Advanced Quantum Field Theory

Roberto CasalbuoniDipartimento di Fisica

Universita di Firenze

Lezioni date all’Universita’ di Firenze nell’a.a. 2004/2005.

Contents

Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1 Notations and Conventions 41.1 Units . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.2 Relativity and Tensors . . . . . . . . . . . . . . . . . . . . . . . . . . 61.3 The Noether’s theorem for relativistic fields . . . . . . . . . . . . . . 61.4 Field Quantization . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

1.4.1 The Real Scalar Field . . . . . . . . . . . . . . . . . . . . . . . 101.4.2 The Charged Scalar Field . . . . . . . . . . . . . . . . . . . . 121.4.3 The Dirac Field . . . . . . . . . . . . . . . . . . . . . . . . . . 141.4.4 The Electromagnetic Field . . . . . . . . . . . . . . . . . . . . 17

1.5 Perturbation Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . 271.5.1 The Scattering Matrix . . . . . . . . . . . . . . . . . . . . . . 271.5.2 Wick’s theorem . . . . . . . . . . . . . . . . . . . . . . . . . . 291.5.3 Feynman diagrams in momentum space . . . . . . . . . . . . . 301.5.4 The cross-section . . . . . . . . . . . . . . . . . . . . . . . . . 33

2 One-loop renormalization 362.1 Divergences of the Feynman integrals . . . . . . . . . . . . . . . . . . 362.2 Higher order corrections . . . . . . . . . . . . . . . . . . . . . . . . . 382.3 The analysis by counterterms . . . . . . . . . . . . . . . . . . . . . . 452.4 Dimensional regularization of the Feynman integrals . . . . . . . . . . 522.5 Integration in arbitrary dimensions . . . . . . . . . . . . . . . . . . . 542.6 One loop regularization of QED . . . . . . . . . . . . . . . . . . . . . 562.7 One loop renormalization . . . . . . . . . . . . . . . . . . . . . . . . . 63

3 Vacuum expectation values and the S matrix 733.1 In- and Out-states . . . . . . . . . . . . . . . . . . . . . . . . . . . . 733.2 The S matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 793.3 The reduction formalism . . . . . . . . . . . . . . . . . . . . . . . . . 81

4 Path integral formulation of quantum mechanics 854.1 Feynman’s formulation of quantum mechanics . . . . . . . . . . . . . 854.2 Path integral in configuration space . . . . . . . . . . . . . . . . . . . 88

1

4.3 The physical interpretation of the path integral . . . . . . . . . . . . 914.4 The free particle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 974.5 The case of a quadratic action . . . . . . . . . . . . . . . . . . . . . . 994.6 Functional formalism . . . . . . . . . . . . . . . . . . . . . . . . . . . 1044.7 General properties of the path integral . . . . . . . . . . . . . . . . . 1074.8 The generating functional of the Green’s functions . . . . . . . . . . . 1134.9 The Green’s functions for the harmonic oscillator . . . . . . . . . . . 116

5 The path integral in field theory 1235.1 The path integral for a free scalar field . . . . . . . . . . . . . . . . . 1235.2 The generating functional of the connected Green’s functions . . . . . 1275.3 The perturbative expansion for the theory λϕ4 . . . . . . . . . . . . . 1295.4 The Feynman’s rules in momentum space . . . . . . . . . . . . . . . . 1355.5 Power counting in λϕ4 . . . . . . . . . . . . . . . . . . . . . . . . . . 1375.6 Regularization in λϕ4 . . . . . . . . . . . . . . . . . . . . . . . . . . . 1425.7 Renormalization in the theory λϕ4 . . . . . . . . . . . . . . . . . . . 145

6 The renormalization group 1526.1 The renormalization group equations . . . . . . . . . . . . . . . . . . 1526.2 Renormalization conditions . . . . . . . . . . . . . . . . . . . . . . . . 1566.3 Application to QED . . . . . . . . . . . . . . . . . . . . . . . . . . . 1606.4 Properties of the renormalization group equations . . . . . . . . . . . 1616.5 The coefficients of the renormalization group equations and the renor-

malization conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . 167

7 The path integral for Fermi fields 1697.1 Fermionic oscillators and Grassmann algebras . . . . . . . . . . . . . 1697.2 Integration over Grassmann variables . . . . . . . . . . . . . . . . . . 1747.3 The path integral for the fermionic theories . . . . . . . . . . . . . . . 178

8 The quantization of the gauge fields 1808.1 QED as a gauge theory . . . . . . . . . . . . . . . . . . . . . . . . . . 1808.2 Non-abelian gauge theories . . . . . . . . . . . . . . . . . . . . . . . . 1828.3 Path integral quantization of the gauge theories . . . . . . . . . . . . 1868.4 Path integral quantization of QED . . . . . . . . . . . . . . . . . . . 1968.5 Path integral quantization of the non-abelian gauge theories . . . . . 2008.6 The β-function in non-abelian gauge theories . . . . . . . . . . . . . . 204

9 Spontaneous symmetry breaking 2139.1 The linear σ-model . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2139.2 Spontaneous symmetry breaking . . . . . . . . . . . . . . . . . . . . . 2199.3 The Goldstone theorem . . . . . . . . . . . . . . . . . . . . . . . . . . 2239.4 The Higgs mechanism . . . . . . . . . . . . . . . . . . . . . . . . . . 2259.5 Quantization of a spontaneously broken gauge theory in the Rξ-gauge 230

2

9.6 ξ-cancellation in perturbation theory . . . . . . . . . . . . . . . . . . 235

10 The Standard model of the electroweak interactions 23910.1 The Standard Model of the electroweak interactions . . . . . . . . . . 23910.2 The Higgs sector in the Standard Model . . . . . . . . . . . . . . . . 24410.3 The electroweak interactions of quarks and the Kobayashi-Maskawa-

Cabibbo matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24910.4 The parameters of the SM . . . . . . . . . . . . . . . . . . . . . . . . 25810.5 Ward identities and anomalies . . . . . . . . . . . . . . . . . . . . . . 259

A 265A.1 Properties of the real antisymmetric matrices . . . . . . . . . . . . . 265

3

Chapter 1

Notations and Conventions

1.1 Units

In quantum relativistic theories the two fundamental constants c e h/ , the lightvelocity and the Planck constant respectively, appear everywhere. Therefore it isconvenient to choose a unit system where their numerical value is given by

c = h/ = 1 (1.1)

For the electromagnetism we will use the Heaviside-Lorentz system, where we takealso

ε0 = 1 (1.2)

From the relation ε0µ0 = 1/c2 it follows

µ0 = 1 (1.3)

In these units the Coulomb force is given by

|F | =e1e24π

1

|x1 − x2|2 (1.4)

and the Maxwell equations appear without any visible constant. For instance thegauss law is

∇ · E = ρ (1.5)

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

α =e2

4πε0h/c(1.6)

is given by

α =e2

4π(1.7)

4

Any physical quantity can be expressed equivalently by using as fundamentalunit energy, mass, lenght or time in an equivalent fashion. In fact from our choicethe following equivalence relations follow

ct ≈ =⇒ time ≈ lenght

E ≈ mc2 =⇒ energy ≈ mass

E ≈ pv =⇒ energy ≈ momentum

Et ≈ h/ =⇒ energy ≈ (time)−1 ≈ (lenght)−1 (1.8)

In practice, it is enough to notice that the product ch/ has dimensions [E·]. Therefore

ch/ = 3 · 108 mt · sec−1 · 1.05 · 10−34 J · sec = 3.15 · 10−26 J · mt (1.9)

Recalling that1 eV = e · 1 = 1.602 · 10−19 J (1.10)

it follows

ch/ =3.15 · 10−26

1.6 · 10−13MeV · mt = 197 MeV · fermi (1.11)

From which1 MeV−1 = 197 fermi (1.12)

Using this relation we can easily convert a quantity given in MeV (the typical unitused in elementary particle physics) in fermi. For instance, using the fact that alsothe elementary particle masses are usually given in MeV , the wave lenght of anelectron is given by

λeCompton =1

me≈ 1

0.5 MeV≈ 200 MeV · fermi

0.5 MeV≈ 400 fermi (1.13)

Therefore the approximate relation to keep in mind is 1 = 200 MeV · fermi. Fur-thermore, using

c = 3 · 1023 fermi · sec−1 (1.14)

we get1 fermi = 3.3 · 10−24 sec (1.15)

and1 MeV−1 = 6.58 · 10−22 sec (1.16)

Also, using1 barn = 10−24 cm2 (1.17)

it follows from (1.12)1 GeV−2 = 0.389 mbarn (1.18)

5

1.2 Relativity and Tensors

Our conventions are as follows: the metric tensor gµν is diagonal with eigenvalues(+1,−1,−1,−1). The position and momentum four-vectors are given by

xµ = (t, x), pµ = (E, p), µ = 0, 1, 2, 3 (1.19)

where x and p are the three-dimensional position and momentum. The scalar prod-uct between two four-vectors is given by

a · b = aµbνgµν = aµbµ = aµbµ = a0b0 − a ·b (1.20)

where the indices have been lowered by

aµ = gµνaν (1.21)

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

∂µ =∂

∂xµ=

(∂

∂t,∂

∂x

)=(∂t, ∇

)(1.22)

The four-momentum operator in position space is

pµ → i∂

∂xµ= (i∂t,−i∇) (1.23)

The following relations will be useful

p2 = pµpµ → − ∂

∂xµ∂

∂xµ= − (1.24)

x · p = Et− p · x (1.25)

1.3 The Noether’s theorem for relativistic fields

We will now review the Noether’s theorem. This allows to relate symmetries of theaction with conserved quantities. More precisely, given a transformation involvingboth the fields and the coordinates, if it happens that the action is invariant underthis transformation, then a conservation law follows. When the transformationsare limited to the fields one speaks about internal transformations. When bothtypes of transformations are involved the total variation of a local quantity F (x)(that is a function of the space-time point) is given by

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

∼= F (x) − F (x) + δxµ∂F (x)

∂xµ(1.26)

6

The total variation ∆ keeps into account both the variation of the reference frameand the form variation of F . It is then convenient to define a local variation δF ,depending only on the form variation

δF (x) = F (x) − F (x) (1.27)

Then we get

∆F (x) = δF (x) + δxµ∂F (x)

∂xµ(1.28)

Let us now start form a generic four-dimensional action

S =

∫V

d4x L(φi, x), i = 1, . . . , N (1.29)

and let us consider a generic variation of the fields and of the coordinates, x′µ =xµ + δxµ

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

∂xµ(1.30)

If the action is invariant under the transformation, then

SV ′ = SV (1.31)

This gives rise to the conservation equation

∂µ

[Lδxµ +

∂L∂φi,µ

∆φi − δxν∂L∂φi,µ

φi,ν

]= 0 (1.32)

This is the general result expressing the local conservation of the quantity in paren-thesis. According to the choice one does for the variations δxµ and ∆φi, and ofthe corresponding symmetries of the action, one gets different kind of conservedquantities.

Let us start with an action invariant under space and time translations. In thecase we take δxµ = aµ with aµ independent on x e ∆φi = 0. From the general resultin eq. (1.32) we get the following local conservation law

T µν =∂L∂φi,µ

φi,ν − Lgµν , ∂µTµν = 0 (1.33)

Tµν is called the energy-momentum tensor of the system. From its local conservationwe get four constant of motion

Pµ =

∫d3xT 0

µ (1.34)

Pµ is the four-momentum of the system. In the case of internal symmetries we takeδxµ = 0. The conserved current will be

Jµ =∂L∂φi,µ

∆φi =∂L∂φi,µ

δφi, ∂µJµ = 0 (1.35)

7

with an associated constant of motion given by

Q =

∫d3xJ0 (1.36)

In general, if the system has more that one internal symmetry, we may have morethat one conserved charge Q, that is we have a conserved charge for any ∆.

The last case we will consider is the invariance with respect to Lorentz transfor-mations. Let us recall that they are defined as the transformations leaving invariantthe norm of a four-vector

x2 = x′2 (1.37)

For an infinitesimal transformation

x′µ = Λµνxν ≈ xµ + εµνx

ν (1.38)

withεµν = −ενµ (1.39)

We see that the number of independent parameters characterizing a Lorentz trans-formation is six. As well known, three of them correspond to spatial rotations,whereas the remaining three correspond to Lorentz boosts. In general, the relativis-tic fields are chosen to belong to a representation of the Lorentz group ( for instancethe Klein-Gordon field belongs to the scalar representation). This means that undera Lorentz transformation the components of the field mix together, as, for instance,a vector field does under rotations. Therefore, the transformation law of the fieldsφi under an infinitesimal Lorentz transformation can be written as

∆φi = −1

2Σijµνε

µνφj (1.40)

where we have required that the transformation of the fields is of first order in theLorentz parameters εµν . The coefficients Σµν (antisymmetric in the indices (µ, ν))define a matrix in the indices (i, j) which can be shown to be the representative ofthe infinitesimal generators of the Lorentz group in the field representation. Usingthis equation and the expression for δxµ we get the local conservation law

0 = ∂µ

[(∂L∂φi,µ

φi,ν −Lgµν)ενρxρ +

1

2

∂L∂φi,µ

Σijνρε

νρφj]

=1

2ενρ∂µ

[(T µν xρ − T µρ xν

)+

∂L∂φi,µ

Σijνρφ

j

](1.41)

and defining (watch at the change of sign)

Mµρν = xρT

µν − xνT

µρ − ∂L

∂φi,µΣijρνφ

j (1.42)

8

it follows the existence of six locally conserved currents (one for each Lorentz trans-formation)

∂µMµνρ = 0 (1.43)

and consequently six constants of motion (notice that the lower indices are antisym-metric)

Mνρ =

∫d3xM0

νρ (1.44)

Three of these constants (the ones with ν and ρ assuming spatial values) are nothingbut the components of the angular momentum of the field.

1.4 Field Quantization

The quantization procedure for a field goes through the following steps:

• construction of the lagrangian density and determination of the canonical mo-mentum density

Πα(x) =∂L

∂φα(x)(1.45)

where the index α describes both the spin and internal degrees of freedom

• quantization through the requirement of the equal time canonical commutationrelations for a spin integer field or through canonical anticommutation relationsfor half-integer fields

[φα(x, t),Πβ(y, t)]± = iδαβδ3(x− y) (1.46)

[φα(x, t), φβ(y, t)]± = 0, [Πα(x, t),Πβ(y, t)]± = 0 (1.47)

In the case of a free field, or for an interacting field in the interaction picture (andtherefore evolving with the free hamiltonian), we can add the following steps

• expansion of φα(x) in terms of a complete set of solutions of the Klein-Gordonequation, allowing the definition of creation and annihilation operators;

• construction of the Fock space through the creation and annihilation operators.

In the following we will give the main results about the quantization of the scalar,the Dirac and the electromagnetic field.

9

1.4.1 The Real Scalar Field

The relativistic real scalar free field φ(x) obeys the Klein-Gordon equation

( +m2)φ(x) = 0 (1.48)

which can be derived from the variation of the action

S =

∫d4xL (1.49)

where L is the lagrangian density

L =1

2

[∂µφ ∂

µφ−m2φ2]

(1.50)

giving rise to the canonical momentum

Π =∂L∂φ

= φ (1.51)

From this we get the equal time 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 (1.52)

By using the box normalization, a complete set of solutions of the Klein-Gordonequation is given by

fk(x) =1√V

1√2ωk

e−ikx, f ∗k(x) =

1√V

1√2ωk

eikx (1.53)

corresponding respectively to positive and negative energies, with

k0 = ωk =

√|k|2 +m2 (1.54)

andk =

2π

Ln (1.55)

withn = n1

i1 + n2i2 + n3

i3, ni ∈ ZZ (1.56)

and L being the side of the quantization box (V = L3). For any two solutions f andg of the Klein-Gordon equation, the following quantity is a constant of motion anddefines a scalar product (not positive definite)

〈f |g〉 = i

∫d3xf ∗ ∂(−)

t g (1.57)

wheref ∗ ∂(−)

t g = f ∗g − f ∗g (1.58)

10

The solutions of eq. (1.53) are orthogonal with respect to this scalar product, forinstance

〈fk|fk′〉 = δk,k′ ≡3∏i=1

δni,n′i

(1.59)

We can go to the continuum normalization by the substitution

1√V

→ 1√(2π)3

(1.60)

and sending the Kronecker delta function into the Dirac delta function. The expan-sion of the field in terms of these solutions is then (in the continuum)

φ(x) =

∫d3k[fk(x)a(

k) + f ∗k(x)a†(k)] (1.61)

or, more explicitly

φ(x) =

∫d3k

1√(2π)32ωk

[a(k)e−ikx + a†(k)eikx] (1.62)

using the orthogonality properties of the solutions one can invert this expression

a(k) = i

∫d3xf ∗

k(x)∂

(−)t φ(x), a†(k) = i

∫d3xφ(x)∂

(−)t fk(x) (1.63)

and obtain the commutation relations

[a(k), a†(k′)] = δ3(k − k′) (1.64)

[a(k), a(k′)] = [a†(k), a†(k′)] = 0 (1.65)

From these equations one constructs the Fock space starting from the vacuum statewhich is defined by

a(k)|0〉 = 0 (1.66)

for any k. The statesa†(k1) · · ·a†(kn)|0〉 (1.67)

describe n Bose particles of equal mass m2 = k2i , and of total four-momentum

k = k1 + · · · kn. This follows by recalling the expression (1.33) for the energy-momentum tensor, that for the Klein-Gordon field is

Tµν = ∂µφ∂νφ− 1

2

(∂ρφ∂

ρφ−m2φ2)gµν (1.68)

from which

P 0 = H =

∫d3xT 00 =

1

2

∫d3x

(Π2 + |∇φ|2 +m2φ2

)(1.69)

11

P i =

∫d3xT 0i = −

∫d3xΠ

∂φ

∂xi, (P = −

∫d3xΠ∇φ) (1.70)

and for the normal ordered operator Pµ we get (the normal ordering is defined byputting all the creation operators to the left of the annihilation operators)

: Pµ :=

∫d3k kµa

†(k)a(k) (1.71)

implying: Pµ : a†(k)|0〉 = kµa

†(k)|0〉 (1.72)

The propagator for the scalar field is given by

〈0|T (φ(x)φ(y))|0〉 = −i∆F (x− y) (1.73)

where

∆F (x) = −∫

d4k

(2π)4e−ikx∆F (k) (1.74)

and

∆F (k) =1

k2 −m2 + iε(1.75)

1.4.2 The Charged Scalar Field

The charged scalar field can be described in terms of two real scalar fields of equalmass

L =1

2

2∑i=1

[(∂µφi)(∂

µφi) −m2φ2i

](1.76)

and we can write immediately the canonical commutation relations

[φi(x, t), φj(y, t)] = iδijδ3(x− y) (1.77)

[φi(x, t), φj(y, t)] = [φi(x, t), φj(y, t)] = 0 (1.78)

The charged field is better understood in a complex basis

φ =1√2(φ1 + iφ2), φ† =

1√2(φ1 − iφ2) (1.79)

The lagrangian density becomes

L = ∂µφ†∂µφ−m2φ†φ (1.80)

The theory is invariant under the phase transformation φ→ eiαφ, and therefore itadmits the conserved current (see eq. (1.35))

jµ = i[(∂µφ)φ† − (∂µφ

†)φ]

(1.81)

12

with the corresponding constant of motion

Q = i

∫d3x φ†∂(−)

t φ (1.82)

The commutation relations in the new basis are given by

[φ(x, t), φ†(y, t)] = iδ3(x− y) (1.83)

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 (1.84)

Let us notice that these commutation relations could have also been obtained directlyfrom the lagrangian (1.80), since

Πφ =∂L∂φ

= φ†, Πφ† =∂L∂φ† = φ (1.85)

Using the expansion for the real fields

φi(x) =

∫d3k

[fk(x)ai(

k) + f ∗k(x)a†i(k)

](1.86)

we get

φ(x) =

∫d3k

[fk(x)

1√2(a1(k) + ia2(k)) + f ∗

k(x)

1√2(a†1(k) + ia†2(k))

](1.87)

Introducing the combinations

a(k) =1√2(a1(k) + ia2(k)), b(k) =

1√2(a1(k) − ia2(k)) (1.88)

it follows

φ(x) =∫d3k

[fk(x)a(

k) + f ∗k(x)b†(k)

]φ†(x) =

∫d3k

[fk(x)b(

k) + f ∗k(x)a†(k)

](1.89)

from which we can evaluate the commutation relations for the creation and annihi-lation operators in the complex basis

[a(k), a†(k′)] = [b(k), b†(k′)] = δ3(k − k′) (1.90)

[a(k), b(k′)] = [a(k), b†(k′)] = 0 (1.91)

13

We get also

: Pµ :=

∫d3kkµ

2∑i=1

a†i(k)ai(k) =

∫d3kkµ

[a†(k)a(k) + b†(k)b(k)

](1.92)

Therefore the operators a†(k) e b†(k) both create particles states with momentumk, as the original operators a†i. For the normal ordered charge operator we get

: Q :=

∫d3k

[a†(k)a(k) − b†(k)b(k)

](1.93)

showing explicitly that a† and b† create particles of charge +1 and −1 respectively.The only non vanishing propagator for the scalar field is given by

〈0|T (φ(x)φ†(y))|0〉 = −i∆F (x− y) (1.94)

1.4.3 The Dirac Field

The Dirac field is described by the action

S =

∫V

d4x ψ(i∂ −m)ψ (1.95)

where we have used the following notations for four-vectors contracted with the γmatrices

v ≡ vµγµ (1.96)

The canonical momenta result to be

Πψ =∂L∂ψ

= iψ†, Πψ† =∂L∂ψ† = 0 (1.97)

The canonical momenta do not depend on the velocities. In principle, this createsa problem in going to the hamiltonian formalism. In fact a rigorous treatmentrequires an extension of the classical hamiltonian approach which was performed byDirac himself. In this particular case, the result one gets is the same as proceedingin a naive way. For this reason we will avoid to describe this extension, and wewill proceed as in the standard case. From the general expression for the energymomentum tensor (see eq. 1.33) we get

T µν = iψγµψ,ν − gµν (ψ(i∂ −m)ψ) (1.98)

and using the Dirac equationT µν = iψγµψ,ν (1.99)

The momentum of the field is given by

P k =

∫d3x T 0k =⇒ P = −i

∫d3x ψ†∇ψ (1.100)

14

and the hamiltonian

H =

∫d3x T 00 =⇒ H = i

∫d3x ψ†∂tψ (1.101)

For a Lorentz transformation, the field variation is defined by

∆φi = −1

2Σijµνε

µνφj (1.102)

In the Dirac case one has

∆ψ(x) = ψ′(x′) − ψ(x) = − i

4σµνε

µνψ(x) (1.103)

from which

Σµν =i

2σµν = −1

4[γµ, γν] (1.104)

Therefore the angular momentum density (see eq. (1.42)) is

Mµρν = iψγµ

(xρ∂ν − xν∂ρ − i

2σρν

)ψ = iψγµ

(xρ∂ν − xν∂ρ +

1

4[γρ, γν]

)ψ

(1.105)By taking the spatial components we obtain

J = (M23,M31,M12) =

∫d3x ψ†

(−ix ∧ ∇ +

1

2σ ⊗ 12

)ψ (1.106)

where 12 is the identity matrix in 2 dimensions, and we have defined

σ ⊗ 12 =

(σ 00 σ

)(1.107)

The expression of J shows the decomposition of the total angular momentum inthe orbital and in the spin part. The theory has a further conserved quantity, the

current ψγµψ resulting from the phase invariance ψ → eiαψ.The decomposition of the Dirac field in plane waves is obtained by using the

spinors u(p,±n) e v(p,±n) solutions of the Dirac equation in momentum space forpositive and negative energies respectively. We write

ψ(x) =∑±n

∫d3 p√(2π)3

√m

Ep

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

](1.108)

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

(1.109)where Ep =

√|p|2 +m2. We will collect here the various properties of the spinors.The four-vector nµ is the one that identifies the direction of the spin quantization inthe rest frame. For instance, by quantizing the spin along the third axis one takesnµ in the rest frame as nµR = (0, 0, 0, 1). Then nµ is obtained by boosting from therest frame to the frame where the particle has four-momentum pµ.

15

• Dirac equation

(p−m)u(p, n) = u(p, n)(p−m) = 0

(p+m)v(p, n) = v(p, n)(p+m) = 0 (1.110)

• 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 (1.111)

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

∑±n

u(p, n)u(p, n) =p+m

2m∑±n

v(p, n)v(p, n) =p−m

2m(1.112)

The Dirac field is quantized by canonical anticommutation relations in order tosatisfy the Pauli principle

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

]+

= 0 (1.113)

and[Πψ(x, t), ψ(y, t)]+ = iδ3(x− y) (1.114)

or [ψ(x, t), ψ†(y, t)

]+

= δ3(x− y) (1.115)

The normal ordered four-momentum is given by

: Pµ :=∑±n

∫d3p pµ[b

†(p, n)b(p, n) + d†(p, n)d(p, n)] (1.116)

If we couple the Dirac field to the electromagnetism through the minimal substitu-tion we find that the free action (1.95) becomes

S =

∫V

d4xψ(i∂ − eA−m)ψ (1.117)

Therefore the electromagnetic field is coupled to the conserved current

jµ = eψγµψ (1.118)

16

This forces us to say that the integral of the fourth component of the current shouldbe the charge operator. In fact, we find

: Q : = e

∫d3x ψ†ψ

=∑±n

∫d3p e

[b†(p, n)b(p, n) − d†(p, n)d(p, n)

](1.119)

The expressions for the momentum angular momentum and charge operators showthat the operators b†(p) and d†(p) create out of the vacuum particles of spin 1/2,four-momentum pµ and charge e and −e respectively.

The propagator for the Dirac field is given by

〈0|T (ψα(x)ψβ(y))|0〉 = iSF (x− y)αβ (1.120)

where

SF (x) = −(i∂ +m)∆F (x) =

∫d4k

(2π)4e−ikxSF (k) (1.121)

and

SF (k) =k +m

k2 −m2 + iε=

1

k −m+ iε(1.122)

1.4.4 The Electromagnetic Field

The lagrangian density for the free electromagnetic field, expressed in terms of thefour-vector potential is

L = −1

4FµνF

µν (1.123)

whereFµν = ∂µAν − ∂νAµ (1.124)

and

E = −∇A0 − ∂ A

∂t, B = ∇∧ A (1.125)

from whichE = (F 10, F 20, F 30), B = (−F 23,−F 31,−F 12) (1.126)

The resulting equations of motion are

Aµ − ∂µ(∂νAν) = 0 (1.127)

The potentials are defined up to a gauge transformation

Aµ(x) → A′µ(x) = Aµ(x) + ∂µΛ(x) (1.128)

In fact, Aµ and A′µ satisfy the same equations of motion and give rise to the same

electromagnetic field. It is possible to use the gauge invariance to require some

17

particular condition on the field Aµ. For instance, we can perform a gauge transfor-mation in such a way that the transformed field satisfies

∂µAµ = 0 (1.129)

This is called the Lorentz gauge. Or we can choose a gauge such that

A0 = ∇ · A = 0 (1.130)

This is called the Coulomb gauge. Since in this gauge all the gauge freedom is com-pletely fixed (in contrast with the Lorentz gauge), it is easy to count the independentdegrees of freedom of the electromagnetic field. From the equations (1.130) we seethat Aµ has only two degrees of freedom. Another way of showing that there areonly two independent degrees of freedom is through the equations of motion. Letus consider the four dimensional Fourier transform of Aµ(x)

Aµ(x) =

∫d4k eikxAµ(k) (1.131)

Substituting this expression in the equations of motion we get

−k2Aµ(k) + kµ(kνAν(k)) = 0 (1.132)

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 (1.133)

The decomposition of Aµ(k) reads

Aµ(k) = aλ(k)eλµ + b(k)kµ + c(k)kµ (1.134)

From the equations of motion we get

−k2(aλeλµ + bkµ + ckµ) + kµ(bk

2 + c(k · k)) = 0 (1.135)

The term in b(k) cancels, therefore it is left undetermined by the equations of motion.For the other quantities we have

k2aλ(k) = c(k) = 0 (1.136)

The arbitrariness of b(k) is a consequence of the gauge invariance. In fact if wegauge transform Aµ(x)

Aµ(x) → Aµ(x) + ∂µΛ(x) (1.137)

thenAµ(k) → Aµ(k) + ikµΛ(k) (1.138)

18

where

Λ(x) =

∫d4k eikxΛ(k) (1.139)

Since the gauge transformation amounts to a translation in b(k) by an arbitraryfunction of k, we can always to choose it equal to zero. Therefore we are left with thetwo degrees of freedom described by the amplitudes aλ(k), λ = 1, 2. Furthermorethese amplitudes are different from zero only if the dispersion relation k2 = 0 issatisfied. This shows that the corresponding quanta will have zero mass. With thechoice b(k) = 0, the field Aµ(k) becomes

Aµ(k) = aλ(k)eλµ(k) (1.140)

showing that kµAµ(k) = 0. Therefore the choice b(k) = 0 is equivalent to the choiceof the Lorentz gauge.

Let us consider now the quantization of this theory. Due to the lack of manifestcovariance of the Coulomb gauge we will discuss here the quantization in the Lorentzgauge, where however we will encounter other problems. If we want to maintain theexplicit covariance of the theory we have to require non trivial commutation relationsfor all the component of the field. That is

[Aµ(x, t),Πν(y, t)] = igνµδ

3(x− y) (1.141)

[Aµ(x, t), Aν(y, t)] = [Πµ(x, t),Πν(y, t)] = 0 (1.142)

with

Πµ =∂L∂Aµ

(1.143)

To evaluate the conjugated momenta is better to write the lagrangian density (seeeq. (1.123) in the following form

L = −1

4[Aµ,ν −Aν,µ][A

µ,ν −Aν,µ] = −1

2Aµ,νA

µ,ν +1

2Aµ,νA

ν,µ (1.144)

Therefore∂L∂Aµ,ν

= −Aµ,ν + Aν,µ = F µν (1.145)

implying

Πµ =∂L∂Aµ

= F µ0 (1.146)

It follows

Π0 =∂L∂A0

= 0 (1.147)

We see that it is impossible to satisfy the condition

[A0(x, t),Π0(y, t)] = iδ3(x− y) (1.148)

19

We can try to find a solution to this problem modifying the lagrangian density insuch a way that Π0 = 0. But doing so we will not recover the Maxwell equation.However we can take advantage of the gauge symmetry, modifying the lagrangiandensity in such a way to recover the equations of motion in a particular gauge. Forinstance, in the Lorentz gauge we have

Aµ(x) = 0 (1.149)

and this equation can be obtained by the lagrangian density

L = −1

2Aµ,νA

µ,ν (1.150)

(just think to the Klein-Gordon case). The minus sign is necessary to recover apositive hamiltonian density. We now express this lagrangian density in terms ofthe gauge invariant one, given in eq. (1.123). To this end we observe that thedifference between the two lagrangian densities is nothing but the second term ofeq. (1.144)

1

2Aµ,νA

ν,µ = ∂µ[1

2Aµ,νA

ν

]− 1

2(∂µAµ,ν)A

ν

= ∂µ[1

2Aµ,νA

ν

]− ∂ν

[1

2(∂µAµ)Aν

]+

1

2(∂µAµ)

2 (1.151)

Then, up to a four divergence, we can write the new lagrangian density in the form

L = −1

4FµνF

µν − 1

2(∂µAµ)

2 (1.152)

One can check that this form gives the correct equations of motion. In fact from

∂L∂Aµ,ν

= −Aµ,ν + Aν,µ − gµν(∂λAλ),∂L∂Aµ

= 0 (1.153)

we get0 = −Aµ + ∂µ(∂νAν) − ∂µ(∂λAλ) = −Aµ (1.154)

the term

−1

2(∂µAµ)

2 (1.155)

which is not gauge invariant, is called the gauge fixing term. More generally, wecould add to the original lagrangian density a term of the form

−λ2(∂µAµ)

2 (1.156)

The corresponding equations of motion would be

Aµ − (1 − λ)∂µ(∂λAλ) = 0 (1.157)

20

In the following we will use λ = 1. From eq. (1.153) we see that

Π0 =∂L∂A0

= −∂µAµ (1.158)

In the Lorentz gauge we find again Π0 = 0. To avoid the corresponding problemwe can ask that ∂µAµ = 0 does not hold as an operator condition, but rather as acondition upon the physical states

〈phys|∂µAµ|phys〉 = 0 (1.159)

The price to pay to quantize the theory in a covariant way is to work in a Hilbertspace much bigger than the physical one. The physical states span a subspace whichis defined by the previous relation. A further bonus is that in this way one has todo with local commutation relations. On the contrary, in the Coulomb gauge, oneneeds to introduce non local commutation relations for the canonical variables. Wewill come back later to the condition (1.159).

Since we don’t have to worry any more about the operator condition Π0 = 0, wecan proceed with our program of canonical quantization. The canonical momentumdensities are

Πµ =∂L∂Aµ

= F µ0 − gµ0(∂λAλ) (1.160)

or, explicitly

Π0 = −∂λAλ = −A0 − ∇ · AΠi = ∂iA0 − ∂0Ai = −Ai + ∂iA0 (1.161)

Since the spatial gradient of the field commutes with the field itself at equal time,the canonical commutator (1.141) gives rise to

[Aµ(x, t), Aν(y, t)] = −igµνδ3(x− y) (1.162)

To get the quanta of the field we look for plane wave solutions of the wave equation.We need four independent four vectors in order to expand the solutions in themomentum space. In a given frame, let us consider the unit four vector whichdefines the time axis. This must be a time-like vector, n2 = 1, and we will choosen0 > 0. For instance, nµ = (1, 0, 0, 0). Then we take two four vectors ε

(λ)µ , λ = 1, 2,in

the plane orthogonal to nµ and kµ. Notice that now k2 = 0,since we are consideringsolutions of the wave equation. Therefore

kµε(λ)µ = nµε(λ)

µ = 0, λ = 1, 2 (1.163)

The four vectors ε(λ)µ , being orthogonal to nµ are space-like, then they will be chosen

orthogonal and normalized in the following way

ε(λ)µ ε(λ

′)µ = −δλλ′ (1.164)

21

Next, we define a unit space-like four vector, orthogonal to nµ and lying in the plane(k, n)

nµε(3).µ = 0 (1.165)

withε(3)µ ε(3)

µ= −1 (1.166)

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) (1.167)

As a last unit four vector we choose nµ

ε(0)µ = nµ (1.168)

These four vectors are orthonormal

ε(λ)µ ε(λ

′)µ = gλλ′

(1.169)

and linearly independent. Then they satisfy the completeness relation

ε(λ)µ ε(λ

′)ν gλλ′ = gµν (1.170)

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) (1.171)

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]

(1.172)

where we have included the hermiticity condition for Aµ(x). For any fixed µ, thisexpansion is the same as the one that we wrote for the Klein-Gordon field, with thesubstitution ε

(λ)µ aλ(k) → a(k). Then, from eq. (1.63)

ε(λ)µ (k)aλ(k) = i

∫d3x f ∗

k(x)∂

(−)t Aµ(x) (1.173)

with the functions fk(x) defined as in eq. (1.53). Using the orthogonality of the

ε(λ′)µ’s we find

aλ(k) = igλλ′

∫d3x ε(λ

′)µ(k)f ∗k(x)∂

(−)t Aµ(x) (1.174)

and analogously

a†λ(k) = igλλ′

∫d3x ε(λ

′)µ(k)Aµ(x)∂(−)t fk(x) (1.175)

22

From these expressions we can evaluate the commutator

[aλ(k), a†λ′(k

′)] =

=

∫d3x d3y

[− f ∗

k(x) ˙fk′(y)

(igµνgλλ′′ε

(λ′′)µ(k)ε(λ′′′)ν(k′)gλ′λ′′′δ3(x− y)

)− f ∗

k(x)fk′(y)

(−igµνgλλ′′ε(λ′′)µ(k)ε(λ′′′)ν(k′)gλ′λ′′′δ3(x− y)

) ]= −

∫d3x f ∗

k(x)i∂

(−)t fk′(x)gλλ′ (1.176)

and using the orthogonality relations for the functions f)k

[aλ(k), a†λ′(k

′)] = −gλλ′δ3(k − k′) (1.177)

Analogously[aλ(k), aλ′(k

′)] = [a†λ(k), a†λ′(k

′)] = 0 (1.178)

The commutation rules we have derived for the operators aλ(k) create some problem.Let us consider a one-particle state

|1, λ〉 =

∫d3k f(k)a†λ(k)|0〉 (1.179)

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 (1.180)

Therefore the states with λ = 0 have negative norm. This problem does not comeout completely unexpected. In fact, our expectation is that only the transversestates (λ = 1, 2), are physical states. For the moment being we have ignored thegauge fixing condition 〈phys|∂µAµ|phys〉 = 0, but its meaning is that only part ofthe total Hilbert space is physical. Therefore the relevant thing is to show that thestates satisfying the Lorentz condition have positive norm. To discuss the gaugefixing condition, let us notice that formulated in the way we did, being bilinear inthe states, it could destroy the linearity of the Hilbert space. So we will try tomodify the condition in a linear one

∂µAµ|phys.〉 = 0 (1.181)

But this would be a too strong requirement. Not even the vacuum state satisfies it.However, if we consider the positive and negative frequency parts 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))† (1.182)

23

it is possible to weaken the condition, and require

∂µA(+)µ (x)|phys.〉 = 0 (1.183)

This allows us to satisfy automatically the original requirement

〈phys.|(∂µA(+)µ + ∂µA(−)

µ )|phys.〉 = 0 (1.184)

To make this condition more explicit let us evaluate the four divergence of A(+)µ

i∂µA(+)µ (x) =

∫d3k√

2ωk(2π)3e−ikx

∑λ=0,3

kµε(λ)µ (k)aλ(k) (1.185)

Using eq. (1.167), we get

kµε(3)µ = −(n · k), kµε(0)µ = (n · k) (1.186)

from which[a0(k) − a3(k)]|phys.〉 = 0 (1.187)

Notice that

[a0(k) − a3(k), a†0(k

′) − a†3(k′)] = −δ3(k − k′) + δ3(k − k′) = 0 (1.188)

Let us denote by Φk(n0, n3) the state with n0 scalar photons (that is with 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 (1.183)

Φ(m)k

=1

m!(a†0(k) − a†3(k))

mΦk(0, 0) (1.189)

These states have vanishing norm

||Φ(m)k

||2 = 0 (1.190)

More generally we can make the following observation. Let us consider the numberoperator for scalar and longitudinal photons

N =

∫d3k (a†3(k)a3(k) − a†0(k)a0(k)) (1.191)

Notice the minus sign that is a consequence of the commutation relations, and itensures 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〉 (1.192)

24

Let us consider a physical state with a total number n of scalar and longitudinalphotons. Then

〈ϕn|N |ϕn〉 = 0 (1.193)

since a0 and a3 act in the same way on a physical state (see eq. (1.187)). It follows

n〈ϕn|ϕn〉 = 0 (1.194)

Therefore all the physical states with a total definite number of scalar and longitu-dinal photons have zero norm, except for the vacuum state (n = 0). Then

〈ϕn|ϕn〉 = δn,0 (1.195)

A generic physical state with zero transverse photons is a linear superposition of theprevious states

|ϕ〉 = c0|ϕ0〉 +∑i=0

ci|ϕi〉 (1.196)

This state has a positive definite norm

〈ϕ|ϕ〉 = |c0|2 ≥ 0 (1.197)

The proof that a physical state has a positive norm can be extended to the case inwhich also transverse photons are present. Of course, the coefficients ci, appearingin the expression of a physical state, are completely arbitrary, but this is not goingto modify the values of the observables. For instance, consider the hamiltonian, wehave

H =

∫d3x : [ΠµAµ − L] :

=

∫d3x :

[F µ0Aµ − (∂λAλ)A0 +

1

4FµνF

µν +1

2(∂λAλ)

2

]: (1.198)

One can easily show the hamiltonian is given by the sum of all the degrees of freedomappearing in Aµ

H =1

2

∫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)

]: (1.199)

Since on the physical states a0 and a3 act in the same way, we get

〈phys.|H|phys.〉 = 〈phys.|∫

d3k ωk

2∑λ=1

a†λ(k)aλ(k)|phys.〉 (1.200)

25

The generic physical state is of the form |ϕT 〉 ⊗ |ϕ〉. with |ϕ〉 defined as in eq.(1.196). Since only |ϕT 〉, contributes to the evaluation of an observable quantity ,we can always choose |ϕ〉 proportional to |ϕ0〉. However, this does not mean thatwe are always working in the restricted physical space, because in a sum over theintermediate states we need to include all the degrees of freedom. This is crucial forthe explicit covariance and locality of the theory.

The arbitrariness in defining the state |ϕ〉 has, in fact, a very simple interpreta-tion. It corresponds to add to Aµ a four gradient, that is it corresponds to performa gauge transformation. Consider the following matrix element

〈ϕ|Aµ(x)|ϕ〉 =∑n,m

cncm〈ϕn|Aµ(x)|ϕm〉 (1.201)

Since Aµ change the occupation number by one unit and all the states |ϕn〉 havezero norm (except for the state with n = 0), the only non vanishing contributionscome from n = 0, m = 1 and n = 1, m = 0

〈ϕ|Aµ(x)|ϕ〉 = c0c1〈0|∫

d3k√2ωk(2π)3

e−ikx[ε(3)µ (k)a3(k) + ε(0)µ (k)a0(k)]|ϕ1〉 + c.c.

(1.202)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〉 (1.203)

and therefore

〈ϕ|Aµ(x)|ϕ〉 =

∫d3k√

2ωk(2π)3[ε(3)µ (k) + ε(0)µ (k)][c0c1e

−ikxf(k) + c.c.] (1.204)

From eqs. (1.167) and (1.168) we have

ε(3)µ + ε(0)µ =kµ

(k · n)(1.205)

from which〈ϕ|Aµ(x)|ϕ〉 = ∂µΛ(x) (1.206)

with

Λ(x) =

∫d3k√

2ωk(2π)3

1

n · k (ic0c1e−ikxf(k) + c.c.) (1.207)

It is important to notice that this gauge transformation leaves Aµ in the Lorentzgauge, since

Λ = 0 (1.208)

26

because the momentum k inside the integral satisfies k2 = 0. From the expansionin eq. (1.172) one gets immediately the expression for the photon propagator

〈0|T (Aµ(x)Aν(y)|0〉 = −igµν∫

d4k

(2π)4

e−ikxk2 + iε

(1.209)

By defining

D(x) =

∫d4k

(2π)4e−ikxD(k) (1.210)

and

D(k) = − 1

k2 + iε(1.211)

we get〈0|T (Aµ(x)Aν(y)|0〉 = igµνD(x− y) (1.212)

1.5 Perturbation Theory

1.5.1 The Scattering Matrix

In order to describe a scattering process we will assign to the vector of state acondition at t = −∞

|Φ(−∞)〉 ≡ |Φi〉 (1.213)

where the state Φi will be specified by assigning the set of incoming free particlesin terms of eigenstates of momentum, spin and so on. For instance, in QED we willhave to specify how many electrons, positrons and photons are in the initial stateand we will have to specify their momenta, the spin projection of fermions and thepolarization of the photons. The equations of motion will tell us how this stateevolves with time and it will be possible to evaluate the state at t = +∞, where,ideally, we will detect the final states. In practice the preparation and the detectionprocesses are made at some finite times. It follows that our ideal description willbe correct only if these times are much bigger than the typical interaction time ofthe scattering process. Once we know Φ(+∞), we are interested to evaluate theprobability amplitude of detecting at t = +∞ a given set of free particles specifiedby a vector state Φf . The amplitude is

Sfi = 〈Φf |Φ(+∞)〉 (1.214)

We will define the S matrix as the operator that give us |Φ(+∞)〉 once we know|Φ(−∞)〉

|Φ(+∞)〉 = S|Φ(−∞)〉 (1.215)

The amplitude Sfi is thenSfi = 〈Φf |S|Φi〉 (1.216)

27

Therefore Sfi is the S matrix element between free states (|Φi〉 and 〈Φf | are calledin and out states respectively). In the interaction representation defined by

|Φ(t)〉 = eiH0St|ΦS(t)〉, O(t) = eiH

0StOS(t)e

−iH0St (1.217)

where the index S identifies the Schrodinger representation and H0S is the free hamil-

tonian, the state vectors satisfy the Schrodinger equation with the interaction hamil-tonian (evaluated in the interaction picture)

i∂

∂t|Φ(t)〉 = HI |Φ(t)〉 (1.218)

whereas the operators evolve with the free hamiltonian. To evaluate the S matrixwe first transform the Schrodinger equation in the interaction representation in anintegral equation

|Φ(t)〉 = |Φ(−∞)〉 − i

∫ t

−∞dt1 H

I(t1)|Φ(t1)〉 (1.219)

One can verify that this is indeed a solution, and that it satisfies explicitly theboundary condition at t = −∞. The perturbative expansion consists in evaluating|Φ(t)〉 by iterating this integral equation. The result in terms of ordered T -productsis

S = 1 +

∞∑n=1

(−i)nn!

∫ +∞

−∞dt1 · · ·

∫ +∞

−∞dtn T

(HI(t1) · · ·HI(tn)

)(1.220)

The T -product of n terms means that the factors have to be written from left toright with decreasing times. For instance, if t1 ≥ t2 ≥ · · · ≥ tn, then

T(HI(t1) · · ·HI(tn)

)= HI(t1) · · ·HI(tn) (1.221)

This result can be written in a more compact form by introducing the T -orderedexponential

S = T(e−i∫ +∞

−∞dt HI(t))

(1.222)

This expression is a symbolic one and it is really defined by its series expansion. Themotivation for introducing the T -ordered exponential is that it satisfies the followingfactorization property

T(e

∫ t3

t1

O(t)dt)= T

(e

∫ t3

t2

O(t)dt)T(e

∫ t2

t1

O(t)dt)(1.223)

If there are no derivative interactions we have

S = T(e−i∫ +∞

−∞dt HI(t))

= T(e+i

∫d4x Lint)

(1.224)

It follows that if the theory is Lorentz invariant, also the S matrix enjoys the sameproperty.

28

1.5.2 Wick’s theorem

The matrix elements of the S matrix between free particle states can be expressedas vacuum expectation values (VEV’s) of T -products. These V EV ′s satisfy animportant theorem due to Wick that states that the T -products of an arbitrarynumber of free fields (the ones we have to do in the interaction representation) canbe expressed as combinations of T -products among two fields, that is in terms ofFeynman propagators. The Wick’s theorem is summarized in the following equation

T(e−i∫

d4x j(x)φ(x))

=: e−i∫

d4x j(x)φ(x): e−1

2

∫d4x d4y j(x)j(y)〈0|T (φ(x)φ(y))|0〉

(1.225)

where φ(x) is a free real scalar field and j(x) an ordinary real function. The previousformula can be easily extended to charged scalar, fermionic and photon fields. TheWick’s theorem is then obtained by expanding both sides of this equation in powersof j(x) and taking the VEV of both sides. Let us now expand both sides of eq.(1.225) in a series of j(x) and compare term by term. We will use the simplifiednotation φi ≡ φ(xi). We get

T (φ) = : φ : (1.226)

T (φ1φ2) = : φ1φ2 : +〈0|T (φ1φ2)|0〉 (1.227)

T (φ1φ2φ3) = : φ1φ2φ3 : +

3∑i=j =k=1

: φi : 〈0|T (φjφk)|0〉 (1.228)

T (φ1φ2φ3φ4) = : φ1φ2φ3φ4 : +4∑

i=j =k =l=1

[: φiφj : 〈0|T (φkφl)|0〉

+ 〈0|T (φiφj)|0〉〈0|T (φkφl)|0〉]

(1.229)

and so on. By taking the VEV of these expression, and recalling that the VEV ofa normal product is zero, we get the Wick’s theorem. The T -product of two fieldoperators is sometimes called the contraction of the two operators. Therefore toevaluate the VEV of a T -product of an arbitrary number of free fields, it is enoughto consider all the possible contractions of the fields appearing in the T -product.For instance, from the last of the previous relations we get

〈0|T (φ1φ2φ3φ4)|0〉 =4∑

i=j =k =l=1

〈0|T (φiφj)|0〉〈0|T (φkφl)|0〉 (1.230)

29

An analogous theorem holds for the photon field. For the fermions one has toremember that the T -product is defined in a slightly different way. This gives aminus sign any time we have a permutation of the fermion fields which is odd withrespect to the original ordering. As an illustration the previous formula becomes

〈0|T (ψ1ψ2ψ3ψ4)|0〉 =4∑

i=j =k =l=1

σP 〈0|T (ψiψj)|0〉〈0|T (ψkψl)|0〉 (1.231)

where σP = ±1 is the sign of the permutation (i, j, k, l) with respect to the funda-mental one (1, 2, 3, 4) appearing on the right 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〉 (1.232)

1.5.3 Feynman diagrams in momentum space

We will discuss here the Feynman rules for the scalar and spinor QED. Let us recallthat the electromagnetic interaction is introduced via the minimal substitution

∂µ → ∂µ + ieAµ (1.233)

For instance, for the charged scalar field we get

Lfree = ∂µφ†∂µφ−m2φ†φ− 1

4FµνF

µν →

→ [(∂µ + ieAµ)φ)]† [(∂µ + ieAµ)φ] −m2φ†φ− 1

4FµνF

µν (1.234)

The interaction part is given by

Lint. = −ie [φ†∂µφ− (∂µφ†)φ]Aµ + e2A2φ†φ (1.235)

Therefore the electromagnetic field is coupled to the current

jµ = ie[φ†∂µφ− (∂µφ

†)φ]

(1.236)

but another interaction term appears (the seagull term). In the spinor case we havea simpler situation

Lfree = ψ(i∂ −m)ψ − 1

4FµνF

µν → ψ(i∂ − eA−m)ψ − 1

4FµνF

µν (1.237)

giving the interaction termLint = −eψγµψAµ (1.238)

30

In a typical experiment in particle physics we prepare beams of particles with def-inite momentum, polarization, etc. At the same time we measure momenta andpolarizations of the final states. For this reason it is convenient to work in a mo-mentum representation. The result for a generic matrix element of the S matrix canbe expressed in the following form

〈f |S|i〉 = (2π)4δ4(∑in

pin −∑out

pout)∏

ext. bosons

√1

2EV

∏ext. fermions

√m

V EM (1.239)

Here f = i, and we are using the box normalization. The Feynman amplitude M(often one defines a matrix T wich differs from M by a factor i) can be obtained bydrawing at a given perturbative order all the connected and topologically inequiv-alent Feynman’s diagrams. The diagrams are obtained by combining together thegraphics elements given in Figures 1.5.1 and 1.5.2. The amplitude M is given by thesum of the amplitudes associated to these diagrams, after extraction of the factor(2π)4δ4(

∑p) in (1.239).

.

Fermion line

Photon line

Scalar line

Fig. 1.5.1 - Graphical representation for both internal and external particle lines.

.

QED vertex

Scalar QED seagull

Fig. 1.5.2 - Graphical representation for the vertices in scalar and spinor QED.

31

At each diagram we associate an amplitude obtained by multiplying together thefollowing factors

• for each ingoing and/or outgoing external photon line a factor ελµ (or its com-plex conjugate for outgoing photons if we consider complex polarizations asthe circular one)

• for each ingoing and/or outgoing external fermionic line a factor u(p) and/oru(p)

• for each ingoing and/or outgoing external antifermionic line a factor v(p)and/or v(p)

• for each internal scalar line a factor i∆F (p) (propagator)

• for each internal photon line a factor igµνD(p) (propagator)

• for each internal fermion line a factor iSF (p) (propagator)

• for each vertex in scalar QED a factor −ie(p + p′)(2π)4δ4(∑

in pi −∑

out pf),where p and p′ are the momenta of the scalar particle (taking p ingoing and p′

outgoing)

• For each seagull term in scalar QED a factor 2ie2gµν(2π)4δ4(∑

in pi−∑

out pf )

• for each vertex in spinor QED a factor −ieγµ(2π)4δ4(∑

ingoing pi−∑

outgoing pf )

• Integrate over all the internal momenta with measure d4p/(2π)4. This givesrise automatically to the conservation of the total four-momentum.

• In a graph containing closed lines (loops), a factor (−1) for each fermionicloop (see Fig. 1.5.3)

.

Fermion loop

Fig. 1.5.3 - A fermion loop can appear inside any Feynman diagram (attaching therest of the diagram by photonic lines).

In the following we will use need also the rules for the interaction with a staticexternal e.m. field. The correspending graphical element is given in Fig. 1.5.4. The

32

rule is to treat it as a normal photon, except that one needs to substitute the wavefunction of the photon (

1

2EV

)1/2

ε(λ)µ (q) (1.240)

with the Fourier transform of the external field

Aextµ (q ) =

∫d3qe−iq · xAext

µ (x ) (1.241)

where q is the momentum in transfer at the vertex where the external field line isattached to. Also, since the static external field breaks translational invariance, weloose the factor (2π)3δ(

∑in pin −

∑out pout).

.

Fig. 1.5.4 - Graphical representation for an external electromagnetic field. Theupper end of the line can be attached to any fermion line.

1.5.4 The cross-section

Let us consider a scattering process with a set of initial particles with four momentapi = (Ei, pi) which collide and produce a set of final particles with four momentapf = (Ef , pf). From the rules of the previous Chapter we know that each externalphoton line contributes with a factor (1/2V E)1/2, whereas each external fermionicline contributes with (m/V E)1/2. Furthermore the conservation of the total fourmomentum gives a term (2π)4δ4(

∑i pi −

∑f pf). If we exclude the uninteresting

case f = i we can write

Sfi = (2π)4δ4

(∑i

pi −∑f

pf

) ∏fermioni

( m

V E

)1/2 ∏bosoni

(1

2V E

)1/2

M (1.242)

where M is the Feynman amplitude which can be evaluated by using the rules ofthe previous Section.

Let us consider the typical case of a two particle collision giving rise to an Nparticles final state. Therefore, the probability per unit time of the transition isgiven by w = |Sfi|2 with

w = V (2π)4δ4(Pf − Pi)∏i

1

2V Ei

∏f

1

2V Ef

∏fermioni

(2m)|M|2 (1.243)

33

where, for reasons of convenience, we have written

m

V E= (2m)

1

2V E(1.244)

w gives the probability per unit time of a transition toward a state with well definedquantum numbers, but we are rather interested to the final states having momentabetween pf and pf + dpf . Since in the volume V the momentum is given by p =2πn/L, the number of final states is given by

(L

2π

)3

d3p (1.245)

The cross-section is defined as the probability per unit time divided by the flux ofthe ingoing particles, and has the dimensions of a length to the square

[t−12t] = [2] (1.246)

In fact the flux is defined as vrelρ = vrel/V , since in our normalization we have aparticle in the volume V , and vrel is the relative velocity of the ingoing particles.For the bosons this follows from the normalization condition of the functions fk(x).For the fermions recall that in the box normalization the wave function is√

m

EVu(p)e−ipx (1.247)

from which ∫V

d3xρ(x) =

∫V

d3xm

V Eu†(p)u(p) = 1 (1.248)

Then the cross-section for getting the final states with momenta between pf e pf+dpfis given by

dσ = wV

vreldNF = w

V

vrel

∏f

V d3pf(2π)3

(1.249)

We obtain

dσ =V

vrel

∏f

V d3pf(2π)3

V (2π)4δ4(Pf − Pi)1

4V 2E1E2

∏f

1

2V Ef

∏fermioni

(2m)|M|2

= (2π)4δ4(Pf − Pi)1

4E1E2vrel

∏fermioni

(2m)∏f

d3pf(2π)32Ef

|M|2 (1.250)

Notice that the dependence on the quantization volume V disappears, as it shouldbe, in the final equation. Furthermore, the total cross-section, which is obtained byintegrating the previous expression over all the final momenta, is Lorentz invariant.

34

In fact, as it follows from the Feynman rules, M is invariant, as the factors d3p/2E.Furthermore

vrel = v1 − v2 =p1

E1

− p2

E2

(1.251)

but in the frame where the particle 2 is at rest (laboratory frame) we have p2 =(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 (1.252)

We see that also this factor is Lorentz invariant.

35

Chapter 2

One-loop renormalization

2.1 Divergences of the Feynman integrals

Let us consider the Coulomb scattering. If we expand the S matrix up to thethird order in the electric charge, and we assume that the external field is weakenough such to take only the first order, we can easily see that the relevant Feynmandiagrams are the ones of Fig. 2.1.1

. .

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. 2.1.1 - The Feynman diagram for the Coulomb scattering at the third orderin the electric charge and at the first order in the external field.

