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 Noethers 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 -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 Wicks 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 + . . . . . . . . . . . . . . . . . . . . . . 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 - Heisenbergs first paper on quantummechanics (matrix mechanics). September - Born and Jordan extend Heisenbergs 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 = 2 [(qn+1 qn) + (qn1 qn)] = 2 [qn+1 + qn1 2qn] (1.1)

5

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

U =1

22

Nn=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

Nn=1

(p2n +

2 (qn qn+1)2)

(1.3)

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

q(j)n = Ajeikjaneijt (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

2j q(j)n = 42q(j)n sin2(kja

2

)(1.6)

from which

2j = 42 sin2

(kja

2

)(1.7)

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

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

kj = kj + 2m

a, m = 1,2, ... (1.8)

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

6

- 3 - 2 - 1 1 2 3akj

1

2

3

4 j2

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)eijt = q(j)n = Ajeikjaneijt (1.9)

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

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

kj =2

aNj =

2

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| /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 = Ajeijteiakjn (1.12)

The most general solution is obtained by a linear superposition

qn =j

ei2

Njn Qj

N(1.13)

pn =j

ei2

Njn Pj

N(1.14)

7

From the reality of qn and pn we get

Qj = Qj, Pj = Pj (1.15)

In the following we will make use of the following relation

Nn=1

ei2

N(j j)n

= Njj (1.16)

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

Nn=1 e

i2

N(j j)n

=N

n=0

ei2N (j j)n

1

=1 e

i2

N(j j)(N + 1)

1 ei2

N(j j)

1 = 0 (1.17)

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

Nn=1

qnei2

Njn

=j

n

ei2

N(j j)n Qj

N=

NQj (1.18)

obtaining

Qj =1N

Nn=1

qnei2

Njn

(1.19)

Pj =1N

Nn=1

pnei2

Njn

(1.20)

Notice that Pj = Qj. Substituting inside the hamiltonian we find

H =N/2j=1

(|Pj|2 + 2j |Qj|2

)(1.21)

This is nothing but the hamiltonian of N decoupled harmonic oscillators each havinga frequency j, 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) = 2 [u(xn+1, t) + u(xn1, t) 2u(xn, t)]

= 2 [(u(xn+1, t) u(xn, t)) (u(xn, t) u(xn1, 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(xn1, t)a

u(x, t) (1.27)

and

(u(xn+1, t) u(xn, t)) (u(xn, t) u(xn1, t)) a(u(xn+1, t) u(xn, t)) a2u(x, t) (1.28)

The equation of motion becomes

u(x, t) = a22u(x, t) (1.29)

Let us recall that the quantity 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 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

lima0

a = 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+1n=1

1a

L0

dx (1.32)

from which

H =1

2a

L0

dx[(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) =a(x, t) (1.34)

getting finally

H =1

2

L0

dx[((x, t)

)2+ v2 ((x, t))

2]

(1.35)

The normal modes decomposition becomes

(x, t) =u(x, t)

a qn(t)

a

j

eikjanQjaN

(1.36)

kj =2

Lj, < j < + (1.37)

giving rise to

(x, t) =1L

+j=

ei2

Ljx

Qj(t) (1.38)

The eigenfrequencies are given by

2j = 42 sin2

(a

Lj) 42

(j

L

)2a2 = (akj)

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 (x, t) can be inverted byusing the following relation L

0dx eix(k k

) = Lk,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/2j=1

(|Pj|2 + 2j |Qj|2

)(2.1)

whereQj = Qj, P

j = Pj (2.2)

2j = 42 sin2

kja

2, kj =

2

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

2(2.3)

whereas in the continuum case

H =j=1

(|Pj|2 + 2j |Qj|2

)(2.4)

withj = 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 =

j2Qj + i

12j

P j , aj =

j2Qj i

12j

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, ak] =

1

2jk

[jQj + iPj , kQ

k iPk] = jk (2.7)

and[aj, ak] = [a

j, a

k] = 0 (2.8)

where we have made use of the canonical commutation relations

[Qj, Pk] = [Qj, P

k ] = ijk (2.9)

