2 and (ψψ)3.

This process never ends.The renormalization requirement restricts in a fantastic way the possi-

ble field theories. However, one could think that this requirement is tootechnical and one could imagine other ways of giving a meaning to La-grangians which do not satisfy this condition. This can be done but at theexpense of limiting the validity of the theory within a well defined rangeof energy. Consider again a theory with a four-fermion interaction. Sincedim g2 = −2, if we consider the scattering ψ + ψ → ψ + ψ in the highenergy limit (where we can neglect all the masses), on dimensional groundwe get that the total cross-section, at the lowest order in the perturbatiove



theory, behaves like

σ ≈ g22E

2. (10.33)

Analogously, in any non renormalizable theory, due to the presence of cou-plings with negative dimensions, the cross-section will increase with theenergy. But the cross-section has to do with the S matrix which must beunitary. Since a unitary matrix has eigenvalues of modulus 1, it followsthat its matrix elements must be bounded. Translating this argument interms of the cross-section one gets the bound

σ ≤ c2

E2, (10.34)

where c is some constant. For the previous example we get

g2E2 ≤ c. (10.35)

This implies a violation of the S matrix unitarity at energies such that

E ≥√

c

g2. (10.36)

It follows that we can give a meaning also to non renormalizable theories,but only for a limited range of values of the energy. This range is fixed bythe value of the non renormalizable couplings. It should be clear that nonrenormalizability and bad behavior of the amplitudes at high energies arestrictly connected.

The idea of considering field theories in a limited range of energy leadsto the modern view of effective quantum field theories (see for example,[Weinberg (1995b)]).

10.2 Dimensional regularization of the Feynman integrals

As we have discussed in the previous Section we need a procedure to givesense to the otherwise divergent Feynman integrals. The simplest of theseprocedures is to introduce a cut-off Λ and define a divergent integral as∫ ∞

0

→ limΛ→∞

∫ Λ

0

. (10.37)

Of course, in the spirit of renormalization, we have first to perform the per-turbative expansion and then take the limit over the cut-off. Although thisprocedure is very simple, it results to be inadequate in situations like gaugetheories. In fact, one can show that the cut-off regularization breaks the



translational invariance and this creates problems in gauge theories. Onekind of regularization, which nowadays is very popular, is the dimensionalregularization. This consists in considering the integration in an arbitrarynumber of space-time dimensions and taking the limit of four dimensions atthe end. Dimensional regularizing is very convenient because it respects allthe usual symmetries. In fact, except for a very limited number of cases,the symmetries do not depend on the number of space-time dimensions.Let us now show what is dimensional regularization about. We want toevaluate integrals of the type

I4(k) =∫d4p F (p, k), (10.38)

with F (p, k) ≈ p−2 or p−4. The idea is that integrating on a lower numberof dimensions improves the convergence properties of the integral in theultraviolet. For instance, if F (p, k) ≈ p−4, the integral is convergent in 2and in 3 dimensions. Therefore, we would like to introduce a quantity

I(ω, k) =∫

d2ωp F (p, k), (10.39)

to be regarded as a function of the complex variable ω. Then, if we candefine a complex function I ′(ω, k) on the entire complex plane, with definitesingularities, and such to coincide with I on some common domain, thenby analytic continuation, I and I ′ define the same analytic function. Asimple example of this procedure is given by Euler’s Gamma function. Thiscomplex function is defined for Re z > 0 by the integral representation

Γ(z) =∫ ∞

0

dt e−ttz − 1. (10.40)

If Re z ≤ 0, the integral diverges asdt

t1+ |Re z| , (10.41)

in the limit t→ 0. However it is easy to get a representation valid also forRe z ≤ 0. Let us divide the integration region in two parts defined by aparameter α

Γ(z) =∫ α

0

dt e−ttz − 1 +∫ ∞α

dt e−ttz − 1. (10.42)

Expanding the exponential in the first integral and integrating term byterm we get

Γ(z) =∞∑n=0

(−1)n

n!

∫ α

0

dt tn+ z − 1 +∫ ∞α

dt e−ttz − 1

=∞∑n=0

(−1)n

n!αn+ z

n+ z+∫ ∞α

dt e−ttz − 1. (10.43)



The second integral converges for any z since α > 0. This expressioncoincides, for Re z > 0, with the representation (10.40) for the Γ function,but it is defined also for Re z < 0 where it has simple poles located atz = −n. Therefore it is a meaningful expression on all the complex z-plane. Notice that in order to isolate the divergences we need to introducean arbitrary parameter α. However the result does not depend on theparticular value of this parameter. This is the Weierstrass representationof the Euler Γ(z). From this example we see that we need to perform thefollowing three steps

• Find a domain where I(ω, k) is convergent. Typically this will be forRe ω < 2.

• Construct an analytic function identical to I(ω, k) in the domain ofconvergence, but defined on a larger domain including the point ω = 2.

• At the end of the calculation take the limit ω → 2.

10.3 Integration in arbitrary dimensions

Let us consider the integral

IN =∫dNpF (p2), (10.44)

with N an integer number and p a vector in an Euclidean N -dimensionalspace. Since the integrand is invariant under rotations of the N -dimensionalvector p, we can perform the angular integration by means of

dNp = dΩNpN−1dp, (10.45)

where dΩN is the element of solid angle in N -dimensions. Therefore∫dΩN = SN , with SN the surface of the unit sphere in N -dimensions.

Then

IN = SN

∫ ∞0

pN−1F (p2)dp. (10.46)

The value of the surface of the sphere, SN , can be evaluated by the followingtrick. Consider

I =∫ +∞

−∞e−x

2dx =

√π. (10.47)

By taking N of these factors we get

IN =∫dx1 · · · dxNe−(x2

1 + · · ·+ x2N ) = πN/2. (10.48)



The same integral can be evaluated in polar coordinates

πN/2 = SN

∫ ∞0

ρN−1e−ρ2dρ. (10.49)

By putting x = ρ2 we have

πN/2 =12SN

∫ ∞0

xN/2−1e−xdx =12SNΓ

(N

2

), (10.50)

where we have used the representation of the Euler Γ function given in theprevious Section. Therefore

SN =2πN/2

Γ(N2

) (10.51)

and

IN =πN/2

Γ(N2

) ∫ ∞0

xN/2−1F (x)dx, (10.52)

with x = p2.The integrals we will be interested in are of the type

I(M)N =

∫dNp

(p2 − a2 + iε)A, (10.53)

with p a vector in a N -dimensional Minkowski space. We have also usedthe index (M) to distinguish explicitly this integral, defined in Minkowskispace, from an analogous integral in the Euclidean space (see later). We canperform an anti-clockwise rotation of 900 (Wick’s rotation) in the complexplane of p0 without hitting any singularity. Then we perform a change ofvariable p0 → ip0 obtaining

I(M)N = i

∫dNp

(−p2 − a2)A= i(−1)AIN , (10.54)

where IN is an Euclidean integral of the kind discussed at the beginning ofthis Section with F (x) given by

F (x) = (x+ a2)−A. (10.55)

Therefore we have to evaluate

IN =πN/2

Γ(N2

) ∫ ∞0

xN/2−1

(x+ a2)Adx. (10.56)

By putting x = a2y we get

IN = (a2)N/2−AπN/2

Γ(N2

) ∫ ∞0

yN/2−1(1 + y)−Adx (10.57)



and recalling the integral representation for the Euler B(x, y) function(valid for Rex, y > 0)

B(x, y) =Γ(x)Γ(y)Γ(x+ y)