36

The Coulomb scattering can be used, in principle, to define the physical electriccharge of the electron. This is done assuming that the amplitude is linear in ephys,from which we get an expansion of the type

ephys = e+ a2e3 + · · · = e(1 + a2e

2 + · · ·) (2.1)

in terms of the parameter e which appears in the original lagrangian. The firstproblem we encounter is that we would like to have the results of our calculation interms of measured quantities as ephys. This could be done by inverting the previousexpansion, but, and here comes the second problem, the coefficient of the expan-sion are divergent quantities. To show this, consider, for instance the self-energycontribution to one of the external photons (as the one in Fig. 2.1.1a). We have

Ma = u(p′)(−ieγµ)Aextµ (p ′ − p)

i

p−m+ iε

[ie2Σ(p)

]u(p) (2.2)

where

ie2Σ(p) =

∫d4k

(2π)4(−ieγµ) −ig

µν

k2 + iε

i

p− k −m+ iε(−ieγν)

= −e2∫

d4k

(2π)4γµ

1

k2 + iε

1

p− k −m+ iεγµ (2.3)

or

Σ(p) = i

∫d4k

(2π)4γµ

p− k +m

(p− k)2 −m2 + iεγµ

1

k2 + iε(2.4)

For large momentum, k, the integrand behaves as 1/k3 and the integral divergeslinearly. Analogously one can check that all the other third order contributionsdiverge. Let us write explicitly the amplitudes for the other diagrams

Mb = u(p′)ie2Σ(p′)i

p ′ −m+ iε(−ieγµ)Aext

µ (p ′ − p)u(p) (2.5)

Mc = u(p′)(−ieγµ) −igµνq2 + iε

ie2Πνρ(q)Aextρ (p ′ − p)u(p), q = p′ − p (2.6)

where

ie2Πµν(q) = (−1)

∫d4k

(2π)4Tr

[i

k + q −m+ iε(−ieγµ) i

k −m+ iε(−ieγν)

](2.7)

(the minus sign originates from the fermion loop) and therefore

Πµν(q) = i

∫d4k

(2π)4Tr

[1

k + q −mγµ

1

k −m+ iεγν]

(2.8)

The last contribution is

Md = u(p′)(−ie)e2Λµ(p′, p)u(p)Aextµ (p ′ − p) (2.9)

37

where

e2Λµ(p′, p) =

∫d4k

(2π)4(−ieγα) i

p ′ − k −m+ iεγµ

i

p− k −m+ iε(−ieγβ) −igαβ

k2 + iε(2.10)

or

Λµ(p′, p) = −i∫

d4k

(2π)4γα

1

p ′ − k −m+ iεγµ

1

p− k −m+ iεγα

1

k2 + iε(2.11)

The problem of the divergences is a serious one and in order to give some sense tothe theory we have to define a way to define our integrals. This is what is called theregularization procedure of the Feynman integrals. That is we give a prescription inorder to make the integrals finite. This can be done in various ways, as introducingan ultraviolet cut-off, or, as we shall see later by the more convenient means ofdimensional regularization. However, we want that the theory does not depend onthe way in which we define the integrals, otherwise we would have to look for somephysical meaning of the regularization procedure we choose. This bring us to theother problem, the inversion of eq. (2.1). Since now the coefficients are finite we canindeed perform the inversion and obtaining e as a function of ephys and obtain allthe observables in terms of the physical electric charge (that is the one measured inthe Coulomb scattering). By doing so, a priori we will introduce in the observablesa dependence on the renormalization procedure. We will say that the theory isrenormalizable when this dependence cancels out. Thinking to the regularization interms of a cut-off this means that considering the observable quantities in terms ofephys, and removing the cut-off (that is by taking the limit for the cut-off going tothe infinity), the result should be finite. Of course, this cancellation is not obviousat all, and in fact in most of the theories this does not happen. However there isa restrict class of renormalizable theories, as for instance QED. We will not discussthe renormalization at all order and neither we will prove which criteria a theoryshould satisfy in order to be renormalizable. We will give these criteria withouta proof but we will try only to justify them On a physical basis. As far QED isconcerned we will study in detail the renormalization at one-loop.

2.2 Higher order corrections

In this section we will illustrate, in a simplified context, the previous considerations.Consider again the Coulomb scattering. At the lowest order the Feynman amplitudein presence of an external static electromagnetic field is simply given by the followingexpression (see Fig. 2.1.1), (f = i)

Sfi = (2π)δ(Ef − Ei)m

V EM(1) (2.12)

whereM(1) = u(p′)(−ieγµ)u(p)Aext

µ (p ′ − p ) (2.13)

38

and

Aextµ (q) =

∫d3x Aext

µ (x)e−iq · x (2.14)

is the Fourier transform of the static e.m. field. In the case of a point-like source,of charge Ze we have

Aextµ (x) = (− Ze

4π|x| ,0) (2.15)

For the following it is more convenient to re-express the previous amplitude in termsof the external current generating the external e.m. field. Since we have

Aµ = jµ (2.16)

we can invert this equation in momentum space, by getting

Aµ(q) = − gµνq2 + iε

jν(q) (2.17)

from which

M(1) = u(p′)(−ieγµ)u(p) −igµν

q2 + iε(−ijext

ν (p ′ − p)) (2.18)

withjextµ (x) = (Zeδ3(x),0) (2.19)

From these expressions one can get easily the differential cross-section dσ/dΩ. Thisquantity could be measured and then we could compare the result with the theoret-ical result obtained from the previous expression. In this way we get a measure ofthe electric charge. However the result is tied to a theoretical equation evaluated atthe lowest order in e. In order to get a better determination one has to evaluate thehigher order corrections, and again extract the value of e from the data. The lowestorder result is at the order e2 in the electric charge. the next-to-leading contribu-tions ar at the order e4. For instance, the diagram in Fig. 2.1.1c gives one of thesecontribution. This is usually referred to as the vacuum polarization contributionand in this Section we will consider only this correction, just to exemplify how therenormalization procedure works. This contribution is given in eqs. (2.6) and (2.7).The total result is

M(1) + Mc = u(p′)(−ieγµ)u(p)[ −igµνq2 + iε

+−igµρq2 + iε

(ie2Πρλ)−igλνq2 + iε

](−ijxt

ν (q))

(2.20)The sum gives a result very similar the to one obtained from M(1) alone, but withthe photon propagator modified in the following way

−igµνq2

→ −igµνq2

+−igµρq2

(ie2Πρλ(q2))

−igλνq2

=−igµνq2

+−ie2Πµν

q4(2.21)

39

As discussed in the previous Section Πµν is a divergent quantity. In fact, at highvalues of the momentum it behaves as∫

d4p

p2(2.22)

and therefore it diverges in a quadratic way. This follows by evaluating the integralby cutting the upper limit by a cut-off Λ. As we shall see later on, the real divergenceis weaker due to the gauge invariance. We will see that the correct behaviour is givenby ∫

d4p

p4(2.23)

The corresponding divergence is only logarithmic, but in any case the integral isdivergent. By evaluating it with a cut-off we get

Πµν(q2) = −gµνq2I(q2) + · · · (2.24)

The terms omitted are proportional to qµqν , and since the photon is always attachedto a conserved current, they do not contribute to the final result. The functions I(q2)can be evaluated by various tricks (we will evaluate it explicitly in the following)and the result is

I(q2) =1

12π2log

Λ2

m2− 1

2π2

∫ 1

0

dz z(1 − z) log

[1 − q2z(1 − z)

m2

](2.25)

For small values of q2 (−q2 << m2) one can expand the logarithm obtaining

I(q2) ≈ 1

12π2log

Λ2

m2+

1

60π2

q2

m2(2.26)

Summarizing the propagator undergoes the following modification

−igλνq2

→ −igµνq2

[1 − e2I(q2)] (2.27)

This relation is shown in graphical terms in Fig. 2.2.1.

.

+

Fig. 2.2.1 - The vacuum polarization correction to the propagator.

The total result is then

M(1) + Mc = u(p′)(−ieγ0)u(p)−iq2

(1 − e2

12π2log

Λ2

m2− e2

60π2

q2

m2

)(−iZe) (2.28)

40

in the limit q → 0 the result looks exactly as at the lowest order, except that weneed to introduce a renormalized charge eR given by

eR = e

[1 − e2

12π2log

Λ2

m2

]1/2

(2.29)

In term of eR and taking into account that we are working at the order e4 we canwrite

M(1) + Mc = uf(ieRγ0)ui−iq2

[1 − e2R

60π2

q2

m2

](−iZeR) (2.30)

Now we compare this result with the experiment where we measure dσ/dΩ in thelimit q → 0. We can extract again from the data the value of the electric charge, butthis time we will measure eR. Notice that in this expression there are no more diver-gent quantities, since they have been eliminated by the definition of the renormalizedcharge. It must be stressed that the quantity that is fixed by the experiments isnot the ”parameter” e appearing in the original lagrangian, but rather eR. Thismeans that the quantity e is not really defined unless we specify how to relate itto measured quantities. This relation is usually referred to as the renormalizationcondition. As we have noticed the final expression we got does not contain anyexplicit divergent quantity. Furthermore, also eR, which is obtained from the exper-imental data, is a finite quantity. It follows that the parameter e, must diverge forΛ → ∞. In other words, the divergence which we have obtained in the amplitude, iscompensated by the divergence implicit in e. Therefore the procedure outlined herehas meaning only if we start with an infinite charge. Said that, the renormalizationprocedure is a way of reorganizing the terms of the perturbative expansion in sucha way that the parameter controlling it is the renormalized charge eR. If this ispossible, the various terms in the expansion turn out to be finite and the theory issaid to be renormalizable. As already stressed this is not automatically satisfied.In fact renormalizability is not a generic feature of the relativistic quantum fieldtheories. Rather the class of renormalizable theories (that is the theories for whichthe previous procedure is meaningful) is quite restricted.

We have seen that the measured electric charge is different from the quantityappearing in the expression of the vertex. It will be convenient in the following to callthis quantity eB (tha bare charge). On the contrary, the renormalized charge will becalled e, rather than eR as done previously. For the moment being we have includedonly the vacuum polarization corrections, however if we restrict our problem to thecharge renormalization, the divergent part coming from the self-energy and vertexcorrections cancel out (see later). Let us consider now a process like e−µ− → e−µ−.Neglecting other corrections the total amplitude is given by Fig. 2.2.2.

We will define the charge by choosing an arbitrary value of the square of themomentum in transfer qµ, such that Q2 = −q2 = µ2. In this way the renormalizedcharge depends on the scale µ. In the previous example we were using q → 0. Thegraphs in Fig. 2.2.2 sum up to give (omitting the contribution of the fermionic

41

.

+ +

e B

e B

e B

e B

e B

eB

e B

e B

e B

e B

e B

e B

µ−

e−e−

e−

µ−µ−

q +

Fig. 2.2.2 - The vacuum polarization contributions to the e−µ− → e−µ− scattering.

lines,)

e2B

[−igµνq2

+−igµρq2

(ie2BΠρλ)−igλνq2

+−igµρq2

(ie2BΠρλ)−igλσq2

(ie2BΠστ )−igτνq2

+ · · ·]Q2=µ2

(2.31)By considering only the gµν contribution to Πµν , we get

e2B−igµνq2

[1 − e2BI(q

2) + e4BI2(q2) + · · ·]

Q2=µ2 (2.32)

This expression looks as the lowest order contribution to the propagator be definingthe renormalized electric charge as

e2 = e2B[1 − e2BI(µ2) + e4BI

2(µ2) + · · ·] (2.33)

Since the fermionic contribution is the same for all the diagrams, by calling it F (Q2),we get the amplitude in the form

M(eB) = e2BF (Q2)[1 − e2BI(Q2) + e4BI

2(Q2) · · ·] (2.34)

where e2BF (Q2) is the amplitude at the lowest order in e2B. As we know I(Q2) is adivergent quantity and therefore the amplitude is not defined. However, followingthe discussion of the previous Section we can try to re-express everything in termsof the renormalized charge (2.33). We can easily invert by series the eq. (2.33)obtaining at the order e4

e2B = e2[1 + e2I(µ2) + · · ·] (2.35)

42

By substituting this expression in M(e0) we get

M(eB) = e2[1 + e2I(µ2) + · · ·]F (Q2)[1 − e2I(Q2) + · · ·] =

= e2F (Q2)[1 − e2(I(Q2) − I(µ2)) + · · ·] ≡ MR(e) (2.36)

From the expression (2.25) for I(Q2) we get

I(Q2) =1

12π2log

Λ2

m2− 1

2π2

∫ 1

0

dz z(1 − z) log

[1 +

Q2z(1 − z)

m2

](2.37)

Since the divergent part is momentum independent it cancels out in the expressionfor MR(e). Furthermore, the renormalized charge e will be measured through thecross-section, and therefore, by definition, it will be finite. Therefore, the chargerenormalization makes the amplitude finite. Evaluating I(Q2) − I(µ2) per large Q2

(and µ2) we get

e2(I(Q2) − I(µ2)) = −2α

π

∫ 1

0

dz z(1 − z) log

[m2 +Q2z(1 − z)

m2 + µ2z(1 − z)

]→

→ −2α

π

∫ 1

0

dz z(1 − z) logQ2

µ2= − α

3πlog

Q2

µ2(2.38)

and in this limit

MR(e) = e2F (Q2)

[1 − α

3πlog

Q2

µ2+ · · ·

](2.39)

At first sight MR(e) depends on µ, but actually this is not the case. In fact, e isdefined through the eq. (2.33) and therefore it depends also on µ. What is goingon is that the implicit dependecnce of e on µ cancels out the explicit dependenceof MR(e). This follows from MR(e) = M(eB) and on the observation that M(eB)does not depend on µ. This can be formalized in a way that amounts to introducethe concept of renormalization group that we will discuss in much more detail lateron. In fact, let us be more explicit by writing

M(eB) = MR(e(µ), µ) (2.40)

By differentiating this expression with respect to µ we get

dM(eB)

dµ=∂MR

∂µ+∂MR

∂e

∂e

∂µ= 0 (2.41)

This equation say formally that the amplitude expressed as a function of the renor-malized charge is indeed independent on µ. This equation is the prototype of therenormalization group equations, and tells us how we have to change the coupling,when we change the definition scale µ.

Going back to the original expression for M(e0), we see that this is given bythe tree amplitude e2BF (Q2) times a geometric series arising from the iteration of

43

the vacuum polarization diagram. Therefore, it comes natural to define a couplingrunning with the momentum of the photon.

e2(Q2) = e2B[1 − e2BI(Q2) + e4BI

2(Q2) + · · ·] =e2B

1 + e2BI(Q2)

(2.42)

Also this expression is cooked up in terms of the divergent quantities eB and I(Q2),but again e2(Q2) is finite. This can be shown explicitly by evaluating the renormal-ized charge, e(µ2), in terms of the bare one

e2(µ2) =e2B

1 + e2BI(µ2)

(2.43)

We can put the last two equations in the form

1

e2(Q2)=

1

e2B+ I(Q2),

1

e2(µ2)=

1

e2B+ I(µ2) (2.44)

and subtracting them

1

e2(Q2)− 1

e2(µ2)= I(Q2) − I(µ2) (2.45)

we get

e2(Q2) =e2(µ2)

1 + e2(µ2)[I(Q2) − I(µ2)](2.46)

By using this result we can write MR(e) as

MR(e) = e2(Q2)F (Q2) (2.47)

It follows that the true expansion parameter in the perturbative series is the runningcoupling e(Q2). By using eq. (2.38) we obtain

α(Q2) =α(µ2)

1 − α(µ2)

3πlog

Q2

µ2

(2.48)

We see that α(Q2) increases with Q2, therefore the perturbative approach loosesvalidity for those values Q2 of such that α(Q2) ≈ 1, that is when

1 − α(µ2)

3πlog

Q2

µ2= α(µ2) (2.49)

or

Q2

µ2= e

3π

(1

α(µ2)− 1

)(2.50)

44

For α(µ2) ≈ 1/137 we getQ2

µ2≈ e1281 ≈ 10556 (2.51)

Actually α(Q2) is divergent for

Q2

µ2= e

3π

α(µ2) ≈ e1291 ≈ 10560 (2.52)

This value of Q2 is referred to as the Landau pole. Therefore in the case of QED theperturbative approach maintains its validity up to gigantic values of the momenta.This discussion shows clearly that the behaviour of the running coupling constantis determined by I(Q2), that is from the vaccum polarization contribution. In par-ticular, α(Q2) increases with Q2 since in the expression e2(µ2)(I(Q2) − I(µ2)) thecoefficient of the logarithm is negative (equal to −α(µ2)/(3π)). We will see that inthe so called non-abelian gauge theories the running is opposite, that is the theorybecomes more and more perturbative, by increasing the energy.

2.3 The analysis by counterterms

The previous way of defining a renormalizable theory amounts to say that the origi-nal parameters in the lagrangian, as e, should be infinite and that their divergencesshould compensate the divergences of the Feynman diagrams. Then one can try toseparate the infinite from the finite part of the parameters (this separation is ambigu-ous, see later). The infinite contributions are called counterterms, and by definitionthey have the same operator structure of the original terms in the lagrangian. Onthe other hand, the procedure of regularization can be performed by adding to theoriginal lagrangian counterterms cooked in such a way that their contribution killsthe divergent part of the Feynman integrals. This means that the coefficients ofthese counterterms have to be infinite. However they can also be regularized in thesame way as the other integrals. We see that the theory will be renormalizable ifthe counterterms we add to make the theory finite have the same structure of theoriginal terms in the lagrangian, in fact, if this is the case, they can be absorbedin the original parameters, which however are arbitrary, because they have to befixed by the experiments (renormalization conditions). We will show now that atone-loop the divergent contributions come only from the three diagrams discussedbefore. In fact, let us define D as the superficial degree of divergence of a one-loopdiagram the difference between 4 (the number of integration over the mimentum)and the number of powers of momentum coming from the propagators (we assumehere that no powers of momenta are coming from the vertices, which is not true inscalar QED). We get

D = 4 − IF − 2IB (2.53)

45

where IF and IB are the number of internal fermionic and bosonic lines. Goingthrough the loop one has to have a number of vertices, V equal to the number ofinternal lines

V = IF + IB (2.54)

Also, if at each vertex there are Ni lines of the particle of type i, then we have

NiV = 2Ii + Ei (2.55)

from which2V = 2IF + EF , V = 2IB + EB (2.56)

From these equations we can get V , IF and IB in terms of EF and EB

IF =1

2EF + EB, IB =

1

2EF , V = EF + EB (2.57)

from which

D = 4 − 3

2EF − EB (2.58)

It follows that the only one-loop superficially divergent diagrams in spinor QED arethe ones in Fig. 2.3.1. In fact, all the diagrams with an odd number of externalphotons vanish. This is trivial at one-loop, since the trace of an odd number of γmatrices is zero. In general (Furry’s theorem) it follows from the conservation ofcharge conjugation, since the photon is an eigenstate of C with eigenvalue equal to-1.

Furthermore, the effective degree of divergence can be less than the superficialdegree. In fact, gauge invariance often lowers the divergence. For instance thediagram of Fig. 2.3.1 with four external photons (light-light scattering) has D = 0(logarithmic divergence), but gauge invariance makes it finite. Also the vacuumpolarization diagram has an effective degree of divergence equal to 2. Therefore,there are only three divergent one-loop diagrams and all the divergences can bebrought back to the three functions Σ(p), Πµν(q) e Λµ(p′, p). This does not meanthat an arbitrary diagram is not divergence, but it can be made finite if the previousfunctions are such. In such a case one has only to show that eliminating these threedivergences (primitive divergences) the theory is automatically finite. In particularwe will show that the divergent part of Σ(p) can be absorbed into the definitionof the mass of the electron and a redefinition of the electron field (wave functionrenormalization). The divergence in Πµν , the photon self-energy, can be absorbedin the wave function renormalization of the photon (the mass of the photon is notrenormalized due to the gauge invariance). And finally the divergence of Λµ(p

′, p)goes into the definition of the parameter e. To realize this program we divide upthe lagrangian density in two parts, one written in terms of the physical parameters,the other will contain the counterterms. We will call also the original parametersand fields of the theory the bare parameters and the bare fields and we will use anindex B in order to distinguish them from the physical quantities. Therefore the

46

.

E EF B D

0 2 2

0 4 0

2 0 1

2 1 0

Fig. 2.3.1 - The superficially divergent one-loop diagrams in spinor QED

two pieces of the lagrangian should look like as follows: the piece in terms of thephysical parametersLp

Lp = ψ(i∂ −m)ψ − eψγµψAµ − 1

4FµνF

µν − 1

2(∂µA

µ)2 (2.59)

and the counter terms piece Lc.t.

Lc.t. = iBψ∂ψ − Aψψ − C

4Fµν − E

2(∂µA

µ)2 − eDψAψ (2.60)

and we have to require that la sum of these two contributions should coincide withthe original lagrangian written in terms of the bare quantities. Adding together Lp

and Lc.t. we get

L = (1 +B)iψ∂ψ− (m+A)ψψ− e(1 +D)ψγµψAµ− 1 + C

4FµνF

µν +gauge − fixing

(2.61)where, for sake of simplicity, we have omitted the gauge fixing term. Defining therenormalization constant of the fields

Z1 = (1 +D), Z2 = (1 +B), Z3 = (1 + C) (2.62)

we write the bare fields as

ψB = Z1/22 ψ, AµB = Z

1/23 Aµ (2.63)

47

we obtain

L = iψB ∂ψB − m+ A

Z2

ψBψB − eZ1

Z2Z1/23

ψBγµψBAµB − 1

4FB,µνF

µνB + gauge − fixing

(2.64)and putting

mB =m+ A

Z2, eB =

eZ1

Z2Z1/23

(2.65)

we get

L = iψB ∂ψB −mBψBψB − eBψBγµψBAµB − 1

4FB,µνF

µνB + gauge − fixing (2.66)

So we have succeeded in the wanted separation. Notice that the division of the pa-rameters in physical and counter term part is well defined, because the finite pieceis fixed to be an observable quantity. This requirement gives the renormalizationconditions. The counter terms A, B, ... are determined recursively at each per-turbative order in such a way to eliminate the divergent parts and to respect therenormalization conditions. We will see later how this works in practice at one-looplevel. Another observation is that Z1 and Z2 have to do with the self-energy of theelectron, and as such they depend on the electron mass. Therefore if we considerthe theory for a different particle, as the muon, which has the same interactions asthe electron and differs only for the value of the mass (mµ ≈ 200me), one wouldget a different bare electric charge for the two particles. Or, phrased in a differentway, one would have to tune the bare electric charge at different values in order toget the same physical charge. This looks very unnatural, but the gauge invarianceof the theory implies that at all the perturbative orders Z1 = Z2. As a consequenceeB = e/Z

1/23 , and since Z3 comes from the photon self-energy, the relation between

the bare and the physical electric charge is universal (that is it does not depend onthe kind of charged particle under consideration).

Summarizing, one starts dividing the bare lagrangian in two pieces. Then weregularize the theory giving some prescription to get finite Feynman integrals. Thepart containing the counter terms is determined, order by order, by requiring thatthe divergences of the Feynman integrals, which come about when removing theregularization, are cancelled out by the counter term contributions. Since the sepa-ration of an infinite quantity into an infinite plus a finite term is not well defined, weuse the renormalization conditions, to fix the finite part. After evaluating a physicalquantity we remove the regularization. Notice that although the counter terms aredivergent quantity when we remove the cut off, we will order them according to thepower of the coupling in which we are doing the perturbative calculation. That is wehave a double limit, one in the coupling and the other in some parameter (regulator)which defines the regularization. The order of the limit is first to work at some orderin the coupling, at fixed regulator, and then remove the regularization.

Before going into the calculations for QED we want to illustrate some generalresults about the renormalization. If one considers only theory involving scalar,

48

fermion and massless spin 1 (as the photon) fields, it is not difficult to constructan algorithm which allows to evaluate the ultraviolet (that is for large momenta)divergence of any Feynman diagram. In the case of the electron self-energy (seeFigs. 2.1.1a and 2.1.1b) one has an integration over the four momentum p and abehaviour of the integrand, coming from the propagators, as 1/p3, giving a lineardivergence (it turns out that the divergence is only logarithmic). From this countingone can see that only the lagrangian densities containing monomials in the fieldswith mass dimension smaller or equal to the number of space-time dimensions havea finite number of divergent diagrams. It turns out also that these are renormalizabletheories ( a part some small technicalities). The mass dimensions of the fields can beeasily evaluated from the observation that the action is dimensionless in our units(h/ = 1). Therefore, in n space-time dimensions, the lagrangian density, defined as∫

dnx L (2.67)

has a mass dimension n. Looking at the kinetic terms of the bosonic fields (twoderivatives) and of the fermionic fields (one derivative), we see that

dim[φ] = dim[Aµ] =n

2− 1, dim[ψ] =

n− 1

2(2.68)

In particular, in 4 dimensions the bosonic fields have dimension 1 and the fermionic3/2. Then, we see that QED is renormalizable, since all the terms in the lagrangiandensity have dimensions smaller or equal to 4

dim[ψψ] = 3, dim[ψγµψAµ] = 4, dim[(∂µAµ)

2] = 4 (2.69)

The condition on the dimensions of the operators appearing in the lagrangian canbe translated into a condition over the coupling constants. In fact each monomialOi will appear multiplied by a coupling gi

L =∑i

giOi (2.70)

thereforedim[gi] = 4 − dim[Oi] (2.71)

The renormalizability requiresdim[Oi] ≤ 4 (2.72)

from whichdim[gi] ≥ 0 (2.73)

that is the couplings must have positive dimension in mass are to be dimensionless.In QED the only couplings are the mass of the electron and the electric charge

49

which is dimensionless. As a further example consider a single scalar field. Themost general renormalizable lagrangian density is characterized by two parameters

L =1

2∂µφ∂

µφ− 1

2m2φ2 − ρφ3 − λφ4 (2.74)

Here ρ has dimension 1 and λ is dimensionless. We see that the linear σ-models arerenormalizable theories.

Giving these facts let us try to understand what makes renormalizable and nonrenormalizable theories different. In the renormalizable case, if we have writtenthe most general lagrangian, the only divergent diagrams which appear are theones corresponding to the processes described by the operators appearing in thelagrangian. Therefore adding to L the counter term

Lc.t. =∑i

δgiOi (2.75)

we can choose the δgi in such a way to cancel, order by order, the divergences. Thetheory depends on a finite number of arbitrary parameters equal to the number ofparameters gi. Therefore the theory is a predictive one. In the non renormalizablecase, the number of divergent diagrams increase with the perturbative order. Ateach order we have to introduce new counter terms having an operator structuredifferent from the original one. At the end the theory will depend on an infinitenumber of arbitrary parameters. As an example consider a fermionic theory with aninteraction of the type (ψψ)2. Since this term has dimension 6, the relative couplinghas dimension -2

Lint = −g2(ψψ)2 (2.76)

At one loop the theory gives rise to the divergent diagrams of Fig. 2.3.2. Thedivergence of the first diagram can be absorbed into a counter term of the originaltype

−δg2(ψψ)2 (2.77)

The other two need counter terms of the type

δg3(ψψ)3 + δg4(ψψ)4 (2.78)

These counter terms originate new one-loop divergent diagrams, as for instance theones in Fig. 2.3.3. The first diagram modifies the already introduced counter term(ψψ)4, but the second one needs a new counter term

δg5(ψψ)5 (2.79)

This process never ends.The renormalization requirement restricts in a fantastic way the possible field

theories. However one could think that this requirement is too technical and onecould imagine other ways of giving a meaning to lagrangians which do not satisfy this

50

.

a) b)

c)

Fig. 2.3.2 - Divergent diagrams coming from the interaction (ψψ)2.

condition. But suppose we try to give a meaning to a non renormalizable lagrangian,simply requiring that it gives rise to a consistent theory at any energy. We will showthat this does not happen. Consider again a theory with a four-fermion interaction.Since dim g2 = −2, if we consider the scattering ψ + ψ → ψ + ψ in the high energylimit (where we can neglect all the masses), on dimensional ground we get that thetotal cross-section behaves like

σ ≈ g22E

2 (2.80)

Analogously, in any non renormalizable theory, being there couplings with negativedimensions, the cross-section will increase with the energy. But the cross-section hasto do with the S matrix which is unitary. Since a unitary matrix has eigenvalues of

.

a) b)

Fig. 2.3.3 - Divergent diagrams coming from the interactions (ψψ)2 and (ψψ)4.

51

modulus 1, it follows that its matrix elements must be bounded. Translating thisargument in the cross-section one gets the bound

σ ≤ c2

E2(2.81)

where c is some constant. For the previous example we get

g2E2 ≤ c (2.82)

This implies a violation of the S matrix unitarity at energies such that

E ≥√

c

g2

(2.83)

It follows that we can give a meaning also to non renormalizable theories, but onlyfor a limited range of values of the energy. This range is fixed by the value of thenon renormalizable coupling. It is not difficulty to realize that non renormalizabilityand bad behaviour of the amplitudes at high energies are strictly connected.

2.4 Dimensional regularization of the Feynman

integrals

As we have discussed in the previous Section we need a procedure to give sense atthe otherwise divergent Feynman diagrams. The simplest of these procedures is justto introduce a cut-off Λ and define our integrals as

∫∞0

→ limΛ→∞∫ Λ

0). Of course,

in the spirit of renormalization we have first to perform the perturbative expansionand then take the limit over the cut-off. Although this procedure is very simple itresults to be inadequate in situations like gauge theories. In fact one can show thatthe cut-off regularization breaks the translational invariance and creates problemsin gauge theories. One kind of regularization which nowadays is very much used isthe dimensional regularization. This consists in considering the integration in anarbitrary number of space-time dimensions and taking the limit of four dimensionsat the end. This way of regularizing is very convenient because it respects all thesymmetries. In fact, except for very few cases the symmetries do not depend on thenumber of space-time dimensions. Let us what is dimensional regularization about.We want to evaluate integrals of the type

I4(k) =

∫d4p F (p, k) (2.84)

with F (p, k) ≈ p−2 or p−4. The idea is that integrating on a lower number ofdimensions the integral improves the convergence properties in the ultraviolet. For

52

instance, if F (p, k) ≈ p−4, the integral is convergent in 2 and in 3 dimensions.Therefore, we would like to introduce a quantity

I(ω, k) =

∫d2ωp F (p, k) (2.85)

to be regarded as a function of the complex variable ω. Then, if we can definea complex function I ′(ω, k) on the entire complex plane, with definite singularity,and as such that it coincides with I on some common domain, then by analyticcontinuation I and I ′ define the same analytic function. A simple example of thisprocedure is given by the Euler’s Γ. This complex function is defined for Re z > 0by the integral representation

Γ(z) =

∫ ∞

0

dt e−ttz−1 (2.86)

If Re z ≤ 0, the integral diverges as

dt

t1+|Re z| (2.87)

in the limit t→ 0.However it is easy to get a representation valid also for Re z ≤ 0.Let us divide the integration region in two parts defined by a parameter α

Γ(z) =

∫ α

0

dt e−ttz−1 +

∫ ∞

α

dt e−ttz−1 (2.88)

Expanding the exponential in the first integral and integrating term by term 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 (2.89)

The second integral converges for any z since α > 0. This expression coincides withthe representation for the Γ function for Re z > 0, but it is defined also for Re z < 0where it has simple poles located at z = −n. Therefore it is a meaningful expressionon all the complex plane z. Notice that in order to isolate the divergences we needto introduce an arbitrary parameter α. However the result does not depend on theparticular value of this parameter. This the Weierstrass representation of the EulerΓ(z). From this example we see that we need the following three steps

• Find a domain where I(ω, k) is convergent. Typically this will be for Re ω < 2.

• Construct an analytic function identical to I(ω, k) in the domain of conver-gence, but defined on a larger domain including the point ω = 2.

• At the end of the calculation take the limit ω → 2.

53

2.5 Integration in arbitrary dimensions

Let us consider the integral

IN =

∫dNpF (p2) (2.90)

with N an integer number and p a vector in an Euclidean N -dimensional space.Since the integrand is invariant under rotations of the N -dimensional vector p, wecan perform the angular integration by means of

dNp = dΩNpN−1dp (2.91)

where dΩN is the solid angle element 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 (2.92)

The value of the sphere surface can be evaluated by the following trick. Consider

I =

∫ +∞

−∞e−x2

dx =√π (2.93)

By taking N of these factors we get

IN =

∫dx1 · · · dxNe−(x2

1 + · · ·+ x2N ) = πN/2 (2.94)

The same integral can be evaluated in polar coordinates

πN/2 = SN

∫ ∞

0

ρN−1e−ρ2dρ (2.95)

By putting x = ρ2 we have

πN/2 =1

2SN

∫ ∞

0

xN/2−1e−xdx =1

2SNΓ

(N

2

)(2.96)

where we have used the representation of the Euler Γ function given in the previousSection. Therefore

SN =2πN/2

Γ(N2

) (2.97)

and

IN =πN/2

Γ(N2

) ∫ ∞

0

xN/2−1F (x)dx (2.98)

with x = p2.

54

The integrals we will be interested in are of the type

I(M)N =

∫dNp

(p2 − a2 + iε)A(2.99)

with p a vector in a Ndimensional Minkowski space. We can perform an anti-clockwise rotation of 900 (Wick’s rotation) in the complex plane of p0 without hittingany singularity . Then we do a change of variables p0 → ip0, and we use the definitionp2 → −p2 =

∑Ni=1 p

2i , obtaining

I(M)N = i

∫dNp

(−p2 − a2)A= i(−1)AIN (2.100)

where IN is an integral of the kind discussed at the beginning of this Section withF (x) given by

F (x) = (x+ a2)−A (2.101)

It follows

IN =πN/2

Γ(N2

) ∫ ∞

0

xN/2−1

(x+ a2)Adx (2.102)

By putting x = a2y we get

IN = (a2)N/2−AπN/2

Γ(N2

) ∫ ∞

0

yN/2−1(1 + y)−Adx (2.103)

and recalling the integral representation for the Euler B(x, y) function (valid forRex, y > 0)

B(x, y) =Γ(x)Γ(y)

Γ(x+ y)=

∫ ∞

0

tx−1(1 + t)−(x+y)dt (2.104)

it follows

IN = πN/2Γ(A−N/2)

Γ(A)

1

(a2)A−N/2(2.105)

We have obtained this representation for N/2 > 0 and Re(A − N/2) > 0. But weknow how to extend the Euler gamma-function to the entire complex plane, andtherefore we can extend this formula to complex dimensions N = 2ω

I2ω = πωΓ(A− ω)

Γ(A)

1

(a2)A−ω(2.106)

This shows that I2ω has simple poles located at

ω = A,A+ 1, · · · (2.107)

Therefore our integral will be perfectly defined at all ω such that ω = A,A+ 1, · · ·.At the end we will have to consider the limit ω → 2. The original integral inMinkowski space is then∫

d2ωp

(p2 − a2)A= iπω(−1)A

Γ(A− ω)

Γ(A)

1

(a2)A−ω(2.108)

55

For the following it will be useful to derive another formula. Let us put in theprevious equation p = p′ + k and b2 = −a2 + k2, then∫

d2ωp′

(p′2 + 2p′ · k + k2 − a2)A= iπω(−1)A

Γ(A− ω)

Γ(A)

1

(a2)A−ω(2.109)

from which∫d2ωp

(p2 + 2p · k + b2)A= iπω(−1)A

Γ(A− ω)

Γ(A)

1

(k2 − b2)A−ω(2.110)

Differentiating with respect to kµ we get various useful relations as∫d2ωp

pµ(p2 + 2p · k + b2)A

= iπω(−1)AΓ(A− ω)

Γ(A)

−kµ(k2 − b2)A−ω

(2.111)

and ∫d2ωp

pµpν(p2 + 2p · k + b2)A

= iπω(−1)A

Γ(A)(k2 − b2)A−ω

×[Γ(A− ω)kµkν − 1

2gµν(k

2 − b2)Γ(A− ω − 1)

](2.112)

Since at the end of our calculation we will have to take the limit ω → 2, it will beuseful to recall the expansion of the Gamma function around its poles

Γ(ε) =1

ε− γ + O(ε) (2.113)

whereγ = 0.5772... (2.114)

is the Euler-Mascheroni constant, and for (n ≥ 1):

Γ(−n+ ε) =(−1)n

n!

[1

ε+ ψ(n+ 1) + O(ε)

](2.115)

where

ψ(n + 1) = 1 +1

2+ · · · + 1

n− γ (2.116)

2.6 One loop regularization of QED

In this Section we will regularize, by using dimensional regularization, the relevantdivergent quantities in QED, that is Σ(p), Πµν(q) e Λµ(p, p′). Furthermore, in orderto define the counter terms we will determine the expressions which become divergentin the limit of ω → 2. It will be also convenient to introduce a parameter µ suchthat these quantities have the same dimensions in the space with d = 2ω as in d = 4.

56

The algebra of the Dirac matrices is also easily extended to arbitrary dimensions d.For instance, starting from

[γµ, γν ]+ = 2gµν (2.117)

we getγµγµ = d (2.118)

andγµγνγ

µ = (2 − d)γν (2.119)

Other relations can be obtained by starting from the algebraic properties of theγ-matrices. Let us start with the electron self-energy which we will require to havedimension 1 as in d = 4. From eq. (2.4) we have

Σ(p) = iµ4−2ω

∫d2ωk

(2π)2ωγµ

p− k +m

(p− k)2 −m2 + iεγµ

1

k2 + iε(2.120)

In order to use the equations of the previous Section it is convenient to combinetogether the denominators of this expression into a single one. This is done by usinga formula due to Feynman

1

ab=

∫ 1

0

dz

[az + b(1 − z)]2(2.121)

which is proven using

1

ab=

1

b− a

[1

a− 1

b

]=

1

b− a

∫ b

a

dx

x2(2.122)

and doing the change of variables

x = az + b(1 − z) (2.123)

We get

Σ(p) = iµ4−2ω

∫ 1

0

dz

∫d2ωk

(2π)2ωγµ

p− k +m

[(p− k)2z −m2z + k2(1 − z)]2γµ (2.124)

The denominator can be written in the following way

[...] = p2z −m2z + k2 − 2p · kz (2.125)

and the term p · k can be eliminated through the following change of variablesk = k′ + pz. We find

[...] = (p2 −m2)z + (k′ + pz)2 − 2p · (k′ + pz)z = k′2 −m2z + p2z(1 − z) (2.126)

57

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 (2.127)

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 (2.128)

and integrating over k

Σ(p) = iµ4−2ω

∫ 1

0

dz1

(2π)2ωiπω

Γ(2 − ω)

Γ(2)γµ

p(1 − z) +m

[m2z − p2z(1 − z)]2−ωγµ (2.129)

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 (2.130)

Contracting the γ matrices

Σ(p) = − 1

16π2Γ(ε/2)

∫ 1

0

dz(4πµ2)ε/2(ε− 2)p(1 − z) + (4 − ε)m

[m2z − p2z(1 − z)]ε/2(2.131)

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(2.132)

Defining

A = 2p(1 − z) − 4m, B = −p(1 − z) +m, C =m2z − p2z(1 − z)

4πµ2(2.133)

and expanding for ε→ 0

Σ(p) =1

16π2

∫ 1

0

dz

[2A

ε+ 2B − γA

] [1 − ε

2logC

]

=1

16π2

∫ 1

0

dz

[2A

ε− A logC + 2B − γA

]

=1

8π2ε(p− 4m) − 1

16π2[p− 2m+ γ(p− 4m)]

− 1

8π2

∫ 1

0

dz[p(1 − z) − 2m] logm2z − p2z(1 − z)

4πµ2

=1

8π2ε(p− 4m) + finite terms (2.134)

58

Consider now the vacuum polarization (see eq. (2.8))

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)(2.135)

Using again the Feynman representation for the denominators

Πµν(q) = iµ4−2ω

∫ 1

0

dz

∫d2ωk

(2π)2ω

Tr[γµ(k +m)γν(k + q +m)]

[(k2 −m2)(1 − z) + ((k + q)2 −m2)z]2(2.136)

We write the denominator as

[...] = k2 + q2z −m2 + 2k · qz (2.137)

through the change of variable k = k′ − qz we cancel the mixed term obtaining

[...] = (k′ − qz)2 + 2(k′ − qz) · qz + q2z −m2 = k′2 + q2z(1 − z) −m2 (2.138)

and therefore (we put again k′ = k)

Πµν(q) = iµ4−2ω

∫ 1

0

dz

∫d2ωk

(2π)2ω

Tr[γµ(k − qz +m)γν(k + q(1 − z) +m)]

[k2 + q2z(1 − z) −m2]2

(2.139)Since the integral of the odd terms in k is zero, it is enough to evaluate the contri-bution of the even term to the trace

Tr[...]even = Tr[γµ(k − qz +m)γν(k + q(1 − z)]pari +m2Tr[γµγν ]

= Tr[γµkγν k] − Tr[γµqγν q]z(1 − z) +m2Tr[γµγν ] (2.140)

If we define the γ as matrices of dimension 2ω × 2ω we can repeat the standardcalculation by obtaining a factor 2ω instead of 4. Therefore

Tr[...]even = 2ω[2kµkν − gµνk2 − (2qµqν − gµνq

2)z(1 − z) +m2gµν ]

= 2ω[2kµkν − 2z(1 − z)(qµqν − gµνq2) − gµν(k

2 −m2 + q2z(1 − z)](2.141)

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]

](2.142)

From the relations of the previous Section we get∫d2ωp

2pµpν[p2 − a2]2

= −igµνπωΓ(1 − ω)

(a2)1−ω (2.143)

59

whereas ∫d2ωp

1

[p2 − a2]= −iπωΓ(1 − ω)

(a2)1−ω (2.144)

Therefore the first and the third contribution to the vacuum polarization cancel outand 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

(2.145)Notice that the original integral was quadratically divergent, but due to the previouscancellation the divergence is only logarithmic. The reason is again gauge invari-ance. In fact it is possible to show that this implies qµΠµν(q) = 0. Performing themomentum 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(2.146)

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 − gµνq

2)

∫ 1

0

dz z(1 − z)Γ(ε/2)

[m2 − q2z(1 − z)

4πµ2

]−ε/2

=2

16π2(4 − 2ε log 2)(qµqν − gµνq

2)

×∫ 1

0

dz z(1 − z)(2

ε− γ)

[1 − ε

2logC

](2.147)

where C is definite in eq. (2.133). Finally

Πµν(q) =1

8π2(qµqν − gµνq

2)

∫dz z(1 − z)

[8

ε− 4γ − 4 log 2

] [1 − ε

2logC

]=

1

2π2(qµqν − gµνq

2)

×[

1

3ε− γ

6−∫dz z(1 − z) log

[m2 − q2z(1 − z)

2πµ2

]](2.148)

or, in abbreviated way

Πµν(q) =1

6π2(qµqν − gµνq

2)1

ε+ finite terms (2.149)

We have now to evaluate Λµ(p′, p). From eq. (2.11) we have

Λµ(p′, p) = −iµ4−2ω

∫d2ωk

(2π)2ωγα

p ′ − k +m

(p′ − k)2 −m2γµ

p− k +m

(p− k)2 −m2γα

1

k2(2.150)

60

the general formula to reduce n denominators to a single one is

n∏i=1

1

ai= (n− 1)!

∫ 1

0

n∏i=1

dβiδ(1 −∑n

i=1 βi)

[∑n

i=1 βiai]n

(2.151)

To show this equation notice that

n∏i=1

1

ai=

∫ ∞

0

n∏i=1

dαie

−n∑i=1

αiai(2.152)

Introducing the identity

1 =

∫ ∞

0

dλδ(λ−n∑i=1

αi) (2.153)

and changing variables αi = λβi

n∏i=1

1

ai=

∫ ∞

0

n∏i=1

dαidλδ(λ−n∑i=1

αi)e

−n∑i=1

αiai

=

∫ ∞

0

n∏i=1

dβiλndλ

λe

−λn∑i=1

βiaiδ(1 −

n∑i=1

βi) (2.154)

The integration over βi can be restricted to the interval [0, 1] due to the deltafunction, and furthermore ∫ ∞

0

dλλn−1e−ρλ =(n− 1)!

ρn(2.155)

In our case 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

(2.156)

The denominator is

[...] = k2 −m2(x+ y) + p2x+ p′2y − 2k · (px+ p′y) (2.157)

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 (2.158)

61

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 (2.159)

The odd term in k is zero after integration, the term in k2 is logarithmically di-vergent, whereas the remaining part is convergent. Separating the divergent piece,Λ

(1)µ , from the convergent one, Λ

(3)µ ,

Λµ = Λ(1)µ + Λ(2)

µ (2.160)

we get for the first term

Λ(1)µ (p′, p) = −2iµ4−2ω

∫ 1

0

dx

∫ 1−x

0

dy

∫d2ωk

(2π)2ω

γαγλγµγσγαkλkσ

[k2 −m2(x+ y) + p2x(1 − x) + p′2y(1 − y) − 2p · p′xy]3

= −2iµ4−2ω

∫ 1

0

dx

∫ 1−x

0

dyiπω(−1)3

Γ(3)

(−1

2

)Γ(2 − ω)

γαγλγµγλγα

[m2(x+ y) − p2x(1 − x) − p′2y(1 − y) + 2p · p′xy]2−ω

=1

2µ4−2ω

(1

4π

)ωΓ(2 − ω)

∫ 1

0

dx

∫ 1−x

0

dy

γαγλγµγλγα

[m2(x+ y) − p2x(1 − x) − p′2y(1 − y) + 2p · p′xy]2−ω (2.161)

Using the relationγαγλγ

µγλγα = (2 − d)2γµ (2.162)

we obtain (ε = 4 − 2ω)

Λ(1)µ (p′, p) =

1

2µε(

1

4π

)2−ε/2Γ(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

=1

32π2

[2

ε− 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

=1

8π2εγµ − 1

16π2(γ + 2)γµ − 1

8π2γµ

∫ 1

0

dx

∫ 1−x

0

dy

62

× log

[m2(x+ y) − p2x(1 − x) − p′2y(1 − y) + 2p · p′xy

4πµ2

](2.163)

and finally

Λ(1)µ (p′, p) =

1

8π2εγµ + finite terms (2.164)

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

= − 1

16π2

∫ 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 (2.165)

2.7 One loop renormalization

We summarize here the results of the previous Section

Σ(p) =1

8π2ε(p− 4m) + Σf (p) (2.166)

Πµν(q) = (qµqν − gµνq2)

[1

6π2ε+ Πf (q)

]≡ (qµqν − gµνq

2)Π(q2) (2.167)

Λµ(p′, p) =

1

8π2εγµ + Λf

µ(p′, p) (2.168)

where the functions with the superscript f represent the finite contributions. Letus start discussing the electron self-energy. As shown in eq. (2.2), the effect of Σ(p)is to correct the electron propagator. In fact we have (see Fig. 2.7.1):

SF (p) =i

p−m+

i

p−mie2Σ(p)

i

p−m+ · · · (2.169)

.

Fig. 2.7.1 - The loop expansion for the electron propagator.

from which, at the same order in the perturbative expansion

SF (p) =i

p−m

(1 +

e2Σ(p)

p−m

)−1

=i

p−m+ e2Σ(p)(2.170)

63

Therefore the effect of the divergent terms is to modify the coefficients of p and m:

iS−1F (p) = p−m+ e2Σ(p) = p

(1 +

e2

8π2ε

)−m

(1 +

e2

2π2ε

)+ finite terms (2.171)

This allows us to define the counter terms to be added to the lagrangian expressedin terms of the physical parameters in such a way to cancel these divergences

(L1)ct = iBψ∂ψ − Aψψ (2.172)

(L1)ct modifies the Feynman rules adding two particles interacting terms. Thesecan be easily evaluated noticing that the expression of the propagator, taking intoaccount (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(2.173)

where, consistently with our expansion we have taken only the first order terms inA and B. We can associate to these two terms the diagrams of Fig. 2.7.2, withcontributions −iA to the mass term, and iBp to p.

.

-iA iBp/

Fig. 2.7.2 - The counter terms for the self-energy (p/ in the figure should be read asp).

The propagator at the second order in the coupling constant is then given by addingthe diagrams of Fig. 2.7.3.At this order we get

SF (p) =i

p−m+

i

p−m

(ie2Σ(p) + iBp− iA

) i

p−m(2.174)

and the correction to the free propagator is given by

e2Σ(p) +Bp− A =

(e2

8π2ε+B

)p−

(e2

2π2εm+ A

)+ finite terms (2.175)

We can now fix the counter terms by choosing

B = − e2

8π2

(1

ε+ F2

(m

µ

))(2.176)

64

.

=

Fig. 2.7.3 - The second order contributions at the electron propagator.

A = −me2

2π2

(1

ε+ Fm

(m

µ

))(2.177)

with F2 andFm finite for ε → 0. Notice also that these two functions are dimen-sionless and arbitrary up to this moment. However they can be determined by therenormalization conditions, that is by fixing the arbitrary constants appearing inthe lagrangian. In fact, given

iS−1F (p) ≡ Γ(2)(p) = p−m+Bp−A + e2Σ(p) (2.178)

we can require that at the physical pole, p = m, the propagator coincides with thefree propagator

SF (p) ≈ i

p−m, for p = m (2.179)

From here we get two conditions. The first one is

0 = Γ(2)(p = m) =e2

8π2ε(p−4m)+e2Σf (p = m)− e2

8π2

(1

ε+ F2

)p+

me2

2π2ε

(1

ε+ Fm

)(2.180)

from which

e2Σf(p = m) − me2

8π2F2 +

me2

2π2Fm = 0 (2.181)

The second condition is∂Γ(2)(p)

∂p

∣∣∣p=m

= 1 (2.182)

65

which gives

e2∂Σf (p)

∂p

∣∣∣p=m

− e2

8π2F2 = 0 (2.183)

One should be careful because these particular conditions of renormalization givessome problems related to the zero mass of the photon. In fact one finds some ill-defined integral in the infrared region. However these are harmless divergences,because giving a small to the photon and letting it to zero at the end of the calcula-tions gives rise to finite results. Notice that these conditions of renormalization havethe advantage of being expressed directly in terms of the measured parameters, asthe electron mass. However, one could renormalize at an arbitrary mass scale, M . Inthis case the parameters comparing in Lp are not the directly measured parameters,but they can be correlated to the actual parameters by evaluating some observablequantity. From this point of view one could avoid the problems mentioned above bychoosing a different point of renormalization.

As far as the vacuum polarization is concerned, Πµν gives rise to the followingcorrection of the photon propagator (illustrated in Fig. 2.7.4)

D′µν(q) =

−igµνq2

+−igµλq2

ie2Πλρ(q)−igρνq2

+ · · · (2.184)

.

Fig. 2.7.4 - The loop expansion for the photon propagator.

from which

D′µν(q) =

−igµνq2

+−igµλq2

[(ie2)(qλqρ − gλρq2)Π(q2)

] −igρνq2

+ · · ·

=−igµνq2

[1 − e2Π(q2)

]− iqµqνq4

e2Π(q2) + · · · (2.185)

We see that the one loop propagator has a divergent part in gµν , and also divergentand finite pieces in the term proportional to the momenta. Therefore the propagatoris not any more in the Feynman gauge. It follows that we should add to Lp

L2 = −1

4FµνF

µν − 1

2(∂µA

µ)2 =1

2AµgµνAν (2.186)

66

the two counterterms

(L2)ct = −C4FµνF

µν − E

2(∂µA

µ)2 = −C(

1

4FµνF

µν +1

2(∂µA

µ)2

)− E − C

2(∂µA

µ)2

(2.187)As for the electron propagator, we can look at these two contributions as pertur-bations to the free lagrangian, and evaluate the corresponding Feynman rules, orevaluate the effect on the propagator, and then expand in the counter terms. Theeffect of these two terms to the equation which defines the propagator in momentumspace is

[(1 + C)q2gµν − (C − E)qµqν ]Dνλ(q) = −igλµ (2.188)

We solve this equation by putting

Dµν(q) = α(q2)gµν + β(q2)qµqν (2.189)

Substituting in the previous equation we determine α e β. The result is

α = − i

q2(1 + C), β = − i

q4

C − E

(1 + C)(1 + E)(2.190)

The free propagator, including the corrections at the first order in C and E is

Dµν(q) =−igµνq2

(1 − C) − iqµqνq4

(C −E) (2.191)

and the total propagator

D′µν(q) =

−igµνq2

[1 − e2Π(q2) − C] − iqµqνq4

[e2Π(q2) + C −E] (2.192)

We can now choose E = 0 and cancel the divergence by the choice

C = − e2

6π2ε+ F3

(m

µ

)(2.193)

In fact we are free to choose the finite term of the gauge fixing, since this choice doesnot change the physics. This is because the terms proportional to qµqν , as it followsfrom the gauge invariance, and the conservation of the electromagnetic current. Forinstance, if we have a vertex with a virtual photon (that is the vertex is connectedto an internal photon line) and two external electrons, the term proportional to qµqνis saturated with

u(p′)γµu(p), q = p′ − p (2.194)

The result is zero, bay taking into account the Dirac equation. Let us see what thefull propagator says about the mass of the photon. We have

D′µν(q) =

−igµνq2

[1 − F3 − e2Πf

]− iqµqνq4

[F3 + e2Πf

] ≡ −igµνq2

[1 − Π] − iqµqνq4

Π

(2.195)

67

whereΠ = e2Πf + F3 (2.196)

At the second order in the electric charge we can write the propagator in the form

D′µν(q) =

−iq2(1 + Π(q))

[gµν +

qµqνq2

Π(q)

](2.197)

and we see that the propagator has a pole at q2 = 0, since Π(q2) is finite for q2 → 0.Therefore the photon remains massless after renormalization. This is part of a rathergeneral aspect of renormalization which says that if the regularization procedurerespects the symmetries of the original lagrangian, the symmetries are preserved atany perturbative order. An exception is the case of the anomalous symmetries, whichare symmetries at the classical level, but are broken by the quantum corrections.All the anomalous symmetries can be easily identified in a given theory.

Also in this case we will require that at the physical pole, q2 → 0, the propagatorhas the free form, that is

Π(0) = 0 (2.198)

from whichF3 = −e2Πf (0) (2.199)

For small momenta we can write

Π(q) ≈ e2q2dΠf(q)

dq2

∣∣∣q2=0

(2.200)

since F3 is a constant. Using the expression for Πf from the previous Section, weget

Πf (q) ≈ − 1

2π2

(γ

6+

1

6log

(m2

2πµ2

))+

1

2π2

∫ 1

0

dz z2(1 − z)2 q2

m2+ · · · (2.201)

and

Π(q) ≈ e2q2

60π2m2(2.202)

from which

D′µν = −igµν

q2

[1 − e2q2

60π2m2

]+ gauge terms (2.203)

The first term, 1/q2, gives rise to the Coulomb potential, e2/4πr. Therefore this ismodified by a constant term in momentum space, or by a term proportional to thedelta function in the real space

∆12 = e2∫

dt

∫d4q

(2π)4e−iq(x1 − x2)

[1

q2− e2

60π2m2

]= − e2

4πr− e4

60π2m2δ3(r)

(2.204)

68

This modification of the Coulomb potential modifies the energy levels of the hydro-gen atom, and it is one of the contributions to the Lamb shift, which produces asplitting of the levels 2S1/2 and 2P1/2. The total Lamb shift is the sum of all theself-energy and vertex corrections, and turns out to be about 1057.9 MHz. Thecontribution we have just calculated is only −27.1 MHz, but it is important sincethe agreement between experiment and theory is of the order of 0.1 MHz.

We have now to discuss the vertex corrections. We have seen that the divergentcontribution is Λ

(1)µ (p′, p), and this is proportional to γµ. The counter term to add

to interacting part of Lp

Lintp = −eψγµψAµ (2.205)

is(L3)ct = −eDψγµψAµ (2.206)

The complete vertex is given by (see Fig. 2.7.5)

−ie [γµ + e2Λµ +Dγµ]

(2.207)

.

(counter-term)

Fig. 2.7.5 - The one loop vertex corrections.

