Quantum Field Theory1

Roberto CasalbuoniDipartimento di Fisica

Universita di Firenze

1Lectures given at the Geneva University during the academic year 1997/98.

Contents

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

1 Introduction 31.1 Major steps in quantum field theory . . . . . . . . . . . . . . . . . . 31.2 Many degrees of freedom . . . . . . . . . . . . . . . . . . . . . . . . . 41.3 Linear atomic string . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

2 Lagrangian formalism for continuum systems and quantization 112.1 String quantization . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112.2 The lagrangian formalism for continuum systems . . . . . . . . . . . 152.3 The canonical quantization of a continuum system . . . . . . . . . . . 19

3 The Klein-Gordon field 253.1 Relativistic quantum mechanics and its problems . . . . . . . . . . . 253.2 Quantization of the Klein-Gordon field . . . . . . . . . . . . . . . . . 283.3 The Noether’s theorem for relativistic fields . . . . . . . . . . . . . . 343.4 Energy and momentum of the Klein-Gordon field . . . . . . . . . . . 383.5 Locality and causality in field theory . . . . . . . . . . . . . . . . . . 413.6 The charged scalar field . . . . . . . . . . . . . . . . . . . . . . . . . 46

4 The Dirac field 514.1 The Dirac equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 514.2 Covariance properties of the Dirac equation . . . . . . . . . . . . . . 544.3 Free particle solutions of the Dirac equation . . . . . . . . . . . . . . 594.4 Wave packets and negative energy solutions . . . . . . . . . . . . . . 654.5 Electromagnetic interaction of a relativistic point-like particle . . . . 674.6 Non relativistic limit of the Dirac equation . . . . . . . . . . . . . . . 734.7 Charge conjugation, time reversal and PCT transformation . . . . . . 764.8 Dirac field quantization . . . . . . . . . . . . . . . . . . . . . . . . . . 81

5 The electromagnetic field 895.1 The quantization of the electromagnetic field . . . . . . . . . . . . . . 89

1

6 Symmetries in field theories 1016.1 The linear σ-model . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1016.2 Spontaneous symmetry breaking . . . . . . . . . . . . . . . . . . . . . 1076.3 The Goldstone theorem . . . . . . . . . . . . . . . . . . . . . . . . . . 1116.4 QED as a gauge theory . . . . . . . . . . . . . . . . . . . . . . . . . . 1136.5 Non-abelian gauge theories . . . . . . . . . . . . . . . . . . . . . . . . 1156.6 The Higgs mechanism . . . . . . . . . . . . . . . . . . . . . . . . . . 119

7 Time ordered products 1257.1 Time ordered products and propagators. . . . . . . . . . . . . . . . . 1257.2 A physical application of the propagators . . . . . . . . . . . . . . . . 131

8 Perturbation theory 1368.1 The electromagnetic interaction . . . . . . . . . . . . . . . . . . . . . 1368.2 The scattering matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . 1388.3 The Wick’s theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . 1468.4 Evaluation of the S matrix at second order in QED . . . . . . . . . . 1498.5 Feynman diagrams in momentum space . . . . . . . . . . . . . . . . . 157

9 Applications 1649.1 The cross-section . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1649.2 The scattering e+e− → µ+µ− . . . . . . . . . . . . . . . . . . . . . . 1669.3 Coulomb scattering . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171

10 One-loop renormalization 17510.1 Divergences of the Feynman integrals . . . . . . . . . . . . . . . . . . 17510.2 Dimensional regularization of the Feynman integrals . . . . . . . . . . 18310.3 Integration in arbitrary dimensions . . . . . . . . . . . . . . . . . . . 18410.4 One loop regularization of QED . . . . . . . . . . . . . . . . . . . . . 18710.5 One loop renormalization . . . . . . . . . . . . . . . . . . . . . . . . . 19310.6 Lamb shift and g − 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . 198

2

Chapter 1

Introduction

1.1 Major steps in quantum field theory

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

1925

• January - Pauli formulates the exclusion principle.

• July - Heisenberg’s first paper on quantummechanics (matrix mechanics).

• September - Born and Jordan extend Heisenberg’s formulation of quan-tum mechanics to electrodynamics.

1926

• January - Schrodinger writes down the wave equation.

• February - Fermi introduces a new statistics (Fermi-Dirac).

• August - Dirac relates statistics and symmetry properties of the wavefunction, and shows that the quantized electromagnetic field is equivalentto a set of harmonic oscillators satisfying the Bose-Einstein statistics.

1927

• March - Davisson and Germer detect the electron diffraction by a crystal.

• October - Jordan and Klein show that quantum fields satisfy commuta-tion rules.

1928

• January - The Dirac equation.

• January - Jordan and Wigner introduce anticommuting fields for describ-ing particles satisfying Fermi-Dirac statistics.

3

• January - Pauli and Heisenberg develop the analog for fields of the La-grangian and Hamiltonian methods of mechanics.

• Klein and Nishina complete the theory of the scattering Compton basedon the Dirac equation.

1929

• March - Weyl formulates gauge invariance and its relation to charge con-servation.

• December - Dirac introduces the notion of hole theory, identifying a holewith a proton.

1931 Dirac proposes the positron to interpret the energy negative solutions of hisequation and Heisenberg introduces the idea of antiparticles.

1932 Anderson detects the positron.

1934 Dirac and Heisenberg evaluate the vacuum polarization of the photon. Firstbattle with infinities in quantum field theory.

1936 Serber introduces the concept of renormalized charge.

1947 Bethe evaluates the Lamb-shift.

1948 Schwinger ends the calculation of the Lamb-shift and the renormalizationprogram starts.

1.2 Many degrees of freedom

Aim of this course is to extend ordinary quantum mechanics, which describes nonrelativistic particles in interaction with given forces, to the relativistic case whereforces are described by fields, as for the electromagnetic case. The most relevantdifferences between the two cases are that the forces become dynamical degrees offreedom, and that one needs a relativistic treatment of the problem. In order to geta consistent description we will need to quantize the field degrees of freedom.

The concept of field is a very general one. A field represents a physical quantitydepending on the space-time point. Examples are the distribution of temperaturesin a room, the distribution of the pressure in the atmosphere, the particle velocitiesinside a fluid, the electric and magnetic fields in a given region of space. The commonphysical feature of these systems is the existence of a fundamental state, for example:

• pressure or temperature =⇒ state with T = constant, or P = constant

• particle velocities in a fluid =⇒ state at rest

• electromagnetic field =⇒ state of vacuum.

4

In most of these cases one is interested in discussing small deviations of thesystem from the fundamental state. By doing so one gets, in a first approximation,linear equations for the fields (these being defined in terms of the deviations). Onecan then improve the situation by adding small corrections and treating the problemthrough some perturbative approximation. This linear approximation is generallyvery similar for many different physical situations. For instance, in many cases onegets the wave equation. The quantization of such a system will lead us to describethe system in terms of particles corresponding to the different classical excitations.

The field quantization is done by considering the equation of motion for the fieldas a hamiltonian system describing an infinite number of degrees of freedom. Inorder to understand this point we will begin with a very simple system. That is astring of N linear oscillators (for instance a one-dimensional string of atoms) in thelimit of N → ∞ with a separation among the atoms going to zero. In this way weget a vibrating string as the continuum limit.

1.3 Linear atomic string

Let us consider a string of N +1 harmonic oscillators, or N +1 atoms (of unit mass)interacting through a harmonic force, as in Fig. (1.1). The length of the string is L

a a a a

qn-1 qn qn+1

Fig. 1.1 - In the upper line the atoms are in their equilibrium position, whereas inthe lower line they are displaced by the quantities qn.

and the inter-atomic distance is a. Therefore L = Na. The equations of motion arethe following

qn = ω2 [(qn+1 − qn) + (qn−1 − qn)] = ω2 [qn+1 + qn−1 − 2qn] (1.1)

5

as it follows immediately from the expression of the potential energy of the system

U =1

2ω2

N∑n=1

(qn − qn+1)2 (1.2)

In order to define the problem one has to specify the boundary conditions, althoughin the N → ∞ limit we do not expect that they play any role. Usually one considerstwo possible boundary conditions

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

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

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

H = T + U =1

2

N∑n=1

(p2n + ω2 (qn − qn+1)

2)

(1.3)

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

q(j)n = Ajeikjane−iωjt (1.4)

where the index j enumerates the possible eigenvalues. Notice that in this equationthe dependence on the original equilibrium position has been made explicit through

q(j)n = q(j)(xn) = q(j)(na) ≈ eikjxn (1.5)

where xn = na is the equilibrium position of the nth atom. By substituting eq. (1.4)into the equations of motion we get

−ω2j q

(j)n = −4ω2q(j)n sin2

(kja

2

)(1.6)

from which

ω2j = 4ω2 sin2

(kja

2

)(1.7)

The relation between kj (wave vector) and ωj (frequency of oscillation), showngraphically in Fig. (1.2), is called a dispersion relation.

We may notice that wave vectors differing for integer multiples of 2π/a, that is,such that

k′j = kj + 2mπ

a, m = ±1,±2, ... (1.8)

correspond to the same ωj. This allows us to restrict kj to be in the so called firstBrillouin zone, that is |kj| ≤ π/a. Let us now take into account the boundary

6

- 3 - 2 - 1 1 2 3akj

1

2

3

4 ωj2

Fig. 1.2 - The first Brillouin zone.

conditions. Here we will choose periodic boundary conditions, that is qN+1 = q1, or,more generally, qn+N = qn. This gives us

q(j)n+N = Aje

ikja(n+N)e−iωjt = q(j)n = Ajeikjane−iωjt (1.9)

from whichkjaN = 2πj (j = integer) (1.10)

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

kj =2π

aNj =

2π

Lj j = 0,±1,±2, . . .± N

2(1.11)

where we have taken N even. The restriction on j follows from considering the firstBrillouin zone (|kj| ≤ π/a). Notice that the possible values of kj are 2(N/2) + 1 =N + 1, and that j = 0 corresponds to a uniform translation of the string (with zerofrequency). Since we are interested only in the oscillatory motions, we will omit thissolution in the following. It follows that we have N independent solutions

q(j)n = Aje−iωjteiakjn (1.12)

The most general solution is obtained by a linear superposition

qn =∑j

ei2π

Njn Qj√

N(1.13)

pn =∑j

e−i2π

Njn Pj√

N(1.14)

7

From the reality of qn and pn we get

Q⋆j = Q−j, P ⋆

j = P−j (1.15)

In the following we will make use of the following relation

N∑n=1

ei2π

N(j′ − j)n

= Nδjj′ (1.16)

This can be proven by noticing that for j = j′:

∑Nn=1 e

i2π

N(j′ − j)n

=∑Nn=0

ei2πN (j′ − j)n

− 1

=1− e

i2π

N(j′ − j)(N + 1)

1− ei2π

N(j′ − j)

− 1 = 0 (1.17)

whereas for j = j′ the sum gives N . By using this equation we can invert theprevious expansions

N∑n=1

qne−i2π

Nj′n

=∑j

∑n

ei2π

N(j′ − j)n Qj√

N=

√NQj′ (1.18)

obtaining

Qj =1√N

N∑n=1

qne−i2π

Njn

(1.19)

Pj =1√N

N∑n=1

pnei2π

Njn

(1.20)

Notice that Pj = Q−j. Substituting inside the hamiltonian we find

H =N/2∑j=1

(|Pj|2 + ω2

j |Qj|2)

(1.21)

This is nothing but the hamiltonian of N decoupled harmonic oscillators each havinga frequency ωj, as it can be seen by putting

Pj = Xj + iYjQj = Zj + iTj (1.22)

The result we have obtained so far shows that the string of N atoms is equivalent toN decoupled harmonic oscillators. The oscillator modes are obtained through the

8

expansion of the displacements from the equilibrium condition in normal modes.We are now in the position to introduce the concept of displacement field. Let usdefine a function of the equilibrium position of the atoms

xn = na, L = Na (1.23)

as the displacement of the nth atom from its equilibrium position

u(xn, t) = qn(t) (1.24)

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

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

= ω2 [(u(xn+1, t)− u(xn, t))− (u(xn, t)− u(xn−1, t))] (1.25)

Let us now consider the continuum limit of this system. Physically this is equivalentto say that we are looking at the system at a scale much bigger than the inter-atomicdistance. We will define the limit by taking a → 0, by keeping fixed the length ofthe string, that is to say

a→ 0, N → ∞, aN = L fisso (1.26)

The quantity u(xn, t) goes to a function of the variable x defined in the interval(0, L). Furthermore

u(xn, t)− u(xn−1, t)

a→ u′(x, t) (1.27)

and

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

→ a2u′′(x, t) (1.28)

The equation of motion becomes

u(x, t) = a2ω2u′′(x, t) (1.29)

Let us recall that the quantity ω appearing in the equation of motion for the fieldis the elastic constant divided by the mass of the atom. In order to give a sense tothe equation of motion in the previous limit we need that ω diverges in the limit.One could say that in order the string has a finite mass in the continuum, the massof each atom must go to zero. That is, we will require

lima→0

aω = v finite (1.30)

where v has the dimensions of a velocity. We see that in the limit we get the equationfor the propagation of waves with velocity given by v

u(x, t) = v2u′′(x, t) (1.31)

9

In the limit we have alsoN+1∑n=1

→ 1

a

∫ L

0dx (1.32)

from which

H =1

2a

∫ L

0dx[(u(x, t))2 + v2 (u′(x, t))

2]

(1.33)

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

u(x, t) =√aϕ(x, t) (1.34)

getting finally

H =1

2

∫ L

0dx[(ϕ(x, t)

)2+ v2 (ϕ′(x, t))

2]

(1.35)

The normal modes decomposition becomes

ϕ(x, t) =u(x, t)√

a≈ qn(t)√

a≈∑j

eikjanQj√aN

(1.36)

kj =2π

Lj, −∞ < j < +∞ (1.37)

giving rise to

ϕ(x, t) =1√L

+∞∑j=−∞

ei2π

LjxQj(t) (1.38)

The eigenfrequencies are given by

ω2j = 4ω2 sin2

(πa

Lj)→ 4ω2

(πj

L

)2

a2 = (aωkj)2 → v2k2j (1.39)

In the continuum limit the frequency is a linear function of the wave vector. Therelation between the normal modes Qj(t) and the field ϕ(x, t) can be inverted byusing the following relation ∫ L

0dx eix(k − k′) = Lδk,k′ (1.40)

which holds for k and k′ of the form (1.37). The hamiltonian is easily obtained as

H =∞∑j=1

(|Qj|2 + v2k2j |Qj|2

)(1.41)

The main result here is that in the continuum limit the hamiltonian of the systemdescribes an infinite set of decoupled harmonic oscillators. In the following we willshow that the quantization of field theories of the type described in this Sectiongives rise, naturally, to a description in terms of particles.

10

Chapter 2

Lagrangian formalism forcontinuum systems andquantization

2.1 String quantization

We have shown that a string of N atoms can be described in terms of a set ofdecoupled harmonic oscillators, and this property holds true also in the continuumlimit (N → ∞). In the discrete case we have shown that the hamiltonian of thesystem can be written as

H =N/2∑j=1

(|Pj|2 + ω2

j |Qj|2)

(2.1)

whereQ†j = Q−j, P †

j = P−j (2.2)

ω2j = 4ω2 sin2 kja

2, kj =

2π

Lj, |j| = 1, 2, . . . ,

N

2(2.3)

whereas in the continuum case

H =∞∑j=1

(|Pj|2 + ω2

j |Qj|2)

(2.4)

withωj = v|kj| (2.5)

and kj given by eq. (2.3). In both cases the quantization is trivially done byintroducing creation and annihilation operators

aj =

√ωj2Qj + i

1√2ωj

P †j , a†j =

√ωj2Q†j − i

1√2ωj

Pj (2.6)

11

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

[aj, a†k] =

1

2√ωjωk

[ωjQj + iP †j , ωkQ

†k − iPk] = δjk (2.7)

and[aj, ak] = [a†j, a

†k] = 0 (2.8)

where we have made use of the canonical commutation relations

[Qj, Pk] = [Q†j, P

†k ] = iδjk (2.9)

Notice that

a−j =

√ωj2Q−j + i

1√2ωj

P †−j =

√ωj2Q†j + i

1√2ωj

Pj (2.10)

implying

a†−j =

√ωj2Qj − i

1√2ωj

P †j = aj (2.11)

We see that aj e a†j are 2N (in the discrete case) independent operators as Qj and

Pj. The previous relations can be inverted to give

Qj =1√2ωj

(aj + a†−j), Pj = −i√ωj2(a−j − a†j) (2.12)

In terms of aj and a†j the hamiltonian is

H =N/2∑j=1

ωj[a†jaj + a†−ja−j + 1] =

N/2∑j=−N/2

ωj

[a†jaj +

1

2

](2.13)

The fundamental state is characterized by the equation

aj|0⟩ = 0 (2.14)

and its energy is

E0 =∑j

ωj2

(2.15)

In the continuum limit the energy of the fundamental state is infinite (we will comeback later on this point). The generic energy eigenstate is obtained by applyingto the fundamental state the creation operators (the space generated in this way iscalled the Fock space)

|n−N/2, · · · , nN/2⟩ =1

(n−N/2! · · ·nN/2!)1/2(a†−N/2)

n−N/2 · · · (a†N/2)nN/2 |0⟩ (2.16)

12

The state given above can be thought of being formed by n−N/2 quanta of type−N/2 of energy ω−N/2, up to nN/2 quanta of type N/2 of energy ωN/2. In this kindof interpretation the nj quanta (or particles) of energy ωj are indistinguishable onefrom each other. Furthermore, in a given state we can put as many particles wewant. We see that we are describing a set of particles satisfying the Bose-Einsteinstatistics. Formally this follows from the commutation relation

[a†i , a†j] = 0 (2.17)

from which the symmetry of the wave-function follows. For instance a two-particlestate is given by

|i, j⟩ = a†ia†j = |j, i⟩ (2.18)

As we have already noticed the energy of the fundamental state becomes infinitein the continuum limit. This is perhaps the most simple of the infinities that we willencounter in our study of field quantization. We will learn much later in this coursehow it is possible to keep them under control. For the moment being let us noticethat in the usual cases only relative energies are important, and then the value ofE0 (see eq. (2.15)) is not physically relevant. However there are situations, as in theCasimir effect (see later) where it is indeed relevant. Forgetting momentarily thesespecial situations we can define a new hamiltonian by subtracting E0. This can bedone in a rather formal way by defining the concept of normal ordering. Given anoperator which is a monomial in the creation and annihilation operators, we defineits normal ordered form by taking all the annihilation operators to the right of thecreation operators. We then extend the definition to polynomials by linearity. Forinstance, in the case of the hamiltonian (2.4) we have

: H : ≡ N(H) =∑j

ωj2N(a†jaj + aja

†j) =

∑j

ωja†jaj (2.19)

Coming back to the discrete case, recalling eqs. (1.13) and (1.14)

qn =∑j

ei2π

Njn Qj√

N, pn =

∑j

e−i2π

Njn Pj√

N(2.20)

and using the canonical commutators (2.9), we get

[qn, pm] =∑jk

ei2π

N(jn− km) 1

N[Qj, Pk] =

i

N

∑j

ei2π

Nj(n−m)

= iδnm (2.21)

In the continuum case we use the analogue expansion

ϕ(x, t) =1√L

+∞∑j=−∞

ei2π

LjxQj, ϕ(x, t) =

1√L

+∞∑j=−∞

e−i2π

LjxPj (2.22)

13

from which

[ϕ(x, t), ϕ(y, t)] =1

L

∑jk

ei2π

L(jx− ky)

iδjk

=i

L

+∞∑j=−∞

ei2π

Lj(x− y)

= iδ(x− y) (2.23)

This relation could have been obtained from the continuum limit by recalling that

ϕ(x, t) ≈ u(x, t)√a

≈ qn√a

(2.24)

implying

[ϕ(xn, y), ϕ(xm, t)] = iδnma

(2.25)

In the limit

lima→0

δnma

= δ(x− y) (2.26)

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

1 =∑n

a

(δnma

)→∫dx

(lima→0

δnma

)(2.27)

showing that the properties of a delta-approximation are indeed satisfied. It followsthat we have the following correspondence between the discrete and the continuumcase

qn → ϕ(x, t), pn → ϕ(x, t) (2.28)

Said in different words, ϕ(x, t) e ϕ(x, t) appear to be the canonical variables in thelimit. This remark suggests a way to approach the quantization of a field theorydifferent from the one followed so far. The way we have used is based upon theconstruction of the normal modes of oscillation, but in the discrete case this is notnecessary at all. In fact, in such a case, the quantization is made starting from thecommutation relations among the canonical variables, [q, p] = i, without having ana priori knowledge of the dynamics of the system. This suggests that one shouldstart directly by the field operators ϕ(x, t) e ϕ(x, t), and quantize the theory byrequiring [ϕ(x, t), ϕ(y, t)] = iδ(x − y). To make this approach a consistent one, weneed to extend the hamiltonian and lagrangian description to a continuum system.Let us recall how we proceed in the discrete case. We start by giving a lagrangianfunction L(qn, qn, t). Then we define the conjugated momenta by the equation

pn =∂L

∂qn(2.29)

14

We then go the hamiltonian formalism by taking the conjugated momenta, pn, asindependent variables. The previous equation is used in order to solve the velocitiesin terms of qn and pn. Next we define the hamiltonian as

H(qn, pn) =∑n

pnqn − L (2.30)

At the classical level the time evolution of the observables is obtained through theequation

A = A,H (2.31)

where the Poisson brackets can be defined starting form the brackets between thecanonical variables qn, pm = δnm. The theory is then quantized through the rule

[., .] → i., . (2.32)

In the next Section we will learn how to extend the lagrangian and hamiltonianformalism to the continuum case.

2.2 The lagrangian formalism for continuum sys-

tems

We will now show how to construct the lagrangian starting from the equations ofmotion. For the string this can be simply done by starting from the kinetic energyand the potential energy. Let us start recalling the procedure in the discrete case.In this case the kinetic energy is given by

T =1

2

N∑n=1

p2n =1

2

N∑n=1

u2(xn, t)

=1

2

N∑n=1

aϕ2(xn, t) →1

2

∫ L

0ϕ2(x, t)dx (2.33)

whereas the potential energy is

U =1

2

N∑n=1

ω2(qn − qn+1)2

=1

2

N∑n=1

ω2(u(xn, t)− u(xn+1, t))2

=1

2

N∑n=1

ω2a(ϕ(xn, t)− ϕ(xn+1, t))2 (2.34)

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

U =1

2

N∑n=1

av2(ϕ(xn, t)− ϕ(xn+1, t)

a

)2

→ v2

2

∫ L

0ϕ′2(x, t)dx (2.35)

15

Therefore the total energy and the lagrangian are respectively

E = T + U =1

2

∫ L

0dx[ϕ2(x, t) + v2ϕ′2(x, t)

](2.36)

and

L = T − U =1

2

∫ L

0dx[ϕ2(x, t)− v2ϕ′2(x, t)

](2.37)

The important result is that in the continuum limit, the lagrangian can be writtenas a spatial integral of a function of the field ϕ and its first derivatives, which willbe called lagrangian density, and having the expression

L =1

2

(ϕ2 − v2ϕ′2

)(2.38)

The total lagrangian is obtained by integrating spatially the lagrangian density

L =∫ L

0Ldx (2.39)

Of course, this is not the most general situation one can envisage, but we willconsider only the case in which the lagrangian density is a local function of the fieldand its derivatives

L =∫

L(ϕ, ϕ, ϕ′, x, t)dx (2.40)

Furthermore, we will consider only theories in which the lagrangian contains atmost the first derivatives of the fields. The reason is that otherwise one can run intoproblems with the conservation of probability.

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

S =∫ t2

t1Ldt =

∫ t2

t1dt∫

dxL(ϕ, ϕ, ϕ′, x, t) (2.41)

We require that S is stationary with respect to those variations that are consistentwith the boundary conditions satisfied by the fields. If Σ is the spatial surfacedelimiting the region of spatial integration (for the string Σ reduces to the endpoints), we will ask that

δϕ(x, t) = 0 on Σ (2.42)

Furthermore we will require that the variations at the times t1 and t2 are zero atany space point x

δϕ(x, t1) = δϕ(x, t2) = 0, at any x (2.43)

In the discrete case we have only boundary conditions of the second type, but herethe first ones are necessary in order to be consistent with the boundary conditions

16

for the field. Let us now require the stationarity of S with respect to variationssatisfying the previous boundary conditions (2.42) and (2.43)

0 = δS =∫ t2

t1dt∫

dxδL =∫ t2

t1dt∫

dx

(∂L∂ϕ

δϕ+∂L∂ϕ

δϕ+∂L∂ϕ′ δϕ

′)

(2.44)

Integrating by parts

0 =∫ t2

t1dt∫

dx[∂L∂ϕ

δϕ+∂

∂t

(∂L∂ϕ

δϕ

)−(∂

∂t

∂L∂ϕ

)δϕ

+∂

∂x

(∂L∂ϕ′ δϕ

)−(∂

∂x

∂L∂ϕ′

)δϕ]

=∫

dx

[∂L∂ϕ

δϕ

]t2t1

+∫ t2

t1dt

[∂L∂ϕ′ δϕ

]L0

+∫ t2

t1dt∫

dx

[∂L∂ϕ

− ∂

∂t

∂L∂ϕ

− ∂

∂x

∂L∂ϕ′

]δϕ (2.45)

The boundary terms are zero due to eqs. (2.42) and (2.43). Then from the arbi-trariness of δϕ within the region of integration, we get the Euler-Lagrange equations

∂L∂ϕ

− ∂

∂t

∂L∂ϕ

− ∂

∂x

∂L∂ϕ′ = 0 (2.46)

In fact δϕ can be chosen to be zero everywhere except for a small region around anygiven point x (see Fig. (2.1).

x

Fig. 2.1 - Here the arbitrary variation δϕ(x) is chosen to be zero all along thestring, except for a small region around the point x.

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

S =∫ t2

t1dt∫VdnxL

(ϕi, ϕi,

∂ϕi∂xα

)(2.47)

17

Here V is the spatial volume of integration. We will require again the stationarity ofthe action with respect to variations of the fields satisfying the boundary conditions

δϕi(xα, t) = 0, on Σ, for any t, t1 ≤ t ≤ t2 (2.48)

where Σ is the boundary of V , and

δϕi(xα, t1) = δϕi(xα, t2) = 0, for any xα ∈ V (2.49)

The first boundary conditions are required because, in the general case, one requiresthe fields to go to zero at the boundary of the spatial region (usually the infinite).The Euler-Lagrange equations one gets in this case are

∂L∂ϕi

− ∂

∂t

∂L∂ϕi

− ∂

∂xα

∂L

∂∂ϕi∂xα

= 0, i = 1, . . . , N, α = 1, . . . , n (2.50)

To go to the hamiltonian description one introduces the momentum densities con-jugated to the fields ϕi:

Πi =∂L∂ϕi

(2.51)

and the hamiltonian densityH =

∑i

Πiϕi − L (2.52)

In the case of the string one gets from eq. (2.37)

∂L∂ϕ

= ϕ,∂L∂ϕ′ = −v2ϕ′,

∂L∂ϕ

= 0 (2.53)

From which one recovers the equations of motion for the field ϕ. Furthermore

Π = ϕ (2.54)

implying

H = Πϕ− L = Π2 −(1

2Π2 − 1

2v2ϕ′2

)=

1

2

(Π2 + v2ϕ′2

)(2.55)

which coincides with the energy density given in eq. (2.36).A big merit of the lagrangian formalism is the possibility to formulate in a simple

way the symmetry properties of the theory. We shall see later on that this is dueto the first theorem of Emmy Noether which allows to put in a direct relationthe symmetry properties of the lagrangian and the conservation laws. Due to thiscorrespondence it is also possible to make use of the theorem in a constructive way,that is to restrict the possible forms of the lagrangian from the requirement of a givenset of symmetries. We will discuss later on the theorem. For the moment being wewill show how the equations of the vibrating string give rise to conservation laws.

18

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

E(a, b) =1

2

∫ b

adx[ϕ2 + v2ϕ′2

](2.56)

We can evaluate its time variation

dE(a, b)

dt=

∫ b

adx[ϕϕ+ v2ϕ′ϕ′

]= v2

∫ b

adx[ϕϕ′′ + ϕ′ϕ′

]= v2

∫ b

adx

∂

∂x

[ϕϕ′

]= v2

[ϕϕ′

]ba

(2.57)

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

P (x, t) = −v2ϕϕ′ (2.58)

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

−dE(a, b)dt

= [P (b, t)− P (a, t)] (2.59)

This is the classical energy conservation law, expressing the fact that if the energydecreases in the segment [a, b], then there must be a flux of energy at the endpoints a and b. The total energy is conserved due to the boundary conditions,P (0, t) = P (L, t). But the previous law says something more, because it gives us alocal conservation law, as it follows by taking the limit b→ a. In fact, in this limit

E(a, b) → (b− a)H (2.60)

with H given by (2.55), and∂H∂t

+∂P

∂x= 0 (2.61)

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

2.3 The canonical quantization of a continuum

system

As we have seen in Section 2.2, in a field theory one defines the density of conjugatedmomenta as

Πi =∂L∂ϕi

(2.62)

it is then natural to assume the following commutation relations

[ϕi(xα, t),Πj(yα, t)] = iδijδn(xα − yα) α = 1, . . . , n, i, j = 1, . . . , N (2.63)

19

and[ϕi(xα, t), ϕj(yα, t)] = 0, [Πi(xα, t),Πj(yα, t)] = 0 (2.64)

In the string case we have Π = ϕ and we reproduce eq. (2.23). Starting from theprevious commutation relations and expanding the field in terms of normal modesone gets back the commutation relations for the creation and annihilation operators.Therefore we reconstruct the particle interpretation. Using the Heisenberg repre-sentation (but omitting from now on the corresponding index for the operators),the expansion of the string field in terms of creation and annihilation operators isobtained through the eqs. (2.22) and (2.12).

ϕ(x, t) =1√L

∑j

ei2π

LjxQj

=1√L

∑j

1√2ωj

ei2π

Ljx (

aj(t) + a†−j(t))

=1√L

∑j

1√2ωj

ei2πL jxaj(t) + e

−i2πLjxa†j(t)

(2.65)

Using the equations of motion of the string

ϕ− v2ϕ′′ = 0 (2.66)

we find from the above expansion of ϕ in terms of Qj

Qj + ω2jQj = 0 (2.67)

Using the decomposition (2.6) of aj in terms of Qj and Pj = Q†j, we get

aj + iωjaj = 0 (2.68)

from which

aj(t) = aj(0)e−iωjt ≡ aje

−iωjt, a†j(t) = a†j(0)eiωjt ≡ a†je

iωjt (2.69)

and

ϕ(x, t) =1√L

∑j

1√2ωj

ei(2π

Ljx− ωjt

)aj + e

−i(2π

Ljx− ωjt

)a†j

(2.70)

From this equation one gets immediately the commutation rules for the creationand annihilation operators, but before doing that let us notice the structure of theprevious expansion. This can be written in the following way

ϕ(x, t) =∑j

[fj(x, t)aj + f ∗

j (x, t)a†j

](2.71)

20

with

fj(x, t) =1√2ωjL

ei(2π

Ljx− iωjt

)(2.72)

or

fj(x, t) =1√2ωjL

ei(kjx− iωjt) (2.73)

where we have made use of the definition of kj (see eq.(1.37))

kj =2π

Lj (2.74)

The functions fj(x, t) and their complex conjugated satisfy the wave equation

∂2fj(x, t)

∂t2− v2

∂2fj(x, t)

∂x2= 0 (2.75)

and the boundary conditions

fj(0, t) = fj(L, t) (2.76)

It is immediate to verify that they are a complete set of orthonormal functions∑j

f ∗j (x, t)i∂

(−)t fj(y, t) = δ(x− y) (2.77)

∫ L

0dxf ∗

j (x, t)i∂(−)t fl(x, t) = δjl (2.78)

whereA∂

(−)t B = A(∂tB)− (∂tA)B (2.79)

Let us consider the first relation. We have

∑j

f ∗j (x, t)i∂

(−)t fj(y, t) =

∑j

2ωjf∗j (x, t)fj(y, t) =

∑j

1

Le−ikj(x− y) = δ(x− y)

(2.80)Evaluating this expression with two fj(x, t)’s or two f

∗j (x, t)’s one gets zero. As far

as the second relation is concerned we get∫ L

0dxf ∗

j (x, t)i∂(−)t fl(x, t) =

∫ L

0dx[ωl + ωj]f

∗j (x, t)fl(x, t)

]=

1

L

ωl + ωj2√ωjωl

∫ L

0dxeix(kl − kj)ei(ωj − ωl)t

=1

L

ωl + ωj2√ωjωl

ei(ωj − ωl)tLδjl = δjl (2.81)

21

Also in this case, by taking two fj(x, t)’s or two f∗j (x, t)’s, the result is zero due to

the factor ωl − ωj.We repeat that the set fj(x, t) is a complete set of orthonormal solutions of the

wave equation with periodic boundary conditions. A legitimate question is whythe operator ∂

(−)t appears in these relations. The reason is that the scalar product

should be time independent (otherwise two orthonormal solutions at a given timecould loose these feature at a later time). For instance, in the case of the Schrodingerequation, we define the scalar product as∫

d3xψ∗(x, t)ψ(x, t) (2.82)

because for hermitian hamiltonians this is indeed time independent, as it can bechecked by differentiating the scalar product with respect to time and using theSchrodinger equation:

d

dt

∫d3xψ∗(x, t)ψ(x, t) =

∫d3x

[ψ∗ψ + ψ∗ψ

]=∫

d3x [i(Hψ)∗ψ − iψ∗(Hψ)] = 0

(2.83)In the present case we can define a time independent scalar product, by consider-ing two solutions f and f of the wave equation, and evaluating the following twoexpressions ∫ L

0dxf

[∂2f

∂t2− v2

∂2f

∂x2

]= 0 (2.84)

∫ L

0dx

[∂2f

∂t2− v2

∂2f

∂x2

]f = 0 (2.85)

Subtracting these two expressions one from the other we get

∫ L

0dx

[∂

∂t

(f∂f

∂t− ∂f

∂tf

)− v2

∂

∂x

(f∂f

∂x− ∂f

∂xf

)](2.86)

If both f and f satisfy periodic boundary conditions, the second term is zero, andit follows that the quantity ∫ L

0dxf∂

(−)t f (2.87)

is a constant of motion. Using eq. (2.78), we can invert the relation between fieldand creation and annihilation operators. We get∫ L

0dxf ∗

j (x, t)i∂(−)t ϕ(x, t) =

∑k

∫ L

0dxf ∗

j (x, t)i∂(−)t

[fk(x, t)ak + f ∗

k (x, t)a†k

]= aj

(2.88)and therefore

aj =∫ L

0dxf ∗

j (x, t)i∂(−)t ϕ(x, t) (2.89)

22

and

a†j =∫ L

0dxϕ(x, t)i∂

(−)t fj(x, t) (2.90)

From the field commutation relations we find

[aj, a†k] =

∫ L

0dxdy[(if ∗

j ϕ− if ∗j ϕ)(x,t), iϕfk − iϕfk)(y,t)]

=∫ L

0dxdy

(−f ∗

j fk(−iδ(x− y))− f∗kfk(iδ(x− y))

)=

∫ L

0dxf ∗

j i∂(−)t fk = δjk (2.91)

In analogous way we get[aj, ak] = [a†j, a

†k] = 0 (2.92)

We have seen that the total energy of the string is a constant of motion. Thereis another constant which corresponds to the total momentum of the string, definedby

P =∫ L

0dxP = −

∫ L

0dx ϕϕ′ (2.93)

We will show in the following that this expression is just the total momentum of thestring, by showing that its conservation derives from the invariance of the theoryunder spatial translations. For the moment being let us check that this is in fact aconserved quantity:

dP

dt= −

∫ L

0dx(ϕϕ′ + ϕϕ′) = −

∫ L

0dx

∂

∂x

1

2(ϕ2 + v2ϕ′2) = 0 (2.94)

where we have used the equations of motion of the string and the boundary condi-tions. By using the field expansion

P = −∑j,l

∫ L

0dx ωjkl[fjaj − f∗

j a†j][flal − f ∗

l a†l ]

= −∑j,l

∫ L

0dx

ωjkl2L

√ωjωl

[ei(kjx− ωjt)aj − e−i(kjx− ωjt)a†j

]

×[ei(klx− ωlt)al − e−i(klx− ωlt)a†l

]= −

∑l

1

2kl[a−lale

−2iωlt + a†−la†l e2iωlt − ala

†l − a†lal] (2.95)

The first two terms in the last step give zero contribution because they are antisym-metric in the index of summation(kl ≈ l). Therefore

P =1

2

∑j

kj[aja†j + a†jaj] =

∑j

kja†jaj (2.96)

23

where we have used ∑j

kj = 0 (2.97)

for the antisymmetry j. We see that P has an expression similar to that of H (seeeq. (2.19). We deduce that the states

(a†−N/2)n−N/2 · · · (a†j)nj · · · |0⟩ = |ψ⟩ (2.98)

have energyH|ψ⟩ = (n−N/2ω−N/2 + · · ·+ njωj + · · ·)|ψ⟩ (2.99)

and a momentum

P |ψ⟩ = (n−N/2k−N/2 + · · ·+ njkj + · · ·)|ψ⟩ (2.100)

as it follows from[H, a†j] = ωja

†j, [P, a†j] = kja

†j (2.101)

In a complete general way, if an operator A can be written as

A =∑j

αja†jaj (2.102)

we have[A, a†j] = αja

†j (2.103)

Then, if |ρ⟩ is an eigenket of A with eigenvalue ρ, a†j|ρ⟩ is an eigenket of A with

eigenvalue ρ+αj. Therefore a†j and aj increase and lower the eigenvalues of A. This

is a trivial consequence of the commutation relations

A(a†j|ρ⟩) = (a†jA+ αja†j)|ρ⟩ = (ρ+ αj)a

†j|ρ⟩ (2.104)

24

Chapter 3

The Klein-Gordon field

3.1 Relativistic quantum mechanics and its prob-

lems

The extension of quantum mechanics to the relativistic case gives rise to numerousproblems. The difficulties originate from the relativistic dispersion relation

E2 = |p|2 +m2 (3.1)

This relation gives rise to two solutions

E = ±√|p|2 +m2 (3.2)

It is not difficult to convince himself that the solutions with negative energy haveunphysical behaviour. For instance, increasing the momentum, the energy decreases!But their presence is not a real problem at a classical level. In fact, we see fromeq. (3.2) that there is a gap of at least 2m between the energies of the two types ofsolutions. At the classical level, the way in which the energy is transferred is alwaysa continuous one. So there is no way to start with an energy positive particle andfinish with a negative energy one. On the contrary, in quantum mechanics one can,through the emission of a quantum of energy E > 2m, go from positive energy tonegative energy states. Since a system behaves in such a way to lower its energy, allthe positive energy states would migrate to negative energy ones, causing a collapseof the usual matter. In fact we shall see that it is not possible to ignore this kindof solutions, but they will be reinterpreted in terms of antiparticles. This will allowus to get rid of the problems connected with the negative energy solutions, but itwill cause another problem. In fact, one of the properties of antiparticles is thatthey may be annihilated or pair created. Let us suppose now to try to localizea particle on a distance of the order of its Compton wave-length, that is of order1/m. By doing that we will allow an uncertainty on the momentum of about m,due to the uncertainty principle. This means that the momentum (and the energy)

25

of the particle could reach values of order 2m, enough to create a pair particle-antiparticle. This will be possible only violating the conservation of energy andmomentum. Again, this is the case if the violation of energy conservation is on atime-scale of order ∆t ≈ ∆x ≈ 1/m. But this is the scale of the Compton wave-length, therefore the attempt of localization will be nullified by the fact that at thesame scale we start pair creating particles and antiparticles, meaning that we willbe unable to define the concept of a localized single particle. At the Compton scalethere is no such a thing as a particle, but the picture we get from the previousconsiderations is the one of a cloud of particles and antiparticles surrounding ourinitial particle, and there is no way to distinguish our particle from the many aroundit.