Notice that

aj =

j2Qj + i

12j

P j =

j2Qj + i

12j

Pj (2.10)

implying

aj =

j2Qj i

12j

P j = aj (2.11)

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

Pj. The previous relations can be inverted to give

Qj =12j

(aj + aj), Pj = i

j2(aj aj) (2.12)

In terms of aj and aj the hamiltonian is

H =N/2j=1

j[ajaj + a

jaj + 1] =

N/2j=N/2

j

[ajaj +

1

2

](2.13)

The fundamental state is characterized by the equation

aj|0 = 0 (2.14)

and its energy is

E0 =j

j2

(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)

|nN/2, , nN/2 =1

(nN/2! nN/2!)1/2(aN/2)

nN/2 (aN/2)nN/2 |0 (2.16)

12

The state given above can be thought of being formed by nN/2 quanta of typeN/2 of energy N/2, up to nN/2 quanta of type N/2 of energy N/2. In this kindof interpretation the nj quanta (or particles) of energy j 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

[ai , aj] = 0 (2.17)

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

|i, j = aiaj = |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

j2N(ajaj + aja

j) =

j

jajaj (2.19)

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

qn =j

ei2

Njn Qj

N, pn =

j

ei2

Njn Pj

N(2.20)

and using the canonical commutators (2.9), we get

[qn, pm] =jk

ei2

N(jn km) 1

N[Qj, Pk] =

i

N

j

ei2

Nj(nm)

= inm (2.21)

In the continuum case we use the analogue expansion

(x, t) =1L

+j=

ei2

Ljx

Qj, (x, t) =1L

+j=

ei2

Ljx

Pj (2.22)

13

from which

[(x, t), (y, t)] =1

L

jk

ei2

L(jx ky)

ijk

=i

L

+j=

ei2

Lj(x y)

= i(x y) (2.23)

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

(x, t) u(x, t)a

qna

(2.24)

implying

[(xn, y), (xm, t)] = inma

(2.25)

In the limit

lima0

nma

= (x y) (2.26)

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

1 =n

a

(nma

)

dx

(lima0

nma

)(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 (x, t), pn (x, t) (2.28)

Said in different words, (x, t) e (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 (x, t) e (x, t), and quantize the theory byrequiring [(x, t), (y, t)] = i(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} = nm. 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

Nn=1

p2n =1

2

Nn=1

u2(xn, t)

=1

2

Nn=1

a2(xn, t) 1

2

L0

2(x, t)dx (2.33)

whereas the potential energy is

U =1

2

Nn=1

2(qn qn+1)2

=1

2

Nn=1

2(u(xn, t) u(xn+1, t))2

=1

2

Nn=1

2a((xn, t) (xn+1, t))2 (2.34)

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

U =1

2

Nn=1

av2((xn, t) (xn+1, t)

a

)2 v

2

2

L0

2(x, t)dx (2.35)

15

Therefore the total energy and the lagrangian are respectively

E = T + U =1

2

L0

dx[2(x, t) + v2

2(x, t)

](2.36)

and

L = T U = 12

L0

dx[2(x, t) v22(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 and its first derivatives, which willbe called lagrangian density, and having the expression

L = 12

(2 v22

)(2.38)

The total lagrangian is obtained by integrating spatially the lagrangian density

L = L0

Ldx (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(, , , 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 = t2t1

Ldt = t2t1

dt

dxL(, , , 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 is the spatial surfacedelimiting the region of spatial integration (for the string reduces to the endpoints), we will ask that

(x, t) = 0 on (2.42)

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

(x, t1) = (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 = S = t2t1

dt

dxL = t2t1

dt

dx

(L

+L

+L

)

(2.44)

Integrating by parts

0 = t2t1

dt

dx[L

+

t

(L

)

(

t

L

)

+

x

(L

)

(

x

L

)

]

=

dx

[L

]t2t1

+ t2t1

dt

[L

]L0

+ t2t1

dt

dx

[L

t

L

x

L

] (2.45)

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

L

t

L

x

L

= 0 (2.46)