We fix the counter term by

D = − e2

8π2

[1

ε+ F2

](2.208)

with F2 the same as in eq. (2.7). The reason is that with this choice we get theequality of the wave function renormalization factors Z1 = Z2. This equality followsfrom the conservation of the current, and therefore we can use the arbitrariness inthe finite part of the vertex counter term to guarantee it. The one-loop vertex isthen

(Λµ)′ =

[γµ + e2(Λf

µ −F2

8π2γµ)

](2.209)

One can see that this choice of F2 is such that for spinors on shell the vertex is thefree one

u(p)(Λµ)′u(p)|p=m = u(p)γµu(p) (2.210)

69

We will evaluate now the radiative corrections to the g − 2 of the electron. Here gis the gyromagnetic ratio, which the Dirac equation predicts to be equal to be two.To this end we first need to prove the Gordon identity for the current of a Diracparticle

u(p′)γµu(p) = u(p′)[pµ + pµ′

2m+

i

2mσµνq

ν

]u(p) (2.211)

con q = p′ − p. For the proof we start from

pγµu(p) = (−mγµ + 2pµ)u(p) (2.212)

andγµpu(p) = mγµ (2.213)

Subtracting these two expressions we obtain

γµu(p) =

(pµm

− i

mσµνp

ν

)u(p) (2.214)

An analogous operation on the barred spinor leads to the result. We observe alsothat the Gordon identity shows immediately that g = 2, because it implies that thecoupling with the electromagnetic field is just

e

2mσµνF

µν(q) (2.215)

To evaluate the correction to this term from the one loop diagrams, it is enough toevaluate the matrix element

e2u(p′)Λ(2)µ (p′, p)u(p) (2.216)

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 the free vertex,as implied by the renormalization condition. Therefore we will ignore all the termsproportional to γµ and we will take the first order in the momentum q. For momenta

on shell, the denominator of Λ(2)µ is given by

[...] = m2(x+ y) −m2x(1 − x) −m2y(1 − y) + 2m2xy = m2(x+ y)2 (2.217)

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) (2.218)

Using pγα = −γαp+2pα, and an analogous equation for p′, we can bring p to act onthe spinor at the right of the expression, and p′ on the spinor at the left, obtaining

Vµ = u(p′)[myγα + 2(1 − y)p′α − γαpx

]γµ [mxγα + 2(1 − x)pα − p′yγα] u(p)

(2.219)

70

Making use ofγαγ

µγα = −2γµ (2.220)

γαγµγνγα = 4gµν (2.221)

γαpγµp′γα = −2p′γµp (2.222)

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) (2.223)

and for the piece which does not contain γµ

Vµ = 4mu(p′)[pµ(y − xy − x2) + p′µ(x− xy − y2)

]u(p) (2.224)

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

dy4

mu(p′)

[pµ

(y − xy − x2)

(x+ y)2

+ p′µ(x− xy − y2)

(x+ y)2

]u(p) (2.225)

By changing variable z = x+ y we have∫ 1

0

dx

∫ 1−x

0

dyy − xy − x2

(x+ y)2=

∫ 1

0

dx

∫ 1

x

dz

[1 − x

z− x

z2

]

=

∫ 1

0

[−(1 − x) log x+ x− 1]

=

[−x log x+ x+

x2

2log x− x2

4+x2

2− x

]1

0

=1

4(2.226)

from which

e2u(p′)Λ(2)µ (p′, p)u(p) → − e2

16π2mu(p′)[pµ + p′µ]u(p) (2.227)

Using the Gordon in this expression, and eliminating the further contribution in γµwe obtain

e2u(p′)Λ(2)µ (p′, p)u(p)|mom. magn. → ie2

16π2mu(p′)σµνqνu(p) (2.228)

Finally we have to add this correction to the vertex part taken at p′ = p, whichcoincides with the free vertex

u(p′)[γµ +ie2

16π2mσµνq

ν ]u(p) = u(p′)[pµ + p′µ

2m+

i

2m

(1 +

e2

8π2

)σµνq

ν ]u(p) (2.229)

71

Therefore the correction is

e

2m→ e

2m

(1 +

α

2π

)(2.230)

Recalling that g is the ratio between S · B and e/2m, we get

g

2= 1 +

α

2π+ O(α2) (2.231)

This correction was evaluated by Schwinger in 1948. Actually we know the firstthree terms of the expansion

ath =1

2(g − 2) =

1

2

α

π− 0.32848

(απ

)2

+ 1.49(α

π)3 + · · · = (1159652.4± 0.4) × 10−9

(2.232)to be compared with the experimental value

aexp = (1159652.4 ± 0.2) × 10−9 (2.233)

72

Chapter 3

Vacuum expectation values andthe S matrix

3.1 In- and Out-states

For the future we will need a formulation of field theory showing that any S-matrixelement can be reduced to the evaluation of the vacuum expectation value of a T -product of field operators. Such a formulation is called LSZ (from the authors l(Lehaman, Symanzik and Zimmermann). This formalism is based on a series ofvery general properties as displacement and Lorentz invariance. It is interestingto see how far one can go in determining the features of a field theory by usingonly invariance arguments. In the following we will work in the Heisenberg picture,that is the operators will be time-dependent and evolving with the full hamiltonian.Formally this representation can be obtained from eq. (1.217) by sending the freehamiltonian into the full hamiltonian. It should be also kept in mind that sinceno exact solution of an interacting field theory is known, we are not sure that theassumptions we are going to make are consistent. Said that, our assumptions are:

1. The eigenvalues of the four-momentum lie within the forward light cone

p2 = pµpµ ≥ 0, p0 ≥ 0 (3.1)

2. There exists a nondegenerate Lorentz invariant states of minimum energy.This is called the vacuum state

Φ0 ≡ |0〉 (3.2)

By convention, we assume that the corresponding energy is zero

P0|0〉 = 0 (3.3)

From eq. (3.1) we get also

P|0〉 = 0 → Pµ|0〉 = 0 (3.4)

73

Since pµ = 0 is a Lorentz invariant condition, it follows that Φ0 appears as thevacuum states in all the Lorentz frames.

3. For each stable particle of mass m, there exists a stable stae of single particle

Φ1 ≡ |p〉 (3.5)

where Φ1 is a momentum eigenstate with eigenvalue pµ such that p2 = m2.

4. Neglecting the complications from the zero mass states we can add a furtherrequirement: the vacuum and the single particle states give a discrete spectrumin pµ. For instance, the case of the π mesons is given in Fig. 3.1.

.

xxxxxxxxx

xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

xxxx

xx

xxxx

E

p

Vacuum

m

2m

Continuum

singl

e par

ticle

state

Fig. 3.1 - Energy-momentum spectrum for particles of mass m.

Remember that the pion is not a stable particle and decays into µ + νµ witha lifetime of the order of ≈ 2 · 10−8 sec., but this number is very large withrespect to the natural decay frequency 1/mπ ≈ 5·10−24 sec. (mπ ≈ 140 MeV ).Therefore, by neglecting the weak interactions (responsible for the decay) wecan assume that the pion is stable. The same argument can be applied toquarks and leptons that will be the main actors in the following. Howeverwe mentioned also the pions, to say that the arguments presented here canbe applied (in the appropriate range of energy) also to composite particles.Notice that this assumption is very much in the spirit of perturbation theory.That is we introduce a field for each particle, assuming that the interactionsdo not modify the spectrum in a violent way.

The main tools of investigation in the field of elementary particle physics arethe scattering experiments. Ideally these experiments are realized by preparing at

74

t = −∞ a system of free particles. That is assuming that they do not interact eachother (this requires that the particles are well separated in space). These particleswill interact with a target at some finite time. After that scattering process weimagine that at t = +∞ the emerging particles will behave again as free particles.In real life the times t = ±∞ correspond to laboratory time. That is we preparethe beam at some finite time −T , the scattering process will occur around the timet = 0 (assuming this as the time scale) and then we will measure the outgoingparticles at a time T . The ideal description considered before can be considered ascorrect if the interaction time ∆t around t = 0 is much smaller that T . For instance,imagine an electron prepared in an eigenstate of momentum p, impinging over anatom. Here there is a further simplification that we are doing since in a finite time ora finite volume we cannot prepare a state of definite momentum. This is one of thereasons why analyzing a scattering process it is always convenient to quantize in abox of the dimension of the experiment itself (that is about the distance between thesource of the beam and the detector). The same consideration holds for the energy,that is one has always to remember that the experiments is taking place between tofinite times −T and T . Having prepared the electron in the initial state at the time−T , the particle will travel since it will reach the target where it will be subject tothe interaction with the atoms of target itself. The interaction time will be of theorder of the time necessary to cross the atomic dimensions for the case of a singlescattering, or, at maximum, of the order of the thickness of the target for a multiplescattering. After that we will let again the electron to move freely since it will reachthe detector where we will measure its properties as momentum and polarization. Inconclusion we can think to describe the particles at t = ±∞ in terms of free fields,except that they will be subject to the self-interaction which, as we know, cannotbe neglected. The free fields that we are going to use will be denoted by φin(x) fort→ −∞ and by φout(x) for t→ +∞. The interacting field φ(x)) can be thought ofas constructed in terms of these free fields operators and such that at t → ±∞ itreduces to φout(x) or φin(x). In order that the in- and out-fields describe correctlythe incoming and outgoing free particles we have to require some properties. Inparticular we will require

1. φin(x) must transform as the corresponding φ(x) with respect to the symme-tries of the theory. In particular, with respect to translation it must satisfy

[Pµ, ] = −i∂φin(x)

∂xµ(3.6)

equivalent to require

φin(x+ a) = eiP · aφin(x)e−iP · a (3.7)

2. φin(x) must satisfy the free Klein-Gordon equation (for a scalar field, or theDirac free equation for a spinor field, etc.) corresponding to the physicalmass m

( +m2)φin(x) = 0 (3.8)

75

Notice that this requirement implies that we have taken into account the self-interactions of the field which are responsible for the mass renormalization (wewill mention the wave function renormalization later).

The same properties are required for the out-fields. By using these two propertiesit follows at once that the in-field creates from the vacuum the physical one particlestate from the vacuum. In fact, let |p〉 an eigenstate ofPµ with eigenvalue pµ. Wehave

−i ∂∂xµ

〈p|φin(x)|0〉 = 〈p|[Pµ, φin(x)]|0〉 = pµ〈p|φin(x)|0〉 (3.9)

By applying −i∂/∂xµ once more, we find

−〈p|φin(x)|0〉 = p2〈p|φin(x)|0〉 (3.10)

and using the Klein-Gordon equation

(p2 −m2)〈p|φin(x)|0〉 = 0 (3.11)

We see that the only state that φin(x) can create out of the vacuum is the one withpµ such that p2 = m2. From our starting hypotheses we have that this is the stateof single particle with mass m. Since the in- and out-fields are free fields we canapply all the corresponding formalism. In particular they can be Fourier expanded

φin(x) =

∫d3k

[ain(k)fk(x) + a†in(k)f

∗k(x)]

(3.12)

with

fk(x) =1√

(2π)32ωke−ikx, ωk =

√k2 +m2 (3.13)

Inverting this relation, we find

ain(k) = i

∫d3x

[f ∗k(x)∂0φin(x) − ∂0f

∗k(x)φin(x)

](3.14)

from which

[Pi, ain(k)] =

∫d3x

[f ∗k(x)

(∂

∂x0

∂φin(x)

∂xi

)−(

∂

∂x0f ∗k(x)

)∂φin(x)

∂xi

]

= −∫d3x

[∂f ∗

k(x)

∂xi

(∂

∂x0φin(x)

)−(

∂

∂x0

∂f ∗k(x)

∂xi

)φin(x)

]

= −kiain(k) (3.15)

Since f ∗k(x) and φin(x) both satisfy the free Klein-Gordon equation, we see that the

operators defined in eq. (3.14) are time independent. This follows immediately, by

76

recalling that if φ1(x) and φ2(x) are both solutions of the Klein-Gordon equation,then the four-vector

jµ = φ1∂φ2

∂xµ− ∂φ1

∂xµφ2 (3.16)

is a conserved current, and therefore the charge

Q =

∫d3xj0 (3.17)

is a constant of motion. Using this, it follows

[P0, ain(k)] =

∫d3x

[f ∗k(x)

(∂

∂x0

∂φin(x)

∂x0

)−(

∂

∂x0f ∗k(x)

)∂φin(x)

∂x0

]

= −∫d3x

[∂f ∗

k(x)

∂x0

(∂

∂x0φin(x)

)−(

∂

∂x0

∂f ∗k(x)

∂x0

)φin(x)

]

= −k0ain(k) (3.18)

In conclusion[Pµ, ain(k)] = −kµain(k) (3.19)

Also, by hermitian conjugation we get

[Pµ, a†in(k)] = kµa

†in(k) (3.20)

It follows that a†in(k) creates states of four-momentum kµ whereas ain(k) annihilatesstates with momentum kµ. For instance,

Pµa†in(k1)a

†in(k2)|0〉 = [Pµ, a

†in(k1)a

†in(k2)]|0〉 = (k1 + k2)µa

†in(k1)a

†in(k2)|0〉 (3.21)

Pµain(k1)a†in(k2)|0〉 = [Pµ, ain(k1)a

†in(k2)]|0〉 = (k2 − k1)µain(k1)a

†in(k2)|0〉 (3.22)

By using the condition on the spectrum of the momentum operator we get also

ain(k)|0〉 = 0 (3.23)

since, otherwise, this state would have a negative eigenvalue of the energy. We getalso

〈p1, · · · , pM ; in|k1, · · · , kN ; in〉 = 0 (3.24)

unless M = N and the set (p1, · · · , pM) is identical to the set (k1, · · · , kN).Let us now look for a relation between the in- and the φ-field. The field φ(x)

will satisfy an equation of the type

( +m2)φ(x) = j(x) (3.25)

where j(x) (the scalar current) contains the self-interactions of the fields and possibleinteractions with other fields in the theory. We need to solve this equation with the

77

boundary conditions at t = ±∞ relating φ(x) with the in- and out-fields. Notice thatin writing the previous equation we have included the effects of mass renormalizationinside the current j(x), but we know that at the same time we have also a wavefunction renormalization. Therefore asymptotically the field φ(x) cannot be thesame as the in- and out-fields but it will differ by a factor

√Z, where

√Z is the

wave function renormalization. Therefore, the required asymptotic conditions are

limt→−∞

φ(x) = limt→−∞

√Zφin(x), lim

t→+∞φ(x) = lim

t→+∞

√Zφout(x) (3.26)

The solution for φ(x) is then

φ(x) =√Zφin(x) +

∫d4y∆ret(x− y;m2)j(y) (3.27)

We will see in a moment the meaning of Z in this context. The function ∆ret(x;m2)

is the retarded Green’s function, defined by

( +m2)∆ret(x;m2) = δ4(x) (3.28)

such that∆ret(x;m

2) = 0, for x0 < 0 (3.29)

The need of Z in this context, is because we want the the in-field properly normalizedin order to produce single particle states. In fact, both 〈1|φin(x)|0〉 e 〈1|φ(x)|0〉 havethe same x-dependence (they must be both proportional to exp(ipx), see eq. (3.6)),but φ(x) may create out of the vacuum states other than the single particle state,and therefore the normalization of the two previous matrix elements cannot be thesame.

It can be shown that the asymptotic condition (3.26), cannot hold as an operatorstatement. This is because it is not possible to isolate the operator φ(x) from thescalar current j(x) at t = −∞, since this includes also the self-interactions. Howeverthe statement can be given in a weak form for the matrix elements. The correctway of doing it is first to smear out the field over a space-like region

φf(t) = i

∫d3x (f ∗(x, t)∂0φ(x, t) − ∂0f

∗(x, t)φ(x, t)) (3.30)

with f(x, t) an arbitrary normalizable solution of the Klein-Gordon equation

( +m2)f(x) = 0 (3.31)

Then the asymptotic condition is formulated as

limt→−∞

〈α|φf(t)|β〉 =√Z limt→−∞

〈α|φf(t)in|β〉 (3.32)

where |α〉 and |β〉 are any two normalizable states.

78

Similar considerations can be made for the out-field. In particular we can find arelation between the in- and out-fields

φ(x) =√Zφout(x) +

∫d4y∆adv(x− y;m2)j(y) (3.33)

where∆adv(x;m

2) = 0, for x0 > 0 (3.34)

( +m2)∆adv(x;m2) = δ4(x) (3.35)

is the advanced Green’s function. By comparing this expression with eq. (3.27) wefind √

Zφin(x) =√Zφout(x) +

∫d4y∆(x− y;m2)j(y) (3.36)

with∆(x− y;m2) = ∆adv(x− y;m2) − ∆ret(x− y;m2) (3.37)

The construction of the out-operators of creation and annihilation goes along thesame lines discussed for the in-operators.

3.2 The S matrix

We have shown in the previous Section that an initial state of n non-interactingparticles can be described by

|p1, · · · , pn; in〉 = a†in(p1) · · ·a†in(pn)|0; in〉 ≡ |α; in〉 (3.38)

where α stands for the all collection of the quantum numbers of the initial state.After the scattering process we are interested to evaluate the probability amplitudefor the final state to be a state of non-interacting particles described by the out-operators

|p′1, · · · , p′n; out〉 = a†out(p′1) · · ·a†out(p

′n)|0; out〉 ≡ |β; out〉 (3.39)

The probability amplitude is given by the S matrix elements

Sβα = 〈β; out|α; in〉 (3.40)

We can also define an operator S transforming the in-states into the out-states

〈β; in|S = 〈β; out| (3.41)

such thatSβα = 〈β; in|S|α; in〉 (3.42)

From the definition we can derive some important properties for the operator S

79

1. Since the vacuum is stable and not degenerate, it follows |S00| = 1, and there-fore

〈0in|S = 〈0out| = eiφ〈0in| (3.43)

By choosing the phase equal to zero we get

|0; out〉 = |0; in〉 ≡ |0〉 (3.44)

2. In an analogous way the stability of the one particle state requires

|p; in〉 = |p; out〉 (3.45)

3. The operator S maps the in- to the out-fields

φin(x) = Sφout(x)S−1 (3.46)

In fact, let us consider the following matrix element

〈β; out|φout|α; in〉 = 〈β; in|Sφout|α; in〉 (3.47)

But 〈β; out|φout is an out-state and therefore

〈β; out|φout = 〈β; in|φinS (3.48)

Then

〈β; out|φout(x)|α; in〉 = 〈β; in|Sφout(x)|α; in〉 = 〈β; in|φin(x)S|α; in〉 (3.49)

which shows the relation (3.46).

4. The S is unitary. This follows from the very definition (3.41) implying

S†|α; in〉 = |α; out〉 (3.50)

from which〈β; in|SS†|α; in〉 = 〈β; out|α; out〉 = δαβ (3.51)

showing thatSS† = 1 (3.52)

From eq. (3.50) and the unitarity of S we get

|α; in〉 = S†−1|α; out〉 = S|α; out〉 (3.53)

5. S is Lorentz invariant, as it follows from the fact that transform in- into out-fields having the same Lorentz properties. For the same reason the S matrixis invariant under any symmetry of the theory.

80

3.3 The reduction formalism

In this Section we will describe a systematic way of evaluating the S matrix elementsin terms of the vacuum expectation values of T -products of field operators. Let usstart considering the following S matrix element

Sβ,αp = 〈β; out|α, p; in〉 (3.54)

where |α, p; in〉 is the state of a set of particles α plus an additional incoming particleof momentum pµ. Here we will consider the case of scalar particles, but the formalismcan be easily extended to other cases. We want to show that, by using the asymptoticcondition, it is possible to extract the particle of momentum pµ from the state byintroducing the corresponding field operator. Let us consider the following identities

〈β; out|α, p; in〉 = 〈β; out|a†in(p)|α; in〉= 〈β; out|a†out(p)|α; in〉 + 〈β; out|a†in(p) − a†out(p)|α; in〉= 〈β − p; out|α; in〉− i

∫d3x[fp(x)

(∂

∂x0〈β; out|φin(x) − φout(x)|α; in〉

)

−(

∂

∂x0fp(x)

)〈β; out|φin(x) − φout(x)|α; in〉

](3.55)

where we have used the adjoint of eq. (3.14). We have denoted |β− p; out〉 the out-state obtained by removing the particle with momentum pµ, if present in the state,otherwise it must be taken as the null vector. As we know the integral appearingin the previous expression is time-independent since fp(x), φin(x) and φout(x) aresolutions of the Klein-Gordon equation. Therefore we are allowed to make thefollowing substitutions (in the weak sense, since we are taking matrix elements)

limx0→−∞

φin(x) = limx0→−∞

1√Zφ(x)

limx0→+∞

φout(x) = limx0→+∞

1√Zφ(x) (3.56)

obtaining

〈β; out|α, p; in〉 = 〈β − p; out|α; in〉+

i√Z

(lim

x0→+∞− lim

x0→−∞

)∫d3x[fp(x)

(∂

∂x0〈β; out|φ(x)|α; in〉

)

−(

∂

∂x0fp(x)

)〈β; out|φ(x)|α; in〉

](3.57)

By using the identity∫ +∞

−∞d4x

∂

∂x0

[g1(x)

(∂

∂x0g2(x)

)−(

∂

∂x0g1(x)

)g2(x)

]=

81

=

(lim

x0→+∞− lim

x0→−∞

)∫d3x

[g1(x)

(∂

∂x0g2(x)

)−(

∂

∂x0g1(x)

)g2(x)

]

=

∫ +∞

−∞d4x [g1(x)g2(x) − g1(x)g2(x)] (3.58)

we can write

〈β; out|α, p; in〉 = 〈β − p; out|α; in〉+

i√Z

∫d4x〈β; out|

[fp(x)φ(x) − fp(x)φ(x)

]|α; in〉 (3.59)

By using the Klein-Gordon equation for fp(x) we get

〈β; out|α, p; in〉 = 〈β − p; out|α; in〉+

i√Z

∫d4x〈β; out|

[fp(x)φ(x) − ((∇2 −m2)fp(x)

)φ(x)

]|α; in〉 (3.60)

Integrating by parts

〈β; out|α, p; in〉 = 〈β − p; out|α; in〉+

i√Z

∫d4xfp(x)( +m2)〈β; out|φ(x)|α; in〉 (3.61)

We can iterate this procedure in order to extract all the in- and out-particles fromthe states. Let us see also the case in which we want to extract from the previousmatrix element an out-particle. This exercise is interesting because it shows how theT -product comes about. Let us suppose that the out state is of the type β = (γ, p′),with p′ the momentum of a single particle. We want to extract this particle fromthe state

〈β; out|φ(x)|α; in〉 = 〈γ, p′; out|φ(x)|α; in〉 = 〈γ; out|aout(p′)φ(x)|α; in〉

= 〈γ; out|[aout(p′)φ(x) − φ(x)ain(p

′)]|α; in〉 + 〈γ; out|φ(x)ain(p′)|α; in〉

= 〈γ; out|φ(x)|α− p′; in〉+ i

∫d3y[fp′∗(y)

(∂

∂y0〈γ; out|(φout(y)φ(x) − φ(x)φin(y))|α; in〉

)

−(

∂

∂y0f ∗p′(y)

)〈γ; out|(φout(y)φ(x) − φ(x)φin(y))|α; in〉

](3.62)

Using again the time independence of the integral we have

〈γ; out|[φout(y)φ(x) − φ(x)φin(y)]|α; in〉=

1√Z

(lim

y0→+∞− lim

y0→−∞

)〈γ; out|T (φ(y)φ(x))|α; in〉 (3.63)

82

and eq. (3.58)

〈β; out|φ(x)|α; in〉 = 〈γ; out|φ(x)|α− p′; in〉+

i√Z

(lim

y0→+∞− lim

y0→−∞

)∫d3y[f ∗p′(y)

(∂

∂y0〈γ; out|T (φ(y)φ(x))|α; in〉

)

−(

∂

∂y0f ∗p′(y)

)〈γ; out|T (φ(y)φ(x))|α; in〉

]= 〈γ; out|φ(x)|α− p′; in〉+

i√Z

∫d4y〈γ; out|

[f ∗p′(y)

∂2

∂y20

T (φ(y)φ(x)) − f ∗p′(y)T (φ(y)φ(x))

]|α; in〉

= 〈γ; out|φ(x)|α− p′; in〉+

i√Z

∫d4yf ∗

p′(y)( +m2)y〈γ; out|T (φ(y)φ(x))|α; in〉 (3.64)

Substituting inside the (3.61)

〈β; out|α, p; in〉= 〈β − p; out|α; in〉+

i√Z

∫d4xfp(x)( +m2)x〈γ; out|φ(x)|α− p′; in〉

+

(i√Z

)2 ∫d4xd4y

fp(x)f∗p′(y)( +m2)x( +m2)y〈γ; out|T (φ(y)φ(x))|α; in〉 (3.65)

More generally if we want to remove all the particles in (q1, · · · , qm) and in (p1, · · · , pn)with the condition pi = qj we get

〈p1, · · · , pm; out|q1, · · · , qn; in〉

=

(i√Z

)m+n ∫ m,n∏i,j=1

d4xid4yjfqi(xi)f

∗pj

(yj)( +m2)xi( +m2)yj

·

〈0|T (φ(y1) · · ·φ(yn)φ(x1) · · ·φ(xm))|0〉 (3.66)

If some of the qi’s are equal to some of the pj’s, we can reduce further the matrixelements of the type 〈β; out|α−p; in〉. Therefore we have shown that it is possible toevaluate all the S matrix elements once one knows the vacuum expectation values ofthe T -products (called also the n-point Green’s function). The result obtained herecan be easily extended to generic fields simply by substituting to the wave functionsfp(x) the corresponding solutions of the wave equation (for instance, for spinor fields,spinors times plane waves) and to the Klein-Gordon operator the corresponding waveoperators (for the spinor case the Dirac operator (i∂/ −m)). In the Dirac case onehalso to change i/

√Z to −i/√Z2 for fermions and to +i/

√Z2 for antifermions,

where Z2 is the wave function renormalization for the spinor field.

83

We will show in the following that a simple technique to evaluate perturbativelythe vacuum expectation value of n fields, a n-point Green’s function, can be obtainedin the context of the path-integral formulation of quantum mechanics.

84

Chapter 4

Path integral formulation ofquantum mechanics

4.1 Feynman’s formulation of quantum mechan-

ics

In 1948 Feynman gave a formulation of quantum mechanics quite different from thestandard one. This formulation did not make use of operators and of Schrodingerequation, but rather it was giving an explicit expression for the quantum-mechanicalamplitude, although using a mathematical formalism which was far from being welldefined, which was obtained directly from the physical idea that the probabilityamplitude to go from one state to another can be obtained by summing togetherall the possible ways with an appropriate weight. This was as apply the ordinarycomposition laws in probability theory, to the probability amplitudes (rememberthat the probability is the square of the amplitude). The key observation to get theappropriate weight was taken by the Dirac’s book where it was shown that, for onedegree of freedom, the probability amplitude

〈q′, t+ ∆t|q, t〉 (4.1)

for ∆t infinitesimal, is given by the eiS where S is the classical action to go from q atthe time t to q′ at the tome t+ ∆t. In fact we will show that the Feynman’s formu-lation can be derived from the usual formulation of quantum mechanics. Althoughthe path integral formulation (as the Feynman’s approach is usually called) is justa different way of formulating quantum mechanics, at the beginning of the 70’s itturned out to be a crucial tool for the quantization of gauge theories. Also it is onethe few methods which can be used for studying situations outside the perturba-tive approach. In order to derive the Feynman’s formulation we will start studyinga quantum system described by one degree of freedom, and we will see later howthis can be extended also to infinite degrees of freedom as needed in quantum fieldtheory. First of all we notice that the quantum mechanical problem is to evaluate

85

the time evolution, which is given by the propagator. This can be evaluated in anybasis, for instance in configuration space, its matrix elements are given by

〈q′, t′|q, t〉, t′ ≥ t (4.2)

It will be convenient to work in the Heisenberg representation, that is using time-dependent operators q(t). The states of the previous equation are defined as eigen-states of the operator q(t) at the time t, that is

q(t)|q, t〉 = |q, t〉q (4.3)

It follows immediately that the relation between these states at different times isgiven by

|q, t〉 = eiHt|q, 0〉 (4.4)

for a time independent hamiltonian. The matrix element (4.2) can be evaluatedby slicing the time interval t′ − t in infinitesimal pieces and using the completenessrelation. we get

〈q′, t′|q, t〉 = limn→∞

∫dq1 · · · dqn−1〈q′, t′|qn−1, tn−1〉〈qn−1, tn−1|qn−2, tn−2〉 ·

· · · 〈q1, t1|q, t〉 (4.5)

wheretk = t+ kε, t0 ≡ t, tn ≡ t′ (4.6)

Therefore

〈q′, t′|q, t〉 = limn→∞

∫ (n−1∏k=1

dqk

)(n−1∏k=0

〈qk+1, tk+1|qk, tk〉)

(4.7)

We have still to evaluate the matrix element (4.2), but now for an infinitesimal timeinterval ε. let us use again the completeness in momentum space with states takenat a time t in between tk e tk+1:

〈qk+1, tk + ε|qk, tk〉 =

∫dp〈qk+1, tk + ε|p, t〉〈p, t|qk, tk〉 (4.8)

The mixed matrix elements can be written as

〈qk+1, tk + ε|p, t〉 = 〈qk+1, 0|e−iH(p,q)(tk + ε− t)|p, 0〉 (4.9)

(remember /h = 1). making use of the canonical commutation relations for p andq, the hamiltonian can be written in such a way that all the operators q are at theleft of all the operators p. The result will be denoted by the symbol H+(q,p). Bydoing so we get

〈qk+1, tk + ε|p, t〉 〈qk+1, 0|p, 0〉(1 − iH+(qk+1, p)(tk + ε− t)

) 〈qk+1, 0|p, 0〉e−iH+(qk+1, p)(tk + ε− t) (4.10)

86

The function H+(q, p) is obtained by substituting in H+(q,p) the operators q andp with their eigenvalues. Notice that although H(q,p) and H+(q,p) are the sameoperator, H(q, p) e H+(q, p) are, in general, different functions. In this way we get

〈qk+1, tk + ε|p, t〉 1√2πei(pqk+1 −H+(qk+1, p)(tk+1 − t)) (4.11)

and

〈p, t|qk, tk〉 1√2πei(−pqk −H−(qk, p)(t− tk)) (4.12)

where H− is the eigenvalue of the operatorH−(p,q) defined by putting the operatorsq to the right of p by using the canonical commutation relations. By choosingt = (tk + tk+1)/2 and the previous two equations, we get

〈qk+1, tk + ε|qk, t〉 ∫

dp

2πeiε(p(qk+1 − qk)/ε−Hc(qk+1, qk, p)) (4.13)

where

Hc(qk+1, qk, p) =1

2(H+(qk+1, p) +H−(qk, p)) (4.14)

Since in the limit ε→ 0 we have

〈qk+1, tk + ε|qk, tk〉 → δ(qk+1 − qk) (4.15)

it is reasonable to define a variable velocity as

qk =qk+1 − qk

ε(4.16)

and writeHc(qk+1, qk, p) ≡ H(qk, p) (4.17)

In the same limit H(qk, p) is nothing but the hamiltonian of the system evaluatedin the phase space. Therefore

〈qk+1, tk + ε|qk, tk〉 ∫

dp

2πeiεL(qk, p) (4.18)

whereL(qk, p) = pqk −H(qk, p) (4.19)

is the lagrangian in phase space. Coming back to (4.7) and using (4.18) we obtain

〈q′, t′|q, t〉 = limn→∞

∫ (n−1∏k=1

dqk

)(n−1∏k=0

dpk2π

)e

i

n−1∑k=0

εL(qk, pk)

(4.20)

87

In the limit, the argument of the exponential gives the integral between t and t′ ofthe lagrangian in the phase space, therefore we get the canonical action

S =

∫ t′

t

dt L(q, p) (4.21)

In the limit n → ∞, the multiple integral in eq. (4.20) is called the functionalintegral and we will denote it symbolically as

〈q′, t′|q, t〉 =

∫ q′,t′

q,t

dµ(q(t))dµ(p(t))ei

∫ t′

t

dtL(q, p)(4.22)

where

dµ(q(t)) =∏

t<t′′<t′dq(t′′), dµ(p(t)) =

∏t≤t′′≤t′

dp(t′′)2π

(4.23)

The asymmetry between the integration in q(t) and p(t) implies that the integral isnot invariant under a generic canonical transformation. Notice that the integrationis made over all the functions q(t′′) such that q(t) = q e q(t′) = q′. On the contrarythere are no limitations on p(t′′), as it required by the uncertainty principle.

4.2 Path integral in configuration space

The path integral in eq. (4.20) can be easily written as a functional integral inconfiguration space when the hamiltonian of the system is of the type

H(q,p) =p2

2m+ V (q) (4.24)

In such a case

H+(q, p) = H−(q, p) = H(q, p) =p2

2m+ V (q) (4.25)

The momentum integration can be done naively

〈q′, t′|q, t〉 = limn→∞

∫ (n−1∏k=1

dqk

)

·n−1∏k=0

dpk2π

ei[pk(qk+1 − qk) − εp2k/2m− ε(V (qk+1) + V (qk))/2] =

= limn→∞

∫ (n−1∏k=1

dqk

)(1

2π

√2mπ

iε

)n n−1∏k=0

e(−2m/iε)(qk+1 − qk)2/4

· e−iε(V (qk+1) + V (qk))/2 =

88

= limn→∞

∫ (√m

2iπε

)n(n−1∏k=1

dqk

)

·n−1∏k=0

eiε[m((qk+1 − qk)/ε)2/2 − (V (qk+1) + V (qk))/2] (4.26)

with the result

〈q′, t′|q, t〉 = limn→∞

∫ (√m

2iπε

)n(n−1∏k=1

dqk

)

· e

iε

n−1∑k=0

[m((qk+1 − qk)/ε)2/2 − (V (qk+1) + V (qk))/2]

(4.27)

We have also

limn→∞

n−1∑k=0

ε

[m

2

(qk+1 − qk

ε

)2

− V (qk+1) + V (qk)

2

]=

∫ t′

t

dtL(q, q) (4.28)

where L(q, q) is the lagrangian in configuration space. Therefore we will write

〈q′, t′|q, t〉 =

∫ q′,t′

q,t

D(q(t))ei

∫ t′

t

dtL(q, q)(4.29)

with the functional measure given by

D(q(t)) = limn→∞

(√m

2iπε

)n n−1∏k=1

dqk (4.30)

It is the expression (4.29) (really defined by (4.27)) which was originally formulatedby Feynman. In fact it has a simple physical. But before discussing the interpre-tation let us notice that for N degrees of freedom and with hamiltonians of thetype

H(qi,pi) =

N∑i=1

p2i

2m+ V (qi) i = 1, . . . , N (4.31)

the expression (4.29) can be easily generalized. However for arbitrary system onehas to start with the formulation in the phase space of the previous Section, whichalso can be easily generalized to an arbitrary number of degrees of freedom.

For discussing the interpretation of the expression (4.29), let us recall the bound-ary conditions

q(t) = q, q(t′) = q′ (4.32)

89

.

q q'

t

t'

Fig. 4.2.1 - The Feynman definition for the path integral.

The expression in the exponent of (4.29) is nothing but the action evaluated for acontinuous path starting from q at time t and ending at q′ at time t′. This path isapproximated in eq. (4.27) by many infinitesimal straight paths as shown in Fig.4.2.1.

Therefore the interpretation is that in order to evaluate the amplitude to go from(q, t) to (q′, t′) is obtained by multiplying together the amplitudes corresponding tothe infinitesimal straight paths. Each of these amplitudes (neglecting the normal-ization factor

√m/2iπε) is given by the exponential of i times the action evaluated

along the path. Furthermore we need to sum (or integrate functionally) over all thepossible paths joining the two given end points. In a more concise way one says thatthe amplitude to from one point to another is obtained by summing over all thepossible paths joining the two points with a weight proportional to the exponentialof iS, where S is the classical action evaluated along the path.

90

4.3 The physical interpretation of the path inte-

gral

Feynman arrived to the path integral formulation in configuration space on a morephysical basis. First of all he knew that the amplitude for infinitesimal time differ-ences is proportional to the exponential of iS, with S the classical action evaluatedalong the infinitesimal path. From that one can derive directly the final formula inthe following way. Let us consider the classical double slit experiment, as illustratedin Fig. 4.3.1, where S is the electron source. The electrons come out of the slit Aand arrive to the screen C, where they are detected by D, going through one of thetwo silts A or B. By counting the electrons with the detectpr D we can determine,as a function of x, how many electrons arrive on C, and therefore the probabilitydistribution of the electrons on the screen C. Under the conditions of Fig. 4.3.1,that is with both slits 1 and 2 open, one gets the distribution given in Fig. 4.3.2.In order to explain this distribution it would seem reasonable to make the followingassumptions:

.

S

D

x

A B C

2

1

Fig. 4.3.1 - The double slit experiment.

.

x

P

Fig. 4.3.2 - The experimental probability distribution, P , with both slits open.

• Each electron must go through the slit 1 or through the slit 2.

91

• As a consequence the probability for the electron to reach the screen at thepoint x must be given by the sum of the probability to reach x through 1 withthe probability to go through 2.

These hypotheses can be easily checked by closing one of the two slits and doingagain the experiment. The results obtained in the two cases are given in Fig. 4.3.3.

.

x

P P2 1

Fig. 4.3.3 - The experimental probability distributions, P1 and P2 obtained keepingopen the sli1 1 or the slit 2 respectively.

We see that P = P1 + P2, therefore one of our hypotheses must be wrong. Infact, quantum mechanics tells us that we can say that an electron went through oneof the two slits with absolute certainty, only if we detect its passage through theslit. Only in this case one would get P = P1 +P2. In fact, this is the result that oneobtains when doing the experiment with a further detector telling us which of theholes have been crossed by the electron. This means that with this experimentalsetting the interference gets destroyed. On the contrary the experiment tells us thatwhen we do not detect the hole crossed by the electron, the total amplitude for theprocess is given by the sum of the amplitudes corresponding to the two differentpossibilities, that is

P = |A|2 = |A1+A2|2 = |A1|2+|A2|2+A1A2+A1A2 = |A1|2+|A2|2 = P1+P2 (4.33)

The two expressions differ for a term which is not positive definite and which is theresponsible for the interference pattern. Therefore, quantum mechanics tells us thatthe usual rules for combining together probabilities are correct, but the result thatwe get follows from our ignorance about the path followed by the particle, unlesswe measure it. That is it does not make sense, under such circumstances, define aconcept as the path followed by a particle. The Feynnman’s interpretation is a littlebit different although at the end the theories turn out to be equivalent. He saysthat it is still possible to speak about trajectories of the electron if one gives up tothe usual rules of combining probabilities. This point of view requires taking intoaccount different type of alternatives according to different experimental settings.In particular one defines the following alternatives

92

1. Exclusive alternative - We will speak about exclusive alternatives when weuse an apparatus to determine the alternative chosen by the system. For in-stance one we use a detector to determine the slit crossed by the electrons. Inthis case the total probability is given by the sum of the probabilities corre-sponding to the different alternatives,

P =∑i

Pi. (4.34)

2. Interfering alternatives - This is the situation where the alternative chosenby the system is not detected by an experimental apparatus. In this casewe associate to each alternative an amplitude Ai and assume that the totalamplitude is given by the sum of all the amplitudes Ai

A =∑i

Ai (4.35)

The total probability for the process is evaluated as

P = |A|2 = |∑i

Ai|2 (4.36)

Therefore the probability rules are strictly related to the experimental setting. Thereare situations in which one may have both type of alternatives. For instance, if wewant to know the total probability of having an electron in C within the interval(a, b), having a detector in C and not looking at the slit crossed by the electrons. Inthis case the total probability will be given by

P =

∫ b

a

dx|A1(x) + A2(x)|2 (4.37)

Since the alternatives of getting the electron at different points in C are exclusiveones.

From this point of view it is well conceivable to speak about trajectories of theelectron, but we have to associate to each possible path an amplitude and then sumall over the possible paths. For instance, suppose that we want to evaluate theamplitude to go from S to the screen C when the screen B is eliminated. This canbe done by increasing the number of holes in the screen B. Then, if we denote byA(x, y) the amplitude for the electron to reach x going through the hole placed ata distance y from the center of B, the total amplitude will be given by

A(x) =

∫dyA(x, y) (4.38)

We can continue by using more and more screen of type B with more and moreholes. That is we can construct a lattice of points between A and C and construct

93

all the possible paths from A to the point x on C joining the sites of the lattice,as shown in Fig. 4.3.4. By denoting with C the generic path we get that the totalamplitude is given by

A(x) =∑CA(x, C) (4.39)

.

S

D

x

A C

C

C

1

2

Fig. 4.3.4 - The lattice for the evaluation of equation (4.39).

From this point it is easy to get the path integral expression for the total am-plitude, since each path is obtained by joining together many infinitesimal paths,and for each of these paths the amplitude is proportional to exp(iS) where S is theaction for the infinitesimal path. Therefore, for the single path C the amplitude willbe given by exp(iS(C). This is the final result, except for a proportionality factor.This is easy to get if one starts from the phase space formulation. This is importantbecause in many cases the simple result outlined above is not the correct one. Forinstance, consider the following lagrangian

L =m

2q2f(q) − V (q) (4.40)

where f(q) is an arbitrary function of q. The canonical momentum is given by

p = mqf(q) (4.41)

and the corresponding hamiltonian is

H =1

2mf(q)p2 + V (q) (4.42)

94

Notice that for qk ≈ qk+1 we have f(qk) ≈ (f(qk) + f(qk+1))/2, where qk = (qk +qk+1)/2. Therefore, form eq. (4.20) we get

〈q′, t′|q, t〉 = limn→∞

∫ (n−1∏k=1

dqk

)

·n−1∏k=0

dpk2π

ei[pk(qk+1 − qk) − εp2k/(2mf(qk)) − ε(V (qk+1) + V (qk))/2] (4.43)

This expression is formally the same as the one in eq. (4.26) after sending m →mf(qk). Therefore we get

〈q′, t′|q, t〉 = limn→∞

∫ (√mf(qk)

2iπε

)n(n−1∏k=1

dqk

)

· e

iεn−1∑k=0

[mf(qk)((qk+1 − qk)/ε)2/2 − (V (qk+1) + V (qk))/2]

(4.44)

We can rewrite this expression in a way similar to the one used in eq. (4.29)

〈q′, t′|q, t〉 =

∫ q′,t′

q,t

D(q(t))ei

∫ t′

t

dtL(q, q)(4.45)

with L(q, q) given in (4.40), but the functional measure is now

D(q(t)) = limn→∞

(√mf(qk)

2iπε

)n n−1∏k=1

dqk (4.46)

The formal expression for the path integral is always the same, but the integrationmeasure must be determined by using the original phase space formulation. In thecontinuum limit we have

limn→∞

n−1∏k=0

√f(qk) = lim

n→∞e

1

2

n−1∑k=0

log f(qk)

=

= limn→∞

e

1

2ε

n−1∑k=0

ε log f(qk)

= e

1

2δ(0)

∫ t′

t

dt log f(q)(4.47)

where we have used

limn→∞

n−1∑k=0

ε =

∫ t′

t

dt , limn→∞

δhkε

= δ(th − tk) (4.48)

95

Therefore the result can be also interpreted by saying that the classical lagrangiangets modified by quantum effects as follows

L→ L− i

2δ(0) log f(q) (4.49)

The reason why we say that this is a quantum modification is because the secondterm must be proportional to /h, as it follows from dimensional arguments.

An interesting question is how one gets the classical limit from this formulation.Let us recall that the amplitude for a single path is exp(iS(C)//h) (where we haveput back the factor /h). To find the total amplitude we must sum over all the possiblepaths C. Since the amplitudes are phase factors the will cancel except for the pathswhich are very close to a stationary points of S. Around this point the phaseswill add coherently, whereas they will cancel going away from this point due to thefast variation of the phase. Roughly we can expect that the coherence effect willtake place for those paths such that their phase differs from the stationary value,S(Cclass)//h, of the order of one. For a macroscopical particle (m ≈ 1 gr and times ofabout 1 sec) the typical action is about 1027/h, and the amplitude gets contributiononly from the classical path. For instance, take a free particle starting from theorigin at t = 0 and arriving at x0 at the time t0. The classical path (the one makingS stationary) is

xclass(t) = x0t

t0(4.50)

The corresponding value for S is

S(xclass) =1

2m

∫ t0

0

q2 =1

2mx2

0

t0(4.51)

Let us now consider another path joining the origin with x0 as, for instance,

x(t) = x0t2

t20(4.52)

The corresponding action is

S(x) =2

3mx2

0

t0(4.53)

and we get

∆S = S(x) − S(xclass) =1

6mx2

0

t0(4.54)

Let us take t0 = 1 sec and x0 = 1 cm. We get ∆S = m/6. For a macroscopic particle(say m = 1 gr) and recalling that /h ≈ 10−27 erg·sec, we get ∆S = 1.6 ·1026/h and thisnon-classical path can be safely ignored. However, for an electron m ≈ 10−27 gr,and ∆S = /h/6. It follows that in the microscopic world, many paths may contributeto the amplitude.

96

4.4 The free particle

We shall now consider the case of the one-dimensional free motion. In this case wecan easily make the integrations going back to the original definition (eq. (4.27).The expression we wish to evaluate is

〈q′, t′|q, t〉 = limn→∞

∫ (√m

2iπε

)n(n−1∏k=1

dqk

)e

iε

n−1∑k=0

m((qk+1 − qk)/ε)2/2

(4.55)

since the lagrangian is

L =1

2mq2 (4.56)

The integrations can be done immediately starting from the gaussian integral∫ ∞

−∞dxeax

2 + bx =

√−πae−b2/4a, Re a ≤ 0 (4.57)

which allows us to evaluate the expression∫dxea(x1 − x)2 + b(x2 − x)2

=

√− π

a + beab(x1 − x2)

2/(a+ b) (4.58)

let us start by integrating q1∫dq1e

im[(q2 − q1)2 + (q1 − q0)

2]/(2ε) =

√iπε

meim(q2 − q0)

2/(2(2ε)

=2πiε

m

(m

2πi(2ε)

) 12

eim(q2 − q0)2/(2(2ε) (4.59)

Multiplying by the exponential depending on q2 and integrating over this variablewe get

2πiε

m

(m

2πi(2ε)

) 12∫dq2e

im[(q3 − q2)2 + (q2 − q0)

2/2]/(2ε)

=2πiε

m

(m

2πi(2ε)

) 12(

2πi(2ε)

3m

) 12

eim(q3 − q0)2/(2(3ε))

=

(2πiε

m

) 32(

m

2πi(3ε)

) 12

eim(q3 − q0)2/(2(3ε)) (4.60)

By repeating this procedure, after n− 1 steps we find

∫ (n−1∏k=1

dqk

)e

iε

n−1∑k=0

m((qk+1 − qk)/ε)2/2

=

(2πiε

m

)n2(

m

2πi(nε)

) 12

eim(qn − q0)2/(2(nε)) (4.61)

97

The first factor in the right hand side of this equation cancels out an analogousfactor in the integration measure (see eq. (4.55))

〈q′, t′|q, t〉 =

[m

2πi(t′ − t)

] 12

eim(q′ − q)2/(2(t′ − t)) (4.62)

This expression can be generalized to a free system with N degrees of freedom

〈q′i, t′|qi, t〉 =

[m

2πi(t′ − t)

]N2

e

im

N∑i=1

(q′i − qi)2/(2(t′ − t))

, i = 1, · · · , N (4.63)

Notice also thatlimt′→t

〈q′, t′|q, t〉 = δ(q′ − q) (4.64)

The classical path connecting (q, t) to (q′, t′) for the free particle is given by

qcl(τ) = q +τ − t

t′ − t(q′ − q) (4.65)

The corresponding classical action is

Scl =

∫ t′

t

m

2q2cldτ =

m

2

(q′ − q)2

(t′ − t)2

∫ t′

t

dτ =m

2

(q′ − q)2

(t′ − t)(4.66)

Therefore our result can be written as

〈q′, t′|q, t〉 =

[m

2πi(t′ − t)

] 12

eiScl (4.67)

We shall see that for all the lagrangians quadratic in the coordinates an in themomenta this result, that is that the amplitude is proportional to the exponentialof iS with S the action evaluated along the classical path, is generally true.

Of course, one could formulate quantum mechanics directly in terms of the pathintegral. It is not difficult to show that it is possible to recover all the results of theordinary formulation and prove that one has complete equivalence. However we willnot do it here, but we will limit ourselves to notice that the formulations agree inthe free case. we take a fixed initial point of propagation, for instance (q, t) = (0, 0).Then we define the wave function as

ψ(q′, t′) = 〈q′, t′|0, 0〉 =

∫ q′,t′

0,0

D(q(t))eiS (4.68)

For the free particle we get

ψ(q, t) =[ m

2πit

] 12eimq2

2t (4.69)

98

This is indeed the wave function for a free particle in configuration space, as it canbe checked by going in the momentum space

ψ(p, t) =[ m

2πit

] 12

∫dq e−ipqei

mq2

2t =[ m

2πit

] 12

[2πit

m

] 12

e−i p

2

2mt

= e−iEt (4.70)

where E = p2/2m.

4.5 The case of a quadratic action

In this Section we will consider the case of a quadratic lagrangian for a single degreeof freedom (it can be easily extended to many degrees of freedom)

L =1

2mq2 + b(τ)qq − 1

2c(τ)q2 + d(τ)q − e(τ)q − f(τ) (4.71)

To evaluate the integral (4.29) let us start by the Euler-Lagrange equations associ-ated to eq. (4.71)

d

dτ[mq + b(τ)q + d(τ)] − [b(τ)q − c(τ)q − e(τ)] = 0 (4.72)

that ismq + bq + d+ cq + e = 0 (4.73)

Let us denote by qcl(τ) the solution of the classical equations of motion satisfyingthe boundary conditions

qcl(t) = q, qcl(t′) = q′ (4.74)

We then perform the change of variables

η(τ) = q(τ) − qcl(τ) (4.75)

which in terms of our original definition (4.29) means

ηk = qk − qclk, k = 1, · · · , n− 1 (4.76)

Since this is a translation the functional measure remains invariant. The new vari-able η(τ) satisfies the boundary condition

η(t) = η(t′) = 0 (4.77)

The lagrangian in terms of the new variables is given by

L(q) = L(qcl + η) = L(qcl) +1

2mη2 + bηη − 1

2cη2

+ dη − eη +mqclη + bqclη + bηqcl − cqclη (4.78)

99

Inside the action we can integrate by parts the terms containing both qcl and η. Byusing the boundary conditions for η we get

L(q) = L(qcl) +1

2mη2 + bηη − 1

2cη2 + dη − eη −

[mqcl + bqcl + cqcl

]η (4.79)

Recalling the equations of motion for qcl and performing a further integration byparts we find

L(q) = L(qcl) +1

2mη2 + bηη − 1

2cη2 + dη − eη −

[−d − e

]η

= L(qcl) +1

2mη2 + bηη − 1

2cη2 (4.80)

Therefore we can write the path integral as follows

〈q′, t′|q, t〉 = eiScl(q, q′)∫ 0,t′

0,t

D(η(τ))ei

∫ t′

t

(1

2mη2 + bηη − 1

2cη2)dτ

(4.81)

Notice that here all the dependence on the variables q and q′ is in the classical action.In fact the path integral which remains to be evaluated depends only on t and t′

due to the boundary conditions satisfied by η. For the future purposes of extensionof this approach to field theory, this term turns out to be irrelevant. However it isimportant for systems with a finite number of degrees of freedom and therefore wewill give here an idea how one can evaluate it in general. First, let us notice that theterm bηη is a trivial one, since we can absorb it into a re-definition of c(τ), throughan integration by part

bηη =d

dτ

(1

2bη2

)− 1

2bη2 (4.82)

and definingc = c+ b (4.83)

In this way we get

i

∫ t′

t

(1

2mη2 − 1

2cη2

)dτ = lim

n→∞in−1∑k=0

(m

2ε(ηk+1 − ηk)

2 − 1

2εckη

2k

)

= limn→∞

n−1∑k=1

(im

2ε

(2η2

k − 2ηkηk+1

)− i

2εckη

2k

)

= limn→∞

iηTση (4.84)

where we have used the boundary conditions on η (η0 = ηn = 0) and we haveintroduced the vector

η =

⎡⎢⎢⎢⎢⎣

η1

η2

··

ηn−1

⎤⎥⎥⎥⎥⎦ (4.85)

100

and the matrix

σ =m

2ε

⎡⎢⎢⎢⎢⎢⎢⎣

2 −1 0 · · · 0 0 0−1 2 −1 · · · 0 0 00 −1 2 · · · 0 0 0· · · · · · · · ·0 0 0 · · · −1 2 −10 0 0 · · · 0 −1 2

⎤⎥⎥⎥⎥⎥⎥⎦− ε

2

⎡⎢⎢⎢⎢⎢⎢⎣

c1 0 · · · 0 00 c2 · · · 0 0· · · · · · ·· · · · · · ·0 0 · · · cn−2 00 0 · · · 0 cn−1

⎤⎥⎥⎥⎥⎥⎥⎦

(4.86)Therefore

G(t′, t) ≡∫ 0,t′

0,t

D(η(τ))ei

∫ t′

t

(1

2mη2 + bηη − 1

2cη2)dτ

= limn→∞

( m

2πiε

)n2

∫ (n−1∏k=1

dηk

)eiη

Tση (4.87)

Since σ is a symmetric matrix, it can be diagonalized by an orthogonal transforma-tion

σ = UTσDU (4.88)

The eigenvectors of σ, ζ = Uη, are real and the Jacobian of the transformation isone, since det|U | = 1. We get∫ (

n−1∏k=1

dηk

)eiη

Tση =

∫ (n−1∏k=1

dζk

)eiζ

TσDζ =

n−1∏k=1

√π

(−iσk) =π

n−12√

det(−iσ)

(4.89)We get

G(t′, t) = limn→∞

[( m

2πiε

)n πn−1

det(−iσ)

] 12

= limn→∞

[m

2πi

1

ε

1

(2iεm

)n−1det(−iσ)

] 12

(4.90)

Let us define

f(t′, t) = limn→∞

[ε

(2iε

m

)n−1

det(−iσ)

](4.91)

then

G(t′, t) =

√m

2πif(t′, t)(4.92)

In order to evaluate f(t′, t), let us define (for j = 1, · · · , n− 1)

Pj =

⎡⎢⎢⎢⎢⎢⎢⎣

2 −1 0 · · · 0 0 0−1 2 −1 · · · 0 0 00 −1 2 · · · 0 0 0· · · · · · · · ·0 0 0 · · · −1 2 −10 0 0 · · · 0 −1 2

⎤⎥⎥⎥⎥⎥⎥⎦− ε2

m

⎡⎢⎢⎢⎢⎢⎢⎣

c1 0 0 · · · 0 00 c2 0 · · · 0 00 0 c3 · · · 0 0· · · · · · · ·0 0 0 · · · cj−1 00 0 0 · · · 0 cj

⎤⎥⎥⎥⎥⎥⎥⎦

(4.93)

101

andpj = det|Pj| (4.94)

Clearly (2iε

m

)n−1

det(−iσ) = pn−1 (4.95)

For the first values of j, pj is given by

p1 = 2 − ε2

mc1

p2 =

(2 − ε2

mc2

)p1 − 1

p3 =

(2 − ε2

mc3

)p2 − p1 (4.96)

Therefore we find the recurrency relation

pj+1 =

(2 − ε2

mcj+1

)pj − pj−1, j = 1, · · · , n− 1 (4.97)

where we define p0 = 1. It is convenient to put this expression in the form

pj+1 − 2pj + pj−1

ε2= − cj+1pj

m(4.98)

It is also convenient to introduce the finite difference operator

∆pj =pj+1 − pj

ε(4.99)

a time τ = t+ jε, and a function φ(τ) such that

limε→0

εpj = φ(τ) (4.100)

in the limit in which we keep τ fixed. It follows

limε→0

∆εpj =dφ(τ)

dτ(4.101)

The equation (4.98) can be written as

∆2pj−1 = − cj+1pjm

(4.102)

therefore we get the following differential equation for φ(τ)

d2φ(τ)

dτ 2= − c(τ)

mφ(τ) (4.103)

102

Furthermore, φ(τ) satisfies the following boundary conditions

φ(t) = limε→0

εp0 = 0

dφ(τ)

dτ|τ=t = lim

ε→0εp1 − p0

ε= lim

ε→02 − ε2

mc1 − 1 = 1 (4.104)

Recalling eq. (4.91) we have

f(t′, t) = limε→0

εpn−1 = φ(t′) (4.105)

Therefore, for quadratic lagrangian we get the general result

〈q′, t′|q, t〉 =

√m

2πif(t′, t)eiScl (4.106)

with f(t′, t) = φ(t′).

Examples:

The free particle.

We need only to evaluate f(t′, t). The function φ(τ) satisfies

d2φ(τ)

dτ 2= 0, φ(t) = 0,

dφ(τ)

dτ|τ=t = 1 (4.107)

with solutionφ(τ) = τ − t (4.108)

from whichf(t′, t) = t′ − t (4.109)

in agreement with the direct calculation.

The harmonic oscillator.

In this case b = 0 and c = mω2. Therefore

d2φ(τ)

dτ 2= −ω2φ(τ), φ(t) = 0,

dφ(τ)

dτ|τ=t = 1 (4.110)

from which

φ(τ) =1

ωsinω(τ − t) (4.111)

by putting T = t′ − t:

f(t′, t) =1

ωsinωT (4.112)

103

To evaluate the classical action we need to solve the classical equations of motion

mq +mω2q = 0 (4.113)

with boundary conditionsq(t) = q, q(t′) = q′ (4.114)

The general solution is given by

q(τ) = A sinω(τ − t) +B cosω(τ − t) (4.115)

From the boundary conditions

A =q′ − q cosωT

sinωT, B = q (4.116)

and substituting inside the lagrangian

L =1

2mq2 − 1

2mω2q2 =

1

2mω2

[(A2 −B2) cos 2ω(τ − t) − 2AB sin 2ω(τ − t)

](4.117)

Recalling the following integrals

∫ t′

t

dτ cos 2ω(τ − t) =sin 2ωT

2ω,

∫ t′

t

dτ sin 2ω(τ − t) =1 − cos 2ωT

2ω(4.118)

we get

Scl =1

2

mω

sinωT

[(q2 + q′2) cosωT − 2qq′

](4.119)

and

〈q′, t′|q, t〉 =

√mω

2πi sinωTe

i

2

mω

sinωT

[(q2 + q′2) cosωT − 2qq′

](4.120)

In the limit ω → 0 we get back correctly the free case.