These considerations imply that the relativistic theories cannot be seen as the-ories at a fixed number of particles, which is the usual way of describing things inordinary quantum mechanics. In this sense a field theory, as far as we have seen tillnow, looks as the most natural way to describe such systems. In fact, it embeds,in a natural way, the possibility of describing situations with variable number ofparticles.

One can look also at different ways leading to the necessity of using field theories.For instance, by looking at the quantization of the electromagnetic field, physicistsrealized that this gives a natural explanation of the particle-wave duality, and thatin the particle description one has to do with a variable number of photons. On thecontrary, physical entities as the electrons, were always described in particle termstill 1927, when Davisson and Germer showed experimentally their wave-like behav-ior. This suggested that the particle-wave duality would be a feature valid for anytype of waves or particles. Therefore, based on the analogy with the electromagneticfield, it is natural to introduce a field for any kind of particle.

Historically, the attempt of making quantum mechanics a relativistic theorywas pursued by looking for relativistic generalizations of the Schrodinger equation.Later it was realized that these equations should be rather used as equations forthe fields describing the corresponding particles. As we shall see, these equationsdescribe correctly the energy dispersion relation and the spin of the various particles.Therefore they can be used as a basis for the expansion of the field in terms ofcreation and annihilation operators. In order to illustrate this procedure, let usstart considering the Schrodinger equation for a free particle

i∂ψ

∂t= Hψ (3.3)

where H is the hamiltonian

H =|p|2

2m= − 1

2m|∇|2 (3.4)

If ψ describes an eigenstate of the energy and of the momentum

ψ ≈ e−iEt+ ip · x (3.5)

26

all the information in the equation is to describe correctly the energy-momentumrelation

E =|p|2

2m(3.6)

In the relativistic case one could try to reproduce the positive energy branch of thedispersion relation (3.1). In that case one could start from the hamiltonian

H =√|p|2 +m2 (3.7)

which gives rise to the following wave equation

i∂ψ

∂t=(√

−|∇|2 +m2

)ψ (3.8)

The two obvious problems of this equation are

• spatial and time derivatives appear in a non symmetric way;

• the equation is non-local, that is it depends on an infinite number of spatialderivatives

(√−|∇|2 +m2

)ψ = m

√√√√1− |∇|2

m2

ψ = m∑k

ck(|∇|2

)kψ (3.9)

Both these difficulties are eliminated by iteration

−∂2ψ

∂t2=(−|∇|2 +m2

)ψ (3.10)

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

∂2 +m2)ψ = 0 (3.11)

where

∂2 =∂2

∂t2− |∇|2 (3.12)

is the D’Alembert operator in (3 + 1) dimensions. Notice that in order to solvethe difficulties we have listed above we have been obliged to consider both types of

solutions: positive energy, E =√|p|2 +m2, and negative energy E = −

√|p|2 +m2.

The equation we have obtained in this way is known as the Klein-Gordon equation.As relativistic extension of the Schrodinger theory it was initially discarded becauseit gives rise to a non definite positive probability. In fact, if ψ and ψ⋆ are twosolutions of such an equation, we can write the following identity

0 = ψ⋆(∂2 +m2)

)ψ − ψ

(∂2 +m2)

)ψ⋆ = ∂µ [ψ

⋆∂µψ − (∂µψ⋆)ψ] (3.13)

27

Therefore the currentJµ = ψ⋆∂µψ − (∂µψ

⋆)ψ (3.14)

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

∫d3x(ψ⋆ψ − ψ⋆ψ) (3.15)

is a constant of motion. But we cannot interpret the time-component of the currentas a probability density, as we do in the Schrodinger case, because it is not positivedefinite.

Let us end this Section by stating our conventions for the relativistic notations.The position and momentum four-vectors are given by

xµ = (t, x), pµ = (E, p), µ = 0, 1, 2, 3 (3.16)

The metric tensor gµν is diagonal with components (+1,−1,−1,−1). The four-momentum operator in coordinate space is given by

pµ → i∂

∂xµ=

(i∂

∂t,−i∇

)(3.17)

We have also the following relations

p2 = pµpµ → − ∂

∂xµ

∂

∂xµ= −∂2 (3.18)

x · p = Et− p · x (3.19)

3.2 Quantization of the Klein-Gordon field

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

• construction of the lagrangian density and determination of the canonical mo-mentum density Π(x);

• quantization through the requirement of canonical commutation relations

[ϕ(x, t),Π(y, t)] = iδ3(x− y), [ϕ(x, t), ϕ(y, t)] = 0, [Π(x, t),Π(y, t)] = 0(3.20)

• expansion of ϕ(x, t) 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.

28

We start by the construction of the lagrangian, requiring that the related Euler-Lagrangian equation gives rise to the Klein-Gordon equation. To this end let usrecall how one proceeds in the discrete case. Suppose to have a system of N degreesof freedom satisfying the following equations of motion

miqi = −∂V∂qi

(3.21)

Multiplying these equations by some arbitrary variations δqi, satisfying the followingboundary conditions

δqi(t1) = δqi(t2) = 0 (3.22)

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

∫ t2

t1dt

[N∑i=1

miqiδqi

]= −

∫ t2

t1dt

N∑i=1

δqi∂V

∂qi(3.23)

Integrating by parts

δ

∫ t2

t1dt

[1

2

N∑i=1

miq2i − V

]−[N∑i=1

miqiδqi

]t2t1

= 0 (3.24)

Using the boundary conditions we see that if the equations of motion are satisfied,than the lagrangian, as defined by

S =∫ t2

t1

[1

2

N∑i=1

miq2i − V

]dt (3.25)

is stationary. Conversely from the requirement that the action is stationary undervariations satisfying eq. (3.22), the equations of motion follow. Analogously, in theKlein-Gordon case, we multiply the equation by arbitrary local variations of thefield δϕ(x) = ϕ(x)− ϕ(x), with boundary conditions

δϕ(x, t1) = δϕ(x, t2) = 0, limx→∞

δϕ(x, t) = 0 (3.26)

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

0 =∫ t2

t1dt∫

d3x

[∂

∂t

(ϕδϕ

)− ϕδϕ− ∇ ·

(∇ϕδϕ

)+ ∇ϕ · ∇δϕ+m2ϕδϕ

](3.27)

Using the boundary conditions we get

0 = δ∫ t2

t1dt∫

d3x[1

2ϕ2 − 1

2∇ϕ · ∇ϕ− 1

2m2ϕ2

](3.28)

Therefore the lagrangian will be given by

L =∫

d3xL (3.29)

29

with

L =1

2

[∂µϕ∂

µϕ−m2ϕ2]

(3.30)

In fact, we have just shown that the quantity (the action)

S =∫ t2

t1dtL (3.31)

is stationary at the point in which the equations of motion are satisfied. We cannow write down the canonical momentum density

Π =∂L∂ϕ

= ϕ (3.32)

and the canonical commutation relations

[ϕ(x, t), ϕ(y, t)] = iδ3(x− y), [ϕ(x, t), ϕ(y, t)] = [ϕ(x, t), ϕ(y, t)] = 0 (3.33)

Let us now construct a complete set of solutions of the Klein-Gordon equation. Firstof all we need a scalar product. But we have already one, because we have shown inthe previous Section that the Klein-Gordon equation admits a conserved quantity(see eq. (3.15)), therefore, if f and g are two solutions, the scalar product is

⟨f |g⟩ = i∫

d3xf ∗∂(−)t g (3.34)

Let us now look for plane-wave solutions

f = A(k)e−ikx = A(k)e−i(k0x0 − k · x) (3.35)

From the wave equation we get

(∂2 +m2)f = (−k2 +m2)f = 0 (3.36)

from whichk2 = m2 =⇒ k20 = |k|2 +m2 (3.37)

To fix the normalization, we proceed as in the one-dimensional case by taking a finitevolume and requiring periodic boundary conditions (normalization in the box). Bytaking a cube of side L we require

ϕ(x+ L, y, z, t) = ϕ(x, y + L, z, t) = ϕ(x, y, z + L, t) = ϕ(x, y, z, t) (3.38)

it follows

k =2π

Ln (3.39)

wheren = n1i1 + n2i2 + n3i3 (3.40)

30

is a vector with integer components (n1, n2, n3). The normalization condition is

⟨fk|fk′⟩ = i∫Vd3xf ∗

k∂(−)t fk′ = δk,k′ (3.41)

where the delta is a Kronecker symbol defined as

δk,k′ =3∏i=1

δni,n′i

(3.42)

with n e n′ are two vectors with integer components, related to k and k′, by therelation (3.39). It follows∫

Vd3xA∗

kAk′e

i(k0 − k0′)x0 − i(k − k′) · x(k′0 + k0) = δk,k′ (3.43)

Using ∫Ldxe

i2π

L(n1 − n′

1)x= Lδn1,n1

′ (3.44)

we get ∫Vd3xei(k − k′) · x = L3δk,k′ (3.45)

from whichi∫Vd3xf ∗

k∂(−)t fk′ = |Ak|

22k0L3δk,k′ (3.46)

where

k20 =(2π

L

)2

|n|2 +m2 (3.47)

By considering, for the moment being, the positive solution of this equation, weobtain

Ak =1

L3/2

1√2ωk

, ωk =

√(2π

L

)2

|n|2 +m2 =√|k|2 +m2 (3.48)

and the normalized solution turns out to be

fk(x) =1

L3/2

1√2ωk

e−ikx (3.49)

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

⟨fk|fk′⟩ = i∫

d3xf ∗k∂(−)t fk′ = δ3(k − k′) (3.50)

In this case the spatial momentum can assume all the possible values in R3. Itfollows ∫

d3xA∗kAk′e

ikx− ik′x(k0 + k′0) = (2π)3δ3(k − k′)|Ak|22k0 (3.51)

31

and the corresponding normalization is

Ak =1√(2π)3

1√2ωk

(3.52)

where

ωk =√|k|2 +m2 (3.53)

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

1√V

→ 1√(2π)3

(3.54)

The wave function in the continuum is

fk(x) =1√(2π)3

1√2ωk

e−ikx (3.55)

In both cases the dispersion relation

k20 = |k|2 +m2 (3.56)

is obviously satisfied. But we have to remember that it has two solutions

k0 = ±√|k|2 +m2 = ±ωk (3.57)

As a consequence we get two kind of wave functions having positive and negative

energy and behaving as e−iωkx0 and eiωkx0 , ωk > 0, respectively. The second kindof solutions has negative norm in the scalar product we have defined. This wouldbe a big problem if this equation had the same interpretation as the Schrodingerequation. In the field theory, no such a problem exists. In fact, the physical Hilbertspace is the Fock space, where the scalar product is between the states build upin terms of creation and annihilation operators. Having two types of solutions themost general expansion for the field operator (in the Heisenberg representation) is

ϕ(x) =1√(2π)3

∫d3k

1√2ωk

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

](3.58)

In the second term we can exchange k → −k, obtaining (kx = ωkx0 − k · x)

ϕ(x) =1√(2π)3

∫d3k

1√2ωk

[a(k)e−ikx + a(−k)eikx

]≡∫

d3k[fka(k) + f ∗ka(−k)]

(3.59)

32

Notice that the energy positive and negative solutions are orthogonal (rememberthe one-dimensional case discussed in Section 2.3). We can then invert the previousexpansion with the result

a(k) = i∫

d3xf ∗k(x)∂

(−)t ϕ(x), a(−k) = i

∫d3xϕ(x)∂

(−)t fk(x) (3.60)

If ϕ(x) is a hermitian Klein-Gordon field, we have

a(−k) = a†(k) (3.61)

and the expansion becomes

ϕ(x) =∫

d3k[fk(x)a(k) + f ∗k(x)a†(k)] (3.62)

From these equations one can evaluate the commutators among the operators a(k)

e a†(k), obtaining

[a(k), a†(k′)] = δ3(k − k′) (3.63)

[a(k), a(k′)] = [a†(k), a†(k′)] = 0 (3.64)

These commutation relations depend on the normalization defined for the fk’s. Forinstance, if we change this normalization by a factor Nk

⟨fk|fk′⟩ = i∫

d3xf ∗k∂(−)t fk′ = Nkδ

3(k − k′) (3.65)

leaving unchanged the expansion for the field

ϕ =∫

d3k[fka(k) + f∗ka†(k)] (3.66)

we get

a(k) =i

Nk

∫d3xf ∗

k∂(−)t ϕ, a†(k) =

i

Nk

∫d3xϕ∂

(−)t fk (3.67)

and therefore

[a(k), a†(k′)] =i

NkNk′

∫d3xf ∗

k∂(−)t fk′ =

1

Nk

δ3(k − k′) (3.68)

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

ϕ(x) =1

(2π)3

∫d3k

1

2ωk[A(k)e−ikx + A†(k)eikx] (3.69)

The name comes from the fact that the factor 1/2ωk makes the integration over thethree-momentum Lorentz invariant. In fact one has

1

(2π)3

∫d3k

1

2ωk=

1

(2π)4

∫d4k(2π)δ(k2 −m2)θ(k0) (3.70)

33

as it follows by noticing that for k0 ≈ ωk

k2 −m2 ≈ 2ωk(k0 −√|k|2 +m2) (3.71)

In this case the basis functions for the expansion are

fk(x) =1

(2π)31

2ωke−ikx (3.72)

with normalization

i∫

d3xf ∗k∂(−)t fk′ =

1

(2π)31

2ωkδ3(k − k′) (3.73)

and therefore[A(k), A†(k′)] = (2π)32ωkδ

3(k − k′) (3.74)

3.3 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, it is convenient to evaluate, in general, thevariation of a local quantity F (x) (that is a function of the space-time point)

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

∼= F (x)− F (x) + δxµ∂F (x)

∂xµ(3.75)

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) (3.76)

Then we get

∆F (x) = δF (x) + δxµ∂F (x)

∂xµ(3.77)

Let us now start form a generic four-dimensional action

S =∫Vd4x L(ϕi, x), i = 1, . . . , N (3.78)

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(3.79)

34

If the action is invariant under the transformation, then

SV ′ = SV (3.80)

The variation of S under the transformation (3.79) is given by (here ∂µ = ∂/∂µ andϕi,µ = ∂ϕi/∂xµ)

δSV =∫V ′

d4x′L(ϕi, x′)−∫Vd4xL(ϕi, x)

=∫Vd4xL(ϕi, x+ δx)

∂(x′)

∂(x)−∫Vd4xL(ϕi, x)

≈∫Vd4xL(ϕi, x+ δx)(1 + ∂µδx

µ)−∫Vd4xL(ϕi, x)

=∫Vd4x[L(ϕi, x+ δx)− L(ϕi, x)] +

∫Vd4xL(ϕi, x)∂µδxµ

≈∫Vd4x

[∂L∂ϕi

δϕi +∂L∂ϕi,µ

δϕi,µ +∂L∂xµ

δxµ]+∫Vd4xL∂µδxµ

=∫Vd4x

[∂L∂ϕi

− ∂µ∂L∂ϕi,µ

]δϕi +

∫Vd4x∂µ

[Lδxµ + ∂L

∂ϕi,µδϕi

](3.81)

The first term in the last line is zero due to the Euler-Lagrange equations of motion

∂L∂ϕi

− ∂µ∂L∂ϕi,µ

= 0 (3.82)

Therefore, if the action in invariant under the transformation under consideration,using eq. (3.79), we get∫

Vd4x∂µ

[Lδxµ + ∂L

∂ϕi,µ∆ϕi − δxν

∂L∂ϕi,µ

ϕi,ν

]= 0 (3.83)

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. (3.83) we get the following local conservation law

T µν =∂L∂ϕi,µ

ϕi,ν − Lgµν , ∂µTµν = 0 (3.84)

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

Pν =∫

d3xT 0ν (3.85)

35

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 (3.86)

with an associated constant of motion given by

Q =∫

d3xJ0 (3.87)

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

(3.88)

For an infinitesimal transformation

x′ = x+ δx (3.89)

it followsx2 ≈ x2 + 2x · δx =⇒ x · δx = 0 (3.90)

Since Lorentz transformations are linear

x′µ = Λµνxν ≈ xµ + ϵµνx

ν (3.91)

we getx · δx = 0 =⇒ xµϵµνx

ν = 0 (3.92)

The most general solution for the parameters ϵµν of the transformation is that theform an antisymmetric second order tensor

ϵµν = −ϵνµ (3.93)

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 (3.94)

36

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

](3.95)

and defining (watch at the change of sign)

Mµρν = xρT

µν − xνT

µρ − ∂L

∂ϕi,µΣijρνϕ

j (3.96)

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

∂µMµνρ = 0 (3.97)

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

Mνρ =∫

d3xM0νρ (3.98)

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

In the case of Klein-Gordon

Tµν = ∂µϕ∂νϕ− 1

2

(∂ρϕ∂

ρϕ−m2ϕ2)gµν (3.99)

from which

T 00 =

1

2ϕ2 +

1

2

(|∇ϕ|2 +m2ϕ2

)(3.100)

This current corresponds to the invariance under time translations, and it must beidentified with the energy density of the field (compare with the equation (2.55 forthe one-dimensional case). In analogous way

T 0i = ϕ

∂ϕ

∂xi(3.101)

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

P 0 = H =∫

d3xT 00 =1

2

∫d3x

(Π2 + |∇ϕ|2 +m2ϕ2

)(3.102)

P i =∫

d3xT 0i = −∫

d3xΠ∂ϕ

∂xi, (P = −

∫d3xΠ∇ϕ) (3.103)

37

3.4 Energy and momentum of the Klein-Gordon

field

It is very easy to verify that the energy density found previously coincides with thehamiltonian density evaluated in the canonical way through the Legendre transfor-mation of the lagrangian density

H = Πϕ− L (3.104)

We will verify now, that the momentum P µ is the generator, as it should be, ofthe space-time translations. Which amounts to say that it satisfies the followingcommutation relation with the field

[ϕ(x), P µ] = i∂ϕ

∂xµ(3.105)

In fact

[ϕ(y, t), H] =1

2

∫d3x[ϕ(y, t),Π2(x, t)] = iΠ(y, t) = iϕ(y, t) (3.106)

Analogously

[ϕ(y, t), P i] = −∫

d3x[ϕ(y, t),Π(x, t)∂ϕ(x, t)

∂xi] = −i∂ϕ(y, t)

∂yi= i

∂ϕ(y, t)

∂yi(3.107)

Therefore the operator

U = eiaµPµ (3.108)

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

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

∂xµ≈ ϕ(x+ a) (3.109)

With a calculation completely analogue to the one done in Section 2.3 we canevaluate the hamiltonian and the momentum in terms of the creation and annihila-tion operators

H =1

2

∫d3kωk[a

†(k)a(k) + a(k)a†(k)] (3.110)

P =∫

d3kka†(k)a(k) (3.111)

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

[H, a†(k)] = ωka†(k), [P, a†(k)] = ka†(k) (3.112)

This shows that the operators a†(k), acting on the vacuum, create states of momen-

tum k and energy ωk =√|k|2 +m2, whereas the annihilation operators a(k) destroy

38

the corresponding states. In the case of the box normalization, for any k = (2π/L)n(that is for any choice of the three integer components of the vector n, (n1, n2, n3)),one can build up a state |nk⟩ such that

|nk⟩ =1√nk!

(a†(k)

)nk |0⟩ (3.113)

contains nk particles of momentum k. The most general state is obtained by tensorproduct of states similar to the previous one. Any of these states is characterizedby a triple of integers defining the momentum k, that is

|nk1 . . . nkα⟩ =∏⊗|nki⟩ =

1√nk1 ! · · ·nkα !

(a†(k1))nk1 · · · (a†(kα))nkα |0⟩ (3.114)

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

|0⟩ =∏⊗|0⟩i (3.115)

where |0⟩i is the fundamental state for the momentum in the cell i. That is

aki|0⟩i = 0 (3.116)

In this normalization the hamiltonian is given by

H =1

2

∑k

ωk[a†(k)a(k) + a(k)a†(k)] (3.117)

and therefore

H|0⟩ = 1

2

∑k

ωk|0⟩ (3.118)

This sum is infinite. Recalling that k = (2π/L)n, it follows that the cell in the

k-space has a volume

∆Vk =(2π)3

L3(3.119)

from which

1

2

∑k

ωk =1

2

∑k

∆Vkωk∆Vk

=⇒ 1

2

L3

(2π)3

∫d3k

√|k|2 +m2 (3.120)

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

normal product. In other words by subtracting the infinite energy of the vacuumfrom the hamiltonian. In the box normalization we have

: H :=∑k

ωka†(k)a(k) (3.121)

39

whereas in the continuum

: H :=∫

d3kωka†(k)a(k) (3.122)

As we see, the energy of the vacuum depends on the quantization volume. Thisimplies that it depends on the boundary conditions of the problem. In the realvacuum this is not a difficulty, but it must be considered when one quantize fieldswhich are inside a finite given volume. In this case this dependence produces mea-surable effects, as it was pointed out theoretically by Casimir in 1948, and thenproved experimentally by Sparnay in 1958.

RL

Fig. 3.1 - The Casimir effect

We will discuss very briefly the Casimir effect arising when we have an electro-magnetic field confined between two large perfectly conducting plates. We idealizethe two plates as two large parallel squares of side L at a distance R ≪ L. Thetheory shows that there is an attractive force per unit surface between the two platesgiven by

p = − π2

240

/hc

R4= − 0.013

(Rµm)4dyn/cm2 (3.123)

We can understand the origin of this force in a very qualitative way by quantizingthe electromagnetic field (that we will take here as a Klein-Gordon field with zeromass,m = 0) in a box of side L. The vacuum energy will be

E0 ≈ L3∫ kmax

1/Lk d3k (3.124)

with the integration between a lower momentum of order 1/L and an arbitrary uppermomentum which is necessary in order to make finite the integral. If we insert twoplates of side L, as shown in Fig. 3.1, at a distance R, the energy of the field in thisregion, before the introduction of the plates is

E ≈ L2R∫ kmax

1/Lk d3k (3.125)

40

When we insert the plates we get an analogous result, but the lower momentum willbe of order 1/R. therefore the variation of the energy results to be

∆E ≈ L2R∫ 1/L

1/Rk d3k = L2R

∫ 1/L

1/Rk3 dk =

L2R

4

[(1

L

)4

−(1

R

)4]

(3.126)

Therefore, for R ≪ L, we get

∆E ≈ −L2

R3(3.127)

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

p ≈ −∂∆E/L2

∂R≈ − 1

R4(3.128)

3.5 Locality and causality in field theory

For a free particle there are generally three conserved quantum numbers, as the spa-tial momentum, or energy, angular momentum and its third component. All thesequantities can be expressed as spatial integrals of local functions of the fields. Thelocality property is a crucial one and is connected with the causality. To understandthis point let us consider the following example. For a Klein-Gordon free field thereis a further constant of motion, the number of particles

N =∫

d3ka†(k)a(k) (3.129)

We will show now that this cannot be written as the spatial integral of a localquantity, and that this implies the non observability of the quantity number ofparticles. We know that, apart the energy momentum tensor, the Klein-Gordontheory admits a further conserved current

Jµ = ϕ†(∂µϕ)− (∂µϕ†)ϕ (3.130)

However this expression vanishes for a hermitian field. But it turns out that theoperator N can be expressed in terms of the positive energy

ϕ(+)(x) =∫

d3k1√

2ωk(2π)3e−i(ωkt− k · x)a(k) (3.131)

and negative energy components of the field

ϕ(−)(x) = ϕ(+)†(x) (3.132)

In fact, it is not difficult to show that

N =∫

d3xϕ(+)i∂(−)t ϕ(−) (3.133)

41

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

jµ = ϕ(−)(∂µϕ(+))− (∂µϕ

(−))ϕ(+) (3.134)

is a conserved current. However this current is not a local expression in the fieldϕ. This is because ϕ(+) and ϕ(−) are not local functions of ϕ. In fact, in order toproject out these components from the field we need a time integration. In fact, bydefining

ϕ(x) =∫d4kϕ(k)e−ikx (3.135)

with

ϕ(k) =

√2ωk(2π)3

δ(k2 −m2)(a(k)θ(k0) + a†(k)θ(−k0)

)(3.136)

one hasϕ(+)(x) =

∫d4kϕ(+)(k)e−ikx (3.137)

withϕ(+)(k) = θ(k0)ϕ(k) (3.138)

Using the convolution theorem for the Fourier transform we get

ϕ(+)(x) =∫d4x′θ(x− x′)ϕ(x′) (3.139)

But

θ(x− x′) =1

(2π)4

∫d4keik(x− x′)θ(k0)

= δ3(x− x′)∫ dk0

2πeik0(x0 − x′0)θ(k0)

= δ3(x− x′)θ(x0 − x′0) (3.140)

Thereforeϕ(+)(x, x0) =

∫dx′0θ(x0 − x′0)ϕ(x, x

′0) (3.141)

To show the implications of having to do with a non local current, let us define aparticle density operator

N (x) = iϕ(+)∂(−)t ϕ(−) (3.142)

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

[N (x, t),N (y, t)] = 0, x = y (3.143)

However, for local operators, O(ϕ), this commutator is automatically zero, due tothe canonical commutation relations

[O(ϕ(x, t)),O(ϕ(y, t))] = 0, ∀ x = y (3.144)

42

We want to argue that the vanishing of this commutator is just the necessary con-dition in order that O represents an observable quantity. In fact, if the commutatorof a local operator with itself is not zero at space-like distances, then the measure ofthe observable at some point, x, would influence the measures done at points withspace-like separation from x, because we cannot measure the operator simultane-ously at two such points. But this would imply the propagation of a signal at avelocity greater than the light velocity, in contrast with the causality principle. Wesee that the vanishing of the commutator of a local observable with itself at space-like distances is a necessary condition in order to satisfy the causality principle. Weshow now that this is automatically satisfied if the operator under consideration isa local function of the fields. We will start showing that the commutator of thefield with itself is a Lorentz invariant function. Therefore, from the vanishing of thecommutator for separations between points of the type xµ = (t, x), and yµ = (t, y),it follows the vanishing for arbitrary space-like separations. Let us evaluate thecommutator

[ϕ(x), ϕ(y)] =

=∫ d3k1d

3k2(2π)3

√2ωk12ωk2

[[a(k1), a

†(k2)]e−ik1x+ ik2y

+ [a†(k1), a(k2)]eik1x− ik2y

]=

∫ d3k

(2π)32ωk

[e−ik(x− y) − eik(x− y)

]

= −2i∫ d3k

(2π)32ωksin(ωk(x0 − y0))e

ik(x− y) (3.145)

Using eq. (3.70), this expression can be written in invariant form

[ϕ(x), ϕ(y)] =∫ d4k

(2π)3θ(k0)δ(k

2 −m2)[e−ik(x− y) − eik(x− y)

]

=∫ d4k

(2π)3ϵ(k0)δ(k

2 −m2)e−ik(x− y) (3.146)

Since the sign of the fourth component of a time-like fourvector is invariant underproper Lorentz transformations, we see that by putting

[ϕ(x), ϕ(y)] = i∆(x− y) (3.147)

the function

∆(x− y) = −i∫ d4k

(2π)3ϵ(k0)δ(k

2 −m2)e−ik(x− y) (3.148)

is Lorentz invariant and, as such, it depends only on (x − y)2. Since ∆(x − y)vanishes at equal times, it follows that it is zero for arbitrary space-like separations.

43

Therefore the canonical commutation relations make sure the observability for theKlein-Gordon field. For the negative and positive energy components we get

[ϕ(+)(x), ϕ(−)(y)] =∫ d3k

(2π)32ωke−ik(x− y)

=∫ d4k

(2π)3θ(k0)δ(k

2 −m2)e−ik(x− y)

≡ ∆(+)(x− y) (3.149)

Also in this case we have a Lorentz invariant function, and therefore it is enough tostudy its equal times behaviour:

∆(+)(0, x) =∫ d3k

(2π)32ωkeik · x

=∫ k2dkd(cos θ)dφ

(2π)32ωkeikr cos θ

= −i 1

4π2r

∫ ∞

0

kdk

2ωk

[eikr − e−ikr

]

= − 1

8π2r

d

dr

∫ +∞

−∞dk

eikr√|k|2 +m2

(3.150)

By putting k = m sinh θ, dk = m cosh θdθ, we get

∆(+)(0, x) = − 1

8π2r

d

dr

∫ +∞

−∞dθeimr sinh θ (3.151)

Since ∫ +∞

−∞dθeimr sinh θ = iπH

(1)0 (imr) (3.152)

where H is a Hankel’s function, and using

d

drH

(1)0 (imr) = −imH(1)

1 (imr) (3.153)

we obtain∆(+)(0, x) = − m

8πrH

(1)1 (imr) (3.154)

The asymptotic behaviour of the Hankel’s function H(1)1 (imr) for large and small

values of r is given by

limr→∞

H(1)1 (imr) ≈ −

√2

πmre−mr, lim

r→0H

(1)1 (imr) ≈ − 2

mr(3.155)

from which

limr→∞

∆(+)(0, x) ≈ m

8πr

√2

πmre−mr, lim

r→0∆(+)(0, x) ≈ 1

4πr2(3.156)

44

We see that for space-like separations this commutator does not vanish. But forspace separations larger than the Compton wave length 1/m, ∆(+) is practicallyzero. Remember that for an electron the Compton wave length is about 3.9 · 10−11

cm. Clearly, an analogous result is obtained for the commutator of the particledensity operator. From this we can derive the impossibility of localize a Klein-Gordon particle (but the result can be extended to any relativistic particle) overdistances of the order of 1/m. We start defining the following operators

N(V ) =∫Vd3xN (x) (3.157)

R

riV0

Vi

x0

Fig. 3.2 - In order to localize a particle inside V0, there should be no otherparticles within a distance ri ≈ R

where V is a sphere in the three dimensional space. Suppose we want to localizethe particle around a point x0 with an uncertainty R. We consider a sphere V0centered at x0 with radius R. We then take other spheres Vi not connected to V0,that is with center separated by x0 by a distance ri > R, as shown in Fig. 3.2. Therequirement to localize the particle within V0, with a radius R < 1/m, is equivalentto ask for the existence of a state with eigenvalue 1 for the operator N(V0) andeigenvalues 0 for all the N(Vi) with ri ≈ 1/m. But such an eigenstate does not existbecause, as we have shown previously [N(V0), N(Vi)] = 0. On the contrary, if wetake volumes Vi at distances ri much bigger than 1/m, the corresponding operatorsN(Vi) commute, and we can construct the desired state. Therefore it is possibleto localize the particle only over distances much bigger than the Compton wavelength. The physical explanation is that to realize the localization over distancesmuch smaller than 1/m, we need energies much bigger than m. But in this casethere is a non zero probability to create particle antiparticle pairs.

To summarize, in order to make a local quantity an observable, it is necessarythat once commuted with itself, the result vanishes at space-like distances, otherwisewe violate the causality principle. If the quantity is a local function of the fields,

45

the previous condition is automatically satisfied due to the canonical commutationrelations. A relativistic particle cannot be localized over distances of the orderof 1/m, because the particle density is not a local function of the fields. Fromthese considerations we see also that the components with negative energy of thefields are essential for the internal consistency of the theory. Otherwise the particleinterpretation of the field (that is the commutation relations among creation andannihilation operators) and the locality properties (vanishing of the commutatorsat space-like distances) would not be compatible.

3.6 The charged scalar field

We have shown that a hermitian Klein-Gordon field describes a set of identical scalarparticles. If we want to describe different kind of particles we need to introducedifferent kind of fields. Let us begin with two different hermitian scalar fields. Thefree lagrangian is a simple sum

L =1

2

2∑i=1

[(∂µϕi)(∂

µϕi)−m2iϕ

2i

](3.158)

and we can write immediately the canonical commutation relations

[ϕi(x, t), ϕj(y, t)] = iδijδ3(x− y) (3.159)

[ϕi(x, t), ϕj(y, t)] = [ϕi(x, t), ϕj(y, t)] = 0 (3.160)

All the considerations done up to now can be easily extended to the case of two fields.However we notice that there are two kind of creation and annihilation operators,a†i(k), i = 1, 2, and as a consequence

a†1(k1)a†2(k2)|0⟩ = a†2(k1)a

†1(k2)|0⟩ (3.161)

The two particle state is not any more symmetric, since it is built up in terms of twodifferent types of creation operators. That means that the two fields correspond todistinguishable particles.