In fact 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 (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 i, i = 1, . . . , N(think, as an example, to the electromagnetic field), and to the case of n spatialdimensions with points labelled by x, = 1, . . . , n. In this case the structure ofthe action will be

S = t2t1

dtV

dnxL(i, i,

ix

)(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

i(x, t) = 0, on , for any t, t1 t t2 (2.48)

where is the boundary of V , and

i(x, t1) = i(x, t2) = 0, for any x 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

Li

t

Li

x

L

ix

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

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

i =Li

(2.51)

and the hamiltonian densityH =

i

ii L (2.52)

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

L

= ,L

= v2, L

= 0 (2.53)

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

= (2.54)

implying

H = L = 2 (1

22 1

2v2

2)=

1

2

(2 + v2

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

ba

dx[2 + v2

2]

(2.56)

We can evaluate its time variation

dE(a, b)

dt=

ba

dx[+ v2

]= v2

ba

dx[ +

]= v2

ba

dx

x

[

]= v2

[

]ba

(2.57)

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

P (x, t) = v2 (2.58)

which is the analogous of the Poyntings 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), andHt

+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

i =Li

(2.62)

it is then natural to assume the following commutation relations

[i(x, t),j(y, t)] = iijn(x y) = 1, . . . , n, i, j = 1, . . . , N (2.63)

19

and[i(x, t), j(y, t)] = 0, [i(x, t),j(y, t)] = 0 (2.64)

In the string case we have = 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).

(x, t) =1L

j

ei2

Ljx

Qj

=1L

j

12j

ei2

Ljx (

aj(t) + aj(t)

)

=1L

j

12j

ei2L jxaj(t) + ei2

Ljx

aj(t)

(2.65)Using the equations of motion of the string

v2 = 0 (2.66)

we find from the above expansion of in terms of Qj

Qj + 2jQj = 0 (2.67)

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

aj + ijaj = 0 (2.68)

from which

aj(t) = aj(0)eijt ajeijt, aj(t) = a

j(0)e

ijt ajeijt (2.69)

and

(x, t) =1L

j

12j

ei(2

Ljx jt

)aj + e

i(2

Ljx jt

)aj

(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

(x, t) =j

[fj(x, t)aj + f

j (x, t)a

j

](2.71)

20

with

fj(x, t) =12jL

ei(2

Ljx ijt

)(2.72)

or

fj(x, t) =12jL

ei(kjx ijt) (2.73)

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

kj =2

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 functionsj

f j (x, t)i()t fj(y, t) = (x y) (2.77)

L0

dxf j (x, t)i()t fl(x, t) = jl (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

2jfj (x, t)fj(y, t) =

j

1

Leikj(x y) = (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 L0

dxf j (x, t)i()t fl(x, t) =

L0

dx[l + j]f

j (x, t)fl(x, t)

]=

1

L

l + j2jl

L0

dxeix(kl kj)ei(j l)t

=1

L

l + j2jl

ei(j l)tLjl = jl (2.81)

21

Also in this case, by taking two fj(x, t)s or two fj (x, t)s, the result is zero due to

the factor l j.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

d3x(x, t)(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

d3x(x, t)(x, t) =

d3x

[ +

]=

d3x [i(H) i(H)] = 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)

L0

dx

[2f

t2 v2

2f

x2

]f = 0 (2.85)

Subtracting these two expressions one from the other we get

L0

dx

[

t

(ff

t f

tf

) v2

x

(ff

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 (x, t) =

k

L0

dxf j (x, t)i()t

[fk(x, t)ak + f

k (x, t)a

k

]= aj

(2.88)and therefore

aj = L0

dxf j (x, t)i()t (x, t) (2.89)

22

and

aj = L0

dx(x, t)i()t fj(x, t) (2.90)

From the field commutation relations we find

[aj, ak] =

L0

dxdy[(if j if j )(x,t), ifk ifk)(y,t)]

= L0

dxdy(f j fk(i(x y)) fkfk(i(x y))

)=

L0

dxf j i()t fk = jk (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 = L0

dxP = L0

dx (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=

L0

dx( + ) = L0

dx

x

1

2(2 + v2

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

L0

dx jkl[fjaj fj aj][flal f l a

l ]