=∫ ∞

0

tx−1(1 + t)−(x+y)dt, (10.58)

it follows

IN = πN/2Γ(A−N/2)

Γ(A)1

(a2)A−N/2. (10.59)

We have obtained this representation for N/2 > 0 and Re(A − N/2) > 0.But we know how to extend the Euler Gamma function to the entire com-plex plane, and therefore we can extend this formula to complex dimensionsN = 2ω

I2ω = πωΓ(A− ω)

Γ(A)1

(a2)A−ω. (10.60)

This shows that I2ω has simple poles located at

ω = A,A+ 1, · · · . (10.61)

Therefore our integral is defined in all the ω complex plane excluding thepoints (10.61). At the end we will have to consider the limit ω → 2. Theoriginal integral in Minkowski space is then∫

d2ωp

(p2 − a2)A= iπω(−1)A

Γ(A− ω)Γ(A)

1(a2)A−ω

. (10.62)

For the following it will be useful to derive another formula. Let us put inthe previous equation p = p′ + k∫

d2ωp′

(p′2 + 2p′ · k + k2 − a2)A= iπω(−1)A

Γ(A− ω)Γ(A)

1(a2)A−ω

. (10.63)

Defining b2 = −a2 + k2 we find∫d2ωp

(p2 + 2p · k + b2)A= iπω(−1)A

Γ(A− ω)Γ(A)

1(k2 − b2)A−ω

. (10.64)

Differentiating this expression with respect to kµ we get various useful re-lations as∫

d2ωppµ

(p2 + 2p · k + b2)A= iπω(−1)A

Γ(A− ω)Γ(A)

−kµ(k2 − b2)A−ω

(10.65)

and ∫d2ωp

pµpν(p2 + 2p · k + b2)A

= iπω(−1)A

Γ(A)(k2 − b2)A−ω

×[Γ(A− ω)kµkν −

12gµν(k2 − b2)Γ(A− ω − 1)

]. (10.66)



Since at the end of our calculations we will have to take the limit ω → 2,it will be useful to recall the expansion of the Gamma function around itspoles

Γ(ε) =1ε− γ +O(ε), (10.67)

where

γ = 0.5772... (10.68)

is the Euler-Mascheroni constant, and for (n ≥ 1):

Γ(−n+ ε) =(−1)n

n!

[1ε

+ ψ(n+ 1) +O(ε)], (10.69)

with

ψ(n+ 1) = 1 +12

+ · · ·+ 1n− γ. (10.70)

10.4 One loop regularization of QED

We will show now that at one-loop the divergent contributions come onlyfrom the three diagrams discussed before. In fact, let us define D, thesuperficial degree of divergence of a one-loop diagram, as the differencebetween 4 (the number of integration over the momentum) and the numberof powers of momentum coming from the propagators (we assume here thatno powers of momenta are coming from the vertices, which is true in QED,but not for scalar electrodynamics. We get

D = 4− IF − 2IB (10.71)

where IF and IB are the number of internal fermionic and bosonic lines.Going through the loop one has to have a number of vertices, V equal tothe number of internal lines

V = IF + IB (10.72)

Always for the one-loop diagrams we have

2V = 2IF + EF , V = 2IB + EB (10.73)

with EF and EB the number of external fermionic and bosonic lines. Fromthese equations we can get V , IF and IB in terms of EF and EB

IF =12EF + EB , IB =

12EF , V = EF + EB (10.74)



from which

D = 4− 32EF − EB (10.75)

It follows that the only one-loop superficially divergent diagrams in QEDare the ones in Fig. 10.4. In fact, all the diagrams with an odd number ofexternal photons vanish. This is trivial at one-loop, since the trace of an oddnumber of γ matrices is zero. In general (Furry’s theorem) it follows fromthe conservation of charge conjugation, since the photon is an eigenstate ofC with eigenvalue equal to −1.

E EF B D

0 2 2

0 4 0

2 0 1

2 1 0

Fig. 10.4 The superficially divergent one-loop diagrams in spinor QED.

Furthermore, the effective degree of divergence can be less than thesuperficial degree. In fact, gauge invariance often lowers the divergence.



For instance the diagram of Fig. 10.4, with four external photons (light-light scattering), has D = 0 (logarithmic divergence), but gauge invariancemakes it finite. Also the vacuum polarization diagram has an effectivedegree of divergence equal to 0. Therefore, there are only three divergentone-loop diagrams and all the divergences can be brought back to the threefunctions Σ(p), Πµν(q) e Λµ(p′, p). This does not mean that an arbitrarydiagram is not divergent, but that it can be made finite if the divergencesare eliminated in the three previous functions. In particular we will showthat the divergent part of Σ(p) can be absorbed into the definition of themass of the electron and a redefinition of the electron field (wave functionrenormalization). The divergence in Πµν , the photon self-energy, can beabsorbed in the wave function renormalization of the photon (the mass ofthe photon is not renormalized due to gauge invariance). And finally thedivergence of Λµ(p′, p) goes into the definition of the electric charge.

We will now use dimensional regularization in order to regularize therelevant divergent quantities in QED, that is Σ(p), Πµν(q) and Λµ(p, p′).Furthermore, in order to define the counter terms we will determine thedivergent parts of these expressions in the limit ω → 2. It will be alsoconvenient to introduce a parameter µ such that these quantities have thesame dimensions in the space with d = 2ω as in d = 4. The algebra ofthe Dirac matrices is also easily extended to arbitrary dimensions d. Forinstance, starting from

[γµ, γν ]+ = 2gµν , (10.76)

we get

γµγµ = d (10.77)

and

γµγνγµ = (2− d)γν . (10.78)

Other relations can be obtained by starting from the algebraic propertiesof the γ-matrices. Let us begin with the electron self-energy which we willrequire to have dimension 1 as in d = 4. From eq. (10.4) we have

Σ(p) = iµ4−2ω

∫d2ωk

(2π)2ωγµ

p− k +m

(p− k)2 −m2 + iεγµ

1k2 + iε

. (10.79)

In order to use the equations of the previous Section it is convenient tocombine together the denominators of this expression into a single one.This is done by using a formula due to Feynman

1ab

=∫ 1

0

dz

[az + b(1− z)]2, (10.80)



which is proven using

1ab

=1

b− a

[1a− 1b

]=

1b− a

∫ b

a

dx

x2. (10.81)

Performing the change of variable

x = az + b(1− z). (10.82)

We get

Σ(p) = iµ4−2ω

∫ 1

0

dz

∫d2ωk

(2π)2ωγµ

p− k +m

[(p− k)2z −m2z + k2(1− z)]2γµ.

(10.83)The denominator can be written in the following way

[...] = p2z −m2z + k2 − 2p · kz. (10.84)

The term p · k can be eliminated through the following change of variablek = k′ + pz. We find

[...] = (p2−m2)z+(k′+pz)2−2p·(k′+pz)z = k′2−m2z+p2z(1−z). (10.85)

It follows (we put k′ = k)

Σ(p) = iµ4−2ω

∫ 1

0

dz

∫d2ωk

(2π)2ωγµ

p(1− z)− k +m

[k2 −m2z + p2z(1− z)]2γµ. (10.86)

The linear term in k has vanishing integral, therefore

Σ(p) = iµ4−2ω

∫ 1

0

dz

∫d2ωk

(2π)2ωγµ

p(1− z) +m

[k2 −m2z + p2z(1− z)]2γµ (10.87)

and integrating over k

Σ(p) = iµ4−2ω

∫ 1

0

dz1

(2π)2ωiπω

Γ(2− ω)Γ(2)

γµp(1− z) +m

[m2z − p2z(1− z)]2−ωγµ.