Something really new comes out when the two fields have the same mass termin the lagrangian

L =1

2[(∂µϕ1)(∂

µϕ1) + (∂µϕ2)(∂µϕ2)]−

1

2m2

[ϕ21 + ϕ2

2

](3.162)

Then the theory acquires a symmetry under rotations in the plane of the two fieldsϕ1 e ϕ2

ϕ′1 = ϕ1 cos θ + ϕ2 sin θ

ϕ′2 = −ϕ1 sin θ + ϕ2 cos θ (3.163)

46

In fact the lagrangian is a function of the norm of the following vectors

ϕ = (ϕ1, ϕ2) ∂µϕ = (∂µϕ1, ∂µϕ2) (3.164)

and the norm is invariant under rotations. For infinitesimal transformations we have

δϕ1 = ϕ2θ, δϕ2 = −ϕ1θ (3.165)

or, in a more compact formδϕi = ϵijϕjθ (3.166)

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

ϵ12 = −ϵ21 = 1 (3.167)

From the Noether’s theorem, we have a conserved current, associated to this sym-metry, given by (see eq. (3.86))

Jµ =∂L∂ϕi,µ

∆ϕi = ϕi,µϵijϕjθ (3.168)

It is convenient to factorize out the angle of the infinitesimal rotation and define anew current

jµ =1

θJµ = ϕi,µϵijϕj = ϕ1,µϕ2 − ϕ2,µϕ1 (3.169)

The conservation of the current follows from the equality of the masses of the twofields, as one can also verify directly

∂µjµ = (∂2ϕ1)ϕ2 − (∂2ϕ2)ϕ1 = −(m2

1 −m22)ϕ1ϕ2 (3.170)

The conserved charge associated to the current is

Q =∫

d3x j0 =∫

d3x(ϕ1ϕ2 − ϕ2ϕ1) (3.171)

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

[Q, ϕ1] = −iϕ2, [Q, ϕ2] = iϕ1 (3.172)

with a finite transformation given by

U = eiQθ (3.173)

In fact

eiQθϕ1e−iQθ = ϕ1 + iθ[Q, ϕ1] +

i2

2!θ2[Q, [Q, ϕ1]] + . . .

= ϕ1 + θϕ2 −1

2ϕ1θ

2 + . . .

= ϕ1 cos θ + ϕ2 sin θ (3.174)

47

In analogous way one can show the transformation properties of ϕ2. The invarianceof L under rotations in the plane (ϕ1, ϕ2) is referred to as the invariance under thegroup O(2). The real basis for the fields used so far is not the most convenient one.In fact, the charge Q mixes the two fields. One can understand better the propertiesof the charge operator in a basis in which the fields are not mixed. This basis is acomplex one and it is given by the combinations

ϕ =1√2(ϕ1 + iϕ2), ϕ† =

1√2(ϕ1 − iϕ2) (3.175)

(the factor 1/√2 has been inserted for a correct normalization of the fields). It

follows

[Q, ϕ] =1√2[Q, ϕ1 + iϕ2] =

1√2(−iϕ2 − ϕ1) = −ϕ (3.176)

and analogously[Q, ϕ†] = ϕ† (3.177)

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

Q(ϕ|q⟩) = ([Q, ϕ] + ϕQ)|q⟩ = (−1 + q)ϕ|q⟩ (3.178)

andϕ|q⟩ ≈ |q − 1⟩ (3.179)

In analogous wayϕ†|q⟩ ≈ |q + 1⟩ (3.180)

Inverting the relations (3.175) we get

ϕ1 =1√2(ϕ+ ϕ†), ϕ2 = − i√

2(ϕ− ϕ†) (3.181)

from which

L =1

4

[(∂µϕ+ ∂µϕ

†)2

−(∂µϕ− ∂µϕ

†)2]

− 1

4m2

[(ϕ+ ϕ†

)2−(ϕ− ϕ†

)2]= ∂µϕ

†∂µϕ−m2ϕ†ϕ (3.182)

and

jµ = ϕ1,µϕ2 − ϕ2,µϕ1

= − i

2(∂µϕ+ ∂µϕ

†)(ϕ− ϕ†) +i

2(∂µϕ− ∂µϕ

†)(ϕ+ ϕ†)

= i[(∂µϕ)ϕ

† − (∂µϕ†)ϕ]

(3.183)

48

The charge results to be

Q = i∫

d3x ϕ†∂(−)t ϕ (3.184)

The commutation relations in the new basis are given by

[ϕ(x, t), ϕ†(y, t)] =1

2[ϕ1(x, t) + iϕ2(y, t), ϕ1(x, t)− iϕ2(y, t)] = iδ3(x− y) (3.185)

and

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

[ϕ(x, t), ϕ(y, t)] = [ϕ†(x, t), ϕ†(y, t)] = 0 (3.186)

Let us notice that these commutation relations could have also been obtained directlyfrom the lagrangian (3.182), since

Πϕ =∂L∂ϕ

= ϕ†, Πϕ† =∂L∂ϕ†

= ϕ (3.187)

The original O(2) symmetry becomes now an invariance of the lagrangian (3.182)under a phase transformation of the fields. This follows from (3.182) but it is seenalso from the change of variables

ϕ → 1√2(ϕ′

1 + iϕ′2)

=1√2(ϕ1 cos θ + ϕ2 sin θ + i(−ϕ1 sin θ + ϕ2 cos θ))

=1√2

(ϕ1e

−iθ + iϕ2e−iθ

)= e−iθϕ (3.188)

andϕ† → eiθϕ† (3.189)

In this basis we speak of invariance under the group U(1) (the group of unitarytransformations on the complex vectors of dimensions d = 1).

Using the expansion for the real fields

ϕi(x) =∫

d3k[fk(x)ai(k) + f ∗

k(x)a†i(k)

](3.190)

we get

ϕ(x) =∫

d3k

[fk(x)

1√2(a1(k) + ia2(k)) + f∗

k(x)

1√2(a†1(k) + ia†2(k))

](3.191)

Introducing the combinations

a(k) =1√2(a1(k) + ia2(k)), b(k) =

1√2(a1(k)− ia2(k)) (3.192)

49

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)

](3.193)

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′) (3.194)

[a(k), b(k′)] = [a(k), b†(k′)] = 0 (3.195)

We get also

: Pµ :=∫

d3kkµ2∑i=1

a†i(k)ai(k) =∫

d3kkµ[a†(k)a(k) + b†(k)b(k)

](3.196)

Therefore the operators a†(k) e b†(k) both create particles states with momentumk, as the original operators a†i. The charge Q is given by

Q = i∫

d3xϕ†∂(−)t ϕ

= i∫

d3x∫

d3k1d3k2

[fk1b(k1) + f ∗

k1a†(k1)

]×

[(∂

(−)t fk2)a(k2) + (∂

(−)t f∗

k2)b†(k2)

]=

∫d3k1d

3k2δ3(k1 − k2)

[a†(k1)a(k1)− b(k1)b

†(k1)]

(3.197)

where we have used the orthogonality relations (3.50). For the normal orderedcharge operator we get

: Q :=∫

d3k[a†(k)a(k)− b†(k)b(k)

](3.198)

showing explicitly that a† and b† create particles of charge +1 and −1 respectively. Itis important to notice that the current density j0 is local in the fields, and therefore

[j0(x), j0(y)] = 0, ∀ (x− y)2 < 0 (3.199)

By the same argument of Section 3.5, we construct the operators

Q(Vi) =∫Vid3x j0(x) (3.200)

which for any non intersecting Vi e Vj commute at equal times. Therefore it ispossible to localize a state of definite charge in an arbitrary spatial region. Thisagrees with the argument we have developed in the case of the number of particles,because the pair creation process does not change the charge of the state.

Finally we notice that the field operator is a linear combination of annihilation,a(k), and creation, b†(k), operators, meaning that a local theory deals in a symmetricway with the annihilation of a particle and the creation of an antiparticle. Forinstance, to annihilate a charge +1 is equivalent to the creation of a charge −1.

50

Chapter 4

The Dirac field

4.1 The Dirac equation

In 1928 Dirac tried to to solve the problem of a non positive probability density,present in the Klein-Gordon case, formulating a new wave equation. Dirac thought,correctly, that in order to get a positive quantity it was necessary to have a waveequation of the first order in the time derivative (as it happens for the Schrodingerequation). Therefore Dirac looked for a way to reduce the Klein-Gordon equation(of the second order in the time derivative) to a first-order differential equation.The Pauli formulation of the electron spin put Dirac on the right track. In fact,Pauli showed that in order to describe the spin, it was necessary to generalize theSchrodinger wave function (a complex number) to a two components object

ψ → ψα =[ψ1

ψ2

](4.1)

modifying also the wave equation to a matrix equation

i∂ψα∂t

=2∑

β=1

Hαβψβ (4.2)

where the hamiltonian H is, in general, a 2 × 2 matrix. The electron spin is thendescribed by a special set of 2× 2 matrices, the Pauli matrices, σ

S =1

2σ (4.3)

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

|k|2 = (σ · k)2 (4.4)

as it follows from[σi, σj]+ = 2δij (4.5)

51

where [A,B]+ = AB +BA.Following this suggestion Dirac tried to write down a first order differential equa-

tion for a many component wave function

i∂ψ

∂t= −iα · ∇ψ + βmψ ≡ Hψ (4.6)

where α and β are matrices. The requirements that this equation should satisfy are

• the wave function ψ, solution of the Dirac equation, should satisfy also theKlein-Gordon equation in order to get the correct dispersion relation betweenenergy and momentum;

• the equation should admit a conserved current with the fourth componentbeing positive definite;

• the equation should be covariant with respect to Lorentz transformations (seelater)

In order to satisfy the first requirement, we iterate the Dirac equation and askthat the resulting second order differential equation coincides with the Klein-Gordonequation

−∂2ψ

∂t2= (−iα · ∇+ βm)2ψ

=

(−αiαj ∂2

∂xi∂xj+ β2m2 − i(βα+ αβ) · ∇

)ψ

=

(−1

2

[αi, αj

]+

∂2

∂xi∂xj+ β2m2 − i(βα+ αβ) · ∇

)ψ (4.7)

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

]+= 2δij,

[αi, β

]+= 0, β2 = 1 (4.8)

In order to get a hamiltonian, H, hermitian, we will require also that α and β arehermitian matrices.Since for any choice of i, (αi)2 = 1, it follows that the eigenvaluesof α and β must be ±1. We can also prove the following relations

Tr(β) = Tr(αi) = 0 (4.9)

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

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

where we have made use of the cyclic property of the trace. The consequence is thatthe matrices αi and β can be realized only in a space of even dimensions. This isperhaps the biggest difficulty that Dirac had to cope with. In fact, the αi’s enjoy

52

the same properties of the Pauli matrices, but in a 2 × 2 matrix space, a furtheranticommuting matrix β does not exist. It required some time to Dirac before herealized that the previous relations could have been satisfied by 4× 4 matrices.

An explicit realization of the Dirac matrices is the following

αi =[0 σiσi 0

], β =

[1 00 −1

](4.11)

as it can be checked [αi, αj

]+=[[σi, σj]+ 0

0 [σi, σj]+

]= 2δij (4.12)

[β, αi

]+=[

0 σi−σi 0

]+[0 −σiσi 0

]= 0 (4.13)

Let us now show that also the second of our requirements is satisfied. We multiplythe Dirac equation by ψ† at the left, and then we consider the equation for ψ†

−i∂ψ†

∂t= i(∇ψ†) · α+mψ†β (4.14)

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

iψ†∂ψ

∂t+ i

∂ψ†

∂tψ = ψ†(−iα · ∇+βm)ψ− (i∇ψ† · α+ψ†βm)ψ = −i∇ · (ψ†αψ) (4.15)

that is

i∂

∂t(ψ†ψ) + i

∂

∂xj(ψ†αjψ) = 0 (4.16)

We see that the currentjµ = (ψ†ψ, ψ†αiψ) (4.17)

is a conserved one.∂jµ

∂xµ= 0 (4.18)

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

d3x ψ†ψ (4.19)

is invariant with respect to Lorentz transformations.

53

4.2 Covariance properties of the Dirac equation

To discuss the properties of transformation of the Dirac equation under Lorentztransformations, it turns out convenient to write the equation in a slightly differentway. Let us multiply the equation by β

iβ∂ψ

∂t= −iβα · ∇ψ +mψ (4.20)

and let us define the following matrices

γ0 = β =[1 00 −1

], γi = βαi =

[0 σi

−σi 0

](4.21)

Then the equation becomes(iγ0

∂

∂x0+ iγi

∂

∂xi−m

)ψ = 0 (4.22)

or, in a compact way(i∂ −m)ψ = 0 (4.23)

where

∂ = γµ∂

∂xµ= γµ∂µ (4.24)

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

]+= βαiβαj + βαjβαi = −

[αi, αj

]+= −2δij (4.25)

[γ0, γi

]+=[β, βαi

]+= αi + βαiβ = 0 (4.26)

or[γµ, γν ]+ = 2gµν (4.27)

Notice that(γi)† = (βαi)† = αiβ = −γi (4.28)

and(γi)2 = −1 (4.29)

The covariance of the Dirac equation means that the following two conditions aresatisfied

• given the Dirac wave function ψ(x) in the Lorentz frame, S, an observer ina different frame, S ′ should be able to evaluate, in terms of ψ(x), the wavefunction ψ′(x′) describing the same physical state as ψ(x) in S;

54

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

iγµ∂

∂x′µ−m

)ψ′(x′) = 0 (4.30)

The matrices γµ should satisfy the same algebra as the matrices γµ, because inboth cases the wave functions should satisfy the Klein-Gordon equation (which isinvariant in form). Therefore, neglecting a possible unitary transformations, the twosets of matrices can be identified. As a consequence, the Dirac equation in S ′ willbe (

iγµ∂

∂x′µ−m

)ψ′(x′) = 0 (4.31)

Since both the Dirac equation and the Lorentz transformations are linear, we willrequire that the wave functions in two different Lorentz frames are linearly correlated

ψ′(x′) = ψ′(Λx) = S(Λ)ψ(x) (4.32)

where S(Λ) is a 4 × 4 matrix operating on the complex vector ψ(x) and Λ is theLorentz transformation. On physical ground, the matrix S(Λ) should be invertible

ψ(x) = S−1(Λ)ψ′(x′) (4.33)

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

ψ(x) = S(Λ−1)ψ′(x′) (4.34)

from whichS−1(Λ) = S(Λ−1) (4.35)

Considering the Dirac equation in the frame S(iγµ

∂

∂xµ−m

)ψ(x) = 0 (4.36)

we can write (iγµ

∂

∂xµ−m

)S−1(Λ)ψ′(x′) = 0 (4.37)

Multiplying to the left by S(Λ) and using

∂

∂xµ=∂x′ν∂xµ

∂

∂x′ν= Λνµ

∂

∂x′ν, x′ν = Λνµx

µ (4.38)

it follows (iS(Λ)γµS−1(Λ)Λνµ

∂

∂x′ν−m

)ψ′(x′) = 0 (4.39)

55

Comparing with eq. (4.30), we get

S(Λ)γµS−1(Λ)Λνµ = γν (4.40)

orS−1(Λ)γνS(Λ) = Λνµγ

µ (4.41)

For an infinitesimal transformation we obtain

Λµν = gµν + ϵµν (4.42)

with ϵµν = −ϵνµ (see eq. (3.93)). By expanding S(Λ) to the first order in ϵµν , weget

S(Λ) = 1− i

4σµνϵ

µν (4.43)

and using (4.41), we find the following condition on σµν(1 +

i

4σρλϵ

ρλ)γν

(1− i

4σαβϵ

αβ)= (gµν + ϵνµγ

µ (4.44)

from whichi

4ϵρλ[σρλ, γν ] = ϵνµγ

µ =1

2ϵρλ(gρνγλ − gλνγρ) (4.45)

and finally[σρλ, γν ] = −2i(gρνγλ − gλνγρ) (4.46)

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

σρλ =i

2[γρ, γλ] (4.47)

In fact

[σρλ, γν ] =i

2[γργλ − γλγρ, γν ] =

i

2[γργλγν − γνγργλ − γλγργν + γνγλγρ]

=i

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

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

= −2i[gρνγλ − gνλγρ] (4.48)

A finite Lorentz transformation is obtained by exponentiation

S(Λ) = e− i

4σµνϵ

µν

(4.49)

with

σµν =i

2[γµ, γν ] (4.50)

56

We can now verify that the current jµ, defined in eq. (4.17) transforms as a fourvector. To this end we introduce the following notation

ψ(x) = ψ†(x)β = ψ†(x)γ0 (4.51)

It followsj0 = ψ†ψ = ψγ0ψ, ji = ψ†αiψ = ψβαiψ = ψγiψ (4.52)

orjµ = ψγµψ (4.53)

The transformation properties of ψ under Lorentz transformations are particularlysimple. By noticing that

γ0㵆γ0 = γµ (4.54)

and

σµν† = − i

2[γµ, γν ]

† =i

2[γµ

†, γν†] (4.55)

it followsγ0σµν

†γ0 = σµν (4.56)

and thereforeγ0S

†(Λ)γ0 = S−1(Λ) (4.57)

from whichψ′(x′) = ψ(x)S−1(Λ) (4.58)

We get

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

νψ(x) = Λµνjν(x)(4.59)

We see that jµ has the desired transformation properties. The representation forthe Lorentz generators, in the same basis used previously for the γµ matrices, is

σ0i =i

2[γ0, γi] =

i

2(β2αi − βαiβ) = −iαi = −i

[0 σiσi 0

](4.60)

σij =i

2[γi, γj] =

i

2(βαiβαj − βαjβαi) = − i

2[αi, αj]

= − i

2

[[σi, σj] 0

0 [σi, σj]

]= ϵijk

[σk 00 σk

](4.61)

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

57

The behaviour of the Dirac wave function under parity x→ −x can be obtainedin analogous way. In this case

ΛPνµ =

1

−1−1

−1

(4.62)

and thereforeS−1(ΛP )γ

µS(ΛP ) = γµ (4.63)

This relation is satisfied by the choice

S(ΛP ) = ηPγ0 (4.64)

where ηP is a non observable arbitrary phase. Then

ψ(x) → ψ′(x′) = ηPγ0ψ(x), x′µ= (x0,−x) (4.65)

We are now in the position to classify the bilinear expressions in the Dirac wavefunction under Lorentz transformations. Let us consider expressions of the typeψAψ, where A is an arbitrary 4× 4 matrix. As a basis for the 4× 4 matrices we cantake the following set of 16 linearly independent matrices

ΓS = 1

ΓVµ = γµ

ΓAµ = γ5γµ

ΓTµν = σµν

ΓP = γ5 (4.66)

where the matrix γ5 is defined as

γ5 = γ5 = iγ0γ1γ2γ3 (4.67)

This matrix has the following properties

γ5† = iγ3γ2γ1γ0 = iγ0γ1γ2γ3 = γ5 (4.68)

γ52 = 1, [γ5, γµ]+ = 0 (4.69)

γ5 =[0 11 0

](4.70)

One can easily verify that the bilinear expressions have the following behaviourunder Lorentz transformations

ψψ ≈ scalar

ψγµψ ≈ four− vector

ψγ5γµψ ≈ axial four− vector

ψσµνψ ≈ 2nd rank antisymmetric tensor

ψγ5ψ ≈ pseudoscalar (4.71)

58

As an example, let us verify the last of these transformation properties

ψ(x)γ5ψ(x) → ψ′(x′)γ5ψ′(x′) = η⋆PηP ψ(x)γ0γ5γ0ψ(x) = −ψ(x)γ5ψ(x) (4.72)

4.3 Free particle solutions of the Dirac equation

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

ψ(t) = ue−imt (4.73)

where u is a four components complex vector (usually called a spinor). This solutionhas positive energy. Substituting inside the Dirac equation we get

(i∂ −m)ψ(t) = (mγ0 −m)ue−imt = 0 (4.74)

that is(γ0 − 1)u = 0 (4.75)

Since γ0 has eigenvalues ±1, we see that the Dirac equation admits also solutions

of the type eimt, corresponding to a negative energy state. More generally we canlook for solutions of the form

ψ(+)(x) = e−ikxu(k), positive energy

ψ(−)(x) = eikxv(k), negative energy (4.76)

Inserting in the Dirac equation

(k −m)u(k) = 0

(k +m)v(k) = 0 (4.77)

In the rest frame we get

(γ0 − 1)u(m, 0) = 0

(γ0 + 1)v(m, 0) = 0 (4.78)

There are two independent spinors of type u and two of type v satisfying theseequations. In the basis where γ0 is a diagonal matrix we can choose the followingsolutions

u(1)(m, 0) =

1000

, u(2)(m, 0) =

0100

v(1)(m, 0) =

0010

, v(2)(m, 0) =

0001

(4.79)

59

In a general Lorentz frame the solutions could be obtained by boosting the solutionsin the rest frame. Or, we can notice that the following expression

(k −m)(k +m) = k2 −m2 (4.80)

vanishes for k2 = m2. Therefore we can solve our problem (except for a normaliza-tion constant), by putting

u(α)(k) = cα(k +m)u(α)(m, 0)

v(α)(k) = dα(−k +m)v(α)(m, 0) (4.81)

with k2 = m2. In order to determine the normalization constants cα and dα wemake use of the orthogonality conditions satisfied by the rest frame solutions (seeeq. (4.79)

u(α)(m, 0)u(β)(m, 0) = δαβ

v(α)(m, 0)v(β)(m, 0) = −δαβu(α)(m, 0)v(β)(m, 0) = 0 (4.82)

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

u(α)(k)u(β)(k) = |cα|2u(α)(m, 0)(k +m)2u(β)(m, 0)

= |cα|2u(α)(m, 0)(2m2 + 2mk)u(β)(m, 0) (4.83)

By taking into account that u(α)(m, 0) and u†(α)

(m, 0) are eigenstates of γ0 witheigenvalue +1, we get

u(α)(m, 0)γµu(β)(m, 0) = u†(α)

(m, 0)γµu(β)(m, 0)

= u†(α)

(m, 0)γ0γµγ0u

(β)(m, 0) = u†(α)

(m, 0)γµu(β)(m, 0) (4.84)

from which

u(α)(m, 0)γµu(β)(m, 0) = u†(α)

(m, 0)γµu(β)(m, 0)

= gµ0u†(α)

(m, 0)γ0u(β)(m, 0) = gµ0δαβ (4.85)

that isu(α)(k)u(β)(k) = |cα|2(2m2 + 2mE)δαβ (4.86)

Than we choose

cα =1√

2m(m+ E), E =

√|k|2 +m2 (4.87)

In analogous way we have

v(α)(k)v(β)(k) = |dα|2v(α)(m, 0)(−k +m)2v(β)(m, 0)

= |dα|2v(α)(m, 0)(2m2 − 2mk)v(β)(m, 0) (4.88)

60

and using the fact that v(α)(m, 0) and v†(α)

(m, 0) are eigenstates of γ0 with eigenvalue−1, we get

v(α)(m, 0)γµv(β)(m, 0) = −v†(α)(m, 0)γµv(β)(m, 0)= −v†(α)(m, 0)γ0γµγ0v(β)(m, 0) = −v†(α)(m, 0)γµv(β)(m, 0) (4.89)

that is

v(α)(m, 0)γµv(β)(m, 0) = −v†(α)(m, 0)γµv(β)(m, 0)= −gµ0v†(α)(m, 0)γ0v(β)(m, 0) = gµ0δαβ (4.90)

Recalling that v(α)(m, 0)v(β)(m, 0) = −δαβ, we obtain

v(α)(k)v(β)(k) = −|dα|2(2m2 + 2mE)δαβ (4.91)

and finally

dα = cα =1√

2m(m+ E)(4.92)

The normalized solutions we have obtained are

u(α)(k) =k +m√

2m(m+ E)u(α)(m, 0), v(α)(k) =

−k +m√2m(m+ E)

v(α)(m, 0) (4.93)

Notice that positive and negative energy spinors are orthogonal. In the following itwill be useful to express our solutions in terms of two component spinors, ϕ(α)(m, 0)and χ(α)(m, 0)

u(α)(m, 0) =[ϕ(α)(m, 0)

0

], v(α)(m, 0) =

[0

χ(α)(m, 0)

](4.94)

From the explicit representation of the γµ matrices we get

k =[E 00 −E

]−[

0 k · σ−k · σ 0

]=

[E −k · σk · σ −E

](4.95)

from which

u(α)(k) =

√m+ E

2mϕ(α)(m, 0)

k · σ√2m(m+ E)

ϕ(α)(m, 0)

, v(α)(k) =

k · σ√

2m(m+ E)χ(α)(m, 0)√

m+ E

2mχ(α)(m, 0)

(4.96)

61

In the following we will need the explicit expression for the projectors of thepositive and negative energy solutions. To this end, let us observe that

2∑α=1

u(α)(m, 0)u(α)(m, 0) =

1000

[ 1 0 0 0 ]+

0100

[ 0 1 0 0 ] =1 + γ0

2(4.97)

and analogously2∑

α=1

v(α)(m, 0)v(α)(m, 0) = −1− γ02

(4.98)

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

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

= 2E(k +m)− (k +m)(−k +m)γ0

= 2E(k +m) (4.99)

Therefore the positive energy projector is given by

Λ+(k) =2∑

α=1

u(α)(k)u(α)(k) =k +m√

2m(m+ E)

1 + γ02

k +m√2m(m+ E)

=1

2m(m+ E)

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

2=k2 +m2 + 2mk + 2E(k +m)

4m(m+ E)

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

4m(m+ E)=k +m

2m(4.100)

Analogously

Λ−(k) = −2∑

α=1

v(α)(k)v(α)(k) =k −m√

2m(m+ E)

1− γ02

k −m√2m(m+ E)

=1

2m(m+ E)

(k −m)2 − 2E(k −m)

2=k2 +m2 − 2mk − 2E(k −m)

4m(m+ E)

=(2E + 2m)(−k +m)

4m(m+ E)=

−k +m

2m(4.101)

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

Λ2± = Λ±, Λ+Λ− = 0, Λ+ + Λ− = 1 (4.102)

In this normalization the density ψ†ψ has the correct Lorentz transformation prop-erties

ψ(+)(α)

†(x)ψ

(+)(β) (x) = u(α)(k)γ0u

(β)(k)

62

=1

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

=1

2m(m+ E)2Eu(α)(m, 0)(k +m)u(β)(m, 0)

=2E(m+ E)

2m(m+ E)δαβ =

E

mδαβ (4.103)

Therefore the density for positive energy solutions transforms as the fourth compo-nent of a four vector. The same is true for the negative energy solutions

ψ(−)(α)

†(x)ψ

(−)(β) = v(α)(k)γ0v

(β)(k)

=1

2m(m+ E)v(α)(m, 0)(k −m)γ0(k −m)v(β)(m, 0)

=1

2m(m+ E)2Ev(α)(m, 0)(k −m)v(β)(m, 0)

=2E(m+ E)

2m(m+ E)δαβ =

E

mδαβ (4.104)

We find also

u†(α)

(k)v(β)(k) = 0, kµ = (E, k), kµ = (E,−k)a (4.105)

In fact

u†(α)

(k)v(β)(k) = u(α)(m, 0)(k +m)γ0(−ˆk +m)

2m(m+ E)v(β)(m, 0)

= u(α)(m, 0)(k +m)(−k +m)γ0

2m(m+ E)v(β)(m, 0) = 0 (4.106)

It follows that solutions with opposite energy and same three momentum are or-thogonal

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

ψ(−) = e+i(Ex0 + k · x)v(k), , kµ = (E,−k) (4.107)

Th positive and negative energy solutions are doubly degenerate. It is possibleto remove the degeneration through the construction of projectors for states withdefinite polarization. Let us consider again the solutions in the frame system. Thegenerator of the rotations along the z-axis is given by

σ12 =[σ3 00 σ3

](4.108)

Clearly u(1)(m, 0) and v(1)(m, 0) are eigenstates of this operator (and therefore ofthe third component of the spin operator) with eigenvalues +1, whereas u(2)(m, 0)

63

and v(2)(m, 0) belong to the eigenvalue −1. The projector for the eigenstates witheigenvalues +1 is given by

1 + σ12n3R

2≡ 1 + σ12

2(4.109)

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

σ12 =i

2[γ1, γ2] = iγ1γ2 = −γ0γ5γ3 = γ5γ3γ0 (4.110)

andσ12n

3R = γ5nRγ0 (4.111)

The presence of γ0 forbids a simple extension of this expression to a generic Lorentzframe. We can avoid this, by changing the definition of the projection in the restframe system. Let us put

Σ(±nR) =1± σ12n

3Rγ0

2=

1

2

[1± σ3 0

0 1∓ σ3

](4.112)

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

Σ(±nR) =1± γ5nR

2(4.113)

In the rest frame we have n2R = −1, nRk = 0. We can go to a generic frame

preserving these conditions

Σ(±n) = 1± γ5n

2, n2 = −1, nk = 0 (4.114)

The projector Σ(±n) projects out energy positive states that in the frame system

have a polarization given by S · n = ±1/2, and the negative energy states with

polarization S · n = ∓1/2.In the following we will use the following notation

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

u(kR,−nR) = u(2)(m, 0)

v(kR,−nR) = v(1)(m, 0)

v(kR, nR) = v(2)(m, 0) (4.115)

These spinors satisfy the following relations

Σ(±nR)u(kR,±nR) = u(kR,±nR), Σ(±nR)v(kR,±nR) = v(kR,±nR) (4.116)

andΣ(±nR)u(kR,∓nR) = Σ(±nR)v(kR,∓nR) = 0 (4.117)

64

All these relations generalize immediately to an arbitrary reference frame (alwaysrequiring n2 = −1 e nk = 0)

Σ(±n)u(k,±n) = u(k,±n), Σ(±n)v(k,±n) = v(k,±n) (4.118)

Σ(±n)u(k,∓n) = Σ(±n)v(k,∓n) = 0 (4.119)

The properties of the spin projectors are

Σ(n) + Σ(−n) = 1, Σ(±n)2 = Σ(±n), Σ(n)Σ(−n) = 0 (4.120)

Let us just verify the second equation(1 + γ5n

2

)2

=1 + (γ5n)

2 + 2γ5n

4=

2 + 2γ5n

4= Σ(n) (4.121)

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

1 + γ5n

2

1− γ5n

2=

1− (γ5n)2

4= 0 (4.122)

4.4 Wave packets and negative energy solutions

As we have shown the Dirac equation leads to a positive probability density. Thissolves the problem one had with the Klein-Gordon equation. On the other hand theDirac equation does not solve the problem of the negative energy solutions (and itshould not, as we have seen their importance in the Klein-Gordon case). In fact,the completeness of the spinors involve all the solutions

2∑α=1

[u(α)(k)u(α)(k)− v(α)(k)v(α)(k)

]= Λ+(k) + Λ−(k) = 1 (4.123)

In the case of a non interacting theory there are no possibilities of transitions amongpositive and negative energy states but, when an interaction is turned on, such apossibility cannot be excluded. In fact, if we try to localize a Dirac particle withindistances of order 1/m the negative energy solutions cannot be ignored. To clarifythis point let us consider the time evolution of a gaussian wave packet, assigned attime t = 0,

ψ(x, 0) =1

(πd2)3/4e−|x|2

2d2 w (4.124)

where w is a fixed spinor, w = (ϕ, 0), with w†w = 1. As one can check, the wavepacket is normalized to one

∫d3x ψ†ψ =

1

(πd2)3/2

∫d3x e

−|x|2

d2

65

=1

(πd2)3/2

3∏i=1

∫dxie

−x2i

d2

=1

(πd2)3/2

3∏i=1

(πd2)1/2 = 1 (4.125)

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

ψ(x, t) =∫

d3k1√(2π)3

√m

E

2∑α=1

[b(k, α)u(α)(k)e−ikx + d⋆(k, α)v(α)(k)eikx

](4.126)

and evaluating the expansion coefficients b(k, α) and d⋆(k, α), by requiring that thesolution coincides with eq. (4.124) at time t = 0. We get

ψ(x, 0) =∫

d3k1√(2π)3

√m

E

2∑α=1

[b(k, α)u(α)(k) + d⋆(k, α)v(α)(k)

]eik · x

=1

(πd2)3/4e−|x|2

2d2 w (4.127)

We Fourier transform both sides of this equation, obtaining

√(2π)3

√m

E

2∑α=1

[b(k, α)u(α)(k) + d⋆(k, α)v(α)(k)

]

=w

(πd2)3/4

∫d3x e

−|x|2

2d2 e−ik · x

=w

(πd2)3/4(2πd2)3/2e

−|k|2d2

2 (4.128)

From which

√m

E

2∑α=1

[b(k, α)u(α)(k) + d⋆(k, α)v(α)(k)

]=

(d2

π

)3/4

e−|k|2d2

2 w (4.129)

Using the orthogonality relations for the spinors we find the amplitudes

b(k, α) =

√m

E

(d2

π

)3/4

e−|k|2d2

2 u†(α)

(k)w (4.130)

d⋆(k, α) =

√m

E

(d2

π

)3/4

e−|k|2d2

2 v†(α)

(k)w (4.131)

66

Expressing u e v in terms of two-component spinors we get

b(k, α) =

√m

E

(d2

π

)3/4

e−|k|2d2

2

√m+ E

2mϕ(α)†(m, 0)ϕ (4.132)

d⋆(k, α) =

√m

E

(d2

π

)3/4

e−|k|2d2

21√

2m(m+ E)χ(α)†(m, 0)k · σϕ (4.133)

from which we can evaluate the ratio of the negative energy amplitudes to thepositive energy ones

d⋆(k, α)

b(k, α)≈ |k|m+ E

(4.134)

The amplitudes (for both signs of the energy) contribute only if |k| ≪ 1/d (due to thegaussian exponential). Suppose that we want to localize the particle over distanceslarger than 1/m, that is we require d ≫ 1/m. Since the negative energy state

amplitudes are important only for |k| > m ≫ 1/d, their contribution is depressedby the gaussian exponential. On the other hand, if we try to localize the particleover distances d ≈ 1/m, the negative energy states contribution becomes important

for values of |k| of order m, or of order 1/d, that is in the momentum region inwhich the corresponding amplitudes are not negligible. We see that the physicalinterpretation is essentially the same following from the uncertainty principle.

4.5 Electromagnetic interaction of a relativistic

point-like particle

Before continuing our discussion about the properties of the Dirac equation, let usdescribe the interaction of a point-like particle with the electromagnetic field in therelativistic formalism.