= j,l

L0

dxjkl

2Ljl

[ei(kjx jt)aj ei(kjx jt)aj

]

[ei(klx lt)al ei(klx lt)al

]=

l

1

2kl[alale

2ilt + alal e2ilt alal 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[ajaj + a

jaj] =

j

kjajaj (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

(aN/2)nN/2 (aj)nj |0 = | (2.98)

have energyH| = (nN/2N/2 + + njj + )| (2.99)

and a momentum

P | = (nN/2kN/2 + + njkj + )| (2.100)

as it follows from[H, aj] = ja

j, [P, a

j] = kja

j (2.101)

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

A =j

jajaj (2.102)

we have[A, aj] = ja

j (2.103)

Then, if | is an eigenket of A with eigenvalue , aj| is an eigenket of A witheigenvalue +j. Therefore a

j and aj increase and lower the eigenvalues of A. This

is a trivial consequence of the commutation relations

A(aj|) = (ajA+ ja

j)| = (+ j)a

j| (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

t= H (3.3)

where H is the hamiltonian

H =|p|2

2m= 1

2m||2 (3.4)

If describes an eigenstate of the energy and of the momentum

eiEt+ 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

t=

(||2 +m2

) (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

) = m

1 ||

2

m2

= mk

ck(||2

)k (3.9)

Both these difficulties are eliminated by iteration

2

t2=

(||2 +m2

) (3.10)

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

2 +m2) = 0 (3.11)

where

2 =2

t2 ||2 (3.12)

is the DAlembert 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 and are twosolutions of such an equation, we can write the following identity

0 = (2 +m2)

)

(2 +m2)

) = [

()] (3.13)

27

Therefore the currentJ =

() (3.14)

has zero four-divergence and the quantityd3x J0 =

d3x( ) (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

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

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

p i x

=

(i

t,i

)(3.17)

We have also the following relations

p2 = pp

x

x= 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 (x);

quantization through the requirement of canonical commutation relations

[(x, t),(y, t)] = i3(x y), [(x, t), (y, t)] = 0, [(x, t),(y, t)] = 0(3.20)

expansion of (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 qi, satisfying the followingboundary conditions

qi(t1) = qi(t2) = 0 (3.22)

summing over i, and integrating in time between t1 and t2, we get t2t1

dt

[Ni=1

miqiqi

]=

t2t1

dtNi=1

qiV

qi(3.23)

Integrating by parts

{ t2t1

dt

[1

2

Ni=1

miq2i V

]}

[Ni=1

miqiqi

]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 = t2t1

[1

2

Ni=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 (x) = (x) (x), with boundary conditions

(x, t1) = (x, t2) = 0, limx

(x, t) = 0 (3.26)

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

0 = t2t1

dt

d3x

[

t

(

)

(

)+ +m2

](3.27)

Using the boundary conditions we get

0 = t2t1

dt

d3x[1

22 1

2 1

2m22

](3.28)

Therefore the lagrangian will be given by

L =

d3xL (3.29)

29

with

L = 12

[

m22]

(3.30)

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

S = t2t1

dtL (3.31)

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

=L

= (3.32)

and the canonical commutation relations

[(x, t), (y, t)] = i3(x y), [(x, t), (y, t)] = [(x, t), (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)eikx = A(k)ei(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

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

it follows

k =2

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 = iV

d3xf k()t fk = k,k (3.41)

where the delta is a Kronecker symbol defined as

k,k =3

i=1

ni,ni (3.42)

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

Vd3xA

kAke

i(k0 k0)x0 i(k k) x(k0 + k0) = k,k (3.43)

Using L

dxei2

L(n1 n1)x

= Ln1,n1 (3.44)

we get V

d3xei(k k) x = L3k,k (3.45)

from whichiV

d3xf k()t fk = |Ak|

22k0L3k,k (3.46)

where

k20 =(2

L

)2|n|2 +m2 (3.47)

