norbury. quantum field theory (wisconsin lecture notes, 2000)(107s)

QUANTUM FIELD THEORY

Professor John W. Norbury

Physics DepartmentUniversity of Wisconsin-Milwaukee

P.O. Box 413Milwaukee, WI 53201

November 20, 2000

Contents

1 Lagrangian Field Theory 71.1 Units . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

1.1.1 Natural Units . . . . . . . . . . . . . . . . . . . . . . . 71.1.2 Geometrical Units . . . . . . . . . . . . . . . . . . . . 10

1.2 Covariant and Contravariant vectors . . . . . . . . . . . . . . 111.3 Classical point particle mechanics . . . . . . . . . . . . . . . . 12

1.3.1 Euler-Lagrange equation . . . . . . . . . . . . . . . . . 121.3.2 Hamiltons equations . . . . . . . . . . . . . . . . . . . 14

1.4 Classical Field Theory . . . . . . . . . . . . . . . . . . . . . . 151.5 Noethers Theorem . . . . . . . . . . . . . . . . . . . . . . . . 181.6 Spacetime Symmetries . . . . . . . . . . . . . . . . . . . . . . 24

1.6.1 Invariance under Translation . . . . . . . . . . . . . . 241.6.2 Angular Momentum and Lorentz Transformations . . 25

1.7 Internal Symmetries . . . . . . . . . . . . . . . . . . . . . . . 261.8 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

1.8.1 Covariant and contravariant vectors . . . . . . . . . . 291.8.2 Classical point particle mechanics . . . . . . . . . . . . 291.8.3 Classical fleld theory . . . . . . . . . . . . . . . . . . . 291.8.4 Noethers theorem . . . . . . . . . . . . . . . . . . . . 30

1.9 References and Notes . . . . . . . . . . . . . . . . . . . . . . . 32

2 Symmetries & Group theory 332.1 Elements of Group Theory . . . . . . . . . . . . . . . . . . . . 332.2 SO(2) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

2.2.1 Transformation Properties of Fields . . . . . . . . . . 342.3 Representations of SO(2) and U(1) . . . . . . . . . . . . . . . 352.4 Representations of SO(3) and SU(1) . . . . . . . . . . . . . . 352.5 Representations of SO(N) . . . . . . . . . . . . . . . . . . . . 36

3

4 CONTENTS

3 Free Klein-Gordon Field 373.1 Klein-Gordon Equation . . . . . . . . . . . . . . . . . . . . . 373.2 Probability and Current . . . . . . . . . . . . . . . . . . . . . 39

3.2.1 Schrodinger equation . . . . . . . . . . . . . . . . . . . 393.2.2 Klein-Gordon Equation . . . . . . . . . . . . . . . . . 40

3.3 Classical Field Theory . . . . . . . . . . . . . . . . . . . . . . 413.4 Fourier Expansion & Momentum Space . . . . . . . . . . . . 423.5 Klein-Gordon QFT . . . . . . . . . . . . . . . . . . . . . . . . 45

3.5.1 Indirect Derivation of a; ay Commutators . . . . . . . 453.5.2 Direct Derivation of a; ay Commutators . . . . . . . . 473.5.3 Klein-Gordon QFT Hamiltonian . . . . . . . . . . . . 473.5.4 Normal order . . . . . . . . . . . . . . . . . . . . . . . 483.5.5 Wave Function . . . . . . . . . . . . . . . . . . . . . . 50

3.6 Propagator Theory . . . . . . . . . . . . . . . . . . . . . . . . 513.7 Complex Klein-Gordon Field . . . . . . . . . . . . . . . . . . 66

3.7.1 Charge and Complex Scalar Field . . . . . . . . . . . 683.8 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70

3.8.1 KG classical fleld . . . . . . . . . . . . . . . . . . . . . 703.8.2 Klein-Gordon Quantum fleld . . . . . . . . . . . . . . 713.8.3 Propagator Theory . . . . . . . . . . . . . . . . . . . . 723.8.4 Complex KG fleld . . . . . . . . . . . . . . . . . . . . 73