Let us recall that the classical expression for the electromagnetic four current isgiven by

jµ = (ρ, ρv) (4.135)

where ρ is the charge density, and v the velocity field. In the case of a point-like particle which follows the world line describe in a parametric form by the fourfunctions xµ(τ), with τ an arbitrary line parameter, the charge density at the timet is localized at the position x(τ), evaluated at the parameter value τ such thatt = x0(τ) (see Fig. 4.1). Therefore

ρ(y, t) = eδ3(y − x(τ))|t=x0(τ) (4.136)

67

t = x ( )τ

τ

0

t

x ( )µ

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

It follows

jµ(y) = edxµ

dx0δ3(y − x(τ))|y0=x0(τ) (4.137)

This expression can be put in a covariant form, after taking into account the followingrelation ∫

dτf(τ)δ(y0 − x0(τ)) =

(dx0

dτ

)−1

f(τ)∣∣∣x0(τ)=y0

(4.138)

From this

jµ(y) = e∫ +∞

−∞dτdxµ

dτδ4(y − x(τ)) (4.139)

This four current is a conserved one:

∂µjµ(y) = −e

∫ +∞

−∞dτdxµ

dτ

∂

∂xµδ4(y − x(τ)) = −e

∫ +∞

−∞dτ

d

dτδ4(y − x(τ)) (4.140)

The expression vanishes at any space-time point y, except at the end points x(±∞).We recall also that the equations of motion for a free relativistic scalar particle

can be derived by the following action

S = −m∫ τf

τidτ

√x2, xµ =

dxµ

dτ(4.141)

We will be interested in deriving the lagrangian describing the interaction betweenour particle and the electromagnetic field (we assume that our particle has chargee). We should be able to derive the following equations of motion

d

dt

mv√1− |v|2

= e(E + v ∧ B) (4.142)

We will show that the lagrangian depends on the four-vector potential Aµ and not

on the fields E and B. In fact we will verify that the following lagrangian reproduces

68

the previous equations of motion

S = −m∫ τf

τidτ

√x2 −

∫d4yAµ(y)j

µ(y)

= −m∫ τf

τidτ

√x2 − e

∫ τf

τidτAµ(x(τ))x

µ(τ) (4.143)

Using

∂L

∂xµ= −e∂Aν

∂xµxν

∂L

∂xµ= −m xµ√

x2− eAµ (4.144)

and the Euler-lagrangian equations

∂L

∂xµ− d

dτ

∂L

∂xµ= 0 (4.145)

we get

−e∂Aν∂xµ

xν +md

dτ

xµ√x2

+ e∂Aµ∂xν

xν = 0 (4.146)

Therefore

md

dτ

xµ√x2

= e(∂µAν − ∂νAµ)xν (4.147)

Since ds = dτ√x2, where ds is the line element measured along the trajectory, we

see that the four-velocity of the particle is

Uµ =xµ√x2

(4.148)

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

md

dsUµ = eFµνU

ν (4.149)

Here we have introduced the electromagnetic tensor

Fµν = ∂µAν − ∂νAµ (4.150)

Since the definition of the fields in terms of the vector and scalar potential is givenby

E = −∇A0 − ∂A

∂t, B = ∇ ∧ A (4.151)

we getE = (F 10, F 20, F 30), B = (−F 23,−F 31,−F 12) (4.152)

69

(we can also write F ij = −ϵijkBk). By choosing τ = x0 in the eq. (4.147) we find

md

dt

−vk√1− |v|2

= eFk0 + eFkidxi

dt= −eEk − ϵkijB

jvi (4.153)

reproducing eq. (4.142).There are various ways to convince oneself about the necessity of the appearance

of the four potential inside the lagrangian. For instance, consider the Maxwellequations

∂µFµν = jν , ∂µFµν = 0 (4.154)

where Fµν =12ϵµνρσF

ρσ is the dual tensor (we define the Ricci tensor in 4 dimensions,through ϵ0123 = +1). In Section 3.2 we have shown how to deduce the expressionfor the lagrangian multiplying the field equations by an infinitesimal variation of thefields. In the actual case, to consider Fµν as the fields to be varied, would createa problem because multiplying both sides of the Maxwell equations by δFµν we donot get a Lorentz scalar. This difficulty is avoided by taking Aµ as the independentdegrees of freedom of the theory (we will show in the following that also this pointof view has its own difficulties). In this case, due to the definition (4.150) of theelectromagnetic tensor, the homogeneous Maxwell equations become identities (infact, it is just solving these equations that one originally introduces the vector andthe scalar potentials)

Fµν = ϵµνρσ∂ρAσ =⇒ ∂µFµν = 0 (4.155)

whereas the inhomogeneous ones give rise to

∂µ(∂µAν − ∂νAµ) = ∂2Aν − ∂ν∂µAµ = jν (4.156)

The previous difficulty disappears because jµδAµ is a Lorentz scalar. By regarding jµ

as a given external current, independent on Aµ, we can now get easily the expressionfor the lagrangian. By multiplying eq. (4.156) by δAν and integrating in d4x we get

0 =∫Vd4xδAν(∂2Aν − ∂ν∂

µAµ − jν)

=∫Vd4x

[− δ(∂µAν)∂

µAν + ∂µ(δAν∂µAν)

+ δ(∂µAν)∂νAµ − ∂µ(δA

ν∂νAµ)− δ(Aµjµ)

]=

∫Vd4x

[− 1

2δ(∂µAν)∂

µAν − 1

2δ(∂νAµ)∂

νAµ

+1

2δ(∂µAν)∂

νAµ +1

2δ(∂νAµ)∂

µAν − δ(Aµjµ)]+ surface terms

= −1

2

∫Vd4x(δFµν)F

µν −∫Vd4xδ(Aµjµ) + surface terms

= δ[∫Vd4x (−1

4FµνF

µν − Aµjµ)]+ surface terms (4.157)

70

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

S = −1

4

∫Vd4x FµνF

µν −∫Vd4x jµA

µ (4.158)

Notice that the interacting term has the same structure we found for the point-likeparticle.

We stress again that the Aµ’s are the canonical variables of the electrodynamics.In principle, one could reintroduce the fields by inverting the relations between fieldsand potentials. However, in this way, one would end up with a non-local action.From these considerations one can argue that the potentials play an importantrole in quantum mechanics, much more than in the classical case, where they areessentially a convenient trick. Recall also that the canonical variables satisfy localcommutation relations (the commutator vanishes at space-like distances), implyingthat local observables should be local functions of the potentials. This is going tocreate us some problem, because the theory is invariant under gauge transformations,whereas the potentials are not

Aµ(x) → Aµ(x) + ∂µΛ(x) (4.159)

(where Λ(x) is an arbitrary function). Therefore, the observables of the theoryshould be gauge invariant. This implies that the potentials cannot be observed. Anexample of observable local in the potentials is the electromagnetic tensor Fµν .

The pure electromagnetic part of the action (4.158) is naively gauge invariant,being a function of Fµν . As far as the interaction term is concerned we have (as-suming that the current is gauge invariant)

jµAµ → jµA

µ + jµ∂µΛ = jµA

µ + ∂µ(jµΛ)− (∂µjµ)Λ (4.160)

Adding a four divergence to the lagrangian density does not change the equationsof motion ∫ t2

t1dt∫

d3x ∂µχµ =

∫ t2

t1dt∂

∂t

∫d3 xχ0 (4.161)

Therefore the invariance of the lagrangian under gauge transformations (neglectinga four divergence) is guaranteed, if the potentials are coupled to a gauge invariantand conserved current

∂µjµ = 0 (4.162)

We have shown that this condition is indeed satisfied for the point-like particle.In order to derive the general prescription to couple the electromagnetic poten-

tials to a charged particle, let us go back to the action for the point-like particle.This prescription is known as the minimal substitution. By choosing x0 = τ in(4.143), we get

L = −m√1− |v|2 − e(A0 − v · A) (4.163)

71

from which

p =∂L

∂v=

mv√1− |v|2

+ eA (4.164)

The hamiltonian is obtained through the usual Legendre transform

H = p · v − L = m|v|2√1− |v|2

+ ev · A+m√1− |v|2 + e(A0 − v · A)

=m√

1− |v|2+ eA0 (4.165)

Therefore, we get the following relations

p− eA = mv√

1− |v|2, H − eA0 =

m√1− |v|2

(4.166)

where the quantities in the left hand sides are the same as in the free case. It followsthat we can go from the free case to the interacting one, by the simple substitution(minimal substitution)

pµ → pµ − eAµ (4.167)

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

|v|2 = |p|2

m2 + |p|2, 1− |v|2 = m2

m2 + |p|2(4.168)

we get the hamiltonian as a function of the canonical momenta

Hfree =√m2 + |p|2 (4.169)

By performing the minimal substitution we get

H − eA0 =√m2 + (p− eA)2 (4.170)

from which

H = eA0 +√m2 + (p− eA)2 (4.171)

which is nothing but eq. (4.165), after using eq. (4.164). From the point of viewof canonical quantization, the minimal substitution corresponds to the followingsubstitution in the derivatives

∂µ → ∂µ + ieAµ (4.172)

72

4.6 Non relativistic limit of the Dirac equation

In order to understand better the role of the spin in the Dirac equation we will studynow the non relativistic limit in presence of an electromagnetic field.

(i∂ −m)ψ(x) = 0 =⇒ (i∂ − eA−m)ψ(x) = 0 (4.173)

Notice that the Dirac equation is invariant under the transformation (4.159)

Aµ(x) → Aµ(x) + ∂µα(x) (4.174)

if we perform also the following local phase transformation on the wave function

ψ(x) → e−ieα(x)ψ(x) (4.175)

Also, eq. (4.173) is invariant under Lorentz transformations, if in going from theframe S to the frame S ′ (x→ x′ = Λx), the field Aµ is transformed as

Aµ(x) → A′µ(x

′) = (Λ−1)νµAν(x) (4.176)

that is if Aµ(x) transforms as ∂µ:

∂

∂xµ→ ∂

∂x′µ=

∂xν

∂x′µ∂

∂xν= (Λ−1)νµ∂ν (4.177)

Eq. (4.176) says simply that Aµ transforms as a four vector under Lorentz transfor-mations.

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

ψ(x) =[ϕ(x)χ(x)

](4.178)

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

π = p− eA (4.179)

and using the representation in 2×2 blocks of the Dirac matrices given in eq. (4.11),we get

i∂

∂t

[ϕ(x)χ(x)

]=[

0 σ · πσ · π 0

] [ϕ(x)χ(x)

]+[m 00 −m

] [ϕ(x)χ(x)

]+ eA0

[ϕ(x)χ(x)

](4.180)

This gives rise to two coupled differential equations

i∂ϕ

∂t= σ · πχ+ (m+ eA0)ϕ

i∂χ

∂t= σ · πϕ− (m− eA0)χ (4.181)

73

In the non relativistic limit, and for weak fields, the mass term is the dominant

one, and the energy positive solution will behave roughly as e−imt. With thisconsideration in mind we put[

ϕ(x)χ(x)

]= e−imt

[ϕ(x)χ(x)

](4.182)

with ϕ and χ functions slowly variable in time. In this way we obtain

i∂ϕ

∂t= σ · πχ+ eA0ϕ

i∂χ

∂t= σ · πϕ− (2m− eA0)χ (4.183)

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

χ ≈ σ · π2m

ϕ (4.184)

from which

i∂ϕ

∂t=

[(σ · π)2

2m+ eA0

]ϕ (4.185)

One must be careful in evaluating (σ · π)2, since the components of the vector π donot commute among themselves. In fact

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

∂xi− ie

∂Ai

∂xj(4.186)

where we have made use of

[pi, f(x)] = −i∂f(x)∂xi

(4.187)

From B = ∇ ∧ A it follows[πi, πj] = ieϵijkB

k (4.188)

and

(σ · π)2 = σiσjπiπj =

(1

2[σi, σj] +

1

2[σi, σj]+

)πiπj

= |π|2 + 1

2[σi, σj]π

iπj = |π|2 + 1

4[σi, σj][π

i, πj]

= |π|2 + i

2ϵijkσk(ie)ϵijlB

l (4.189)

Finally(σ · π)2 = |π|2 − eσ · B (4.190)

74

The equation for ϕ becomes

i∂ϕ

∂t=

(p− eA)2

2m− e

2mσ · B + eA0

ϕ (4.191)

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

µ =e

2mσ =

e

mS (4.192)

where we have introduced the spin matrices S = σ/2. This shows that the Diracequation predicts a gyromagnetic ratio equal to two. We may see this also in aslightly different way, by considering the interaction with an uniform and weakmagnetic field. In this case the vector potential is given by

A =1

2B ∧ x (4.193)

Neglecting the quadratic term in the fields we have

(p− eA)2 ≈ |p|2 − e(p · A+ A · p) (4.194)

Using ∑i

[pi, Ai] = −i∇ · A = 0 (4.195)

it follows

(p− eA)2 ≈ |p|2 − 2ep · A = |p|2 − ep · (B ∧ x)= |p|2 − epiϵijkB

jxk = |p|2 − eϵkijxkpiBj

= |p|2 − e(x ∧ p) · B = |p|2 − eL · B (4.196)

and finally

i∂ϕ

∂t=

[|p|2

2m− e

2m(L+ 2S) · B + eA0

]ϕ (4.197)

which shows explicitly the value of the gyromagnetic ratio. Experimentally this isvery close to two, and we shall see, in the following, that the difference is explained bythe quantum electrodynamics (QED). This is in fact, one of the biggest successes ofthis theory. However, let us notice that, from the point of view of the Dirac equation,to find a value of the gyromagnetic ratio so close to the experimental value is not areal prediction. In fact, one could think to add to the theory a further interactionterm of the kind Fµνψσ

µνψ. This term is both Lorentz and gauge invariant. It isalso possible to show that such a term gives a contribution to the magnetic momentof the electron, and therefore it changes the gyromagnetic ratio. We shall see thatthe requirement that QED is a finite theory forbids, in fact, the appearance of sucha term.

75

4.7 Charge conjugation, time reversal and PCT

transformation

Dirac equation had a great success in explaining the fine structure of the hydro-gen atom, but the problem of negative energy solutions, that, in principle, makethe theory unstable, was still there. Dirac looked for a solution to this problemby taking advantage of the exclusion Pauli principle, which applies to half-integerspin particles. Dirac made the hypothesis that all the negative energy states wereoccupied by electrons. In such a situation, the Pauli principle forbids to any elec-tron in a positive energy state to make a transition to a negative energy state. Thissolves the stability problem, but at the same time new phenomena may happen.For, instance, an electron in a negative energy state could get enough energy (biggerthan 2m which is the minimal energy gap between the negative and positive energystates) to make a transition to a state of positive energy. If we imagine that in thestate of energy −E are present N electrons (we are simplifying things, because dueto the momentum degeneracy there actually an infinite number of electrons), andthat one of these electrons undergoes the transition, the energy of the state changesas follows

E −NE → E − (N − 1)E = E −NE + E (4.198)

where E is the energy of all the other electrons (with energy different from −E) inthe fundamental state. Notice that in the Dirac theory the fundamental state is theone with all the negative energy states occupied and zero electrons in the energypositive states. In a sense this is the physical explanation of the infinite energy ofthe vacuum that we found in the case of the Klein-Gordon field, and one finds alsofor the Dirac case (see later). In a complete analogous way, also the charge of thevacuum is infinite and its variation in the previous transition is given by

Q+Ne→ Q+ (N − 1)e = Q+Ne− e (4.199)

where e is the charge of the electron (e < 0). We see that the vacuum energy andthe charge increase respectively by E and −e in the transition. We can interpretthis by saying that the hole left in the vacuum by the electron has charge −e andenergy E. That is we can think to the hole being a particle of positive energy andpositive charge. This is the way in which the idea of antiparticles came around.That the hole is thought as the antiparticle of the electron. The transition of anelectron of negative energy to a state of positive energy is then seen as the creationof particle antiparticle (the hole) pair. In the same way, once we have a hole in thevacuum, it may happen that a positive energy electron makes a transition to thehole state. In this case both the electron and the hole disappears. This is the pairannihilation phenomenon. Of course this happens with some energy released, thatusually is under electromagnetic form.

The hole theory is nowadays reinterpreted in terms of antiparticles, but this wayof thinking has been extremely fruitful in many fields, as the study of electrons inmetals, in nuclear physics and so on.

76

Assuming seriously the hole theory means that the Dirac equation should admit,beyond the positive energy solutions corresponding to an electron, other positiveenergy solutions with the same mass of the electron, but with opposite charge. Tosee this point in a formal way, we look for a transformation of the electron wavefunction, ψ(x), to the antielectron (positron) wave function ψC(x), such that, if ψsatisfies

(i∂ − eA−m)ψ(x) = 0 (4.200)

then ψC satisfies(i∂ + eA−m)ψC(x) = 0 (4.201)

We will require that the transformation is a local one, and that the transformed of theantiparticle wave function gives back, except for a possible phase factor, the electronwave function. To build up ψC we will start by taking the complex conjugate of ψ.This is clearly the only possibility to change a negative energy solution, described

by eiEt, in a positive energy solution, described by e−iEt. By taking the hermitianconjugate, multiplying by γ0 and transposing, we get

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

→ [γµT (−i∂µ − eAµ)−m]ψT = 0 (4.202)

whereψT = γ0

Tψ⋆ (4.203)

If there exists a matrix, C, such that

CγµTC−1 = −γµ (4.204)

multiplying eq. (4.202) by C, we get

(i∂ + eA−m)CψT = 0 (4.205)

This describes a particle with charge −e, therefore, apart a phase factor ηC , we canidentify ψC with CψT :

ψC = ηCCψT (4.206)

In the representation where γ0 is diagonal we have

γ0T = γ0, γ1

T = −γ1, γ2T = γ2, γ3

T = −γ3 (4.207)

It is enough to choose C such to commute with γ1 and γ2 and anticommute with γ0and γ3. It follows that C must be proportional to γ2γ0. Let us choose

C = iγ2γ0 =(

0 −iσ2−iσ2 0

)(4.208)

In this way, C satisfies−C = C−1 = CT = C† (4.209)

77

To understand how the transformation works let us consider, in the rest frame, anegative energy solution with spin down

ψ(−) = eimt

0001

(4.210)

Then

ψ(−)C = ηCCψT = ηCCγ0ψ

(−)⋆ = ηCiγ2ψ(−)⋆

= ηCe−imt

0 0 0 10 0 −1 00 −1 0 01 0 0 0

0001

(4.211)

and

ψ(−)Cdown = ηCe

−imt

1000

= ηCψ(+)up (4.212)

That is, given an energy negative wave function describing an electron with spindown, its charge conjugated is a positive energy wave function describing a positronwith spin up. For an arbitrary solution with defined energy and spin, by using theprojectors of Section 4.3, we write

ψ =ϵp+m

2m

1 + γ5n

2ψ (4.213)

where p0 > 0. ϵ = ±1 selects the energy sign. Noticing that C commutes with γ5,that γ5

⋆ = γ5, and thatγ0γµ⋆γ0 = γµT (4.214)

following fromγ0㵆γ0 = γµ (4.215)

we obtain

ψC = ηCCγ0ϵp⋆ +m

2m

1 + γ5n⋆

2ψ⋆ = ηCC

ϵpT +m

2m

1− γ5nT

2γ0ψ⋆

=−ϵp+m

2m

1 + γ5n

2ψC (4.216)

We see that ψC is described by the same four vectors pµ and nµ appearing in ψ, butwith opposite sign of the energy. Then

u(p, n) = ηCvC(p, n), v(p, n) = ηCu

C(p, n) (4.217)

78

Since the spin projector selects the states of spin ±1/2 along n according to thesign of the energy, it follows that the charge conjugation inverts the spin projectionof the particle. Notice also that, being ψC a solution of the Dirac equation withe→ −e, it follows that the following transformation

ψ → ψC , Aµ → −Aµ (4.218)

is a symmetry of the Dirac equation. Because we change sign to the four-potential,we say also that the photon has charge conjugation -1.

The last discrete transformation we will consider here is the time reversal. Thephysical meaning of this transformation can be illustrated in terms of a movie werewe record all the observations made on the state described by the wave functionψ(x). If we run the movie backward and we see a series of observations which arephysically doable, we say that the theory in invariant under time reversal. From amathematical point of view we have the time reversal symmetry if, sending t→ t′ =−t, it is possible to transform the wave function in such a way that it satisfies theoriginal Dirac equation. If this happens, the transformed wave function describesan electron propagating backward in time. To build up explicitly the time reversaltransformation, let us consider the electron in interaction with the electromagneticfield. It is convenient to write the Dirac equation in hamiltonian form (see eq. (4.6))

i∂ψ(x, t)

∂t= Hψ(x, t) (4.219)

withH = eA0 + γ0γ · (−i∇ − eA) + γ0m (4.220)

Let us define our transformation through the following equation

ψ′(x, t′) = Kψ(x, t), t′ = −t (4.221)

From eq. (4.219), omitting the spatial argument

i∂

∂tK−1ψ′(t′) = HK−1ψ′(t′) (4.222)

Multiplying this equation by K we get

∂

∂t′K(−i)K−1ψ′(t′) = KHK−1ψ′(t′) (4.223)

The invariance can be realized in two ways

K(−i)K−1 = i; KHK−1 = H (4.224)

orK(−i)K−1 = −i; KHK−1 = −H (4.225)

79

The second possibility can be excluded immediately, since under time reversal wehave

∇ → ∇, A→ −A, A0 → A0 (4.226)

As it follows recalling that the vector potential is generated by a distribution of cur-rents (changing sign under time reversal), whereas the scalar potential is generatedby a distribution of charges. Let us put

K = T (4× 4 matrix)× (complexconjugation) (4.227)

Then, we get from eq. (4.223)

i∂

∂t′ψ′(t′) = TH⋆T−1ψ′(t′) (4.228)

withTH⋆T−1 = H (4.229)

By taking into account the transformation properties of the potentials we get

TH⋆T−1 = T (eA′0 + (γ0γ)

⋆ · (i∇+ eA′) + γ0⋆m)T−1 (4.230)

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

Tγ0(γ)⋆T−1 = −γ0γ, Tγ0T

−1 = γ0 (4.231)

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

T γ⋆T−1 = −γ (4.232)

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

Tγ0T−1 = γ0, Tγ1T−1 = −γ1, Tγ2T−1 = γ2, Tγ3T−1 = −γ3 (4.233)

By choosing arbitrary the phase, we put

T = iγ1γ3 (4.234)

With this choice T satisfies

T † = T, T 2 = 1 (4.235)

To understand the correspondence with the classical results, where momentum andangular momentum change sign under time reversal, let us study how a positiveenergy solution transforms:

K

[p+m

2m

1 + γ5n

2ψ(t)

]= T

[p⋆ +m

2m

1 + γ5n⋆

2

]ψ⋆(t)

= T

[p⋆ +m

2m

]T−1T

[1 + γ5n

⋆

2

]T−1Tψ⋆(t)

=ˆp+m

2m

1 + γ5 ˆn

2ψ′(t′) (4.236)

80

where, again t′ = −t, and

p = (p0,−p), n = (n0,−n) (4.237)

The three discrete symmetry operations described so far, parity, P , charge con-jugation, C, and time reversal, (T ), can be combined together into a symmetrytransformation called PCT . Omitting all the phases, this transformation is rathersimple

ψPCT (−x) = PC[Kψ(x)]T= PCγ0(Kψ(x))

⋆ = iγ0γ2(−iγ1⋆γ3⋆)ψ(x) = iγ5ψ(x)(4.238)

and it suggests a simple correspondence between the wave function of a positronmoving backward in time (ψPCT (−x)), and the electron wave function. For a freeparticle of negative energy we have

ψPCT (−x) = iγ5−p+m

2m

1 + γ5n

2ψ(x)

=p+m

2m

1− γ5n

2(iγ5ψ(x)) =

p+m

2m

1− γ5n

2ψPCT (−x) (4.239)

Comparison with eq. (4.216), giving the charge conjugated of an energy negativestate

ψC =p+m

2m

1 + γ5n

2ψC (4.240)

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

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

= (i∂x′ + eA(x′)−m)ψPCT (x′) (4.241)

showing that a positron moving backward in time satisfies the same equation asan electron moving forward. Eq. (4.241) tells us that the PCT transformationon ψ, combined with the PCT transformation on the four-vector potential, that isAµ(x) → −Aµ(−x), is a symmetry of the theory.

The interpretation of the positrons as negative energy electrons moving backwardin time is the basis of the positron theory formulated by Stuckelberg and Feynman.In this approach it is possible to formulate the scattering theory without using fieldtheory. In fact, the pair creation and pair annihilation processes can be reinterpretedin terms of scattering processes among electrons moving forward and backward intime.

4.8 Dirac field quantization

In this Section we abandon the study of the Dirac wave equation thought as a gen-eralization of the Scrhodinger equation, due to its difficulties to cope with many

81

particle states. We will adopt here the point of view of the quantum field the-ory. That is the relativistic equation is the equation describing the field operator.However, we have shown in the Klein-Gordon case, that after quantization we geta many particle system satisfying Bose-Einstein statistics. On the other hand wehave seen that the Dirac equation describes spin 1/2 particles, which should satisfythe Fermi-Dirac statistics. It is quite clear that we will run into troubles insisting inquantizing the Dirac field as we did for the Klein-Gordon case. However, in order tounderstand the problems and the way to deal with them, we will follow the canonicalway of quantization, showing that this leads to problems with the positivity of theenergy. Looking for a solution of this problem we will find also the solution to theproblem of the wrong statistics.

We will begin our study by looking for the action giving rise to the Dirac equation.We will take the quantities ψ and ψ as independent ones. Following the usualprocedure, we multiply the Dirac equation by δψ (in such a way to form a Lorentzscalar) and integrate over the space-time volume V

0 =∫Vd4x δψ(i∂ −m)ψ = δ

∫Vd4x ψ(i∂ −m)ψ (4.242)

We will the assume the following action

S =∫Vd4x ψ(i∂ −m)ψ (4.243)

It is simply verified that this action gives rise to the correct equation of motion forψ. In fact,

∂L∂ψ

= −mψ, ∂L∂ψ,µ

= ψiγµ (4.244)

from which−mψ − i∂µψγ

µ = 0 (4.245)

The canonical momenta result to be

Πψ =∂L∂ψ

= iψ†, Πψ† =∂L∂ψ†

= 0 (4.246)

The canonical momenta do not depend on the velocities. In principle, this creates aproblem in going to the hamiltonian formalism. In fact a rigorous treatment requiresan extension of the classical hamiltonian treatment, which was performed by Dirachimself. In this particular case, the result one gets is the same as proceeding ina naive way. For this reason we will avoid to describe this extension, and we willproceed as in the standard case. Then the hamiltonian density turns out to be

H = Πψψ − L = iψ†ψ − ψ(iγ0∂0 + iγk∂k −m)ψ = ψ†(−iα · ∇+ βm)ψ (4.247)

If one makes use of the Dirac equation, it is possible to write the hamiltonian densityas

H = ψ†i∂ψ

∂t(4.248)

82

Contrarily to the Klein-Gordon case (see eq. (3.100), the hamiltonian density isnot positive definite. Let us recall the general expression for the energy momentumtensor (see eq. 3.84)

T µν =∂L∂ϕi,µ

ϕi,ν − gµνL (4.249)

In our case we getT µν = iψγµψ,ν − gµν (ψ(i∂ −m)ψ) (4.250)

and using the Dirac equationT µν = iψγµψ,ν (4.251)

We verify immediately that this expression has vanishing four divergence. Also

T 0k = iψ†∂kψ (4.252)

from which we get the momentum of the field

P k =∫

d3x T 0k =⇒ P = −i∫

d3x ψ†∇ψ (4.253)

In analogous way, by using the general expression for the angular momentum density(see eq. (3.96))

Mµρν = xρT

µν − xνT

µρ − ∂L

∂ϕi,µΣijρνϕ

j (4.254)

The matrices Σijµν are defined in terms of the transformation properties of the

field(see eq. (3.94))

∆ϕi = −1

2Σijµνϵ

µνϕj (4.255)

In our case from eq. (4.32), and from eq. (4.43) in the case of an infinitesimalLorentz transformation, we get

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

4σµνϵ

µνψ(x) (4.256)

from which

Σµν =i

2σµν = −1

4[γµ, γν ] (4.257)

Our result is then

Mµρν = iψγµ

(xρ∂ν − xν∂ρ −

i

2σρν

)ψ = iψγµ

(xρ∂ν − xν∂ρ +

1

4[γρ, γν ]

)ψ (4.258)

By taking the spatial components we obtain

J = (M23,M31,M12) =∫

d3x ψ†(−ix ∧ ∇+

1

2σ ⊗ 12

)ψ (4.259)

83

where 12 is the identity matrix in 2 dimensions, and using eq. (4.61) we have defined

σ ⊗ 12 =(σ 00 σ

)(4.260)

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, thecurrent ψγµψ.

We will need the decomposition of the Dirac field in plane waves. To this endwe will make use of the spinors u(p,±n) e v(p,±n) that we have defined at the endof the Section 4.3. The expansion is similar to the one used in eq. (4.126), but nowb(k) and d(k) are operators

ψ(x) =∑±n

∫ d3 p√(2π)3

√m

Ep

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

](4.261)

ψ†(x) =∑±n

∫d3 p√(2π)3

√m

Ep

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

]γ0 (4.262)

where Ep =√|p|2 +m2. We will collect here the various properties of the previous

spinors, which are nothing but a trivial extension of the case in which, in the restframe, the spin is quantized along the z axis:

• Dirac equation

(p−m)u(p, n) = u(p, n)(p−m) = 0

(p+m)v(p, n) = v(p, n)(p+m) = 0 (4.263)

• Orthogonality

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

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

v(p, n)u(p, n′) = v†(p, n)u(p, n′) = 0 (4.264)

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

• Completeness

∑±nu(p, n)u(p, n) =

p+m

2m∑±nv(p, n)v(p, n) =

p−m

2m(4.265)

84

We are now in the position to express the hamiltonian in terms of the operatorsb(p, n) e d(p, n). Using eq. (4.248) we find

H =∫

d3x∑

±n,±n′

∫ d3 p√(2π)3

√m

Ep

d3p′√(2π)3

√m

Ep′Ep′

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

]×[b(p′, n′)u(p′, n′)e−ip

′x − d†(p′, n′)v(p′, n′)eip′x]

=∑

±n,±n′

∫d3p d3p′

Ep′m√EpEp′

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

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

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

−b†(p, n)d(p, n′)e+i(Ep + Ep′)tu†(p, n)v(p, n′)δ3(p+ p′)]

(4.266)

Performing one of the integrations and using the orthogonality relations, we get

H =∑±n

∫d3p Ep[b

†(p, n)b(p, n)− d(p, n)d†(p, n)] (4.267)

In analogous way we get

P =∑±n

∫d3p p[b†(p, n)b(p, n)− d(p, n)d†(p, n)] (4.268)

If we try to interpret these expressions as we did in the Klein-Gordon case, we wouldassume that the operator d(p, n) creates from the vacuum a state of energy −Ep andmomentum −p. Dirac tried to solve the problem assuming that the vacuum wasfilled up by the negative energy solutions. Due to the Pauli principle this wouldmake impossible for any other negative energy state to be created. Let us calledthe vacuum filled up by the negative energy solutions the Dirac vacuum. Thenthe operator d(p, n) should give zero when acting upon this state. We then defineas the true vacuum of the theory the Dirac vacuum and require

d(p, n)|0⟩Dirac = 0 (4.269)

That is, in the Dirac vacuum, the operator d(p, n) behaves as an annihilation opera-tor (thing that we have anticipated in writing). Since the Dirac vacuum is obtainedby the original vacuum acting with d(p, n), it follows that these operators shouldsatisfy the following algebraic identity

(d(p, n))2 = 0 (4.270)

85

We can satisfy this relation in a uniform algebraic way by requiring that the opera-tors d(p, n) anticommute among themselves

[d(p, n), d(p′, n′)]+ = 0 (4.271)

This leads to the Jordan and Wigner idea of quantizing the Dirac field in terms ofanticommutators[

b(p, n), b†(p′, n′)]+=[d(p, n), d†(p′, n′)

]+= δnn′δ3(p− p′) (4.272)

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

P µ =∑±n

∫d3p pµ

[b†(p, n)b(p, n) + d†(p, n)d(p, n)−

[d(p, n), d†(p, n)

]+

](4.273)

Due to the anticommutation relations, the last term turns out to be an infinitenegative constant, which, physically, can be associated to the energy of the infiniteelectrons filling up the Dirac vacuum (called also the Dirac sea). If we ignore thisconstant (as we did in the Klein-Gordon case, and with the same warnings), theenergy operator is positive definite. The use of the anticommutators solves also theproblem of the wrong statistics. In fact, the wave functions are now antisymmetricin the exchange of two Dirac particles (from now on we will put |0⟩Dirac = |0⟩):

b†(p1, n1)b†(p2, n2)|0⟩ = −b†(p2, n2)b

†(p1, n1)|0⟩ (4.274)

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

In the Klein-Gordon case we interpreted the conserved current as the electromag-netic current. We shall show now that in the Dirac case the expression ψγµψ, hasthe same interpretation. Let us start evaluating the spatial integral of the density∫

d3x ψ†ψ (4.275)

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

∑±n

∫d3p

[b†(p, n)b(p, n) + d(p, n)d†(p, n)

](4.276)

As we know, this expression is formally positive definite. However, if we couple theDirac field to the electromagnetism through the minimal substitution we find thatthe free action (4.243) becomes

S =∫Vd4xψ(i∂ − eA−m)ψ (4.277)

86

Therefore the electromagnetic field is coupled to the conserved current

jµ = eψγµψ (4.278)

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

Q = e∫

d3x ψ†ψ

=∑±n

∫d3p e

[b†(p, n)b(p, n)− d†(p, n)d(p, n) +

[d(p, n), d†(p, n)

]+

](4.279)

The subtraction of the infinite charge associated to the Dirac sea leaves us withan operator which is not anymore positive definite. We see also that the operatorsb† create particles of charge e (electrons) whereas d† create particles of charge −e(positrons). Notice that the interpretation of Q as the charge operator would nothave worked by using commutation relations.

A further potential problem is connected with the causality. We have seen inthe Klein-Gordon case that the causality properties is guaranteed, for the localobservable, by the canonical commutation relations for the fields. But this is justthe property we have given up in the Dirac case. In order to discuss this point, letus start evaluating the equal time anticommutator for the Dirac field

[ψ(x, t), ψ†(y, t)

]+=

∑±n,±n′

∫ d3 p√(2π)3

√m

Ep

∫ d3 p′√(2π)3

√m

Ep′δnn′δ3(p− p′)

×[u(p, n)u(p, n)γ0e

ip · (x− y) + v(p, n)v(p, n)γ0e−ip · (x− y)

]=

∫ d3p

(2π)3m

Ep

[(p+m

2m

)eip · (x− y) +

(p−m

2m

)e−ip · (x− y)

]γ0

=∫

d3p

(2π)3m

Ep

[(p+m

2m

)+

(ˆp−m

2m

)]γ0e

ip · (x− y)

=∫ d3p

(2π)3m

Ep

2Ep2m

e−ip · (x− y) = δ3(x− y) (4.280)

In analogous way we get

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

]+= 0 (4.281)

By using eq. (4.246). the anticommutator between ψ and ψ† can be written as

[Πψ(x, t), ψ(y, t)]+ = iδ3(x− y) (4.282)

This shows that also in the Dirac case one can use the canonical formalism, but withanticommutators in place of the commutators. For arbitrary space-time separations

87

we get [ψ(x), ψ†(y)

]+=

=∫ d3p

(2π)3m

Ep

[(p+m

2m

)e−ip(x− y) +

(p−m

2m

)eip(x− y)

]γ0

=[(i∂ +m

)xγ0] ∫ d3p

(2π)31

2Ep

[e−ip(x− y) − eip(x− y)

]=

[(i∂ +m

)xγ0]i∆(x− y) (4.283)

where ∆(x) is the invariant function defined in eq. (3.148), in the evaluation ofcommutator for he Klein-Gordon field. From the properties of the ∆(x) function, itfollows that the anticommutator of the Dirac fields vanishes at space-like distances.It follows that the Dirac field cannot be an observable quantity. In fact one could domore, by evaluating the commutator of the Dirac field. It is easy to show that theresult does not vanish at space-like distances. This by itself would put in a serioustrouble the idea of quantizing Dirac field via commutation relations. But how dowe solve the causality problem. The crucial observation is in the following identity

[AB,C] = A[B,C] + [A,C]B = A[B,C]+ − [A,C]+B (4.284)

which holds for arbitrary operators. The identity shows that AB commutes withC if A and B separately commute or anticommute with C. An immediate conse-quence is that a local quantity containing an even number of Dirac fields commuteswith itself at space-like distances. So, in order to reconcile the causality with thequantization of the Dirac field we have to give up with its property of being anobservable. However, all the important physical quantities, as energy-momentumtensor and electromagnetic current are bilinear in the Fermi fields, and thereforethey are observable quantities.

What we have shown here is that, in order to give a sense to the quantizationof the Dirac field, we have been forced to use anticommutation relations, which, inturn, imply that the corresponding quanta obey the Fermi-Dirac statistics. Thisis nothing but an example of the celebrated spin statistics theorem that wasproved by Pauli in 1940. This theorem says that in a Lorentz invariant local fieldtheory, integer and half-integer particles must satisfy respectively Bose-Einstein andFermi-Dirac statistics.

88

Chapter 5

The electromagnetic field

5.1 The quantization of the electromagnetic field

In Section 4.5 we have shown that the action for the electromagnetic field must beexpressed in terms of the four-vector potential Aµ. We recall that the lagrangiandensity for the free case is given by (see. eq. (4.158)

L = −1

4FµνF

µν (5.1)

whereFµν = ∂µAν − ∂νAµ (5.2)

The resulting equations of motion are

∂2Aµ − ∂µ(∂νAν) = 0 (5.3)

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

Aµ(x) → A′µ(x) = Aµ(x) + ∂µΛ(x) (5.4)

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

electromagnetic field). In other words the action of electrodynamics is gauge invari-ant. Up to now, we have considered only symmetries depending on a finite numberof parameters. For instance, in the case of the O(2) symmetry for the charged scalarfield, the transformation symmetry depends on a single parameter, the rotation an-gle. In the case of the gauge symmetry one deals with a continuous number ofparameters, given by the function Λ(x). In fact, in each space-time point, we canchange the definition of Aµ by adding the gradient of Λ evaluated at that point.The main consequence of this type of invariance is to reduce the effective degrees offreedom of the theory from 4 to 2. Let us start from the classical theory. We recallthat it is possible to use the gauge invariance to require some particular condition

89

on the field Aµ. For instance, we can perform a gauge transformation in such a waythat the transformed field satisfies

∂µAµ = 0 (5.5)

In fact, given an arbitrary Aµ, let us make a gauge transformation by choosing Λ(x)such that

∂2Λ + ∂µAµ = 0 (5.6)

Then, the transformed field A′µ = Aµ+∂µΛ has vanishing four divergence. When Aµ

satisfies the condition ∂µAµ = 0, we say that the potential is in the Lorentz gauge.