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

Ak =1

L3/212k

, k =

(2

L

)2|n|2 +m2 =

|k|2 +m2 (3.48)

and the normalized solution turns out to be

fk(x) =1

L3/212k

eikx (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 =

3(k k) (3.50)

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

d3xAkAke

ikx ikx(k0 + k0) = (2)33(k k)|Ak|22k0 (3.51)

31

and the corresponding normalization is

Ak =1(2)3

12k

(3.52)

where

k =|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

1V

1(2)3

(3.54)

The wave function in the continuum is

fk(x) =1(2)3

12k

eikx (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 = k (3.57)

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

energy and behaving as eikx0 and eikx0 , k > 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

(x) =1(2)3

d3k

12k

[a(k)eikx0 + ik x + a(k)eikx0 + ik x

](3.58)

In the second term we can exchange k k, obtaining (kx = kx0 k x)

(x) =1(2)3

d3k

12k

[a(k)eikx + 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 (x), a(k) = i

d3x(x)

()t fk(x) (3.60)

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

a(k) = a(k) (3.61)

and the expansion becomes

(x) =

d3k[fk(x)a(k) + fk(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)] = 3(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 fks. Forinstance, if we change this normalization by a factor Nk

fk|fk = i

d3xf k()t fk = Nk

3(k k) (3.65)

leaving unchanged the expansion for the field

=

d3k[fka(k) + fka(k)] (3.66)

we get

a(k) =i

Nk

d3xf

k()t , a

(k) =i

Nk

d3x

()t fk (3.67)

and therefore

[a(k), a(k)] =i

NkNk

d3xf

k()t fk =

1

Nk3(k k) (3.68)

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

(x) =1

(2)3

d3k

1

2k[A(k)eikx + A(k)eikx] (3.69)

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

1

(2)3

d3k

1

2k=

1

(2)4

d4k(2)(k2 m2)(k0) (3.70)

33

as it follows by noticing that for k0 k

k2 m2 2k(k0 |k|2 +m2) (3.71)

In this case the basis functions for the expansion are

fk(x) =1

(2)31

2keikx (3.72)

with normalization

i

d3xf k()t fk =

1

(2)31

2k3(k k) (3.73)

and therefore[A(k), A(k)] = (2)32k

3(k k) (3.74)

3.3 The Noethers theorem for relativistic fields

We will now review the Noethers 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+ x) F (x)

= F (x) F (x) + xF (x)

x(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 F ,depending only on the form variation

F (x) = F (x) F (x) (3.76)

Then we get

F (x) = F (x) + xF (x)

x(3.77)

Let us now start form a generic four-dimensional action

S =V

d4x L(i, x), i = 1, . . . , N (3.78)

and let us consider a generic variation of the fields and of the coordinates, x =x + x

i(x) = i(x) i(x) i(x) + x x

(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 = / andi, =

i/x)

SV =V

d4xL(i, x)V

d4xL(i, x)

=V

d4xL(i, x+ x)(x)

(x)

V

d4xL(i, x)

V

d4xL(i, x+ x)(1 + x)V

d4xL(i, x)

=V

d4x[L(i, x+ x) L(i, x)] +V

d4xL(i, x)x

V

d4x

[Li

i +Li,

i, +Lx

x]+

V

d4xLx

=V

d4x

[Li

Li,

]i +

V

d4x

[Lx + L

i,i

](3.81)

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

Li

Li,

= 0 (3.82)

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

Vd4x

[Lx + L

i,i x L

i,i,

]= 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 x and i, 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 x = a with a independent on x e i = 0. From the general resultin eq. (3.83) we get the following local conservation law

T =Li,

i, Lg , T = 0 (3.84)

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

P =

d3xT 0 (3.85)

35

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

J =Li,

i =Li,

i, J = 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 = x2

(3.88)

For an infinitesimal transformation

x = x+ x (3.89)

it followsx2 x2 + 2x x = x x = 0 (3.90)

Since Lorentz transformations are linear

x = x x + x (3.91)

we getx x = 0 = xx = 0 (3.92)

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

= (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 fieldsi under an infinitesimal Lorentz transformation can be written as

i = 12ij

j (3.94)