4.6 Functional formalism

The most natural mathematical setting to deal with the path integral is the theoryof functionals. The concept of functional is nothing but an extension of the conceptof function. Recall that a function is simply a mapping from a manifold M toR1, that a law which associates a number to each point of the manifold. A realfunctional is defined as an application from a space of functions (and therefore an∞-dimensional space) to R1. There are many spaces of functions. For instance,the space of the square integrable real functions of a real variable. Therefore a real

104

functional associate a real number to a real function. If we denote the space offunctions by L, a real functional is the mapping

F : L → R1 (4.121)

We will make use of the notation

F [η], F, F [·], η ∈ L (4.122)

to denote the functional of defined in eq. (4.121). Examples of functionals are

• L=L1[−∞,+∞]= the space of the integrable functions in R1 with respect tothe measure w(x)dx:

F1[η] =

∫ +∞

−∞dxw(x)η(x) (4.123)

• L=L2[−∞,+∞]= space of the square integrable functions in R1:

F2[η] = e−1

2

∫dxη2(x)

(4.124)

• L=C= space of the continuous functions:

F3[η] = η(x0) (4.125)

F3 is the functional that associates to each continuous function its value at x0.

The next step is to define the derivative of a functional. We need to understandhow a functional varies when we vary the function. A simple way of understandthis point is to consider some path on the definition manifold of the functions. Forinstance consider functions from R1 to R1, and take an interval (t, t′) of R1 madeof many infinitesimal pieces, as we did for defining the path integral. Then we canapproximate a function η(t) with a convenient limit of its discrete approximation(η1, η2, · · · , ηn) obtained by evaluating the functions at the discrete points in whichthe interval has been divided. In this approximation we can think of the functionalF as an application from Rn → R1. That is

F [η] → F (η1, · · · , ηn) ≡ F (ηi), i = 1, · · · , n (4.126)

Then we can consider the variation of F (ηi) when we vary the quantities ηi

F (ηi + δηi) − F (ηi) =

n∑j=1

∂F (ηi)

∂ηjδηj (4.127)

Let us put

Ki =1

ε

∂F

∂ηi(4.128)

105

whereε is the infinitesimal interval where we evaluate the discrete apoproximationto the function. Then, for ε→ 0

F (ηi + δηi) − F (ηi) =n∑j=1

Kjδηjε→∫Kδηdx (4.129)

that is

δF [η] =

∫K(x)δη(x)dx (4.130)

Since δη(x) is the variation of the function evaluated at x, we will define the func-tional derivative as

δF [η] =

∫δF

δη(x)δη(x)dx (4.131)

In other words. to evaluate the functional derivative of a functional we first eval-uate its infinitesimal variation and then we can get the functional derivative bycomparison with the previous formula. Looking at the eq. (4.128) we get

δF [η]

δη(x)= lim

ε→0

1

ε

∂F (ηi)

∂ηi

ηi = η(xi), xi = x1 + (i− 1)ε, ε =xn − x1

n− 1, i = 1, · · · , n (4.132)

let us now evaluate the functional derivatives of the previous functionals:

δF1[η] =

∫w(x)δη(x)dx (4.133)

givingδF1[η]

δη(x)= w(x) (4.134)

Then

δF2[η] = e−1

2

∫dxη2(x)

(−∫η(x)δη(x)) (4.135)

and thereforeδF2[η]

δη(x)= −η(x)F2[η] (4.136)

In order to evaluate the derivative of F3 let us write

F3[η] =

∫η(x)δ(x− x0)dx (4.137)

In this way F3 looks formally as a functional of type F1, and

δF3[η]

δη(x)= δ(x− x0) (4.138)

106

Let us consider also

F4[η] = e−1

2

∫dxdyK(x, y)η(x)η(y)

(4.139)

then

δF4[η] = e−1

2

∫dxdyK(x, y)η(x)η(y)

(−∫dxdyK(x, y)δη(x)η(y)) (4.140)

where we have used the symmetry of the integral with respect to x↔ y. We obtain

δF4[η]

δη(x)= −

∫dyK(x, y)η(y)F4[η] (4.141)

A further example is the action functional. In fact the action is a functional of thepath

S[q] =

∫ t′

t

L(q, q)dt (4.142)

By varying it

δS[q] =

∫ t′

t

(∂L

∂qδq +

∂L

∂qδq

)dt =

∫ t′

t

(∂L

∂q− d

dt

∂L

∂q

)δqdt (4.143)

Since we consider variations with respect to functions with fixed boundary condi-tions, we have δq(t) = δq(t′) = 0. We get

δS

δq(t)=

∂L

∂q(t)− d

dt

∂L

∂q(t)(4.144)

Starting from the functional derivative one can generalize the Taylor expansionto the functional case. This generalization is called the Volterra expansion of afunctional

F [η0 + η1] =∞∑k=0

1

k!

∫dx1 · · · dxk δkF [η]

δη(x1) · · · δη(xk)∣∣∣∣η=η0

η1(x1) · · ·η1(xk) (4.145)

All these definitions are easily generalized to the case of η a function from Rm → Rn,or from a manifold M to a manifold N .

4.7 General properties of the path integral

In this Section we would like to stress some important properties of the path integral,namely

107

• Invariance of the functional measure under translations - It followsimmediately from the definition that

q(τ) → q(τ) + η(τ), η(t) = η(t′) = 0

p(τ) → p(τ) + π(τ), for any π(t), π(t′) (4.146)

• Factorization of the path integral. From the very definition of the path itfollows that in order to evaluate the amplitude for going from q at time t to q′

at time t′ we can first go to q” to a time t ≤ t” ≤ t′, sum up over all the pathsfromq a q” and from q” to q′, and eventually integrating over all the possibleintermediate points q” (see Fig. 4.7.1)).

.

q q"

t

t"

q'

t'

Fig. 4.7.1 - The factorization of the path integral.

In equations we get

〈q′, t′|q, t〉 =

∫ q′t′

q,t

dµ(q(τ))dµ(p(τ))ei

∫ t”

t

Ldτ + i

∫ t′

t”

Ldτ

=

∫dq”

∫ q”,t”

q,t

dµ(q(τ))dµ(p(τ))ei

∫ t”

t

Ldτ

×∫ q′,t′

q”,t”

dµ(q(τ))dµ(p(τ))ei

∫ t′

t”

Ldτ

=

∫dq”〈q′, t′|q”, t”〉〈q”, t”|q, t〉 (4.147)

as it follows from the factorization properties of the measure

dµ(q(τ)) =∏

t<τ<t′dq(τ) = dq”

∏t<τ<t”

dq(τ)∏

t”<τ<t′dq(τ)

108

dµ(p(τ)) =∏

t≤τ≤t”dp(τ) =

∏t≤τ≤t”

dp(τ)∏

t”≤τ≤t′dp(τ) (4.148)

It is clear that the factorization property is equivalent to the completeness of thestates.

Using these properties it is possible to show that the wave function, as definedin (4.68) satisfies the Schrodinger equation.

The next important property of the path integral is that it gives an easy way ofevaluating expectation values. To this end we will consider expressions of the type(which will be called the average value of F )

∫ q′,t′

q,t

dµ(q(τ))dµ(p(τ))F [q(τ), p(τ)]eiS (4.149)

where F is an arbitrary functional of q, p and of their time derivatives. Let us startwith ∫ q′,t′

q,t

dµ(q(τ))dµ(p(τ))eiSq(t”), t ≤ t” ≤ t′ (4.150)

We begin by the observation that

∫ q′,t′

q,t

dµ(q(τ))dµ(p(τ))eiSq(t′) = q′∫ q′,t′

q,t

dµ(q(τ))dµ(p(τ))eiS = q′〈q′, t′|q, t〉(4.151)

since q(t) satisfies the boundary conditions q(t) = q, q(t′) = q′. By using thefactorization we get

∫ q′,t′

q,t

dµ(q(τ))dµ(p(τ))eiSq(t”) =

∫dq”

∫ q”,t”

q,t

dµ(q(τ))dµ(p(τ))eiSq(t”)

·∫ q′,t′

q”,t”

dµ(q(τ))dµ(p(τ))eiS

=

∫dq”〈q′, t′|q”, t”〉q”〈q”, t”|q, t〉 (4.152)

Since |q, t〉 is an eigenstate of q(t) it follows

∫ q′,t′

q,t

dµ(q(τ))dµ(p(τ))eiSq(t”) =

∫dq”〈q′, t′|q(t”)|q”, t”〉〈q”, t”|q, t〉

= 〈q′, t′|q(t”)|q, t〉 (4.153)

Therefore the average value of q(t”) evaluated with exp(iS) coincides with the ex-pectation value of the operator q(t”).

Now let us study the average value of q(t1)q(t2) con t ≤ t1, t2 ≤ t′. By proceedingas before we will get the expectation value of q(t1)q(t2) if t1 ≥ t2, or q(t2)q(t1) if

109

t2 ≥ t1. In terms of the T -product we get∫ q′,t′

q,t

dµ(q(τ))dµ(p(τ))eiSq(t1)q(t2) = 〈q′, t′|T (q(t1)q(t2)|q, t〉 (4.154)

Analogously∫ q′,t′

q,t

dµ(q(τ))dµ(p(τ))eiSq(t1) · · · q(tn) = 〈q′, t′|T (q(t1) · · ·q(tn))|q, t〉 (4.155)

Let us now consider a functional of q(t)such to admit an expansion in a Volterra’sseries ∫ q′,t′

q,t

dµ(q(τ))dµ(p(τ))F [q]eiS

=∞∑k=0

1

k!

∫dt1 · · · dtk δkF [q]

δq(t1) · · · δq(tk)∣∣∣∣q=0

·∫ q′,t′

q,t

dµ(q(τ))dµ(p(τ))eiSq(t1) · · · q(tk)

=

∞∑k=0

1

k!

∫dt1 · · · dtk δkF [q]

δq(t1) · · · δq(tk)∣∣∣∣q=0

〈q′, t′|T (q(t1) · · ·q(tk))|q, t〉

≡ 〈q′, t′|T (F [q])|q, t〉 (4.156)

where

T (F [q]) ≡∞∑k=0

1

k!

∫dt1 · · · dtk δkF [q]

δq(t1) · · · δq(tk)∣∣∣∣q=0

T (q(t1) · · ·q(tk)) (4.157)

If the functional depends also on the time derivatives a certain caution is necessary.For instance, let us study the average value of q(t1)q(t2) (t ≤ t1, t2 ≤ t′):∫ q′,t′

q,t

dµ(q(τ))dµ(p(τ))eiSq(t1)q(t2)

= limε→0

∫ q′,t′

q,t

dµ(q(τ))dµ(p(τ))eiSq(t1)q(t2 + ε) − q(t2)

ε

= limε→0

1

ε〈q′, t′|T

(q(t1)q(t2 + ε)

)− T

(q(t1)q(t2)

)|q, t〉

=d

dt2〈q′, t′|T

(q(t1)q(t2)

)|q, t〉 (4.158)

Let us define an ordered T-product (or covariant T-product) as

T (q(t1)q(t2)

)=

d

dt2T(q(t1)q(t2)

)(4.159)

110

Then it is not difficult to show that for an arbitrary functional of q(t) e q(t) one has

∫ q′,t′

q,t

dµ(q(τ))dµ(p(τ))F [q, q]eiS = 〈q′, t′|T (F [q, q])|q, t〉 (4.160)

The proof can be done by expanding F [q, q] in a Volterra’s series of q and q, inte-grating by parts and using (4.156). At the end one has to integrate again by parts.The T product is obtained as in (4.158),

The power of this formulation of quantum mechanics comes about when we wantto get the fundamental properties of quantum mechanics as the equations of motionfor the operators, the commutation relations and so on so forth. The fundamentalrelation follows from the following property of the path integral∫ q′,t′

q,t

dµ(q(τ))dµ(p(τ))δ

δq(τ)

[F [q(τ), p(τ)]eiS

]= 0 (4.161)

This is a simple consequence of the definition of the path integral and of the func-tional derivative

limn→∞

∫ (n−1∏k=1

dqk

)(n−1∏k=0

dpk2π

)1

ε

∂

∂qk

⎡⎢⎢⎢⎢⎣F (q1, · · · , qn−1)e

i

n−1∑k=0

εL(qk, pk)

⎤⎥⎥⎥⎥⎦ = 0 (4.162)

(of course we are assuming that the integrand goes to zero for large values of the qi’s.To this end one has to define the integral in an appropriate way as it will be shownlater on). On the other hand the previous equation follows from the invarianceunder translation of the functional measure. In fact, let us introduce the followingnotation

dµq,p = dµ(q(τ))dµ(p(τ)) (4.163)

It follows for an infinitesimal function η such that η(t) = η(t′) = 0

∫ q′,t′

q,t

dµq,p A[q] =

∫ q′,t′

q,t

dµq,p A[q+η] =

∫ q′,t′

q,t

dµq,p A[q]+

∫ q′,t′

q,t

dµq,p

∫η(t)

δA[q]

δq(t)dt

(4.164)Proving the validity of eq. (4.161). By performing explicitly the functional derivativein eq. (4.161)we get a relation which is the basis of the Schwinger’s formulation ofquantum mechanics, that is

i〈q′, t′|T (F [q]

δS

δq(τ)

)|q, t〉 = −〈q′, t′|T

(δF [q]

δq(τ)

)|q, t〉 (4.165)

The importance of this relation follows from the fact that we can use this relationto get all the properties of quantum mechanics by selecting the functional F in a

111

convenient way. For instance, by choosing F [q] = 1 we get

〈q′, t′|T (

δS

δq(t1)

)|q, t〉 = 0 (4.166)

that is the quantum equations of motion. With the choice F [q] = q we get

〈q′, t′|T (q(t1)

δS

δq(t2)

)|q, t〉 = i〈q′, t′|q, t〉δ(t1 − t2) (4.167)

and since this relation holds for arbitrary states

T (q(t1)

δS

δq(t2)

)= iδ(t1 − t2) (4.168)

For simplicity, let us consider the following action

S =

∫ t′

t

(1

2mq2 − V (q)

)dt (4.169)

thenδS

δq(t2)= −

(mq(t2) +

∂V (q(t2))

∂q(t2)

)(4.170)

from which

T [q(t1)

(mq(t2) +

∂V (q)

∂q(t2)

)]= −iδ(t1 − t2) (4.171)

By using the definition of the T -product we get

T (q(t1)q(t2) =d2

dt22T (q(t1)q(t2))

=d2

dt22[θ(t1 − t2)q(t1)q(t2) + θ(t2 − t1)q(t2)q(t1)]

=d

dt2

[− δ(t1 − t2)q(t1)q(t2) + δ(t1 − t2)q(t2)q(t1)

+ theta(t1 − t2)q(t1)q(t2) + θ(t2 − t1)q(t2)q(t1)]

= δ(t1 − t2)q(t1)q(t2) + δ(t1 − t2)q(t2)q(t1) + T (q(t1)q(t2))

= δ(t1 − t2)[q(t1), q(t1)] + T (q(t1)q(t2)) (4.172)

Inserting this inside the eq. (4.171

−mδ(t1 − t2)[q(t1), q(t2)] + T

[q(t1)

(mq(t2) +

∂V

∂q(t2)

)]= −iδ(t1 − t2) (4.173)

and using the equations of motion, we get

[q(t), mq(t)] = i (4.174)

Therefore we get the canonical commutator

[q(t),p(t)] = i (4.175)

112

4.8 The generating functional of the Green’s func-

tions

In field theory the relevant amplitudes are the ones evaluated between t = −∞ andt = +∞. We have seen that using the reduction formulas we are lead to evaluatethe vacuum expectation value of T -products of local operators. Therefore we willconsider matrix elements of the type (limiting ourselves, for the moment being, toa single degree of freedom)

〈0|T (q(t1) · · ·q(tn))|0〉 (4.176)

Here |0〉 is the ground state of system. Notice that more generally one can considerthe matrix element

〈q′, t′|T (q(t1) · · ·q(tn))|q, t〉 =

∫ q′,t′

q,t

dµ(q(τ))dµ(p(τ))eiSq(t1) · · · q(tn) (4.177)

These matrix elements can be derived in a more compact form, by introducing thegenerating functional

〈q′, t′|q, t〉J =

∫ q′,t′

q,t

dµ(q(τ))dµ(p(τ))eiSJ (4.178)

where

SJ =

∫ t′

t

dτ [pq −H + Jq] (4.179)

and J(τ) is an arbitrary external source. By differentiating this expression withrespect to the external source and evaluating the derivatives at J = 0 we get

〈q′, t′|T (q(t1) · · ·q(tn))|q, t〉 =

(1

i

)nδn

δJ(t1) · · · δJ(tn)〈q′, t′|q, t〉J

∣∣∣J=0

(4.180)

Now we want to see how it is possible to bring back the evaluation of the matrixelement in (4.176) to the knowledge of the generating functional. Let us introducethe wave function of the ground state

Φ0(q, t) = e−iE0t〈q|0〉 = 〈q|e−iHt|0〉 = 〈q, t|0〉 (4.181)

then

〈0|T (q(t1) · · ·q(tn))|0〉 =

∫dq′dq〈0|q′, t′〉〈q′, t′|T (q(t1) · · ·q(tn))|q, t〉〈q, t|0〉

=

∫dq′dqΦ

0(q′, t′)〈q′, t′|T (q(t1) · · ·q(tn))|q, t〉Φ0(q, t) (4.182)

113

Next we define the functional Z[J ], the generating functional of the vacuum ampli-tudes, that is

Z[J ] =

∫dq′dqΦ

0(q′, t′)〈q′, t′|q, t〉JΦ0(q, t) = 〈0|0〉J (4.183)

It follows

〈0|T (q(t1) · · ·q(tn))|0〉 =

(1

i

)nδn

δJ(t1) · · · δJ(tn)Z[J ]

∣∣∣J=0

(4.184)

We want to show that it is possible evaluate Z[J ] as a path integral with arbitraryboundary conditions on q and q′. That is we want to prove the following relation

Z[J ] = limt→+i∞, t′→−i∞

eiE0(t′ − t)

Φ0(q)Φ0(q′)

〈q′, t′|q, t〉J (4.185)

wit q and q′ arbitrary and with the ground state wave functions taken at t = 0. Tothis end let us consider an external source J(t) vanishing outside the interval (t′′, t′′′)with t′ > t′′′ > t′′ > t. Then, we can write

〈q′, t′|q, t〉J =

∫dq′′dq′′′〈q′, t′|q′′′, t′′′〉〈q′′′, t′′′|q′′, t′′〉J〈q′′, t′′|q, t〉 (4.186)

with

〈q′, t′|q′′′, t′′′〉 = 〈q′|e−iH(t′ − t′′′)|q′′′〉 =∑n

Φn(q′)Φ

n(q′′′)e−iEn(t′ − t′′′) (4.187)

Since E0 is the lowest eigenvalue we get

limt′→−i∞

eiE0t′〈q′, t′|q′′′, t′′′〉 = lim

t′→−i∞

∑n

Φn(q′)Φ

n(q′′′)eiEnt

′′′e−i(En − E0)t

′=

= Φ0(q′)Φ

0(q′′′)eiE0t

′′′= Φ

0(q′′′, t′′′)Φ0(q

′) (4.188)

Analogously

〈q′′, t′′|q, t〉 =∑n

Φn(q′′)Φ

n(q)e−iEn(t′′ − t) (4.189)

and

limt→+i∞

e−iE0t〈q′′, t′′|q, t〉 = limt→+i∞

∑n

Φn(q′′)Φ

n(q)e−iEnt′′ei(En − E0)t =

= Φ0(q′′)Φ

0(q)e−iE0t

′′= Φ0(q

′′, t′′)Φ0(q) (4.190)

And finally we get the result

imt→+i∞,t′→−∞eiE0(t′ − t)〈q′, t′|q, t〉J

=

∫dq′′dq′′′Φ

0(q)Φ0(q′)Φ

0(q′′′, t′′′)〈q′′′, t′′′|q′′, t′′〉JΦ0(q

′′, t′′) =

= Φ0(q)Φ0(q

′)Z[J ] (4.191)

114

Let us notice again that the values q and q′ are completely arbitrary, and furthermorethat the coefficient in front of the matrix element is J-independent. Therefore it isphysically irrelevant. In fact, the relevant physical quantities are the ratios

〈0|T (q(t1) · · ·q(tn))|0〉〈0|0〉 =

(1

i

)n1

Z[0]

δn

δJ(t1) · · · δJ(tn)Z[J ]

∣∣∣J=0

(4.192)

where the J-independent factor cancels out. Then we can write

Z[J ] = limt→+i∞,t′→−i∞

N

∫ q′,t′

q,t

D(q(τ))ei

∫ t′

t

(L+ Jq)dτ(4.193)

where N is a J-independent normalization factor. This expression suggests the defi-nition of an euclidean generating functional ZE[J ]. This is obtained by introducingan euclidean time τ = it (Wick’s rotation)

ZE [J ] = limτ ′→+∞,τ→−∞

N

∫ q′,τ ′

q,τ

D(q(τ))e

∫ τ ′

τ

(L(q, idq

dτ) + Jq)dτ

(4.194)

Consider for instance the case

L =1

2m(

dq

dt)2 − V (q) (4.195)

from which

L(q, idq

dτ) = −1

2m

(dq

dτ

)2

− V (q) (4.196)

It is then convenient to define the euclidean lagrangian and euclidean action

LE =1

2m

(dq

dτ

)2

+ V (q) (4.197)

SE =

∫ τ ′

τ

LEdτ (4.198)

Notice that the euclidean lagrangian coincides formally with the hamiltonian. Moregenerally

LE = −L(q, idq

dτ) (4.199)

Therefore

ZE[J ] = limτ ′→+∞,τ→−∞

N

∫ q′,τ ′

q,τ

D(q(τ))e−SE +

∫ τ ′

τ

Jqdτ(4.200)

115

Since e−SE is a positive definite functional, the path integral turns out to be welldefined and convergent if SE is positive definite.the vacuum expectation values ofthe operators q(t) can be evaluated by using ZE and continuing the result for realtimes

1

Z[0]

δnZ[J ]

δJ(t1) · · · δJ(tn)

∣∣∣J=0

= (i)n1

ZE[0]

δnZE[J ]

δJ(τ1) · · · δJ(τn)

∣∣∣J=0,τi=iti

(4.201)

Since the values of q and q′ are arbitrary a standard choice is to set them at zero.

4.9 The Green’s functions for the harmonic oscil-

lator

It is interesting to evaluate the generating functional for the harmonic oscillatorstarting from both the equations (4.185) and (4.200). Let us start from eq. (4.183),where we do not need to specify the boundary conditions

Z[J ] =

∫D(q(τ))Φ

0(q′, t′)eiSJΦ0(q, t)dqdq

′ (4.202)

with

Φ0(q, t) =(mωπ

) 14e−1

2mωq2

e− i

2ωt

(4.203)

Φ0(q

′, t′) =(mωπ

) 14e−1

2mωq′2

e

i

2ωt′

(4.204)

and

SJ =

∫ t′

t

[1

2mq2 − 1

2mω2q2 + Jq

]dτ (4.205)

In this calculation we are not interested in the normalization factors but only onthe J dependence. Then, let us consider the argument of the exponential in eq.(4.9 and, for the moment being, let us neglect the time dependent terms from theoscillator wave functions

arg = −1

2mω(q2 + q′2) + i

∫ t′

t

[1

2mq2 − 1

2mω2q2 + Jq

]dτ (4.206)

let us perform the change of variable

q(τ) = x(τ) + x0(τ) (4.207)

where we will try to arrange x0(τ) in such a way to extract all the dependence fromthe source J out of the integral. We have

arg = −1

2mω[x(t)2 + x2(t′)] − 1

2mω[x0(t)

2 + x20(t

′)] −mω[x(t)x0(t) + x(t′)x0(t′)]

116

+ i

∫ t′

t

[1

2mx2 − 1

2mω2x2

]dτ + i

∫ t′

t

[1

2mx2

0 −1

2mω2x2

0 + Jx0

]dτ

+ i

∫ t′

t

[mxx0 −mω2xx0 + Jx

]dτ (4.208)

Integrating by parts the last integral can be written as

im[xx0]t′t − i

∫ t′

t

[mx0 +mω2x0 − J ]xdτ (4.209)

and choosing x0 as a solution of the wave equation in presence of the source J

x0 + ω2x0 =1

mJ (4.210)

we getim[x(t′)x0(t

′) − x(t)x0(t)] (4.211)

Let us take the following boundary conditions on x0:

ix0(t′) = ωx0(t

′), ix0(t) = −ωx0(t) (4.212)

In this way, the term (4.211) cancels the mixed terms

arg = −1

2mω[x2(t) + x2(t′)] − 1

2mω[x2

0(t) + x20(t

′)]

+ i

∫ t′

t

[1

2mx2 − 1

2mω2x2]dτ

+ i

∫ t′

t

[1

2mx2

0 −1

2mω2x2

0 + Jx0]dτ (4.213)

Integrating by parts the last term we get

i

∫ t′

t

[1

2mx2

0 −1

2mω2x2

0 + Jx0]dτ =

=

∫ t′

t

[−1

2mx0(x0 + ω2x0 − 1

mJ) +

1

2Jx0]dτ +

i

2m[x0x0]

t′t =

=1

2mω[x2

0(t′) + x2

0(t)] +i

2

∫ t′

t

Jx0dτ (4.214)

Keeping in mind the boundary conditions (4.212), the term coming from the inte-gration by parts cancels the second term in eq. (4.213), and therefore

arg = −1

2mω[x2(t) + x2(t′)] + i

∫ t′

t

[1

2mx2 − 1

2mω2x2]dτ +

i

2

∫ t′

t

Jx0dτ (4.215)

117

The J dependence is now in the last term, and we can extract it from the integral.The first two term give rise to the functional at J = 0. We obtain

Z[J ] = e

i

2

∫ t′

t

J(τ)x0(τ)dτZ[0] (4.216)

with x0(τ) solution of the equations of motion (4.210) with boundary conditions(4.212). For t < τ < t′ this solution can be written as

x0(τ) = i

∫ t′

t

∆(τ − s)J(s)ds (4.217)

with [d2

dτ 2+ ω2

]∆(τ − s) = − i

mδ(τ − s) (4.218)

andi∆(t′ − s) = ω∆(t′ − s), i∆(t− s) = −ω∆(t− s) (4.219)

These equations are easily solved fort t′ > s and t < s

∆(τ) = Ae−iωτ , τ > 0

∆(τ) = Be+iωτ , τ < 0 (4.220)

Therefore we have

∆(τ) = Aθ(τ)e−iωτ +Bθ(−τ)e+iωτ (4.221)

The constants A e B are fixed by the requirement that ∆(τ) satisfies eq. (4.218).We have

∆(τ) = (A−B)δ(τ) − iAωθ(τ)e−iωτ + iBωθ(−τ)e+iωτ (4.222)

and∆(τ) = (A−B)δ(τ) − iω(A+B)δ(τ) − ω2∆(τ) (4.223)

from which

A = B, A =1

2ωm(4.224)

and finally

∆(τ) =1

2ωm

[θ(τ)e−iωτ + θ(−τ)e+iωτ

](4.225)

The functional we were looking for is

Z[J ] = e−1

2

∫ t′

t

dsds′J(s)∆(s− s′)J(s′)Z[0] (4.226)

118

We see that ∆(s − s′) is the vacuum expectation value of the T -product of theposition operators at the times s and s′:

∆(s− s′) = − 1

Z[0]

δ2Z[J ]

δJ(s)δJ(s′)

∣∣∣J=0

=〈0|T (q(s)q(s′))|0〉

〈0|0〉 (4.227)

The analogue calculation in the euclidean version is simpler since the path integralis of the type already considered (quadratic action)

ZE[J ] ≈ e−S(cl)E (4.228)

Here S(cl)E is th euclidean classical action. We have already noticed that here the

boundary conditions are arbitrary and therefore we choose q(τ) → 0 for τ → ±∞.we have to solve the equations of motion in presence of the external source

δ

δq(τ)

[SE −

∫ +∞

−∞Jqdτ

]= −mq(τ) +mω2q(τ) − J = 0 (4.229)

that is

q − ω2q = − 1

mJ (4.230)

Let us now define the euclidean Green’s function[d2

dτ 2− ω2

]DE(τ) = − 1

mδ(τ) (4.231)

with the boundary conditionlim

τ→±∞DE(τ) = 0 (4.232)

then, the solution of the eq. (4.230) is

q(τ) =

∫ +∞

−∞DE(τ − s)J(s)ds (4.233)

We can evaluate DE(τ) starting from the Fourier transform

DE(τ) =

∫dνe−iντDE(ν) (4.234)

Substituting inside (4.231) we get

−∫dνe−iντ [ν2 + ω2]DE(ν) = − 1

mδ(τ) (4.235)

from which

DE(ν) =1

2πm

1

ν2 + ω2(4.236)

119

and

DE(τ) =1

2πm

∫ +∞

−∞dν

1

ν2 + ω2e−iντ (4.237)

To evaluate this integral we close the integral with the circle at the ∞ in the lowerν-plane for τ > 0 or in the upper ν-plane for τ < 0. The result is

DE(τ) =1

2πm

e−ωτ(−2iω)

(−2πi) =1

2mωe−ωτ (4.238)

By doing the same calculation for τ < 0 we finally get

DE(τ) =1

2mωe−ω|τ | (4.239)

The classical action is given by

SE −∫ +∞

−∞Jqdτ =

∫ +∞

−∞

[−1

2m

(q − ω2q +

1

mJ

)q − 1

2Jq

]dτ = −1

2

∫ +∞

−∞Jqdτ

(4.240)and therefore

ZE[J ] = e

1

2

∫ +∞

−∞Jqdτ

Z[0] = e

1

2

∫ +∞

−∞J(s)DE(s− s′)J(s′)dsds′

Z[0] (4.241)

It follows1

ZE [0]

δ2ZE [J ]

δJ(s)δJ(s′)

∣∣∣J=0

= DE(s− s′) (4.242)

and using the eq. (4.201)

1

Z[0]

δ2Z[J ]

δJ(t1)δJ(t2)

∣∣∣J=0

= − 1

ZE [0]

δ2ZE [J ]

δJ(τ1)δJ(τ2)

∣∣∣J=0,τi=iti

(4.243)

we get∆(t) = DE(it) (4.244)

To compare these two expressions let us introduce the Fourier transform of ∆(t).Since it satisfies eq. (4.218) we obtain

∆(t) =

∫dνe−iνt∆(ν) (4.245)

with

(−ν2 + ω2)∆(ν) = − i

2πm(4.246)

that is

∆(ν) =i

2πm

1

ν2 − ω2(4.247)

120

Whereas eq. (4.237) does not present any problem (ν2 + ω2 > 0), in eq. (4.246) weneed to define the singularities present in the denominator. In fact, this expressionhas two poles at ν = ±ω, and therefore we need to specify how the integral isdefined. In order to reproduce the euclidean result, it is necessary to specify theintegration path as in Fig. 4.9.1. For instance, for t > 0 by closing the integral inthe lower plane we get

i

2πm(−2πi)

1

2ωe−iωt =

1

2mωe−iωt (4.248)

.

Im

Re

ν

ν− ω + ω

Fig. 4.9.1 - The integration path for equation (4.245).

We would get the same result by integrating along the real axis but translatingthe denominator of +iε (ε > 0) in such a way to move the pole at ω in the lowerplane and the one at −ω in the upper plane. We get

∆(t) = limε→0+

i

2πm

∫ +∞

−∞

e−iνtν2 − ω2 + iε

dν (4.249)

With this prescription we can safely rotate the integration path anti-clockwise by900, since we do not encounter any singularity (see Fig. 4.9.2)

∆(t) =i

2πm

∫ +i∞

−i∞

e−iνtν2 − ω2

dν (4.250)

We have omitted here the limit since the expression is now regular. Performingthe change of variable ν = iν ′

∆(t) = − 1

2πm

∫ +∞

−∞

e+ν′t

−ν ′2 − ω2dν ′ = DE(it) (4.251)

Therefore we have shown how it works the analytic continuation implicit in eq.(4.244). By doing so we have also seen that the Feynman prescription (the +iε term)is equivalent to the euclidean formulation. One could arrive to this connection also

121

.

Im

Re

− ω +

ε

ν

νi

εi

ω −

Fig. 4.9.2 - The Wick rotation in the ν-plane.

by noticing that the Feynman prescription is equivalent to substitute in the pathintegral the term

e− i

2mω2

∫q2(t)dt

(4.252)

with

e− i

2m(ω2 − iε)

∫q2(t)dt

(4.253)

Therefore the Feynman prescription is there to make the integral convergent.

e−1

2mε

∫q2(t)dt

(4.254)

The same convergence is ensured by the euclidean formulation. Therefore we haveshown that, in the case of quadratic actions, the path integral can be formulatedas an euclidean integral, making it a well defined mathematical expression. Thisfact is enough to guarantee that all the manipulations that we make with the pathintegral in the perturbative expansion are well defined, since we will be expandingaround the quadratic case.

122

Chapter 5

The path integral in field theory

5.1 The path integral for a free scalar field

We will extend now the previous approach to the case of a scalar field theory. Wewill show later how to extend the formulation to the case of Fermi fields. Let us startwith a free scalar field. We have denoted the quantum operator by φ(x). Here wewill denote the classical field by ϕ(x). For a neutral particle this is a real function,and its dynamics is described by the lagrangian

S =

∫V

d4x1

2

[∂µϕ∂

µϕ−m2ϕ2] ≡ ∫ t′

t

dt

∫d3x

1

2

[∂µϕ∂

µϕ−m2ϕ2]

(5.1)

A free field theory can be seen as a continuous collection of non-interacting harmonicoscillators. As it is well known this can be seen by looking for the normal modes.To this end, let us consider the Fourier transform of the field

ϕ(x, t) =1

(2π)3

∫d3kei

k · xq(k, t) (5.2)

By substituting inside eq. (5.1) we get

S =1

2

∫V

d4x

∫d3k

(2π)3

d3k′

(2π)3[q(k, t)q(k′, t)

+ (k · k′ −m2)q(k, t)q(k′, t)]eik · x+ ik′ · x

=1

2

∫ t′

t

dt

∫d3k

(2π)3[q(k, t)q(−k, t) − (|k|2 +m2)q(k, t)q(−k, t)] (5.3)

Since ϕ(x, t) is a real field

q∗(k, t) = q(−k, t) (5.4)

and

S =

∫ t′

t

dt

∫d3k

(2π)3

1

2

[|q(k, t)|2 − ω2

k|q(k, t)|2

], ω2

k= |k|2 +m2 (5.5)

123

Therefore the system is equivalent to a continuous set of complex oscillators q(k, t)(or a pair of real oscillators)). The equations of motion are

q(k, t) + ω2kq(k, t) = 0 (5.6)

We are now in the position of evaluating the generating functional Z[J ] by addingan external source term to the action

S =

∫V

d4x

1

2

[∂µϕ∂

µϕ−m2ϕ2]+ Jϕ

(5.7)

From the Fourier decomposition we get

S =

∫ t′

t

dt

∫d3k

(2π)3

1

2

[|q(k, t)|2 − ω2

k|q(k, t)|2

]+ J(−k, t)q(k, t)

(5.8)

where we have also expanded J(x) in terms of its Fourier components. Let us

separate q(k, t) and J(k, t) into their real and imaginary parts

q(k, t) = x(k, t) + iy(k, t)

J(k, t) = l(k, t) + im(k, t) (5.9)

It follows from the reality conditions that the real parts are even functions of kwhereas the imaginary parts are odd functions

x(k, t) = x(−k, t), l(k, t) = l(−k, t)y(k, t) = −y(−k, t), m(k, t) = −m(−k, t) (5.10)

Now we can use the result found in the previous Section for a single oscillator. it willbe enough to take the mass equal to one and sum over all the degrees of freedom.In particular we get for the exponent in eq. (4.226)

− 1

2

∫ t′

t

dsds′J(s)∆(s− s′)J(s′) →

→ −1

2

∫ t′

t

dsds′∫

d3k

(2π)3

[l(k, s)∆(s− s′;ωk)l(k, s

′)

+ m(k, s)∆(s− s′;ωk)m(k, s′)]

= −1

2

∫ t′

t

dsds′∫

d3k

(2π)3J(−k, s)∆(s− s′;ωk)J(k, s′) (5.11)

where we have made use of eq. (5.9). Going back to J(x, t), we get

− 1

2

∫d4xd4yJ(x)

∫d3k

(2π)3∆(s− s′;ωk)e

ik · (x− y)J(y)

≡ − i

2

∫d4xd4yJ(x)∆F (x− y;m2)J(y) (5.12)

124

where we have defined xµ = (x, s), yµ = (y, s′) and

i∆F (x,m2) =

∫d3k

(2π)3∆(s;ωk)e

ik · x = limε→0+

∫d4k

(2π)4

i

k2 −m2 + iεe−ikx (5.13)

Again we have put kµ = (ν,k). Therefore, the generating functional is

Z[J ] = e− i

2

∫d4xd4yJ(x)∆F (x− y;m2)J(y)

Z[0] (5.14)

In particular we get

1

〈0|0〉〈0|T (φ(x1)φ(x2))|0〉 = − 1

Z[0]

δ2Z[J ]

δJ(x1)δJ(x2)

∣∣∣J=0

= i∆F (x1 − x2;m2) (5.15)

We could repeat the exercise in the euclidean formulation, with the result

ZE[J ] = e

1

2

∫d4xd4yJ(x)G0(x− y;m2)J(y)

ZE [0] (5.16)

where

G0(x;m2) =

∫d4k

(2π)4

1

k2 +m2e−ikx (5.17)

We go from Minkowskian to the Euclidean description through the substitution

−i∆F → G0 (5.18)

For the future it will be convenient to use the following notation for the N -pointsGreen’s functions

G(N)(x1, · · · , xN) =〈0|T (φ(x1) · · ·φ(xN))|0〉

〈0|0〉 (5.19)

In the free case we can get evaluate easily the generic G(N) In fact G(N)0 (the in-

dex recalls that we are in the free case) is obtained by developing the generatingfunctional

Z[J ]

Z[0]=

∞∑n=0

∫d4x1 · · · d4xnJ(x1) · · ·J(xn)

(i)n

n!

〈0|T (φ(x1) · · ·φ(xn))|0〉〈0|0〉 ≡

≡∞∑n=0

∫d4x1 · · · d4xnJ(x1) · · ·J(xn)

(i)n

n!G

(n)0 (x1, · · · , xn) (5.20)

By comparison with the expansion of eq. (5.14)

Z[J ]

Z[0]= 1 − i

2

∫d4x1d

4x2J(x1)J(x2)∆F (x1 − x2)

− 1

8

∫d4x1d

4x2J(x1)J(x2)∆F (x1 − x2)

·∫d4x3d

4x4J(x3)J(x4)∆F (x3 − x4) + · · · (5.21)

125

Since Z[J ] is even in J we have

G(2k+1)0 (x1, · · · , x2k+1) = 0 (5.22)

For G(2)0 we get back eq. (5.13). To evaluate G

(4)0 we have first to notice that

all the G(N)0 ’s are symmetric in their arguments. therefore in order to extract the

coefficients we need to symmetrize. Therefore we write∫d4x1 · · · d4x4J(x1) · · ·J(x4)∆F (x1 − x2)∆F (x3 − x4)

=1

3

∫d4x1 · · · d4x4J(x1) · · ·J(x4)

[∆F (x1 − x2)∆F (x3 − x4)

+ ∆F (x1 − x3)∆F (x2 − x4) + ∆F (x1 − x4)∆F (x2 − x3)]

(5.23)

It follows

G(4)0 (x1, · · · , x4) = −

[∆F (x1 − x2)∆F (x3 − x4) + ∆F (x1 − x3)∆F (x2 − x4)

+ ∆F (x1 − x4)∆F (x2 − x3)]

(5.24)

If we represent G(2)0 (x, y) by a line as in Fig. 5.1.1, G

(4)0 will be given by Fig.

5.1.2 since it is obtained in terms of products of G(2)0 . Therefore G

(2n)0 is given by

a combination of products of n two point functions. We will call disconnected adiagram in which we can isolate two subdiagrams with no connecting lines. Wesee that in the free case all the non vanishing Green’s functions are disconnectedexcept for the two point function. Therefore it is natural to introduce a functionalgenerating only the connected Green’s functions. In the free case we can put

.

x y

Fig. 5.1.1 - The two point Green’s function G(2)0 (x, y).

.

x x

x x

1 2

3 4

x1 x1x2x2

x3

x3x4

x4

++

Fig. 5.5.2 - The expansion of the four point Green’s function G(4)0 (x1, x2, x3, x4).

Z[J ] = eiW [J ] (5.25)

126

with

W [J ] = −1

2

∫d4x1d

4x2J(x1)J(x2)∆F (x1 − x2) +W [0] (5.26)

and we have

1

Z[0]

δ2Z[J ]

δJ(x1)δJ(x2)

∣∣∣J=0

= iδ2W [J ]

δJ(x1)δJ(x2)

∣∣∣J=0

= −i∆F (x1 − x2) (5.27)

Therefore the derivatives of W [J ] give the connected diagrams with the propernormalization(that is divided by 1/Z[0]). In the next Section we will see that thisproperty generalizes to the interacting case. That is W [J ] is defined through eq.(5.25), and its Volterra expansion gives rise to the connected Green’s functions

iW [J ] =

∞∑n=0

(i)n

n!

∫d4x1 · · · d4xnJ(x1) · · ·J(xn)G

(n)c (x1, · · · , xn) (5.28)

where

G(n)c (x1, · · · , xn) =

1

〈0|0〉〈0|T (φ(x1) · · ·φ(xn))|0〉conn (5.29)

The index ”conn” denotes the connected Green’s functions. Notice that once we findthe connected Green’s functions the theory is completely solved, since the generatingfunctional is recovered by exponentiation of W [J ].

5.2 The generating functional of the connected

Green’s functions

Let us start by defining recursively the connected Green’s functions.

G(1)(x1) = G(1)c (x1) ≡ G(1)

c ;

G(2)(x1, x2) = G(2)c (x1, x2) +G(1)

c (x1)G(1)c (x2) ≡ G(2)

c +G(1)c

2;

G(3)(x1, x2, x3) = G(3)c (x1, x2, x3) +G(1)

c (x1)G(2)c (x2, x3)

+G(1)c (x2)G

(2)c (x1, x3) +G(1)

c (x3)G(2)c (x1, x2) +G(1)

c (x1)G(1)c (x2)G

(1)c (x3) ≡

≡ G(3)c + 3G(1)

c G(2)c +G(1)

c

3(5.30)

On the right hand side we have used a symbolic notation by adding together thetopological equivalent configurations. To define G

(n)c (x1, · · · , xn) we decompose the

set (x1, · · · , xn) in all the possible subsets, and then we write

G(n)(x1, · · · , xn) =∑

G(n1)c (xp1, · · · , xpn1

)G(n2)c (xq1 , · · · , xqn2

) · · ·G(nk)c (xr1 , · · · , xrnk

)

(5.31)where we sum over all the indices ni such that n1 + · · · + nk = n and over all thepermutations from 1 to n of the indices in the set p1, · · · , pn1, etc. A graphical

127

.

xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

= c

xxxxxxxxxxxxxxxxxxxxxxxxx

=c

c+

xxxxxxxxxxxxxxxxxxxxxxxxx

=

c

+cc

c3 +

c

c

c

xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

= c +c

c3 + 4

c

c+

+ 6

c

c

c +cc

cc

Fig. 5.2.1 - The connected Green’s functions.

description is given in Fig. 5.2.1. By using these definitions it is not difficult tocheck that W [J ] is the generating functional of the connected Green’s functions. It

is enough to compare the expansion of Z[J ] in series of G(n)c with the expansion from

the G(n). This can be done by induction. We will limit ourselves to an explicit checkup to the 4th order. By defining G(0) = 1 and using the same symbolic notation asbefore, we get

Z[J ]

Z[0]=

∞∑n=0

(i)n

n!JnG(n) = ei(W [J ] −W [0])

= exp

[iG(1)

c J +i2

2!G(2)c J2 +

i3

3!G(3)c J3 +

i4

4!G(4)c J4 + · · ·

]

≈ 1 + iG(1)c J +

i2

2!G(2)c J2 +

i3

3!G(3)c J3 +

i4

4!G(4)c J4

+1

2!

[i2G(1)

c

2J2 +

i4

4!G(2)c

2J4 + i3G(1)

c G(2)c J3 +

i4

3G(1)c G(3)

c J4

]

+1

3!

[i3G(1)

c

3J3 +

3

2i4G(1)

c

2G(2)c J4

]

+1

4!

[i4G(1)

c

4J4]

+ · · ·

128

= 1 + iJG(1)c +

i2

2!

[G(2)c +G(1)

c

2]

+i3

3!J3[G(3)c + 3G(1)

c G(2)c +G(1)

c

3]

+i4

4!J4[G(4)c + 3G(2)

c

2+ 4G(1)

c G(3)c + 6G(1)

c

2G(2)c +G(1)

c

4]

+ · · · (5.32)

5.3 The perturbative expansion for the theory λϕ4

Let us consider an interacting scalar field. The action will be of the type

S =

∫V

d4x

(1

2

[∂µϕ∂

µϕ−m2ϕ2]− V (ϕ)

)(5.33)

where V (ϕ) is the interaction potential. A typical example is the theory λϕ4 withthe potential given by

V (ϕ) =λ

4!ϕ4 (5.34)

The derivation of the perturbative expansion is very simple in this formalism. Wetake advantage of the following identity valid for any functional F

∫D(ϕ(x))F [ϕ]e

iS + i

∫d4xJ(x)ϕ(x)

= F

[1

i

δ

δJ

] ∫D(ϕ(x))e

iS + i

∫d4xJ(x)ϕ(x)

(5.35)

We separate in eq. (5.33) the quadratic part of the action, S0, from the interactingpart and then we use the previous identity

Z[J ] = N

∫D(ϕ(x))e

iS + i

∫d4xJ(x)ϕ(x)

= N

∫D(ϕ(x))e

−i∫d4xV (ϕ)

eiS0 + i

∫d4xJ(x)ϕ(x)

= e−i∫d4xV

(1

i

δ

δJ

)Z0[J ] (5.36)

Notice that we have suppressed the temporal limits in the expression for the gen-erating functional. Z0[J ] is the generating functional for the free case, evaluated ineq. (5.14). Therefore we get

Z[J ] = eiW [J ] = Z[0]e−i∫d4xV

(1

i

δ

δJ

)e− i

2

∫d4xd4yJ(x)∆F (x− y)J(y)

(5.37)

129

In this expression we have suppressed the mass in the argument of ∆F . For thefollowing calculation it will be useful to introduce the following shorthand notation∫

d4x1 · · · d4xnF (x1, · · · , xn) ≡ 〈F (x1, · · · , xn)〉 (5.38)

Then, eq. (5.37) can be rewritten in the following way

Z[J ] = eiW [J ] = e−i〈V

(1

i

δ

δJ

)〉e− i

2〈J(1)∆F (1, 2)J(2)〉

Z[0]

= e−i〈V

(1

i

δ

δJ

)〉eiW0[J ] (5.39)

This gives rise to the following expression for the generating functional of the con-nected Green’s functions

W [J ] = −i log

⎡⎢⎣eiW0[J ] +

⎛⎜⎝e−i〈V

(1

i

δ

δJ

)〉− 1

⎞⎟⎠ eiW0[J ]

⎤⎥⎦

= W0[J ] − i log

⎡⎢⎣1 + e−iW0[J ]

⎛⎜⎝e−i〈V

(1

i

δ

δJ

)〉− 1

⎞⎟⎠ eiW0[J ]

⎤⎥⎦(5.40)

This expression can be easily expanded in V . In fact, by defining

δ[J ] = e−iW0[J ]

⎛⎜⎝e−i〈V

(1

i

δ

δJ

)〉− 1

⎞⎟⎠ eiW0[J ] (5.41)

we see that we can expand W [J ] in a series of δ. Notice also that the expansion isin term of the interaction lagrangian. At the second order in δ we get

W [J ] = W0[J ] − i

(δ − 1

2δ2

)+ · · · (5.42)

Let us now consider the case of λϕ4. the functional δ[J ] can be in turn expanded ina series of the dimensionless coupling λ

δ = λδ1 + λ2δ2 + · · · (5.43)

It follows

W [J ] = W0[J ] − iλδ1 − iλ2

(δ2 − 1

2δ21

)+ · · · (5.44)

130

Then, from eq. (5.43) we get

δ1 = − i

4!e−iW0[J ]〈

(1

i

δ

δJ

)4

〉eiW0[J ]

δ2 = − 1

2(4!)2e−iW0[J ]〈

(1

i

δ

δJ

)4

〉〈(

1

i

δ

δJ

)4

〉eiW0[J ] (5.45)

By using the following list of functional derivatives

δ

δJ(x)eiW0[J ] = −i∆(x, 1)J(1)eiW0[J ]

δ2

δJ(x)2eiW0[J ] = (−i∆(x, x) − ∆(x, 1)∆(x, 2)J(1)J(2)) eiW0[J ]

δ3

δJ(x)3eiW0[J ] =

(− 3∆(x, x)∆(x, 1)J(1)

+ i∆(x, 1)∆(x, 2)∆(x, 3)J(1)J(2)J(3))eiW0[J ]

δ4

δJ(x)4eiW0[J ] = (−3∆(x, x)2 + 6i∆(x, x)∆(x, 1)∆(x, 2)J(1)J(2)

+∆(x, 1)∆(x, 2)∆(x, 3)∆(x, 4)J(1)J(2)J(3)J(4))eiW0[J ] (5.46)

where the integration on the variables 1, 2, 3, 4 is understood, we obtain

δ1 = − i

4!

[〈∆(y, 1)∆(y, 2)∆(y, 3)∆(y, 4)J(1)J(2)J(3)J(4)〉

+ 6i〈∆(y, y)∆(y, 1)∆(y, 2)J(1)J(2)〉− 3〈∆2(y, y)〉]

(5.47)

The quantity δ2 can be made in terms of δ1 by writing

δ2 = − 1

2(4!)2e−iW0[J ]〈

(1

i

δ

δJ

)4

〉(eiW0[J ]e−iW0[J ]〈

(1

i

δ

δJ

)4

〉eiW0[J ]

)

= − i

2(4!)e−iW0[J ]〈

(1

i

δ

δJ

)4

〉(eiW0[J ]δ1

)(5.48)

Then we have

〈(

1

i

δ

δJ

)4

〉(eiW0[J ]δ1

)= 〈δ

4eiW0[J ]

δJ(x)4〉δ1 + 4〈δ

3eiW0[J ]

δJ(x)3

δδ1δJ(x)

〉

+ 6〈δ2eiW0[J ]

δJ(x)2

δ2δ1δJ(x)2

〉 + 4〈δeiW0[J ]

δJ(x)

δ3δ1δJ(x)3

〉

+ eiW0[J ]〈 δ4δ1δJ(x)4

〉 (5.49)

131

and therefore

δ2 =1

2δ21 −

i

2 · 4!e−iW0[J ]

[4〈δ

3eiW0[J ]

δJ(x)3

δδ1δJ(x)

〉

+ 6〈δ2eiW0[J ]

δJ(x)2

δ2δ1δJ(x)2

〉 + 4〈δeiW0[J ]

δJ(x)

δ3δ1δJ(x)3

〉

+ eiW0[J ]〈 δ4δ1δJ(x)4

〉]

(5.50)

The δ21 gives a disconnected contribution since there are no propagators connecting

the two terms. In fact the previous expression shows that δ2 contains a term whichcancels precisely the term δ2

1 in eq. (5.44). By using the expression for δ1 and eqs.(5.46) we get

δ2 − 1

2δ21 =

=

(− i

2 · 4!

)(− i

4!

)〈4(− 3∆(x, x)∆(x, 1)J(1)

+ i∆(x, 1)∆(x, 2)∆(x, 3)J(1)J(2)J(3))

·(4∆(y, x)∆(y, 4)∆(y, 5)∆(y, 6)J(4)J(5)J(6) + 12i∆(y, y)∆(y, x)∆(y, 4)J(4)

)+ 6

(− i∆(x, x) − ∆(x, 1)∆(x, 2)J(1)J(2)

)(12∆2(y, x)∆(y, 3)∆(y, 4)J(3)J(4)

+ 12i∆(y, y)∆2(y, x))

+ 4(− i∆(x, 1)J(1)

)(24∆3(y, x)∆(y, 2)J(2)

)+ 24∆4(y, x)〉 (5.51)

By omitting the terms independent on the external source we have

δ2 − 1

2δ21 =

=i

2〈J(1)∆(1, x)

(1

6∆3(x, y) +

1

4∆(x, x)∆(x, y)∆(y, y)

)∆(y, 2)J(2)〉

+i

8〈J(1)∆(1, x)∆2(x, y)∆(y, y)∆(x, 2)J(2)〉

+1

12〈J(1)∆(1, x)∆(x, x)∆(x, y)∆(y, 2)∆(y, 3)∆(y, 4)J(2)J(3)J(4)〉

+1

16〈J(1)J(2)∆(1, x)∆(2, x)∆2(x, y)∆(y, 3)∆(y, 4)J(3)J(4)〉

− i

72〈J(1)J(2)J(3)∆(1, x)∆(2, x)∆(3, x)∆(x, y)∆(y, 4)

· ∆(y, 5)∆(y, 6)J(4)J(5)J(6)〉 (5.52)

132

Now, by differentiating functionally eq. (5.44) and putting J = 0 we can evaluatethe connected Green’s functions. For instance, recalling that

iG(2)c (x1, x2) =

δ2W [J ]

δJ(x1)δJ(x2)

∣∣∣J=0

(5.53)

we get

G(2)c (x1, x2) = i∆F (x1 − x2) − λ

2

∫d4x∆F (x1 − x)∆F (x− x)∆F (x− x2)

− iλ2

∫d4xd4y∆F (x1 − x)

[16∆3F (x− y)

+1

4∆F (x− y)∆F (x− x)∆F (y − y)

]∆F (y − x2)

− iλ2

4

∫d4xd4y∆F (x1 − x)∆2

F (x− y)∆F (y − y)∆F (x− x2) (5.54)

.

xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

xxxx

xxxx

x x1 2= x

xxx

xxxx

x1 x2xx

xxxx

x1 x2+

xxxx

x +

+ xx

xxxx

x1 x2xxxx

xxxx

x y+ x

xxxxx

xxxx

x1 x2x +

xxxx y

xxxx

xxxx

x1 x2xx

x y

Fig. 5.3.1 - The graphical expansion for the two point connected Green’s functionG

(2)c (x, y).

In Fig. 5.3.1) we give a graphical form to this expression. as we see the resulthas an easy interpretation. The diagrams of Fig. 5.3.1) are given in terms oftwo elements, the propagator ∆F (x− y), corresponding to the lines and the vertex

λϕ4 corresponding to a point joining four lines. In order to get G(n)c we have to

draw all the possible connected diagrams containing a number of interaction verticescorresponding to the fixed perturbative order. The analytic expression is obtainedassociating a factor i∆F (x−y) to each line connecting the point x with the point y,and a factor (−iλ/4!) for each interaction vertex. The numerical factors appearingin eq. (5.54) can be obtained by counting the possible ways to draw a given diagram.For instance let us consider the diagram of Fig. 5.3.2. This diagram contains thetwo external propagators, the internal one, and the vertex at x. To get the numericalfactor we count the number of ways to attach the external propagators to the vertex,after that we have only a possibility to attach the vertex at the internal propagator.There are four ways to attach the the vertex at the first propagator and three forthe second. Therefore (

1

4!

)· 4 · 3 =

1

2(5.55)

133

.

= xx

xxxx

xxxx

x1 xxxxx

xxxx

x1 x2x xxxx

xxxx

x x2

xxxx

x

Fig. 5.3.2 - How to get the numerical factor for the first order contributions toG

(2)c (x1, x2).

where the factor 1/4! comes from the vertex. We proceed in the same way for thediagram of Fig. 5.3.3.

.

= xx

xx

x1

xxxxx

xxxx

x1 x2xxxx

xxxx

x yxxxx

xxxx

x2

yx y

Fig. 5.3.3 - How to get the numerical factor for one of the second order contribu-tions to G

(2)c (x1, x2).

.

= xxxx

xxxx

xx

xxxx

xxxx

x xx

xx1

2 3

4

xx

xxxx

xxxx

xx xxxx

x x

xx1

2 3

4

xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

xxxx

xxxxxx

xxxx

xxxx

x xx

xx1

2 3

4

xx

xxxx

xxxx

xxxx

xxxx

x xx

xx1

2 3

4

xxxx

xx

xx

xx

xxxx

x x

xxx1

2 3

4

xxxx

xxxx

xxxx

xx xx

x x

xxx1

2 3

4xxxx

xx xxxx

xxxx

y

yy y+

+ +

+

+

+

xx xxxx

xxxx

xxxx

xxxx x

x1

x3x2

x4

x y

xxxx

xxxx

xxxx

xxxx xx

xx

xxxx

y

x

+

+ +

x1

x2

x4

x3

+ xxxx

xxxx

xx xx

xx xx

x y

x1

x2

x4

x3

Fig. 5.3.4 - The graphical expansion up to the second order of G(4)c (x1, x2, x3, x4).