Notice also that there is still a freedom in the potential. In fact by performing afurther gauge transformation

Aµ → A′µ = Aµ + ∂µΛ

′ (5.7)

we can stay in the Lorentz gauge, that is with ∂µA′µ = 0, if

∂2Λ′ = 0 (5.8)

In the Lorentz gauge the equations of motion simplify and reduce to the wave equa-tion, or to the Klein-Gordon equation with m = 0. This, together the fact, thatin the Lorentz gauge the covariance of the theory is explicitly preserved (the gaugecondition is Lorentz invariant), makes this gauge a very popular one. However, thecounting of the effective degrees of freedom is not so evident. From this point ofview a better choice is the Coulomb gauge, which is defined as the gauge wherethe scalar potential and the spatial divergence of the vector potential vanish. To seethat such a gauge exists, let us perform the following gauge transformation

A′µ(x) = Aµ(x)− ∂µ

∫ t

0A0(x, t

′)dt′ (5.9)

ClearlyA′

0 = 0 (5.10)

Then we perform a second gauge transformation

A′′µ = A′

µ − ∂µΛ (5.11)

in such a way to have ∇ · A′′ = 0. To this end we choose Λ(x) such that

∇ · A′′ = ∇ · A′ + ∇2Λ = 0 (5.12)

This equation can be solved by recalling that

∇2 1

|x|= −4πδ3(x) (5.13)

90

Then

Λ(x, t) =1

4π

∫ d3x′

|x− x′|∇ · A′(x′, t) (5.14)

from which∂Λ(x, t)

∂t=

1

4π

∫ d3x′

|x− x′|∇ · A′(x′, t) (5.15)

From the Gauss equation for the electric field we get

∇ · E = −∇2A0 − ∇ · A = 0 (5.16)

and, in terms of A′µ

∇ · E = ∇ · A′ = 0 (5.17)

Therefore∂Λ(x, t)

∂t= 0 (5.18)

from which A′′0 = A′

0 = 0 showing that the second gauge transformation does notdestroy the vanishing of the scalar potential. In conclusion, we have shown that itis possible to choose a gauge such that

A0 = ∇ · A = 0 (5.19)

It follows that the independent degrees of freedom are only two. Unfortunately inthis gauge the explicit Lorentz covariance is lost. Another way of showing that Aµhas only two degrees of freedom is through the equations of motion. Let us considerthe four dimensional Fourier transform of Aµ(x)

Aµ(x) =∫

d4k eikxAµ(k) (5.20)

Substituting this expression in the equations of motion we get

−k2Aµ(k) + kµ(kνAν(k)) = 0 (5.21)

Let us now decompose Aµ(k) in terms of four independent four vectors, which can

be chosen as kµ = (E, k), kµ = (E,−k), and two further four vectors eλµ(k), λ = 1, 2,orthogonal to kµ

kµeλµ = 0, λ = 1, 2 (5.22)

The decomposition of Aµ(k) reads

Aµ(k) = aλ(k)eλµ + b(k)kµ + c(k)kµ (5.23)

From the equations of motion we get

−k2(aλeλµ + bkµ + ckµ) + kµ(bk2 + c(k · k)) = 0 (5.24)

91

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 (5.25)

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) (5.26)

thenAµ(k) → Aµ(k) + ikµΛ(k) (5.27)

whereΛ(x) =

∫d4k eikxΛ(k) (5.28)

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) (5.29)

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. If we would like to require theexplicit covariance of the theory we would require non trivial commutation relationsfor all the component of the field. That is

[Aµ(x, t),Πν(y, t)] = igνµδ

3(x− y) (5.30)

[Aµ(x, t), Aν(y, t)] = [Πµ(x, t),Πν(y, t)] = 0 (5.31)

with

Πµ =∂L∂Aµ

(5.32)

To evaluate the conjugated momenta is better to write the lagrangian density (seeeq. (5.1) 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 (5.33)

Therefore∂L∂Aµ,ν

= −Aµ,ν + Aν,µ = F µν (5.34)

92

implying

Πµ =∂L∂Aµ

= F µ0 (5.35)

It follows

Π0 =∂L∂A0

= 0 (5.36)

We see that it is impossible to satisfy the condition

[A0(x, t),Π0(y, t)] = iδ3(x− y) (5.37)

We can try to find a solution to this problem modifying the 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

∂2Aµ(x) = 0 (5.38)

and this equation can be obtained by the lagrangian density

L = −1

2Aµ,νA

µ,ν (5.39)

(just think to the Klein-Gordon case). We will see that the minus sign is necessaryto recover a positive hamiltonian density. We now express this lagrangian densityin terms of the gauge invariant one, given in eq. (5.1). To this end we observe thatthe difference between the two lagrangian densities is nothing but the second termof eq. (5.33)

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 (5.40)

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

L = −1

4FµνF

µν − 1

2(∂µAµ)

2 (5.41)

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 (5.42)

we get0 = −∂2Aµ + ∂µ(∂νAν)− ∂µ(∂λAλ) = −∂2Aµ (5.43)

93

the term

−1

2(∂µAµ)

2 (5.44)

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 (5.45)

The corresponding equations of motion turn out to be

∂2Aµ − (1− λ)∂µ(∂λAλ) = 0 (5.46)

These equations are the same as the Maxwell equations in the Lorentz gauge. There-fore, in the following we will use λ = 1. From eq. (5.42) we see that

Π0 =∂L∂A0

= −∂µAµ (5.47)

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 (5.48)

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 (5.48).

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) (5.49)

or, explicitly

Π0 = −∂λAλ = −A0 − ∇ · AΠi = ∂iA0 − ∂0Ai = −Ai + ∂iA0 (5.50)

Since the spatial gradient of the field commutes with the field itself at equal time,the canonical commutator (5.30) gives rise to

[Aµ(x, t), Aν(y, t)] = −igµνδ3(x− y) (5.51)

94

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,inthe 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 (5.52)

The four vectors ϵ(λ)µ , being orthogonal to nµ are space-like, then they will be chosenorthogonal and normalized in the following way

ϵ(λ)µ ϵ(λ′)µ = −δλλ′ (5.53)

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

nµϵ(3).µ = 0 (5.54)

withϵ(3)µ ϵ(3)

µ= −1 (5.55)

By construction ϵ(3)µ is orthogonal to ϵ(λ)µ . This four vector is completely fixed bythe previous conditions, and we get

ϵ(3)µ =kµ − (n · k)nµ

(n · k)(5.56)

A last unit four vector we choose nµ

ϵ(0)µ = nµ (5.57)

These four vector are orthonormal and we can write

ϵ(λ)µ ϵ(λ′)µ = gλλ

′(5.58)

and being linearly independent, they satisfy the completeness relation

ϵ(λ)µ ϵ(λ′)

ν gλλ′ = gµν (5.59)

In the frame where nµ = (1, 0) and kµ = (k, 0, 0, k), we have

ϵ(1)µ= (0, 1, 0, 0), ϵ(2)

µ= (0, 0, 1, 0), ϵ(3)

µ= (0, 0, 0, 1) (5.60)

The plane wave expansion of Aµ is

Aµ(x) =∫

d3k√2ωk(2π)3

3∑λ=0

ϵ(λ)µ (k)[aλ(k)e

−ikx + a†λ(k)eikx

](5.61)

95

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. (3.60)

ϵ(λ)µ (k)aλ(k) = i∫

d3x f ∗k(x)∂

(−)t Aµ(x) (5.62)

with the functions fk(x) defined as in Section 3.2. 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) (5.63)

and analogously

a†λ(k) = igλλ′∫

d3x ϵ(λ′)µ(k)Aµ(x)∂

(−)t fk(x) (5.64)

Comparison with the calculation done in eq. (2.91) we get

[aλ(k), a†λ′(k

′)] =

=∫

d3x d3y[− f ∗

k(x) ˙fk′(y)

(igµνgλλ′′ϵ

(λ′′)µ(k)ϵ(λ′′′)ν(k′)gλ′λ′′′δ

3(x− y))

− f∗k(x)fk′(y)

(−igµνgλλ′′ϵ(λ

′′)µ(k)ϵ(λ′′′)ν(k′)gλ′λ′′′δ

3(x− y)) ]

= −∫

d3x f ∗k(x)i∂

(−)t fk′(x)gλλ′ (5.65)

and using the orthogonality relations of eq. (3.50)

[aλ(k), a†λ′(k

′)] = −gλλ′δ3(k − k′) (5.66)

Analogously[aλ(k), aλ′(k

′)] = [a†λ(k), a†λ′(k

′)] = 0 (5.67)

Again from the comparison of the Klein-Gordon commutators we have

[Aµ(x), Aν(y)] = −igµν∆(x− y) (5.68)

with the invariant function ∆(x) defined as in eq. (3.148), but with m = 0. Thecommutation 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⟩ (5.69)

its norm is given by

⟨1, λ|1, λ⟩ =∫

d3k d3k′ f⋆(k)f(k′)⟨0|aλ(k)a†λ(k′)|0⟩

=∫

d3k d3k′ f⋆(k)f(k′)⟨0|[aλ(k), a†λ(k′)|0⟩

= −gλλ∫

d3k |f(k)|2 (5.70)

96

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 (5.71)

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))† (5.72)

it is possible to weaken the condition, and require

∂µA(+)µ (x)|phys.⟩ = 0 (5.73)

This allows us to satisfy automatically the original requirement

⟨phys.|(∂µA(+)µ + ∂µA(−)

µ )|phys.⟩ = 0 (5.74)

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

i∂µA(+)µ (x) =

∫ d3k√2ωk(2π)3

e−ikx∑λ=0,3

kµϵ(λ)µ (k)aλ(k) (5.75)

Using eq. (5.56), we get

kµϵ(3)µ = −(n · k), kµϵ(0)µ = (n · k) (5.76)

from which[a0(k)− a3(k)]|phys.⟩ = 0 (5.77)

Notice that

[a0(k)− a3(k), a†0(k

′)− a†3(k′)] = −δ3(k − k′) + δ3(k − k′) = 0 (5.78)

Let us denote by Φk(n0, n3) the state with n0 scalar photons (that is with polarizationϵ(0)µ (k)), and with n3 longitudinal photons (that is with polarization ϵ(3)µ (k)). Thenthe following states satisfy the condition (5.73)

Φ(m)

k=

1

m!(a†0(k)− a†3(k))

mΦk(0, 0) (5.79)

97

These states have vanishing norm

||Φ(m)

k||2 = 0 (5.80)

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)) (5.81)

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⟩ (5.82)

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

⟨φn|N |φn⟩ = 0 (5.83)

since a0 and a3 act in the same way on a physical state (see eq. (5.77)). It follows

n⟨φn|φn⟩ = 0 (5.84)

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 (5.85)

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

|φ⟩ = c0|φ0⟩+∑i=0

ci|φi⟩ (5.86)

This state has a positive definite norm

⟨φ|φ⟩ = |c0|2 ≥ 0 (5.87)

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]: (5.88)

98

One can easily show the hamiltonian is given by the sum of all the degrees of freedomappearing in Aµ (see the Klein-Gordon case, eq. (3.102))

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)

]: (5.89)

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

⟨phys.|H|phys.⟩ = ⟨phys.|∫

d3k ωk2∑

λ=1

a†λ(k)aλ(k)|phys.⟩ (5.90)

The generic physical state is of the form |φT ⟩ ⊗ |φ⟩. with |φ⟩ defined as in eq.(5.86). 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

c⋆ncm⟨φn|Aµ(x)|φm⟩ (5.91)

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⟩ = c⋆0c1⟨0|∫ d3k√

2ωk(2π)3e−ikx[ϵ(3)µ (k)a3(k) + ϵ(0)µ (k)a0(k)]|φ1⟩+ c.c.

(5.92)In order to satisfy the gauge condition the state |φ1⟩ is of the form

|φ1⟩ =∫

d3q f(q)[a†3(q)− a†0(q)]|0⟩ (5.93)

and therefore

⟨φ|Aµ(x)|φ⟩ =∫ d3k√

2ωk(2π)3[ϵ(3)µ (k) + ϵ(0)µ (k)][c⋆0c1e

−ikxf(k) + c.c.] (5.94)

From eqs. (5.56) and (5.57) we have

ϵ(3)µ + ϵ(0)µ =kµ

(k · n)(5.95)

99

from which⟨φ|Aµ(x)|φ⟩ = ∂µΛ(x) (5.96)

with

Λ(x) =∫ d3k√

2ωk(2π)3

1

n · k(ic⋆0c1e

−ikxf(k) + c.c.) (5.97)

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

∂2Λ = 0 (5.98)

because the momentum k inside the integral satisfies k2 = 0.

100

Chapter 6

Symmetries in field theories

6.1 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

(6.1)

This lagrangian is invariant under linear transformations acting upon the vectorϕ = (ϕ1, · · · , ϕN) and leaving invariant its norm

|ϕ|2 =N∑i=1

ϕiϕi (6.2)

Consider an infinitesimal transformation (from now on we will omit the index ofsum over the indices which are repeated)

δϕi = ϵijϕj (6.3)

The condition for letting the norm invariant gives

|ϕ+ δϕ|2 = |ϕ|2 (6.4)

from whichϕ · δϕ = 0 (6.5)

or, in components,ϕiϵijϕj = 0 (6.6)

This is satisfied byϵij = −ϵji (6.7)

101

showing that the rotations in N dimensions depend on N(N−1)/2 parameters. Fora finite transformation we have

|ϕ′|2 = |ϕ|2 (6.8)

withϕ′i = Sijϕj (6.9)

implyingSST = 1 (6.10)

In fact, by exponentiating the infinitesimal transformation one gets

S = eϵ (6.11)

withϵT = −ϵ (6.12)

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 (6.13)

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 ) (6.14)

It is not difficult to show that these matrices satisfy the algebra

[TAB, TCD] = −iδACTBD + iδADTBC − iδBDTAC + iδBCTAD (6.15)

This is nothing but the Lie algebra of the group O(N), and the TAB are 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 (6.16)

since the N(N − 1)/2 parameters ϵAB are linearly independent, we find the N(N −1)/2 conserved currents

JABµ = −iϕi,µTABij ϕj (6.17)

In the case of N = 2 the symmetry is the same that we have studied in Section 3.6,and the only conserved current is given by

J12µ = −J21

µ = ϕ1,µϕ2 − ϕ2,µϕ1 (6.18)

102

in agreement with (3.169). One can easily check that the charges associated to theconserved currents close the same Lie algebra as the generators TAB. More generally,if we have conserved currents given by

jAµ = −iϕi,µTAij ϕj (6.19)

with[TA, TB] = ifABCTC (6.20)

then, using the canonical commutation relations, we get

[QA, QB] = ifABCQC (6.21)

withQA =

∫d3x jA0 (x) = −i

∫d3x ϕiT

Aij ϕj (6.22)

A particular example is the case N = 4. We parameterize our fields in the form

ϕ = (π1, π2, π3, σ) = (π, σ) (6.23)

These fields can be arranged into a 2× 2 matrix

M = σ + iτ · π (6.24)

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 (6.25)

Furthermore we have the relation

|ϕ|2 = σ2 + |π|2 = 1

2Tr(M †M) (6.26)

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(6.27)

This lagrangian is invariant under the following transformation of the matrix M

M → LMR† (6.28)

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 (6.25). In fact, if A isa 2× 2 matrix with detA = 1, then

τ2AT τ2 = A−1 (6.29)

103

Therefore, for M ′ = LMR† we get

τ2M′∗τ2 = τ2L

∗M∗RT τ2 = τ2L∗τ2(τ2M

∗τ2)τ2RT τ2 (6.30)

and from (6.29)

τ2L∗τ2 = τ2L

†T τ2 = L†−1= L

τ2RT τ2 = R−1 = R† (6.31)

since 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 (6.26) 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 (6.32)

then we get

δLM = (− i

2θL · τ)M = (− i

2θL · τ)(σ+ iπ · π) = 1

2θL · π+

i

2(θL ∧ π− θLσ) · τ (6.33)

where we have usedτiτj = δij + iϵijkτk (6.34)

SinceδLM = δLσ + iδLπ · τ (6.35)

we get

δLσ =1

2θL · π, δLπ =

1

2(θL ∧ π − θLσ) (6.36)

Analogously we obtain

δRσ = −1

2θR · π, δRπ =

1

2(θR ∧ π + θRσ) (6.37)

The combined transformation is given by

δσ =1

2(θL − θR) · π, δπ =

1

2[(θL + θR) ∧ π − (θL − θR)σ] (6.38)

104

and we can check immediately that

σδσ + π · δπ = 0 (6.39)

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† (6.40)

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 (6.41)

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) (6.42)

and dividing by θL/2

JLµ = σ,µπ − π,µσ − π,µ ∧ π (6.43)

and analogouslyJRµ = −σ,µπ + π,µσ − π,µ ∧ π (6.44)

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 (6.45)

By taking the following combinations of the currents

JVµ =1

2(JLµ + JRµ ), JAµ =

1

2(JLµ − JRµ ) (6.46)

one hasJVµ = π ∧ π,µ (6.47)

andJAµ = σ,µπ − π,µσ (6.48)

The corresponding algebra of charges is

[QVi , Q

Vj ] = iϵijkQ

Vk , [QV

i , QAj ] = iϵijkQ

Ak , [QA

i , QAj ] = iϵijkQ

Vk (6.49)

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 (6.50)

105

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)

](6.51)

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 (6.52)

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 (6.53)

The lagrangian density becomes

L =1

2∂µϕ

′i∂µϕ′

i − V (|ϕ′ + v|2) (6.54)

Expanding V in series of ϕ′i we get

V = V (|v|2) + 1

2

∂2V

∂ϕi∂ϕj

∣∣∣ϕ=v

ϕ′iϕ

′j + · · · (6.55)

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 (6.56)

Therefore∂V

∂ϕi= µ2ϕi + λϕi|ϕ|2 (6.57)

In order to have a solution to the stationary condition we must have ϕi = 0, or

|ϕ|2 = −µ2

λ(6.58)

This equation has real solutions only if µ2/λ < 0. However, in order to have apotential bounded from below one has to require λ > 0, therefore we may have non

106

zero 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.

6.2 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 (6.59)

The lagrangian density will be of the type

L =1

2∂µϕ∂

µϕ− V (ϕ2) (6.60)

If the vacuum state is non degenerate, barring a phase factor, we must have

P |0⟩ = |0⟩ (6.61)

since P commutes with the hamiltonian. It follows

⟨0|ϕ|0⟩ = ⟨0|P−1PϕP−1P |0⟩ = ⟨0|PϕP−1|0⟩ = −⟨0|ϕ|0⟩ (6.62)

from which⟨0|ϕ|0⟩ = 0 (6.63)

107

Things change if the fundamental state is degenerate. This would be the case in theexample (6.60), if

V (ϕ2) =µ2

2ϕ2 +

λ

4ϕ4 (6.64)

with µ2 < 0. In fact, this potential has two minima located at

ϕ = ±v, v =

√−µ

2

λ(6.65)

By denoting with |R⟩ e |L⟩ the two states corresponding to the classical configura-tions ϕ = ±v, we have

P |R⟩ = |L⟩ = |R⟩ (6.66)

Therefore⟨R|ϕ|R⟩ = ⟨R|P−1PϕP−1P |R⟩ = −⟨L|ϕ|L⟩ (6.67)

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 (6.68)

whereϕ · ϕ = ϕ2

1 + ϕ22 (6.69)

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 (6.70)

with v defined as in (6.65). 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⟩ (6.71)

and⟨0|ϕ|0⟩ = ⟨0|R−1RϕR−1R|0⟩ = ⟨0|ϕθ|0⟩ = 0 (6.72)

since ϕθ = ϕ. In the case µ2 < 0 (degenerate case), we have

R(θ)|0⟩ = |θ⟩ (6.73)

where |θ⟩ is one of the infinitely many degenerate fundamental states 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⟩ (6.74)

108

withϕθi = R(θ)ϕiR

−1(θ) = ϕi (6.75)

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⟩ (6.76)

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

](6.77)

109

The eigenvalues of H are(ϵ0 + ϵ1, ϵ0 − ϵ1) (6.78)

We have no more degeneracy and the eigenstates are

|S⟩ = 1√2(|R⟩+ |L⟩) (6.79)

with eigenvalue ES = ϵ0 + ϵ1, and

|A⟩ = 1√2(|R⟩ − |L⟩) (6.80)

with eigenvalue EA = ϵ0 − ϵ1. One can show that ϵ1 < 0 and therefore 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⟩) (6.81)

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⟩

)(6.82)

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(6.83)

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 (6.84)

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) = −∫Vd3x

[µ2

2v2 +

λ

4v4]=µ4

4λ

∫Vd3x =

µ4

4λV (6.85)

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6.3 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 (6.86)

The conditions that V must satisfy in order to have a minimum are

∂V

∂ϕl= µ2ϕl + λϕl|ϕ|2 = 0 (6.87)

with solutions

ϕl = 0, |ϕ|2 = v2, v =

√−µ2

λ(6.88)

The character of the stationary points can be studied by evaluating the secondderivatives

∂2V

∂ϕl∂ϕm= δlm(µ

2 + λ|ϕ|2) + 2λϕlϕm (6.89)

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 (6.90)

• 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 (6.91)

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Expanding the potential around this minimum we get

V (ϕ) ≈ V∣∣∣minimum

+1

2

∂2V

∂ϕl∂ϕm

∣∣∣minimum

(ϕl − vδlN)(ϕm − vδmN) (6.92)

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

(6.93)

Therefore the masses of the fields ϕa, a = 1, · · · , N − 1 , and χ = ϕN − v, are givenby

m2ϕa = 0, m2

χ = −2µ2 (6.94)

By definingm2 = −2µ2 (6.95)

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

(6.96)

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 (6.97)

since a, b = 1, · · · , N − 1, and

T aNij ϕj|min = i(δai δNj − δNi δ

aj )vδjN = ivδai = 0 (6.98)

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

112

(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.

6.4 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) (6.99)

which is invariant under the global phase transformation

ψ(x) → ψ′(x) = e−iQαψ(x) (6.100)

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 (6.101)

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 (6.99) cannotbe gauge invariant due to the derivative coming from the kinetic term. The idea issimply 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) (6.102)

113

In this case the termψDµψ (6.103)

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 (6.104)

The transformation law of Aµ is obtained from eq. (6.102)

[(∂µ + ieQAµ)ψ]′ = (∂µ + ieQA′

µ)ψ′(x)

= (∂µ + ieQA′µ)e

−iQα(x)ψ

= e−iQα(x)[∂µ + ieQ(A′

µ −1

e∂µα)

]ψ (6.105)

from which

A′µ = Aµ +

1

e∂µα (6.106)

The lagrangian density

Lψ = ψ[iD/ −m]ψ = ψ[iγµ(∂µ + ieQAµ)−m)ψ = L0 − eψQγµψAµ (6.107)

is then invariant under gauge transformations, or under the local 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 thateq. (6.102) implies that under a gauge transformation, the covariant derivativeundergoes a unitary transformation

Dµ → Dµ′ = e−iQα(x)Dµe

iQα(x) (6.108)

Then, also the commutator of two covariant derivatives

[Dµ, Dν ] = [∂µ + ieQAµ, ∂ν + ieQAν ] = ieQFµν (6.109)

withFµν = ∂µAν − ∂νAµ (6.110)

transforms in the same way

Fµν → e−iQα(x)FµνeiQα(x) = Fµν (6.111)

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 (6.112)

114

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 (6.113)

from which, eliminating the infinitesimal parameter α,

Jµ = ψγµQψ (6.114)

6.5 Non-abelian gauge theories

The approach of the previous section can be easily extended to local non-abeliansymmetries. We will consider the case of N Dirac fields. The free lagrangian

L0 =N∑a=1

ψa(i∂/ −m)ψa (6.115)

is invariant under the global transformation

Ψ(x) → Ψ′(x) = AΨ(x) (6.116)

where A is a unitary N ×N matrix, and we have denoted by Ψ the 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 (6.117)

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 (6.118)

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

](6.119)

we have

TA =σA

2, A = 1, 2, 3 (6.120)

115

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 (6.121)

To make local the transformation (6.117), means to promote the parameters αA tospace-time functions

αA → αA(x) (6.122)

Notice, that now the group does not need to be abelian, and therefore, in general

e−iαATAe−iβAT

A= e−iβAT

Ae−iαAT

A(6.123)

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)] (6.124)

We will put againDµ = ∂µ + igBµ (6.125)

where Bµ is a N ×N matrix acting upon Ψ(x). In components

Dµab = δab∂

µ + ig(Bµ)ab (6.126)

The eq. (6.124) implies

DµΨ → (∂µ + igB′µ)U(x)Ψ

= U(x)∂µΨ+ U(x)[U−1(x)igB′µU(x)]Ψ + (∂µU(x))Ψ

= U(x)[∂µ + U−1(x)igB′µU(x) + U−1(x)∂µU(x)]Ψ (6.127)

thereforeU−1(x)igB′

µU(x) + U−1(x)∂µU(x) = igBµ (6.128)

and

B′µ(x) = U(x)Bµ(x)U

−1(x) +i

g(∂µU(x))U

−1(x) (6.129)

For an infinitesimal transformation

U(x) ≈ 1− iαA(x)TA (6.130)

we get

δBµ(x) = −iαA(x)[TA, Bµ(x)] +1

g(∂µαA(x))T

A (6.131)

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 (6.132)

116

The transformation law for Aµ becomes

δAµC = fABC αAAµB +

1

g∂µαC (6.133)

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 (6.134)

in virtue of the eq. (6.124), 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 (6.135)

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) (6.136)

The invariant kinetic term will be assumed as

LA = −1

2Tr[FµνF

µν ] (6.137)

Let us now evaluate Fµν

igFµν = [Dµ, Dν ] = [∂µ + igBµ, ∂ν + igBν ]

= ig(∂µBν − ∂νBµ)− g2[Bµ, Bν ] (6.138)

orFµν = (∂µBν − ∂νBµ) + ig[Bµ, Bν ] (6.139)

in componentsF µν = F µν

C TC (6.140)

withF µνC = ∂µAνC − ∂νAµC − gfABC AµAA

νB (6.141)

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 (6.142)

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.

117

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 ∫

Vd4x

[Ψ(i∂/ −m)Ψ− gΨγµB

µΨ]+ SA (6.143)

where

SA = −1

2

∫Vd4xTr(FµνF

µν) (6.144)

and the variation of SA

δSA = −∫Vd4xTr(FµνδF

µν) (6.145)

Using the definition (6.139) for the field strength we get

δFµν = ∂µδBν + ig(δBµ)Bν + igBµδ(Bν)− (µ↔ ν) (6.146)

from which

δSA = −2∫Vd4xTr[F µν(∂µδBν + ig(δBµ)Bν + igBµ(δBν))] (6.147)

where we have taken into account the antisymmetry properties of Fµν . Integratingby parts we obtain

δSA = −2∫Vd4xTr[−(∂µF

µν)δBν − igBµFµνδBν + igF µνBµδBν ]

= 2∫Vd4xTr [(∂µF

µν + ig[Bµ, Fµν ]) δBν ]

=∫Vd4x (∂µF

µν + ig[Bµ, Fµν ])A δAνA (6.148)

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 (6.149)

From the first equation we see that the currents ΨγµTAΨ are not conserved. In fact

the conserved currents turn out to be

JAν = ΨγνTAΨ− i[Bµ, Fµν ]

A (6.150)

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 respect

118

to 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 (6.151)

and we get

jµ =∂L∂Ψ,µ

δΨ+∂L

∂Aν,µCδAνC (6.152)

from whichjµ = ΨγµαAT

AΨ− F µνC fABC αAAνB (6.153)

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 (6.154)

Thereforejµ = ΨγµαAT

AΨ− i[Bν , Fνµ]AαA (6.155)

After division by αA we get the Noether’s currents (6.150). 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 (6.156)

which vanishes for abelian gauge fields, whereas it is compensated by the gaugefields contribution in the non abelian case.

6.6 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, the

119

theory 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 (6.157)

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(6.158)

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 (6.159)

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 (6.160)

from whichδχ = −αϕ2, δϕ2 = αχ+ αv (6.161)

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 (6.162)

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 =√ϕ21 + ϕ2

2, sin θ =ϕ2√

ϕ21 + ϕ2

2

(6.163)

120

Under a finite rotation, the new fields transform as

ρ→ ρ, θ → θ + α (6.164)

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 =√ϕ22 + χ2 + 2χv + v2 ≈ v + χ, θ ≈ ϕ2

v + χ≈ ϕ2

v(6.165)

Again, if we make the theory invariant under a local transformation, we will haveinvariance under

θ(x) → θ(x) + α(x) (6.166)

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) (6.167)

The O(2) transformations become phase transformations on ϕ

ϕ→ eiαϕ (6.168)

and the lagrangian (6.157) can be written as

L = ∂µϕ†∂µϕ− µ2ϕ†ϕ− λ(ϕ†ϕ)2 (6.169)

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 (6.170)

from which

L = (∂µ − igAµ)ϕ†(∂µ + igAµ)ϕ− µ2ϕ†ϕ− λ(ϕ†ϕ)2 − 1

4FµνF

µν (6.171)

In terms of the polar coordinates (ρ, θ) we have

ϕ =1√2ρeiθ, ϕ† =

1√2ρe−iθ (6.172)

By performing the following gauge transformation on the scalars

ϕ→ ϕ′ = ϕe−iθ (6.173)

121

and the corresponding transformation on the gauge fields

Aµ → A′µ = Aµ +

1

g∂µθ (6.174)

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 (6.175)

In this way the Goldstone boson disappears. We have now to translate the field ρ

ρ = χ+ v, ⟨0|χ|0⟩ = 0 (6.176)

and we see that this generates a bilinear term in Aµ, coming from the covariantderivative, given by

1

2g2v2AµA

µ (6.177)

Therefore the gauge field acquires a mass

m2A = g2v2 (6.178)

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 (6.179)

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 (6.180)

where(Dµ)ij = δij∂µ + i

g

2(TAB)ijW

ABµ (6.181)

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 (6.158)

ϕi = vδiN (6.182)

122

Recalling that

T abij ϕj∣∣∣min

= 0, T aNij ϕj∣∣∣min

= ivδai , a, b = 1, · · · , N − 1 (6.183)

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

4g2v2W aN

µ W µbN (6.184)

Therefore, the fields W aNµ associated to the broken directions T aN acquire a mass

g2v2/2, whereasW 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. 6.1.We can now show how to eliminate the Goldstone bosons. In fact we can define

new fields ξa and χ as

ϕi =(e−iT

aNξa)iN

(χ+ v) (6.185)

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 fieldsϕ and (ξa, χ) can be seen easily by expanding around the vacuum(

e−iTaNξa

)iN

≈ δiN − i(T aN)iNξa = δiN + δai ξa (6.186)

from whichϕi ≈ (vξa, χ+ v) (6.187)

showing that the ξa’s are really the Goldstone fields. The unitary gauge is definedthrough the transformation

ϕi →(eiT

aNξa)ijϕj = δiN(χ+ v) (6.188)

123

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.

Wµ → eiTaNξaWµe

−iT aNξa − i

g

(∂µe

iT aNξa)e−iT

aNξa (6.189)

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 fieldW 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.

124

Chapter 7

Time ordered products

7.1 Time ordered products and propagators.

One of the most relevant quantities in perturbative field theory is the propagator,that is the vacuum expectation value of a time ordered product of two fields. Tointroduce the propagator from a physical point of view we will consider a chargedKlein-Gordon. As we know from Section 3.6, the field ϕ destroys a particle of charge+1 and creates a particle of charge −1. In any case the net variation of the chargeis −1. In analogous way the field ϕ† gives rise to a net variation of the charge equalto +1. Let us now construct a state with charge +1 applying ϕ† to the vacuum

|ψ(y, t)⟩ = ϕ†(y)|0⟩ =∫ d3k√

2ωk(2π)3eiky |k,m, 1⟩ (7.1)

where |k,m, 1⟩ is the single particle state with charge +1, momentum k and mass m.We want to evaluate the probability amplitude for the state, |ψ(y, t)⟩, to propagateto the same state, |ψ(x, t′)⟩, at a later time t′ > t. This is given by the matrixelement

θ(t′ − t)⟨ψ(x, t′)|ψ(y, t)⟩ = θ(t′ − t)⟨0|ϕ(x, t′)ϕ†(y, t)|0⟩ (7.2)

It turns out to be convenient to think to this matrix element as the one correspondingto the creation of a charge +1 at the point y and time t, and to its annihilation atthe point x and time t′. This interpretation is a correct one, since the state |ψ(y, t)⟩is eigenstate of the charge density operator

ρ(z, t)|ψ(y, t)⟩ = [ρ(z, t), ϕ†(y, t)]|0⟩ = +δ3(z − y)ϕ†(y, t)|0⟩ = δ3(z − y)|ψ(y, t)⟩(7.3)

where we have used eq. (3.197)

ρ =: iϕ†∂(−)t ϕ : (7.4)

and the canonical commutation relations.

125

.

y x y x

p n p n

pn pn

t t'

t' t

+1 -1

t

Fig. 7.1 - The two probability amplitudes contributing to the process np→ np.

However for t′ < t we could reach the same result by creating a particle of charge-1 at (x, t′), and annihilating it at (y, t). The corresponding amplitude is

θ(t− t′)⟨0|ϕ†(y, t)ϕ(x, t′)|0⟩ (7.5)

The situation is represented in Fig. 7.1, where we have considered the case of acharged particle exchanged between a proton (charge +1) and a neutron(charge 0).The total amplitude is obtained by adding the two contributions together. We definethe vacuum expectation value of the time ordered product (T product) of two fieldsas

⟨0|T (ϕ(x)ϕ†(y)|0⟩ = ⟨0|T (ϕ†(y)ϕ(x))|0⟩= θ(x0 − y0)⟨0|ϕ(x)ϕ†(y)|0⟩+ θ(y0 − x0)⟨0|ϕ(y)†ϕ(x)|0⟩≡ −i∆F (x− y) (7.6)

The function ∆F (x − y) is called the Feynman propagator, and we will showimmediately that it depends indeed on the difference of the two coordinates x andy. Using the expressions (3.193) for the fields, we find

−i∆F (x− y) =∫

d3k d3k′[θ(x0 − y0)f

∗k(y)fk′(x)⟨0|a(k

′)a†(k)|0⟩

+ θ(y0 − x0)fk(y)f∗k′(x)⟨0|b(k′)b†(k)|0⟩

]=

∫ d3k

(2π)31

2ωk

[θ(x0 − y0)e

−ik(x− y)

+ θ(y0 − x0)eik(x− y)

](7.7)

where, we recall that k0 = ωk. This expression can be written in a more convenientway by using the following integral representation of the step function

θ(t) = limη→0+

i

2π

∫dω

e−iωt

ω + iη(7.8)

126

This representation can be verified immediately by noticing that for t < 0 the inte-gral is convergent in the upper complex half-plane of ω. Since there no singularitiesin this region (the integral has a pole at ω = −iη)), we see that the integral vanishes.In the case t > 0 the integral is convergent in the lower half-plane. Then we pickup the contribution of the pole (in a clockwise direction) and we find

−i∆F (x− y) = i∫ d3k

(2π)4

∫dω

1

2ωk

[e−iω(x0 − y0)

ω + iη e−ik(x− y)

+ eiω(x0 − y0)ω + iη eik(x− y)

](7.9)

By the following change of variable k0 = ω + ωk, we get

−i∆F (x− y) = i∫ d4k

(2π)41

2ωk

e−ik(x− y)

k0 − ωk + iη+

eik(x− y)

k0 − ωk + iη

= i

∫ d4k

(2π)41

2ωke−ik(x− y)

[1

k0 − ωk + iη− 1

k0 + ωk − iη

]

= i∫ d4k

(2π)4e−ik(x− y)

k2 −m2 + iϵ(7.10)

where we have defined ϵ = 2ηωk. Notice that ϵ is a positive quantity. Then

∆F (x− y) = −∫ d4k

(2π)4e−ik(x− y)

k2 −m2 + iϵ(7.11)

From this representation it follows that ∆F (x) is a Green function for the Klein-Gordon operator

(∂2 +m2)∆F (x) = δ4(x) (7.12)

That the T -product is a Green function for the Klein-Gordon operator is a simpleconsequence of its very definition, by using the canonical commutators

(∂2 +m2)x⟨0|T (ϕ(x)ϕ†(y))|0⟩ = ∂20⟨0|T (ϕ(x)ϕ†(y))|0⟩+ ⟨0|T ((−∇2

x +m2)ϕ(x)ϕ†(y))|0⟩= ∂0⟨0|δ(x0 − y0)[ϕ(x), ϕ

†(y)]|0⟩+ ∂0⟨0|T (ϕ(x)ϕ†(y))|0⟩+ ⟨0|T ((−∇2

x +m2)ϕ(x)ϕ†(y))|0⟩= ⟨0|δ(x0 − y0)[ϕ(x), ϕ

†(y)]|0⟩+ ⟨0|T ((∂2x +m2)ϕ(x)ϕ†(y))|0⟩= −iδ4(x− y) (7.13)

It is easily seen that an analogous result holds for the hermitian Klein-Gordon field,that is

∆F (x− y) = i⟨0|T (ϕ(x)ϕ(y))|0⟩ (7.14)

127

.