(10.88)By defining ε = 4− 2ω we get

Σ(p) = −µε∫ 1

0

dz1

(2π)4−επ(4−ε)/2Γ(ε/2)γµ

p(1− z) +m

[m2z − p2z(1− z)]ε/2γµ.

(10.89)Contracting the γ matrices

Σ(p) = − 116π2

Γ(ε/2)∫ 1

0

dz(4πµ2)ε/2(ε− 2)p(1− z) + (4− ε)m

[m2z − p2z(1− z)]ε/2, (10.90)



we obtain

Σ(p) =1

16π2Γ(ε/2)

∫ 1

0

dz[2p(1− z)− 4m− ε(p(1− z)−m)]

×[m2z − p2z(1− z)

4πµ2

]−ε/2. (10.91)

Defining

A = 2p(1−z)−4m, B = −p(1−z)+m, C =m2z − p2z(1− z)

4πµ2(10.92)

and expanding for ε→ 0

Σ(p) =1

16π2

∫ 1

0

dz

[2Aε

+ 2B − γA] [

1− ε

2logC

]≈ 1

16π2

∫ 1

0

dz

[2Aε−A logC + 2B − γA

]=

18π2ε

(p− 4m)− 116π2

[p− 2m+ γ(p− 4m)]

− 18π2

∫ 1

0

dz[p(1− z)− 2m] logm2z − p2z(1− z)

4πµ2

=1

8π2ε(p− 4m) + finite terms. (10.93)

Consider now the self-energy of the photon (vacuum polarization) (seeeq. (10.8))

Πµν(q) = iµ4−2ω

∫d2ωk

(2π)2ωTr

[1

k + q −mγµ

1

k −mγν]

= iµ4−2ω

∫d2ωk

(2π)2ω

Tr[γµ(k +m)γν(k + q +m)](k2 −m2)((k + q)2 −m2)

. (10.94)

We use again the Feynman representation


∫ 1

0

dz

∫d2ωk

(2π)2ω

Tr[γµ(k +m)γν(k + q +m)][(k2 −m2)(1− z) + ((k + q)2 −m2)z]2

.

(10.95)Then we write the denominator as

[...] = k2 + q2z −m2 + 2k · qz. (10.96)

We then perform the change of variable k = k′ − qz in order to cancel themixed term 2k · qz. In this way we get

[...] = (k′−qz)2 +2(k′−qz) ·qz+q2z−m2 = k′2 +q2z(1−z)−m2 (10.97)



and therefore (we put again k′ = k)


∫ 1

0

dz

∫d2ωk

(2π)2ω

×Tr[γµ(k − qz +m)γν(k + q(1− z) +m)][k2 + q2z(1− z)−m2]2

. (10.98)

Since the integral of the odd terms in k is zero, it is enough to evaluate thecontribution of the even terms to the trace

Tr[...]even = Tr[γµkγν k]− Tr[γµqγν q]z(1− z) +m2Tr[γµγν ]. (10.99)

If we define the γ’s as matrices of dimension 2ω × 2ω we can repeat thecalculation of Section 9.2 obtaining a factor 2ω instead of 4. Therefore

Tr[...]even = 2ω[2kµkν − gµνk2

−(2qµqν − gµνq2)z(1− z) +m2gµν ]

= 2ω[2kµkν − 2z(1− z)(qµqν − gµνq2)− gµν(k2 −m2

+q2z(1− z)] (10.100)

and we find

Πµν(q) = iµ4−2ω2ω∫ 1

0

dz

∫d2ωk

(2π)2ω

[ 2kµkν[k2 + q2z(1− z)−m2]2

−2z(1− z)(qµqν − gµνq2)[k2 + q2z(1− z)−m2]2

− gµν[k2 + q2z(1− z)−m2]

]. (10.101)

From the relations of the previous Section we get∫d2ωp

2pµpν[p2 − a2]2

= −igµνπωΓ(1− ω)(a2)1−ω , (10.102)

whereas ∫d2ωp

1[p2 − a2]

= −iπω Γ(1− ω)(a2)1−ω . (10.103)

Therefore the first and the third contribution to the vacuum polarizationcancel out and we are left with

Πµν(q) = −iµ4−2ω2ω(qµqν − gµνq2)∫ 1

0

dz 2z(1− z)

×∫

d2ωk

(2π)2ω

1[k2 + q2z(1− z)−m2]2

. (10.104)



Notice that the original integral was quadratically divergent, but in thefinal result there is only a logarithmic divergence. The reason is gaugeinvariance. In fact, the photon self-energy is coupled to the photon field.Gauge invariance (in momentum space) implies that the theory should beinvariant under the replacement Aµ(q)→ Aµ(q)+λqµ. But this means thatwe must have qµΠµν(q) = qνΠµν = 0. In order to satisfy this condition thetensor Πµν must be necessarily of the form

Πµν = (qµqν − gµνq2)Π(q). (10.105)

Since the photon self-energy has dimension 2, it follows that Π(q) has di-mension zero. This quantity is an integral over d4k, therefore the integrandmust have dimension −4 and it is logarithmically divergent. With the sameargument one shows that the one-loop diagram with 4 external photons,which is logarithmically divergent by the power counting made in Section10.4, is finite.

Performing the momentum integration we have

Πµν(q) = −i2µ4−2ω2ω(qµqν − gµνq2)

×∫ 1

0

dz z(1− z) iπω

(2π)2ω

Γ(2− ω)[m2 − q2z(1− z)]2−ω

. (10.106)

By putting again ε = 4− 2ω and expanding the previous expression

Πµν(q) = 2µε22−ε/2(qµqν − gµνq2)

×∫ 1

0

dz z(1− z) π2−ε/2

(2π)4−εΓ(ε/2)

[m2 − q2z(1− z)]ε/2

=2

16π222−ε/2(qµqν − gµνq2)

×∫ 1

0

dz z(1− z)Γ(ε/2)[m2 − q2z(1− z)

4πµ2

]−ε/2≈ 2

16π2(4− 2ε log 2)(qµqν − gµνq2)

×∫ 1

0

dz z(1− z)(2ε− γ)

[1− ε

2logC

], (10.107)

where C is definite in eq. (10.92). Finally

Πµν(q) =1

8π2(qµqν−gµνq2)

∫ 1

0

dz z(1−z)[

8ε− 4γ − 4 log 2

] [1− ε

2logC

](10.108)



and

Πµν(q) ≈ 12π2

(qµqν − gµνq2)[ 1

3ε− γ

6

−∫ 1

0

dz z(1− z) log[m2 − q2z(1− z)

2πµ2

] ], (10.109)

from which

Πµν(q) =1

6π2(qµqν − gµνq2)

1ε

+ finite terms. (10.110)

We have now to evaluate Λµ(p′, p). From eq. (10.11) we have

Λµ(p′, p) = −iµ4−2ω

∫d2ωk

(2π)2ωγα

p ′ − k +m

(p′ − k)2 −m2γµ

p− k +m

(p− k)2 −m2γα

1k2.