There are four ways to attach the first vertex to the first propagator and four forattaching the second vertex to the second propagator. Three ways for attaching thesecond leg of the first vertex to the second vertex and two ways of attaching the thethird leg of the first vertex to the second one. Therefore we get(

1

4!

)· 4 · 4 · 3 · 2 =

1

6(5.56)

134

The four point function can be evaluated in analogous way and the result is givenin Fig. 5.3.4. Of course we can get it by functional differentiation of W [J ]. Thefirst order contribution is

G(4)c (x1, x2, x3, x4) = −iλ

∫d4x∆F (x1−x)∆F (x2−x)∆F (x3−x)∆F (x4−x) (5.57)

where the numerical coefficient is one, since there are 4 · 3 · 2 of attaching the vertexto the four external lines.

5.4 The Feynman’s rules in momentum space

The rules for the perturbative expansion of the Green’s functions in space-time thatwe got in the previous Section can be easily extended to the momentum space. Wewill consider again the two point function given in eq. (5.54) where we will take

separately the various contributions in λ that we will denote by G(2)(i)c , where i

stands for the power of λ. let us also define the Fourier transform of G(2)c (x1, x2) as

G(2)c (p1, p2) =

∫d4x1d

4x2G(2)c (x1, x2)e

ip1x1 + ip2x2 (5.58)

Then we get

G(2)(0)c (p1, p2) = (2π)4δ4(p1 + p2)

i

p21 −m2 + iε

(5.59)

and

G(2)(1)c (p1, p2) =

= −iλ2

∫d4x1d

4x2d4xeip1x1 + ip2x2

∫d4p

(2π)4e−ip(x1 − x) i

p2 −m2 + iε

·∫

d4q

(2π)4

i

q2 −m2 + iε

∫d4k

(2π)4e−ik(x− x2) i

k2 −m2 + iε

= −iλ2

∫d4p

(2π)4

d4q

(2π)4

d4k

(2π)4(2π)4δ4(p− k)(2π)4δ4(p1 − p)(2π)4δ4(p2 + k)

· i

p2 −m2 + iε

i

q2 −m2 + iε

i

k2 −m2 + iε(5.60)

The product of the δ4 functions may be rewritten in such a way to pull out theoverall four-momentum conservation

(2π)4δ4(p1 + p2)(2π)4δ4(p1 − p)(2π)4δ4(p− k) (5.61)

By integrating over p and k we get

G(2)(1)c (p1, p2) = −iλ

2(2π)4δ4(p1+p2)

(i

p21 −m2 + iε

)2 ∫d4q

(2π)4

i

q2 −m2 + iε(5.62)

Also in momentum space we can make use of a diagrammatic expansion with thefollowing rules:

135

• For each propagator draw a line with associated momentum p (see Fig. 5.4.1).

.

:εp m2 2_ + i

___________i

Fig. 5.4.1 - The propagator in momentum space.

• For each factor (−iλ/4!) draw a vertex with the convention that the momentumflux is zero (see Fig. 5.4.2).

.

λi4 = 0

p

p

p

p

p p

1

2 3

4

: _4!__ p p1 2 3 ++ +,

Fig. 5.4.2 - The vertex in momentum space.

• To get G(n)c draw all the topological inequivalent diagrams after having fixed

the external legs. The number of ways of drawing a given diagram is itstopological weight. The contribution of such a diagram is multiplied by itstopological weight.

• After the requirement of the conservation of the four-momentum at each vertexwe need to integrate over all the independent internal four-momenta withweight ∫

d4q

(2π)4(5.63)

A more systematic way is to associate a factor

−iλ4!

(2π)4δ4(

4∑i=1

pi) (5.64)

to each vertex and integrate over all the momenta of the internal lines. Thisgives automatically a factor

(2π)4δ4(

n∑i=1

pi), (n = numero linee esterne) (5.65)

corresponding to the conservation of the total four-momentum of the diagram.

By using these rules we can easily evaluate the 2nd order contribution of Fig.5.4.3.

136

.

1

2

3q

q

q

p2

p1

Fig. 5.4.3 - One of the second order contributions to G(2)c .

We get

4 × 4 × 3 × 2 ×(−iλ

4!

)2i

p22 −m2 + iε

∫ (3∏i=1

d4qi(2π)4

i

q2i −m2 + iε

)

·(2π)4δ4(p1 − q1 − q2 − q3)(2π)4δ4(q1 + q2 + q3 − p2)i

p21 −m2 + iε

=λ2

6

1

(p21 −m2 + iε)2

(2π)4δ4(p1 + p2)

·∫ (

3∏i=1

d4qi(2π)4

i

q2i −m2 + iε

)(2π)4δ4(p1 − q1 − q2 − q3) (5.66)

5.5 Power counting in λϕ4

We start by going to the euclidean formulation. Let us denote the time by tR. Therelation with the euclidean time is tE = itR. Recalling from the Section 4.9 thatfor the harmonic oscillator of unit mass one has (we have defined xR = (tR, x),xE = (tE , x))

i∆F (xR) =

∫d3p

(2π)3∆(tR, ω

2k)eip · x =

∫d3p

(2π)3DE(tE, ω

2k)eip · x

=

∫d4pE(2π)4

e−ipExEp2E +m2

(5.67)

where pE = (EE , p), and we have used eq. (4.237) for unit mass and with ν = EE.From this we get the following modifications to Feynman’s rules in the euclidean

version

• Associate to each line a propagator

1

p2 +m2(5.68)

137

• To each vertex associate a factor

− λ

4!(5.69)

This rule follows because the weight in the euclidean path-integral is given by

e−SE instead of eiS .

The other rules of the previous Section are unchanged. The reason to use theeuclidean version is because in this way the divergences of the integrals becomemanifest by simply going in polar coordinates.

We will now in the position to make a general analysis of the divergences. Forsimplicity we will consider only scalar particles. First of all consider the numberof independent momenta in a given diagram, or the number of ”loops”, L. Thisis given, by definition, by the number of momenta upon which we are performingthe integration, since they are not fixed by the four-momentum conservation at thevertices. Since one of this conservation gives rise to the overall four-momentumconservation we see that the number of loops is given by

L = I − (V − 1) = I − V + 1 (5.70)

where I is the number of internal lines and V the number of vertices. As in Section2.3 we will be interested in evaluating the superficial degree of divergence of adiagram. To this end notice that

• The L independent integrations give a factor∏L

k=1 ddpk where d is the number

of space-time dimensions.

• Each internal line gives rise to a propagator containing two inverse powers inmomenta

I∏i=1

1

p2i +m2

(5.71)

Given that, the superficial degree of divergence is defined as

D = dL− 2I (5.72)

Notice that D has to do with a common scaling of all the integration variables,therefore it may well arise the situation where D < 0 (corresponding to a convergentsituation), but there are divergent sub-integrations. Therefore we will need a moreaccurate analysis that we will do at the end of this Section. For the moment beinglet us stay with this definition. We will try now to get a relation between D andthe number of external legs in a diagram. By putting VN equal to the total numberof vertices with N lines we have (see Section 2.3)

NVN = E + 2I (5.73)

138

with E the number of external legs and I the number of internal ones (recall weare dealing with scalar particles only). Eliminating I between this relation and theprevious one, we get

I =1

2(NVN −E) (5.74)

from which

D = d(I−VN+1)−2I = (d−2)I−d(VN−1) =d− 2

2(NVN−E)−d(VN−1) (5.75)

This relation can be rewritten as

D = d− 1

2(d− 2)E +

(N − 2

2d−N

)VN (5.76)

In particular, for d = 4 we have

D = 4 − E + (N − 4)VN (5.77)

For the λϕ4 theory this relation becomes particularly simple since then D dependsonly on the number of external legs

D = 4 − E (5.78)

Therefore the only diagrams superficially divergent, D ≥ 0, are the ones with E = 2and E = 4.

As noticed before, this does not prove that a diagram with E > 4 or D < 0 isconvergent. Let us discuss more in detail this point. Let us consider the diagram ofFig. 5.5.1, and let us suppose that the diagrams 1 and 2 have degrees of superficialdivergence D1 and D2 respectively. Since they are connected by n lines, we haven− 1 independent momenta, and therefore

.

1 2

nD D1 2

Fig. 5.5.1 - Two diagrams connected by n-lines.

D = D1 +D2 + 4(n− 1) − 2n = D1 +D2 + 2n− 4 (5.79)

Therefore we may have D < 0 with D1 and D2 positive, that is both divergent.However for n > 1 we have

D ≥ D1 +D2 (5.80)

139

and therefore in order to have D < 0 with 1 or 2 being divergent diagrams, we needD1 or D2 to be negative enough to overcome the other. To exclude the case n = 1means to consider only the one-particle irreducible (1PI) Feynman’s diagrams. InFig. 5.5.2 we give an example of a one-particle reducible diagram.

.

xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

Fig. 5.5.2 - One-particle reducible Feynman’s diagram.

The reason of excluding these diagrams from our considerations is that the mo-mentum of the internal line connecting the two diagrams is fixed, and therefore itcannot give rise to divergences. Furthermore, the one-particle reducible diagramscan be easily reconstructed from the 1PI’s. After that we look for the diagramshaving D < 0 but yet divergent, in order to have a complete classification of thedivergences. The idea is simply to look for the reducibility of the diagram. We willnow stay with the λϕ4 theory. Then, we can consider the 2-, 3-particle reduciblediagrams. In fact, for the nature of the vertex it is impossible to have a connected4-particle irreducible diagram. We begin with the 3-particle reducibility. In thiscase we look for diagrams of the type of Fig. 5.5.3.

.

xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

1

N

Fig. 5.5.3 - Three-particle reducible Feynman’s diagram.

In this case we have D1 = 0 and D2 = 1 −N , therefore

D = D1 +D2 + 2 = 3 −N (5.81)

In order the superficial degree of divergence is negative, we must have N > 4. Inthis diagram we have a primitively divergent four point function and a diagram withD < 0 which can be again analyzed in the same way. In analogous way we lookfor two-particle reducible diagrams with D < 0 and we extract contributions of theforms given in Fig 5.5.4.

140

.

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xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

1

N

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xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

1

N

Fig. 5.5.4 - Two-particle reducible Feynman’s diagram.

In the first case we have D1 = 2 and D2 = 2 −N , therefore

D = D1 +D2 = 4 −N (5.82)

Again this requires N > 4. The second diagram has again D2 < 0 and we can iteratethe procedure. In the second case we have D1 = 0 and D2 = 2 −N , therefore

D = D1 +D2 = 2 −N (5.83)

requiring N > 2. By proceeding in this way we see that truly convergent diagramsdo not contain hidden two and four point divergent functions. What we have shownis that for the λϕ4 theory in 4 dimensions a 1PI Feynman diagram is convergentif D < 0 and it cannot be split up into 3- or 2-particle reducible parts containingtwo-and four-point divergent functions. This turns out to be a general theoremdue to Weinberg, valid in any field theory and it says that a Feynman diagram isconvergent if its superficial degree of divergence and that of all its subgraphs arenegative.

In the case of the λφ4 theory the Weinberg’s theorem tells us that all the diver-gences rely in the two and four point functions. If it happens that all the divergentdiagrams (in the primitive sense specified above) correspond to terms which arepresent in the original lagrangian, we say that the theory is renormalizable. Ingeneral we can look at this question by studying the equation for the superficialdivergence that we got before

D = d− 1

2(d− 2)E +

(N − 2

2d−N

)VN (5.84)

141

For d = 4 we recallD = 4 − E + (N − 4)VN (5.85)

We see that D increases with VN for N > 4. Therefore we have infinite divergentdiagrams and the theory is not renormalizable. We can have a renormalizable scalartheory only for N ≤ 4, meaning that the dimension N of the vertex operator, ϕN ,must be less or equal to 4. An interesting case is the one of the theory λϕ3 in 6dimensions. For d = 6 we get

D = 6 − 2E + (2N − 6)VN (5.86)

and for N = 3D = 6 − 2E (5.87)

and the only divergent functions are the one- two- and three-point ones. In d = 6we have dim[ϕ] = 2 and again in order to have renormalizability we must have thatthe dimensions of the vertex operator must be less or equal to 6.

5.6 Regularization in λϕ4

In this Section we will evaluate the one-loop contributions to the two- and four-point Green’s functions by using the dimensional regularization. To this end, let usconsider the action in dimensions d = 2ω

S2ω =

∫d2ωx

[1

2∂µϕ∂

µϕ+1

2m2ϕ2 +

λ

4!ϕ4

](5.88)

In our unit system we have (/h = c = 1), and therefore the action S2ω should bedimensionless. As discussed in Section 2.3 this allows to evaluate the dimensions ofthe fields, and we recall that for the scalar field we have

dim[ϕ] =d

2− 1 = ω − 1 (5.89)

From this it follows at once that

dim[m] = 1, dim[λ] = 4 − 2ω (5.90)

As we did in QED for the electric charge, here too it is convenient to introduce adimensionful parameter µ in order to be able to define a dimensionless coupling

λnew = λold(µ2)ω−2 (5.91)

Then the action will be (λ = λnew)

S2ω =

∫d2ωx

[1

2∂µϕ∂

µϕ+1

2m2ϕ2 +

λ

4!(µ2)2−ωϕ4

](5.92)

The Feynman’s rules are slightly modified

142

• The scalar product among two four-vectors becomes a sum over 2ω compo-nents.

• In the loop integrals we will have a factor∫d2ωp

(2π)2ω(5.93)

and the δ function in 2ω-dimensions will be defined by∫d2ωpδ(2ω)(

∑i

pi) = 1 (5.94)

• The coupling λ goes into λ(µ2)2−ω.

Also, we will do all the calculation in the euclidean version. Remember thatin Section 2.5 we gave the relevant integrals both in the euclidean and in theminkowskian version.

We start our calculation from the first order contribution to the two-point func-tion corresponding to the diagram (tadpole) of Fig. 5.6.1. This is given by

(2π)2ωδ(2ω)(p1 + p2)1

p21 +m2

T21

p21 +m2

(5.95)

.

p

p

2p1

Fig. 5.6.1 - The one-loop contribution to the two-point function in the λϕ4 theory.

with

T2 =1

2(−λ)(µ2)2−ω

∫d2ωp

(2π)2ω

1

p2 +m2= −λ

2(µ2)2−ω πω

(2π)2ωΓ(1 − ω)

1

(m2)1−ω

= −λ2

m2

(4π)2

(4πµ2

m2

)2−ωΓ(1 − ω) (5.96)

For ω → 2 we get

T2 ≈ −λ2

m2

(4π)2

[1 + (2 − ω) log

4πµ2

m2+ · · ·

]× (−1)

[1

2 − ω+ ψ(2) + · · ·

]

=λ

32π2m2

[1

2 − ω+ ψ(2) + log

4πµ2

m2+ · · ·

](5.97)

143

.

3

4p1

p2

p

p

p - k

k

Fig. 5.6.2 - The one-loop contribution to the four-point function in the λϕ4 theory.

This expression has a simple pole at ω = 2 and a finite part depending explicitly onµ2.

Let us consider now the one-loop contribution to the four-point function. Thecorresponding diagram is given in Fig. 5.6.2, with an amplitude

(2π)2ωδ(2ω)(p1 + p2 + p3 + p4)

4∏k=1

1

p2k +m2

T4 (5.98)

with

T4 =1

2(−λ)2(µ2)4−2ω

∫d2ωk

(2π)2ω

1

k2 +m2

1

(k − p)2 +m2, p = p1 + p2 (5.99)

We reduce the two denominator to a single one, through the equation (see Section2.6)

1

k2 +m2

1

(k − p)2 +m2=

∫ 1

0

dx1

[x(k2 +m2) + (1 − x) ((k − p)2 +m2)]2

=

∫ 1

0

dx1

[x(k2 +m2) + (1 − x) (k2 +m2 − 2k · p+ p2)]2

=

∫ 1

0

dx1

[(k2 +m2) + (1 − x)(p2 − 2k · p)]2 (5.100)

the denominator can be written as

(k2 +m2) + (1 − x)(p2 − 2k · p) = [k − (1 − x)p]2 +m2 + p2(1 − x) − p2(1 − x)2

= [k − (1 − x)p]2 +m2 + p2x(1 − x) (5.101)

Since our integral is a convergent one, it is possible to translate k into k − p(1 − x)obtaining

T4 =λ2

2(µ2)4−2ω

∫ 1

0

dx

∫d2ωk

(2π)2ω

1

[k2 +m2 + p2x(1 − x)]2(5.102)

144

By performing the momentum integration we get

T4 =λ2

2(µ2)4−2ω

∫ 1

0

dx

(4π)ωΓ(2 − ω)

1

[m2 + p2x(1 − x)]2−ω(5.103)

In the limit ω → 2

T4 ≈ λ2

2(µ2)2−ω [1 + (2 − ω) logµ2

] ∫ 1

0

dx

(4π)2

[1

2 − ω+ ψ(1)

]

×[1 − (2 − ω) log

m2 + p2x(1 − x)

4π

]

≈ (µ2)2−ω λ2

32π2

[1

2 − ω+ ψ(1) −

∫ 1

0

dx logm2 + p2x(1 − x)

4πµ2

](5.104)

Also the x integration can be done by using the formula∫ 1

0

dx log

[1 +

4

ax(1 − x)

]= −2 +

√1 + a log

√1 + a + 1√1 + a− 1

, a > 0 (5.105)

The final result is

T4 ≈ (µ2)2−ω λ2

32π2

[ 1

2 − ω+ ψ(1) + 2 + log

4πµ2

m2

−√

1 +4m2

p2log

√1 + 4m2

p2+ 1√

1 + 4m2

p2− 1

+ · · ·]

(5.106)

5.7 Renormalization in the theory λϕ4

In order to renormalize the theory λϕ4 we proceed as in QED. That is we split theoriginal lagrangian, expressed in terms of bare couplings and bare fields in a part

Lp =1

2∂µϕ∂

µϕ+1

2m2ϕ2 +

λ

4!µ4−2ωϕ4 (5.107)

depending on the renormalized parameters m and λ and field ϕ, and in the counter-terms contribution

Lct =1

2δZ∂µϕ∂

µϕ+1

2δm2ϕ2 +

δλ

4!µ4−2ωϕ4 (5.108)

The sum of these two expressions

L =1

2(1 + δZ)∂µϕ∂

µϕ+1

2(m2 + δm2)ϕ2 +

(λ+ δλ)

4!µ4−2ωϕ4 (5.109)

145

gives back the original lagrangian, if we redefine couplings and fields as follows

ϕB = (1 + δZ)1/2ϕ ≡ Z1/2ϕ ϕ (5.110)

m2B =

m2 + δm2

1 + δZ= (m2 + δm2)Z−1

ϕ (5.111)

λB = µ4−2ω λ+ δλ

(1 + δZ)2= µ4−2ω(λ+ δλ)Z−2

ϕ (5.112)

In fact

L =1

2∂µϕB∂

µϕB +1

2m2Bϕ

2B +

λ0

4!ϕ4B (5.113)

The counter-terms are then evaluated by the requirement of removing the di-vergences arising in the limit ω → 2, except for a finite part which is fixed by therenormalization conditions. From the results of the previous Section, by summing up

.

xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

xx xx = xx xx xxxx xx+

xxxx

xxxx

xxxx

xxxx+ + xx

xxxxxx

xxxx

+

Fig. 5.7.1 - The contributions to the teo-point connected Green’s function.

all the 1PI contributions we get the following expression for the two-point connectedGreen’s function (see Fig. 5.7.1)

G(2)c (p1, p2) = (2π)2ωδ2ω(p1 + p2)

[1

p21 +m2

+1

p21 +m2

T21

p21 +m2

+ · · ·]

≈ (2π)2ωδ2ω(p1 + p2)1

p21 +m2 − T2

(5.114)

Therefore, the effect of the first order contribution is to produce a mass correctiondetermined by T2. The divergent part can be taken in care by the counterterm δm2

which produces an additional contribution to the one-loop diagram, represented inFig. 5.7.2. We choose the counterterm as

1

2δm2ϕ2 =

λ

64π2m2

[2

ε+ F1

(ω,m2

µ2

)]ϕ2 (5.115)

with ε = 4−2ω and F1 an arbitrary dimensionless function Adding the contributionof the counterterm to the previous result we get

G(2)c (p1, p2) = (2π)2ωδ2ω(p1 + p2)

[1

p21 +m2 − T2

+1

p21 +m2

(−δm2)1

p21 +m2

](5.116)

146

.

xx x_ m 2δ=

Fig. 5.7.2 - The Feynman’rule for the counterterm δm2.

at the order λ we have

G(2)c (p1, p2) = (2π)2ωδ2ω(p1 + p2)

1

p21 +m2 + δm2 − T2

(5.117)

Of course the same result could have been obtained by noticing that the countertermmodifies the propagator by shifting m2 into m2+δm2. Now the combination δm2−T2

is finite and given by

δm2 − T2 =λ

32π2m2

[F1 − ψ(2) + log

m2

4πµ2

](5.118)

The finite result at the order λ is then

G(2)c (p1, p2) = (2π)2ωδ2ω(p1 +p2)

1

p21 +m2

(1 + λ

32π2

[F1 − ψ(2) + log m2

4πµ2

]) (5.119)

This expression has a pole in Minkowski space, and we can define the renormalizationcondition by requiring that the inverse propagator at the physical mass is

p2 +m2phys (5.120)

This choice determines uniquely the F1 term. However we will discuss more in detailthe renormalization conditions in the next Chapter when we will speak about therenormalization group. Notice that at one loop there is no wave function renormal-ization, but this comes about at two loops, as we shall see later.

Consider now the four-point connected Green’s function. Its diagrammatic ex-pansion at one loop is given in Fig. 5.7.3 (again we have included only 1PI diagrams).

From the results of the previous Section and introducing the Mandelstam invari-ants

s = (p1 + p2)2, t = (p1 + p3)

2, u = (p1 + p4)2 (5.121)

we get

G(4)c (p1, p2, p3, p4) = (2π)2ωδ(2ω)(p1 + p2 + p3 + p4)

4∏k=1

1

p2k +m2

× (−µ4−2ω)λ

[1 − 3λ

32π2

(2

ε+ ψ(1) + 2 − log

m2

4πµ2− 1

3A(s, t, u)

)](5.122)

147

.

= xx

xxxx

xxxx

xxxx

xxxx

xxxx

xx

xxxx xx

xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

+ xx x

xxxxxx

xxxx

xx

xxxx

xx

xxxx

xxxx

xxxx

xx

+

+ +xxxx

xxxx

xx

xxxx

xxxx

xxxx

Fig. 5.7.3 - The one-loop expansion for the four-point Green’s function.

with

A(s, t, u) =∑z=s,t,u

√1 +

4m2

zlog

√1 + 4m2

z+ 1√

1 + 4m2

z− 1

(5.123)

The divergent part can be disposed off by the counterterm

δλ

4!µ4−2ωϕ4 =

1

4!µ4−2ω 3λ2

32π2

[2

ε+G1

(ω,m2

µ2

)]ϕ4 (5.124)

with G1 an arbitrary dimensionless function. This counterrtem produces an addi-tional contribution with a Feynman rule represented in Fig. 5.7.4.

.

_ 22_

µω

δλ( )=

Fig. 5.7.4 - The Feynman’s rule for the couterterm δλ.

By adding the counterterm we get

G(4)c (p1, p2, p3, p4) = (2π)2ωδ(2ω)(p1 + p2 + p3 + p4)

4∏k=1

1

p2k +m2

(−µ2−4ω)λ

×[1 − 3λ

32π2

(−G1 + ψ(1) + 2 − log

m2

4πµ2− 1

3A(s, t, u)

)](5.125)

A possible renormalization condition is to fix the scattering amplitude at some valueof the invariants (for instance, s = 4m2

phys, t = u = 0) to be equal to the lowestorder value

G(4)c (p1, p2, p3, p4) = (2π)2ωδ(2ω)(p1 + p2 + p3 + p4)

4∏k=1

1

p2k +m2

× (−µ2−4ω)λ (5.126)

148

In any case G1 is completely fixed by this condition.Without entering into the calculation we want now show how the renormalization

program works at two loops. We will examine only the two-point function. Itscomplete expansion up to two loops, including also the counterterm contributions isgiven in Fig. 5.7.5. By omitting the external propagators and the delta-function of

.

xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

xxxx

xxxx = xx

xxxxxx

xxxx

xxxx

xxxx+ +

+ xx xxxx xx + xxxx

xxxx

xx

xxxx

xx xx+ +

xx xx+ xxx x+

a) b)

c) d)

Fig. 5.7.5 - The two-loop expansion for the two-point Green’s function.

conservation of the four-momentum, the various two-loop contributions are (noticethat a counterterm inside a loop gives a second order contribution as a pure two-loops diagram):

• Contribution from the diagram a) of Fig 5.7.5

− λ2

6(16π2)2

[1

2εp2 +

6m2

ε2+

6m2

ε

(3

2+ ψ(1) − log

m2

4πµ2

)](5.127)

• Contribution from the diagram b) of Fig 5.7.5

− λ2

4(16π2)2

[4

ε2+

2

ε

(ψ(2) + ψ(1) − 2 log

m2

4πµ2

)](5.128)

• Contribution from the diagram c) of Fig 5.7.5

3λ2

4(16π2)2m2

[4

ε2+

2

ε

(ψ(2) +G1 − log

m2

4πµ2

)](5.129)

• Contribution from the diagram d) of Fig 5.7.5

λ2

4(16π2)2m2

[4

ε2+

2

ε

(ψ(1) + F1 − log

m2

4πµ2

)](5.130)

149

Adding together these terms we find

− λ2

12(16π2)2

p2

ε+m2 2λ2

(16π2)2

1

ε2+m2 λ2

2(16π2)2

1

ε[F1 + 3G1 − 1] (5.131)

where we have used ψ(2) − ψ(1) = 1. This expression is still divergent but at theorder λ2. Therefore we can make finite the diagram by modifying at the secondorder the δm2 counterterm

1

2

(δm2

1 + δm22

)ϕ2 (5.132)

with δm21 given in eq. (5.115) and

δm22 = m2 λ2

2(16π2)2

[4

ε2+

1

ε(F1 + 3G1 − 1) + F2

(ω,m2

µ2

)](5.133)

and F2 a new dimensionless function. The further divergence in p2 can be eliminatedthrough the wave function renormalization counterterm, that is

1

2δZ∂µϕ∂

µϕ (5.134)

with the choice

δZ = − λ2

24(16π2)2

[2

ε+H2(ω,

m2

µ2)

](5.135)

with H2 dimensionless. All the functions F1, F2, G1 H2 are finite in the limitω → 2. Notice also that we are not introducing more and more freedom, becauseonly particular combinations of these functions appear in the counterterms, and areprecisely these combinations which are fixed by the renormalization conditions. Also,the function H2 is fixed by the renormalization condition on the inverse propagator,which are really two conditions: a) the position of the pole, b) the residuum at thepole.

.

Fig. 5.7.6 - Insertion of counterterms in multi-loop diagrams.

The previous procedure can be easily extended to higher and higher orders inthe expansion in the coupling constant. What makes the theory renormalizableis the fact that the new divergences that one encounters can all be disposed offby modifying the counterterms δm2 and λ. This depends from the fact that thedivergent terms are constant or quadratic in the external momentum, and thereforethey can be reproduced by operators as ϕ2, ∂µϕ∂

µϕ and ϕ4. The non trivial part of

150

the renormalization program is just the last one. In fact consider diagrams of thetype represented in Fig. 5.7.6, where we have taken out all the possible externallines. These diagrams contain 1/ε terms coming from the counterterms times log p2

terms from the loop integration. In the previous two-loop calculation we had nosuch terms but we got their strict relatives log(m2/4πµ2). If present the log p2

terms would be a real disaster since they are non-local in configuration space andas such they cannot be absorbed by local operators. However if one makes all therequired subtractions, that is one subtracts correctly also the diagrams of the typein Fig. 5.7.6, it is possible to show that the log p2 terms cancel out. For instance,in the expression for δm2

2 there are not such terms. In fact they cancel in adding upthe four two-loop contributions.

151

Chapter 6

The renormalization group

6.1 The renormalization group equations

For the following considerations it will be convenient to introduce the one-particleirreducible truncated n-point functions Γ(n)(p1, p2, · · · , pn) defined by removing thepropagators on the external legs and omitting the delta-function expressing theconservation of the four-momentum

(2π)2ωδ(2ω)(

n∑i=1

pi)Γ(n)(p1, p2, · · · , pn) = G

(n)1PI(p1, p2, · · · , pn)

n∏i=1

DF (pi)−1 (6.1)

where

DF (p) =

∫d2ωxeipx〈0|T (ϕ(x)ϕ(0))|0〉 (6.2)

is the exact euclidean propagator. We have two ways of evaluating these functions,we can start from the original lagrangian and evaluate them in terms of the barecouplings constants (in the case of λϕ4, λB and mB). On the other hand we canexpress the bare parameters in terms of the renormalized ones, and, at the sametime, renormalizing the operators through the wave function renormalization (ϕB =

Z1/2ϕ ϕ), obtaining the relation

Γ(n)B (p1, p2, · · · , pn;λB, mB, ω) = Z−n/2

ϕ Γ(n)(p1, p2, · · · , pn;λ,m, µ, ω) (6.3)

The factor Z−n/2ϕ originates from a factor Z

n/2ϕ coming from the renormalization

factors in G(n)c and from n factors Z−1

ϕ from the inverse propagators. Of course, in a

renormalizable theory the Γ(n)’s are finite by construction. In the previous relationwe must regard the renormalized parameters as function of the bare ones. In theλϕ4 case, this means λ and m depending on λB and mB. However the right handside of the previous equation does not depend on the choice of the parameter µ.This implies that also the functions Γ(n) do not depend on µ except for the fieldrescaling factor Z

−n/2ϕ . These considerations can be condensed into a differential

152

equation obtained by requiring that the total derivative with respect to µ of theright hand side of eq. (6.3) vanishes[

µ∂

∂µ+ µ

∂λ

∂µ

∂

∂λ+ µ

∂m

∂µ

∂

∂m− n

2µ∂ logZϕ∂µ

]Γ(n)(p1, p2, · · · , pn;λ,m, µ, ω) = 0

(6.4)The content of this relation is that a change of the scale µ can be compensated by aconvenient change of the couplings and of the field normalization. Notice that sinceΓ(n) is finite in the limit ω → 2, it follows that also the derivative of logZϕ is finite.It is convenient to define the following dimensionless coefficients

µ∂λ

∂µ= β

(λ,m

µ, ω

)1

2µ∂ logZϕ∂µ

= γd

(λ,m

µ, ω

)µ

m

∂m

∂µ= γm

(λ,m

µ, ω

)(6.5)

obtaining[µ∂

∂µ+ β

∂

∂λ+ γmm

∂

∂m− nγd

]Γ(n)(p1, p2, · · · , pn;λ,m, µ, ω) = 0 (6.6)

We can try to eliminate from this equation the term ∂/∂µ. To this end let us considerthe dimensions of Γ(n). In dimensions d = 2ω the delta function has dimensions −2ωand dim[ϕ] = ω − 1, therefore

G(n)c (p1, · · · , pn) =

∫ n∏i=1

d2ωxiei(p1x1 + · · ·+ pnxn)〈0|T (ϕ(x1) · · ·ϕ(xn)|0〉conn.

(6.7)has dimensions n(ω−1)−2nω = −n(ω+1). The propagator DF (p) has dimensions2(ω − 1) − 2ω = −2, and we get

dim[Γ(n)

]− 2ω = −n(ω + 1) + 2n→ dim[Γ(n)

]= n+ ω(2 − n) (6.8)

and putting ε = 4 − 2ω

dim[Γ(n)

]= 4 − n+

ε

2(n− 2) (6.9)

Therefore Γ(n) must be a homogeneous function of degree 4 − n+ ε(n− 2)/2 in thedimensionful parameters pi, m e µ. Introducing a common scale s for the momentawe get (from the Euler theorem)[

µ∂

∂µ+ s

∂

∂s+m

∂

∂m− (4 − n +

ε

2(n− 2))

]Γ(n)(spi;λ,m, µ, ε) = 0 (6.10)

153

Since everything is finite for ε→ 0 we can take the limit and eliminate ∂/∂µ betweenthis equation and eq. (6.6)[

−s ∂∂s

+ β∂

∂λ+ (γm − 1)m

∂

∂m− nγd + 4 − n

]Γ(n)(spi;λ,m, µ, ε) = 0 (6.11)

This equation tells us how the Γ(n) scale with the momenta, and allows us to evaluatethem at momenta spi once we know them at momenta pi. To this end one needsto know the coefficients β, γm and γd. For this purpose one needs to specify therenormalization conditions since these coefficients depend on the arbitrary functionsF1, G1, · · ·.

In order to use the renormalization group equations we need to know more aboutthe structure of the functions β, γd and γm. We have seen that the counterterms aresingular expressions in ε,therefore it will be possible to express them as a Laurentseries in the renormalized parameters

λB = µε

[a0 +

∞∑k=1

akεk

](6.12)

m2B = m2

[b0 +

∞∑k=1

bkεk

](6.13)

Zϕ =

[c0 +

∞∑k=1

ckεk

](6.14)

with the coefficients functions of the dimensionless parameters λ and m/µ. Noticethat a0, b0 and c0 may depend on ε but only in an analytical way. Recalling fromthe previous Section the expressions for the counterterms at the second order in λ

δλ =3λ2

32π2

[2

ε+G1

](6.15)

δm2 =λ

32π2m2

[2

ε+ F1

]+

λ2

2(16π2)2m2

[4

ε2+

1

ε(F1 + 3G1 − 1) + F2

](6.16)

δZ = − λ2

24(16π2)2

[2

ε+H2

](6.17)

we get for the bare parameters

λB = µε[λ+

3λ2

32π2

(2

ε+G1

)]+ O(λ3) (6.18)

154

m2B = m2

[1 +

λ

32π2

(2

ε+ F1

)

+λ2

2(16π2)2

(4

ε2+

1

ε(F1 + 3G1 − 1) + F2

)

+λ2

24(16π2)2

(2

ε+H2

)]+ O(λ3) (6.19)

Zϕ = 1 − λ2

24(16π2)2

(2

ε+H2

)+ O(λ3) (6.20)

Therefore the coefficients are given by

a0 = λ+3λ2

32π2G1 + O(λ3) (6.21)

a1 =3λ2

16π2+ O(λ3) (6.22)

b0 = 1 +λ

32π2F1 +

λ2

2(16π2)2(F2 +

1

12H2) + O(λ3) (6.23)

b1 =λ

16π2+

λ2

2(16π2)2(F1 + 3G1 − 1) +

λ2

12(16π2)2+ O(λ3) (6.24)

b2 =2λ2

(16π2)2+ O(λ3) (6.25)

c0 = 1 − λ2

24(16π2)2H2 + O(λ3) (6.26)

c1 = − λ2

12(16π2)2+ O(λ3) (6.27)

Notice that the dependence on m/µ comes only from the arbitrary functions definingthe finite parts of the counterterms. This is easily understood since the divergentterms originate from the large momentum dependence where the masses can beneglected. This is an essential feature, since it is very difficult to solve the grouprenormalization equations when the coefficients β, γm e γd depend on both variablesλ andm/µ. On the other hand since these coefficients depend on the renormalizationconditions, it is possible to define a scheme where they do not depend on m/µ.Another option would be to try to solve the equations in the large momentum limitwhere the mass dependence can be neglected.

Callan and Szymanzik have considered a different version of the renormalizationgroup equations. They have considered the dependence of Γ(n) on the physicalmass. In that case one can show that β and γd depend only on λ. Furthermorein their equation the terms γm and µ/∂µ do not appear. In their place there is ainhomogeneous term in the large momentum limit.

155

6.2 Renormalization conditions

Let us consider again the renormalization conditions. The renormalization procedureintroduces in the theory a certain amount of arbitrariness in terms of the scale µ andof the arbitrary functions F1, F2 · · ·. Recall that the lagrangian has been decomposedas

L = Lp + Lct (6.28)

therefore, the finite part of Lct can be absorbed into a redefinition of the parametersappearing in L since both have the same operatorial structure. This is another wayof saying that the finite terms in Lct are fixed once we assign λ and m. In turn theseare fixed by relating them to some observable quantity as a scattering cross-section.Furthermore we have to fix a scale µ at which we define the parameters λ and m.The renormalization group equations then tell us how to change the couplings whenwe change µ, in order to leave invariant the physical quantities. As we have saidthe procedure for fixing m and λ is largely arbitrary and we will exhibit variouspossibilities. To do that it is convenient to write explicitly the expression for thetwo- and four-point truncated amplitudes

Γ(2)(p) =

(1 − λ2

24(16π2)2H2

)p2

+ m2(1 +

λ

32π2

(F1 − ψ(2) + log

m2

4πµ2

)

+λ2

2(16π2)2F2 + · · ·

)(6.29)

Γ(4)(p1, p2, p3, p4) = −µελ(

1 − 3λ

32π2

(ψ(1) + 2 − log

m2

4πµ2− 1

3A(s, t, u) −G1

))(6.30)

In the two point function we have omitted the finite terms coming from the secondorder contributions for which we have given only the divergent part. let us nowdiscuss various possibilities of renormalization conditions (sometimes one speaks ofsubtraction conditions)

• A first possibility is to renormalize at zero momenta, that is requiring

Γ(2)(p)∣∣∣p2∼0

= p2 +m2A (6.31)

Γ(4)(p1, p2, p3, p4)∣∣∣pi=0

= −µελA (6.32)

where m2A and λA are quantities to be fixed by comparing some theoretical

cross-section with the experimental data. Then, comparing with eqs. (6.29)and (6.30) and using A(s, t, u) = 6 for pi = 0, we get

HA2 = 0, FA

1 = ψ(2) − logm2A

4πµ2, GA

1 = ψ(1) − logm2A

4πµ2(6.33)

156

Another possibility would be to fix the momenta at some physical value in theMinkowski space. For instance we could require Γ(−p2 = m2) = 0 where m isthe physical mass and then fix Γ(4) with all the momenta on shell.

• To choose the subtraction point at zero momenta is generally dangerous formassless particles (in this way one would introduce spurious infrared diver-gences). As a consequence a popular choice is to subtract at an arbitraryvalue of the momenta. For instance one requires

Γ(2)(p) = p2 +m2B, p2 = M2 (6.34)

Γ(4)(p1, p2, p3, p4) = −µελB, s = t = u = M2 (6.35)

Notice that the point chosen for the four-point function is a non-physical one.In this case we get

HB2 = 0, FB

1 = ψ(2)−logm2B

4πµ2, GB

1 = ψ(1)+2−logm2B

4πµ2−1

3A(M2,M2,M2)

(6.36)This way of renormalize has however the problem of leaving an explicit depen-dence on the mass in the coefficients of the renormalization group equations.

• The last way of renormalizing we will consider is due to Weinberg and ’tHooft and the idea is simply to put equal to zero all the arbitrary functionsF1, F2, · · · at each order in the expansion. In this way the coefficients ai, biand ci appearing in the Laurent series defining the bare couplings turn out tobe mass independent.

The Weinberg and ’t Hooft prescription allows to evaluate the renormalizationgroup equations coefficients in a very simple way. For instance, let us start with thebare coupling

λB = µε

[λ+

∞∑k=1

ak(λ)

εk

](6.37)

Differentiating with respect to µ and recalling that λB does not depend on thisvariable, we get

0 = ε

[λ+

∞∑k=1

ak(λ)

εk

]+ µ

∂λ

∂µ

[1 +

∞∑k=1

a′k(λ)

εk

](6.38)

where

a′k(λ) =dak(λ)

dλ(6.39)

Since β = µ∂λ/∂µ is analytic for ε → 0, for small ε we can write β = A + Bε,obtaining

0 = ε

[λ +

∞∑k=1

ak(λ)

εk

]+ (A+Bε)

[1 +

∞∑k=1

a′k(λ)

εk

](6.40)

157

and identifying the coefficients of the various powers in ε

B + λ = 0 (6.41)

a1 + A+Ba′1 = 0 (6.42)

ak+1 + Aa′k +Ba′k+1 = 0 (6.43)

from whichB = −λ (6.44)

A = −(

1 − λd

dλ

)a1 (6.45)

We then get

β(λ) = −(

1 − λd

dλ

)a1 − λε (6.46)

and for ε→ 0

β(λ) = −(

1 − λd

dλ

)a1 (6.47)

and (1 − λ

d

dλ

)ak+1 =

dakdλ

(1 − λ

d

dλ

)a1 (6.48)

From the expression of the previous Section for a1 at the second order in λ, we get

µ∂λ

∂µ= β(λ) =

3λ2

16π2(6.49)

This equation can be integrated in order to know how we have to change λ when wechange the value of the scale µ. The function λ(µ) is known as the running couplingconstant. Its knowledge, together with the knowledge of m(µ) allows us to integrateeasily the renormalization group equations. In fact, let us define

t = log s (6.50)

and let us consider the following partial differential equation (we shall see that therenormalization group equations can be brought back to this form)

∂

∂tF (xi, t) = βi(xi)

∂

∂xiF (xi, t) (6.51)

The general solution to this equation is obtained by using the method of the ”char-acteristics”. That is one considers the integral curves (the characteristic curves)defined by the ordinary differential equations

dxi(t)

dt= βi(xi(t)) (6.52)

158

with given boundary conditions xi(0) = xi. Then, the general solution is

F (xi, t) = ψ(xi(t)) (6.53)

whereψ(xi) = F (xi, 0) (6.54)

is an arbitrary function to be determined by the condition at t = 0 required for F .In fact we can check that ψ satisfies the differential equation

∂

∂tψ(xi(t)) =

dxi(t)

dt

∂ψ

∂xi= βi

∂ψ

∂xi(6.55)

In our case the equation contains a non-derivative term

∂

∂tG(xi, t) = βi(xi)

∂

∂xiG(xi, t) + γ(xi)G(xi, t) (6.56)

with γ(xi) a given function. By putting

G(xi, t) = e

∫ t

0

dt′γ(xi(t′))F (xi, t) (6.57)

we see that F (xi, t) satisfies eq. (6.51) and therefore the general solution will be

G(xi, t) = e

∫ t

0

dt′γ(xi(t′))ψ(xi(t)) (6.58)

with xi(t) satisfying eq. (6.52). For the Γ(n) we get (t = log s)

Γ(n)(spi;m,λ, µ) = Γ(n)(pi;m(s), λ(s), µ)s4−ne−n

∫ s

1

γd(λ(s′))ds′

s′ (6.59)

with

s∂λ(s)

∂s= β(λ(s)) (6.60)

s∂m(s)

∂s= m(s)(γm(λ(s)) − 1) (6.61)

and λ(s = 1) = λ, m(s = 1) = m.This result tells us that when we perform a re-scaling on the momenta, the

amplitudes Γ(n) do not scale only with the trivial factor s4−n due to their dimensions,but that they undergo also a non trivial scaling, due to γd = 0, which is necessaryto compensate the variations of λ and m with the scale.

159

6.3 Application to QED

In this Section we will apply the previous results to QED. In particular, we want tocheck that the running coupling obtained via the renormalization group equationsis the same that we have defined in Section 2.2. First of all let us start with theQED lagrangian in d = 2ω dimensions (neglecting the gauge fixing terms)

S2ω =

∫d2ωx

[ψ(i∂ −m)ψ − eψAψ − 1

4FµνF

µν

](6.62)

Since

dim[ψAψ] = ω − 1 + 2

(ω − 1

2

)= 3ω − 2 (6.63)

it followsdim[e] = 2 − ω (6.64)

Therefore we defineenew = eold(µ)ω−2 = eoldµ

−ε/2 (6.65)

As before we write the lagrangian as Lp+Lct, with (we ignore here the gauge fixingterm)

Lp = ψ(i∂ −m)ψ − eµε/2ψAψ − 1

4FµνF

µν (6.66)

and

Lct = iδZ2ψ∂ψ − δmψψ − eδZ1µε/2ψAψ − δZ3

4FµνF

µν (6.67)

giving

L = i(1 + δZ2)ψ∂ψ − (m+ δm)ψψ − e(1 + δZ1)µε/2ψAψ − 1 + δZ3

4FµνF

µν (6.68)

After re-scaling of the fields one gets back the original lagrangian via, in particular,the identification

eB = µε/2e(1 + δZ1)

(1 + δZ2)(1 + δZ3/2)= µε/2e(1 − δZ3/2) (6.69)

where we have used δZ1 = δZ2. From Section 2.7 we recall that

δZ3 = C = − e2

6π2ε(6.70)

and therefore

eB = µε/2e

(1 +

e2

12π2ε

)(6.71)

We proceed as in the previous Section by defining

eB = µε/2

(e+

∑k

akεk

)(6.72)

160

with

a1 =e3

12π2(6.73)

Requiring that eB is µ-independent, and in the limit ε→ 0 we get

β(e) = µ∂e

∂µ= −1

2

(1 − e

d

de

)a1 (6.74)

from which

β(e) =e3

12π2(6.75)

The differential equation defining the running coupling constant is

µ∂e(µ)

∂µ= β(e) =

e3

12π2(6.76)

Integrating this equation we obtain

1

e2(µ2)= − 1

12π2log µ2 +K (6.77)

where K is an integration constant which can be eliminated by evaluating e atanother momentum scale Q2. We get

1

α(Q2)− 1

α(µ2)= − 1

3πlog

Q2

µ2(6.78)

or

α(Q2) =α(µ2)

1 − α(µ2)3π

log Q2

µ2

(6.79)

which agrees with the result obtained in Section 2.2.

6.4 Properties of the renormalization group equa-

tions

Integrating eq. (6.49)

µdλ

dµ=

3λ2

16π2+ O(λ3) (6.80)

we get

− 1

λ(µ)=

3

16π2log

µ

µs− 1

λs(6.81)

where µs is a reference scale and λs = λ(µs). Therefore

λ(µ) = λs1

1 − 3λs

16π2 log µµs

(6.82)

161

From this expression we see that starting from µ = µs, λ increases with µ. Supposethat we start with a small λ in such a way that the perturbative expansion is sensible.However increasing µ we will need to add more and more terms to the expansiondue to the increasing of λ. Therefore the λϕ4 theory allows a perturbative expansiononly at small mass scales, or equivalently at large distances. In this theory we cangive a definite sense to the asymptotic states. The increasing of λ is related to thesign of the β(λ) function. For the λϕ4 theory and for QED the β function is positive,however if it would be negative the theory would become perturbative at large massscales or at small distances. In this case one could solve the renormalization groupequations at large momenta by using the perturbative expansion at a lower orderto evaluate the β and the other coefficients. However in this case the coupling isincreasing at large distances and this would be a problem for our approach based onth in- and out- states. As we shall see this is the situation in QCD. here the fieldsthat are useful to describe the dynamics at short distances (the quark fields) cannotdescribe the asymptotic states, which can be rather described by bound states ofquarks as mesons and baryons.

Going back to the eq. (6.82) we see that, assuming the validity of the expressionfor the β function up to large momenta, the coupling constant becomes infinite atthe value of µ given by

µ = µse

16π2

3λs (6.83)

which is a very big scale if λs is small. The pole in eq. (6.82) is called the Landaupole (see Section 2.2 in the case of QED). Of course there are no motivations whyeq. (6.82) should be valid for large values of λ.

Let us now discuss several possible scenarios starting from a theory where β = 0for λ = 0.

• If β(λ) > 0 for any λ, than the running coupling is always increasing, and itwill become infinite at some value of µ. If this happens for a finite µ we saythat we have a Landau pole at that scale.

• Suppose that β(λ) is positive at small λ and that it becomes negative goingthrough a zero at λ = λF (see Fig. 6.4.1)).

The point λF is called a fixed point since if we start with the initial conditionλ = λF than λ remains fixed at that value. To study the behaviour of λ aroundthe fixed point, let us expand the β(λ) around its zero

β(λ) ≈ (λ− λF )β ′(λF ) (6.84)

It follows

µdλ

dµ= (λ− λF )β ′(λF ) + · · · (6.85)

162

.

F

β(λ)

λλ

Sλ SλSλ

Fig. 6.4.1 - The behaviour of β(λ) giving rise to an ultraviolet fixed point.

from whichdλ

λ− λF= β ′(λF )

dµ

µ(6.86)

Integrating this equation we get

λ− λFλs − λF

=

(µ

µs

)β′(λF )

(6.87)

Notice that the sign of β ′(λF ) plays here a very important role. In the presentcase the sign is negative since β(λ) > 0 for λ < λF and β(λ) < 0 for λ > λF .For large values of µ we have that λ → λF independently on the initial valueλs. For this reason λF is referred to as a ultraviolet (UV) fixed point (seeFig. 6.4.2)). The large scale (or small distance) behaviour of a field theory of

.

S

Fλ

λ

µ µ

Fig. 6.4.2 - The behaviour of λ close to a UV fixed point.

this type would be dictated by the value of λF . In particular if λF 1 andλs < λF we would be always in the perturbative regime. Otherwise, startingfrom λs > λF the theory would become perturbative at large scales (see Fig.6.4.1). About this point, notice that λ = 0 is a fixed point since β(0) = 0.However this is an infrared (IR) fixed point, since the coupling goes to zero for

163

µ → 0 independently on the initial value λs. We say also that the fixed pointis repulsive since λ goes away from the fixed point for increasing µ, whereas aUV fixed point is said to be attractive.

• Let us consider now the case of β(λ) < 0 for small values of λ and decreasingmonotonically. For instance, suppose β(λ) = −aλ2 with a > 0. Integratingthe equation for the running coupling we get

λ(µ) = λs1

1 + aλs log µµs

(6.88)

In this case λ is a decreasing function of µ and goes to zero for µ→ ∞ (that isλ = 0 is a UV fixed point). This property is known as ”asymptoyic freedom”and it holds for non-abelian gauge theories that we will describe later on inthe course. these are the only 4-dimensional theories to share this property.In higher dimensions other theories enjoy this property. For instance in d = 6the theory λϕ3 is asymptotically free. Notice that also this time the runningcoupling has a pole, but at a smaller scale than µs

µ

µs= e

− 1

aλs (6.89)

Therefore the coupling increases at large distances.

• Finally let us consider the case of β(λ) < 0 for small values of λ, becomes zeroat λF and then positive. Then β ′(λF ) > 0. By expanding β(λ) around λF weget (by the same analysis we did before)

λ− λFλs − λF

=

(µ

µs

)β′(λF )

(6.90)

and we see that λ → λF µ → 0 in a way independent on λs, whereas it goesaway from λs for increasing µ. Therefore λF is an IR fixed point (see Fig.6.4.3)).

Let us now derive the perturbative expressions for γm and γd using the samemethod followed for β(λ). e possibile ricavare. We differentiate the eq. (6.13) withrespect to µ (notice that in this renormalization scheme b0 = 1)

0 = 2m∂m

∂µ

[1 +

∞∑k=1

bkεk

]+m2∂λ

∂µ

∞∑k=1

b′kεk

(6.91)

from which

0 = 2γm

[1 +

∞∑k=1

bkεk

]+ β(λ)

∞∑k=1

b′kεk

(6.92)

164

.

S

λ

µµ

Fλ

Fig. 6.4.3 - The behaviour of λ close to an IR fixed point.

where we have made use of the definition γm = (µ/m)(∂m/∂µ). Using eq. (6.46)we find

2γm − λdb1dλ

= 0

2γmbk − dbkdλ

(1 − λ

d

dλ

)a1 − λ

dbk+1

dλ(6.93)

or

γm =1

2λdb1dλ

(6.94)

and

λdb1dλ

bk − dbkdλ

(1 − λ

d

dλ

)a1 − λ

dbk+1

dλ= 0 (6.95)

Using the eq. (6.24) for b1 we get for γm

γm(λ) =λ

32π2− 5

12

(λ

16π2

)2

+ O(λ3) (6.96)

Let us now consider Zϕ (recall that although this quantity is infinite in the limitε→ 0, γd is finite). By differentiating eq. (6.14) we find (c0 = 1)

Zϕγd =1

2µ∂Zϕ∂µ

=1

2µdλ

dµ

∞∑k=1

c′kεk

=1

2β(λ)

∞∑k=1

c′kεk

(6.97)

or (1 +

∞∑k=1

ckεk

)γd =

1

2(A+Bε)

∞∑k=1

c′kεk

(6.98)

Using again the expression for β(λ) (see eq. (6.46)

γd = −1

2λdc1dλ

(6.99)

165

λdck+1

dλ= λck

dc1dλ

− dckdλ

(1 − λ

d

dλ

)a1 (6.100)

At the lowest order, using eq. (6.27) we get

γd =1

12

(λ

16π2

)2

+ O(λ3) (6.101)

To conclude this Section, let us study the behaviour of Γ(n) when the theory hasa UV fixed point at λ = λF . Since for large scales λ→ λF we may suppose that γmand γd go to γm(λF ) e γd(λF ) respectively. Then, integrating the following equation(see eq. (6.61))

sdm

ds= m(γm(λF ) − 1) (6.102)

we havem(s) = m(1)sγm(λF )−1 (6.103)

Analogously ∫ s

1

ds′

s′γd(λF ) = γd(λF ) log s (6.104)

Then, for large values of s (see eq. (6.59)

Γ(n)(spi;m,λ, µ) → s(4−n−nγd(λF ))Γ(n)(pi;msγm(λF )−1, λF , µ) (6.105)

Therefore, if 1 − γm(λF ) > 0 we can safely neglect the mass on the right hand sideof this equation. From this equation we understand also the reason why γd is calledthe anomalous dimension.

In non abelian gauge theories there is a UV fixed point at λ = 0 where γm = 0and therefore one can neglect the masses at large momenta.

To justify why we have assumed that in the evaluation of γm and γd the onlyimportant contribution comes from λ = λF we notice that the integral∫ s

1

ds′

s′γm(λ(s′)) (6.106)

can be rewritten (using ds/s = dλ/β)

∫ λ(s)

λ

dλ′γm(λ′)β(λ′)

(6.107)

Therefore, barring the case in which γm and β have simultaneous zeroes, the integralis dominated by the points where there is a zero of the β function.

166

6.5 The coefficients of the renormalization group

equations and the renormalization conditions

We have already noticed that the coefficients of the renormalization group equationsdepend on the subtraction procedure. In particular, using the first two renormal-ization conditions considered in Section 6.2 a mass dependence of the coefficients isintroduced. To study better this problem, let us consider, for generic renormaliza-tion conditions, the expression (6.12) for the bare coupling

λB = µε

(a0 +

∞∑k=1

akεk

)(6.108)

Differentiating with respect to µ we find

ε

(a0 +

∞∑k=1

akεk

)+ µ

∂λ

∂µ

(a′0 +

∞∑k=1

a′kεk

)+ µ

∂(m/µ)

∂µ

(a0 +

∞∑k=1

akεk

)= 0 (6.109)

where the dot stands for the derivative with respect to m/µ. Notice also that (seeeq. (6.5)

µ∂

∂µ

(m

µ

)=m

µ(γm − 1) (6.110)

With a familiar procedure, we put β = A+Bε, obtaining

a0 +Ba′0 = 0 (6.111)

a1 + Aa′0 +Ba′1 +m

µ(γm − 1)a0 = 0 (6.112)

At the second order in λ

B = −λ(

1 +3λ

32π2G1

)(1 +

3λ

16π2G1

)−1

≈ −λ(

1 − 3λ

32π2G1

)(6.113)

and

Aa′0 = − 3λ2

16π2+ λ

(1 − 3λ

32π2G1

)3λ

8π2− m

µ(γm − 1)

3λ2

32π2G1 (6.114)

Since the right hand side is at least of the second order in λ, neglecting higher orderterms we get

A =3λ2

16π2+m

µ

3λ2

32π2G1 (6.115)

and for ε→ 0

β =3λ2

16π2+m

µ

3λ2

32π2G1 (6.116)

We see directly how the β function depends on the subtraction procedure. Similarexpression can be obtained for γm e γd. Therefore, when using subtraction procedure

167

such that the coefficients of the renormalization group equations depend explicitlyon the masses, it is convenient to look for solutions in a region of momenta wherethe masses can be neglected. Another possibility is to choose the arbitrary scale µsuch that m/µ can be neglected.

Of course, changing the renormalization procedure will induce a finite renormal-ization in the couplings. For instance we get expressions of the type

λ′ = f

(λ,m

µ

)= λ+ O(λ2) (6.117)

Therefore we obtain

β ′ = µ∂λ′

∂µ= µ

∂λ

∂µf ′ + µ

∂(m/µ)

∂µf = βf ′ +

m

µ(γm − 1)f (6.118)

Working in a region where we can neglect the mass we have

β ′(λ′) = β(λ)f ′(λ) (6.119)

and we see that a fixed point at λ = λF implies a fixed point in λ′. Thereforethe presence of a fixed point does not depend on the scheme used (at least in thisapproximation). It is possible to show that also the sign of the β function is scheme-independent.

168

Chapter 7

The path integral for Fermi fields

7.1 Fermionic oscillators and Grassmann algebras

The Feynman’s formulation of the path integral can be extended to field theoriesfor fermions. However, since the path integral is strictly related to the classicaldescription of the theory, it could be nice to have such a ”classical” description forfermionic systems. In fact it turns out that it is possible to have a ”classical” (calledpseudoclassical) description in a way such that its quantization leads naturally toanticommutation relations. In this theory one introduces anticommuting variablesalready at the classical level. This type of mathematical objects form the so calledGrassmann algebras. Before proceeding further let us introduce some concepts aboutthese algebras which will turn out useful in the following.