ωIm( )

Re( )

ωω

kωk-

C+

Fig. 7.2 - The integration path C+.

.

ωIm( )

Re( )

ωω

kωk-

C_

Fig. 7.3 - The integration path C−.

All the invariant functions we have encountered so far can be obtained by specifyingin a convenient way the integration path of the following integral

∆C(x) =∫C

d4k

(2π)4e−ikx

k2 −m2(7.15)

In fact the integrand has two poles in k0, which are located at k0 = ±ωk, withωk =

√|k|2 +m2. To define the integral we need to specify how the integration

path goes around the poles. In particular, we can define the two integration pathsC± given in Figs. 7.2 and 7.3. Correspondingly we have the following integrals

∆(±)(x) = −i∫C±

d4k

(2π)4e−ikx

k2 −m2(7.16)

128

.

ωIm( )

Re( )

ω

ω

kωk-

C

Fig. 7.4 - The integration path C

We find

∆(+)(x) = −i∫C+

d4k

(2π)4e−ikx

(k0 − ωk)(k0 + ωk)=∫ d3k

(2π)3e−ikx

2ωk= [ϕ(+)(x), ϕ(−)(y)]

(7.17)where we have used eq. (3.149). Also

∆(−)(x) = −∫ d3k

(2π)3eikx

2ωk= −∆(+)†(x) = [ϕ(−)(x), ϕ(+)(y)] (7.18)

It follows∆(+)(x) + ∆(−)(x) = i∆(x) (7.19)

and using eq. (3.147), i∆(x − y) = [ϕ(x), ϕ(y)]. Therefore the commutator can berepresented as

∆(x) = −∫C

d4k

(2π)4e−ikx

k2 −m2(7.20)

with C given in Fig. 7.4.In the case of ∆F , the poles position is the one in Fig. 7.5. Then it can be also

defined by taking the poles on the real axis and choosing an integration path CF ,as specified in Fig. 7.6. That is

∆F (x) = −∫CF

d4k

(2π)4e−ikx

k2 −m2(7.21)

129

.

ωIm( )

Re( )ω ωk

ωk

-

Fig. 7.5 - The position of the poles in the definition of ∆F (x)..

ωIm( )

Re( )

ω

ω

kωk-

CF

Fig. 7.6 - The integration path CF .

Using eqs. (7.7), (7.17) and (7.18), we see that

∆F (x) = iθ(x0)∆(+)(x)− iθ(−x0)∆(−)(x) (7.22)

Let us also notice that all the ∆C invariant functions defined on a close path C,satisfy the homogeneous Klein-Gordon equation. In fact, the action of the Klein-Gordon operator removes the singularities from the integrand, leaving the integralof an analytic function, which vanishes due to the Cauchy theorem.

For a free Dirac field, the Feynman propagator has a similar definition

SF (x− y)αβ = −i⟨0|T (ψα(x)ψβ(y))|0⟩ (7.23)

but with the T -product defined as follows

T (ψα(x)ψβ(y)) = θ(x0 − y0)ψα(x)ψβ(y)− θ(y0 − x0)ψβ(y)ψα(x) (7.24)

Notice thatT (ψα(x)ψβ(y)) = −T (ψβ(y)ψα(x)) (7.25)

130

(an analogous property holds for the Klein-Gordon case, but with a plus sign). Theminus sign introduced in the definition of the T -product for the Dirac field, is neededbecause only in this way it may represent the Green function for the Dirac operator.In fact

(i∂x −m)αβT (ψβ(x)ψγ(y))

= i(γ0)αβδ(x0 − y0)[ψβ(x), ψγ(y)

]+= iδ(x0 − y0)(γ0)αβ(γ0)βγδ

3(x− y)

= iδ4(x− y)δαγ (7.26)

that is(i∂ −m)SF (x) = δ4(x) (7.27)

It is clear that the choice of sign is related to the way in which we perform thecanonical quantization. From the property of SF of being the Green function of theDirac operator, we can see that

SF (x) = −(i∂ +m)∆F (x) (7.28)

and

SF (x) =∫ d4k

(2π)4e−ikx k +m

k2 −m2 + iϵ=∫CF

d4k

(2π)4e−ikx k +m

k2 −m2(7.29)

Finally we consider the photon propagator. The only difference with the Klein-Gordon case is that the polarization vector give an extra factor −gµν , and therefore

⟨0|T (Aµ(x)Aν(y)|0⟩ = −igµν∫ d4k

(2π)4e−ikx

k2 + iϵ(7.30)

Defining

D(x) = −∫ d4k

(2π)4e−ikx

k2 + iϵ(7.31)

we get⟨0|T (Aµ(x)Aν(y)|0⟩ = +igµνD(x− y) (7.32)

7.2 A physical application of the propagators

The choice of the integration paths in eq. (7.15) allows us to define various types ofGreen’s functions according to the boundary conditions we require. Suppose thatwe want to solve the Klein-Gordon in a given external source

(∂2 +m2)ϕ(x) = j(x) (7.33)

The solution can be given in terms of the Green’s function defined by

(∂2 +m2)G(x) = δ4(x) (7.34)

131

.

ωIm( )

Re( )

ω

ω

kωk

-

C

C

ret

adv

Fig. 7.7 - The integration paths for Gret and Gadv.

In fact,

ϕ(x) = ϕ(0)(x) +∫

d4y G(x− y)j(y) (7.35)

where ϕ(0)(x) satisfy the homogeneous Klein-Gordon equation and it is chosen insuch a way that ϕ(x) satisfies the boundary conditions of the problem. For instance,if we give the the function ϕ(x) at t = −∞ we may require

limt→−∞

ϕ(x) = limt→−∞

ϕ(0)(x) (7.36)

Then, to satisfy the boundary conditions it is enough to choose for G(x) the retardedsolution defined by

Gret(x, x0 < 0) = 0 (7.37)

Such a solution can be found easily by applying the method we have illustratedin the previous Section. By choosing the integration path as in Fig. 7.7, that is,leaving both poles below the path we get

Gret(x) = −∫Cret

d4k

(2π)4e−ikx

k2 −m2(7.38)

Clearly Gret(x) vanishes for x0 < 0. In fact, in this case, we can close the pathon the half-plane Im ω > 0 without hitting any singularity, therefore the functionvanishes. In analogous way we can define a function Gadv(x) vanishing for x0 > 0, bychoosing a path below the poles (see Fig. 7.7). By integrating explicitly over ω onesees easily that the retarded solution propagates forward in time both the positiveand negative energy solutions, whereas the advanced one propagates both solutionsbackward in time. By using the expression (7.7) for the Feynman propagator we seethat it propagates the positive energy solutions forward in time and the negativeenergy ones backward in time. In fact, Feynman and Stueckelberg showed that thebackward propagation of the negative energy solutions is equivalent to the forwardpropagation of the anti-particles. The Feynman propagator acquires its full meaningonly in the quantum theory, where. as we shall prove, it represents the centralelement of the perturbation theory.

132

In order to stress the relevance of the Feynman propagator we will consider nowa simple application of what we have learned so far. Let us consider two staticpoint-like electric charges placed at x1 and x2. They can be described by the chargedensity

j0(x) =∑m=1,2

emδ3(x− xm) (7.39)

where em are the values of the electric charges. The current density vanishes becausewe have supposed the charges to be static. Therefore the interaction hamiltonian is

Hint = −Lint =∫

d3x jµ(x)Aµ(x) =

∑m=1,2

emA0(xm, 0) (7.40)

In this equation we have taken the electromagnetic field operator A0 at the timet = 0, because in this case it will be more convenient to use the Schrodinger rep-resentation. We have worked so far with the Heisenberg representation because itturns out to be more convenient from the point of view of the relativistic covarianceof the formalism, but the problem we will be interested here is the evaluation of theinteraction energy between the two charges. We recall that the Heisenberg and theSchrodinger representation, as far as the operators are concerned, are related by

AH(t) = eiHtASe−iHt (7.41)

where AH(t) is the operator in the Heisenberg representation and AS is the operatorin the Schrodinger one. Therefore the two operators are the same at t = 0. Toevaluate the interaction energy we can do a perturbative calculation by evaluatingthe energy shift induced by the interaction hamiltonian. Since the electric chargesare classical we are quantizing only the photon field A0(x, 0), and the state that weare perturbing is the vacuum state (the state without photons). Since

⟨0|Hint|0⟩ = 0 (7.42)

we must evaluate the energy shift at the second order in the perturbative theory.We will have

∆E =∑n

⟨0|Hint|n⟩⟨n|Hint|0⟩E0 − En

=∫ d3k

−ωk⟨0|Hint |k⟩⟨k|Hint|0⟩ (7.43)

The only state which contributes in the sum is the state with a single photon ofenergy ωk. By using the expression for Hint and the following representation for1/ωk

1

ωk= lim

ϵ→0+i∫ ∞

0dt e−i(ωk − iϵ)tdt (7.44)

we can write ∆E in the form (notice that we have suppressed the limit because theresulting expression is regular at for ϵ→ 0+).

∆E = −i∑m,n

emen

∫d3k

∫ ∞

0dt e−iωkt⟨0|A0(xn, 0)|k⟩⟨k|A0(xm, 0)|0⟩

133

.

Fig. 7.8 - The graphical description of eq. (7.46).

= −i∑m,n

emen

∫d3k

∫dt θ(t)⟨0|eiH0tA0(xn, 0)e

−iH0t|k⟩⟨k|A0(xm, 0)|0⟩

= −i∑m,n

∫dt θ(t)⟨0|A0(xn, t)A0(xm, 0)|0⟩ (7.45)

where H0 is the free hamiltonian for the electromagnetic field. To get this expressionwe have used the completeness (recalling again that A0 couples only states that differby a photon), and now the operators appearing in the last line can be interpreted asoperators in the Heisenberg representation. By writing explicitly the various termsin the sum we get

∆E = −i∫

dtθ(t)[e21⟨0|A0(x1, t)A0(x1, 0)|0⟩+ e1e2⟨0|A0(x1, t)A0(x2, 0)|0⟩

+ e1e2⟨0|A0(x2, t)A0(x1, 0)|0⟩+ e22⟨0|A0(x2, t)A0(x2, 0)|0⟩]

(7.46)

The intermediate states description of these four contributions is given in Fig. 7.8.Notice that the first and the fourth diagram describe a correction to the intrinsicproperties of the charges, and since we are interested in the evaluation of the inter-action energy we can omit them from our calculation. Then, the energy interaction∆E12 is given by

∆E12 = −ie1e2∫

dt θ(t)[⟨0|A0(x1, t)A0(x2, 0)|0⟩

+ ⟨0|A0(x2, t)A0(x1, 0)|0⟩]

(7.47)

Sending t→ −t in the second term, we get

∆E12 = −ie1e2∫

dt[θ(t)⟨0|A0(x1, t)A0(x2, 0)|0⟩

+ θ(−t)⟨0|A0(x2,−t)A0(x1, 0)|0⟩]

(7.48)

and using H0|0⟩ = 0,

⟨0|A0(x2,−t)A0(x1, 0)|0⟩ = ⟨0|e−iH0tA0(x2, 0)eiH0tA0(x1, 0)e

−iH0t|0⟩= ⟨0|A0(x2, 0)A0(x1, t)|0⟩ (7.49)

134

Therefore our final result is

∆E12 = −ie1e2∫

dt ⟨0|T (A0(x1, t)A0(x2, 0))|0⟩ (7.50)

We see that the energy interaction is expressed in terms of the Feynman propagator.Recalling the eqs.(7.31) and (7.32) we get

∆E12 = −e1e2∫

dt∫ d4k

(2π)4e−ik(x1 − x2)

k2 + iϵ

= e1e2

∫ d3k

(2π)3eik(x1 − x2)

k2=e1e24π

1

|x1 − x2|(7.51)

To evaluate the integration over

∇2∫ d3k

(2π)3eik · x

|k|2= −δ3(x) (7.52)

and

∇2 1

|x|= −4πδ3(x) (7.53)

from which ∫ d3k

(2π)3eik · x

|k|2=

1

4π

1

|x|(7.54)

135

Chapter 8

Perturbation theory

8.1 The electromagnetic interaction

As we already discussed the electromagnetic interaction can be introduced for anarbitrary charged particle via the minimal substitution or invoking the gauge prin-ciple

∂µ → ∂µ + ieAµ (8.1)

Fore instance, in the Klein-Gordon case one gets the lagrangian density for a chargedfield given by

Lfree = ∂µϕ†∂µϕ−m2ϕ†ϕ− 1

4FµνF

µν →

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

4FµνF

µν (8.2)

In this case the interacting part is given by

Lint. = −ie[ϕ†∂µϕ− (∂µϕ

†)ϕ]Aµ + e2A2ϕ†ϕ (8.3)

We see that the gauge field is coupled to the current

jµ = ie[ϕ†∂µϕ− (∂µϕ

†)ϕ]

(8.4)

but another interacting term appears. This term is a straight consequence of thegauge invariance. In fact the current jµ which was conserve in absence of the inter-action is now neither conserved, neither gauge invariant. Consider the infinitesimalgauge transformation

δϕ(x) = −ieΛ(x)ϕ(x), δAµ(x) = ∂µΛ(x) (8.5)

thenδLfree = ieΛ,µϕ

†∂µϕ− ie∂µϕ†Λ,µϕ = jµ∂µΛ (8.6)

136

and writing Lint in the form

Lint = −jµAµ + e2A2ϕ†ϕ (8.7)

we findδLint = −jµΛ,µ − (δjµ)A

µ + 2e2Λ,µAµϕ†ϕ (8.8)

The first term cancels with the variation of Lfree, whereas the other two terms cancelamong themselves

δjµ = ie[ϕ†(−ie)Λ,µϕ− ieΛ,µϕ

†ϕ]= 2e2Λ,µϕ

†ϕ (8.9)

This shows that the A2 term is necessary to compensate the fact that δjµ = 0 sincethe current is not gauge invariant. In fact the conserved and gauge invariant currentcomes by using the Noether’s theorem

Jµ = ie[ϕ†(∂µ + ieAµ)ϕ− (∂µ − ieAµ)ϕ

†)ϕ]= jµ − 2e2Aµϕ

†ϕ (8.10)

The situation is far more simple in the case of the Dirac equation where

Lfree = ψ(i∂ −m)ψ − 1

4FµνF

µν → ψ(i∂ − eA−m)ψ − 1

4FµνF

µν (8.11)

giving the interaction termLint = −eψγµψAµ (8.12)

Here the gauge field is coupled to a conserved and gauge invariant current. As aconsequence −jµAµ is the only interaction term. In fact,

δLfree = eψγµψΛ,µ = jµΛ

,µ (8.13)

andδLint = −jµδAµ = −jµΛ,µ (8.14)

The canonical quantization for an interacting system follows the same procedureas in the non interacting case. We require canonical commutation and/or anti-commutation relations at equal times for the various fields. For different fields werequire equal time vanishing commutation (anticommutation) relations for spin inte-ger (half-integer) fields, whereas we require zero commutation relations among fieldsof integer spin and fields of half-integer spin. Usually the canonical commutationrelations among the fields are not changed by the interactions with respect to thefree case. However this is not the case if the interaction term involves derivatives ofthe fields. This follows from the definition of the canonical momentum densities

Πi =∂L∂ϕi

=∂Lfree

∂ϕi+∂Lint

∂ϕi(8.15)

137

For instance, for the charged scalar field we get

Π ≡ Πϕ =∂L∂ϕ

= ϕ† − ieϕ†A0, Π† = Πϕ† = ϕ+ ieϕA0 (8.16)

Since the canonical momenta contain the time component of the gauge field, one canverify that the canonical commutators among the scalar fields and their derivativesare changed by the interaction. Also the propagators are modified. However wewill not insist on this point, because in practice it has no consequences on theperturbation theory (see later). When derivative interactions are not present thecanonical momentum densities coincide with the free ones, and we get

H = Πϕ− L = Πϕ− Lfree − Lint = Hfree − Lint (8.17)

and thereforeHint = −Lint (8.18)

This is what happens for the interaction between a Dirac and the electromagneticfields. The corresponding theory is called QED (Quantum Electro Dynamics). Werecall also that in general the hamiltonian and the electromagnetic current are nor-mal ordered in such a way that the vacuum is an eigenstate of these operators withvanishing eigenvalues. Therefore the interaction term is written as

Lint = −e : ψγµψ : Aµ (8.19)

We can verify that this is equivalent to write

Lint = −e2[ψ, γµψ]A

µ (8.20)

For instance, if we consider the electric charge, we get

Q =e

2

∫d3x(ψ†ψ − γ0ψψ

†γ0)

=e

2

∑±n

∫d3p

[b†(p, n)b(p, n) + d(p, n)d†(p, n)− b(p, n)b†(p, n)− d†(p, n)d(p, n)

]=e

2

∑±n

∫d3p

[2b†(p, n)b(p, n)− 2d†(p, n)d(p, n)

−[b†(p, n), b(p, n)

]++[d(p, n), d†(p, n)

]+

]=: Q : (8.21)

since the two anticommutators cancel out among themselves.

8.2 The scattering matrix

The scattering processes are a central element in the study of the elementary parti-cles, since they are the only experimental technique available. In the typical scat-tering process the incoming particles are prepared in a state of definite momentum,

138

after that the scattering process os some target has taken place, one looks at the finalstates. In ordinary quantum mechanics this situation is well described by using freewave functions for the initial and final states. This description is certainly correctif one has to do with short-range potentials. In field theory this representation isnot really correct, since also in absence of reciprocal interactions the particles haveself interactions as we have already noticed. For instance, a real electron can bethought of as if it would have a surrounding cloud of photons which can be emit-ted and absorbed also when very far from other electrons. A rigorous treatment ofthese problems is highly non trivial and it is outside of the scopes of this course.Therefore we will confine ourselves to a rather intuitive treatment of the problem.On the other side the imitations of the method will be rather obvious so it may wellconstitute the basis for a more refined approach. To simplify the matter we willmake use of the adiabatic hypothesis. This consists in looking at a scatteringprocess in the following way. At time t = −∞ we will suppose that our systemcan be described in terms of free particles, that is with the interaction turned off.Between t = −∞ and a time t = −T , much before the scattering process takesplace, we let the coupling describing the interaction grow from zero to its actualvalue. In the interval −T < t < +T , the coupling stays at this value, and then fromt = +T and t = +∞ the coupling goes again to zero (see Fig. 8.1)

.g(t)

T-T t

Fig. 8.1 - The adiabatic switching of the coupling constant.

In practice this can be realized by defining the interacting part of the hamiltonianas

Hint(t, ϵ) = e−ϵ|t|Hint (8.22)

performing all the calculations and taking the limit ϵ→ 0+ at the end. The consis-tency of this procedure has been shown by various authors and a detailed discussion

139

can be found, for instance, in the book by Jauch and Rohrlich, Theory of Photonsand Electrons.

By using the adiabatic hypothesis we can now discuss the perturbative calcula-tion of the scattering amplitudes. The perturbative expansion will be possible onlyif the interaction term is small. For in stance in QED one gets a series of powersin the fine structure constant e2/4π ≈ 1/137, therefore, if the coefficients of theexpansion do not grow too much, the expansion is justified. Let us start with theequation of motion for the states in the Schrodinger representation

i∂|ΦS(t)⟩

∂t= HS|ΦS(t)⟩ (8.23)

Suppose also that we have two interacting fields A and B. Then we can write

HS = H0S +HI

S (8.24)

withH0S = H0

S(A) +H0S(B) (8.25)

andHIS ≡ HI

S(A,B) (8.26)

where H0S(A) and H0

S(B) are the free hamiltonians for the fields A and B, andHIS is the interaction hamiltonian. It turns out convenient to introduce a new

representation for the vectors of state, the interaction representation. This is definedby the following unitary transformation upon the states and on the operators in theSchrodinger representation

|Φ(t)⟩ = eiH0St|ΦS(t)⟩, O(t) = eiH

0StOSe

−iH0St (8.27)

Of course the matrix elements of any operator in the interaction representation arethe same as in the Schrodinger representation

⟨Φ′(t)|O(t)|Φ(t)⟩ = ⟨Φ′S(t)|OS|ΦS(t)⟩ (8.28)

We have also H0S = H0, where H0 is the free hamiltonian in the interaction represen-

tation. Notice also that the interaction representation coincides with the Heisenbergrepresentation when we switch off the interaction. In the interaction representationthe time evolution of the states is dictated by the interaction hamiltonian

i∂|Φ(t)⟩∂t

= i∂

∂t

(eiH

0St|ΦS(t)⟩

)= −H0

SeiH0

St|ΦS(t)⟩+ eiH0St(H0

S +HIS)|ΦS(t)⟩

= eiH0StHI

Se−iH0

St|Φ(t)⟩ (8.29)

from which

i∂|Φ(t)⟩∂t

= HI |Φ(t)⟩ (8.30)

140

where HI is the interaction hamiltonian in the interaction representation. On theother side the operators evolve with the free hamiltonian. Therefore, in the in-teraction representation they coincide with the Heisenberg operators of the non-interacting case.

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

|Φ(−∞)⟩ ≡ |Φi⟩ (8.31)

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 (see theadiabatic hypothesis) specified by a vector state Φf . This amplitude is

Sfi = ⟨Φf |Φ(+∞)⟩ (8.32)

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

|Φ(+∞)⟩ = S|Φ(−∞)⟩ (8.33)

The amplitude Sfi is thenSfi = ⟨Φf |S|Φi⟩ (8.34)

ThereforeSfi is the S matrix element between free states. 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)⟩ (8.35)

One can verify that this indeed a solution, and furthermore it satisfies explicitly theboundary condition at t = −∞. The perturbative expansion consists in evaluating|Φ(t)⟩ by iterating this integral equation

|Φ(t)⟩ = |Φ(−∞)⟩− i∫ t

−∞dt1 H

I(t1)[|Φ(−∞)⟩ − i

∫ t1

−∞dt2 H

I(t2)|Φ(t2)⟩](8.36)

Continuing the iteration we get

|Φ(t)⟩ =[1− i

∫ t

−∞dt1 H

I(t1) + (−i)2∫ t

−∞dt1

∫ t1

−∞dt2 H

I(t1)HI(t2)

+ · · ·]|Φ(−∞)⟩ (8.37)

141

Of course this is meaningful only if the expansion is a convergent one. By takingthe limit for t = +∞ we get the perturbative expansion of the S matrix

S = 1 +∞∑n=1

(−i)n∫ +∞

−∞dt1

∫ t1

−∞dt2 · · ·

∫ tn−1

−∞dtn

[HI(t1)H

I(t2) · · ·HI(tn)]

(8.38)

We can rewrite this expression in terms of T -products

S = 1 +∞∑n=1

(−i)n

n!

∫ +∞

−∞dt1 · · ·

∫ +∞

−∞dtn T

(HI(t1) · · ·HI(tn)

)(8.39)

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

T(HI(t1) · · ·HI(tn)

)= HI(t1) · · ·HI(tn) (8.40)

The equality of the two expressions (8.38) and (8.39) holds term by term. As anexample, consider n = 2. The term in eq. (8.39) can be written as

A =∫ t2

t1

∫ t2

t1dt ds T

(HI(t)HI(s)

)=

∫ t2

t1dt HI(t)

(∫ t

t1ds HI(s)

)+∫ t2

t1dt(∫ t2

tds HI(s)

)HI(t) (8.41)

By looking at Figs. 8.2 and 8.3 one sees easily that exchanging the integrations ons and t one gets ∫ t2

t1dt∫ t2

tds =

∫ t2

t1ds∫ s

t1dt (8.42)

.

t1

t1

t2

t2 t

s

t

t2

Fig. 8.2 - The figure represents schematically the integral∫ t2t1

dt∫ t2t ds.

142

.

t1

t1

t2

t2 t

s

t1 s

Fig. 8.3 - The figure represents schematically the integral∫ t2t1

ds∫ st1dt.

Therefore

A =∫ t2

t1dt∫ t

t1ds HI(t)HI(s) +

∫ t2

t1ds∫ s

t1dt HI(s)HI(t) (8.43)

and exchanging s↔ t in the second integral

A = 2∫ t2

t1dt∫ t

t1ds HI(t)HI(s) (8.44)

The result for the nth term in the series can be obtained in a completely analogousway.

Since the S matrix connects the set of free states at t = −∞ with a set of freestates at t = +∞, it should represent simply a change of basis and as such it shouldbe unitary. From this point of view the unitarity property of the S matrix is avery fundamental one because it has to do with the very fundamental properties ofquantum mechanics. So it is important to check that at least formally (this refersto the fact that we don’t really know if the series which we have found for the Smatrix is a convergent one) the expression (8.39) represents a unitary operator. Inorder to do that we start rewriting S in the form

S = T(e−i∫ +∞

−∞dt HI(t))

(8.45)

This expression is a symbolic one and it is really defined by its series expansion

S =∞∑n=0

(−i)n

n!T(∫ +∞

−∞dt HI(t)

)n=

∞∑n=0

(−i)n

n!

∫ +∞

−∞dt1 · · · dtnT

(HI(t1) · · ·HI(tn)

)(8.46)

143

The motivation for introducing the T -ordered exponential is that it satisfies thefollowing factorization property

T(e

∫ t3

t1O(t)dt)

= T(e

∫ t3

t2O(t)dt)

T(e

∫ t2

t1O(t)dt)

(8.47)

To prove this relation we first consider the following expression (t1 ≤ t2 ≤ t3)

T( ∫ t3

t1O(t)dt

)n=∫ t3

t1· · ·

∫ t3

t1ds1 · · · dsn T (O(s1) · · ·O(sn))

=(∫ t3

t2+∫ t2

t1

)(∫ t3

t2+∫ t2

t1

)· · ·

(∫ t3

t2+∫ t2

t1

)ds1 · · · dsnT (O(s1) · · ·O(sn))

=n∑k=0

n!

(n− k)!k!

∫ t3

t2· · ·

∫ t3

t2ds1 · · · dsn−k

×∫ t2

t1· · ·

∫ t2

t1dz1 · · · dzk T (O(s1) · · ·O(sn−k)O(z1) · · ·O(zk))

=n∑k=0

n!

(n− k)!k!

∫ t3

t2· · ·

∫ t3

t2ds1 · · · dsn−kT (O(s1) · · ·O(sn−k))

×∫ t2

t1· · ·

∫ t2

t1dz1 · · · dzk T (O(z1) · · ·O(zk)) (8.48)

In the last term we have used the fact that all the times zi are smaller than thetimes si. What we have proved is the relation

T(∫ t3

t1dt O(t)

)n=

n∑k=0

n!

(n− k)!k!T(∫ t3

t2dt O(t)

)n−kT(∫ t2

t1dt O(t)

)k(8.49)

The factorization property (8.47) follows immediately if we remember that the anal-ogous property for the ordinary exponential

ea+ b = eaeb (8.50)

just follows from the binomial expansion

ea+ b =∞∑n=0

1

n!(a+ b)n =

∞∑n=0

1

n!

n∑k=0

n!

(n− k)!k!an−kbk (8.51)

using∞∑n=0

n∑k=0

=∞∑k=0

∞∑n=k

(8.52)

and putting h = n − k. Since eq. (8.49) generalizes the binomial formula to T -products of powers of time integrals of operators, by the same token we get theformula (8.47). With this property we can now prove the unitarity of any operatorof the form

U = T(e−i∫ tf

tidt O(t))

(8.53)

144

with O(t)a hermitian operator. To this end let us divide the time interval (ti, tf ) inN infinitesimal intervals ∆t with

ti ≡ t1 ≤ t2 ≤ · · · ≤ tN = tf (8.54)

then we can write

U = limN→∞

e−i∆tO(tN)e−i∆tO(tN−1) · · · e−i∆tO(t1) (8.55)

from whichU † = lim

N→∞e+i∆tO(t1)e+i∆tO(t2) · · · e+i∆tO(tN) (8.56)

and the unitarity follows immediately.Let us notice that if there are no derivative interactions we have

S = T(e−i∫ +∞

−∞dt HI(t))

= T(e+i∫

d4x Lint)(8.57)

It follows that if the theory is Lorentz invariant, also the S matrix enjoys the sameproperty. One could think that for theories with derivative interactions the Lorentzinvariance is lost. However it is possible to show that also in these theories theS matrix is given by the same equation. To see that the S matrix is Lorentzinvariant, notice that the statement would be trivial but for the presence of the T -product. However this is invariant under proper Lorentz transformations for time-like separations (remember that a proper Lorentz transformation cannot change thesign of the time component of a four-vector). In the case of space-like separationsconsider, for instance, the second order term in the series for S∫

d4x1 d4x2

[θ(x01 − x02)Lint(x1)Lint(x2) + θ(x02 − x01)Lint(x2)Lint(x1)

](8.58)

Since we integrate over x1 and x2, the separation between the two points can be ei-ther space-like either time-like. However, if the lagrangian density is a local functionof the fields, it follows that

[Lint(x1),Lint(x2)] = 0 per (x1 − x2)2 < 0 (8.59)

and therefore for (x1 − x2)2 < 0 we get

T (Lint(x1)Lint(x2)) = Lint(x1)Lint(x2) (8.60)

This shows that the T -product of local invariant Lorentz operators is Lorentz in-variant. We had an example of this property when we evaluated the propagator fora scalar field.

145

8.3 The Wick’s theorem

In the previous Section we have shown that the S matrix can be evaluated in termsof matrix elements of T -products. As we shall see in the applications, the matrixelements of the S matrix between free particle states can in turn 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. In order to prove the theorem we will use the technique ofgenerating functionals. That is we we will start by proving the following identity

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⟩

(8.61)

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 start by expanding the left handside of the formula, by using the factorization property in eq. (8.47). Let us alsodefine

O(t) =∫

d3x j(x)ϕ(x) (8.62)

and notice that for free fields [O(t), O(t′)] is just an ordinary number (called a c-number, to be contrasted with operators which are called q-numbers). Dividingagain the interval (ti, tf ) in N pieces of amplitude ∆t as in the previous Section, weget

T(e−i∫ tf

tidt O(t))

= limN→∞

e−i∆tO(tN)e−i∆tO(tN−1) · · · e−i∆tO(t1) (8.63)

and using

eAeB = eA+Be+1

2[A,B]

(8.64)

valid if [A,B] commutes con A e B, we get

T(e−i∫ tf

tidt O(t))

= limN→∞

N∏i=3

e−i∆tO(ti)e−i∆t(O(t2) +O(t1))e−1

2∆t2[O(t2), O(t1)]

146

= limN→∞

N∏i=4

e−i∆tO(ti)e−i∆t(O(t3) +O(t2) +O(t1))

×e−1

2∆t2[O(t3), O(t2) +O(t1)]−

1

2∆t2[O(t2), O(t1)]

= limN→∞

e−i∆t

N∑i=1

O(ti)e

−1

2∆t2

∑1≤i≤j≤N

[O(tj), O(ti)]

= e−i∫ tf

tidt O(t)

e−1

2

∫ tf

tidt1 dt2 θ(t1 − t2)[O(t1), O(t2)]

(8.65)

from which

T(e−i∫

d4x j(x)ϕ(x))

= e−i∫

d4x j(x)ϕ(x)e−1

2

∫d4x d4y j(x)j(y)θ(x0 − y0)[ϕ(x), ϕ(y)]

(8.66)

The next step is to expand the first exponential on the right hand side of thisequation in normal products. We have

e−i∫

d4x j(x)ϕ(x)= e

−i∫

d4x j(x)(ϕ(+)(x) + ϕ(−)(x))

= e−i∫

d4x j(x)ϕ(−)(x)e−i∫

d4x j(x)ϕ(+)(x)

× e+1

2

∫d4x d4y j(x)j(y)[ϕ(−)(x), ϕ(+)(y)]

(8.67)

where we have used again the equation (8.64). Therefore

e−i∫

d4x j(x)ϕ(x)= : e

−i∫

d4x j(x)ϕ(x):

× e+1

2

∫d4x d4y j(x)j(y)[(ϕ(x)(−), ϕ(+)(y)]

(8.68)

Substituting in eq. (8.66)

T(e−i∫

d4x j(x)ϕ(x))=: e

−i∫

d4x j(x)ϕ(x):

×e+1

2

∫d4x d4y j(x)j(y)([ϕ(−)(x), ϕ(+)(y)]− θ(x0 − y0)[ϕ(x), ϕ(y)])

(8.69)

The argument of the second exponential is a c-number because it proportional tocommutators of free fields, and therefore we can evaluate it just by taking its VEV

A ≡ [ϕ(−)(x), ϕ(+)(y)]− θ(x0 − y0)[ϕ(x), ϕ(y)]

= ⟨0|[ϕ(−)(x), ϕ(+)(y)]− θ(x0 − y0)[ϕ(x), ϕ(y)]|0⟩ (8.70)

147

Since ϕ(+)|0⟩ = 0, we can write

A = −⟨0| [ϕ(y)ϕ(x) + θ(x0 − y0)ϕ(x)ϕ(y)− θ(x0 − y0)ϕ(y)ϕ(x)] |0⟩= −⟨0| [θ(y0 − x0)ϕ(y)ϕ(x) + θ(x0 − y0)ϕ(x)ϕ(y)| 0⟩= −⟨0|T (ϕ(x)ϕ(y))|0⟩ (8.71)

This proves our identity (8.61). Let us now expand both sides of eq. (8.61) in a seriesof j(x) and compare term by term. We will use the simplified notation ϕi ≡ ϕ(xi).We get

T (ϕ) = : ϕ : (8.72)

T (ϕ1ϕ2) = : ϕ1ϕ2 : +⟨0|T (ϕ1ϕ2)|0⟩ (8.73)

T (ϕ1ϕ2ϕ3) = : ϕ1ϕ2ϕ3 : +3∑

i=j =k=1

: ϕi : ⟨0|T (ϕjϕk)|0⟩ (8.74)

T (ϕ1ϕ2ϕ3ϕ4) = : ϕ1ϕ2ϕ3ϕ4 : +4∑

i =j =k =l=1

[: ϕiϕj : ⟨0|T (ϕkϕl)|0⟩

+ ⟨0|T (ϕiϕj)|0⟩⟨0|T (ϕkϕl)|0⟩]

(8.75)

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⟩ (8.76)

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⟩ (8.77)

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⟩ (8.78)

148

8.4 Evaluation of the S matrix at second order in

QED

In the case of QED the S matrix is given by

S = 1 +∞∑n=1

(+i)n

n!

∫· · ·

∫d4x1 · · · d4xnT (LI(x1) · · · LI(xn)) (8.79)

with (see eq. (8.19)LI = −e : ψAψ : (8.80)

We have now to understand how to use the Wick’s theorem in the actual situation.Consider, for simplicity, two scalar fields. From eq. (8.73) we get (at equal timesthe T -product and the usual product coincides)

ϕa(x)ϕb(x) =: ϕa(x)ϕb(x) : +⟨0|T (ϕa(x)ϕb(x))|0⟩ (8.81)

from which: ϕa(x)ϕb(x) := ϕa(x)ϕb(x)− ⟨0|T (ϕa(x)ϕb(x))|0⟩ (8.82)

Therefore

T (: ϕa(x)ϕb(x) : ϕ1(x1) · · ·ϕn(xn)) = T (ϕa(x)ϕb(x)ϕ1(x1) · · ·ϕn(xn))−⟨0|T (ϕa(x)ϕb(x))|0⟩T (ϕ1(x1) · · ·ϕn(xn)) (8.83)

Since the second term subtracts the contraction between the two operators taken atthe same point, we can generalize the Wick expansion by saying that when normalproduct are contained inside a T -product the Wick’s expansion applies with thefurther rule that the contractions of operators at the same point, inside the normalproduct, are vanishing. With this convention we can write

T (: A(x1)B(x1) · · · : · · · : A(xn)B(xn) · · · :) = T (A(x1)B(x1) · · ·A(xn)B(xn) · · ·)(8.84)

In the case of QED one can get convinced more easily by recalling that

: ψγµψ :=1

2[ψ, γµψ] (8.85)

and noticing that inside a T -product the fields can be freely commuted except fortaking into account of their statistics. For instance

T (: ψγµψ :) =1

2T ([ψ, γµψ]) = T (ψγµψ) (8.86)

For the following analysis it is useful to remember how the various field operatorsact on the kets

ψ+ annihilates e− ψ− creates e+

ψ+ annihilates e+ ψ− creates e−

A+ annihilates γ A− creates γ (8.87)

149

Decomposing LI in positive and negative frequency components

LI = −e : (ψ+ + ψ−)(A+µ + A−

µ )γµ(ψ+ + ψ−) : (8.88)

we get 8 terms with non vanishing matrix elements. For instance

: ψ+α A

−αβψ

−β := −ψ−

β A−αβψ

+α (8.89)

has the following non vanishing matrix element

⟨e+γ|ψ−A−ψ+|e+⟩ (8.90)

This process corresponds to a positron emitting a photon. This and the other sevenprocesses described by the S matrix at the first order

S(1) = −ie∫

d4x : ψ(x)A(x)ψ(x) : (8.91)

are represented by the diagrams of Figs. 8.4 and 8.5.