(10.111)The general formula to reduce n denominators to a single one is

n∏i=1

1ai

= (n− 1)!∫ 1

0

n∏i=1

dβiδ(1−

∑ni=1 βi)

[∑ni=1 βiai]n

. (10.112)

To show this equation notice that

n∏i=1

1ai

=∫ ∞

0

n∏i=1

dαie

−n∑i=1

αiai. (10.113)

Introducing the identity

1 =∫ ∞

0

dλδ(λ−n∑i=1

αi) (10.114)

and changing variables αi = λβi

n∏i=1

1ai

=∫ ∞

0

n∏i=1

dαidλδ(λ−n∑i=1

αi)e−

n∑i=1

αiai

=∫ ∞

0

n∏i=1

dβiλn dλ

λe

−λn∑i=1

βiaiδ(1−

n∑i=1

βi). (10.115)

The integration over βi can be restricted to the interval [0, 1] due to thedelta function. We get∫ ∞

0

dλλn−1e−ρλ =(n− 1)!ρn

. (10.116)



proving eq. (10.112). In the case of the vertex function we get

Λµ(p′, p) = −2iµ4−2ω

∫d2ωk

(2π)2ω

∫ 1

0

dx

∫ 1−x

0

dy

× γα(p ′ − k +m)γµ(p− k +mγα[k2(1− x− y) + (p− k)2x−m2x+ (p′ − k)2y −m2y]3

. (10.117)

The denominator is

[...] = k2 −m2(x+ y) + p2x+ p′2y − 2k · (px+ p′y). (10.118)

Changing variable, k = k′ + px+ py

[...] = (k′ + px+ p′y)2 −m2(x+ y) + p2x

+p′2y − 2(k′ + px+ p′y) · (px+ p′y)

= k′ 2 −m2(x+ y) + p2x(1− x) + p′2y(1− y)

−2p · p′xy. (10.119)

Letting again k′ → k

Λµ(p′, p) = −2iµ4−2ω

∫ 1

0

dx

∫ 1−x

0

dy

∫d2ωk

(2π)2ω

×γα(p′(1− y)− px− k +m)γµ(p(1− x)− p′y − k +m)γα

[k2 −m2(x+ y) + p2x(1− x) + p′2y(1− y)− 2p · p′xy]3. (10.120)

The odd term in k is zero after integration. The term in k2 is logarithmi-cally divergent, whereas the remaining part is convergent. Separating thedivergent piece, Λ(1)

µ , from the convergent one, Λ(2)µ ,

Λµ = Λ(1)µ + Λ(2)

µ , (10.121)

we get for the first term

Λ(1)µ (p′, p) = −2iµ4−2ω

∫ 1

0dx∫ 1−x

0dy

∫d2ωk

(2π)2ω

× γαγλγµγσγαkλkσ

[k2 −m2(x+ y) + p2x(1− x) + p′2y(1− y)− 2p · p′xy]3

= −2iµ4−2ω∫ 1

0dx∫ 1−x

0dy

iπω(−1)3

(2π)2ωΓ(3)

(−1

2

)Γ(2− ω)

× γαγλγµγλγα

[m2(x+ y)− p2x(1− x)− p′2y(1− y) + 2p · p′xy]2−ω

= 12µ

4−2ω(

14π

)ω Γ(2− ω)∫ 1

0dx∫ 1−x

0dy

× γαγλγµγλγα

[m2(x+ y)− p2x(1− x)− p′2y(1− y) + 2p · p′xy]2−ω. (10.122)



Using the relation

γαγλγµγλγα = (2− d)2γµ, (10.123)

we obtain

Λ(1)µ (p′, p) =

12µε(

14π

)2−ε/2

Γ(ε/2)(ε− 2)2γµ

∫ 1

0

dx

∫ 1−x

0

dy

× 1[m2(x+ y)− p2x(1− x)− p′2y(1− y) + 2p · p′xy]ε/2

≈(

132π2

)[2ε− γ]

[4− 4ε]γµ∫ 1

0

dx

∫ 1−x

0

dy

×

[m2(x+ y)− p2x(1− x)− p′2y(1− y) + 2p · p′xy

4πµ2

]−ε/2

≈ 18π2ε

γµ −1

16π2(γ + 2)γµ −

18π2

γµ

∫ 1

0

dx

∫ 1−x

0

dy

× log

[m2(x+ y)− p2x(1− x)− p′2y(1− y) + 2p · p′xy

4πµ2

](10.124)

and finally

Λ(1)µ (p′, p) =

18π2ε

γµ + finite terms. (10.125)

In the convergent part we can put directly ω = 2

Λ(2)µ (p′, p) = − i

8π4

∫ 1

0

dx

∫ 1−x

0

dyiπ2(−1)3

Γ(3)

×γα(p′(1− y)− px+m)γµ(p(1− x)− p′y +m)γαm2(x+ y)− p2x(1− x)− p′2y(1− y) + 2p · p′xy

= − 116π2

∫ 1

0

dx

∫ 1−x

0

dy

×γα(p′(1− y)− px+m)γµ(p(1− x)− p′y +m)γαm2(x+ y)− p2x(1− x)− p′2y(1− y) + 2p · p′xy

. (10.126)

10.5 One loop renormalization

We summarize here the results of the previous Section

Σ(p) =1

8π2ε(p− 4m) + Σf (p). (10.127)



Πµν(q) = (qµqν − gµνq2)[

16π2ε

+ Πf (q)]≡ (qµqν − gµνq2)Π(q2). (10.128)

Λµ(p′, p) =1

8π2εγµ + Λfµ(p′, p), (10.129)

where the functions with the superscript f represent the finite contribu-tions. Let us start discussing the electron self-energy. As shown in eq.(10.2), the effect of Σ(p) is to correct the electron propagator. In fact wehave (see Fig. 10.5):

SF (p) =i

p−m+

i

p−mie2Σ(p)

i

p−m+ · · · , (10.130)

from which, at the same order in the perturbative expansion

Fig. 10.5 The loop expansion for the electron propagator.

SF (p) =i

p−m

(1 +

e2Σ(p)p−m

)−1

=i

p−m+ e2Σ(p). (10.131)

Therefore the effect of the divergent terms is to modify the coefficients ofp and m:

iS−1F (p) = p−m+ e2Σ(p) = p

(1 +

e2

8π2ε

)−m

(1 +

e2

2π2ε

)+ finite terms.

(10.132)This allows us to define the counterterms to be added to the Lagrangianexpressed in terms of the physical parameters in such a way to cancel thesedivergences

(L1)ct = iBψ∂ψ −Aψψ, (10.133)

(L1)ct modifies the Feynman rules adding two operators parameterized byA and B. These coefficients can be evaluated noticing that the expressionof the propagator, taking into account (L1)ct, is

i

(1 +B)p− (m+A)≈ i

p−m

(1− Bp−A

p−m

)≈ i

p−m+

i

p−m(iBp− iA)

i

p−m, (10.134)



where, consistently with our expansion we have taken only the first orderterms in A and B. We can associate to these two terms the diagrams ofFig. 10.6, with contributions −iA to the mass term, and iBp to p.

-iA iBp/

Fig. 10.6 The counter terms for the self-energy (p/ in the figure should be read as p).

The propagator at the second order in the coupling constant is thenobtained by adding the diagrams of Fig. 10.7 to the free part. We get

x+ +

Fig. 10.7 The second order contributions at the electron propagator.