36

where we have required that the transformation of the fields is of first order in theLorentz parameters . The coefficients (antisymmetric in the indices (, ))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 x we get the local conservation law

0 =

[(Li,

i, Lg

)x +

1

2

Li,

ijj

]

=1

2

[(T x T x

)+

Li,

ijj

](3.95)

and defining (watch at the change of sign)

M = xT xT Li,

ijj (3.96)

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

M = 0 (3.97)and consequently six constants of motion (notice that the lower indices are antisym-metric)

M =

d3xM0 (3.98)

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

In the case of Klein-Gordon

T = 1

2

(

m22)g (3.99)

from which

T 00 =1

22 +

1

2

(||2 +m22

)(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 =

xi(3.101)

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

P 0 = H =

d3xT 00 =1

2

d3x

(2 + ||2 +m22

)(3.102)

P i =

d3xT 0i =

d3x

xi, (P =

d3x) (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 = L (3.104)

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

[(x), P ] = i

x(3.105)

In fact

[(y, t), H] =1

2

d3x[(y, t),2(x, t)] = i(y, t) = i(y, t) (3.106)

Analogously

[(y, t), P i] =

d3x[(y, t),(x, t)(x, t)

xi] = i(y, t)

yi= i

(y, t)

yi(3.107)

Therefore the operator

U = eiaP (3.108)

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

eia P(x)eia P (x) + ia[P, (x)] = (x) + a(x)

x (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

d3kk[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)] = ka(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 k =|k|2 +m2, whereas the annihilation operators a(k) destroy

38

the corresponding states. In the case of the box normalization, for any k = (2/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 =1nk!

(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 . . . nk =|nki =

1nk1 ! nk !

(a(k1))nk1 (a(k))nk |0 (3.114)

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

|0 =|0i (3.115)

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

aki|0i = 0 (3.116)

In this normalization the hamiltonian is given by

H =1

2

k

k[a(k)a(k) + a(k)a(k)] (3.117)

and therefore

H|0 = 12

k

k|0 (3.118)

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

k-space has a volume

Vk =(2)3

L3(3.119)

from which

1

2

k

k =1

2

k

VkkVk

= 12

L3

(2)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

ka(k)a(k) (3.121)

39

whereas in the continuum

: H :=

d3kka(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 = 2

240

/hc

R4= 0.013

(Rm)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 kmax1/L

k 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 kmax1/L

k 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/L1/R

k d3k = L2R 1/L1/R

k3 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

J = () () (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

(+)(x) =

d3k1

2k(2)3ei(kt k x)a(k) (3.131)

and negative energy components of the field

()(x) = (+)(x) (3.132)

In fact, it is not difficult to show that

N =

d3x(+)i()t

() (3.133)

41

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

j = ()(

(+)) (())(+) (3.134)

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

(x) =

d4k(k)eikx (3.135)

with

(k) =

2k(2)3

(k2 m2)(a(k)(k0) + a

(k)(k0))

(3.136)

one has(+)(x) =

d4k(+)(k)eikx (3.137)

with(+)(k) = (k0)(k) (3.138)

Using the convolution theorem for the Fourier transform we get

(+)(x) =d4x(x x)(x) (3.139)

But

(x x) = 1(2)4

d4keik(x x

)(k0)

= 3(x x) dk0

2eik0(x0 x

0)(k0)

= 3(x x)(x0 x0) (3.140)

Therefore(+)(x, x0) =

dx0(x0 x0)(x, x0) (3.141)

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

N (x) = i(+)()t () (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(), this commutator is automatically zero, due tothe canonical commutation relations

[O((x, t)),O((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 x = (t, x), and y = (t, y),it follows the vanishing for arbitrary space-like separations. Let us evaluate thecommutator

[(x), (y)] =

= d3k1d3k2

(2)32k12k2

[[a(k1), a

(k2)]eik1x+ ik2y

+ [a(k1), a(k2)]eik1x ik2y

]=

d3k(2)32k

[eik(x y) eik(x y)

]

= 2i d3k

(2)32ksin(k(x0 y0))eik(x y) (3.145)