4 Dirac Field 754.1 Probability & Current . . . . . . . . . . . . . . . . . . . . . . 774.2 Bilinear Covariants . . . . . . . . . . . . . . . . . . . . . . . . 784.3 Negative Energy and Antiparticles . . . . . . . . . . . . . . . 79

4.3.1 Schrodinger Equation . . . . . . . . . . . . . . . . . . 794.3.2 Klein-Gordon Equation . . . . . . . . . . . . . . . . . 804.3.3 Dirac Equation . . . . . . . . . . . . . . . . . . . . . . 82

4.4 Free Particle Solutions of Dirac Equation . . . . . . . . . . . 834.5 Classical Dirac Field . . . . . . . . . . . . . . . . . . . . . . . 87

4.5.1 Noether spacetime current . . . . . . . . . . . . . . . . 874.5.2 Noether internal symmetry and charge . . . . . . . . . 874.5.3 Fourier expansion and momentum space . . . . . . . . 87

4.6 Dirac QFT . . . . . . . . . . . . . . . . . . . . . . . . . . . . 884.6.1 Derivation of b; by; d; dy Anticommutators . . . . . . . 88

4.7 Pauli Exclusion Principle . . . . . . . . . . . . . . . . . . . . 884.8 Hamiltonian, Momentum and Charge in terms of creation and

annihilation operators . . . . . . . . . . . . . . . . . . . . . . 88

CONTENTS 5

4.8.1 Hamiltonian . . . . . . . . . . . . . . . . . . . . . . . . 884.8.2 Momentum . . . . . . . . . . . . . . . . . . . . . . . . 884.8.3 Angular Momentum . . . . . . . . . . . . . . . . . . . 884.8.4 Charge . . . . . . . . . . . . . . . . . . . . . . . . . . 88

4.9 Propagator theory . . . . . . . . . . . . . . . . . . . . . . . . 884.10 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88

4.10.1 Dirac equation summary . . . . . . . . . . . . . . . . . 884.10.2 Classical Dirac fleld . . . . . . . . . . . . . . . . . . . 884.10.3 Dirac QFT . . . . . . . . . . . . . . . . . . . . . . . . 884.10.4 Propagator theory . . . . . . . . . . . . . . . . . . . . 88


5 Electromagnetic Field 895.1 Review of Classical Electrodynamics . . . . . . . . . . . . . . 89

5.1.1 Maxwell equations in tensor notation . . . . . . . . . . 895.1.2 Gauge theory . . . . . . . . . . . . . . . . . . . . . . . 895.1.3 Coulomb Gauge . . . . . . . . . . . . . . . . . . . . . 895.1.4 Lagrangian for EM fleld . . . . . . . . . . . . . . . . . 895.1.5 Polarization vectors . . . . . . . . . . . . . . . . . . . 895.1.6 Linear polarization vectors in Coulomb gauge . . . . . 915.1.7 Circular polarization vectors . . . . . . . . . . . . . . 915.1.8 Fourier expansion . . . . . . . . . . . . . . . . . . . . . 91

5.2 Quantized Maxwell fleld . . . . . . . . . . . . . . . . . . . . . 915.2.1 Creation & annihilation operators . . . . . . . . . . . 91

5.3 Photon propagator . . . . . . . . . . . . . . . . . . . . . . . . 915.4 Gupta-Bleuler quantization . . . . . . . . . . . . . . . . . . . 915.5 Proca fleld . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91

6 S-matrix, cross section & Wicks theorem 936.1 Schrodinger Time Evolution Operator . . . . . . . . . . . . . 93

6.1.1 Time Ordered Product . . . . . . . . . . . . . . . . . . 956.2 Schrodinger, Heisenberg and Dirac (Interaction) Pictures . . 96

6.2.1 Heisenberg Equation . . . . . . . . . . . . . . . . . . . 976.2.2 Interaction Picture . . . . . . . . . . . . . . . . . . . . 97