SF (p) =i

p−m+

i

p−m(ie2Σ(p) + iBp− iA

) i

p−m(10.135)

and the correction is given by

e2Σ(p)+Bp−A =(

e2

8π2ε+B

)p−(

e2

2π2εm+A

)+finite terms. (10.136)

We can now fix the counter terms by choosing

B = − e2

8π2

(1ε

+ F2

(m

µ

)), (10.137)

A = −me2

2π2

(1ε

+ Fm

(m

µ

)), (10.138)

with F2 and Fm finite for ε → 0. Notice that these two functions aredimensionless and for the moment being completely arbitrary. Howeverthey can be determined by the renormalization conditions, that is by fixingthe arbitrary constants appearing in the Lagrangian. In fact, given

iS−1F (p) ≡ Γ(2)(p) = p−m+Bp−A+ e2Σ(p)

=(

1− e2

8π2F2

)p−m+

me2

2π2Fm + e2Σf , (10.139)



we can require that at the physical pole, p = m, the propagator coincideswith the free propagator

SF (p) ≈ i

p−m, for p = m. (10.140)

From here we get two conditions. The first one is

Γ(2)(p = m) = 0 (10.141)

from which

e2Σf (p = m)− me2

8π2F2 +

me2

2π2Fm = 0. (10.142)

The second condition is

∂Γ(2)(p)∂pµ

∣∣∣p=m

= γµ, (10.143)

giving

e2 ∂Σf (p)∂pµ

∣∣∣p=m

− e2

8π2F2γµ = 0. (10.144)

One should be careful because these particular conditions of renormaliza-tion give some problem related to the zero mass of the photon. In fact,one finds some ill-defined integral in the infrared region. However theseare harmless divergences, and can be eliminated giving a small mass to thephoton and sending this mass to zero at the end of the calculations. Noticethat these conditions of renormalization have the advantage of being ex-pressed directly in terms of the measured parameters, as the electron mass.However, one could renormalize at an arbitrary mass scale, M . In this casethe parameters comparing in Lp are not the directly measured parameters,but they can be correlated to the actual parameters by evaluating someobservable quantity. From this point of view one could avoid the problemsmentioned above by choosing a different point of renormalization.

As far as the vacuum polarization is concerned, Πµν gives rise to thefollowing correction to the photon propagator (illustrated in Fig. 10.8)

D′µν(q) =−igµνq2

+−igµλq2

ie2Πλρ(q)−igρνq2

+ · · · , (10.145)

from which


+−igµλq2

[(ie2)(qλqρ − gλρq2)Π(q2)

] −igρνq2

+ · · ·

=−igµνq2

[1− e2Π(q2)

]− i qµqν

q4e2Π(q2) + · · · . (10.146)



Fig. 10.8 The loop expansion for the photon propagator.

We see that the one loop propagator has a divergent part in gµν , as wellin the term proportional to the momenta. Therefore the propagator doesnot correspond any more to the Lorenz gauge and we need to add to thefollowing terms in Lp

L2 = −14FµνF

µν − 12

(∂µAµ)2, (10.147)

the two counterterms

(L2)ct = −C4FµνF

µν − E

2(∂µAµ)2

= −C(

14FµνF

µν +12

(∂µAµ)2

)− E − C

2(∂µAµ)2. (10.148)

As for the electron propagator, we can look at these two contributionsas perturbations to the free Lagrangian, and evaluate the correspondingFeynman rules, or evaluate the effect on the propagator. The modificationin the equation defining the propagator due to these two terms is

[(1 + C)q2gµν − (C − E)qµqν ]Dνλ(q) = −igλµ. (10.149)

We solve this equation by putting

Dµν(q) = α(q2)gµν + β(q2)qµqν . (10.150)

Substituting in the previous equation we determine α and β. The result is

α = − i

q2(1 + C), β = − i

q4

C − E(1 + C)(1 + E)

. (10.151)

The free propagator, including the corrections at the first order in C andE is

Dµν(q) =−igµνq2

(1− C)− i qµqνq4

(C − E) (10.152)

and the total propagator


[1− e2Π(q2)− C]− i qµqνq4

[e2Π(q2) + C − E]. (10.153)



We can now choose E = 0 and cancel the divergence by the choice

C = − e2

6π2ε+ F3

(m

µ

). (10.154)

In fact, we are free to choose the finite term in the gauge fixing, since thischoice does not change the physics. This is because the terms proportionalto qµqν , as it follows from gauge invariance and the consequent conservationof the electromagnetic current, do not contribute to the physical amplitude.For instance, if we have a vertex with a virtual photon (that is a vertexconnected to an internal photon line) and two external electrons, the termproportional to qµqν is saturated with

u(p′)γµu(p), q = p′ − p. (10.155)

The result is zero, by taking into account the Dirac equation. Let us nowconsider the mass of the photon. We have


[1− F3 − e2Πf

]− i qµqν

q4

[F3 + e2Πf

]≡ −igµν

q2[1− Π]− i qµqν

q4Π, (10.156)

where

Π = e2Πf + F3. (10.157)

At the second order in the electric charge we can write the propagator inthe form

D′µν(q) =−i

q2(1 + Π(q))

[gµν +

qµqνq2

Π(q)]

(10.158)

and we see that the propagator has a pole at q2 = 0, since Π(q2) is finitefor q2 → 0. Therefore the photon remains massless after renormalization.This is part of a rather general aspect of renormalization which says that,if the regularization procedure respects the symmetries of the original La-grangian, the symmetries are preserved at any perturbative order. However,there are cases where it is not possible to devise a regularization proceduresuch to preserve a given symmetry. This the case of the anomalous sym-metries which are symmetries only at the classical level but broken byquantum corrections.

Also for the photon we will require that at the physical pole, q2 → 0,the propagator has the free form, that is

Π(0) = 0, (10.159)

from which

F3 = −e2Πf (0). (10.160)



10.6 Lamb shift and g − 2

In this Section we will use the technology developed in this Chapter toevaluate two important physical quantities. The first one is a contributionto the Lamb shift. The levels 2S1/2 and 2P1/2 of the hydrogen atom are de-generate except for corrections coming from QED. Among these correctionsthere is one coming from the radiative correction to the photon propagatorand this is the one that we will evaluate here. The second quantity thatwe will evaluate comes from the vertex corrections and it is a contributionto the gyromagnetic factor of the electron which is equal to 2 in the Diractheory. Let us start with the vacuum polarization.

We will consider the expression (10.157) for small momenta:

Π(q) ≈ e2q2 dΠf (q)dq2

∣∣∣q2=0

, (10.161)

where use has been made of F3 being constant. Using the expression forΠf from the previous Section, we get

Πf (q) ≈ − 12π2

(γ

6+

16

log(m2

2πµ2

))+

12π2

∫ 1

0

dz z2(1− z)2 q2

m2+ · · ·

(10.162)and

Π(q) ≈ e2q2

60π2m2. (10.163)

It follows

D′µν = −igµνq2

[1− e2q2

60π2m2

]+ gauge terms. (10.164)

The first term, 1/q2, gives rise to the Coulomb potential, e2/4πr. Thesecond gives a correction by a term proportional to a delta function in thereal space (we are using the same notations of Section 7.2)

∆12 = e2

∫ +∞

−∞dt

∫d4q

(2π)4e−iq(x1 − x2)

[1q2− e2

60π2m2

]= − e2

4πr− e4

60π2m2δ3(~r). (10.165)

This modification of the Coulomb potential changes the energy levels ofthe hydrogen atom, and it is one of the contributions to the Lamb shift,which produces a splitting of the levels 2S1/2 and 2P1/2. The total Lambshift is the sum of all the self-energy and vertex corrections, and turns outto be about 1057.9 MHz. The contribution we have just evaluated is only



−27.1 MHz, but it is important since the agreement between experimentand theory is at the level of 0.1 MHz.