First of all we recall that an algebra V is nothing but a vector space where itis defined a bilinear mapping V × V → V (the algebra product). A Grassmannalgebra Gn is a vector space of dimensions 2n and it can be described in terms of ngenerators θi (i = 1, · · · , n) satisfying the relations

θiθj + θjθi ≡ [θi, θj ]+ = 0, i, j = 1, · · · , n (7.1)

Then, the elements constructed by all the possible products among the generatorsform a basis of Gn:

1, θi, θiθj , θiθjθk, · · · , θ1θ2 · · · θn (7.2)

Notice that there are no other possible elements since

θ2i = 0 (7.3)

from eq. (7.1). The subset of elements of Gn generated by the product of an even

number of generators is called the even part of the algebra, G(0)n , whereas the subset

generated by the product of an odd number of generators is the odd part, G(1)n . It

followsGn = G(0)

n ⊕ G(1)n (7.4)

169

A decomposition of this kind is called a grading of the vector space, if it is possibleto assign a grade to each element of the subspaces. In this case the grading is asfollows

1. If x ∈ G(0)n , deg(x) = +1.

2. If x ∈ G(1)n , deg(x) = −1.

Notice that this grading is also consistent with the algebra product, since if x, y arehomogeneous elements of Gn (that is to say even or odd), then

deg(xy) = deg(x)deg(y) (7.5)

Since we have no time to enter into the details of the classical mechanics ofthe Grassmann variables, we will formulate the path integral starting directly fromquantum mechanics. Let us recall that in the Feynman’s formulation one introducesstates which are eigenstates of a complete set of observables describing the configu-ration space of the system. In the case of field theories one takes eigenstates of thefield operators in the Heisenberg representation

φ(x, t)|ϕ(x), t〉 = ϕ(x)|ϕ(x), t〉 (7.6)

this is possible since the fields commute at equal times

[φ(x, t), φ(y, t)]− = 0 (7.7)

and as a consequence it is possible to diagonalize simultaneously at fixed time theset of operators φ(x, t). For Fermi fields such a possibility does not arise since theyanticommute at equal times. Therefore it is not possible to define simultaneouseigenstates of Fermi fields. To gain a better understanding of this point let us stayfor the moment with free fields, and let us recall that this is nothing but a set ofharmonic oscillators. This is true also for Fermi fields, except for the fact that theseoscillators are of Fermi type, that is they are described by creation and annihilationoperators satisfying the anticommutation relations

[ai, aj]+ =[a†i , a

†j

]+

= 0,[ai, a

†j

]+

= δij (7.8)

For greater simplicity take the case of a single Fermi oscillator, and let us try todiagonalize the corresponding annihilation operator

a|θ〉 = θ|θ〉 (7.9)

Then, from a2 = 0, as it foillows from eqs. (7.8), we must have θ2 = 0. Thereforethe eigenvalue θ cannot be an ordinary number. However it can be thought as anodd element of a Grassmann algebra. If we have n Fermi oscillators and we require

ai|θ1, · · · , θn〉 = θi|θ1, · · · , θn〉 (7.10)

170

we getai (aj |θ1, · · · , θn〉) = θj (ai|θ1, · · · , θn〉) = θjθi|θ1, · · · , θn〉 (7.11)

But aiaj = −ajai and therefore

[θi, θj ]+ = 0 (7.12)

We see that the equations (7.10) are meaningful if we identify the eigenvalues θiwith the generators of a Grassmann algebra Gn. In this case the equation (7.9) canbe solved easily

|θ〉 = eθa† |0〉 (7.13)

where |0〉 is the ground state of the oscillator. In fact

a|θ〉 = a(1 + θa†)|0〉 = θaa†|0〉 = θ(1 − a†a)|0〉= θ|0〉 = θ(1 + θa†|0〉 = θ|θ〉 (7.14)

where we have used

eθa†

= 1 + θa† (7.15)

since θ2 = a2 = 0. We have also

a†|θ〉 = a†(1 + θa†)|0〉 = a†|0〉 =∂

∂θ(1 + θa†)|0〉 =

∂

∂θ|θ〉 (7.16)

Here we have used the left-derivative on the Gn defined by

∂

∂θiθi1θi2 · · · θik = δi1,iθi2θi3 · · · θik − δi2,iθi1θi3 · · · θik

+ · · ·+ (−1)k−1δik,iθi1θi3 · · · θik−1(7.17)

Suppose we want to define a scalar product in the vector space of the eigenstatesof a (notice that these are vector spaces with coefficients in a Grassmann algebra.mathematically such an object is called a modulus over the algebra). Then it isconvenient to extend the algebra G1 generated by θ to a comples Grassmann algebra,G2, generated by θ and θ∗. The operation ”” is defined as the complex conjugationon the ordinary numbers and it is an involution (automorphism of the algebra withsquare one) with the property

(θi1θi2 · · · θik)∗ = θ∗ik · · · θ∗i2θ∗i1 (7.18)

In the algebra G2 we can define normalized eigenstates

|θ〉 = eθa† |0〉N(θ, θ∗) (7.19)

where N(θ, θ∗) ∈ G(0)2 and the phase is fixed by the convention

N∗(θ, θ∗) = N(θ, θ∗) (7.20)

171

The corresponding bra will be

〈θ| = N(θ, θ∗)〈0|eθ∗a (7.21)

the normalization factor N(θ, θ∗) is determined by the condition

1 = 〈θ|θ〉 = N2(θ, θ∗)〈0|eθ∗aeθa† |0〉= N2(θ, θ∗)〈0|(1 + θ∗a)(1 + θa†)|0〉= N2(θ, θ∗)〈0|(1 + θ∗θaa†)|0〉= N2(θ, θ∗)(1 + θ∗θ) = N2(θ, θ∗)e−θθ∗ (7.22)

It followsN(θ, θ∗) = eθθ

∗/2 (7.23)

Let us recall some simple properties of a Fermi oscillator. the occupation numberoperator has eigenvalues 1 and 0. In fact

(a†a)2 = a†aa†a = a†(1 − a†a)a = a†a (7.24)

from whicha†a(a†a− 1) = 0 (7.25)

Starting from the state with zero eigenvalue for a†a, we can build up only theeigenstate with eigenvalue 1, a†|0〉, since a†2|0〉 = 0. Therefore the space of theeigenstates of a†a is a two-dimensional space. By defining the ground state as

|0〉 =

[01

](7.26)

one sees easily that

a =

[0 01 0

], a† =

[0 10 0

](7.27)

(one writes down a as the most general 2 × 2 matrix and then requires a|0〉 = 0,a2 = 0 and

[a, a†

]+

= 1). The state with occupation number equal to 1 is

|1〉 =

[10

](7.28)

In this basis the state |θ〉 is given by

|θ〉 = eθθ∗/2eθa

† |0〉 = eθθ∗/2(1 + θa†)|0〉

= eθθ∗/2 (|0〉 + θ|1〉) = eθθ

∗/2[θ1

](7.29)

whereas〈θ| = eθθ

∗/2 [ θ∗ 1 ] (7.30)

172

We see that in order to introduce eigenstates of the operator a has forced us togeneralize the ordinary notion of vector space on the real or on the complex numbers,to a vector space on a Grassmann algebra (a modulus). In this space the linearcombinations of the vectors are taken with coefficients in G2 (or more generally inG2n).

Let us now come to the definition of the integral over a Grassmann algebra.To this end we recall that the crucial element in order to derive the path integralformalism from quantum mechanics is the completeness relation for the eigenstatesof the position operator ∫

dq|q〉〈q| = 1 (7.31)

What we are trying to do here is to think to the operators a as generalized posi-tion operators, and therefore we would like its eigenstates to satisfy a completenessrelation of the previous type. Therefore we will ”define” the integration over theGrassmann algebra G2 by requiring that the states defined in eqs. (7.29) and (7.30)satisfy ∫

dθ∗dθ|θ〉〈θ| =

(1 00 1

)≡ 12 (7.32)

where the identity on the right hand side must be the identity on the two-dimensionalvector space. Using the explicit representation for the states∫

dθ∗dθ|θ〉〈θ| =

∫dθ∗dθeθθ

∗(θ1

)( θ∗ 1 )

=

∫dθ∗dθeθθ

∗(θθ∗ θθ∗ 1

)

=

∫dθ∗dθ

(θθ∗ θθ∗ 1 + θθ∗

)=

(1 00 1

)(7.33)

Therefore the integration must satisfy∫dθ∗dθ · θθ∗ = 1 (7.34)

∫dθ∗dθ · θ =

∫dθ∗dθ · θ∗ =

∫dθ∗dθ · 1 = 0 (7.35)

These relations determine the integration in a unique way, since they fix the valueof the integral over all the basis elements of G2. The integral over a generic elementof the algebra is then defined by linearity. the previous integration rules can berewritten as integration rules over the algebra G1 requiring

[dθ, dθ∗]+ = [θ, dθ∗]+ = [θ, dθ]+ = [θ∗, dθ]+ = [θ∗, dθ∗]+ = 0 (7.36)

Then, we can reproduce the rules (7.34) and (7.35) by requiring∫dθ · θ = 1,

∫dθ · 1 = 0 (7.37)

173

For n Fermi oscillators we get

|θ1, · · · , θn〉 = e

n∑i=1

θiθ∗i /2

e

n∑i=1

θia†i

|0〉 (7.38)

with ∫ (n∏i=1

dθ∗i dθi

)|θ1, · · · , θn〉〈θ1, · · · , θn| = 1 (7.39)

and integration rules ∫dθiθj = δij (7.40)

Assuming also[dθi, θj ]+ = [dθi, dθj]+ = 0 (7.41)

Notice that the measure dθ is translationally invariant. In fact if f(θ) = a+ bθ and

α ∈ G(1)1 ∫

dθf(θ) =

∫dθf(θ + α) (7.42)

since’ ∫dθbα = 0 (7.43)

7.2 Integration over Grassmann variables

Let us consider the Grassmann algebra Gn. Its generic element can be expandedover the basis elements as

f(θ1, · · · , θn) = f0+∑i1

f1(i1)θi1 +∑i1,12

f2(11.i2)θi1θi2 +· · ·∑i1,···,in

fn(i1, · · · , in)θi1 · · · θin(7.44)

Of course the coefficients of the expansion are antisymmetric. If we integrate fover the algebra Gn only the last element of the expansion gives a non vanishingcontribution. Noticing also that for the antisymmetry we have∑

i1,···,infn(i1, · · · , in)θi1 · · · θin = n!fn(1, · · · , n)θ1 · · · θn (7.45)

it follows ∫dθn · · · dθ1f(θ1, · · · , θn) = n!fn(1, · · · , n) (7.46)

An interesting property of the integration rules for Grassmann variables is the inte-gration by parts formula. This can be seen either by the observation that a derivativedestroys a factor θi in each element of the basis and therefore∫

dθn · · · dθ1 ∂

∂θjf(θ1, · · · , θn) = 0 (7.47)

174

either by the property that the integral is invariant under translations (see the pre-vious Section and Section 4.7). Let us now consider the transformation propertiesof the measure under a change of variable. For simplicity consider a linear transfor-mation

θi =∑k

aikθ′k (7.48)

The eq. (7.40) must be required to hold in each basis, since it can be seen easilythat the previous transformation leaves invariant the multiplication rules. In fact,

θ′iθ′j = cikcjhθkθh = −cikcjhθhθk == −cjhcikθhθk = −θ′jθ′i (7.49)

where cij is the inverse of the matrix aij . Therefore we require∫dθiθj =

∫dθ′iθ

′j = δij (7.50)

By putting

dθi =∑k

dθ′kbki (7.51)

we obtain ∫dθiθj =

∫ ∑h,k

dθ′kbkiajhθ′h =

∑h,k

bkiajhδhk = (ab)ij = δij (7.52)

from whichbij = (a−1)ij (7.53)

The transformation properties of the measure are

dθn · · ·dθ1 = det|a−1|dθ′n · · · dθ′1 (7.54)

where we have used the anticommutation properties of dθi among themselves. Wesee also that

det|a| = det

∣∣∣∣ ∂θ∂θ′∣∣∣∣ (7.55)

and

dθ′n · · · dθ′1 = det−1

∣∣∣∣∂θ′∂θ

∣∣∣∣ dθn · · · dθ1 (7.56)

Therefore the rule for a linear change of variables is∫dθ′n · · · dθ′1f(θ′i) =

∫dθn · · · dθ1det−1

∣∣∣∣∂θ′∂θ

∣∣∣∣ f(θ′i(θi)) (7.57)

This rule should be compared with the one for commuting variables∫dx′1 · · · dx′nf(x′i) =

∫dx1 · · · dxndet

∣∣∣∣∂x′∂x

∣∣∣∣ f(x′i(xi)) (7.58)

175

We see that the Grassmann measure transforms according to the inverse jacobianrather than with the jacobian itself. As in the commuting case, the more importantintegral to be evaluated for the path integral approach is the gaussian one

I =

∫dθn · · · dθ1e

∑i,j

Aijθiθj/2

≡∫dθn · · · dθ1eθ

TAθ/2 (7.59)

The matrix A is supposed to be real antisymmetric

AT = −A (7.60)

As shown in the Appendix a real and antisymmetric matrix can be reduced to theform

As =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎣

0 λ1 0 0 · · ·−λ1 0 0 0 · · ·0 0 0 λ2 · · ·0 0 −λ2 0 · · ·· · · · · · ·· · · · · · ·· · · · · · ·

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎦

(7.61)

by means of an orthogonal transformation S. Then, let us do the following changeof variables

θi =∑j

(ST )ijθ′j (7.62)

we getθTAθ = (ST θ′)AST θ′ = θ′TSAST θ′ = θ′TAsθ′ (7.63)

and using det|S| = 1

I =

∫dθ′n · · · dθ′1eθ

′TAsθ′/2 (7.64)

The expression in the exponent is given by

1

2θ′TAsθ′ = λ1θ

′1θ

′2 + · · ·+ λn

2θ′n−1θ

′n, for n even

1

2θ′TAsθ′ = λ1θ

′1θ

′2 + · · ·+ λn−1

2θ′n−2θ

′n−1, for n odd (7.65)

From eq. (7.46) we get contribution to the integral only from the term in theexpansion of the exponential (θ′TAsθ′)

n2 for n even. In the case of n odd we get

I = 0 since θ′TAsθ′ does not contain θ′n. Therefore, for n even

I =

∫dθ′n · · ·dθ′1eθ

′TAsθ′/2 =

∫dθ′n · · · dθ′1

1

(n2)!

(1

2θ′TAsθ′)

n2

=

∫dθ′n · · ·dθ′1

1

(n2)!

(n

2)!λ1 · · ·λn

2θ′1 · · · θ′n

= λ1 · · ·λn2

=√

detA (7.66)

176

where we have used det|As| = det|A|. Then∫dθn · · ·dθ1eθ

TAθ/2 =√

detA (7.67)

This result is valid also for n odd since in this case det|A| = 0. A similar resultholds for commuting variables. In fact, let us define

J =

∫dx1 · · ·dxne−x

TAx/2 (7.68)

with A a real and symmetric matrix. A can be diagonalized by an orthogonaltransformation with the result

J =

∫dx′1 · · · dx′ne−x

′TAdx′/2 = (2π)n2

1√detA

(7.69)

Again, the result for fermionic variables is opposite to the result of the commutingcase. Another useful integral is

I(A;χ) =

∫dθn · · ·dθ1eθ

TAθ/2 + χT θ (7.70)

with χi Grassmann variables. To evaluate this integral let us put

θ = θ′ + A−1χ (7.71)

We get

1

2θTAθ + χT θ =

1

2(θ′T − χTA−1)A(θ′ + A−1χ) + χT (θ′ + A−1χ)

=1

2θ′TAθ′ +

1

2χTA−1χ (7.72)

where we have used the antisymmetry of A and χT θ′ = −θ′Tχ. Since the measureis invariant under translations we find

I(A;χ) =

∫dθ′n · · · dθ′1eθ

′TAθ′/2 + χTA−1χ/2 = eχTA−1χ/2√detA (7.73)

again to be compared with the result for commuting variables

I(A; J) =

∫dx1 · · · dxne−x

TAx/2 + JTx = eJTA−1J/2 (2π)

n2√

detA(7.74)

The previous equations can be generalized to complex variables∫dθ1dθ

∗1 · · · dθndθ∗neθ

†Aθ = det|A| (7.75)

177

and ∫dθ1dθ

∗1 · · · dθndθ∗neθ

†Aθ + χ†θ + θ†χ = e−χ†A−1χdet|A| (7.76)

The second of these equations follows from the first one as in the real case. we willprove the first one for the case n = 1. Starting from∫

dθ2dθ1eθTAθ/2 = A12 (7.77)

and putting

θ1 =1√2(η + η∗), θ2 = − i√

2(η − η∗) (7.78)

we get1

2θTAθ = −iη∗A12η (7.79)

and recalling that (remember that here one has to use the inverse of the jacobian)

dθ2dθ1 = idηdη∗ (7.80)

we get ∫dθ2dθ1e

θTAθ/2 = i

∫dηdη∗e−iη∗A12η (7.81)

and ∫dηdη∗eη

∗(−iA12)η = −iA12 (7.82)

7.3 The path integral for the fermionic theories

In this Section we will discuss the path integral formulation for a system describedby anticommuting variables. Considering a Fermi oscillator, one could start fromthe ordinary quantum mechanical description and then by using the completeness interms of Grassmann variable, it would be possible to derive the corresponding pathintegral formulation. We will give here only the result of this analysis. It turns outthat the amplitude among eigenstates of the Fermi annihilation operator is given by

〈θ′, t′|θ, t〉 =

∫ θ∗(t′)=θ′∗,t′

θ(t)=θ,t

D(θ, θ∗)eθ′∗θ(t′)/2 + θ∗(t)θ/2 + iS (7.83)

whereD(θ, θ∗) =

∏dθdθ∗ (7.84)

and

S =

∫ t′

t

[i

2(θ∗θ − θ∗θ) −H(θ, θ∗)

]dt (7.85)

178

where, for the Fermi oscillator H = θ∗θ. The expression is very similar to theone obtained in the commuting case if we realize that the previous formula is theanalogue in the phase space using complex coordinates of the type q± ip. It shouldbe noticed the particular boundary conditions for the θ variables originating fromthe fact that the equations of motion are of the first order. Correspondingly wecan assign only the coordinates at a particular time. In any case, since at the endwe are interested in field theory where only the generating functional is relevantand where the boundary conditions are arbitrary (for instance we can take the fieldsvanishing at t = ±∞), and in any case do not affect the dependence of the generatingfunctional on the external source. The extension to the Fermi field is done exactlyas we have done in the bosonic case, and correspondingly the generating functionalfor the free Dirac theory will be

Z[η, η] =

∫D(ψ, ψ)e

i

∫[ψ(i∂/ −m)ψ + ηψ + ψη]

(7.86)

where η and η are the external sources (of Grassmann character). Using the inte-gration formula (7.73) we get formally

Z[η, η] = Z[0]e−(iη)

−ii∂/ −m

(iη)= Z[0]e

−iη 1

i∂/ −mη

(7.87)

where the determinant of the Dirac operator has been included in the term at zerosource. In this expression we have to specify how to take the inverse operator. Weknow that this is nothing but the Dirac propagator defined by

(i∂/ −m)SF (x− y) = δ4(x− y) (7.88)

It follows

Z[η, η] = Z[0]e−i∫d4xd4yη(x)SF (x− y)η(y)

(7.89)

with

SF (x) = limε→0+

∫d4k

(2π)4

1

/k −m+ iεe−ikx = lim

ε→0+

∫d4k

(2π)4

/k +m

k2 −m2 + iεe−ikx

(7.90)The Green’s functions are obtained by differentiating the generating functional withrespect to the external sources (Grassmann differentiation). In particular, the twopoint function is given by

1

Z[0]

δZ

δη(x)δη(y)

∣∣∣η=η=0

=1

〈0|0〉〈0|T (ψ(x)ψ(y))|0〉 = iSF (x− y) (7.91)

The T -product in this expression is the one for the Fermi fields, as it follows fromthe derivation made in the bosonic case, taking into account that the classical fieldof this case are anticommuting.

We will not insist here further on this discussion, but all we have done in thebosonic case can be easily generalized for Fermi fields.

179

Chapter 8

The quantization of the gaugefields

8.1 QED as a gauge theory

Many field theories possess global symmetries. These are transformations leavinginvariant the action of the system and are characterized by a certain number ofparameters which are independent on the space-time point. As a prototype we canconsider the free Dirac lagrangian

L0 = ψ(x)[i∂/ −m]ψ(x) (8.1)

which is invariant under the global phase transformation

ψ(x) → ψ′(x) = e−iQαψ(x) (8.2)

If one has more than one field, Q is a diagonal matrix having as eigenvalues thecharges of the different fields measured in unit e. For instance, a term as ψ2ψ1φ,with φ a scalar field, is invariant by choosing Q(ψ1) = Q(φ) = 1, and Q(ψ2) = 2.This is a so called abelian symmetry since

e−iαQe−iβQ = e−i(α + β)Q = e−iβQe−iαQ (8.3)

It is also referred to as a U(1) symmetry. The physical meaning of this invariancelies in the possibility of assigning the phase to the fields in an arbitrary way, withoutchanging the observable quantities. This way of thinking is in some sort of contradic-tion with causality, since it requires to assign the phase of the fields simultaneouslyat all space-time points. It looks more physical to require the possibility of assigningthe phase in an arbitrary way at each space-time point. This invariance, formulatedby Weyl in 1929, was called gauge invariance. The free lagrangian (8.1) cannot begauge invariant due to the derivative coming from the kinetic term. The idea is

180

simply to generalize the derivative ∂µ to a so called covariant derivative Dµ havingthe property that Dµψ transforms as ψ, that is

Dµψ(x) → [Dµψ(x)]′ = e−iQα(x)Dµψ(x) (8.4)

In this case the termψDµψ (8.5)

will be invariant as the mass term under the local phase transformation. To con-struct the covariant derivative, we need to enlarge the field content of the theory,by introducing a vector field, the gauge field Aµ, in the following way

Dµ = ∂µ + ieQAµ (8.6)

The transformation law of Aµ is obtained from eq. (8.4)

[(∂µ + ieQAµ)ψ]′ = (∂µ + ieQA′µ)ψ

′(x)

= (∂µ + ieQA′µ)e

−iQα(x)ψ

= e−iQα(x)[∂µ + ieQ(A′

µ −1

e∂µα)

]ψ (8.7)

from which

A′µ = Aµ +

1

e∂µα (8.8)

The lagrangian density

Lψ = ψ[iD/ −m]ψ = ψ[iγµ(∂µ + ieQAµ) −m)ψ = L0 − eψQγµψAµ (8.9)

is then invariant under gauge transformations, or under the local group U(1). We seealso that by requiring local invariance we reproduce the electromagnetic interactionas obtained through the minimal substitution we discussed before.

In order to determine the kinetic term for the vector field Aµ we notice that eq.(8.4) implies that under a gauge transformation, the covariant derivative undergoesa unitary transformation

Dµ → Dµ′ = e−iQα(x)Dµe

iQα(x) (8.10)

Then, also the commutator of two covariant derivatives

[Dµ, Dν ] = [∂µ + ieQAµ, ∂ν + ieQAν ] = ieQFµν (8.11)

withFµν = ∂µAν − ∂νAµ (8.12)

transforms in the same way

Fµν → e−iQα(x)FµνeiQα(x) = Fµν (8.13)

181

The last equality follows from the commutativity of Fµν with the phase factor. Thecomplete lagrangian density is then

L = Lψ + LA = ψ[iγµ(∂µ + ieQAµ) −m]ψ − 1

4FµνF

µν (8.14)

The gauge principle has automatically generated an interaction between the gaugefield and the charged field. We notice also that gauge invariance prevents anymass term, 1

2M2AµAµ. Therefore, the photon field is massless. Also, since the

local invariance implies the global ones, by using the Noether’s theorem we find theconserved current as

jµ =∂L∂ψ,µ

δψ = ψγµ(Qα)ψ (8.15)

from which, eliminating the infinitesimal parameter α,

Jµ = ψγµQψ (8.16)

8.2 Non-abelian gauge theories

The approach of the previous section can be easily extended to local non-abeliansymmetries. We will consider the case of N Dirac fields. The free lagrangian

L0 =

N∑a=1

ψa(i∂/ −m)ψa (8.17)

is invariant under the global transformation

Ψ(x) → Ψ′(x) = AΨ(x) (8.18)

where A is a unitary N ×N matrix, and we have denoted by Ψ the column vectorwith components ψa. In a more general situation the actual symmetry could be asubgroup of U(N). For instance, when the masses are not all equal. So we willconsider here the gauging of a subgroup G of U(N). The fields ψa(x) will belong, ingeneral, to some reducible representation of G. Denoting by U the generic elementof G, we will write the corresponding matrix Uab acting upon the fields ψa as

U = e−iαATA, U ∈ G (8.19)

where TA denote the generators of the Lie algebra associated to G, Lie(G), (that isthe vector space spanned by the infinitesimal generators of the group) in the fermionrepresentation. The generators TA satisfy the algebra

[TA, TB] = ifABC TC (8.20)

182

where fABC are the structure constants of Lie(G). For instance, if G = SU(2), andwe take the fermions in the fundamental representation,

Ψ =

[ψ1

ψ2

](8.21)

we have

TA =σA

2, A = 1, 2, 3 (8.22)

where σA are the Pauli matrices. In the general case the TA’s are N ×N hermitianmatrices that we will choose normalized in such a way that

Tr(TATB) =1

2δAB (8.23)

To make local the transformation (8.19), means to promote the parameters αA tospace-time functions

αA → αA(x) (8.24)

Notice, that now the group does not need to be abelian, and therefore, in general

e−iαATAe−iβATA = e−iβATAe−iαATA (8.25)

Let us now proceed to the case of the local symmetry by defining again the conceptof covariant derivative

DµΨ(x) → [DµΨ(x)]′ = U(x)[Dµψ(x)] (8.26)

We will put againDµ = ∂µ + igBµ (8.27)

where Bµ is a N ×N matrix acting upon Ψ(x). In components

Dµab = δab∂

µ + ig(Bµ)ab (8.28)

The eq. (8.26) 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 (8.29)

thereforeU−1(x)igB′

µU(x) + U−1(x)∂µU(x) = igBµ (8.30)

and

B′µ(x) = U(x)Bµ(x)U

−1(x) +i

g(∂µU(x))U−1(x) (8.31)

183

For an infinitesimal transformation

U(x) ≈ 1 − iαA(x)TA (8.32)

we get

δBµ(x) = −iαA(x)[TA, Bµ(x)] +1

g(∂µαA(x))TA (8.33)

Since Bµ(x) acquires a term proportional to TA, the transformation law is consistentwith a Bµ linear in the generators of the Lie algebra, that is

(Bµ)ab ≡ AµA(TA)ab (8.34)

The transformation law for Aµ becomes

δAµC = fABC αAAµB +

1

g∂µαC (8.35)

The difference with respect to the abelian case is that the field undergoes also ahomogeneous transformation.

The kinetic term for the gauge fields is constructed as in the abelian case. Infact the quantity

[Dµ, Dν ]Ψ ≡ igFµνΨ (8.36)

in virtue of the eq. (8.26), transforms as Ψ under gauge transformations, that is

([Dµ, Dν ]Ψ)′ = igF ′µνΨ

′ = igF ′µνU(x)Ψ

= U(x)([Dµ, Dν ]Ψ) = U(x) (igFµν)Ψ (8.37)

This time the tensor Fµν is not invariant but transforms homogeneously, since itdoes not commute with the gauge transformation as in the abelian case

F ′µν = U(x)FµνU

−1(x) (8.38)

The invariant kinetic term will be assumed as

LA = −1

2Tr[FµνF

µν ] (8.39)

Let us now evaluate Fµν

igFµν = [Dµ, Dν ] = [∂µ + igBµ, ∂ν + igBν]

= ig(∂µBν − ∂νBµ) − g2[Bµ, Bν ] (8.40)

orFµν = (∂µBν − ∂νBµ) + ig[Bµ, Bν ] (8.41)

in componentsF µν = F µν

C TC (8.42)

184

withF µνC = ∂µAνC − ∂νAµC − gfABC AµAA

νB (8.43)

The main feature of the non-abelian gauge theories is the bilinear term in the pre-vious expression. Such a term comes because fABC = 0, expressing the fact that G isnot abelian. The kinetic term for the gauge field, expressed in components, is givenby

LA = −1

4

∑A

FµνAFµνA (8.44)

Therefore, whereas in the abelian case LA is a free lagrangian (it contains onlyquadratic terms), now it contains interaction terms cubic and quartic in the fields.The physical motivation lies in the fact that the gauge fields couple to everythingwhich transforms in a non trivial way under the gauge group. Therefore they couplealso to themselves (remember the homogeneous piece of transformation).

To derive the equations of motion for the gauge fields, let us consider the totalaction ∫

V

d4x[Ψ(i∂/ −m)Ψ − gΨγµB

µΨ]+ SA (8.45)

where

SA = −1

2

∫V

d4xTr(FµνFµν) (8.46)

and the variation of SA

δSA = −∫V

d4xTr(FµνδFµν) (8.47)

Using the definition (8.41) for the field strength we get

δFµν = ∂µδBν + ig(δBµ)Bν + igBµδ(Bν) − (µ↔ ν) (8.48)

from which

δSA = −2

∫V

d4xTr[F µν(∂µδBν + ig(δBµ)Bν + igBµ(δBν))] (8.49)

where we have taken into account the antisymmetry properties of Fµν . Integratingby parts we obtain

δSA = −2

∫V

d4xTr[−(∂µFµν)δBν − igBµF

µνδBν + igF µνBµδBν ]

= 2

∫V

d4xTr [(∂µFµν + ig[Bµ, F

µν ]) δBν ]

=

∫V

d4x (∂µFµν + ig[Bµ, F

µν ])A δAνA (8.50)

185

where we used the cyclic property of the trace. By taking into account also the freeterm for the Dirac fields and the interaction we find the equations of motion

∂µFµνA + ig[Bµ, F

µν ]A = gΨγνTAΨ

(i∂/ −m)Ψ = gγµBµΨ (8.51)

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 (8.52)

The reason is that under a global transformation of the symmetry group, the gaugefields are not invariant, said in different words they are charged fields with respectto the gauge fields. In fact we can verify immediately that the previous currents areprecisely the Noether’s currents. Under a global variation we have

δAµC = fABC αAAµB, δΨ = −iαATAΨ (8.53)

and we get

jµ =∂L∂Ψ,µ

δΨ +∂L

∂Aν,µCδAνC

(8.54)

from whichjµ = ΨγµαAT

AΨ − F µνC fABC αAAνB (8.55)

In the case of simple compact Lie groups one can define fABC = fABC with theproperty fABC = fBCA. It follows

F µνC fABCAνBT

A = i[Bν , Fνµ] ≡ i[Bν , Fνµ]ATA (8.56)

Thereforejµ = ΨγµαAT

AΨ − i[Bν , Fνµ]AαA (8.57)

After division by αA we get the Noether’s currents (8.52). The contribution of thegauge fields to the currents is also crucial in order they are conserved quantities. Infact, the divergence of the fermionic contribution is given by

∂µ(ΨγµTAΨ) = −igΨγµTABµΨ + igΨγµB

µTAΨ = −igΨγµ[TA, Bµ]Ψ (8.58)

which vanishes for abelian gauge fields, whereas it is compensated by the gaugefields contribution in the non abelian case.

8.3 Path integral quantization of the gauge theo-

ries

As we have seen at the beginning of this course that the quantization of the elec-tromagnetic field is far from being trivial, the difficulties being related to the gauge

186

invariance of the theory. In this Section we would like to discuss how these difficul-ties arise also in the path integral quantization. To this end let us recall the generalscheme of field quantization in this approach. First of all we will deal only with theperturbative approach, meaning that the only thing that one needs to evaluate isthe generating functional for the free case. The free case is defined by the quadraticpart of the action. Therefore the evaluation of the path integral reduces to calculatea Gaussian integral of the type given in eqs. (7.73) and (7.73) for the fermionic andbosonic case respectively. For instance, in the bosonic case one evaluates

I(A; J) =

∫dx1 · · · dxne−x

TAx/2 + JTx = eJTA−1J/2 (2π)

n2√

detA(8.59)

In our case the operator A appearing in the gaussian factor is nothing but the waveoperator. The result of the integration is meaningful only if the operator A is non-singular (it needs to have an inverse) otherwise the integral is not well defined. Thisis precisely the point where we run into problems in the case of a gauge theory.Remember from eq. (1.132) that the wave operator for the electromagnetic field inmomentum space is given by

(−k2gµν + kµkν)Aν(k) = 0 (8.60)

We see that the operator has a null eigenvector kν, since

(−k2gµν + kµkν)kν = 0 (8.61)

and therefore it is a singular operator. This is strictly related to the gauge invariance,which in momentum space reads

Aµ(k) → Aµ(k) + kµΛ(k) (8.62)

In fact, we see that the equations of motion are invariant under a gauge transforma-tion precisely because the wave operator has kµ as a null eigenvector. Another wayof seeing this problem is the following: in order to get physical results, as to evaluateS matrix elements, it is necessary to integrate over gauge-invariant functionals∫

D(Aµ)F [Aµ]eiS (8.63)

with F [Aµ] = F [AΩµ ] (here AΩ

µ is the gauge transformed of Aµ). But being the actionand the functional F gauge invariant, it follows that the integrand is invariant alongthe gauge group orbits. The gauge group orbits are defined as the set of points, inthe field space, which can be reached by a given Aµ via a gauge transformation. Inother words, given Aµ, its orbit is given by all the fields AΩ

µ obtained by varying theparameters of the gauge group (that is varying Ω), see Fig. 8.3.1. For instance, inthe abelian case

AΩµ = Aµ + ∂µΩ (8.64)

187

.

A

A'

µ

µ

Ω

ΩA

A'

µ

µ

Fig. 8.3.1 - The orbits of the gauge group in the field space.

Since the orbits define equivalence classes (or cosets) we can imagine the field spaceas the set of all the orbits. Correspondingly we can divide our functional integralin a part parallel and in one perpendicular to the orbits. Then, the reason why ourfunctional integral is not well defined depends from the fact that the integrand alongthe space parallel to the orbits is infinite (the integrand being gauge invariant doesnot depend on the corresponding variables). However this observation suggests asimple solution to our problem. We could define the integral by dividing it by theintegral along the part parallel to the orbit (notice that is the same as the volumeof the gauge group). In order to gain a clear understanding of this problem let usconsider the following very simple example in two dimensions

Z =

∫ +∞

−∞dxdye

−1

2a2(x− y)2

(8.65)

This is as reducing the space-time two two points with x and y the possible values ofthe field in those points. At the same time the exponent can be seen as an euclideanaction with the (gauge) symmetry

xΩ = x+ Ω, yΩ = y + Ω (8.66)

The integral (8.65) is infinite due to this invariance. In fact this implies that theintegrand depends only on x − y, and therefore the integration over the remainingvariable gives rise to an infinite result. In order to proceed along the lines we havedescribed above, let us partition the space of the field, that is the two-dimensionalspace (x, y) in the gauge group orbits. Given a point (x0, y0), the group orbit isgiven by the set of points (xΩ = x + Ω, yΩ = y + Ω) corresponding to the linex− y = x0 − y0, as shown in Fig. 8.3.2 We can perform the integral by integrating

188

.

x + y = c

f(x,y) = 0

(x , y )

(x , y )

0 0

Ω Ω

x

y

Fig. 8.3.2 - The field space (x, y) partitioned in the orbits of the gauge group. Weshow also the line x+ y = c and the arbitrary line f(x, y) = 0 (dashed).

along the perpendicular space, the line x + y = c (c is a constant), and then alongthe parallel space. It is this last integral which gives a divergent result, since we areadding up the same contribution an infinite number of times. therefore the obviousway to do is to introduce as integration variables x− y and x+ y, integrating onlyon x − y. Or, said in a different way, perform both integrals and dividing up bythe integral over x+ y, which cancels the divergence at the numerator. This can bedone in a more systematic way by using the following identity

1 =

∫ +∞

−∞dΩδ

(1

2(x+ y) + Ω

)(8.67)

Then we multiply Z per by this factor obtaining

Z =

∫dΩ

⎡⎣∫ dxdyδ

(1

2(x+ y) + Ω

)e−1

2a2(x− y)2

⎤⎦ (8.68)

The integral inside the parenthesis is what we are looking for since it is evaluatedalong the line x+y = constant. Of course the result should not depend on the choiceof this constant (it corresponds to translate the integration line in a way parallel tothe orbits). In fact, by doing the change of variables

x→ x+ Ω, y → y + Ω (8.69)

189

we get

Z =

∫dΩ

⎡⎣∫ dxdyδ

(1

2(x+ y)

)e−1

2a2(x− y)2

⎤⎦ (8.70)

As promised the infinite factor is now factorized out and we see also clearly thatthis nothing but the volume of the gauge group (Ω parameterize the gauge group,which in the present case is isomorphic to R1). Then, we define our integral as

Z ′ =Z

VG=

∫dxdyδ

(1

2(x+ y)

)e−1

2a2(x− y)2

(8.71)

with

VG =

∫dΩ (8.72)

In this way the integral is defined by integrating over the particular surface x+y = 0.Of course we would get the same result integrating over any other line (or surfacein the general case) having the property of intersecting each orbit only once. Forinstance, let us choose the line (see Fig. 8.3.2)

f(x, y) = 0 (8.73)

Then, proceeding as in eq. (8.67) we consider the following identity, which allowsus to define a gauge invariant function ∆f (x, y)

1 = ∆f (x, y)

∫dΩδ

[f(xΩ, yΩ)

]= ∆f(x, y)

∫dΩδ [f(x+ Ω, y + Ω)] (8.74)

In the previous case we had f(x, y) = (x + y)/2 = 0 giving rise to f(xΩ, yΩ) =(x+ y)/2+Ω = 0, from which ∆f(x, y) = 1. Multiplying Z by the expression (8.74)we get

Z =

∫dΩ

⎡⎣∫ dxdy∆f(x, y)δ [f(x+ Ω, y + Ω)] e

−1

2a2(x− y)2

⎤⎦ (8.75)

Let us show that ∆f(x, y) is gauge invariant. In fact by the transformation x →x+ Ω′, y → y + Ω′ we obtain

1 = ∆f(x+ Ω′, y + Ω′)∫dΩδ [f(x+ Ω + Ω′, y + Ω + Ω′)] (8.76)

and changing the integration variable Ω → Ω + Ω′

1 = ∆f(x+ Ω′, y + Ω′)∫dΩδ [f(x+ Ω, y + Ω)] (8.77)

190

Comparison with eq. (8.74) gives

∆f(x, y) = ∆f(x+ Ω′, y + Ω′) (8.78)

The integral inside the parenthesis in eq. (8.75) does not depend on Ω (that is isgauge invariant). This can be shown exactly as we did before in the case f(x, y) =(x + y)/2. Therefore we can again factorize the integration over Ω, being the sumof infinite gauge copies

Z =

∫dΩ

⎡⎣∫ dxdy∆f(x, y)δ [f(x, y)] e

−1

2a2(x− y)2

⎤⎦ (8.79)

and define

Z ′ =Z

VG=

∫dxdy∆f(x, y)δ [f(x, y)] e

−1

2a2(x− y)2

(8.80)

The function ∆f (x, y) (called the Faddeev-Popov determinant) can be calculateddirectly from its definition. To this end notice that in (8.79), ∆f (x, y) appearsmultiplied by δ[f(x, y)]. As a consequence, we need to know ∆f only at points veryclose at the surface f(x, y) = 0. But in this case, the equation f(xΩ, yΩ) = 0 hasthe only solution Ω = 0, since by hypothesis the surface f(x, y) crosses the gaugeorbits only once. Therefore we can write

δ[f(xΩ, yΩ)] =1[

∂fΩ

∂Ω

]Ω=0

δ[Ω] (8.81)

giving rise to

∆f (x, y) =∂fΩ

∂Ω

∣∣∣Ω=0

(8.82)

where fΩ = f(xΩ, yΩ). Clearly ∆f represents the answer of the gauge fixing f = 0to an infinitesimal gauge transformation. As a last remark, notice that our originalintegral can be written as

Z =

∫d2xe

−1

2xiAijxj

(8.83)

with

A = a2

[1 −1−1 1

](8.84)

a singular matrix. As we have already noticed the singularity of A is related to thegauge invariance, and the solution we have given here solves also the problem of thesingularity of A. In fact, integrating over x− y only, means to integrate only on theeigenvectors of A corresponding to non zero eigenvalues.

191

Since the solution we have found to our problem is of geometrical type, it is veryeasy to generalize it to the case of a gauge field theory, abelian or not. We start bychoosing a surface in the gauge field space given by

fB(AµB) = 0, B = 1, · · · , n (8.85)

where n is the number of parameters of LieG. Also, this surface should intersect thegauge orbits only once. To write down the analogue of equation (8.74) we need alsoa measure on the gauge group. This can be obtained as follows. We have seen thatwe are interested in the solutions of

fB(AΩµ ) = 0 (8.86)

around fB(Aµ) = 0, therefore it is enough to know the measure around the identityof the gauge group. In such a case the transformation Ω can be written as

Ω ≈ 1 + iαA(x)TA (8.87)

Since at the first order in the group parameters

ΩΩ′ ≈ 1 + i(αA + α′A)TA (8.88)

we see that the invariant measure of the group, around the identity, is given by

dµ(Ω) =∏A,x

dαA(x) (8.89)

It follows that for infinitesimal transformations

dµ(ΩΩ′) = dµ(Ω′Ω) = dµ(Ω) (8.90)

next we define the Faddeev-Popov determinant as in (8.74)

1 = ∆f [Aµ]

∫dµ(Ω)δ[fA(AΩ

µ )] (8.91)

In this expression we have introduced a functional delta-function, defined by theidentity

1 =

∫dµ(Ω)δ[Ω] (8.92)

having the usual properties extended to the functional case. We can see that ∆f isa gauge invariant functional, since

∆−1f [AΩ′−1

µ ] =

∫dµ(Ω)δ[fA(A(Ω′−1Ω)

µ ] (8.93)

and changing variables Ω′Ω = Ω′′ we get

∆−1f (aΩ′−1

µ ) =

∫dµ(Ω′′Ω′)δ[fA(AΩ′′

µ )] =

∫dµ(Ω)δ[fA(AΩ

µ )] (8.94)

192

Consider now the naive path integral

Z =

∫D(Aµ)F [Aµ]e

i

∫LAd4x

(8.95)

where F is a gauge invariant functional. Since in the following F will not play anyparticular role we will discuss only the case F = 1, but all the following consider-ations are valid for an arbitrary gauge invariant F . We multiply Z by the identity(8.91)

Z =

∫dµ(Ω)

⎡⎢⎣∫ D(Aµ)e

i

∫LAd4x

δ[fA(AΩµ )]∆f [Aµ]

⎤⎥⎦ (8.96)

The functional measure is invariant under gauge transformations

Aµ → ΩAµΩ−1 +

i

g(∂µΩ)Ω−1 (8.97)

since the homogeneous part has determinant one, and the inhomogeneous part givesrise to a translation. Then by performing the change of variables Aµ → AΩ

µ , andusing the gauge invariance of D(Aµ), LA and ∆f [Aµ], we get

Z =

∫dµ(Ω)

⎡⎢⎣∫ D(Aµ)e

i

∫LAd4x

δ[fA(Aµ)]∆f [Aµ]

⎤⎥⎦ (8.98)

Again, we define the functional integral as

Z → Z

VG=

Z∫dµ(Ω)

=

∫D(Aµ)e

i

∫LAd4x

δ[fA(Aµ)]∆f [Aµ] (8.99)

The evaluation of ∆f goes as before. Since in the previous equation it appearsmultiplied by δ[fA(Aµ)], the only solution of fA(AΩ

µ ) = 0 in this neighborhood isΩ = 1, or αA = 0. therefore

δ[fA(AΩµ )] = det−1

∥∥∥∥∥δfA(AΩµ )

δαB

∣∣∣αA=0

∥∥∥∥∥ δ[αA] (8.100)

from which

∆f [Aµ] = det

∥∥∥∥∥δfA(AΩµ )

δαB

∣∣∣αA=0

∥∥∥∥∥ (8.101)

Once again, the Faddeev-Popov determinant δf [Aµ] is the answer of the gauge-fixingto an infinitesimal gauge transformation. The final expression that we get for Z is

Z =

∫D(Aµ)δ[fA(Aµ)]det

∥∥∥∥∥δfA(AΩµ )

δαB

∣∣∣αA=0

∥∥∥∥∥ ei

∫LAd4x

(8.102)

193

This expression gives the right measure to be used in the path integral quantizationof the gauge fields. In particular, the vacuum expectation value (v.e.v.) of a gaugefield functional will be defined as

〈0|T ∗(F [A])|0〉f =

∫D(Aµ)δ[f ]∆fe

iSF [A] (8.103)

where the index f specifies the gauge in which we evaluate the v.e.v. of F . Aboutthis point, we can prove an important result, that is: if F [A] is a gauge invariantfunctional

F [AΩ] = F [A] (8.104)

than its v.e.v. does not depend on the gauge fixing fA = 0. We will prove thisresult by showing the relation between the v.e.v.’s of F [A] in due different gauges.Consider the identity

〈0|T ∗(F [A])|0〉f1 =

∫D(Aµ)δ[f1]∆f1e

iSF [A]

∫dµ(Ω)δ[fΩ

2 ]∆f2 (8.105)

where we have used eq. (8.91)and defined fΩ2 [A] ≡ f2[A

Ω]. By change of variableAµ → AΩ

µ , we get

〈0|T ∗(F [A])|0〉f1 =

∫dµ(Ω)

∫D(Aµ)δ[f

Ω−1

1 ]∆f1eiSF [AΩ−1

]δ[f2]∆f2

=

∫dµ(Ω)

∫D(Aµ)e

iSδ[f2]∆f2

[δ[fΩ

1 ]∆f1F [AΩ]]

(8.106)

where in the second line we have made a further change of variables Ω → Ω−1 andused the invariance of dµ(Ω). Then

〈0|T ∗(F [A])|0〉f1 =

∫dµ(Ω)〈0|T ∗(δ[fΩ

1 ]∆f1F [AΩ])|0〉f2 (8.107)

This is the relation we were looking for. If F is gauge invariant we get at once

〈0|T ∗(F [A])|0〉f1 = 〈0|T ∗(F [A])|0〉f2 (8.108)

For the following we will define a generating functional

Zf [ηAµ ] =

∫D(Aµ)δ[f ]∆fe

iSei

∫d4xηAµA

µA

(8.109)

which is not gauge invariant, but in terms of which we can easily construct thev.e.v.’s of gauge invariant quantities, since it is built up in terms of the correctfunctional measure.

For the purpose of building up the perturbative theory it is convenient to writethe integrand of eq. (8.109) in the form exp(iSeff), where Seff is an effective action

194

taking into account the gauge fixing f = 0 and the Faddeev-Popov determinant.We will discuss later the exponentiation of ∆f . As far as δ[f ] it is concerned, wecan exponentiate it by using its Fourier representation, or, more conveniently bychoosing the gauge condition in the form fA(A) − BA(x) = 0, with BA(x) a set ofarbitrary functions independent on the gauge fields. Then

∆(f−B)[A] = det

∥∥∥∥δ(fA[AΩ] −BA)

δαB

∣∣∣αA=0

∥∥∥∥ = det

∥∥∥∥δfA[AΩ]

δαB

∣∣∣αA=0

∥∥∥∥ = ∆f [A] (8.110)

and therefore

1 = ∆f [A]

∫dµ(Ω)δ[fΩ

A − BA] (8.111)

We can further multiply this identity by the following expression which does notdepend on the gauge fields

constant =

∫D(B)e

− i

2β

∫d4x

∑A

B2A(x)

(8.112)

We get

constant = ∆f [A]

∫dµ(Ω)D(B)e

− i

2β

∫d4x

∑A

(fΩA (x))2

δ(fA − BA)

= ∆f [A]

∫dµ(Ω)e

− i

2β

∫d4x

∑A

(fΩA (x))2

(8.113)

Multiplying this expression by the functional (8.95)we can show as before that thegauge volume factor factorizes, allowing us to define a class of generating functionals

Zβ[ηAµ ] =

∫D(Aµ)∆f [A]e

i

∫d4x(LA − 1

2β

∑A

(fA)2)

ei

∫d4xηAµA

µA

(8.114)

Finally, defining

Sf = − 1

2β

∫d4x

∑A

(fA(x))2 (8.115)

we find the analogue of eq. (8.107) is

〈0|T ∗(F [A])|0〉f1 =

∫dµ(Ω)〈0|T ∗(∆f1e

iSΩf1F [AΩ])|0〉f2 (8.116)

showing again that the v.e.v. a gauge invariant functional gives rise to a gaugeinvariant v.e.v. for the corresponding operator.

195

8.4 Path integral quantization of QED

Let us now discuss the path integral quantization of QED in a covariant gauge. Westart from the lagrangian

L = ψ [iγµ(∂µ + ieAµ) −m]ψ − 1

4FµνF

µν (8.117)

withFµν = ∂µAν − ∂νAµ (8.118)

invariant under

ψ(x) → ψ′(x) = e−ieα(x)ψ(x)

Aµ(x) → A′µ(x) = Aµ(x) +

1

e∂µα(x) (8.119)

The Lorentz gauge is defined by

f(Aµ) = ∂µAµ(x) = 0 (8.120)

and

f(AΩµ ) = ∂µAµ(x) +

1

eα(x) (8.121)

withδf(AΩ

µ )

δα(y)

∣∣∣α=0

=1

exδ

4(x− y) (8.122)

Then the Faddeev-Popov determinant turns out to be field-independent and it canbe absorbed into the normalization of the generating functional, which is given by

Z[η, η, ηµ] = N

∫D(ψ, ψ)D(Aµ)e

i

∫d4xL

δ[∂µAµ]ei

∫[ηψ + ψη + ηµA

µ]d4x

(8.123)We extract the interaction term in the usual way

Z[η, η, ηµ] = e−ie

∫d4x

(−1

i

δ

δη

)γµ(

1

i

δ

δηµ

)(1

i

δ

δη

)Z0[η, η, ηµ] (8.124)

with

Z0[η, η, ηµ] = N

∫D(ψ, ψ)D(Aµ)e

i

∫d4xL0

δ[∂µAµ]ei

∫[ηψ + ψη + ηµA

µ]d4x

(8.125)Writing

SA = −1

4

∫FµνF

µνd4x =1

2

∫d4x Aµ(g

µν − ∂µ∂ν)Aν (8.126)

196

using δ[∂µAµ] and integrating over the Fermi fields we get

Z0[η, η, ηµ] = e−i〈η(x)SF (x− y)η(y)〉

·N∫

D(Aµ)δ[∂µAµ]e

i

2〈Aµ(x)Aµ(x)〉

ei〈ηµ(x)Aµ(x)〉(8.127)

Then we exponentiate the δ[∂µAµ] through its Fourier transform

δ[∂µAµ] =

∫D(C)ei〈C(x)∂µA

µ(x)〉 =

∫D(C)e−i〈∂µC(x)Aµ(x)〉 (8.128)

obtaining the following expression for Z0

Z0[η, η, ηµ] = e−i〈η(x)SF (x− y)η(y)〉

·N∫

D(Aµ)D(C)ei〈1

2Aµ(x)Aµ(x) + (ηµ(x) − ∂µC(x))Aµ(x)〉

(8.129)

Doing the gaussian integral over Aµ we get

Z0[η, η, ηµ] = e−i〈η(x)SF (x− y)η(y)〉

·N∫

D(C)e

1

2〈i(ηµ(x) − ∂µC(x))

(i

)x,y

i(ηµ(y) − ∂µC(y))〉

= e−i〈η(x)SF (x− y)η(y)〉e−i

2〈ηµ 1

ηµ〉

·∫

D(C)ei〈∂µC 1

ηµ〉 − i

2〈∂µC 1

∂µC〉

(8.130)

After integrating by parts, we can rewrite the integral over C(x) as

∫D(C)e

−i〈C 1

∂µηµ〉 +

i

2〈CC〉

(8.131)

and doing again this gaussian integral we find

e

1

2〈(− i

∂µηµ

)(+i)

(− i

∂νην

)〉 = e

i

2〈ηµ∂µ∂ν2

ην 〉(8.132)

The final result is

Z0[η, η, ηµ] = Z[0]e−i〈η(x)SF (x− y)η(y)〉e− i

2〈ηµ

(gµν − ∂µ∂ν

2

)ην〉

(8.133)

197

Introducing the function

Gµν = limε→0+

∫d4k

(2π)4

[− gµνk2 + iε

+kµkν

(k2 + iε)2

]e−ikx (8.134)

we write

Z0[η, η, ηµ] = Z[0]e−i〈η(x)SF (x− y)η(y)e− i

2〈ηµ(x)Gµν(x− y)ην(y)〉

(8.135)

The photon propagator in this gauge is given by

1

〈0|0〉〈0|T (Aµ(x)Aν(x))|0〉 = iGµν(x− y) (8.136)

Notice that this propagator is transverse, that is it satisfies the Lorentz condition

∂µGµν(x) = 0 (8.137)

The Feynman’s rules in this gauge are the same discussed in Section 1.5.3, exceptfor th photon propagator which is

−i[

gµνk2 + iε

− kµkν(k2 + iε)2

](8.138)

Of course, the additional term in the propagator proportional to kµkν does not affectthe physics since the photon is coupled to a conserved current.

We can perform an analogous calculation but using the gauge fixing in the formf − B discussed at the end of the previous Section, that is using the generatingfunctional (8.114)

Zβ[η, η, ηµ] = N

∫D(ψ, ψ)D(Aµ)e

i

∫d4x

[L − 1

2β(∂µA

µ)2

]

· e

∫d4x

[ηψ + ψη + ηµA

µ]

(8.139)

In this case the only effect of the gauge fixing is to change the lagrangian L toL′ = L − (∂µA

µ)2/(2β). Of course, the addition of this term removes the problemof the singular wave operator. In fact we can write∫

d4xL′ =1

2

∫d4xAµ

(gµν − (1 − 1

β)∂µ∂ν

)Aν (8.140)

Therefore we have to evaluate the functional integral

Z0[η, η, ηµ] = Ne−i〈η(x)SF (x− y)η(y)〉∫

D(Aµ)ei

∫d4x

(1

2AµKµνA

ν + ηµAµ

)

(8.141)

198

with

Kµν = gµν − β − 1

β∂µ∂ν (8.142)

obtaining

Zβ[η, η, ηµ] = Z[0]e−i〈η(x)SF (x− y)η(y)〉e−i

2〈ηµ(x)Gµν(x− y)ην(y)〉

(8.143)

whereKµνG

νρ(x) = δρµδ4(x) (8.144)

We can see that Kµν is non singular for β = ∞. Going to momentum space, eq.(8.144) gives∫

d4k

(2π)4

(−gµνk2 +

β − 1

βkµkν

)Gνρ(k)e−ikx = δρµ

∫d4k

(2π)4e−ikx (8.145)

or (−gµνk2 +

β − 1

βkµkν

)Gνρ(k) = δρν (8.146)

This can be solved by writing the most general second rank symmetric tensor func-tion of kµ

Gµν(k) = Agµν +Bkµkν (8.147)

with the coefficients such that

−Ak2 = 1,β − 1

βA− 1

βBk2 = 0 (8.148)

Solving these equations

Gµν(k) = −gµνk2

+ (1 − β)kµkνk4

(8.149)

The singularities are defined by the usual iε term, and we get

Gµν(x) = limε→0+

∫d4k

(2π)4

[− gµνk2 + iε

+ (1 − β)kµkν

(k2 + iε)2

]e−ikx (8.150)

The discussion made at the beginning of this Section corresponds to β = 0 (the socalled Landau gauge). In fact for β → 0

limβ→0

e− i

2β

∫d4x(∂µA

µ)2

→ δ[∂µAµ] (8.151)

The choice corresponding to the quantization made in Section 1.5.3 is β = 1. Inthis case

Gµν(x) = −gµν∆F (x;m2 = 0) (8.152)

199

This is called the Feynman’s gauge. As we see the parameter β selects differentcovariant gauges. As already stressed the physics does not depend on β since thephoton couples to the fermionic current ψγµψ, which is conserved in the abelian case,and as a consequence the β dependent part of the propagator, being proportionalto kµkν does not contribute. The situation is quite different in the non-abelian casewhere the fermionic current is not conserved since the gauge fields are not neutral.Therefore we need further contributions in order to cancel the β dependent terms.Such contributions arise from the Faddeev-Popov determinant.