γ

e+

e-e+

γ

e+

e+

e-

γ

e+

γ

e+

Fig. 8.4 -Diagrams for the processes described by the S matrix at the first order.

γ

e+

e-e-

γ

e-

γe+

e-e-

γ

e-

Fig. 8.5 -Diagrams for the processes described by the S matrix at the first order.

However none of these contributions corresponds to a physically possible processsince the four momentum is not conserved. We will show later that the conservationof the four momentum is a consequence of the theory. For the moment we willassume it and we will show that for real particles (that is for particles on the massshell p2 = m2) these processes cannot happen. For instance consider

e−(p) → e−(p′) + γ(k) (8.92)

if the four momentum is conserved

p = p′ + k (8.93)

150

from whichm2 = m2 − 2p · k (8.94)

where we have used k2 = 0 for the photon. In the rest frame of the electron we getmk0 = 0. Therefore the process is possible only for a photon with vanishing fourmomentum. Let us now consider the 20 order contribution

S(2) =(−ie)2

2!

∫d4x1 d

4x2 T(ψ(x1)A(x1)ψ(x1)ψ(x2)A(x2)ψ(x2)

)(8.95)

We can expand S(2) with the Wick’s theorem, and classify the various contributionsaccording to the number of contractions. If we associate to a contraction of twofields at the points x1 and x2 a line, we see that the terms originating from S(2) canbe obtained by connecting among them the diagrams depicted in Figs. 8.4 and 8.5in all the possible ways. In this case the only non vanishing contractions are theones between ψ and ψ, and between Aµ andAν . Recalling from Section 7.1

⟨0|T (ψα(x)ψβ(y))|0⟩ = iSF (x− y)αβ

⟨0|T (Aµ(x)Aν(y))|0⟩ = igµνD(x− y) (8.96)

we get

S(2) =6∑i=1

S(2)i (8.97)

where

S(2)1 =

(−ie)2

2!

∫d4x1 d

4x2 : ψ(x1)A(x1)ψ(x1)ψ(x2)A(x2)ψ(x2) : (8.98)

S(2)2 =

(−ie)2

2!

∫d4x1 d

4x2 : ψ(x1)A(x1)iSF (x1 − x2)A(x2)ψ(x2) :

+(−ie)2

2!

∫d4x1 d

4x2 : ψ(x2)A(x2)iSF (x2 − x1)A(x1)ψ(x1) :

= (−ie)2∫

d4x1 d4x2 : ψ(x1)A(x1)iSF (x1 − x2)A(x2)ψ(x2) : (8.99)

S(2)3 =

(−ie)2

2!

∫d4x1 d

4x2 : ψ(x1)γµψ(x1)igµνD(x1 − x2)ψ(x2)γνψ(x2) : (8.100)

S(2)4 =

(−ie)2

2!

∫d4x1 d

4x2 : ψ(x1)γµiSF (x1 − x2)igµνD(x1 − x2)γνψ(x2) :

+(−ie)2

2!

∫d4x1 d

4x2 : ψ(x2)γµiSF (x2 − x1)igµνD(x2 − x1)γνψ(x1) :

= (−ie)2∫

d4x1 d4x2 : ψ(x1)γµiSF (x1 − x2)ig

µνD(x1 − x2)γνψ(x2) : (8.101)

151

S(2)5 =

(−ie)2

2!

∫d4x1 d

4x2 (−1) : Tr[iSF (x1−x2)A(x2)iSF (x2−x1)A(x1)] : (8.102)

S(2)6 =

(−ie)2

2!

∫d4x1 d

4x2 (−1) : Tr[iSF (x1−x2)γµiSF (x2−x1)γνigµνD(x2−x1)] :

(8.103)

The term S(2)1 is nothing but the product of two processes of type S(1) and it does not

give rise to real processes. The term S(2)2 is obtained by contracting a two fermionic

fields, and this means to connect with a fermionic line two of the vertices of Figs.8.4 and 8.5. Th possible external particles are two γ, two e−, two e+, or a pair e+e−.Selecting the external states we can get different physical processes. One of theseprocesses is the Compton scattering γ+ e− → γ+ e−. In this case we must select inS(2)2 ψ+(x2) to destroy the initial electron and ψ− to create the final electron. As for

the photons are concerned, since Aµ is a real field, we can destroy the initial photonboth in x2 and x1 and create the final photon in the other point. Therefore we gettwo contributions

S(2)2 (γe− → γe−) = Sa + Sb (8.104)

with

Sa = (−ie)2∫

d4x1 d4x2 ψ

−(x1)γµiSF (x1 − x2)γ

νA−µ (x1)A

+ν (x2)ψ

+(x2) (8.105)

and

Sb = (−ie)2∫

d4x1 d4x2 ψ

−(x1)γµiSF (x1 − x2)γ

νA−ν (x2)A

+µ (x1)ψ

+(x2) (8.106)

The corresponding diagrams are given in Fig. 8.6

e-

γ

x2 x1

e-

γ

Sa

e-

γ

x2 x1

e-

γ

Sb

Fig. 8.6 -Diagrams for the Compton scattering.

The terms corresponding to the Compton scattering for a positron are obtainedfrom the previous ones by substituting ψ+ (annihilates an electron) with ψ− (createsa positron) and ψ− (creates an electron) with ψ+ (annihilates a positron). The other

two processes coming from S(2)2 are 2γ → e+e− (pair creation) and e+e− → 2γ (pair

annihilation). The S matrix element for the pair creation is given by

S(2)2 (2γ → e+e−) = (−ie)2

∫d4x1 d

4x2

ψ−(x1)γµiSF (x1 − x2)γ

νA+µ (x1)A

+ν (x2)ψ

−(x2) (8.107)

152

Notice that in evaluating

A+µ (x1)A

+ν (x2)|γ(k1)γ(k2)⟩ (8.108)

we get two contributions since one of the fields A+µ can annihilate any one of the two

external photons. The diagrams for these two contributions are given in Fig. 8.7.

γ e-

γ e+

γ

e+γ

e-

x1

x2

x1

x2

Fig. 8.7 -Diagrams for the pair creation.

We have analogous contributions and diagrams for the pair annihilation process.The next processes we consider are the ones generated by S

(2)3 in which we have

contacted the photon fields. They are: electron scattering e−e− → e−e−, positronscattering e+e+ and e+e− → e+e−. For the electron scattering we have

S(2)3 (2e− → 2e−) =

(−ie)2

2!

∫d4x1 d

4x2

: ψ−(x1)γµψ+(x1)ig

µνD(x1 − x2)ψ−(x2)γνψ

+(x2) : (8.109)

The termψ+(x1)ψ

+(x2)|e−(p1)e−(p2)⟩ (8.110)

gives rise to two contributions, and other two come from the final state. The corre-sponding diagrams are given in Fig. 8.8.

The terms a) and d) differ only for the exchange x1 ↔ x2 and therefore theyare equal after having exchanged the integration variables. The same is true for theterms b) and c). In this way we get a factor 2 which cancels the 2! in the denom-

inator. This the same phenomenon as for the term S(2)2 . That is there diagrams

giving rise to the same contribution to the scattering amplitude. At the order n onhas n! equivalent diagrams which cancel the factor n! coming from the expansion ofthe S matrix. This means that it is enough to draw all the inequivalent diagrams.For the electron scattering we have two such diagrams differing for a minus signdue to the exchange of the fermionic lines. This reflects the fact that field theorytakes automatically into account the statistics of the particles. In the present caseby giving rise to a properly antisymmetrized amplitude.

For the process e+e− → e+e− we get the diagrams the diagrams of Fig. 8.9.The inequivalent diagrams are those of Fig. 8.10 and correspond to the followingcontributions

S(2)3 (e+e− → e+e−) = Sa(e

+e− → e+e−) + Sb(e+e− → e+e−) (8.111)

153

e- e-

e- e-

a)

e-

e-e-

e-

b)

e-

e-e-

e-

c)

e- e-

e-e-

d)

Fig. 8.8 -Diagrams for electron scattering.

with

Sa(e+e− → e+e−) = (−ie)2

∫d4x1 d

4x2

: ψ−(x1)γµψ+(x1)ig

µνD(x1 − x2)ψ+(x2)γνψ

−(x2) : (8.112)

and

Sb(e+e− → e+e−) = (−ie)2

∫d4x1 d

4x2

: ψ−(x1)γµψ−(x1)ig

µνD(x1 − x2)ψ+(x2)γνψ

+(x2) : (8.113)

The term S(2)4 gives rise to the two possibilities e− → e− and e+ → e+. These

are not scattering processes, but the exchange of the photon change the intrinsicproperty of the electron, in particular the mass¿ They are called self-energy con-tributions. For the electron we have the diagram of Fig. 8.11 with a contributiongiven by

S(2)4 = (−ie)2

∫d4x1 d

4x2 ψ−(x1)γµiSF (x1 − x2)ig

µνD(x1 − x2)γνψ+(x2) (8.114)

In analogous way the term S(5)2 contributes to the self-energy of the photon. However,

as we shall show in the following, it cannot change the mass of the photon which

154

e- e-

e+ e+

a)

e+ e+

e-e-

b)

e+

e- e-

e+

c)

e+

e- e-

e+

d)

Fig. 8.9 -Diagrams for the process e+e− → e+e−.

is fixed to zero by the gauge invariance of the theory. We get the two equivalentdiagrams of Fig. 8.12, which contribute by

S(2)5 = (−ie)2

∫d4x1 d

4x2

(−1)Tr[iSF (x1 − x2)γµ(x2)iSF (x2 − x1)γ

ν(x1)]A−µ (x1)A

+ν (x2) (8.115)

Notice the minus sign which is due to the fact that in a fermionic loop we have toinvert two fermionic fields inside the T -product. The last term is the one whereall the fields are contracted. There are no external particles and the correspondingdiagram of Fig. 8.13 is called a vacuum diagram. These diagrams can be generallyignored because they contribute to a simple phase factor for the vacuum state.Thesediagrams can be generally ignored because they contribute to a phase factor for thevacuum state.

155

e- e-

e+ e+

Sa

x2

x1

e+

e- e-

e+

Sb

x2 x1

Fig. 8.10 -Inequivalent diagrams for the process e+e− → e+e−.

e-

γ

e-

x2 x1

Fig. 8.11 -Electron self-energy.

156

γ γ

x2 x1

γγ

x2 x1

Fig. 8.12 -Photon self-energy.

x2 x1

Fig. 8.13 -Vacuum diagram.

8.5 Feynman diagrams in momentum space

As we noticed already, in a typical experiment in particle physics we prepare beamsof particles with definite momentum, polarization, etc. At the same time we measuremomenta and polarizations of the final states. For this reason it is convenient towork in a momentum representation. Let us recall the expressions for the fermionand the photon propagators (see eqs. (7.29) and (7.31)

SF (x) =1

(2π)4

∫d4p e−ipxSF (p) (8.116)

and

D(x) =1

(2π)4

∫d4p e−ipxD(p) (8.117)

where

SF (p) =1

p−m+ iϵ(8.118)

and

D(p) = − 1

p2 + iϵ(8.119)

From the fields expansion in terms of creation and annihilation operators, one canevaluate the action of the positive frequency part of ψ, ψ e Aµ on the one particlestates. We get

ψ+(x)|e−(p, r)⟩ =∑±s

∫d3k

√m

(2π)3Eke−ikxu(k, s)b(k, s)b†(p, r)|0⟩ (8.120)

157

and using [b(k, s), b†(p, r)

]+= δrsδ

3(p− k) (8.121)

it follows

ψ+(x)|e−(p, r)⟩ =√

m

(2π)3Eku(p, r)e−ipx|0⟩ (8.122)

For later applications it is more convenient to use the normalization in a box thanthe continuous one. This amounts to the substitution

1√(2π)3

→ 1√V

(8.123)

and to use discrete momenta p = (2π/L)3n. We get

ψ+(x)|e−(p, r)⟩ =√

m

V Epu(p, r)e−ipx|0⟩ (8.124)

ψ+(x)|e+(p, r)⟩ =√

m

V Epv(p, r)e−ipx|0⟩ (8.125)

and for the photon

A+µ (x)|γ(k, λ)⟩ =

√1

2V Ekϵ(λ)µ (k)e−ikx|0⟩ (8.126)

By conjugating these expressions we obtain the action of the negative frequencyoperators on the bras.

As an example let us consider a process associated to S(1)

|i⟩ = |e−(p)⟩ → |f⟩ = |e−(p′), γ(k)⟩ (8.127)

From (8.91) we get

⟨f |S(1)|i⟩ = −ie∫

d4x ⟨e−(p′), γ(k)|ψ−(x)A−µ (x)γ

µψ+(x)|e−(p)⟩

= −ie∫

d4x

(√m

V Ep′u(p′)eip

′x)

×(√

1

2V Ekϵ(k)eikx

)(√m

V Epu(p)e−ipx

)

= −ie m√V 3EpEp′2Ek

(2π)4δ4(p′ + k − p)u(p′)ϵ(k)u(p) (8.128)

This expression can be written as

⟨f |S(1)|i⟩ = (2π)4δ4(p′ + k − p)

√m

V Ep

√m

V Ep′

√1

2V EkM (8.129)

158

whereM = −ieu(p′)ϵ(k)u(p) (8.130)

is called the Feynman amplitude for the process. Notice that M is a Lorentz in-variant quantity. The term (2π)4δ4(p′ + k − p) gives the conservation of the fourmomentum in the process, whereas the other factors are associated to the variousexternal particles (incoming and outgoing). As we said previously this process isnot physically possible since it does not respect the four momentum conservation.

This structure is quite general and for any process we will have the delta-function

expressing the four momentum conservation, and factors as√m/V Ep for each ex-

ternal fermion and√1/2V Ek for each boson. Now we have to investigate the rules

for evaluating M. To this end let us consider the Compton scattering

|i⟩ = |e−(p), γ(k)⟩ → |f⟩ = |e−(p′), γ(k′)⟩ (8.131)

The S matrix element for the Compton scattering is given in eqs. (8.104)-(8.105)

S(2)2 (γe− → γe−) = Sa + Sb (8.132)

and

⟨f |Sa|i⟩ = (−ie)2∫

d4x1 d4x2

√m

V Ep

√m

V Ep′

√1

2V Ek

√1

2V Ek′

× u(p′)eip′x1 ϵ(k′)eik

′x1 i

(2π)4

∫d4qe−iq(x1 − x2)SF (q)

ϵ(k)e−ikx2u(p)e−ipx2

= (−ie)2∫

d4x1 d4x2

∫ d4q

(2π)4

√m

V Ep

√m

V Ep′

√1

2V Ek

√1

2V Ek′

× ei(p′ + k′ − q)x1e−i(p+ k − q)x2ϵµ(k

′)ϵν(k)u(p′)γµ

i

q −m+ iϵγνu(p)

= (2π)4δ4(p′ + k′ − p− k)

√m

V Ep

√m

V Ep′

√1

2V Ek

√1

2V Ek′

× ϵµ(k′)ϵν(k)u(p

′)(−ieγµ) i

p+ k −m+ iϵ(−ieγν)u(p) (8.133)

Also this expression can be written in the form

⟨f |Sa|i⟩ == (2π)4δ4(p′ + k′ − p− k)

√m

V Ep

√m

V Ep′

√1

2V Ek

√1

2V Ek′Ma (8.134)

with

Ma = ϵµ(k′)ϵν(k)u(p

′)(−ieγµ) i

q −m+ iϵ(−ieγν)u(p), q = p+ k (8.135)

159

.

p

k

q=p+k

k’

p’

Fig. 8.13 - Contribution to the Compton scattering.

We will associate to this expression the diagram in Fig 8.13. The four momentum q isdetermined by the conservation of the four momentum at the vertices: q = p+ k =p′ + k′. However notice that in general q2 = m2, that is the exchanged particle(described by the propagator) is not a real particle but a virtual one. Lookingat the previous expression one understands immediately how the various pieces areconnected to the graphical elements in the diagram. In fact we have the followingrules for any given diagram (Feynman diagram)

• for each vertex there is a factor −ieγµ

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

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

• for each ingoing and/or outgoing photon line there is a factor ϵλµ (or its complexconjugate for outgoing photons if we consider complex polarizations as thecircular one).

Notice also that the spinorial factor start from the external states and end upwith the ingoing ones. The further contribution to the Compton scattering (givenby Sb) corresponds to the diagram in Fig. 8.14, and using the previous rules we get

Mb = u(p′)(−ieγµ)i

q −m+ iϵ(−ieγν)u(p)ϵµ(k)ϵν(k′), q = p− k′ (8.136)

If we write down the Compton amplitude for the positrons we see that we mustassociate v(p), v(p) to the initial and final states respectively. This is seen from eq.(8.125) which shows that the annihilation operator for the positrons is associatedto v(p). For this reason, when drawing the diagrams in momentum space, is oftenconvenient to invert the direction of the positron lines, in such a way that the barredspinors are always written to the left and the unbarred to the right (see Fig. 8.15).

160

.

p

k’

q=p−k’

k

p’

Fig. 8.14 - The crossed contribution to the Compton scattering.

In this case one has to be careful with the direction of the momenta which flowin direction opposite to the arrow in the case of the antiparticles. As for the internallines, there is no distinction between particles and antiparticles, therefore, in generalone draws the arrows in a way consistent with the flow of the momenta. Considernow e−e− scattering. The diagrams for this process are in Fig. 8.16, and the matrixelement is given by

⟨f |S(2)3 |i⟩ = (2π)4δ4(p′1 + p′2 − p1 − p2)

4∏i=1

√m

V EiM (8.137)

where M = Ma +Mb, with

Ma = u(p′1)(−ieγµ)u(p1)igµνD(p2 − p′2)u(p′2)(−ieγν)u(p2) (8.138)

andMb = −u(p′2)(−ieγµ)u(p1)igµνD(p2 − p′1)u(p

′1)(−ieγν)u(p2) (8.139)

The relative minus sign comes fro the exchange of the two electrons in the initialstate, that is from the Fermi statistics.

From this example we get the further rule for the Feynman diagrams

• - for each internal photon line there is a factor igµνD(p) (propagator).

As a last example let us consider the electron self-energy

|i⟩ = |e−(p)⟩ → |f⟩ = |e−(p′)⟩ (8.140)

the S matrix element is

⟨f |S(2)4 |i⟩ = −e2

∫d4x1 d

4x2

√m

V Ep′

√m

V Epu(p′)eip

′x1

161

.

e−

u(p)

e+

v−(q’)

e+

v(q)

e−

u−(p’)

initial state final state

Fig. 8.15 - The spinor conventions for antifermions.

.

p2 p2’

p2−p2’

p1 p1’

a)

p2

p1’

p2−p1’

p1

p2’

b)

Fig. 8.16 - The Feynman diagrams for the scattering e−e− → e−e−.

× γµ

∫ d4q1(2π)4

e−iq1(x1 − x2)iSF (q1)

× γν

∫ d4q2(2π)4

e−iq2(x1 − x2)iD(q2)u(p)e−ipx2

=

√m

V Ep′

√m

V Ep

∫ d4q1(2π)4

d4q2(2π)4

(2π)4δ4(p′ − q1 − q2)(2π)4δ4(p− q1 − q2)

× u(p′)(−ieγµ)iSF (q1)(−ieγν)u(p)igµνD(q2)

= (2π)4δ4(p′ − p)

√m

V Ep′

√m

V EpM (8.141)

162

where in the last term we have integrated over q1. M is given by

M =∫ d4q2

(2π)4u(p′)(−ieγµ)iSF (p− q2)(−ieγν)u(p)igµνD(q2) (8.142)

which correspond to the diagram in Fig. 8.17.

.

p

q2

q1=p-q2

p’=p

Fig. 8.16 - The Feynman diagram for the electron self-energy.

We see that the rule of conservation of the four momentum is always valid andalso, that we have to integrate with measure d4q/(2π)4 over all the momenta whichare not determined by this conservation law. For the general case it is more conve-nient to formulate this rule in the following way:

• for each vertex there is an explicit factor (2π)4δ4(∑i pi), where pi are the

momenta entering the vertex

• integrate all the internal momenta with a measure d4pi/(2π)4

In this way the factor (2π)4δ4(∑pext) is automatically produced. We can verify

the previous rule in the self-energy case

∫ d4q1(2π)4

∫ d4q2(2π)4

(2π)4δ4(q1 + q2 − p)(2π)4δ4(p′ − q1 − q2)

= (2π)4δ4(p′ − p)∫ d4q2

(2π)4(8.143)

As a last rule we recall

• for each fermionic loop there is a factor (-1).

163

Chapter 9

Applications

9.1 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 separate in the S matrix the

term 1 corresponding at no scattering events we can write

Sfi = δfi + (2π)4δ4

∑i

pi −∑f

pf

∏fermioni

(m

V E

)1/2 ∏bosoni

(1

2V E

)1/2

M (9.1)

where M is the Feynman amplitude which can be evaluated by drawing the corre-sponding Feynman diagrams and using the rules of the previous Chapter.

Let us consider the typical case of a two particle collision giving rise to an Nparticles final state. Since we are interested to a situation with the final statedifferent from the initial one, the probability for the transition will be the modulussquare of the second term in eq. (9.1). In doing this operation we encounter thesquare of the Dirac delta which is not a definite quantity. However we should recallthat we are quantizing the theory in a box, and really considering the system in afinite, although large, time interval that we parameterize as (−T/2, T/2). Thereforewe have not really to do with the delta function but rather with (Pi,f =

∑i,f pi,f )

(2π)4δ4 (Pf − Pi) →∫Vd3x

∫ T/2

−T/2dtei(Pf − Pi)x (9.2)

Consider one of the factors appearing in this equation, for instance the time integral.By performing the integration we get (∆E = Ef − Ei))∫ T/2

−T/2dtei∆Et =

2 sin(T∆E/2)

∆E(9.3)

164

And evaluating the modulus square

|(2π)δ(Ef − Ei)|2 →4 sin2(T∆E/2)

(∆E)2(9.4)

On the right hand side we have a function of ∆E, whose integral holds 2πT , andhas a peak at ∆E = 0. Therefore in the T → ∞ limit we have a delta-convergentsequence

limT→∞

4 sin2(T∆E/2)

(∆E)2= 2πTδ(Ef − Ei) (9.5)

By doing the same operations also for the space integrals we get∣∣∣(2π)4δ4 (Pf − Pi)∣∣∣2 → (2π)4L3Tδ4(Pf − Pi) (9.6)

where L is the side of the volume V = L3. Therefore, the probability per unit timeof the transition is

w = V (2π)4δ4(Pf − Pi)∏i

1

2V Ei

∏f

1

2V Ef

∏fermioni

(2m)|M|2 (9.7)

where, for reasons of convenience, we have written

m

V E= (2m)

1

2V E(9.8)

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 (9.9)

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−1ℓ2t] = [ℓ2] (9.10)

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. Forthe bosons this follows from the normalization condition (3.41). For the fermionsrecall that in the box normalization the wave function is√

m

EVu(p)e−ipx (9.11)

from which ∫Vd3xρ(x) =

∫Vd3x

m

V Eu†(p)u(p) = 1 (9.12)

165

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

(9.13)

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 (9.14)

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

vrel = v1 − v2 =p1E1

− p2E2

(9.15)

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 (9.16)

We see that also this factor is Lorentz invariant.

9.2 The scattering e+e− → µ+µ−

In order to exemplify the previous techniques we will study the process e+e− →µ+µ−. The interaction lagrangian density is

LI = −e[ψeγ

λψe + ψµγλψµ

]Aλ (9.17)

Fig. 9.1 describes the Feynman diagram for this process at the second order (in thisdiagram the arrows are oriented according to the direction of the momenta). Noticethat in contrast to the process e+e− → e+e− the diagram in the crossed channel ofFig. 9.2 is now missing.The Feynman amplitude is

M = u(p4, r4)(−ieγµ)v(p3, r3)−igµν

(p1 + p2)2v(p1, r1)(−ieγν)u(p2, r2)

= ie2u(p4, r4)γµv(p3, r3)1

(p1 + p2)2v(p1, r1)γ

µu(p2, r2) (9.18)

166

.

e+

p1

e-

p2

µ-

p4

µ+

p3

Fig. 9.1 - The Feynman diagram for the scattering e+e− → µ+µ−..

e+ e+

e− e−

Fig. 9.2 - The crossed diagram for the scattering e+e− → e+e−.

where we have introduced the polarization of the fermions ri (the direction of thespin in the rest frame). Often one is interested in unpolarized cross-sections. In thatcase one has to sum the cross-section over the final polarizations and to average theinitial ones. That is we need the following quantity

X =1

4

∑ri

|M|2 (9.19)

where

M⋆ = −ie2v(p3, r3)γµu(p4, r4)1

(p1 + p2)2u(p2, r2)γ

µv(p1, r1) (9.20)

This expression can be written in the following form

M =ie2

(p1 + p2)2Amuonsµ (r3, r4)A

µelectrons(r1, r2) (9.21)

with

Amuonsµ = u(p4, r4)γµv(p3, r3)

167

Aelectronsµ = v(p1, r1)γµu(p2, r2) (9.22)

We get

X =1

4

e4

(p1 + p2)4∑r1,r2

(Aelectronsµ A⋆ electrons

ν

) ∑r3,r4

(AµmuonsA⋆ νmuons) (9.23)

Defining the quantity

Aelectronsµν =

∑r1,r2

(Aelectronsµ A⋆ electrons

ν

)=∑r1,r2

v(p1, r1)γµu(p2, r2)u(p2, r2)γνv(p1, r1)

(9.24)and using the eqs. (4.100) and (4.101) for the positive and negative energy projec-tors, we obtain

Aelectronsµν = Tr

[p1 −me

2me

γµp2 +me

2me

γν

](9.25)

and the analogous quantity for the muons

Amuonsµν = Tr

[p4 +mµ

2mµ

γµp3 −mµ

2mµ

γν

](9.26)

To evaluate the trace of Dirac matrices we may use several theorems. Let usstart showing that the trace of an odd number of gamma matrices is zero. In factfor odd n

Tr [a1 · · · an] = Tr [a1 · · · anγ5γ5] = Tr [γ5a1 · · · anγ5]= (−1)nTr [a1 · · · an] (9.27)

where we have used the cyclic property of the trace and the anticommutativity ofγ5 and γµ. Obviously we have

Tr[1] = 4 (9.28)

Furthermore

Tr[ab] =1

2Tr[ab+ ba] =

1

2aµbνTr([γ

µ, γν ]+) = 4aµbνgµν = 4a · b (9.29)

Then, using ab = −ba+ 2a · b we can evaluate

Tr[a1a2a3a4] = Tr[(−a2a1 + 2a1 · a2)a3a4]= −Tr[a2(−a3a1 + 2a1 · a3)a4] + 8(a1 · a2)(a3 · a4)= Tr[a2a3(−a4a1 + 2a1 · a4)]− 8(a1 · a3)(a2 · a4) + 8(a1 · a2)(a3 · a4)= −Tr[a2a3a4a1] + 8(a1 · a4)(a2 · a3)−8(a1 · a3)(a2 · a4) + 8(a1 · a2)(a3 · a4) (9.30)

that is

Tr[a1a2a3a4] = 4[(a1 · a2)(a3 · a4)− (a1 · a3)(a2 · a4) + (a1 · a4)(a2 · a3)] (9.31)

168

This relation can be easily extended by induction. Other useful properties are

γµγµ = 4 (9.32)

γµaγµ = (−aγµ + 2aµ)γ

µ = −2a (9.33)

andγµabγ

µ = 4a · b (9.34)

Let us go back to our process. Evaluating the trace we get

Aµνelectrons =1

4m2e

Tr[(p1γµp2γ

ν)−m2eγ

µγν ]

=1

m2e

[pµ1pν2 − gµν(p1 · p2) + pν1p

µ2 −m2

egµν ]

=1

m2e

[pµ1pν2 + pν1p

µ2 − gµν(p1 · p2 +m2

e)] (9.35)

Analogously we get

Aµνmuons =1

m2µ

[pµ3pν4 + pν3p

µ4 − gµν(p3 · p4 +m2

µ)] (9.36)

Substituting into X we get

X =1

4

e4

(p1 + p2)41

m2em

2µ

[2(p1 · p3)(p2 · p4) + 2(p1 · p4)(p2 · p3)− 2(p1 · p2)(p3 · p4 +m2µ)

− 2(p3 · p4)(p1 · p2 +m2e) + 4(p1 · p2 +m2

e)(p3 · p4 +m2µ]

=e4

2m2em

2µ(p1 + p2)4

[(p1 · p3)(p2 · p4) + (p1 · p4)(p2 · p3)

+ m2µ(p1 · p2) +m2

e(p3 · p4) + 2m2em

2µ] (9.37)

This process is studied in the circular colliders where two beams, one of electronsand the other of positron with same energy are made to collide, and looking for afinal pair µ+ − µ−. Therefore it is convenient to use the frame of the center of massfor the pair e+ − e−. We will choose the momentum variables as in Fig. 9.3 with

p1 = (E, p), p2 = (E,−p), p3 = (E, p ′), p4 = (E,−p ′) (9.38)

In this frame the various scalar products are

p1 · p3 = p2 · p4 = E2 − pp′ cos θ, p1 · p4 = p2 · p3 = E2 + pp′ cos θ (9.39)

p1 · p2 = E2 + p2, p3 · p4 = E2 + p′2, (p1 + p2)

2 = 4E2 (9.40)

169

.

p1 p2

p3

p4

Fig. 9.3 - The Feynman diagram for the scattering e+e− → µ+µ−.

In order the process is kinematically possible we must have E > mµ ≈ 200 me. Wecan then neglect me with respect to E and mµ, obtaining

X =e4

2m2em

2µ

1

16E4

[(E2 − pp′ cos θ)2 + (E2 + pp′ cos θ)2 +m2

µ(E2 + p2)

]=

e4

32m2em

2µE

4[2E4 + 2E2p′

2cos2 θ + 2E2m2

µ]

=e4

16m2em

2µ

1

E2[E2 + p′

2cos2 θ +m2

µ] (9.41)

from which

dσ =e4

16m2em

2µ

1

E2[E2 + p′

2cos2 θ +m2

µ](2π)4δ4(p1 + p2 − p3 − p4)

1

4(E2 + p2)

× (2me)2(2mµ)

2 d3p3d

3p4(2π)64E2

= δ4(p1 + p2 − p3 − p4)e4

128π2

1

E6[E2 + p′

2cos2 θ +m2

µ]d3p3d

3p4 (9.42)

where we have used E2 = p2, since we are neglecting the electron mass. We can inte-grate this expression over four variables using the conservation of the four momentumgiven by the delta function. We will integrate over p4 e |p3|. Using d3p3 = p′2dp′dΩwe get

dσ

dΩ=

∫p′

2dp′δ(E1 + E2 − E3 − E4)

e4

128π2

1

E6(E2 + p′

2cos2 θ +m2

µ)

170

=

[∂(E3 + E4)

∂p′

]−1e4

128π2

p′2

E6(E2 + p′

2cos2 θ +m2

µ) (9.43)

The derivative can be evaluated by noticing that E23 = E2

4 = m2µ + p′2

∂(E3 + E4)

∂p′= 2

p′

E(9.44)

Using e2 = 4πα we get the differential cross-section

dσ

dΩ=α2

8

p′2

E6

E

2p′(E2 + p′

2cos2 θ +m2

µ) =α2

16E4

p′

E(E2 + p′

2cos2 θ +m2

µ) (9.45)

In the extreme relativistic limit E >> mµ (p′ ≈ E) we get the expression

dσ

dΩ=

α2

16E2(1 + cos2 θ) (9.46)

from which we obtain the total cross-section

σ =α2

16E2

∫dΩ(1 + cos2 θ) =

α2

16E22π∫ 1

−1dw(1 + w2) =

α2π

3E2(9.47)

In the general case we get

σ =α2

16E4

p′

E

∫dΩ(E2 + p′

2cos2 θ +m2

µ) =α2

16E4

p′

E2π(2E2 +

2

3p′

2+ 2m2

µ

)=

πα2

4E4

p′

E

(E2 +

1

3p′

2+m2

µ

)(9.48)

In the high energy limit we can easily estimate the total cross-section. Recallingthat

1 GeV −2 = 0.389 mbarn (9.49)

we get

σ(mbarn) ≈ 5.6 · 10−5

(E(GeV ))2· 0.389 ≈ 2.17 · 10−5

(E(GeV ))2(9.50)

or in terms of nanobarns, 1 nbarn = 10−6 mbarn,

σ ≈ 20 (nbarn)

(E(GeV ))2(9.51)

9.3 Coulomb scattering

Sometimes one can think to the electromagnetic field as an assigned quantity, inthat case it will be described by a classical function rather than by an operator.This is the case for the scattering of electrons and positrons from an external field

171

as the Coulomb field of a heavy nucleus. The full electromagnetic will be of course,the sum of the classical part and of the quantized part. The expansion of the Smatrix is still given by

S = 1 +∞∑n=1

(i)n

n!

∫· · ·

∫d4x1 · · · d4xnT (LI(x1) · · · LI(xn)) (9.52)

withLI(x) = −e : ψ(x)γµψ(x)[Aµ(x) + Aext

µ (x)] : (9.53)

. .

Ze

p’−p

p’p

Fig. 9.4 - The Feynman diagram for the Coulomb scattering of an electron.