We have now to discuss the vertex corrections. We have seen that thedivergent contribution is Λ(1)

µ (p′, p), and this is proportional to γµ. Thecounter term to add to the interacting part of Lp,

Lintp = −eψγµψAµ, (10.166)

is

(L3)ct = −eDψγµψAµ. (10.167)

The complete vertex (omitting the factor −ie) is given by (see Fig. 10.9)

ΛTµ = γµ(1 +D) + e2Λµ. (10.168)

(counter-term)

Fig. 10.9 The one loop vertex corrections.

Due to the conservation of the electromagnetic current, it is possible toprove the validity of the Ward-Takahashi identity [Ward (1950); Takahashi(1957)] between the total self-energy and the total vertex correction

∂ΣT (p)∂p

= ΛTµ (p, p), (10.169)

where ΣT and ΛTµ are defined as

Γ(2) = p−m+ ΣT , Γµ = γµ + ΛTµ . (10.170)

This identity applied to the expressions (10.139) and (10.168) gives

Bγµ + e2 ∂Σ(p)∂pµ

= Dγµ + e2Λµ(p, p), (10.171)

where Σ and Λ are the self-energy and vertex corrections evaluated fromthe Feynman diagrams associated to Lp. Therefore, from the eq. (10.201)in the first Exercise of this Section, it follows

B = D, (10.172)



at this order in perturbation theory. This equality is certainly satisfied bythe divergent parts of these counterterms, since they are devised in order toeliminate the terms in 1/ε in Σ and Λµ which are equal (see eqs. (10.127)and 10.129)). However, in order to satisfy (10.172) also the finite parts ofB and D should be equal. As a consequence we fix the counter term D by

D = − e2

8π2

[1ε

+ F2

], (10.173)

with F2 the same as in eq. (10.144). Notice that in this way the wavefunction renormalization terms Z1 and Z2 are made equal, and the chargerenormalization depends only on Z3, that is from the vacuum polarization.

The complete one-loop vertex is then

ΛTµ =[γµ + e2(Λfµ −

F2

8π2γµ)]. (10.174)

By using again the identity (10.201), valid at any order in 1/ε and, therefore,also for the finite parts, and the condition (10.144)

Λfµ∣∣∣p=m

=∂Σf

∂pµ

∣∣∣p=m

=F2

8π2γµ, (10.175)

we see that for spinors on shell we have

u(p)ΛTµu(p)|p=m = u(p)γµu(p). (10.176)

It is interesting to notice that we have been able to satisfy the Ward-Takahashi identity since we had the freedom to choose the finite part ofthe counterterm D. There are situations where this is not possible and thisis the case of the anomalies, that is transformations that are symmetriesat the classical level but that are broken at quantum level. A celebratedexample is the one of the axial-vector anomaly. As we know from theExercises in Chapter 6, the Lagrangian for a massless Dirac field is invariantunder the chiral transformation

ψ(x)→ eiαγ5ψ(x), (10.177)

at the classical level, besides being invariant under a phase transformation.Both symmetries are preserved by making the Dirac field to interact withthe electromagnetic field. However, it turns out that it is impossible todefine a renormalization procedure (choice of the counterterms) in order tosatisfy both the Ward-Takahashi identities following from the conservationsof the two currents associated to the two symmetries. Usually one makesthe choice of satisfying the Ward-Takahashi identity associated to the phase



symmetry and the chiral symmetry is broken at the quantum level (axialanomaly).

We will now evaluate the radiative corrections to the g−2 of the electron.Here g is the gyromagnetic ratio, which is predicted to be equal to 2 bythe Dirac equation (see Section 4.7). To this end we first need to prove theGordon identity for the current of a Dirac particle

u(p′)γµu(p) = u(p′)[pµ + pµ′

2m+

i

2mσµνq

ν

]u(p), (10.178)

with q = p′ − p. For the proof we start from

pγµu(p) = (−mγµ + 2pµ)u(p) (10.179)

and

γµpu(p) = mγµu(p). (10.180)

Subtracting these two expressions we obtain

γµu(p) =(pµm− i

mσµνp

ν

)u(p). (10.181)

An analogous operation on the barred spinor leads to the result. We ob-serve also that the Gordon identity shows immediately that the value ofthe gyromagnetic ratio is 2, because it implies that the coupling with theelectromagnetic field is just

e

2mσµνF

µν(q). (10.182)

To evaluate the correction to this term from the one loop diagrams, it isenough to evaluate the matrix element

e2u(p′)Λ(2)µ (p′, p)u(p) (10.183)

in the limit p′ → p and for on-shell momenta. In fact Λ(1)µ contributes only

to the terms in γµ and, in the previous limit, they have to build up thefree vertex, as implied by the renormalization condition. Therefore we willignore all the terms proportional to γµ and we will take the first order inthe momentum q. For momenta on shell, the denominator of Λ(2)

µ is givenby

[...] = m2(x+y)−m2x(1−x)−m2y(1−y)+2m2xy = m2(x+y)2. (10.184)

In order to evaluate the numerator, let us define

Vµ = u(p′)γα [p′(1− y)− px+m] γµ [p(1− x)− p′y +m] γαu(p).(10.185)



Using pγα = −γαp+ 2pα, and an analogous equation for p′, we can bring pto act on the spinor to the right of the expression, and p′ on the spinor tothe left, obtaining

Vµ = u(p′)[myγα + 2(1− y)p′α − γαpx

]γµ

× [mxγα + 2(1− x)pα − p′yγα]u(p). (10.186)

Making use of

γαγµγα = −2γµ, (10.187)

γαγµγνγα = 4gµν , (10.188)

γαpγµp′γα = −2p′γµp, (10.189)

we get

Vµ = u(p′)[− 2m2xyγµ + 2my(1− x)(−mγµ + 2pµ)

−4my2p′µ + 2mx(1− y)(−mγµ + 2p′µ) + 4(1− x)(1− y)m2γµ

−2y(1− y)m2γµ − 4mx2pµ

−2m2x(1− x)γµ − 2xym2γµ]u(p) (10.190)

and for the piece which does not contain γµ

Vµ = 4mu(p′)[pµ(y − xy − x2) + p′

µ(x− xy − y2)]u(p). (10.191)

Therefore the relevant part of the vertex contribution is

e2u(p′)Λ(2)µ (p′, p)u(p)→ − e2

16π2

∫ 1

0

dx

∫ 1−x

0

dy4mu(p′)

[pµ

(y − xy − x2)(x+ y)2

+p′µ(x− xy − y2)

(x+ y)2

]u(p). (10.192)

By changing variable, z = x+ y, we obtain∫ 1

0

dx

∫ 1−x

0

dyy − xy − x2

(x+ y)2=∫ 1

0

dx

∫ 1

x

dz

[1− xz− x

z2

]=∫ 1

0

[−(1− x) log x+ x− 1]

=[−x log x+ x+

x2

2log x− x2

4+x2

2− x]1

0

=14, (10.193)

from which

e2u(p′)Λ(2)µ (p′, p)u(p)→ − e2

16π2mu(p′)[pµ + p′µ]u(p). (10.194)



Using the Gordon identity in this expression, and eliminating the furthercontribution in γµ, we obtain the correction to the magnetic moment

e2u(p′)Λ(2)µ (p′, p)u(p)|magn. mom. →

ie2

16π2mu(p′)σµνqνu(p). (10.195)