8.5 Path integral quantization of the non-abelian

gauge theories

In the case of a non-abelian gauge theory there are several differences with respect toQED. The pure gauge lagrangian is not quadratic, it contains three-linear and four-linear interaction terms and furthermore the Faddeev-Popov determinant typicallydepends on the gauge fields. In this Section we will consider the pure gauge case,since the fermionic coupling is completely analogue to the abelian case. Separatingthe lagrangian in the quadratic part plus the rest, we get

LA = −1

4

∑C

FµνCFµνC

= −1

4

(∂µAνC − ∂νAµC − gfABC AµAAνB

) (∂µAνC − ∂νAµC − gfDEC AµDA

νE

)= L(2)

A + gfABC AµAAνB∂µAνC − 1

4g2fABC fDEC AµAA

νBAµDAνE

≡ L(2)A + LIA (8.153)

where L(2)A and L(I)

A are the quadratic and the interacting part respectively. Theinteraction terms can be extracted by the functional integration by the usual trick

Z[ηµ] = eiS

(I)A

[1

i

δ

δηAµ

]Z0[η

Aµ ] (8.154)

with S(I)A =

∫d4xL(I)

A and

Z0[ηµ] = N

∫D(Aµ)e

i

∫d4x

[L(2)A − 1

2β

∑A

(∂µAµA)2

]ei

∫d4xηµAA

Aµ∆f [Aµ]

(8.155)Consider now ∆f [Aµ]. In the Lorentz gauge

fA[Aµ] = ∂µAµA (8.156)

200

Under an infinitesimal gauge transformation (for later convenience we have changedthe definition of the gauge parameters, sending αA → gαA)

δAµC = gfABC αAAµB + ∂µαC (8.157)

we havefC [AΩ

µ ] = ∂µAµC + gfABC ∂µ(αAA

µB) + αC (8.158)

Then

δfC [AΩµ (x)]

δαB(y)

∣∣∣αA=0

= δBC2xδ

4(x− y)+ gfBAC ∂µ(δ4(x− y)AµA) = MCB(x− y) (8.159)

We see that ∆f [A] depends explicitly on the gauge fields. In order to get an ex-pression for the generating functional which is convenient fir deriving the Feynmanrules, it is useful to give to it an exponential form. This can be done, recalling fromeq. (7.76)) that a determinant can be written as a gaussian integral over Grassmannvariables.

∆f [Aµ] = det

∥∥∥∥∥δfA[AΩµ (x)]

δαB(y)

∣∣∣αA=0

∥∥∥∥∥ = N

∫D(c, c∗)e

i

∫c∗A(x)MAB(x− y)cB(y)d4xd4y

(8.160)The Grassmann fields cA(x) are called the ghost fields. It should be noticed thatthe ghost carry zero spin and therefore in their particle interpretation lead to aviolation of the spin-statistics theorem. However this fact has no consequence, sinceour generating functional do not generate amplitudes with external ghosts. Theaction for the ghost fields can be read from the previous equation and we get

SFPG =

∫d4xc∗A(x)

[δAB − gfACB ∂µA

µC

]cB(x) (8.161)

The ghosts are zero mass particles interacting with the gauge fields through theinteraction term

S(I)FPG = −g

∫d4xc∗Af

ACB ∂µA

µCcB (8.162)

In order to deal with this interaction it turns out convenient to define a new gener-ating functional where we introduce also external sources for the ghost fields

Z[η, η∗, ηµ] = eiS

(I)A

[1

i

δ

δηµ

]eiS

(I)FPG

[−1

i

δ

δη,1

i

δ

δη∗

]Z0[η, η

∗, ηµ] (8.163)

with

Z0[η, η∗, ηµ] =

∫D(Aµ)D(c, c∗)

·ei

∫d4x

[L(2)A − 1

2β

∑A

(∂µAµA)2 + ηAµA

µA

]

· ei

∫d4x

[c∗AcA + c∗Aη

A + ηA∗cA]

(8.164)

201

The integration over AµA like in the abelian case. For the ghost field we use eq.(7.76) with the result

∫D(c, c∗)ei〈c∗AcA + c∗Aη

A + ηA∗cA〉 = e

−〈(iηA∗)

(− i

)(iηA)〉

= e−i〈ηA∗ 1

ηA〉

(8.165)Defining

∆AB(x) = −δAB limε→0+

∫d4k

(2π)4

e−ikxk2 + iε

(8.166)

we get

Z0[η, η∗, ηµ] = e

− i

2〈ηAµ (x)Gµν

AB(x− y)ηνB(y)〉e−i〈ηA

∗(x)∆AB(x− y)ηB(y)〉 (8.167)

withGµνAB = δABG

µν(x− y) (8.168)

It is then easy to read the Feynman’s rules. For the propagators we get

.

µ, ν,A B

k

Fig. 8.5.1 - Gauge field propagator:

−iδAB(gµν − (1 − β)

kµkνk2 + iε

)1

k2 + iε(8.169)

.

.

A B

k

Fig. 8.5.2 - Ghost field propagator:

iδAB1

k2 + iε(8.170)

.The Feynman’s rules for the vertices are obtained by taking the Fourier transform

of the interaction parts (here we have written fABC ≡ fABC).

202

.

µ,

ν,

A

B

C

k

k

k

1

2

3

λ,

Fig. 8.5.3 - Three-linear gluon vertex:

gfABC(2π)4δ4(k1 + k2 + k3) [gµν(k1 − k2)λ + gνλ(k2 − k3)µ + gλµ(k3 − k1)ν ] (8.171)

.

µ,

ν,

A

B C

k

k

k

1

2

3

λ,

ρ, D

k4

Fig. 8.5.4 - Four-linear gluon vertex:

−ig2(2π)4δ4(

4∑i=1

ki)[fABEfCDE(gµλgνρ − gνλgµρ)

+ fACEfBDE(gµνgλρ − gλνgµρ) + fADEfCBE(gµλgρν − gρλgµν)]

(8.172)

.

µ, A

B

k

C

p q

Fig. 8.5.5 - Ghost-gluon vertex:

−gfABC(2π)4δ4(k + p− q)pµ (8.173)

203

When we have fermions coupled to the gauge fields we need to add the rulefor the vertex (the propagator is the usual one). By a simple comparison with thecoupling in QED we see that in this case the rule for the vertex is

.

µ, A

k

p p'

b cβ γ, ,

Fig. 8.5.6 - Fermion-gluon vertex:

−igTAbc(γµ)βγ(2π)4δ4(p− p′ − k) (8.174)

8.6 The β-function in non-abelian gauge theories

We will skip a careful treatment of the renormalization in non-abelian gauge the-ories. However, since all the interaction terms have dimensions ≤ 4, the theory isrenormalizable by power counting. On the other hand we will evaluate the betafunction, since in these theories there is the possibility of having β < 0, giving riseto asymptotic freedom. As it is known, QCD the theory of strong interactions is infact a gauge theory based on the gauge group SU(3), and it is precisely ths featureof being an asymptotically free theory that has made it very attractive phenomeno-logically. In non-abelian gauge theories there are several ways to define the coupling,in fact we can look at the fermionic coupling, at the trilinear or at the four-linearcouplings. If the theory is gauge invariant (and the renormalization should preservethis feature), all these ways should give the same result. This is in fact ensured bya set of identities among the renormalization constants (Ward identities), which arethe analogue of the equality Z1 = Z2 in QED. Therefore we will study the defini-tion coming from the fermionic coupling. In this case the relation between the barecoupling and the renormalized one is the same as in QED (see Section 6.3))

gB = µε/2gZ1

Z2Z1/23

(8.175)

where Z1 is the vertex renormalization constant, and Z@ and Z3 are the wave func-tion renormalization constants for the fermion and the gauge particle respectively.

204

As we shall see, one of the differences with QED is that now Z1 = Z2. As a conse-quence we will need to evaluate all the three renormalization constants. Let us startwith the fermion self-energy. The one-loop contribution is given in Fig. 8.6.1.

.

k

ppp k_

Fig. 8.6.1 - One-loop fermion self-energy.

We will work in the Feynman-’t Hooft gauge, corresponding to the choice β = 1 forthe gauge field propagator (see eq. (8.169)). The Feynman rules for this diagramare the same as in QED except for the group factors. In order to avoid group theorycomplications, we will consider the case of a gauge group SU(2) with fermions in thespinor representation, where all the group factors are easily evaluated. At the sametime we will exhibit also the general results for a theory SU(N) with fermions inthe fundamental representation (that is the one of dimensions N). Then, recallingthe expression for the electron self-energy in QED (see eq. (2.166)), we get

Σ(p) =∑A

(TATA)ΣQED(p) =∑A

(TATA)

[1

8π2ε(p− 4m) + Σf

QED(p)

](8.176)

For SU(2) we have

∑A

(TATA)ab =∑A

(τA

2

τA

2

)ab

=3

4δab (8.177)

In the case of a group SU(N) and fermions in the fundamental representation it ispossible to show that ∑

A

(TATA)ab = C2(F )δab (8.178)

with

C2(F ) =N2 − 1

2N(8.179)

This result is justified by the observation that the operator∑

A TATA (called the

Casimir operator) commutes with all the generators and therefore it must be propor-tional to the identity matrix. From the previous expression for Σ we find immediately(see Section 2.7)

Z2 = 1 − g2

8π2εC2(F ) (8.180)

205

.

= + +

+ + + +

+ +

a)

b) c) d)

e) f)

Fig. 8.6.2 - One-loop expansion of the vacuum polarization diagram in non-abeliangauge theories.

The next task is the evaluation of Z3. To this end we need the one-loop expansionof the gauge particle propagator (the vacuum polarization diagram). This is givenin Fig. 8.6.2.Since we are using dimensional regularization we can see immediately that the lastthree diagrams (diagrams d), e) and f)) of Fig. 8.6.2 are zero (this is not true inother regularizations). In fact all these diagrams involve the integral∫

d2ωk

k2(8.181)

which, in dimensional regularization gives (including a mass term)∫d2ωk

k2 − a2= −iπωΓ(1 − ω)(a2)ω−1 (8.182)

By taking the limit a→ 0, we see that for ω > 1 the integral vanishes.Let us begin our calculation from the gauge particle contributions to the vacuum

polarization. The kinematics is specified in Fig. 8.6.3. We will evaluate the trun-cated Green’s function which differs by a factor ig2 from the vacuum polarizationvacuum Πµν as defined in Section 2.1. We have

ig2Π(a)µν,AB =

1

2(−g)2fADCfDBC

∫d2ωk

(2π)2ωEµν

−i(p + k)2

−ik2

(8.183)

206

The factor 1/2 is a symmetry factor associated to the diagram and the tensor Eµνcomes from the trilinear vertices. It is given by

Eµν =[gσµ(p− k)ρ + gσρ(2k + p)µ − gρµ(2p+ k)σ

]× [gσν(p− k)ρ − gνρ(2p+ k)σ + gρσ(2k + p)ν ]

= gµν[(p− k)2 + (2p+ k)2

]+ 2ω(2k + p)µ(2k + p)ν

− [(2p+ k)µ(2k + p)ν − (2k + p)µ(p− k)ν + (p− k)µ(2p+ k)ν ](8.184)

.

p p

p + k

k

A, B,

C,

D,

µ νρ

σ

Fig. 8.6.3 - One-loop gauge particle contribution to the vacuum polarization.

The group factor is easily evaluated for SU(2), since in this case fABC = εABC .Then we get

εADCεDBC = −2δAB (8.185)

The general result for SU(N) is (taking into account that fABC are completelyantisymmetric)

fADCfDBC = −C2(G)δAB (8.186)

withC2(G) = N (8.187)

This result can be understood by noticing that (TA)BC = ifABC gives a represen-tation of the Lie algebra of the group (the adjoint representation), and thereforeC2(G) is nothing but the Casimir of the adjoint. By using the standard trick wereduce the two denominators in the integral to a single one

1

k2

1

(k + p)2=

∫ 1

0

dz1

(k′2 + z(1 − z)p2)2(8.188)

withk = k′ − pz (8.189)

Operating this substitution into the tensor Eµν , taking into account that the inte-gration over the odd terms in k′ gives zero, and that for symmetry we can make thesubstitution

k′µk′ν →

gµν2ω

k′2 (8.190)

207

we get (by calling k′2 with k2)

Eµν = gµνk26

(1 − 1

2ω

)+ gµνp

2[(1 + z)2 + (2 − z)2

]+ pµpν

[2ω(1 − 2z)2 − 2

((1 + z)2 + (2 − z)(1 − 2z)

)]≡ aµνk

2 + bµν (8.191)

Performing the integral over k with the formulas of Section 2.5 and putting every-thing together, we get

ig2Π(a)µν,AB = −1

2g2(−C2(G))δAB

iπω

(2π)2ω

∫ 1

0

dz[ωz(1 − z)p2Γ(1 − ω)aµν + Γ(2 − ω)bµν

](8.192)

By taking only the 1/ε term (ε = 4 − 2ω) in the expansion and integrating over zwe get

ig2Π(a)µν,AB = i

g2

16π2εC2(G)δAB

[19

6gµνp2 − 11

3pµpν

](8.193)

As we see this contribution is not transverse, that is it is not proportional to(pµpν − gµνp

2), as it should be. in fact, we will see that adding to this diagramthe contribution of the ghost loop we recover the transverse factor. The ghost con-tribution is the part b) of the expansion of Fig. 8.6.2. The relevant kinematics isgiven in Fig. 8.6.4.

.

p p

p + k

k

A, B,µ ν

C

D

Fig. 8.6.4 - One-loop ghost contribution to the vacuum polarization.

We have

ig2Π(b)µν,AB = (−1)g2fACDfBDC

∫d2ωk

(2π)2ω

i

(p+ k)2

i

k2(p+ k)µkν (8.194)

By the standard procedure we get (k = k′ − pz))

ig2Π(b)µν,AB = −g2C2(G)δAB

∫ 1

0

dz

∫d2ωk

(2π)2ω

k′µk′ν − pµpνz(1 − z)

(k′2 + z(1 − z)p2)2(8.195)

208

where we have eliminated the terms linear in k′. Then, we perform the integral overk′

ig2Π(b)µν,AB = −ig2 πω

(2π)2ωC2(G)δAB

∫ 1

0

dz z(1 − z)

[1

2p2Γ(1 − ω) − pµpνΓ(2 − ω)

](8.196)

We perform now the integral in z. Then taking the limit ω → 2, and keeping onlythe leading term, we get

ig2Π(b)µν,AB = i

g2

16π2εC2(G)δAB

[1

6p2gµν +

1

3pµpν

](8.197)

Putting together this contribution with the former one we get

ig2(Π(a)µν,AB + Π

(b)µν,AB) = −i g

2

8π2ε

5

3C2(G)δAB

[pµpν − gµνp

2]

(8.198)

The fermionic loop contribution is given in Fig. 8.6.5, and except for the groupfactor is the same evaluated in Section 2.6..

p p

p + k

k

A, B,µ ν

Fig. 8.6.5 - One-loop fermion contribution to the vacuum polarization.

The leading contribution is

ig2Π(c)µν,AB = ig2Tr(TATB)ΠQED

µν = ig2

12π2εnF δAB

[pµpν − gµνp

2]

(8.199)

where we have used the conventional normalization for the group generators in thefermionic representation

Tr(TATB) =1

2nF δAB (8.200)

where nF is the number of fundamental representations. For instance, in the caseof QCD, we have three different kind of quarks corresponding to different flavors.Then, the one-loop leading term contribution to the vacuum polarization is givenby

ig2Πµν,AB = ig2(Π(a)µν,AB + Π

(b)µν,AB + Π

(c)µν,AB)

= −i g2

8π2ε

(5

3C2(G) − 2

3nF

)[pµpν − gµνp

2]

(8.201)

209

By comparison with Section 2.7 we get

Z3 = 1 +g2

8π2ε

(5

3C2(G) − 2

3nF

)(8.202)

We are now left with the vertex corrections. The fermion contribution is given bythe diagram of Fig. 8.6.6.

.

k pp'

p + kp' + k

A

BB

Fig. 8.6.6 - Fermion one-loop vertex correction.

Again, this is the same as in QED except for the group factors. the leading contri-bution is

−ig3Λµ,Af = −i g

3

8π2ε(TBTATB)γµ (8.203)

The group factor can be elaborated as follows

TBTATB = TB(TBTA + ifABCTC) = C2(F )TA + ifABCTBTC

= C2(F )TA − 1

2fABCfBCDTD =

(C2(F ) − 1

2C2(G)

)TA(8.204)

Therefore we obtain

−ig3Λµ,Af = −i g

3

8π2ε

(C2(F ) − 1

2C2(G)

)TAγµ (8.205)

The final diagram to be evaluated is the vertex correction from the trilinear gaugecoupling given in Fig. 8.6.7.We get (q = p′ − p)

−ig3Λµ,Ag = (−igγνTB)

∫d2ωk

(2π)2ω

ik

k2

−i(p′ − k)2

· (−g)fABC [gµν(q + p′ − k)ρ − gνρ(p+ p′ − 2k)µ + gρµ(p− k − q)ν ]

· −i(p− k)2

(−igγρTC) (8.206)

210

.

k

pp'

B

p - kp' - k

C

A,µ

Fig. 8.6.7 - One-loop vertex correction from the trilinear gauge coupling.

Since we are interested only in the divergent piece, the calculation can be simplifieda lot by the observation that the denominator goes as (k2)3, whereas the numeratorgoes at most as k2. Therefore the divergent piece is obtained by neglecting all theexternal momenta. Therefore the expression simplifies to

−ig3Λµ,Ag = ig3fABCTBTC

∫d2ωk

(2π)2ω

γν k(gµνkρ − 2gνρkµ + gµρkν)γρ

(k2)3(8.207)

The group factor gives

fABCTBTC =i

2fABCfBCDTD =

i

2C2(G)TA (8.208)

whereas the numerator gives simply −3k2γµ. Then we have

−ig3Λµ,Ag = − g3

(2π)2ω

3

2C2(G)TAγµ

∫d2ωk

(k2)2= −i g

3

8π2ε

3

2C2(G)TAγµ (8.209)

Adding up the two vertex contributions we obtain

−ig3Λµ,A = −ig3(Λµ,Af + Λµ,A

g ) = −i g3

8π2ε(C2(F ) + C2(G))TAγµ (8.210)

Again, comparison with Section 2.7 gives

Z1 = 1 − g2

8π2ε(C2(F ) + C2(G)) (8.211)

Using eq. (8.175) we have

gB = µε/2g(1 + ∆Z1 − ∆Z2 − 1

2∆Z3) = µε/2

(g +

a1

ε

)(8.212)

and collecting everything together we get

a1 = − g3

16π2

(11

3C2(G) − 2

3nF

)(8.213)

211

The evaluation of β(g) goes as in the QED case (see Section 6.3), that is

β(g) = µ∂g(µ)

∂µ= −1

2

(1 − g

d

dg

)a1 = −1

2(a1 − 3a1) = a1 (8.214)

that is

β(g) = − g3

16π2

(11

3C2(G) − 2

3nF

)(8.215)

In the case of QCD where the gauge group is SU(3) we have

C2(G) = 3 (8.216)

and therefore

βQCD(g) = − g3

16π2

(11 − 2

3nF

)(8.217)

We see that QCD is asymptotically free (β(g) < 0) for

nF <33

2(8.218)

212

Chapter 9

Spontaneous symmetry breaking

9.1 The linear σ-model

In this Section and in the following we will study, from a classical point of view,some field theory with particular symmetry properties. We will start examining thelinear σ-model. This is a model for N scalar fields, with a symmetry O(N). Thelagrangian is given by

L =1

2

N∑i=1

∂µφi∂µφi − 1

2µ2

N∑i=1

φiφi − λ

4

(N∑i=1

φiφi

)2

(9.1)

This lagrangian is invariant under linear transformations acting upon the vectorφ = (φ1, · · · , φN) and leaving invariant its norm

|φ|2 =N∑i=1

φiφi (9.2)

Consider an infinitesimal transformation (from now on we will omit the index ofsum over the indices which are repeated)

δφi = εijφj (9.3)

The condition for letting the norm invariant gives

|φ+ δφ|2 = |φ|2 (9.4)

from whichφ · δφ = 0 (9.5)

or, in components,φiεijφj = 0 (9.6)

This is satisfied byεij = −εji (9.7)

213

showing that the rotations in N dimensions depend on N(N−1)/2 parameters. Fora finite transformation we have

|φ′|2 = |φ|2 (9.8)

withφ′i = Sijφj (9.9)

implyingSST = 1 (9.10)

In fact, by exponentiating the infinitesimal transformation one gets

S = eε (9.11)

withεT = −ε (9.12)

implying that S is an orthogonal transformation. The matrices S form the rotationgroup in N dimensions, O(N).

The Noether’s theorem implies a conserved current for any symmetry of thetheory. In this case we will get N(N − 1)/2 conserved quantities. It is useful forfurther generalizations to write the infinitesimal transformation in the form

δφi = εijφj = − i

2εABT

ABij φj, i, j = 1, · · · , N, A,B = 1, · · · , N (9.13)

which is similar to what we did in Section 3.3 when we discussed the Lorentz trans-formations. By comparison we see that the matrices TAB are given by

TABij = i(δAi δBj − δAj δ

Bi ) (9.14)

It is not difficult to show that these matrices satisfy the algebra

[TAB, TCD] = −iδACTBD + iδADTBC − iδBDTAC + iδBCTAD (9.15)

This is nothing but the Lie algebra of the group O(N), and the TAB are the in-finitesimal generators of the group. Applying now the Noether’s theorem we findthe conserved current

jµ =∂L

∂(∂µφi)δφi = − i

2φi,µεABT

ABij φj (9.16)

since the N(N − 1)/2 parameters εAB are linearly independent, we find the N(N −1)/2 conserved currents

JABµ = −iφi,µTABij φj (9.17)

In the case of N = 2 the symmetry is the same as for the charged field in a realbasis, and the only conserved current is given by

J12µ = −J21

µ = φ1,µφ2 − φ2,µφ1 (9.18)

214

One can easily check that the charges associated to the conserved currents closethe same Lie algebra as the generators TAB. More generally, if we have conservedcurrents given by

jAµ = −iφi,µTAij φj (9.19)

with[TA, TB] = ifABCTC (9.20)

then, using the canonical commutation relations, we get

[QA, QB] = ifABCQC (9.21)

with

QA =

∫d3x jA0 (x) = −i

∫d3x φiT

Aij φj (9.22)

A particular example is the case N = 4. We parameterize our fields in the form

φ = (π1, π2, π3, σ) = (π, σ) (9.23)

These fields can be arranged into a 2 × 2 matrix

M = σ + iτ · π (9.24)

where τ are the Pauli matrices. Noticing that τ2 is pure imaginary, τ1 and τ3 real,and that τ2 anticommutes with τ1 and τ3, we get

M = τ2M∗τ2 (9.25)

Furthermore we have the relation

|φ|2 = σ2 + |π|2 =1

2Tr(M †M) (9.26)

Using this it is easy to write the lagrangian for the σ-model in the form

L =1

4Tr(∂µM

†∂µM) − 1

4µ2Tr(M †M) − 1

16λ(Tr(M †M)

)2(9.27)

This lagrangian is invariant under the following transformation of the matrix M

M → LMR† (9.28)

where L and R are two special (that is with determinant equal to 1) unitary matrices,that is L,R ∈ SU(2). The reason to restrict these matrices to be special is thatonly in this way the transformed matrix satisfy the condition (9.25). In fact, if A isa 2 × 2 matrix with detA = 1, then

τ2AT τ2 = A−1 (9.29)

215

Therefore, for M ′ = LMR† we get

τ2M′∗τ2 = τ2L

∗M∗RT τ2 = τ2L∗τ2(τ2M∗τ2)τ2RT τ2 (9.30)

and from (9.29)

τ2L∗τ2 = τ2L

†T τ2 = L†−1= L

τ2RT τ2 = R−1 = R† (9.31)

since the L and R are independent transformations, the invariance group in thisbasis is SU(2)L ⊗ SU(2)R. In fact this group and O(4) are related by the followingobservation: the transformation M → LMR† is a linear transformation on thematrix elements of M , but from the relation (9.26) we see that M → LMR† leaves

the norm of the vector φ = (π, σ) invariant and therefore the same must be truefor the linear transformation acting upon the matrix elements of M , that is on σand π. Therefore this transformation must belongs to O(4). This shows that thetwo groups SU(2) ⊗ SU(2) and O(4) are homomorphic (actually there is a 2 to 1relationship, since −L and −R define the same S as L and R).

We can evaluate the effect of an infinitesimal transformation. To this end wewill consider separately left and right transformations. We parameterize the trans-formations as follows

L ≈ 1 − i

2θL · τ , R ≈ 1 − i

2θR · τ (9.32)

then we get

δLM = (− i

2θL ·τ)M = (− i

2θL ·τ)(σ+ iπ ·τ) =

1

2θL ·π+

i

2(θL ∧π− θLσ) ·τ (9.33)

where we have usedτiτj = δij + iεijkτk (9.34)

SinceδLM = δLσ + iδLπ · τ (9.35)

we get

δLσ =1

2θL · π, δLπ =

1

2(θL ∧ π − θLσ) (9.36)

Analogously we obtain

δRσ = −1

2θR · π, δRπ =

1

2(θR ∧ π + θRσ) (9.37)

The combined transformation is given by

δσ =1

2(θL − θR) · π, δπ =

1

2[(θL + θR) ∧ π − (θL − θR)σ] (9.38)

216

and we can check immediately that

σδσ + π · δπ = 0 (9.39)

as it must be for a transformation leaving the form σ2 + |π|2 invariant. Of particular

interest are the transformations with θL = θR ≡ θ. In this case we have L = R and

M → LML† (9.40)

These transformations span a subgroup SU(2) of SU(2)L⊗SU(2)R called the diag-onal subgroup. In this case we have

δσ = 0, δπ = θ ∧ π (9.41)

We see that the transformations corresponding to the diagonal SU(2) are the rota-tions in the 3-dimensional space spanned by π. These rotations define a subgroupO(3) of the original symmetry group O(4). From the Noether’s theorem we get theconserved currents

jLµ =1

2σ,µθL · π +

1

2π,µ · (θL ∧ π − θLσ) (9.42)

and dividing by θL/2JLµ = σ,µπ − π,µσ − π,µ ∧ π (9.43)

and analogouslyJRµ = −σ,µπ + π,µσ − π,µ ∧ π (9.44)

Using the canonical commutation relations one can verify that the correspondingcharges satisfy the Lie algebra of SU(2)L ⊗ SU(2)R

[QLi , Q

Lj ] = iεijkQ

Lk , [QR

i , QRj ] = iεijkQ

Rk , [QL

i , QRj ] = 0 (9.45)

By taking the following combinations of the currents

JVµ =1

2( JLµ + JRµ ), JAµ =

1

2( JLµ − JRµ ) (9.46)

one hasJVµ = π ∧ π,µ (9.47)

andJAµ = σ,µπ − π,µσ (9.48)

The corresponding algebra of charges is

[QVi , Q

Vj ] = iεijkQ

Vk , [QV

i , QAj ] = iεijkQ

Ak , [QA

i , QAj ] = iεijkQ

Vk (9.49)

These equations show that QVi are the infinitesimal generators of a subgroup SU(2)

of SU(2)L ⊗ SU(2)R which is the diagonal subgroup, as it follows from

[QVi , πj ] = iεijkπk, [QV

i , σ] = 0 (9.50)

217

In the following we will be interested in treating the interacting field theoriesby using the perturbation theory. As in the quantum mechanical case, this is welldefined only when we are considering the theory close to a minimum the energy ofthe system. In fact if we are going to expand around a maximum the oscillation ofthe system can become very large leading us outside of the domain of perturbationtheory. In the case of the linear σ-model the energy is given by

H =

∫d3xH =

∫d3x

[N∑i=1

∂L∂φi

φi −L]

=

∫d3x

[1

2

N∑i=1

(φ2i + |∇φi|2) + V (|φ|2)

]

(9.51)Since in the last member of this relation the first two terms are positive definite,it follows that the absolute minimum is obtained for constant field configurations,such that

∂V (|φ|2)∂φi

= 0 (9.52)

Let us call by vi the generic solution to this equation (in general it could happen thatthe absolute minimum is degenerate). Then the condition for getting a minimum isthat the eigenvalues of the matrix of the second derivatives of the potential at thestationary point are definite positive. In this case we define new fields by shiftingthe original fields by

φi → φ′i = φi − vi (9.53)

The lagrangian density becomes

L =1

2∂µφ

′i∂µφ′

i − V (|φ′ + v|2) (9.54)

Expanding V in series of φ′i we get

V = V (|v|2) +1

2

∂2V

∂φi∂φj

∣∣∣φ=v

φ′iφ

′j + · · · (9.55)

This equation shows that the particle masses are given by the eigenvalues of thesecond derivative of the potential at the minimum. In the case of the linear σ-modelwe have

V =1

2µ2|φ|2 +

λ

4(|φ|2)2 (9.56)

Therefore∂V

∂φi= µ2φi + λφi|φ|2 (9.57)

In order to have a solution to the stationary condition we must have φi = 0, or

|φ|2 = −µ2

λ(9.58)

218

This equation has real solutions only if µ2/λ < 0. However, in order to have apotential bounded from below one has to require λ > 0, therefore we may have nonzero solutions to the minimum condition only if µ2 < 0. But, notice that in thiscase, µ2 cannot be identified with a physical mass, these are given by the eigenvaluesof the matrix of the second derivatives of the potential at the minimum and theyare positive definite by definition. We will study this case in the following Sections.In the case of µ2 > 0 the minimum is given by φi = 0 and one can study thetheory by taking the term λ(|φ|2)2 as a small perturbation (that is requiring thatboth λ and the values of φi, the fluctuations, are small). The free theory is givenby the quadratic terms in the lagrangian density, and they describe N particles ofcommon mass m. Furthermore, both the free and the interacting theories are O(N)symmetric.

9.2 Spontaneous symmetry breaking

In this Section we will see that the linear σ-model with µ2 < 0, is just but an exampleof a general phenomenon which goes under the name of spontaneous symmetrybreaking of symmetry. This phenomenon lies at the basis of the modern descriptionof phase transitions and it has acquired a capital relevance in the last years in allfield of physics. The idea is very simple and consists in the observation that atheory with hamiltonian invariant under a symmetry group may not show explicitlythe symmetry at the level of the solutions. As we shall see this may happen whenthe 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 nontrivial way under the symmetry group.

Just as an example consider a scalar field described by a lagrangian invariant underparity

P : φ→ −φ (9.59)

The lagrangian density will be of the type

L =1

2∂µφ∂

µφ− V (φ2) (9.60)

If the vacuum state is non degenerate, barring a phase factor, we must have

P |0〉 = |0〉 (9.61)

since P commutes with the hamiltonian. It follows

〈0|φ|0〉 = 〈0|P−1PφP−1P |0〉 = 〈0|PφP−1|0〉 = −〈0|φ|0〉 (9.62)

219

from which〈0|φ|0〉 = 0 (9.63)

Things change if the fundamental state is degenerate. This would be the case in theexample (9.60), if

V (φ2) =µ2

2φ2 +

λ

4φ4 (9.64)

with µ2 < 0. In fact, this potential has two minima located at

φ = ±v, v =

√−µ

2

λ(9.65)

By denoting with |R〉 e |L〉 the two states corresponding to the classical configura-tions φ = ±v, we have

P |R〉 = |L〉 = |R〉 (9.66)

Therefore〈R|φ|R〉 = 〈R|P−1PφP−1P |R〉 = −〈L|φ|L〉 (9.67)

which now does not imply that the expectation value of the field vanishes. In thefollowing we will be rather interested in the case of continuous symmetries. So letus consider two scalar fields, and a lagrangian density with symmetry O(2)

L =1

2∂µφ · ∂µφ− 1

2µ2φ · φ− λ

4(φ · φ)2 (9.68)

whereφ · φ = φ2

1 + φ22 (9.69)

For µ2 > 0 there is a unique fundamental state (minimum of the potential) φ = 0,whereas for µ2 < 0 there are infinite degenerate states given by

|φ|2 = φ21 + φ2

2 = v2 (9.70)

with v defined as in (9.65). By denoting with R(θ) the operator rotating the fieldsin the plane (φ1, φ2), in the non-degenerate case we have

R(θ)|0〉 = |0〉 (9.71)

and〈0|φ|0〉 = 〈0|R−1RφR−1R|0〉 = 〈0|φθ|0〉 = 0 (9.72)

since φθ = φ. In the case µ2 < 0 (degenerate case), we have

R(θ)|0〉 = |θ〉 (9.73)

where |θ〉 is one of the infinitely many degenerate fundamental states lying on the

circle |φ|2 = v2. Then

〈0|φi|0〉 = 〈0|R−1(θ)R(θ)φiR−1(θ)R(θ)|0〉 = 〈θ|φθi |θ〉 (9.74)

220

withφθi = R(θ)φiR

−1(θ) = φi (9.75)

Again, the expectation value of the field (contrarily to the non-degenerate state)does not need to vanish. The situation can be described qualitatively saying that theexistence of a degenerate fundamental state forces the system to choose one of theseequivalent states, and consequently to break the symmetry. But the breaking is onlyat the level of the solutions, the lagrangian and the equations of motion preservethe symmetry. One can easily construct classical systems exhibiting spontaneoussymmetry breaking. For instance, a classical particle in a double-well potential.This system has parity invariance x → −x, where x is the particle position. Theequilibrium positions are around the minima positions, ±x0. If we put the particleclose to x0, it will perform oscillations around that point and the original symmetryis lost. A further example is given by a ferromagnet which has an hamiltonianinvariant under rotations, but below the Curie temperature exhibits spontaneousmagnetization, breaking in this way the symmetry. These situations are typical forthe so called second order phase transitions. One can describe them through theLandau free-energy, which depends on two different kind of parameters:

• Control parameters, as µ2 for the scalar field, and the temperature for theferromagnet.

• Order parameters, as the expectation value of the scalar field or as themagnetization.

The system goes from one phase to another varying the control parameters, and thephase transition is characterized by the order parameters which assume differentvalues in different phases. In the previous examples, the order parameters were zeroin the symmetric phase and different from zero in the broken phase.

The situation is slightly more involved at the quantum level, since spontaneoussymmetry breaking cannot happen in finite systems. This follows from the existenceof the tunnel effect. Let us consider again a particle in a double-well potential, andrecall that we have defined the fundamental states through the correspondence withthe classical minima

x = x0 → |R〉x = −x0 → |L〉 (9.76)

But the tunnel effect gives rise to a transition between these two states and as aconsequence it removes the degeneracy. In fact, due to the transition the hamiltonianacquires a non zero matrix element between the states |R〉 and |L〉. By denotingwith H the matrix of the hamiltonian between these two states, we get

H =

[ε0 ε1ε1 ε0

](9.77)

221

The eigenvalues of H are(ε0 + ε1, ε0 − ε1) (9.78)

We have no more degeneracy and the eigenstates are

|S〉 =1√2(|R〉 + |L〉) (9.79)

with eigenvalue ES = ε0 + ε1, and

|A〉 =1√2(|R〉 − |L〉) (9.80)

with eigenvalue EA = ε0 − ε1. One can show that ε1 < 0 and therefore the funda-mental state is the symmetric one, |S〉. This situation gives rise to the well knowneffect of quantum oscillations. We can express the states |R〉 and |L〉 in terms ofthe energy eigenstates

|R〉 = 1√2(|S〉 + |A〉)

|L〉 = 1√2(|S〉 − |A〉) (9.81)

Let us now prepare a state, at t = 0, by putting the particle in the right minimum.This is not an energy eigenstate and its time evolution is given by

|R, t〉 =1√2

(e−iESt|S〉 + e−iEAt|A〉

)=

1√2e−iESt

(|S〉 + e−it∆E |A〉

)(9.82)

with ∆E = EA − ES. Therefore, for t = π/∆E the state |R〉 transforms into thestate |L〉. The state oscillates with a period given by

T =2π

∆E(9.83)

In nature there are finite systems as sugar molecules, which seem to exhibit sponta-neous symmetry breaking. In fact one observes right-handed and left-handed sugarmolecules. The explanation is simply that the energy difference ∆E is so small thatthe oscillation period is of the order of 104 − 106 years.

The splitting of the fundamental states decreases with the height of the potentialbetween two minima, therefore, for infinite systems, the previous mechanism doesnot work, and we may have spontaneous symmetry breaking. In fact, coming backto the scalar field example, its expectation value on the vacuum must be a constant,as it follows from the translational invariance of the vacuum

〈0|φ(x)|0〉 = 〈0|eiPxφ(0)e−iPx|0〉 = 〈0|φ(0)|0〉 = v (9.84)

and the energy difference between the maximum at φ = 0, and the minimum atφ = 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 (9.85)

222

9.3 The Goldstone theorem

From our point of view, the most interesting consequence of spontaneous symmetrybreaking is the Goldstone theorem. This theorem says that for any continuoussymmetry spontaneously broken, there exists a massless particle (the Goldstoneboson). The theorem holds rigorously in a local field theory, under the followinghypotheses

• 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 classical scalar fieldtheory. Let us start considering the lagrangian for the linear σ-model with invarianceO(N)

L =1

2∂µφ · ∂µφ− µ2

2φ · φ− λ

4(φ · φ)2 (9.86)

The conditions that V must satisfy in order to have a minimum are

∂V

∂φl= µ2φl + λφl|φ|2 = 0 (9.87)

with solutions

φl = 0, |φ|2 = v2, v =

√−µ2

λ(9.88)

The character of the stationary points can be studied by evaluating the secondderivatives

∂2V

∂φl∂φm= δlm(µ2 + λ|φ|2) + 2λφlφm (9.89)

We have two possibilities

• µ2 > 0, we have only one real solution given by φ = 0, which is a minimum,since

∂2V

∂φl∂φm= δlmµ

2 > 0 (9.90)

• µ2 < 0, there are infinite solutions, among which φ = 0 is a maximum. Thepoints of the sphere |φ|2 = v2 are degenerate minima. In fact, by choosingφl = vδlN as a representative point, we get

∂2V

∂φl∂φm= 2λv2δlNδmN > 0 (9.91)

223

Expanding the potential around this minimum we get

V (φ) ≈ V∣∣∣minimum

+1

2

∂2V

∂φl∂φm

∣∣∣minimum

(φl − vδlN)(φm − vδmN ) (9.92)

If we are going to make a perturbative expansion, the right fields to be used areφl − vδlN , and their mass is just given by the coefficient of the quadratic term

M2lm =

∂2V

∂φl∂φm

∣∣∣minimo

= −2µ2δlNδmN =

⎡⎢⎢⎣

0 0 · 00 0 · 0· · · ·0 0 · −2µ2

⎤⎥⎥⎦ (9.93)

Therefore the masses of the fields φa, a = 1, · · · , N − 1 , and χ = φN − v, are givenby

m2φa

= 0, m2χ = −2µ2 (9.94)

By definingm2 = −2µ2 (9.95)

we can write the potential as a function of the new fields

V = −m4

16λ+

1

2m2χ2 +

√m2λ

2χ

(N−1∑a=1

φ2a + χ2

)+λ

4

(N−1∑a=1

φ2a + χ2

)2

(9.96)

In this form the original symmetry O(N) is broken. However a residual symmetryO(N − 1) is left. In fact, V depends only on the combination

∑N−1a=1 φ

2a, and it is

invariant under rotations around the axis we have chosen as representative for thefundamental state, (0, · · · , v). It must be stressed that this is not the most generalpotential invariant under O(N − 1). In fact the most general potential (up to thefourth order in the fields) describing N scalar fields with a symmetry O(N − 1)would depend on 7 coupling constants, whereas the one we got depends only on thetwo parameters m and λ. Therefore spontaneous symmetry breaking puts heavyconstraints on the dynamics of the system. We have also seen that we have N − 1massless scalars. Clearly the rotations along the first N − 1 directions leave thepotential invariant, whereas the N − 1 rotations on the planes a − N move awayfrom the surface of the minima. This can be seen also in terms of generators. Sincethe field we have chosen as representative of the ground state is φi|min = vδiN , wehave

T abij φj|min = i(δai δbj − δbi δ

aj )vδjN = 0 (9.97)

since a, b = 1, · · · , N − 1, and

T aNij φj |min = i(δai δNj − δNi δ

aj )vδjN = ivδai = 0 (9.98)

Therefore we have N − 1 broken symmetries and N − 1 massless scalars. Thegenerators of O(N) divide up naturally in the generators of the vacuum symmetry

224

(here O(N−1)), and in the so called broken generators, each of them correspondingto a massless Goldstone boson. In general, if the original symmetry group G of thetheory is spontaneously broken down to a subgroup H (which is the symmetry ofthe vacuum), the Goldstone bosons correspond to the generators of G which are leftafter subtracting the generators of H . Intuitively one can understand the origin ofthe massless particles noticing that the broken generators allow transitions from apossible vacuum to another. Since these states are degenerate the operation doesnot cost any energy. From the relativistic dispersion relation this implies that wemust have massless particles. One can say that Goldstone bosons correspond to flatdirections in the potential.

9.4 The Higgs mechanism

We have seen that the spontaneous symmetry breaking mechanism, in the case ofcontinuous symmetry leads to massless scalar particles, the Goldstone bosons. Alsogauge theories lead to massless vector bosons, in fact, as in the electromagnetic case,gauge invariance forbids the presence in the lagrangian of terms quadratic in thefields. Unfortunately in nature the only massless particles we know are the photonand perhaps the neutrinos, which however are fermions. But once one couplesspontaneous symmetry breaking to a gauge symmetry, things change. In fact, if welook back at the hypotheses underlying a gauge theory, it turns out that Goldstonetheorem does not hold in this context. The reason is that it is impossible to quantizea gauge theory in a way which is at the same time manifestly covariant and has aHilbert space with positive definite metric. This is well known already for theelectromagnetic field, where one has to choose the gauge before quantization. Whathappens is that, if one chooses a physical gauge, as the Coulomb gauge, in orderto have a Hilbert space spanned by only the physical states, than the theory loosesthe manifest covariance. If one goes to a covariant gauge, as the Lorentz one, thetheory is covariant but one has to work with a big Hilbert space, with non-definitepositive metric, and where the physical states are extracted through a supplementarycondition. The way in which the Goldstone theorem is evaded is that the Goldstonebosons disappear, and, at the same time, the gauge bosons corresponding to thebroken symmetries acquire mass. This is the famous Higgs mechanism.

Let us start with a scalar theory invariant under O(2)

L =1

2∂µφ · ∂µφ− µ2

2φ · φ− λ

4(φ · φ)2 (9.99)

and let us analyze the spontaneous symmetry breaking mechanism. If µ2 < 0 thesymmetry is broken and we can choose the vacuum as the state

φ = (v, 0), v =

√−µ2

λ(9.100)

225

After the translation φ1 = χ+v, with 〈0|χ|0〉 = 0, we get the potential (m2 = −2µ2)

V = −m4

16λ+

1

2m2χ2 +

√m2λ

2χ(φ2

2 + χ2) +λ

4(φ2

2 + χ2)2 (9.101)

In this case the group O(2) is completely broken (except for the discrete symmetryφ2 → −φ2). The Goldstone field is φ2 . This has a peculiar way of transformingunder O(2). In fact, the original fields transform as

δφ1 = −αφ2, δφ2 = αφ1 (9.102)

from whichδχ = −αφ2, δφ2 = αχ+ αv (9.103)

We see that the Goldstone field undergoes a rotation plus a translation, αv. Thisis the main reason for the Goldstone particle to be massless. In fact one can haveinvariance under translations of the field, only if the potential is flat in the corre-sponding direction. This is what happens when one moves in a way which is tangentto the surface of the degenerate vacuums (in this case a circle). How do things changeif our theory is gauge invariant? In that case we should have invariance under atransformation of the Goldstone field given by

δφ2(x) = α(x)χ(x) + α(x)v (9.104)

Since α(x) is an arbitrary function of the space-time point, it follows that we canchoose it in such a way to make φ2(x) vanish. In other words it must be possibleto eliminate 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

(9.105)

Under a finite rotation, the new fields transform as

ρ→ ρ, θ → θ + α (9.106)

It should be also noticed that the two coordinate systems coincide when we are closeto the vacuum, as when we are doing perturbation theory. In fact, in that case wecan perform the following expansion

ρ =√φ2

2 + χ2 + 2χv + v2 ≈ v + χ, θ ≈ φ2

v + χ≈ φ2

v(9.107)

Again, if we make the theory invariant under a local transformation, we will haveinvariance under

θ(x) → θ(x) + α(x) (9.108)

226

By choosing α(x) = −θ(x) we can eliminate this last field from the theory. The onlyremaining degree of freedom in the scalar sector is ρ(x).

Let us study the gauging of this model. It is convenient to introduce complexvariables

φ =1√2(φ1 + iφ2), φ† =

1√2(φ1 − iφ2) (9.109)

The O(2) transformations become phase transformations on φ

φ→ eiαφ (9.110)

and the lagrangian (9.99) can be written as

L = ∂µφ†∂µφ− µ2φ†φ− λ(φ†φ)2 (9.111)

We know that it is possible to promote a global symmetry to a local one by intro-ducing the covariant derivative

∂µφ→ (∂µ − igAµ)φ (9.112)

from which

L = (∂µ + igAµ)φ†(∂µ − igAµ)φ− µ2φ†φ− λ(φ†φ)2 − 1

4FµνF

µν (9.113)

In terms of the polar coordinates (ρ, θ) we have

φ =1√2ρeiθ, φ† =

1√2ρe−iθ (9.114)

By performing the following gauge transformation on the scalars

φ→ φ′ = φe−iθ (9.115)

and the corresponding transformation on the gauge fields

Aµ → A′µ = Aµ − 1

g∂µθ (9.116)

the lagrangian will depend only on the fields ρ and A′µ (we will put again A′

µ = Aµ)

L =1

2(∂µ + igAµ)ρ(∂

µ − igAµ)ρ− µ2

2ρ2 − λ

4ρ4 − 1

4FµνF

µν (9.117)

In this way the Goldstone boson disappears. We have now to translate the field ρ

ρ = χ+ v, 〈0|χ|0〉 = 0 (9.118)

227

and we see that this generates a bilinear term in Aµ, coming from the covariantderivative, given by

1

2g2v2AµA

µ (9.119)

Therefore the gauge field acquires a mass

m2A = g2v2 (9.120)

It is instructive to count the degrees of freedom before and after the gauge trans-formation. Before we had 4 degrees of freedom, two from the scalar fields and twofrom the gauge field. After the gauge transformation we have only one degree offreedom from the scalar sector, but three degrees of freedom from the gauge vector,because now it is a massive vector field. The result looks a little bit strange, butthe reason why we may read clearly the number of degrees of freedom only after thegauge transformation is that before the lagrangian contains a mixing term

Aµ∂µθ (9.121)

between the Goldstone field and the gauge vector which makes complicate to readthe mass of the states. The previous gauge transformation realizes the purpose ofmaking that term vanish. The gauge in which such a thing happens is called theunitary gauge.

We will consider now the further example of a symmetry O(N). The lagrangianinvariant under local transformations is

L =1

2(Dµ)ijφj(D

µ)ikφk − µ2

2φiφi − λ

4(φiφi)

2 (9.122)

where(Dµ)ij = δij∂µ + i

g

2(TAB)ijW

ABµ (9.123)

where (TAB)lm = i(δAl δBm − δAmδ

Bl ). In the case of broken symmetry (µ2 < 0), we

choose again the vacuum along the direction N , with v defined as in (9.100)

φi = vδiN (9.124)

Recalling that

T abij φj

∣∣∣min

= 0, T aNij φj

∣∣∣min

= ivδai , a, b = 1, · · · , N − 1 (9.125)

the mass term for the gauge field is given by

−1

8g2TABij φj

∣∣∣min

(TCD)ikφk

∣∣∣minWABµ W µCD

= −1

2g2T aNij φj

∣∣∣min

(T bN)ikφk

∣∣∣minW aNµ W µbN

=1

2g2v2δai δ

biW

aNµ W µbN =

1

2g2v2W aN

µ W µaN (9.126)

228

Therefore, the fields W aNµ associated to the broken directions T aN acquire a mass

g2v2/2, whereas W abµ , associated to the unbroken symmetry O(N−1), remain mass-

less.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 in two categories. In the firstcategory fall the broken generators lying in GW ; they have associated massive vectorbosons. In the second category fall the other broken generators; they have associatedmassless Goldstone bosons. Finally the gauge fields associated to generators of GW

lying in H remain massless. From the previous derivation this follows noticing thatthe generators of H annihilate the minimum of the fields, leaving the correspondinggauge bosons massless, whereas the non zero action of the broken generators generatea mass term for the other gauge fields.

The situation is represented in Fig. 9.4.1.

G

H G W

Fig. 6.1 -This figure shows the various groups, G, the global symmetry of thelagrangian, H ∈ G, the symmetry of the vacuum, and GW , the group of local sym-metries. The broken generators in GW correspond to massive vector bosons. Thebroken generators do not belonging to GW correspond to massless Goldstone bosons.Th unbroken generators in GW correspond to massless vector bosons.

We can now show how to eliminate the Goldstone bosons. In fact we can definenew fields ξa and χ as

φi =

(e−iT aNξa

)iN

(χ+ v) (9.127)

where a = 1, · · · , N − 1, that is the sum is restricted to the broken directions. Theother degree of freedom is in the other factor. The correspondence among the fields

229

φ and (ξa, χ) can be seen easily by expanding around the vacuum(e−iT aN ξa

)iN

≈ δiN − i(T aN )iNξa = δiN + δai ξa (9.128)

from whichφi ≈ (vξa, χ+ v) (9.129)

showing that the ξa’s are really the Goldstone fields. The unitary gauge is definedthrough the transformation

φi →(eiT

aNξa)ij

φj = δiN(χ + v) (9.130)

Wµ → eiTaNξaWµe

−iT aN ξa − i

g

(∂µe

iT aNξa)e−iT aN ξa (9.131)

This transformation eliminates the Goldstone degrees of freedom and the resultinglagrangian depends on the field χ, on the massive vector fields W aN

µ and on themassless field W ab

µ . Notice again the counting of the degrees of freedom N+2N(N−1)/2 = N2 in a generic gauge, and 1 + 3(N − 1) + 2(N − 1)(N − 2)/2 = N2 in theunitary gauge.

9.5 Quantization of a spontaneously broken gauge

theory in the Rξ-gauge

In the previous Section we have seen that in the unitary gauge some of the gaugefields acquire a mass. The corresponding quadratic part of the lagrangian will be(we consider here the abelian case corresponding to a symmetry U(1) or O(2))

L = −1

4FµνF

µν +m2

2W 2 (9.132)

where Wµ is the massive gauge field (called also a Yang-Mills massive field). Thefree equations of motion are

( +m2)Wµ − ∂µ(∂νWν) = 0 (9.133)

Let us study the solutions of this equation in momentum space. We define

Wµ(x) =

∫d4kWµ(k)e

−ikx (9.134)

It is convenient to introduce a set of four independent four-vectors. By puttingkµ = (E,k) we introduce

εµ(i) = (0, ni), i = 1, 2

εµ(3) =1

m

(|k|, E

k

|k|

)(9.135)

230

withk · ni = 0, ni · nj = δij (9.136)

Then we can decompose Wµ(k) as follows

Wµ(k) = εµ(λ)aλ(k) + kµb(k) (9.137)

Since kµεµ(λ) = 0, from the equation of motion we get

(−k2 +m2)(εµ(λ)aλ(k) + kµb(k)) + kµk2b(k) = 0 (9.138)

implying(k2 −m2)aλ(k) = 0, m2b(k) = 0 (9.139)

Therefore, for m = 0 the field has three degrees of freedom with k2 = m2, corre-sponding to transverse and longitudinal polarization. When the momentum is onshell, we can define a fourth unit vector

εµ(0) =kµ

m(9.140)

In this way we have four independent unit vectors which satisfy the completenessrelation

3∑λ,λ′=0

εµ(λ)εν(λ′)gλλ′ = gµν (9.141)

Therefore the sum over the physical degrees of freedom gives

3∑λ=1

εµ(λ)εν(λ) = −gµν +kµkν

m2(9.142)

The propagator is defined by the equation[( +m2) − ∂µ∂ν

]Gνλ(x) = igλµδ

4(x) (9.143)

Solving in momentum space we find

Gµν(k) =

∫d4k

(2π)4e−ikx i

k2 −m2 + iε

(−gµν +

kµkν

m2

)(9.144)

Therefore, the corresponding Feynman’s rule for the massive gauge field propagatoris

− i

k2 −m2 + iε

(gµν − kµkν

m2

)(9.145)

Notice that the expression in parenthesis is nothing but the projector over the phys-ical states given in eq. (9.142). This propagator has a very bad high-energy be-haviour. In fact it goes to constant for k → ∞. As a consequence the argument

231

for the renormalizability of the theory based on power counting does not hold anymore. Although in the unitary gauge a spontaneously broken gauge theory does notseem to be renormalizable, we have to take into account that the gauge invarianceallows us to make different choices of the gauge. In fact we will show that thereexist an entire class of gauges where the theory is renormalizable by power counting.We will discuss this point only for the abelian case, with gauge group O(2), but theargument cab be easily extended to the non-abelian case. Let us consider the O(2)theory as presented in Section 9.4. Let us recall that under a transformation of O(2)the two real scalar field of the theory transform as

δφ1 = −αφ2, δφ2 = αφ1 (9.146)

Therefore, in order to make it a gauge theory we introduce covariant derivatives

Dµφ1 = ∂µφ1 + gWµφ2

Dµφ2 = ∂µφ2 − gWµφ1 (9.147)

with Wµ transforming under a local transformation as

δWµ(x) =1

g∂µα(x) (9.148)

Assuming that the spontaneous breaking occurs along the direction φ1, we introducethe fields

φ1 = v + χ, φ2 = φ (9.149)

with

v2 = −µ2

λ(9.150)

Remember that χ is the Higgs field, whereas φ is the Goldstone field. In terms ofthe new variables the lagrangian is

L =1

2(∂µχ+ gWµφ)2 +

1

2(∂µφ− gWµχ− gvWµ)

2 − V (χ, φ) − 1

4FµνF

µν (9.151)

where, the potential up to quadratic terms is given by (see eq. (9.101))

V (χ, φ) = −m4

16λ+

1

2m2χ2 + · · · (9.152)

with m2 = −2µ2. We see that the effect of the shift on φ1 is to produce the terms

1

2g2v2W 2 − gvWµ∂

µφ+ g2vW 2χ (9.153)

The last term is an interaction term and it does not play any role in the followingconsiderations. The first term is the mass term for the W

m2W = g2v2 (9.154)

232

whereas the second term is the mixing between the W and the Goldstone field, thatwe got rid of by choosing the unitary gauge. However this choice is not unique. Infact let us consider a gauge fixing of the form

f − B = 0, f = ∂µWµ + gvξφ (9.155)

where ξ is an arbitrary parameter and B(x) is an arbitrary function. By proceedingas in Section 8.3 we know that the effect of such a gauge fixing is to add to thelagrangian a term

− 1

2βf 2 = − 1

2β(∂µWµ + gvξφ)2 = − 1

2β(∂µWµ)

2 − ξ

βgv∂µWµφ− ξ2

2βg2v2φ2 (9.156)

The first term is the usual covariant gauge fixing. The second term, after integrationby parts reads

+ξ

βgvWµ∂

µφ (9.157)

and we see that with the choiceβ = ξ (9.158)

it cancels the unwanted mixing term. Finally the third contribution

−1

2ξg2v2φ2 (9.159)

is a mass term for the Goldstone field

m2φ = ξg2v2 = ξm2

W (9.160)

However this mass is gauge dependent (it depends on the arbitrary parameter ξ)and this is a signal of the fact that in this gauge the Goldstone field is fictitious.In fact remember that it disappears in the unitary gauge. We will see later, in anexample, that evaluating a physical amplitude the contribution of the Goldstonedisappears, as it should be. Finally we have to consider the ghost contribution fromthe Faddeev-Popov determinant. For this purpose we need to know the variation ofthe gauge fixing f under an infinitesimal gauge transformation

δf(x) =1

gα(x) + gvξα(x)(χ(x) + v) (9.161)

givingδf(x)

δα(y)=

1

gδ4(x− y) + gvξ(χ(x) + v)δ4(x− y) (9.162)

Therefore the ghost lagrangian can be written as

c∗c + ξg2v2c∗c + ξg2vχc∗c (9.163)

Notice that in contrast with QED, the ghosts cannot be ignored, since they are cou-pled to the Higgs field. We are now in the position of reading the various propagatorsfrom the quadratic part of the lagrangian

233

• Higgs propagator (χ):i

k2 −m2

• Goldstone propagator (φ):i

k2 − ξm2W

• Ghost propagator (c): − i

k2 − ξm2W

The propagator for the vector field is more involved. let us collect the variousquadratic terms in Wµ

−1

4FµνF

µν +1

2m2WW

2 − 1

2ξ(∂µWµ)

2 (9.164)

In momentum space wave operator is

(gµνk2 − kµkν) −m2W g

µν +1

ξkµkν (9.165)

which can be rewritten as(gµν − kµkν

k2

)(k2 −m2

W ) +kµkν

k2

1

ξ(k2 − ξm2

W ) (9.166)

The propagator in momentum space is the inverse of the wave operator (except fora factor −i). To invert the operator we notice that it is written as a combination oftwo orthogonal projection operators P1 and P2

αP1 + βP2 (9.167)

The inverse of such an operator is simply

1

αP1 +

1

βP2 (9.168)

Therefore the W propagator is given by

− i

k2 −m2W

(gµν − kµkν

k2

)− iξ

k2 − ξm2W

kµkν

k2

= − i

k2 −m2W

(gµν − kµkν

k2 − ξm2W

(1 − ξ)

)(9.169)

We see that for this class of gauges the asymptotic behaviour of the propagator is1/k2 and we can apply the usual power counting. Notice that the unitary gaugeis recovered in the limit ξ → ∞. In this limit the W -propagator becomes the onediscussed at the beginning of the Section. Of course, the singularity of the limitdestroys the asymptotic behaviour that one has for finite values of ξ. Also, forξ → ∞ the ghost and the Goldstone fields decouple since their mass goes to infinity.