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

[(x), (y)] = d4k

(2)3(k0)(k

2 m2)[eik(x y) eik(x y)

]

= d4k

(2)3(k0)(k

2 m2)eik(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

[(x), (y)] = i(x y) (3.147)

the function

(x y) = i d4k

(2)3(k0)(k

2 m2)eik(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

[(+)(x), ()(y)] = d3k

(2)32keik(x y)

= d4k

(2)3(k0)(k

2 m2)eik(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

(2)32keik x

= k2dkd(cos )d

(2)32keikr cos

= i 142r

0

kdk

2k

[eikr eikr

]

= 182r

d

dr

+

dkeikr

|k|2 +m2(3.150)

By putting k = m sinh , dk = m cosh d, we get

(+)(0, x) = 182r

d

dr

+

deimr sinh (3.151)

Since +

deimr sinh = iH(1)0 (imr) (3.152)

where H is a Hankels function, and using

d

drH

(1)0 (imr) = imH

(1)1 (imr) (3.153)

we obtain(+)(0, x) = m

8rH

(1)1 (imr) (3.154)

The asymptotic behaviour of the Hankels function H(1)1 (imr) for large and small

values of r is given by

limr

H(1)1 (imr)

2

mremr, lim

r0H

(1)1 (imr)

2

mr(3.155)

from which

limr

(+)(0, x) m8r

2

mremr, lim

r0(+)(0, x) 1

4r2(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 1011cm. 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 = 12

2i=1

[(i)(

i)m2i2i]

(3.158)

and we can write immediately the canonical commutation relations

[i(x, t), j(y, t)] = iij3(x y) (3.159)

[i(x, t), j(y, t)] = [i(x, t), j(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,ai(k), i = 1, 2, and as a consequence

a1(k1)a2(k2)|0 = a2(k1)a1(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 = 12[(1)(

1) + (2)(2)]

1

2m2

[21 +

22

](3.162)

Then the theory acquires a symmetry under rotations in the plane of the two fields1 e 2

1 = 1 cos + 2 sin

2 = 1 sin + 2 cos (3.163)

46

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

= (1, 2) = (1, 2) (3.164)

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

1 = 2, 2 = 1 (3.165)

or, in a more compact formi = ijj (3.166)

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

12 = 21 = 1 (3.167)

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

J =Li,

i = i,ijj (3.168)

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

j =1

J = i,ijj = 1,2 2,1 (3.169)

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

j = (21)2 (22)1 = (m21 m22)12 (3.170)

The conserved charge associated to the current is

Q =

d3x j0 =

d3x(12 21) (3.171)

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

[Q, 1] = i2, [Q, 2] = i1 (3.172)

with a finite transformation given by

U = eiQ (3.173)

In fact

eiQ1eiQ = 1 + i[Q, 1] +

i2

2!2[Q, [Q, 1]] + . . .

= 1 + 2 1

21

2 + . . .

= 1 cos + 2 sin (3.174)

47

In analogous way one can show the transformation properties of 2. The invarianceof L under rotations in the plane (1, 2) 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

=12(1 + i2),

=12(1 i2) (3.175)

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

follows

[Q, ] =12[Q, 1 + i2] =

12(i2 1) = (3.176)

and analogously[Q, ] = (3.177)

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

Q(|q) = ([Q, ] + Q)|q = (1 + q)|q (3.178)

and|q |q 1 (3.179)

In analogous way|q |q + 1 (3.180)

Inverting the relations (3.175) we get

1 =12(+ ), 2 =

i2( ) (3.181)

from which

L = 14

[(+

)2

(

)2] 1

4m2

[(+

)2

(

)2]=

m2 (3.182)

and

j = 1,2 2,1

= i2(+

)( ) + i2( )(+ )

= i[()

()]

(3.183)

48

The charge results to be

Q = i

d3x ()t (3.184)

The commutation relations in the new basis are given by

[(x, t), (y, t)] =1

2[1(x, t) + i2(y, t), 1(x, t) i2(y, t)] = i3(x y) (3.185)

and

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

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

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

=L

= , =L

= (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

12(1 + i

2)