We will consider here the scattering of an electron from nucleus, that will bethought as infinitely heavy. Therefore it will give rise to a static Coulomb potential.Let us introduce the Fourier transform of this field

Aextµ (x) =

∫ d3q

(2π)3eiq · xAext

µ (q) (9.54)

At the first order in the external field we have

S(1) = −ie∫

d4x : ψ(x)γµψ(x)Aextµ (x) : (9.55)

and the transition we consider is

|i⟩ = |e−(p, r)⟩ → |f⟩ = |e−(p′, s)⟩ (9.56)

This is described by the diagram of Fig. 9.4 with contribution given by

⟨f |S(1)|i⟩ = −ie∫

d4x ⟨e−(p′, s)|ψ−(x)γµψ+(x)Aextµ (x)|e−(p, r)⟩

172

= −ie(

m

EpV

)1/2 (m

Ep′V

)1/2 ∫d4x eip

′xu(p′, s)

×∫ d3q

(2π)3eiq · xAext

µ (q)γµu(p, s)e−ipx

= −ie(

m

EpV

)1/2 (m

Ep′V

)1/2

(2π)δ(E ′ − E)

×∫ d3q

(2π)3(2π)3δ3(p+ q − p ′)u(p′, s)γµAext

µ (q)u(p, s)

= (2π)δ(E ′ − E)(m

EV

)M (9.57)

withM = u(p′, s)(−ieγµ)Aext

µ (p ′ − p)u(p, s) (9.58)

Notice that in this case we have only the conservation of the energy, whereas thespatial momentum is not conserve. In fact the external field violates the translationalinvariance of the theory, and, as a consequence, the nucleus absorbs the momentump′ − p from the electron. From the [previous expression we see also that when thereare external fields the Feynman rules are modified, and we have to substitute thewave function of a photon (

1

2EqV

)1/2

ϵ(λ)µ (q) (9.59)

with the Fourier transform of the external field

Aextµ (q) (9.60)

The probability per unit time is given by

w =1

T

∣∣∣⟨f |S(1)|i⟩∣∣∣2 = 2πδ(E ′ − E)

(m

EV

)2

|M|2 (9.61)

The expression for the density of final states is still (p′ = |p ′|)

dNf =V d3p′

(2π)3=V p′2dp′dΩ

(2π)3(9.62)

Since E ′2 = E2 = p′2 + m2 = p2 + m2. we have p = |p| = p′, and E ′dE ′ = p′dp′.Therefore

dNf = Vp′E ′dE ′

(2π)3dΩ (9.63)

The incoming flux is v/V = p/V E, and we get

dσ =V wdNf

v=(m

2π

)2

dE′δ(E ′ − E)|M|2dΩ (9.64)

173

The differential cross-section is obtained by integrating over the final energy of theelectron

dσ

dΩ=(m

2π

)2

|M|2 =(me

2π

)2 ∣∣∣u(p′, s)γµAextµ (q)u(p, r)

∣∣∣2 (9.65)

where q = p ′ − p. Averaging over the initial polarizations and summing over thefinal ones we obtain

dσ

dΩ=(me

2π

)2 1

2Aextµ (q)Aext

ν (q)Tr

[p′ +m

2mγµp+m

2mγν]

(9.66)

From the evaluation of the trace we get

Tr[...] =1

m2

[p′µpν − gµν(p

′ · p) + p′νpµ + gµνm2]

(9.67)

Assuming now that the external field is of Coulomb type, we have

Aextµ (x) =

(− Ze

4π|x|, 0

)(9.68)

and

Aext0 (q) = − Ze

|q|2(9.69)

We see that we need only the terms with µ = ν = 0 from the trace

Tr[...] =1

m2[E2 +m2 + p · p ′] (9.70)

and since p · p ′ = p2 cos θ

Tr[...] =1

m2[E2 +m2 + p2 cos θ] (9.71)

we getdσ

dΩ= 2

(Zα)2

|q|4(E2 +m2 + p2 cos θ) (9.72)

Using

|q|2 = |p ′ − p|2 = 4p2 sin2 θ

2(9.73)

m2 = E2 − |p|2, and v = p/E, we finally obtain

dσ

dΩ= 2

(Zα)2

(4p2 sin2(θ/2)2(E2 +m2 + p2 cos θ) =

(Zα)2

4E2v4 sin4(θ/2)[1− v2 sin2(θ/2)]

(9.74)In the non relativistic limit v << 1, E ≈ m

dσ

dΩ=

(Zα)2

4m2 sin4(θ/2)

1

v4=

(Zα)2

16T 2 sin4(θ/2)(9.75)

which is the classical Rutherford formula for the Coulomb scattering with T =mv2/2.

174

Chapter 10

One-loop renormalization

10.1 Divergences of the Feynman integrals

Let us consider again the Coulomb scattering. If we expand the S matrix up tothe third 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. 10.1

. .

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

175

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 + · · ·) (10.1)

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. 10.1a). We have

Ma = u(p′)(−ieγµ)Aextµ (p ′ − p)

i

p−m+ iϵ

[ie2Σ(p)

]u(p) (10.2)

where

ie2Σ(p) =∫ d4k

(2π)4(−ieγµ)

−igµν

k2 + iϵ

i

p− k −m+ iϵ(−ieγν)

= −e2∫ d4k

(2π)4γµ

1

k2 + iϵ

1

p− k −m+ iϵγµ (10.3)

or

Σ(p) = i∫

d4k

(2π)4γµ

p− k +m

(p− k)2 −m2 + iϵγµ

1

k2 + iϵ(10.4)

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) (10.5)

Mc = u(p′)(−ieγµ) −igµνq2 + iϵ

ie2Πνρ(q)Aextρ (p ′ − p)u(p), q = p′ − p (10.6)

where

ie2Πµν(q) = (−1)∫ d4k

(2π)4Tr

[i

k + q −m+ iϵ(−ieγµ) i

k −m+ iϵ(−ieγν)

](10.7)

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

Πµν(q) = i∫ d4k

(2π)4Tr

[1

k + q −mγµ

1

k −m+ iϵγν]

(10.8)

The last contribution is

Md = u(p′)(−ie)e2Λµ(p′, p)u(p)Aextµ (p ′ − p) (10.9)

176

where

e2Λµ(p′, p) =∫ d4k

(2π)4(−ieγα) i

p ′ − k −m+ iϵγµ

i

p− k −m+ iϵ(−ieγβ) −igαβ

k2 + iϵ(10.10)

or

Λµ(p′, p) = −i∫ d4k

(2π)4γα

1

p ′ − k −m+ iϵγµ

1

p− k −m+ iϵγα

1

k2 + iϵ(10.11)

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. (10.1). Since now the coefficients are finite wecan indeed 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 is arestrict class of renormalizable theories, as for instance QED. We will not discuss therenormalization at all order and neither we will prove which criteria a theory shouldsatisfy in order to be renormalizable. We will give these criteria without a proof butwe will try only to justify them in a physical basis. As far QED is concerned we willstudy in detail the renormalization at one-loop.

The previous way of defining a renormalizable theory amounts to say that theoriginal parameters in the lagrangian, as e, should be infinite and that their diver-gences should compensate the divergences of the Feynman diagrams. Then one cantry to separate the infinite from the finite part of the parameters (this separationis ambiguous, see later). The infinite contributions are called counterterms, andby definition they have the same operator structure of the original terms in thelagrangian. On the other hand, the procedure of regularization can be performedby adding to the original lagrangian counterterms cooked in such a way that theircontribution kills the divergent part of the Feynman integrals. This means thatthe coefficients of these counterterms have to be infinite. However they can also beregularized in the same way as the other integrals. We see that the theory will be

177

renormalizable if the counterterms we add to make the theory finite have the samestructure of the original terms in the lagrangian, in fact, if this is the case, they canbe absorbed in the original parameters, which however are arbitrary, because theyhave to be fixed by the experiments (renormalization conditions).

In the case of QED all the divergences can be brought back to the three functionsΣ(p), Πµν(q) e Λµ(p′, p). This does not mean that an arbitrary diagram is notdivergence, but it can be made finite if the previous functions are such. In sucha case one has only to show that eliminating these three divergences (primitivedivergences) the theory is automatically finite. In particular we will show thatthe divergent part of Σ(p) can be absorbed into the definition of the mass of theelectron and a redefinition of the electron field (wave function renormalization).The divergence in Πµν , the photon self-energy, can be absorbed in the wave functionrenormalization of the photon (the mass of the photon is not renormalized due tothe gauge invariance). And finally the divergence of Λµ(p

′, p) goes into the definitionof the parameter e. To realize this program we divide up the lagrangian density intwo parts, one written in terms of the physical parameters, the other will contain thecounterterms. We will call also the original parameters and fields of the theory thebare parameters and the bare fields and we will use an index B in order to distinguishthem from the physical quantities. Therefore the two pieces of the lagrangian shouldlook like as follows: the piece in terms of the physical parametersLp

Lp = ψ(i∂ −m)ψ − eψγµψAµ − 1

4FµνF

µν − 1

2(∂µA

µ)2 (10.12)

and the counter terms piece Lc.t.

Lc.t. = iBψ∂ψ − Aψψ − C

4FµνF

µν − E

2(∂µA

µ)2 − eDψAψ (10.13)

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

(10.14)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) (10.15)

we write the bare fields as

ψB = Z1/22 ψ, AµB = Z

1/23 Aµ (10.16)

we obtain

L = iψB∂ψB − m+ A

Z2

ψBψB − eZ1

Z2Z1/23

ψBγµψBAµB − 1

4FB,µνF

µνB + gauge− fixing

(10.17)

178

and putting

mB =m+ A

Z2

, eB =eZ1

Z2Z1/23

(10.18)

we get

L = iψB∂ψB −mBψBψB − eBψBγµψBAµB − 1

4FB,µνF

µνB + gauge− fixing (10.19)

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,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. 10.1a and 10.1b) one has an integration over the four momentum p and a

179

behaviour 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 (10.20)

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(10.21)

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 (10.22)

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 (10.23)

thereforedim[gi] = 4− dim[Oi] (10.24)

The renormalizability requiresdim[Oi] ≤ 4 (10.25)

from whichdim[gi] ≥ 0 (10.26)

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 chargewhich 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 (10.27)

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

180

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 (10.28)

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 (10.29)

.

a) b)

c)

Fig. 10.2 - Divergent diagrams coming from the interaction (ψψ)2.

At one loop the theory gives rise to the divergent diagrams of Fig. 10.2. Thedivergence of the first diagram can be absorbed into a counter term of the originaltype

−δg2(ψψ)2 (10.30)

The other two need counter terms of the type

δg3(ψψ)3 + δg4(ψψ)

4 (10.31)

181

.

a) b)

Fig. 10.3 - Divergent diagrams coming from the interactions (ψψ)2 and (ψψ)4.

These counter terms originate new one-loop divergent diagrams, as for instance theones in Fig. 10.3. The first diagram modifies the already introduced counter term(ψψ)4, but the second one needs a new counter term

δg5(ψψ)5 (10.32)

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 thiscondition. 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 ≈ g22E2 (10.33)

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

σ ≤ c2

E2(10.34)

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

g2E2 ≤ c (10.35)

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

E ≥√c

g2(10.36)

182

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.

10.2 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→∞

∫ Λ0 ). 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) (10.37)

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. Forinstance, 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) (10.38)

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) =∫ ∞

0dt e−ttz−1 (10.39)

If Re z ≤ 0, the integral diverges as

dt

t1+|Re z| (10.40)

183

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

0dt e−ttz−1 +

∫ ∞

αdt e−ttz−1 (10.41)

Expanding the exponential in the first integral and integrating term by term we get

Γ(z) =∞∑n=0

(−1)n

n!

∫ α

0dt tn+z−1 +

∫ ∞

αdt e−ttz−1

=∞∑n=0

(−1)n

n!

αn+z

n+ z+∫ ∞

αdt e−ttz−1 (10.42)

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.

10.3 Integration in arbitrary dimensions

Let us consider the integral

IN =∫dNpF (p2) (10.43)

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 (10.44)

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

∫ ∞

0pN−1F (p2)dp (10.45)

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

I =∫ +∞

−∞e−x

2dx =

√π (10.46)

184

By taking N of these factors we get

IN =∫dx1 · · · dxNe−(x21 + · · ·+ x2N) = πN/2 (10.47)

The same integral can be evaluated in polar coordinates

πN/2 = SN

∫ ∞

0ρN−1e−ρ

2dρ (10.48)

By putting x = ρ2 we have

πN/2 =1

2SN

∫ ∞

0xN/2−1e−xdx =

1

2SNΓ

(N

2

)(10.49)

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

SN =2πN/2

Γ(N2

) (10.50)

and

IN =πN/2

Γ(N2

) ∫ ∞

0xN/2−1F (x)dx (10.51)

with x = p2.The integrals we will be interested in are of the type

I(M)N =

∫ dNp

(p2 − a2 + iϵ)A(10.52)

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 obtaining

I(M)N = i

∫ dNp

(−p2 − a2)A= i(−1)AIN (10.53)

where IN is an Euclidean integral of the kind discussed at the beginning of thisSection with F (x) given by

F (x) = (x+ a2)−A (10.54)

It follows

IN =πN/2

Γ(N2

) ∫ ∞

0

xN/2−1

(x+ a2)Adx (10.55)

By puttingx = a2y we get

IN = (a2)N/2−AπN/2

Γ(N2

) ∫ ∞

0yN/2−1(1 + y)−Adx (10.56)

185

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)=∫ ∞

0tx−1(1 + t)−(x+y)dt (10.57)

it follows

IN = πN/2Γ(A−N/2)

Γ(A)

1

(a2)A−N/2(10.58)

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(10.59)

This shows that I2ω has simple poles located at

ω = A,A+ 1, · · · (10.60)

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(10.61)

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(10.62)

from which∫d2ωp

(p2 + 2p · k + b2)A= iπω(−1)A

Γ(A− ω)

Γ(A)

1

(k2 − b2)A−ω(10.63)

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

(10.64)

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)]

(10.65)

186

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) (10.66)

whereγ = 0.5772... (10.67)

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

Γ(−n+ ϵ) =(−1)n

n!

[1

ϵ+ ψ(n+ 1) +O(ϵ)

](10.68)

where

ψ(n+ 1) = 1 +1

2+ · · ·+ 1

n− γ (10.69)

10.4 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 Λ

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

[γµ, γν ]+ = 2gµν (10.70)

we getγµγµ = d (10.71)

andγµγνγ

µ = (2− d)γν (10.72)

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. (10.4) we have

Σ(p) = iµ4−2ω∫ d2ωk

(2π)2ωγµ

p− k +m

(p− k)2 −m2 + iϵγµ

1

k2 + iϵ(10.73)

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(10.74)

187

which is proven using

1

ab=

1

b− a

[1

a− 1

b

]=

1

b− a

∫ b

a

dx

x2(10.75)

and doing the change of variables

x = az + b(1− z) (10.76)

We get

Σ(p) = iµ4−2ω∫ 1

0dz∫ d2ωk

(2π)2ωγµ

p− k +m

[(p− k)2z −m2z + k2(1− z)]2γµ (10.77)

The denominator can be written in the following way

[...] = p2z −m2z + k2 − 2p · kz (10.78)

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) (10.79)

It follows (we put k′ = k)

Σ(p) = iµ4−2ω∫ 1

0dz∫ d2ωk

(2π)2ωγµ

p(1− z)− k +m

[k2 −m2z + p2z(1− z)]2γµ (10.80)

The linear term in k has vanishing integral, therefore

Σ(p) = iµ4−2ω∫ 1

0dz∫ d2ωk

(2π)2ωγµ

p(1− z) +m

[k2 −m2z + p2z(1− z)]2γµ (10.81)

and integrating over k

Σ(p) = iµ4−2ω∫ 1

0dz

1

(2π)2ωiπω

Γ(2− ω)

Γ(2)γµ

p(1− z) +m

[m2z − p2z(1− z)]2−ωγµ (10.82)

By defining ϵ = 4− 2ω we get

Σ(p) = −µϵ∫ 1

0dz

1

(2π)4−ϵπ(4−ϵ)/2Γ(ϵ/2)γµ

p(1− z) +m

[m2z − p2z(1− z)]ϵ/2γµ (10.83)

Contracting the γ matrices

Σ(p) = − 1

16π2Γ(ϵ/2)

∫ 1

0dz(4πµ2)ϵ/2

(ϵ− 2)p(1− z) + (4− ϵ)m

[m2z − p2z(1− z)]ϵ/2(10.84)

188

we obtain

Σ(p) =1

16π2Γ(ϵ/2)

×∫ 1

0dz[2p(1− z)− 4m− ϵ(p(1− z)−m)]

[m2z − p2z(1− z)

4πµ2

]−ϵ/2(10.85)

Defining

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

4πµ2(10.86)

and expanding for ϵ→ 0

Σ(p) =1

16π2

∫ 1

0dz[2A

ϵ+ 2B − γA

] [1− ϵ

2logC

]=

1

16π2

∫ 1

0dz[2A

ϵ− A logC + 2B − γA

]=

1

8π2ϵ(p− 4m)− 1

16π2[p− 2m+ γ(p− 4m)]

− 1

8π2

∫ 1

0dz[p(1− z)− 2m] log

m2z − p2z(1− z)

4πµ2

=1

8π2ϵ(p− 4m) + finite terms (10.87)

Consider now the vacuum polarization (see eq. (10.8))

Πµν(q) = iµ4−2ω∫ d2ωk

(2π)2ωTr

[1

k + q −mγµ

1

k −mγν]

= iµ4−2ω∫

d2ωk

(2π)2ωTr[γµ(k +m)γν(k + q +m)]

(k2 −m2)((k + q)2 −m2)(10.88)

Using again the Feynman representation for the denominators

Πµν(q) = iµ4−2ω∫ 1

0dz∫ d2ωk

(2π)2ωTr[γµ(k +m)γν(k + q +m)]

[(k2 −m2)(1− z) + ((k + q)2 −m2)z]2(10.89)

We write the denominator as

[...] = k2 + q2z −m2 + 2k · qz (10.90)

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 (10.91)

and therefore (we put again k′ = k)

Πµν(q) = iµ4−2ω∫ 1

0dz∫ d2ωk

(2π)2ωTr[γµ(k − qz +m)γν(k + q(1− z) +m)]

[k2 + q2z(1− z)−m2]2(10.92)

189

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 ] (10.93)

If we define the γ as matrices of dimension 2ω × 2ω we can repeat the calculation ofSection 9.2 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)](10.94)

and we find

Πµν(q) = iµ4−2ω2ω∫ 1

0dz∫ d2ωk

(2π)2ω

[ 2kµkν[k2 + q2z(1− z)−m2]2

− 2z(1− z)(qµqν − gµνq2)

[k2 + q2z(1− z)−m2]2− gµν

[k2 + q2z(1− z)−m2]

](10.95)

From the relations of the previous Section we get∫d2ωp

2pµpν[p2 − a2]2

= −igµνπωΓ(1− ω)

(a2)1−ω(10.96)

whereas ∫d2ωp

1

[p2 − a2]= −iπωΓ(1− ω)

(a2)1−ω(10.97)

Therefore the first and the third contribution to the vacuum polarization cancel outand we are left with

Πµν(q) = −iµ4−2ω2ω(qµqν − gµνq2)∫ 1

0dz 2z(1− z)

∫ d2ωk

(2π)2ω1

[k2 + q2z(1− z)−m2]2

(10.98)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

0dz z(1− z)

iπω

(2π)2ωΓ(2− ω)

[m2 − q2z(1− z)]2−ω

(10.99)By putting again ϵ = 4− 2ω and expanding the previous expression

Πµν(q) = 2µϵ22−ϵ/2(qµqν − gµνq2)∫ 1

0dz z(1− z)

π2−ϵ/2

(2π)4−ϵΓ(ϵ/2)

[m2 − q2z(1− z)]ϵ/2

190

=2

16π222−ϵ/2(qµqν − gµνq

2)∫ 1

0dz z(1− z)Γ(ϵ/2)

[m2 − q2z(1− z)

4πµ2

]−ϵ/2

=2

16π2(4− 2ϵ log 2)(qµqν − gµνq

2)

×∫ 1

0dz z(1− z)(

2

ϵ− γ)

[1− ϵ

2logC

](10.100)

where C is definite in eq. (10.86). 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

]](10.101)

or, in abbreviated way

Πµν(q) =1

6π2(qµqν − gµνq

2)1

ϵ+ finite terms (10.102)

We have now to evaluate Λµ(p′, p). From eq. (10.11) we have

Λµ(p′, p) = −iµ4−2ω∫ d2ωk

(2π)2ωγα

p ′ − k +m

(p′ − k)2 −m2γµ

p− k +m

(p− k)2 −m2γα

1

k2(10.103)

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)

[∑ni=1 βiai]

n(10.104)

To show this equation notice that

n∏i=1

1

ai=∫ ∞

0

n∏i=1

dαie−

n∑i=1

αiai(10.105)

Introducing the identity

1 =∫ ∞

0dλδ(λ−

n∑i=1

αi) (10.106)

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−λ

n∑i=1

βiai

(10.107)

191

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

0dλλn−1e−ρλ =

(n− 1)!

ρn(10.108)

In our case we get

Λµ(p′, p) = −2iµ4−2ω∫ d2ωk

(2π)2ω

∫ 1

0dx∫ 1−x

0dy

× γα(p ′ − k +m)γµ(p− k +mγα[k2(1− x− y) + (p− k)2x−m2x+ (p′ − k)2y −m2y]3

(10.109)

The denominator is

[...] = k2 −m2(x+ y) + p2x+ p′2y − 2k · (px+ p′y) (10.110)

Changing variable, k = k′ + px+ py

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

= k′ 2 −m2(x+ y) + p2x(1− x) + p′2y(1− y)− 2p · p′xy (10.111)

Letting again k′ → k

Λµ(p′, p) = −2iµ4−2ω∫ 1

0dx∫ 1−x

0dy∫ d2ωk

(2π)2ω

×γα(p ′(1− y)− px− k +m)γµ(p(1− x)− p′y − k +m)γα[k2 −m2(x+ y) + p2x(1− x) + p′2y(1− y)− 2p · p′xy]3

(10.112)

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 (10.113)

we get for the first term

Λ(1)µ (p′, p) = −2iµ4−2ω

∫ 1

0dx∫ 1−x

0dy∫ d2ωk

(2π)2ω

γαγλγµγσγαkλkσ

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

= −2iµ4−2ω∫ 1

0dx∫ 1−x

0dyiπω(−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

0dx∫ 1−x

0dy

γαγλγµγλγα

[m2(x+ y)− p2x(1− x)− p′2y(1− y) + 2p · p′xy]2−ω(10.114)

192

Using the relationγαγλγ

µγλγα = (2− d)2γµ (10.115)

we obtain (ϵ = 4− 2ω)

Λ(1)µ (p′, p) =

1

2µϵ(

1

4π

)2−ϵ/2Γ(ϵ/2)(ϵ− 2)2γµ

∫ 1

0dx∫ 1−x

0dy

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

0dx∫ 1−x

0dy[

m2(x+ y)− p2x(1− x)− p′2y(1− y) + 2p · p′xy4πµ2

]−ϵ/2

=1

8π2ϵγµ −

1

16π2(γ + 2)γµ −

1

8π2γµ

∫ 1

0dx∫ 1−x

0dy

× log

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

4πµ2

](10.116)

and finally

Λ(1)µ (p′, p) =

1

8π2ϵγµ + finite terms (10.117)

In the convergent part we can put directly ω = 2

Λ(2)µ (p′, p) = − i

8π4

∫ 1

0dx∫ 1−x

0dyiπ2(−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

0dx∫ 1−x

0dy

γα(p′(1− y)− px+m)γµ(p(1− x)− p′y +m)γαm2(x+ y)− p2x(1− x)− p′2y(1− y) + 2p · p′xy

(10.118)

10.5 One loop renormalization

We summarize here the results of the previous Section

Σ(p) =1

8π2ϵ(p− 4m) + Σf (p). (10.119)

Πµν(q) = (qµqν − gµνq2)[

1

6π2ϵ+Πf (q)

]≡ (qµqν − gµνq

2)Π(q2). (10.120)

Λµ(p′, p) =

1

8π2ϵγµ + Λfµ(p

′, p), (10.121)

193

where the functions with the superscript f represent the finite contributions. Let usstart discussing the electron self-energy. As shown in eq. (10.2), the effect of Σ(p)is to correct the electron propagator. In fact we have (see Fig. 10.4):

SF (p) =i

p−m+

i

p−mie2Σ(p)

i

p−m+ · · · , (10.122)

from which, at the same order in the perturbative expansion

.

Fig. 10.4 - The loop expansion for the electron propagator.

SF (p) =i

p−m

(1 +

e2Σ(p)

p−m

)−1

=i

p−m+ e2Σ(p). (10.123)

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. (10.124)

This allows us to define the counterterms 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, (10.125)

(L1)ct modifies the Feynman rules adding two operators parameterized by A and B.These coefficients can be evaluated noticing that the expression of the propagator,taking into account (L1)ct, is

i

(1 +B)p− (m+ A)≈ i

p−m

(1− Bp− A

p−m

)

≈ i

p−m+

i

p−m(iBp− iA)

i

p−m, (10.126)

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. 10.5, withcontributions −iA to the mass term, and iBp to p.

The propagator at the second order in the coupling constant is then obtained byadding the diagrams of Fig. 10.6 to the free part. We get

SF (p) =i

p−m+

i

p−m

(ie2Σ(p) + iBp− iA

) i

p−m(10.127)

194

.

-iA iBp/

Fig. 10.5 - The counter terms for the self-energy (p/ in the figure should be read asp)..

=

Fig. 10.6 - The second order contributions at the electron propagator.

and the correction is given by

e2Σ(p) +Bp− A =

(e2

8π2ϵ+B

)p−

(e2

2π2ϵm+ A

)+ finite terms. (10.128)

We can now fix the counter terms by choosing

B = − e2

8π2

(1

ϵ+ F2

(m

µ

)), (10.129)

A = −me2

2π2

(1

ϵ+ Fm

(m

µ

)), (10.130)

with F2 and Fm finite for ϵ → 0. Notice that these two functions are dimensionlessand for the moment being completely arbitrary. However they can be determined bythe renormalization conditions, that is by fixing the arbitrary constants appearingin the Lagrangian. In fact, given

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

=

(1− e2

8π2F2

)p−m+

me2

2π2Fm + e2Σf , (10.131)

195

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

SF (p) ≈i

p−m, for p = m. (10.132)

From here we get two conditions. The first one is

Γ(2)(p = m) = 0 (10.133)

from which

e2Σf (p = m)− me2

8π2F2 +

me2

2π2Fm = 0. (10.134)

The second condition is∂Γ(2)(p)

∂pµ

∣∣∣p=m

= γµ, (10.135)

giving

e2∂Σf (p)

∂pµ

∣∣∣p=m

− e2

8π2F2γµ = 0. (10.136)

One should be careful because these particular conditions of renormalization givesome problem 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, andcan be eliminated giving a small mass to the photon and sending this mass to zero atthe end of the calculations. Notice that these conditions of renormalization have theadvantage of being expressed directly in terms of the measured parameters, as theelectron 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 to the photon propagator (illustrated in Fig. 10.7)

D′µν(q) =

−igµνq2

+−igµλq2

ie2Πλρ(q)−igρνq2

+ · · · , (10.137)

.

Fig. 10.7 - The loop expansion for the photon propagator.

196

from which

D′µν(q) =

−igµνq2

+−igµλq2

[(ie2)(qλqρ − gλρq2)Π(q2)

] −igρνq2

+ · · ·

=−igµνq2

[1− e2Π(q2)

]− i

qµqνq4

e2Π(q2) + · · · . (10.138)

We see that the one loop propagator has a divergent part in gµν , as well in the termproportional to the momenta. Therefore the propagator does not correspond anymore to the Lorenz gauge and we need to add to the following terms in Lp

L2 = −1

4FµνF

µν − 1

2(∂µA

µ)2, (10.139)

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. (10.140)

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. The modification in the equation defining thepropagator due to these two terms is

[(1 + C)q2gµν − (C − E)qµqν ]Dνλ(q) = −igλµ. (10.141)

We solve this equation by putting

Dµν(q) = α(q2)gµν + β(q2)qµqν . (10.142)

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

α = − i

q2(1 + C), β = − i

q4C − E

(1 + C)(1 + E). (10.143)

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

Dµν(q) =−igµνq2

(1− C)− iqµqνq4

(C − E) (10.144)

and the total propagator

D′µν(q) =

−igµνq2

[1− e2Π(q2)− C]− iqµqνq4

[e2Π(q2) + C − E]. (10.145)

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

C = − e2

6π2ϵ+ F3

(m

µ

). (10.146)

197

In fact, we are free to choose the finite term in the gauge fixing, since this choicedoes not change the physics. This is because the terms proportional to qµqν , as itfollows from gauge invariance and the consequent conservation of the electromagneticcurrent, do not contribute to the physical amplitude. For instance, if we have avertex with a virtual photon (that is a vertex connected to an internal photon line)and two external electrons, the term proportional to qµqν is saturated with

u(p′)γµu(p), q = p′ − p. (10.147)

The result is zero, by taking into account the Dirac equation. Let us now considerthe mass of the photon. We have

D′µν(q) =

−igµνq2

[1− F3 − e2Πf

]− i

qµqνq4

[F3 + e2Πf

]≡ −igµν

q2[1− Π]− i

qµqνq4

Π, (10.148)

whereΠ = e2Πf + F3. (10.149)

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)

](10.150)

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. However, there are cases where it is not possible to devisea regularization procedure such to preserve a given symmetry. This the case ofthe anomalous symmetries which are symmetries only at the classical level butbroken by quantum corrections.

Also for the photon we will require that at the physical pole, q2 → 0, the prop-agator has the free form, that is

Π(0) = 0, (10.151)

from whichF3 = −e2Πf (0). (10.152)

10.6 Lamb shift and g − 2

In this Section we will use the technology developed in this Chapter to evaluate twoimportant physical quantities. The first one is a contribution to the Lamb shift. Thelevels 2S1/2 and 2P1/2 of the hydrogen atom are degenerate except for corrections

198

coming from QED. Among these corrections there is one coming from the radiativecorrection to the photon propagator and this is the one that we will evaluate here.The second quantity that we will evaluate comes from the vertex corrections and itis a contribution to the gyromagnetic factor of the electron which is equal to 2 inthe Dirac theory. Let us start with the vacuum polarization.

We will consider the expression (10.149) for small momenta:

Π(q) ≈ e2q2dΠf (q)

dq2

∣∣∣q2=0

, (10.153)

where use has been made of F3 being constant. Using the expression for Πf fromthe previous Section, we get

Πf (q) ≈ − 1

2π2

(γ

6+

1

6log

(m2

2πµ2

))+

1

2π2

∫ 1

0dz z2(1− z)2

q2

m2+ · · · (10.154)

and

Π(q) ≈ e2q2

60π2m2. (10.155)

It follows

D′µν = −igµν

q2

[1− e2q2

60π2m2

]+ gauge terms. (10.156)

The first term, 1/q2, gives rise to the Coulomb potential, e2/4πr. The second givesa correction by a term proportional to a delta function in the real space (we areusing the same notations of Section 7.2)

∆12 = e2∫ +∞

−∞dt

∫ d4q

(2π)4e−iq(x1 − x2)

[1

q2− e2

60π2m2

]

= − e2

4πr− e4

60π2m2δ3(r). (10.157)

This modification of the Coulomb potential changes the energy levels of the hydrogenatom, and it is one of the contributions to the Lamb shift, which produces a splittingof the levels 2S1/2 and 2P1/2. The total Lamb shift is the sum of all the self-energyand vertex corrections, and turns out to be about 1057.9 MHz. The contributionwe have just evaluated is only −27.1 MHz, but it is important since the agreementbetween experiment and theory is at the level 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 addto the interacting part of Lp,

Lintp = −eψγµψAµ, (10.158)

is(L3)ct = −eDψγµψAµ. (10.159)

199

.

(counter-term)

Fig. 10.8 - The one loop vertex corrections.

The complete vertex (omitting the factor −ie) is given by (see Fig. 10.8)

ΛTµ = γµ(1 +D) + e2Λµ. (10.160)

Due to the conservation of the electromagnetic current, it is possible to provethe validity of the Ward-Takahashi identity between the total self-energy and thetotal vertex correction

∂ΣT (p)

∂p= ΛTµ (p, p), (10.161)

where ΣT and ΛTµ are defined as

Γ(2) = p−m+ ΣT , Γµ = γµ + ΛTµ . (10.162)

This identity applied to the expressions (10.131) and (10.160) gives

Bγµ + e2∂Σ(p)

∂pµ= Dγµ + e2Λµ(p, p), (10.163)

where Σ and Λ are the self-energy and vertex corrections evaluated from the Feyn-man diagrams associated to Lp. One can check that the Ward-Takahashi identityholds for the the regularized one-loop expressions for Σ and Λµ in equations (10.73)and (10.103), and therefore it must be satisfied order by order in the ϵ expansion.Using this result one gets

B = D, (10.164)

at this order in perturbation theory. This equality is certainly satisfied by thedivergent parts of these counterterms, since they are devised in order to eliminatethe terms in 1/ϵ in Σ and Λµ which are equal (see eqs. (10.119) and 10.121)).However, in order to satisfy (10.164) also the finite parts of B and D should beequal. As a consequence we fix the counter term D by

D = − e2

8π2

[1

ϵ+ F2

], (10.165)

with F2 the same as in eq. (10.136). Notice that in this way the wave functionrenormalization terms Z1 and Z2 are made equal, and the charge renormalizationdepends only on Z3, that is from the vacuum polarization.

200

The complete one-loop vertex is then

ΛTµ =[γµ + e2(Λfµ −

F2

8π2γµ)

]. (10.166)

By using again the Ward-Takahashi identity for the regularized expressions, valid atany order in 1/ϵ and, therefore, also for the finite parts, and the condition (10.136)

Λfµ∣∣∣p=m

=∂Σf

∂pµ

∣∣∣p=m

=F2

8π2γµ, (10.167)

we see that for spinors on shell we have

u(p)ΛTµu(p)|p=m = u(p)γµu(p). (10.168)

It is interesting to notice that we have been able to satisfy the Ward-Takahashiidentity since we had the freedom to choose the finite part of the counterterm D.There are situations where this is not possible and this is the case of the anomalies,that is transformations that are symmetries at the classical level but that are brokenat quantum level. A celebrated example is the one of the axial-vector anomaly. Thelagrangian for a massless Dirac field is invariant under the chiral transformation

ψ(x) → eiαγ5ψ(x), (10.169)

at the classical level, besides being invariant under a phase transformation. Bothsymmetries are preserved by making the Dirac field to interact with the electromag-netic field. However, it turns out that it is impossible to define a renormalizationprocedure (choice of the counterterms) in order to satisfy both the Ward-Takahashiidentities following from the conservations of the two currents associated to the twosymmetries. Usually one makes the choice of satisfying the Ward-Takahashi identityassociated to the phase symmetry and the chiral symmetry is broken at the quantumlevel (axial anomaly).

We will now evaluate the radiative corrections to the g−2 of the electron. Here gis the gyromagnetic ratio, which is predicted to be equal to 2 by the Dirac equation(see Section 4.6). To this end we first need to prove the Gordon identity for thecurrent of a Dirac particle

u(p′)γµu(p) = u(p′)

[pµ + pµ′

2m+

i

2mσµνq

ν

]u(p), (10.170)

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

pγµu(p) = (−mγµ + 2pµ)u(p) (10.171)

andγµpu(p) = mγµu(p). (10.172)

201

Subtracting these two expressions we obtain

γµu(p) =(pµm

− i

mσµνp

ν)u(p). (10.173)

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

e

2mσµνF

µν(q). (10.174)

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) (10.175)

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 momentaon shell, the denominator of Λ(2)

µ is given by

[...] = m2(x+ y)−m2x(1− x)−m2y(1− y) + 2m2xy = m2(x+ y)2. (10.176)

In order to evaluate the numerator, let us define

Vµ = u(p′)γα [p′(1− y)− px+m] γµ [p(1− x)− p′y +m] γαu(p). (10.177)

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

Vµ = u(p′)[myγα + 2(1− y)p′

α − γαpx]γµ

× [mxγα + 2(1− x)pα − p′yγα] u(p). (10.178)

Making use ofγαγ

µγα = −2γµ, (10.179)

γαγµγνγα = 4gµν , (10.180)

γαpγµp′γα = −2p′γµp, (10.181)

we get

Vµ = u(p′)[− 2m2xyγµ + 2my(1− x)(−mγµ + 2pµ)

−4my2p′µ+ 2mx(1− y)(−mγµ + 2p′

µ) + 4(1− x)(1− y)m2γµ

−2y(1− y)m2γµ − 4mx2pµ

−2m2x(1− x)γµ − 2xym2γµ]u(p) (10.182)

202

and for the piece which does not contain γµ

Vµ = 4mu(p′)[pµ(y − xy − x2) + p′

µ(x− xy − y2)

]u(p). (10.183)

Therefore the relevant part of the vertex contribution is

e2u(p′)Λ(2)µ (p′, p)u(p) → − e2

16π2

∫ 1

0dx∫ 1−x

0dy

4

mu(p′)

[pµ

(y − xy − x2)

(x+ y)2

+p′µ (x− xy − y2)

(x+ y)2

]u(p). (10.184)

By changing variable, z = x+ y, we obtain∫ 1

0dx∫ 1−x

0dyy − xy − x2

(x+ y)2=∫ 1

0dx∫ 1

xdz[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

]10

=1

4, (10.185)

from which

e2u(p′)Λ(2)µ (p′, p)u(p) → − e2

16π2mu(p′)[pµ + p′µ]u(p). (10.186)

Using the Gordon identity in this expression, and eliminating the further contribu-tion in γµ, we obtain the correction to the magnetic moment

e2u(p′)Λ(2)µ (p′, p)u(p)|magn. mom. →

ie2

16π2mu(p′)σµνq

νu(p). (10.187)

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)∣∣∣p′≈p

≈ u(p′)

[pµ + p′µ2m

+i

2m

(1 +

e2

8π2

)σµνq

ν

]u(p). (10.188)

Therefore the correction is

e

2m→ e

2m

(1 +

α

2π

). (10.189)

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

g

2= 1 +

α

2π+O(α2). (10.190)

203

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

(α

π

)2

+ 1.49(α

π)3 + · · ·

= (1159652.4± 0.4)× 10−9, (10.191)

to be compared with the experimental value

aexp = (1159652.4± 0.2)× 10−9. (10.192)

204