Finally we have to add this correction to the vertex part taken at p′ = p,which coincides with the free vertex

u(p′)[γµ +ie2

16π2mσµνq

ν ]u(p)∣∣∣p′≈p

≈ u(p′)[pµ + p′µ

2m+

i

2m

(1 +

e2

8π2

)σµνq

ν

]u(p). (10.196)

Therefore the correction ise

2m→ e

2m

(1 +

α

2π

). (10.197)

Recalling that g is the ratio between ~S · ~B and e/2m, we getg

2= 1 +

α

2π+O(α2). (10.198)

This correction was evaluated by Schwinger in 1948. Actually we know thefirst three terms of the expansion

ath =12

(g − 2) =12α

π− 0.32848

(απ

)2

+ 1.49(α

π)3 + · · ·

= (1159652.4± 0.4)× 10−9, (10.199)

to be compared with the experimental value

aexp = (1159652.4± 0.2)× 10−9. (10.200)

10.7 Exercises

(1) Using the two regularized expressions (10.79) and (10.111) prove theidentity

∂Σ(p)∂p

= Λµ(p, p). (10.201)

This relation must hold order by order in the expansion in ε = 4− 2ω.Check that this is true for the terms of order 1/ε.

(2) Shows that the Lagrangian of a classical massless Dirac field interactingwith an electromagnetic field is invariant under the chiral transforma-tion

ψ → eiαγ5ψ. (10.202)



(3) Verify the relations (10.187), (10.188) and (10.189) satisfied by the γ-matrices.

(4) Another regularization used in QED is the Pauli-Villars, consisting inchanging the photon propagator

1k2→ 1

k2− 1k2 −M2

(10.203)

where M is a large mass. Consider the divergent integral

I =∫d4k

1k2[(k + p)2 −m2]

(10.204)

Show that this integral is finite using Pauli-Villars regularization, andevaluate it. Show that the result diverges logarithmically for M →∞.

(5) Given a scalar and a Dirac field, assume that they are interacting viathe interaction

LI = gψγ5ψφ (10.205)

Although this theory is similar to spinor QED, and its couplings satisfy[gi] ≥ 0, it is not strictly renormalizable. Explain why and which kindof modification is necessary in order to make it strictly renormalizable.


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Index

Abeliansymmetry, 147tensor strength, 148

Absorptionprobability, 224process, 221

Action, 29N fields, 30particle interacting with

electromagnetic field, 89Dirac, 105electromagnetic field, 92free relativistic point-like particle,

89Klein-Gordon, 44non-abelian fields, 152

Adiabatic hypothesis, 181Anderson, 5Angular momentum

Dirac, 106Angular momentum conservation, 51Annihilation operators, 24

Dirac, 109electromagnetic field, 125Klein-Gordon, 47massive vector field, 130

Anomalous symmetries, 255, 258Anticommutation relations

Dirac, 110Anticommutator

Dirac, 109Dirac matrices, 70

gamma matrices, 72Antiparticles, 66

Dirac, 98Atomic systems, 221Axial anomaly, 258, 259Axial currents, 138

Barefields, 230parameters, 230

Bethe, 6Black body radiation, 224Born, 4Bose, 4Bose-Einstein statistics, 25Boundary conditions

PeriodicKlein-Gordon, 45

Boundary conditions for thevariations of the fields, 29N fields, 30Klein-Gordon, 43

Broken generators, 146, 158

Casimir effect, 26, 54–56Cauchy problem

Klein-Gordon, 67Charge conjugation, 99

Dirac spinor, 99matrix, 99symmetry, 101

Chiral symmetry, 160, 258

267



Chirality operator, 112Colliders, 215Color, 217Commutation relation

electromagnetic field, 121, 123Commutation relations, 27, 33, 42O(N) charges, 135charge density, 66charged scalar field, 64, 65electromagnetic field, 125Klein-Gordon, 44, 47, 59massive vector field, 130string, 26

Compton scattering, 195, 196, 202Compton wavelength, 3, 40, 60, 61Conserved chargesO(N), 135charged scalar field, 63–66definition, 50Dirac, 110, 180

Conserved currentO(2), 135definition, 50Dirac, 71, 75, 110gauge invariant, 178Klein-Gordon, 56

Conserved currentsO(4) model, 137abelian fields, 149non-abelian gauge fields, 153

Contraction of two operators, 191Control parameters, 142Correspondence rule, 28Coulomb field, 218Coulomb force, 2Coulomb gauge, 118, 119Coulomb potential, 175

correction, 256Coulomb scattering, 227Counterterms, 230, 231, 251, 254,

255, 258Covariant derivative, 147, 148, 150Covariant normalization, 48CP operation, 113Creation operators, 24

Dirac, 109

electromagnetic field, 125Klein-Gordon, 47massive vector field, 130

Cross-section, 209–211e+e− → hadrons, 217e+e− → µ+µ−, 216, 217Coulomb scattering, 220

Curie temperature, 142Current conservation

charged scalar field, 63, 64

Davisson, 4Dimensional regularization, 236Dipole approximation, 223Dirac, 4, 5

action, 105bilinears, 77equation, 70, 72, 77, 107

solutions, 79hole theory, 98negative energy solutions, 79, 86,

98positive energy solutions, 79, 98representation, 11, 77

DivergencesQED, 230

Effective quantum field theories, 236Einstein, 4Electromagnetic current

Klein-Gordon, 178point-like particle, 88

Electromagnetic fielddegrees of freedom, 119

Electromagnetic interaction, 177charged particle, 88Dirac field, 178–180Klein-Gordon field, 177

Electromagnetic strength tensor, 90dual, 91

e+e− → µ+µ−, 212Electron Positron scattering, 198Electron scattering, 197, 205Emission

probability, 223process, 221


Bibliography 269

Energy interaction, 174Energy-momentum tensor

definition, 50Dirac, 105Klein-Gordon, 52

Euler’s Γ, 239Euler’s γ, 237Euler’s Γ expansion, 241Euler’s Beta, 240Euler-Lagrange equations, 31Euler-Mascheroni constant, 241Expansion of the field

charged scalar field, 65Dirac, 107electromagnetic, 124Klein-Gordon, 47massive vector, 130string, 33

Fermi, 4, 5Fermi-Dirac statistics, 109Fermion self-energy, 198, 206Ferromagnet, 142Feynman amplitude, 202, 209e+e− → µ+µ−, 213Compton scattering, 203, 204Coulomb scattering, 219electron scattering, 205fermion self-energy, 206Photon self-energy, 228vertex, 229

Feynman diagrams, 200e+e− → e+e−, 212e+e− → µ+µ−, 212Compton scattering, 203, 204Coulomb scattering, 218, 227electron scattering, 205electron self-energy, 206rules, 204–206

Feynman formula, 243n denominators, 248

Feynman integralsdivergence, 227

Feynman propagator, 164–166, 168fermions, 168, 169, 200

radiative corrections, 251, 252

photons, 170, 200Field transformationsO(4) model, 136

FieldsDirac, 104electromagnetic field, 121Klein-Gordon, 42massive vector, 129massless spin 1/2, 112Proca, 129

Fine structure constant, 2Flux, 210, 211, 220Fock space, 25, 43

Klein-Gordon, 53Foldy-Wouthuysen transformation,

115Four-momentum

fields, 50

g − 2, 256, 259, 261γ-matricesγµ, 72γ5, 76σµν , 75d dimensions, 243chiral basis, 79covariance, 74identities, 214list, 76trace formulas, 214