=12(1 cos + 2 sin + i(1 sin + 2 cos ))

=12

(1e

i + i2ei)= ei (3.188)

and ei (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

i(x) =

d3k[fk(x)ai(k) + f

k(x)ai(k)

](3.190)

we get

(x) =

d3k

[fk(x)

12(a1(k) + ia2(k)) + f

k(x)

12(a1(k) + ia

2(k))

](3.191)

Introducing the combinations

a(k) =12(a1(k) + ia2(k)), b(k) =

12(a1(k) ia2(k)) (3.192)

49

it follows

(x) =

d3k[fk(x)a(k) + f

k(x)b(k)

](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)] = 3(k k) (3.194)

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

We get also

: P :=

d3kk2

i=1

ai(k)ai(k) =

d3kk[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 ai. The charge Q is given by

Q = i

d3x()t

= i

d3x

d3k1d3k2

[fk1b(k1) + f

k1a(k1)

]

[(

()t fk2)a(k2) + (

()t f

k2)b(k2)

]=

d3k1d

3k23(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) =Vi

d3x 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

=[12

](4.1)

modifying also the wave equation to a matrix equation

it

=2

=1

H (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,

S =1

2 (4.3)

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

|k|2 = ( k)2 (4.4)

as it follows from[i, j]+ = 2ij (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

t= i + m H (4.6)

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

the wave function , 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

2

t2= (i + m)2

=

(ij

2

xixj+ 2m2 i(+ )

)

=

(12

[i, j

]+

2

xixj+ 2m2 i(+ )

) (4.7)

We see that it is necessary to require the following matrix relations[i, j

]+= 2ij,

[i,

]+= 0, 2 = 1 (4.8)

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

Tr() = Tr(i) = 0 (4.9)

For instance, from i = i, we get i = i, and therefore

Tr(i) = Tr(i) = Tr(i) = 0 (4.10)

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

52

the same properties of the Pauli matrices, but in a 2 2 matrix space, a furtheranticommuting matrix 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

i =[0 ii 0

], =

[1 00 1

](4.11)

as it can be checked [i, j

]+=

[[i, j]+ 0

0 [i, j]+

]= 2ij (4.12)

[, i

]+=

[0 i

i 0

]+

[0 ii 0

]= 0 (4.13)

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

i

t= i() +m (4.14)

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

i

t+ i

t = (i +m) (i +m) = i () (4.15)

that is

i

t() + i

xj(j) = 0 (4.16)

We see that the currentj = (, i) (4.17)

is a conserved one.j

x= 0 (4.18)

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

d3x (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

i

t= i +m (4.20)

and let us define the following matrices

0 = =[1 00 1

], i = i =

[0 i

i 0

](4.21)

Then the equation becomes(i0

x0+ ii

xim

) = 0 (4.22)

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

where

=

x= (4.24)

The matrices satisfy the following anticommutation relations[i, j

]+= ij + ji =

[i, j

]+= 2ij (4.25)

[0, i

]+=

[, i

]+= i + i = 0 (4.26)

or[, ]+ = 2g

(4.27)

Notice that(i) = (i) = i = i (4.28)

and(i)2 = 1 (4.29)

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

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

54

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

i

xm

)(x) = 0 (4.30)

The matrices should satisfy the same algebra as the matrices , 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 (

i

xm

)(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

(x) = (x) = S()(x) (4.32)

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

(x) = S1()(x) (4.33)

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

(x) = S(1)(x) (4.34)

from whichS1() = S(1) (4.35)

Considering the Dirac equation in the frame S(i

xm

)(x) = 0 (4.36)

we can write (i

xm

)S1()(x) = 0 (4.37)

Multiplying to the left by S() and using

x=

xx

x=

x, x = x

(4.38)

it follows (iS()S1()

xm

)(x) = 0 (4.39)

55

Comparing with eq. (4.30), we get

S()S1() = (4.40)

orS1()S() =

(4.41)

For an infinitesimal transformation we obtain

= g + (4.42)

with = (see eq. (3.93)). By exp