234

To end this Section we observe that with this approach one is able to prove at thesame time that the theory is renormalizable and that it does not contain spuriousstates. In fact the theory is renormalizable for finite values of ξ and it does notcontain spurious states for ξ → ∞. But, since the gauge invariant quantities do notdepend on the gauge fixing, they do not depend in particular on ξ, and thereforeone gets the wanted result.

9.6 ξ-cancellation in perturbation theory

In this Section we will discuss the cancellation of the ξ dependent terms in thecalculation of a simple amplitude at tree level. To this end we consider again theO(2) model of the previous Section, and we add to the theory a fermion field. Wewill also mimic the standard model by making transform under the gauge grouponly the left-handed part of the fermion field. That is we decompose ψ as

ψ = ψL + ψR (9.170)

where

ψL =1 − γ5

2ψ, ψR =

1 + γ5

2ψ (9.171)

and we assume the following transformation under the gauge group

ψL → eiαψL, ψR → ψR (9.172)

A fermion mass term would destroy the gauge invariance, since it couples left-handedand right-handed fields. In fact

ψL =

(1 − γ5

2

)ψ = ψ† 1 − γ5

2γ0 = ψ

1 + γ5

2(9.173)

and therefore

ψψ = ψ

(1 + γ5

2+

1 − γ5

2

)ψ = ψLψR + ψRψL (9.174)

Therefore we will introduce a Yukawa coupling between the scalar field and thefermions, in such a way to be gauge invariant and to gives rise to a mass term forthe fermion when the symmetry is spontaneously broken. We will write

Lf = ψLiDψL + ψRi∂ψR − λf(ψLψRΦ + ψRψLΦ∗) (9.175)

withDµ = ∂µ − igWµ (9.176)

Here we have used again the complex notation for the scalar field, as in eq. (9.109),and we remind that it transforms as

Φ → eiαΦ (9.177)

235

Therefore Lf is gauge invariant. The lagrangian can be rewritten in terms of thefield ψ by writing explicitly the chiral projectors (1 ± γ5)/2. We get

Lf = ψi∂1 − γ5

2ψ + ψi∂

1 + γ5

2ψ + gψγµ

1 − γ5

2ψW µ

− λf√2

(ψ

1 + γ5

2ψ(φ1 + iφ2) + ψ

1 − γ5

2ψ(φ1 − iφ2)

)(9.178)

from which

L = ψi∂ψ + gψγµ1 − γ5

2ψW µ − λf√

2

(ψψ(v + χ) + iψγ5ψφ

)(9.179)

where χ+ v = φ1 and φ = φ2. This expression shows that the fermion field acquiresa mass through the Higgs mechanism given by

mf =λfv√

2(9.180)

The previous expression shows also that the Higgs field is a scalar (parity +1),whereas the Goldstone field is a pseudoscalar (parity= -1). Consider now thefermion-fermion scattering in this model at tree level. The amplitude gets con-tribution from the diagrams of Fig. 9.6.1.

.

p

p' k

k'

W

q

φ χ+ +

Fig. 9.6.1 - The fermion-fermion scattering in the U(1) Higgs model.

We want to show that the ξ contribution cancels out. The ξ dependence arises fromthe gauge field and the Goldstone field exchange. Therefore we will consider onlythese two contributions. Let us start with the Goldstone exchange. The Feynman’srule for the vertex is given in Fig. 9.6.2.

.

φ

Fig. 9.6.2 - Goldstone-fermion vertex:λf√

2γ5

236

The corresponding amplitude is

Mφ =

(λf√

2

)2

u(p′)γ5u(p)i

q2 − ξm2W

u(k′)γ5u(k) (9.181)

where q = p− p′ = k′ − k. The Feynman rule for the W-fermion vertex is given inFig. 9.6.3.

.

Fig. 9.6.3 - W-fermion vertex: igγµ1 − γ5

2

The corresponding amplitude is

MW = (ig)2u(p′)γµ1 − γ5

2u(p)

× −iq2 −m2

W

(gµν − qµqν

q2 − ξm2W

(1 − ξ)

)u(k′)γν

1 − γ5

2u(k) (9.182)

It is convenient to isolate the ξ dependence by rewriting the gauge field propagatoras

−iq2 −m2

W

[gµν − qµqν

m2W

+ qµqν(

1

m2W

− 1 − ξ

q2 − ξm2W

)]

= =−i

q2 −m2W

(gµν − qµqν

m2W

)+

−iq2 − ξm2

W

qµqν

m2W

(9.183)

The first term is ξ independent and it is precisely the W propagator in the unitarygauge. The second term can be simplified by using the identity

qµu(p′)γµ1 − γ5

2u(p) = (p− p′)µu(p′)γµ

1 − γ5

2u(p) =

= u(p′)(

1 + γ5

2p− p′

1 − γ5

2

)u(p) = mf u(p

′)γ5u(p) (9.184)

By using this relation we have

qµqν u(p′)γµ1 − γ5

2u(p)u(k′)γµ

1 − γ5

2u(k) =

= (mf)u(p′)γ5u(p)(−mf)u(k

′)γ5u(k)

= −(λf√

2

)2

v2u(p′)γ5u(p)u(k′)γ5u(k) (9.185)

237

where we have made use of eq. (9.180). Substituting this result in MW we get

MW = (ig)2u(p′)γµ1 − γ5

2u(p)

−iq2 −m2

W

(gµν − qµqν

m2W

)u(k′)γµ

1 − γ5

2u(k)

+−i

q2 − ξm2W

g2v2

m2W

(λf√

2

)2

u(p′)γ5u(p)u(k′)γ5u(k) (9.186)

Finally, using m2W = g2v2 we see that the last term cancels exactly the Mφ ampli-

tude. Therefore the result is the same that we would have obtained in the unitarygauge where there is no Goldstone contribution.

Now let us make some comment about the result. In QED we have used thefact that the qµqν terms in the propagator do not play a physical role,since theyare contracted with the fermionic current which is conserved. In the present casethis does not happens, but we get a term proportional to the fermion mass. Infact the fermionic current ψγµ(1 − γ5)/2ψ it is not conserved because the fermionacquires a mass through the spontaneous breaking of the symmetry. On the otherhand, the term that one gets in this way it is just the one necessary to give rise tothe cancellation with the Goldstone boson contribution. It is also worth to comparewhat happens for different choices of ξ. For ξ = 0 the gauge propagator is transverse

−ik2 −m2

W

(gµν − kµkν

k2

)(9.187)

This corresponds to the Landau gauge. The propagator has a pole term at k2 = 0,which however it is cancelled by the Goldstone pole which also is at k2 = 0. For thechoice ξ = 1 the gauge propagator has the very simple form

−igµνk2 −m2

W

(9.188)

This is the Feynman-’t Hooft gauge. Here the role of the Goldstone boson, whichhas a propagator

i

k2 −m2W

(9.189)

it is to cancel the contribution of the time-like component of the gauge field, whichis described by the polarization vector ε

(0)µ . In fact the gauge propagator can be

rewritten as

−igµνk2 −m2

W

=−i

k2 −m2W

3∑λ,λ′=0

ε(λ)µ ε(λ

′)ν gλλ′ =

i

k2 −m2W

3∑λ=1

ε(λ)µ ε(λ)

ν +−i

k2 −m2W

kµkν

m2W

(9.190)The first piece is the propagator in the unitary gauge, whereas the second piece iscancelled by the Goldstone boson contribution.

238

Chapter 10

The Standard model of theelectroweak interactions

10.1 The Standard Model of the electroweak in-

teractions

We will describe now the Weinberg Salam model for the electroweak interactions.Let us recall that the Fermi theory of weak interactions is based on a lagrangianinvolving four Fermi fields. This lagrangian can be written as the product of twocurrents which are the sum of various pieces.

LF =GF√

2J (+)µ Jµ(−) (10.1)

Limiting for the moment our considerations to the charged current for the electronand its neutrino (here ν ≡ νe)), the current is given by

J (+)µ = ψeγµ(1 − γ5)ψν , J (−)

µ = J (+)µ

†= ψνγµ(1 − γ5)ψe (10.2)

To understand the symmetry hidden in this couplings it is convenient to introducethe following spinors with 2 × 4 components

L =

((ψν)L(ψe)L

)=

1 − γ5

2

(ψνψe

)(10.3)

where we have introduced left-handed fields. The right-handed spinors, projectedout with (1 + γ5)/2, don’t feel the weak interaction. This is, in fact, the physicalmeaning of the V − A interaction. By using

L =((ψν)L, (ψe)L

)=(ψν , ψe

) 1 + γ5

2(10.4)

239

we can write the weak currents in the following way

J (+)µ = 2ψe

1 + γ5

2γµ

1 − γ5

2ψν = 2

((ψν)L, (ψe)L

)γµ

(0

(ψν)L

)

= 2((ψν)L, (ψe)L

)γµ

(0 01 0

)((ψν)L(ψe)L

)= 2Lγµτ−L (10.5)

where we have defined

τ± =τ1 ± iτ2

2(10.6)

the τi’s being the Pauli matrices. Analogously

J (−)µ = 2Lγµτ+L (10.7)

In the fifties the hypothesis of the intermediate vector bosons was advanced. Ac-cording to this idea the Fermi theory was to be regarded as the low energy limitof a theory where the currents were coupled to vector bosons, very much as QED.However, due to the short range of the weak interactions these vector bosons had tohave mass. The interaction among the vector bosons and the fermions is given by

Lint =g

2√

2

[J (+)µ W−µ + J (−)

µ W+µ]

=g√2Lγµ(τ−W−µ + τ+W

+µ)L (10.8)

The coupling g can be related to the Fermi constant through the relation

GF√2

=g2

8m2W

(10.9)

Introducing real fields in place of W±

W± =W1 ∓ iW2√

2(10.10)

we get

Lint =g

2Lγµ(τ1W

µ1 + τ2W

µ2 )L (10.11)

Let us now try to write the electromagnetic interaction within the same formalism

Lem = eψeQγµψeAµ = eψeQ

(1 + γ5

2γµ

1 − γ5

2+

1 − γ5

2γµ

1 + γ5

2

)ψeA

µ

= e[(ψe)LQγµ(ψe)L + (ψe)RQγµ(ψe)R]Aµ (10.12)

where

(ψe)R =1 + γ5

2ψe, (ψe)R = ψe

1 − γ5

2(10.13)

where Q = −1 is the electric charge of the electron, in unit of the electric charge ofthe proton, e. It is also convenient to denote the right-handed components of theelectron as

R = (ψe)R (10.14)

240

We can write

(ψe)Lγµ(ψe)L = Lγµ

(0 00 1

)L = Lγµ

1 − τ32

L (10.15)

and(ψe)Rγµ(ψe)R = RγµR (10.16)

from which, using Q = −1,

Lem = e

(−1

2LγµL− RγµR+ Lγµ

τ32L

)Aµ = ejem

µ Aµ (10.17)

From this equation we can write

jemµ = j3

µ +1

2jYµ (10.18)

withj3µ = Lγµ

τ32L (10.19)

andjYµ = −LγµL− 2RγµR (10.20)

These equations show that the electromagnetic current is a mixture of j3µ, which is

a partner of the charged weak currents, and of jYµ , which is another neutral current.The following notations unify the charged currents and j3

µ

jiµ = Lγµτi2L (10.21)

Using the canonical anticommutators for the Fermi fields, it is not difficult to provethat the charges associated to the currents

Qi =

∫d3x ji0(x) (10.22)

span the Lie algebra of SU(2),

[Qi, Qj ] = iεijkQk (10.23)

The charge QY associated to the current jYµ commutes with Qi, as it can be easilyverified noticing that the right-handed and left-handed fields commute with eachother. Therefore the algebra of the charges is

[Qi, Qj ] = iεijkQk, [Qi, QY ] = 0 (10.24)

This is the Lie algebra of SU(2) ⊗ U(1), since the Qi’s generate a group SU(2),whereas QY generates a group U(1), with the two groups commuting among them-selves. To build up a gauge theory from these elements we have to start with aninitial lagrangian possessing an SU(2) ⊗ U(1) global invariance. This will produce

241

4 gauge vector bosons. Two of these vectors are the charged one already introduce,whereas we would like to identify one of the the other two with the photon. Byevaluating the commutators of the charges with the fields we can read immediatelythe quantum numbers of the various particles. We have

[Lc, Qi] =

∫d3x [Lc, L

†a

(τi2

)abLb] =

(τi2

)cbLb (10.25)

and[R,Qi] = 0 (10.26)

[Lc, QY ] = −Lc, [R,QY ] = −2R (10.27)

These relations show that L transforms as an SU(2) spinor, or, as the representation2, where we have identified the representation with its dimensionality. R belongs tothe trivial representation of dimension 1. Putting everything together we have thefollowing behaviour under the representations of SU(2) ⊗ U(1)

L ∈ (2,−1), R ∈ (1,−2) (10.28)

The relation (10.18) gives the following relation among the different charges

Qem = Q3 +1

2QY (10.29)

As we have argued, if SU(2) ⊗ U(1) has to be the fundamental symmetry ofelectroweak interactions, we have to start with a lagrangian exhibiting this globalsymmetry. This is the free Dirac lagrangian for a massless electron and a left-handedneutrino

L0 = Li∂L+ Ri∂R (10.30)

Mass terms, mixing left- and right-handed fields, would destroy the global symmetry.For instance, a typical mass term gives

mψψ = mψ†[1 + γ5

2γ0

1 − γ5

2+

1 − γ5

2γ0

1 + γ5

2

]ψ = m(ψRψL + ψLψR) (10.31)

We shall see later that the same mechanism giving mass to the vector bosons can beused to generate fermion masses. A lagrangian invariant under the local symmetrySU(2) ⊗ U(1) is obtained through the use of covariant derivatives

L = Liγµ(∂µ − ig

τi2W iµ + i

g′

2Yµ

)L+ Riγµ (∂µ + ig′Y µ)R (10.32)

The interaction term is

Lint = gLγµ

(τ

2· W µ

)L− g′

2LγµLY

µ − g′RγµRY µ (10.33)

242

In this expression we recognize the charged interaction with W±. We have also, asexpected, two neutral vector bosons W 3 and Y . The photon must couple to theelectromagnetic current which is a linear combination of j3 and jY . The neutralvector bosons are coupled to these two currents, so we expect the photon field,Aµ, to be a linear combination of W 3

µ and Yµ. It is convenient to introduce twoorthogonal combination of these two fields, Zµ, and Aµ. The mixing angle can beidentified through the requirement that the current coupled to Aµ is exactly theelectromagnetic current with coupling constant e. Let us consider the part of Lint

involving the neutral couplings

LNC = gj3µW

µ3 +

g′

2jYµ Y

µ (10.34)

By putting (θ is called the Weinberg angle)

W3 = Z cos θ + A sin θ

Y = −Z sin θ + A cos θ (10.35)

we get

LNC =

[g sin θ j3

µ + g′ cos θ1

2jYµ

]Aµ +

[g cos θ j3

µ − g′ sin θ1

2jYµ

]Zµ (10.36)

The electromagnetic coupling is reproduced by requiring the two conditions

g sin θ = g′ cos θ = e (10.37)

These two relations allow us to express both the Weinberg angle and the electriccharge e in terms of g and g′.

In the low-energy experiments performed before the LEP era (< 1990) the fun-damental parameters used in the theory were e (or rather the fine structure constantα) and sin2 θ. We shall see in the following how the Weinberg angle can be elim-inated in favor of the mass of the Z, which is now very well known. By using eq.(10.37), and jem

µ = j3µ + 1

2jYµ , we can write

LNC = ejemµ Aµ + [g cos θj3

µ − g′ sin θ(jemµ − j3

µ)]Zµ (10.38)

Expressing g′ through g′ = g tan θ, the coefficient of Zµ can be put in the form

[g cos θ + g′ sin θ]j3µ − g′ sin θ jem

µ =g

cos θj3µ −

g

cos θsin2 θ jem

µ (10.39)

from which

LNC = ejemµ Aµ +

g

cos θ[j3µ − sin2 θ jem

µ ]Zµ ≡ ejemµ Aµ +

g

cos θjZµZ

µ (10.40)

where we have introduced the neutral current coupled to the Z

jZµ = j3µ − sin2 θ jem

µ (10.41)

243

The value of sin2 θ was evaluated initially from processes induced by neutral currents(as the scattering ν − e) at low energy. The approximate value is

sin2 θ ≈ 0.23 (10.42)

This shows that both g and g′ are of the same order of the electric charge. Noticethat the eq. (10.37) gives the following relation among the electric charge and thecouplings g and g′

1

e2=

1

g2+

1

g′2(10.43)

10.2 The Higgs sector in the Standard Model

According to the discussion made in the previous Section, we need three brokensymmetries in order to give mass to W± and Z. Since SU(2) ⊗ U(1) has four gen-erators, we will be left with one unbroken symmetry, that we should identify withthe group U(1) of the electromagnetism (U(1)em), in such a way to have the corre-sponding gauge particle (the photon) massless. Therefore, the group U(1)em mustbe a symmetry of the vacuum (that is the vacuum should be electrically neutral).To realize this aim we have to introduce a set of scalar fields transforming in a con-venient way under SU(2) ⊗ U(1). The simplest choice turns out to be a complexrepresentation of SU(2) of dimension 2 (Higgs doublet). As we already noticed thevacuum should be electrically neutral, so one of the components of the doublet mustalso be neutral. Assume that this component is the lower one, than we will write

Φ =

[φ+

φ0

](10.44)

To determine the weak hypercharge, QY , of Φ, we use the relation

Qem = Q3 +1

2QY (10.45)

for the lower component of Φ. We get

QY (φ0) = 2 ×[0 −

(−1

2

)]= 1 (10.46)

Since QY and Qi commute, both the components of the doublet must have the samevalue of QY , and using again eq. (10.45), we get

Qem(φ+) =1

2+

1

2= 1 (10.47)

which justifies the notation φ+ for the upper component of the doublet.

244

The most general lagrangian for Φ with the global symmetry SU(2)⊗U(1) andcontaining terms of dimension lower or equal to four (in order to have a renormal-izable theory) is

LHiggs = ∂µΦ†∂µΦ − µ2Φ†Φ − λ(Φ†Φ)2 (10.48)

where λ > 0 and µ2 < 0. The potential has infinitely many minima on the surface

|Φ|2minimum = −µ2

2λ=v2

2(10.49)

with

v2 = −µ2

λ(10.50)

Let us choose the vacuum as the state

〈0|Φ|0〉 =

[0

v/√

2

](10.51)

By putting

Φ =

[0

v/√

2

]+

[φ+

(h + iη)/√

2

]≡ Φ0 + Φ′ (10.52)

with〈0|Φ′|0〉 = 0 (10.53)

we get

|Φ|2 =1

2v2 + vh+ |φ+|2 +

1

2(h2 + η2) (10.54)

from which

VHiggs = −1

4

µ4

λ+ λv2h2 + 2λvh

(|φ+|2 +

1

2(h2 + η2)

)+ λ

(|φ+|2 +

1

2(h2 + η2)

)2

(10.55)From this expression we read immediately the particle masses (φ− = (φ+)†)

m2η = m2

φ+ = m2φ− = 0 (10.56)

andm2h = 2λv2 = −2µ2 (10.57)

It is also convenient to express the parameters appearing in the original form of thepotential in terms of m2

h e v (µ2 = −2m2h, λ = m2

h/2v2). In this way we get

VHiggs = −1

8m2hv

2+m2h

2h2+

m2h

vh

(|φ+|2 +

1

2(h2 + η2)

)+m2h

2v2

(|φ+|2 +

1

2(h2 + η2)

)2

(10.58)Summarizing we have three massless Goldstone bosons φ± and η, and a massivescalar h. This is called the Higgs field.

245

As usual, by now, we promote the global symmetry to a local one by introducingthe covariant derivative. Recalling that Φ ∈ (2, 1) of SU(2) ⊗ U(1), we have

DHµ = ∂µ − i

g

2τ · Wµ − i

g′

2Yµ (10.59)

andLHiggs = (DH

µ Φ)†DHµΦ − µ2Φ†Φ − λ(Φ†Φ)2 (10.60)

To study the mass generation it is convenient to write Φ in a way analogous to theone used in eq. (9.127)

Φ = eiξ · τ/v

[0

(v + h)/√

2

](10.61)

According to our rules the exponential should contain the broken generators. In factthere are τ1, and τ2 which are broken. Furthermore it should contain the combinationof 1 and τ3 which is broken (remember that (1 + τ3)/2 is the electric charge of thedoublet Φ, and it is conserved). But, at any rate, τ3 is a broken generator, so theprevious expression gives a good representation of Φ around the vacuum. In fact,expanding around this state

eiξ · τ/v

[0

(v + h)/√

2

]≈

[1 + iξ3/v i(ξ1 − iξ2)/v

i(ξ1 + iξ2)/v 1 − iξ3/v

] [0

(v + h)/√

2

]

≈ 1√2

[i(ξ1 − iξ2)

(v + h− iξ3)

](10.62)

and introducing real components for Φ (φ+ = (φ1 − iφ2)/√

2)

Φ =1√2

[φ1 − iφ2

v + h+ iη

](10.63)

we get the relation among the two sets of coordinates

(φ1, φ2, η, h) ≈ (ξ2,−ξ1,−ξ3, h) (10.64)

By performing the gauge transformation

Φ → e−iξ · τ/vΦ (10.65)

Wµ · τ2→ e−iξ · τ/v Wµ · τ

2eiξ · τ/v +

i

g

(∂µe

−iξ · τ/v)eiξ · τ/v (10.66)

Yµ → Yµ (10.67)

the Higgs lagrangian remains invariant in form, except for the substitution of Φ with(0, (v+h)/

√2). We will also denote old and new gauge fields with the same symbol.

Defining

Φd =

[01

], Φu =

[10

](10.68)

246

the mass terms for the scalar fields can be read by substituting, in the kinetic term,Φ with its expectation value

Φ → v√2Φd (10.69)

We get

−i(g

2τ · W +

g′

2Y

)v√2Φd = −i

(g√2(τ−W− + τ+W

+) +g

2τ3W

3 +g′

2Y

)v√2Φd

= −i v√2

[g√2W+Φu − 1

2(gW 3 − g′Y )Φd

](10.70)

Since Φd and Φu are orthogonals, the mass term is

v2

2

[g2

2|W+|2 +

1

4(gW 3 − g′Y )2

](10.71)

From eq. (10.37) we have tan θ = g′/g, and therefore

sin θ =g′√

g2 + g′2, cos θ =

g√g2 + g′2

(10.72)

allowing us to write the mass term in the form

v2

2

[g2

2|W+|2 +

g2 + g′2

4(W 3 cos θ − Y sin θ)2

](10.73)

From eq. (10.35) we see that the neutral fields combination is just the Z field,orthogonal to the photon. In this way we get

g2v2

4|W+|2 +

1

2

g2 + g′2

4v2Z2 (10.74)

Finally the mass of the vector bosons is given by

M2W =

1

4g2v2, M2

Z =1

4(g2 + g′2)v2 (10.75)

Notice that the masses of the W± and of the Z are not independent, since theirratio is determined by the Weinberg angle, or from the gauge coupling constants

M2W

M2Z

=g2

g2 + g′2= cos2 θ (10.76)

Let us now discuss the parameters that we have so far in the theory. The gaugeinteraction introduces the two gauge couplings g and g′, which can also be expressedin terms of the electric charge (or the fine structure constant), and of the Weinbergangle. The Higgs sector brings in two additional parameters, the mass of the Higgs,

247

mh, and the expectation value of the field Φ, v. The Higgs particle has not been yetdiscovered, and at the moment we have only an experimental lower bound on mh,from LEP, which is given by mh > 60 GeV . The parameter v can be expressed interms of the Fermi coupling constant GF . In fact, from eq. (10.9)

GF√2

=g2

8M2W

(10.77)

using the expression for M2W , we get

v2 =1√2GF

≈ (246 GeV )2 (10.78)

Therefore, the three parameters g, g′ and v can be traded for e, sin2 θ and GF .Another possibility is to use the mass of the Z

M2Z =

1

4

1√2GF

e2

sin2 θ cos2 θ=

πα√2GF sin2 θ cos2 θ

(10.79)

to eliminate sin2 θ

sin2 θ =1

2

[1 −

√1 − 4πα√

2GFM2Z

](10.80)

The first alternative has been the one used before LEP. However, after LEP1, themass of the Z is very well known

MZ = (91.1867 ± 0.0021) GeV (10.81)

Therefore, the parameters that are now used as input in the SM are α, GF and MZ .

The last problem we have to solve is how to give mass to the electron, since it isimpossible to construct bilinear terms in the electron field which are invariant underthe gauge group. The solution is that we can build up trilinear invariant terms inthe electron field and in Φ. Once the field Φ acquires a non vanishing expectationvalue, due to the breaking of the symmetry, the trilinear term generates the electronmass. Recalling the behaviour of the fields under SU(2) ⊗ U(1)

L ∈ (2,−1), R ∈ (1,−2), Φ ∈ (2,+1) (10.82)

we see that the following coupling (Yukawian coupling) is invariant

LY = geLΦR + h.c. = ge [ (ψν)L, (ψe)L ]

[φ+

φ0

](ψe)R + h.c. (10.83)

In the unitary gauge, which is obtained by using the transformation (10.65) bothfor Φ and for the lepton field L, we get

LY = ge [ (ψν)L, (ψe)L ]

[0

(v + h)/√

2

](ψe)R + h.c. (10.84)

248

from whichLY =

gev√2(ψe)L(ψe)R +

ge√2(ψe)L(ψe)Rh + h.c. (10.85)

Since

[(ψe)L(ψe)R]† = (ψe1 + γ5

2ψe)

† = ψ†e

1 + γ5

2γ0ψe = ψe

1 − γ5

2ψe = (ψe)R(ψe)L

(10.86)we get

LY =gev√

2ψeψe +

ge√2ψeψeh (10.87)

Therefore the symmetry breaking generates an electron mass given by

me = −gev√2

(10.88)

In summary, we have been able to reproduce all the phenomenological featuresof the V −A theory and its extension to neutral currents. This has been done with atheory that, before spontaneous symmetry breaking is renormalizable. In fact, alsothe Yukawian coupling has only dimension 3. The proof that the renormalizabilityof the theory holds also in the case of spontaneous symmetry breaking (µ2 < 0)is absolutely non trivial. In fact the proof was given only at the beginning of theseventies by ’t Hooft.

10.3 The electroweak interactions of quarks and

the Kobayashi-Maskawa-Cabibbo matrix

So far we have formulated the SM for an electron-neutrino pair, but it can beextended in a trivial way to other lepton pairs. The extension to the quarks is alittle bit less obvious and we will follow the necessary steps in this Section. Firstof all we recall that in the quark model the nucleon is made up of three quarks,with composition, p ≈ uud, n ≈ udd. Then, the weak transition n→ p+ e− + νe isobtained through

d→ u+ e− + νe (10.89)

and the assumption is made that quarks are point-like particles as leptons. There-fore, it is natural to write quark current in a form analogous to the leptonic current(see eqs. (10.5) and (10.7))

Jµ(±)d = 2ΨLγ

µτ∓ΨL (10.90)

with

ΨL =

(uLdL

)=

1 − γ5

2

(ud

)(10.91)

249

Phenomenologically, from the hyperon decays it was known that it was necessary afurther current involving the quark s, given by

Jµ(±)s = 2Ψ′

Lγµτ∓Ψ′

L (10.92)

with

Ψ′L =

(uLsL

)=

1 − γ5

2

(us

)(10.93)

Furthermore the total current contributing to the Fermi interaction had to be of theform

Jµh = Jµd cos θC + Jµs sin θC (10.94)

where θC is the Cabibbo angle, and

sin θC = 0.21 ± 0.03 (10.95)

The presence of the angle in the current was necessary in order to explain why theweak decays strangeness violating (that is the ones corresponding to the transitionu ↔ s, as K+ → µ+νµ) where suppressed with respect to the decays strangenessconserving. Collecting together the two currents we get

Jµ(±)h = 2QLγ

µτ∓QL (10.96)

with

QL =

(uL

dL cos θC + sL sin θC

)≡(uLdCL

)(10.97)

anddCL = dL cos θC + sL sin θC (10.98)

This allows us to assign QL to the representation (2, 1/3) of SU(2)⊗U(1). In fact,from

Qem = Q3 +1

2QY (10.99)

we get

QY (uL) = 2(2

3− 1

2) =

1

3(10.100)

and (Qem(s) = Qem(d) = −1/3)

QY (dCL) = 2(−1

3+

1

2) =

1

3(10.101)

As far as the right-handed components are concerned, we assign them to SU(2)singlets, obtaining

uR ∈ (1,4

3) dCR ∈ (1,−2

3) (10.102)

250

Therefore, the hadronic neutral currents contribution from quarks u, d, s are givenby

j3µ = QLγµ

τ32QL (10.103)

jYµ =1

3QLγµQL +

4

3uRγµuR − 2

3dCRγµd

CR (10.104)

Since the Z is coupled to the combination j3µ − sin2 θjem

µ , it is easily seen that thecoupling contains a bilinear term in the field dC given by

−(

1

2− 1

3sin2 θ

)ZµdCLγµd

CL +

1

3sin2 θZµdCRγµd

CR (10.105)

This gives rise to terms of the type ds and sd which produce strangeness changingneutral current transitions . However the experimental result is that these transitionsare strongly suppressed. In a more general way we can get this result from the simpleobservation that W3 is coupled to the current j3

µ which has an associated charge Q3

which can be obtained by commuting the charges Q1 and Q2

[Q1, Q2] =

∫d3xd3y [Q†

Lτ1QL, Q†Lτ2QL] =

∫d3xQ†

L[τ1, τ2]QL

= i

∫d3xQ†

Lτ3QL = i

∫d3x(u†LuL − dC†

L dCL) (10.106)

The last term is just the charge associated to a Flavour Changing Neutral Current(FCNC). We can compare the strength of a FCNC transition, which from

dC†dC = d†d cos2 θC + s†s sin2 θC + (d†s+ s†d) sin θC cos θC (10.107)

is proportional to sin θC cos θC , with the strength of a flavour changing chargedcurrent transition as u → s, which is proportional to sin θC . From this comparisonwe expect the two transitions to be of the same order of magnitude. But if wecompare K0 → µ+µ− induced by the neutral current , with K+ → µ+νµ induced bythe charged current (see Fig. 2.3), using BR(K0 → µ+µ−), we find

Γ(K0 → µ+µ−)

Γ(K+ → µ+νµ)≤ 6 × 10−5 (10.108)

This problem was solved in 1970 by the suggestion (Glashow Illiopoulos andMaiani, GIM) that another quark with charge +2/3 should exist, the charm. Then,one can form two left-handed doublets

Q1L =

(uLdCL

), Q2

L =

(cLsCL

)(10.109)

wheresCL = −dL sin θC + sL cos θC (10.110)

251

K0d

s–

Z µ+

µ–

K+u

s–

W+ µ+

νµ

Fig. 10.3.1 - The neutral current flavour changing process K0 → µ+µ− and hecharged current flavour changing process K+ → µ+νµ.

is the combination orthogonal to dCL . As a consequence the expression of Q3 getsmodified

Q3 =

∫d3x (u†LuL + c†LcL − dC†

L dCL − sC†

L sCL) (10.111)

ButdC†L d

CL + sC†

L sCL = d†LdL + s†LsL (10.112)

since the transformation matrix(dCLsCL

)=

(cos θC sin θC− sin θC cos θC

)(dLsL

)(10.113)

is orthogonal. The same considerations apply to QY . This mechanism of cancellationturns out to be effective also at the second order in the interaction, as it shouldbe, since we need a cancellation at least at the order G2

F in order to cope withthe experimental bounds. In fact, one can see that the two second order graphscontributing to K0 → µ+µ− cancel out except for terms coming from the massdifference between the quarks u and c. One gets

A(K0 → µ+µ−) ≈ G2F (m2

c −m2u) (10.114)

allowing to get a rough estimate of the mass of the charm quark, mc ≈ 1 ÷ 3 GeV .The charm was discovered in 1974, when it was observed the bound state cc, withJPC = 1−−, known as J/ψ. The mass of charm was evaluated to be around 1.5 GeV .In 1977 there was the observation of a new vector resonance, the Υ, interpreted asa bound state bb, where b is a new quark, the bottom or beauty, with mb ≈ 5 GeV .Finally in 1995 the partner of the bottom, the top, t, was discovered at Fermi Lab.The mass of the top is around 175 GeV . We see that in association with three

252

leptonic doublets (νee−

)L

,

(νµµ−

)L

,

(νττ−

)L

(10.115)

there are three quark doublets(ud

)L

,

(cs

)L

,

(tb

)L

(10.116)

We will show later that experimental evidences for the quark b to belong to a doubletcame out from PEP and PETRA much before the discovery of top. Summarizingwe have three generations of quarks and leptons. Each generation includes twoleft-handed doublets (

νAe−A

)L

,

(uAdA

)L

A = 1, 2, 3 (10.117)

and the corresponding right-handed singlets (except for the neutrinos if we assumethat they are massless). Inside each generation the total charge is equal to zero. Infact

Qtoti =

∑f∈gen

Qf = 0 − 1 + 3 ×(

2

3− 1

3

)= 0 (10.118)

where we have also taken into account that each quark comes in three colors. Thisis very important because it guarantees the renormalizability of the SM. In fact, ingeneral there are quantum corrections to the divergences of the currents destroyingtheir conservation. However, the conservation of the currents coupled to the Yang-Mills fields is crucial for the renormalization properties. In the case of the SM forthe electroweak interactions one can show that the condition for the absence of thesecorrections (Adler, Bell, Jackiw anomalies, see later) is that the total electric chargeof the fields is zero.

We will complete now the formulation of the SM for the quark sector, showingthat the mixing of different quarks arises naturally from the fact that the quarkmass eigenstates are not necessarily the same fields which couple to the gauge fields.We will denote the last ones by

′AL =

(ν ′Ae′A

)L

, q′AL =

(u′Ad′A

)L

, A = 1, 2, 3 (10.119)

and the right-handed singlets by

e′AR, u′AR, d′AR (10.120)

We assume that the neutrinos are massless and that they do not have any right-handed partner. The gauge interaction is then

L =∑fL

fAL

(i∂µ + g

τ

2· Wµ + g′

QY (fL)

2Yµ

)γµfAL

+∑fR

fAR

(i∂µ + g′

QY (fR)

2

)γµfAR (10.121)

253

where fAL = ′AL, q′AL, and fAR = e′AR, u

′AR, d

′AR. We recall also the weak hyper-

charge assignments

QY (′AL) = −1, QY (q′AL) =1

3

QY (e′AR) = −2, QY (u′AR) =4

3

QY (d′AR) = −2

3(10.122)

Let us now see how to give mass to quarks. We start with up and down quarks, anda Yukawian coupling given by

gdqLΦdR + h.c. (10.123)

where Φ is the Higgs doublet. When this takes its expectation value, 〈Φ〉 =(0, v/

√2), a mass term for the down quark is generated

gdv√2

(dLdR + dRdL

)=gdv√

2dd (10.124)

corresponding to a mass md = −gdv/√

2. In order to give mass to the up quark weneed to introduce the conjugated Higgs doublet defined as

Φ = iτ2Φ =

((φ0)

−(φ+)

)(10.125)

Observing that QY (Φ) = −1, another invariant Yukawian coupling is given by

guqLΦuR + h.c. (10.126)

giving mass to the up quark, mu = −guv/√

2. The most general Yukawian couplingamong quarks and Higgs field is then given by

LY =∑AB

(geAB

′ALΦe

′BR + guAB q

′ALΦu

′BR + gdAB q

′ALΦd

′BR

)+ h.c. (10.127)

where giAB, with i = e, u, d, are arbitrary 3 × 3 matrices. When we shift Φ by itsvacuum expectation value we get a mass term for the fermions given by

Lfermion mass =v√2

∑AB

(geABe

′ALe

′BR + guABu

′ALu

′BR + gdABd

′ALd

′BR

)+ h.c. (10.128)

Therefore we obtain three 3 × 3 mass matrices

M iAB = − v√

2giAB, i = e, u, d (10.129)

254

These matrices give mass respectively to the charged leptons and to the up anddown of quarks. They can be diagonalized by a biunitary transformation

M id = S†M iT (10.130)

where S and T are unitary matrices (depending on i) and M id are three diagonal

matrices. Furthermore one can take the eigenvalues to be positive. This followsfrom the polar decomposition of an arbitrary matrix

M = HU (10.131)

where H† = H =√MM † is a hermitian definite positive matrix and U † = U−1 is a

unitary matrix. In fact, by defining

H2 = MM †, U = H−1M (10.132)

we see that H is hermitian and positive definite, and that U is unitary, since

U †U = M †H−2M = M †M †−1M−1M = 1

UU † = H−1MM †H−1 = H−1H2H−1 = 1 (10.133)

Therefore, if the unitary matrix S diagonalizes H we have

Hd = S†HS = S†MU−1S = S†MT (10.134)

with T = U−1S is a unitary matrix. Each of the mass terms has the structureψ′LMψ′

R. Then we get

ψ′LMψ′

R = ψ′LS(S†MT )T †ψ′

R ≡ ψLMdψR (10.135)

where we have defined the mass eigenstates

ψ′L = SψL, ψ′

R = TψR (10.136)

We can now express the charged current in terms of the mass eigenstates. We have

J+(h)µ = 2q′ALγµτ−q

′AL = 2u′ALγµd

′AL = 2uALγµ((S

u)−1Sd)ABdBL (10.137)

DefiningV = (Su)−1Sd, V † = V −1 (10.138)

we getJh(+)µ = 2uALγµd

(V )AL (10.139)

withd

(V )AL = VABdBL (10.140)

The matrix V is called the Cabibbo-Kobayashi-Maskawa (CKM) matrix, and itdescribes the mixing among the down type of quarks (this is conventional we could

255

have chosen to mix the up quarks as well). We see that its physical origin is that,in general, there are no relations between the mass matrix of the up quarks, Mu,and the mass matrix of the down quarks, Md. The neutral currents are expressionswhich are diagonal in the ”primed” states, therefore their expression is the same alsoin the basis of the mass eigenstates, due to the unitarity of the various S matrices.In fact

jh(3)µ ≈ u′Lu

′L − d′Ld

′L = uL(S

u)−1SuuL − dL(Sd)−1SddL = uLuL − dLdL (10.141)

For the charged leptonic current, since we have assumed massless neutrinos we get

J (+)µ = 2νALγµ(S

e)ABeBL ≡ 2ν ′ALγµeAL (10.142)

where ν ′L = (Se)†νL is again a massless eigenstate. Therefore, there is no mixingin the leptonic sector, among different generations. However this is tied to ourassumption of massless neutrinos. By relaxing this assumption we may generate amixing in the leptonic sector producing a violation of the different leptonic numbers.

It is interesting to discuss in a more detailed way the structure of the CKMmatrix. In the case we have n generations of quarks and leptons, the matrix V ,being unitary, depends on n2 parameters. However one is free to choose in anarbitrary way the phase for the 2n up and down quark fields. But an overall phasedoes not change V . Therefore the CKM matrix depends only on

n2 − (2n− 1) = (n− 1)2 (10.143)

parameters. We can see that, in general, it is impossible to choose the phases in sucha way to make V real. In fact a real unitary matrix is nothing but an orthogonalmatrix which depends on

n(n− 1)

2(10.144)

real parameters. This means that V will depend on a number of phases given by

number of phases = (n− 1)2 − n(n− 1)

2=

(n− 1)(n− 2)

2(10.145)

Then, in the case of two generations V depends only on one real parameter (theCabibbo angle). For three generations V depends on three real parameters andone phase. Since the invariance under the discrete symmetry CP implies that allthe couplings in the lagrangian must be real, it follows that the SM, in the caseof three generations, implies a CP violation. A violation of CP has been observedexperimentally by Christensen et al. in 1964. These authors discovered that theeigenstate of CP , with CP = −1

|K02〉 =

1√2

(|K0〉 − |K0〉) (10.146)

256

decays into a two pion state with CP = +1 with a BR

BR(K02 → π+π−) ≈ 2 × 10−3 (10.147)

It is not clear yet, if the SM is able to explain the observed CP violation in quanti-tative terms.

To complete this Section we give the experimental values for some of the matrixelements of V . The elements Vud and Vus are determined through nuclear β-, K−and hyperon-decays, using the muon decay as normalization. One gets

|Vud| = 0.9736 ± 0.0010 (10.148)

and|Vus| = 0.2205 ± 0.0018 (10.149)

The elements Vcd and Vcs are determined from charm production in deep inelasticscattering νµ +N → µ+ c+X. The result is

|Vcd| = 0.224 ± 0.016 (10.150)

and|Vcs| = 1.01 ± 0.18 (10.151)

The elements Vub and Vcb are determined from the b → u and b → c semi-leptonicdecays as evaluated in the spectator model, and from the B semi-leptonic exclusivedecay B → Dν. One gets

|Vub||Vcb| = 0.08 ± 0.02 (10.152)

and|Vcb| = 0.041 ± 0.003 (10.153)

More recently from the branching ratio t→ Wb, CDF (1996) has measured |Vtb|

|Vtb| = 0.97 ± 0.15 ± 0.07 (10.154)

A more recent comprehensive analysis gives the following 90 % C.L. range for thevarious elements of the CKM matrix

V =

⎛⎝ 0.9745 − 0.9757 0.219 − 0.224 0.002 − 0.005

0.218 − 0.224 0.9736 − 0.9750 0.036 − 0.0460.004 − 0.014 0.034 − 0.046 0.9989 − 0.9993

⎞⎠ (10.155)

257

10.4 The parameters of the SM

It is now the moment to count the parameters of the SM. We start with the gaugesector, where we have the two gauge coupling constants g and g′. Then, the Higgssector is specified by the parameters µ and the self-coupling λ. We will rather usev =

√−µ2/λ and m2h = −2µ2. Then we have the Yukawian sector, that is that

part of the interaction between fermions and Higgs field giving rise to the quarkand lepton masses. If we assume three generations and that the neutrinos are allmassless we have three mass parameters for the charged leptons, six mass parametersfor the quarks (assuming the color symmetry) and four mixing angles. In order totest the structure of the SM the important parameters are those relative to thegauge and to the Higgs sectors. As a consequence one assumes that the mass matrixfor the fermions is known. This was not the case till two years ago, before thetop discovery, and its mass was unknown. The masses of the vector bosons can beexpressed in terms of the previous parameters. However, the question arises aboutthe better choice of the input parameters. Before LEP1, the better choice was touse quantities known with great precision related to the parameters (g, g′, v). Thechoice was to use the fine structure constant

α =1

137.0359895(61)(10.156)

and the Fermi constant

GF = 1.166389(22)× 10−5 GeV −2 (10.157)

as measured from the β-decay of the muon. Then, most of the experimental researchof the seventies and eighties was centered about the determination of sin2 θ. Theother parameters as (mt, mh) affect only radiative corrections, which, in this type ofexperiments can be safely neglected due to the relatively large experimental errors.Therefore, in order to relate the set of parameters (α,GF , sin

2 θ) to (g, g′, v) we canuse the tree level relations. The situation has changed a lot after the beginning ofrunning of LEP1. In fact, the mass of the Z has been measured with great precision,δMZ/MZ ≈ 2 × 10−5. In this case the most convenient set is (α,GF ,MZ), with

MZ = (91.1867 ± 0.0021) GeV (10.158)

Also the great accuracy of the experimental measurements requires to take intoaccount the radiative corrections. We will discuss this point in more detail at the endof these lectures. However we notice that one gets informations about parametersas (mt, mh) just because they affect the radiative corrections. In particular, theradiative corrections are particularly sensitive to the top mass. In fact, LEP1 wasable to determine the top mass, in this indirect way, before the top discovery.

258

10.5 Ward identities and anomalies

Up to now we have assumed that the symmetries for the classical lagrangian arestill valid at the quantum level. This is not always so and the motivations can beeasily understood. The problem has to do with the invariance of the functionalintegration measure with respect to the symmetry transformation. In fact, we willsee that there are situations in which the measure is not invariant and therefore thesymmetry is broken at the quantum level. Before dealing with this problem it willbe convenient to rephrase the Noether’s theorem in a way more suitable to derivethe Ward identities in a quantum field theory. These identities are nothing but theexpression of the symmetry in the quantum context. Let us consider a field theorywith a lagrangian invariant under the following global transformation

φi → φi + δφi (10.159)

withδφi(x) = −iεA(TA)ijφj(x) (10.160)

Then, let us consider the transformation of the action under the same transformationof the fields but with εA → εA(x) a space-time function. We find

δS =

∫d4x

[∂L∂φi

δφi +∂L∂φi,µ

δφi,µ

](10.161)

whereδφi,µ = −iεA(TA)ij∂µφj − i(∂µεA)(TA)ijφj (10.162)

Substituting we find

δS =

∫d4x[− i

∂L∂φi

εA(TA)ijφj − i∂L∂φi,µ

εA(TA)ij∂µφj

− i∂L∂φi,µ

(TA)ijφj∂µεA

](10.163)

Since for εA independent on x we have δS = 0 by hypothesis, it follows that the firsttwo terms must vanish. Therefore we are left with

δS =

∫d4x ∂µεAj

Aµ (10.164)

where

jAµ = −i ∂L∂φi,µ

(TA)ijφj (10.165)

are the Noether’s currents after factorizing the infinitesimal parameters εA. However,if we take εA(x) to describe a variation around the classical solutions, that is if the

259

fields inside the currents satisfy the equations of motion, then we must have δS = 0.Then, writing

δS = −∫d4x εA∂µj

µA (10.166)

we get∂µj

µA = 0 (10.167)

Let us now consider the generating functional associated to this theory

Z[η] =

∫D(φ)e

iS[φ] + i

∫d4x ηiφi

(10.168)

If we perform the change of variables

φi → φi + δφi = φi − iεA(x)(TA)ijφj (10.169)

and assume that the integration measure is invariant, we get

Z[η] =

∫D(φ)e

iS[φ] + i

∫d4x ηiφi

ei

∫d4x( ∂µεAj

Aµ + εAηi(T

A)ijφj)(10.170)

By expanding this expression at the first order in εA(x) and using the fact that theprevious expression must agree with eq. (10.168) for any εA(x), we get

0 =

∫D(φ)e

iS[φ] + i

∫d4x ηiφi [−i∂µjAµ + ηi(T

A)ijφj]

(10.171)

From this equation we can generate the Ward identities by differentiating withrespect to ηi and then putting ηi = 0. For instance, for ηi = 0 we get

∂µ〈0|jAµ (x)|0〉 = 0 (10.172)

By differentiating once (remember that one gets the T ∗ product) we have

∂µx 〈0|T (jAµ (x)φi(y))|0〉 = −δ4(x− y)〈0|(TA)ijφj(y)|0〉 (10.173)

and, differentiating N times

∂µx 〈0|T (jAµ (x)φi1(x1) · · ·φiN (xN ))|0〉

=

N∑p=1

δ4(x− xp)〈0|T (φi1(x1) · · · (−(TA)ipjφj(xp)) · · ·φiN (xN))|0〉(10.174)

This derivation is correct only if the integration measure is invariant under thechange of variables (10.169). We will illustrate a simple situation where the measure

260

is not invariant. Consider a massless fermion in a gauge field. For the followingconsiderations the gauge field can be taken as an external field, but the same appliesalso to the quantized case. In fact for an external gauge field we define the generatingfunctional as

Z =

∫DψDψe

i

∫d4x ψiDψ

(10.175)

whereDµ = ∂µ + igAµ (10.176)

As we will be interested only in checking the current conservation we can put tozero the external sources for the fermion fields (see the previous derivation). If thegauge field Aµ is quantized we have to insert a further functional integration in Aµbut this will not play any role in the following, so for simplicity we will take Aµ asan external field. The lagrangian is invariant under the global transformation

ψ → eiαγ5ψ (10.177)

giving rise to the classically conserved current

Jµ = ψγµγ5ψ (10.178)

We will show that the quantum corrections destroy the conservation law. To thisend we proceed as before by performing the change of variables inside Z

ψ(x) → ψ′(x) = (1 + iα(x)γ5)ψ(x)

ψ(x) → ψ′(x) = ψ(x)(1 + iα(x)γ5) (10.179)

The variation of the action is given by

δS =

∫d4xα(x)∂µ(ψγ

µγ5ψ) (10.180)

In order to evaluate the change of the integration measure it is convenient to expandthe fermion field in a basis of eigenvectors of D

iDφm(x) = λmφm(x)

−iDµφm(x)γµ = λmφm(x) (10.181)

where we have used a discrete notation for the eigenvalues. In the case of a zerogauge field Aµ the eigenvalues λm are given by

λ2m = k2

0 − |k|2 (10.182)

Notice that they become negative definite after being Wick rotated

λ2m → −k2

4 − |k|2 = −k2E (10.183)

261

By hypothesis the eigenvectors φm span an orthonormal basis and therefore we canexpand ψ and ψ as

ψ(x) =∑m

amφm(x), ψ(x) =∑m

bmφm(x) (10.184)

where am and bm are Grassmann coefficients. Then, except for a possible constantwe have

DψDψ =∏m

damdbm (10.185)

We can evaluate the effect of the change of variables on the Grassmann coefficientsby starting from

ψ′(x) = (1 + iα(x)γ5)ψ(x) =∑m

a′mφm(x) (10.186)

Then, from the orthogonality relation∫d4xφ†

mφn = δmn (10.187)

we get

a′m =

∫d4xφ†

m(x)ψ′(x) = am +∑n

∫d4xφ†

m(x)iα(x)γ5φn(x)an = am +∑n

Cmnan

(10.188)with

Cmn =

∫d4xφ†

m(x)iα(x)γ5φn(x) (10.189)

A similar transformation holds for bm. Therefore, from the transformation rules forthe Grassmann measure we get

Dψ′Dψ′ =1

det|I|2DψDψ (10.190)

where I = 1 + C is the jacobian of the transformation. Since we are interested inthe infinitesimal transformation, we have

det|I| = eTr log(1 + C) ≈ eTr C (10.191)

that is

log det|I| ≈ TrC = i

∫d4xα(x)

∑n

φ†n(x)γ5φn(x) (10.192)

The coefficient of α(x) is the trace of γ5 taken over all the Hilbert space. Althoughthe trace of γ5 is zero on the spinor space it gets multiplied by an infinite factorcoming from the total Hilbert space. Therefore this expression is ill defined and it

262

needs to be regularized. Furthermore, we would like to maintain the gauge invari-ance, and therefore we should regulate the expression in a gauge invariant way. Anatural choice for the regularization is the following

∑n

φ†n(x)γ5φn(x) = lim

M→∞

∑n

φ†n(x)γ5φn(x)e

λ2n/M

2(10.193)

This is in fact a convergence factor in the euclidean space. The previous expressioncan be rewritten as∑

n

φ†n(x)γ5φn(x) = lim

M→∞

∑n

φ†n(x)γ5e

(iD)2/M2φn(x)

= limM→∞

〈x|tr[γ5e

(iD)2/M2]|x〉 (10.194)

where tr is the trace over the Dirac matrices. The evaluation of D)2 gives

(D)2 = γµγνDµDν =

1

2[γµ, γν ]+D

µDν +1

2[γµ, γν ]−DµDν

= D2 − i

2σµν [D

µ, Dν]− = D2 +g

2σµνF

µν (10.195)

In order to get a contribution from the trace over the Dirac indices we need atleast four γ matrices. Therefore the leading term is obtained by expanding theexponential up to the order (σµνF

µν)2 and neglecting Aµ in all the other places.Therefore we get

∑n

φ†n(x)γ5φn(x) = lim

M→∞tr

[γ5

1

2!

(− g

2M2σµνF

µν)2]〈x|e−/M2 |x〉 (10.196)

The matrix element can be easily evaluated by inserting momentum eigenstates andperforming a Wick’s rotation

〈x|e−/M2 |x〉 = limx→y

∫d4k

(2π)4ek

2/M2e−ik(x − y) = i

∫d4kE(2π)4

e−k2E/M

2= i

M4

16π2

(10.197)Therefore ∑

n

φ†nγ5φn =

ig2

8 · 16π2tr[γ5σµνσρλ]F

µνF ρλ (10.198)

In our definitionsγ5 = iγ0γ1γ2γ3 (10.199)

The trace is easily evaluated by calculating first

tr[γ5σ01σ23] = −itr[γ0γ1γ2γ3γ0γ1γ2γ3] = −4i (10.200)

263

and using ε0123 = 1 we gettr[γ5σµνσρλ] = 4iεµνρλ (10.201)

Then ∑n

φ†nγ5φn = − g2

32π2εµνρλF

µνF ρλ (10.202)

Therefore the determinant we are looking for is given by

det |I| = e−i∫d4xα(x)

g2

32π2εµνρλF

µνF ρλ

(10.203)

and after the change of variables the generating functional is given by

Z =

∫DψDψe

i

∫d4x ψiDψ

ei

∫d4xα(x)(∂µ(ψγ

µγ5ψ) +g2

16π2εµνρλF

µνF ρλ)

(10.204)leading to the anomalous conservation law

∂µjµ5 = − g2

16π2εµνρλF

µνF ρλ (10.205)

This is called the Adler-Bell-Jackiw (ABJ) anomaly.The previous result can be extended to the case of non-abelian chiral symmetries

as, for instance, in the standard model where the SU(2) symmetry acts only on theleft-handed fermions and therefore the corresponding transformations involve γ5.The ABJ anomaly is modified by a factor

Tr[TA[TB, TC]+] (10.206)

where the term TA comes in the chiral transformation in conjunction with γ5,whereas the other two matrices arise from the regulator. It is a crucial point inthe renormalization of the gauge theories that the gauge symmetry is preservedat the quantum level. It follows that the standard model is consistent only if theanomaly vanishes. One has to check that the factor in eq. (10.206) vanishes for allthe generators of SU(2) ⊗ U(1). It is not difficult to show that this is indeed thecase only if the sum of the electric charges of the fermions is zero∑

fermions

Q = 0 (10.207)

This condition holds for any generation, in fact

Qelectron +Qneutrino + 3(Qup +Qdown) = −1 + 0 + 3

(2

3− 1

3

)= 0 (10.208)

264

Appendix A

A.1 Properties of the real antisymmetric matri-

ces

Given a real and antisymmetric matrix n× n A, we want to show that it is possibleto put it in the form (7.61) by means of an orthogonal transformation. To thisend, let us notice that iA is a hermitian matrix and therefore it can be diagonalizedthrough a unitary transformation U :

U(iA)U † = Ad (A.1)

where Ad is diagonal and real. The eigenvalues of iA satisfy the equation

det|iA− λ · 1| = 0 (A.2)

Since AT = −A it follows

det|iA− λ · 1| = det|(iA− λ · 1)T | = det| − iA− λ · 1| = 0 (A.3)

Therefore, if λ is an eigenvalue, the same is for −λ. lo e’. It follows that Ad can bewritten in the form

Ad =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎣

λ1 0 0 0 · · ·0 −λ1 0 0 · · ·0 0 λ2 0 · · ·0 0 0 −λ2 · · ·· · · · · · ·· · · · · · ·· · · · · · ·

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎦

(A.4)

Notice that if n is odd, Ad has necessarily a zero eigenvalue, that we will write as(Ad)nn. We will assume also to order the matrix in such a way that all the λi arepositive.Each submatrix of the type[

λi 00 −λi

](A.5)

265

can be put in the form [0 iλi

−iλi 0

](A.6)

through the unitary 2 × 2 matrix

V2 =1√2

[i 11 i

](A.7)

In fact

V2

[λi 00 −λi

]V †

2 =

[0 iλi

−iλi 0

](A.8)

Defining

V =

⎡⎢⎢⎢⎢⎣V2 0 · · ·0 V2 · · ·· · · · ·· · · · ·· · · · ·

⎤⎥⎥⎥⎥⎦ (A.9)

with Vnn = 0 for n odd, we get

V AdV† =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎣

0 iλ1 0 0 · · ·−iλ1 0 0 0 · · ·

0 0 0 iλ20 · · ·0 0 −iλ1 0 · · ·· · · · · · ·· · · · · · ·· · · · · · ·

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎦

= iAs (A.10)

with As defined in (7.61). Therefore

(V U)(iA)(V U)† = iAs (A.11)

or(V U)(A)(V U)† = As (A.12)

But A and As are real matrices, and therefore also V U must be real. However beingV U real and unitary it must be orthogonal.

266