Gauge fieldsabelian, 148equations of motion, 153non-abelian, 151

Gauge fixing, 118, 122, 123, 126Gauge invariance, 147

electromagnetic field, 117, 120Gauge theories

QED, 147Gauge transformations

abelian fields, 148electromagnetic field, 92, 117, 120,

129, 178non-abelian fields, 151

Germer, 4Goldstone bosons, 144, 154



Goldstone fields, 159Goldstone theorem, 144, 154Gordon identity, 259Green functions, 166

advanced, 171retarded, 170–172

Ground statedegenerate, 140, 142

Ground state energy, 25GroupO(2), 63, 64, 117, 141, 155, 156O(3), 137O(3, 1), 7, 10, 136O(4), 136, 137O(N), 63, 134, 139, 146, 158O(N − 1), 146, 158SL(2, C), 9–12, 136SO(3, 1), 7SU(2), 137, 138SU(2)L ⊗ SU(2)R, 136–138U(1), 65, 147U(N), 149complete Lorentz, 77Lorentz, 6, 7, 9–12, 42, 48, 51, 58,

59, 71, 73, 75–77Poincare, 15, 16

Gyromagnetic ratio, 259Gyromagnetic ratio for a Dirac

particle, 96

HamiltonianO(N) model, 138Dirac, 70, 108electromagnetic field, 127Klein-Gordon, 52, 53

Hamiltonian density, 31, 179Dirac, 105Klein-Gordon, 52

Heaviside-Lorentz units, 2Heisenberg, 4, 5Helicity operator, 112Higgs mechanism, 154, 155Hydrogen atom, 256

Interaction energy, 175Interaction representation, 181, 183

Internal transformations, 48Invariant amplitude

fermion self-energy, 227Invariant functions

∆, 167∆(±), 167∆C , 166, 168

Jordan, 4, 5

Klein, 4, 5Klein-Gordon

equation, 42solutions, 44

normalized in a box, 46normalized in the continuum,

46

Lagrangian, 29Lagrangian densityO(2), 140, 141, 155O(4), 135O(N), 133, 144O(N) model, 139abelian field, 148broken U(1), 157charged scalar field, 62, 64Coulomb interaction, 218Dirac, 105, 147electromagnetic field, 117, 122gauged O(N), 158gauged U(1), 156interacting charged scalar field, 177interacting Dirac field, 178Klein-Gordon, 44massive vector field, 129non-abelian fields, 148, 152QED, 191

Lamb, 5Lamb shift, 5, 256Landau free-energy, 142Lie algebra, 149, 151O(N), 135SU(2)L ⊗ SU(2)R, 138

Local gauge transformation, 147, 150Lorentz


Bibliography 271

Casimir operator, 16generators, 9, 16, 75transformations, 6–10, 15, 51, 72,

80, 942× 2 representation, 10Dirac, 73, 75, 106electromagnetic field, 94improper, 7proper, 7

Lorenz gauge, 118, 120–123, 227

Massbroken gauge fields, 157, 158

Mass dimensions, 233bosons, 233fermions, 233

Massive vector fielddegrees of freedom, 130equation, 129

Maxwell equations, 91electromagnetic potentials, 117

Minimal substitution, 93, 94, 177Minkowski space

distance, 15infinitesimal distance, 6

MomentumDirac, 106, 108Klein-Gordon, 52, 53, 65

Momentum density, 31, 32charged scalar field, 64Dirac, 105electromagnetic field, 121, 123interacting case, 179Klein-Gordon, 44

Muon, 212

Natural units, 1Negative energy solutions, 39Negative norm states, 125Neutrinos, 154Nishina, 5Noether’s theoremO(N), 134charged scalar field, 178general formulation, 48

Non renormalizable theory, 234

Non-abeliangauge symmetries, 147gauge theories, 149tensor strength, 151, 152

Normal modes, 23Normal ordering, 26, 188–190, 192Null states, 126Number of particles

electromagnetic field, 127Klein-Gordon, 56

O(2)potential, 140, 155

O(N)conserved currents, 134generators, 134Lie algebra, 134mass matrix, 145potential, 146

One-looprenormalization, 227

One-loop divergent diagrams, 234One-loop regularization in QED, 241One-loop renormalization, 250Order parameters, 142Orthogonal transformations, 134

Pair annihilation, 39, 196, 197Pair creation, 39, 196Pais, 4Parity, 7, 11, 76, 77, 140Particle density operator, 57Particle-wave duality, 40Pauli, 4, 5Pauli equation, 96Pauli matrices, 9, 69, 135, 150Pauli-Lubanski four-vector, 15Pauli-Villars regularization, 262PCT symmetry, 103PCT transformation, 103Perturbation theory, 163, 173, 177Phase transitions, 142Physical electric charge, 227Physical states

electromagnetic field, 123, 126energy, 128



PoincareCasimir operators, 16generators, 15transformations, 15

Poisson brackets, 28Polarization vectors

electromagnetic field, 123, 124massive vector field, 131sum for massive vector field, 131sum for the electromagnetic field,

124, 223Projectors

negative energy, 83positive energy, 83spin, 85

QCD, 217QED, 179, 191Quantum oscillations, 143Quarks, 217

Ratio R, 217Regularization, 229

cut-off, 229dimensional, 229

Relativistic dispersion relation, 39Renormalizability

criterium, 233Renormalizable

theories, 229, 233Renormalization, 229Renormalization conditions, 231, 252,

253, 255Rest frame, 80

solutions, 80Retherford, 5Ricci tensor, 2Rutherford formula, 220

S-matrix, 183, 185, 187, 209Coulomb scattering, 218Lorentz invariance, 187QED, 191

first order, 193, 201second order, 194–199, 202,

205, 206

third order, 227

unitarity, 185

Scattering

amplitude, 182, 183

atom-photon, 221

Coulomb, 218

matrix, 180

process, 180, 182

Schrodinger, 4

Schrodinger equation

integral form, 183

Schwinger, 6

Self-energy

electrons, 227, 228, 243–245, 250

photon, 198, 245–247, 251

radiative corrections, 253–255

Serber, 5

Sigma-model, 133

O(4), 135, 160

O(N), 133, 144

potential, 139

Space-time metric tensor, 1

Space-time translations, 50

Sphere surface in N dimensions, 239

Spin average, 213

Spin sum, 213

Spin-statistics theorem, 112

Spinors, 12, 13

u, 81, 82

v, 81, 82

bispinor, 77

completeness, 107

Dirac, 71

dotted, 11, 15

four-component, 79

orthogonality, 107

undotted, 11, 14

Spontaneous symmetry breaking, 140,154

finite systems, 142

Step function, 165

String, 22

Lagrangian, 28

Sugar molecules, 144

Symmetries, 133


Bibliography 273

Time ordered exponential, 185Time ordered product, 163, 164, 166,

183, 188, 190, 192fermions, 168Lorentz invariance, 188

Time reversal, 7, 101transformation, 101, 102

Trace propertiesDirac matrices, 70

Tunnel effect, 142

Unitarity violation, 236Unitary gauge, 157, 159

Vacuum contribution, 199Vacuum state

Dirac, 108Klein-Gordon, 54

Vacuum symmetry, 146Variation

local, 49total, 49

Vector currents, 138Vertex corrections, 229, 248–251, 257Virtual particles, 204

Ward-Takahashi identity, 257, 258Weierstrass representation, 238Weinberg, 4Weyl, 5Weyl equation, 78, 113Wick’s rotation, 239Wick’s theorem, 188, 191, 194, 195

fermions, 191Wigner, 5

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