6.3 Cross section and S-matrix . . . . . . . . . . . . . . . . . . . 996.4 Wicks theorem . . . . . . . . . . . . . . . . . . . . . . . . . . 101

6.4.1 Contraction . . . . . . . . . . . . . . . . . . . . . . . . 1016.4.2 Statement of Wicks theorem . . . . . . . . . . . . . . 101

6 CONTENTS

7 QED 1037.1 QED Lagrangian . . . . . . . . . . . . . . . . . . . . . . . . . 1037.2 QED S-matrix . . . . . . . . . . . . . . . . . . . . . . . . . . 103

7.2.1 First order S-matrix . . . . . . . . . . . . . . . . . . . 1037.2.2 Second order S-matrix . . . . . . . . . . . . . . . . . . 1047.2.3 First order S-matrix elements . . . . . . . . . . . . . . 1067.2.4 Second order S-matrix elements . . . . . . . . . . . . . 1077.2.5 Invariant amplitude and lepton tensor . . . . . . . . . 107

7.3 Casimirs trick & Trace theorems . . . . . . . . . . . . . . . . 1077.3.1 Average over initial states / Sum over flnal states . . . 1077.3.2 Casimirs trick . . . . . . . . . . . . . . . . . . . . . . 107

Chapter 1

Lagrangian Field Theory

1.1 Units

We start with the most basic thing of all, namely units and concentrateon the units most widely used in particle physics and quantum fleld the-ory (natural units). We also mention the units used in General Relativity,because these days it is likely that students will study this subject as well.

Some useful quantities are [PPDB]:h h2 = 1:055 1034J sec = 6:582 1022MeV secc = 3 108 msec .1 eV = 1:6 1019Jhc = 197MeV fm1fm = 1015m1barn = 1028m2

1mb = :1fm2

1.1.1 Natural Units

In particle physics and quantum fleld theory we are usually dealing withparticles that are moving fast and are very small, i.e. the particles are bothrelativistic and quantum mechanical and therefore our formulas have lotsof factors of c (speed of light) and h (Plancks constant). The formulasconsiderably simplify if we choose a set of units, called natural units wherec and h are set equal to 1.

In CGS units (often also called Gaussian [Jackson appendix] units), thebasic quantities of length, mass and time are centimeters (cm), gram (g),seconds (sec), or in MKS units these are meters (m), kilogram (kg), seconds.In natural units the units of length, mass and time are all expressed in GeV.

7

8 CHAPTER 1. LAGRANGIAN FIELD THEORY

Example With c 1, show that sec = 3 1010cm.Solution c = 3 1010cm sec1. If c 1

) sec = 3 1010cm

We can now derive the other conversion factors for natural units, in which his also set equal to unity. Once the units of length and time are established,one can deduce the units of mass from E = mc2. These are

sec = 1:52 1024GeV 1m = 5:07 1015GeV 1kg = 5:61 1026GeV

(The exact values of c and h are listed in the [Particle Physics Booklet] asc = 2:99792458 108m=sec and h = 1:05457266 1034Jsec= 6:58212201025GeV sec. )

1.1. UNITS 9

Example Deduce the value of Newtons gravitational constantG in natural units.

Solution It is interesting to note that the value of G is one of theleast accurately known of the fundamental constants. Whereas,say the mass of the electron is known as [Particle Physics Book-let] me = 0:51099906MeV=c2 or the flne structure constant asfi = 1=137:0359895 and c and h are known to many decimalplaces as mentioned above, the best known value of G is [PPDB]G = 6:67259 1011m3kg1sec2, which contains far fewer dec-imal places than the other fundamental constants.

Lets now get to the problem. One simply substitutes the con-version factors from before, namely

G = 6:67 1011m3kg1sec2

=6:67 1011(5:07 1015GeV 1)3

(5:61 1026GeV )(1:52 1024GeV 1)2= 6:7 1039GeV 2

=1

M2Pl

where the Planck mass is deflned as MPl 1:22 1019GeV .

Natural units are also often used in cosmology and quantum gravity[Guidry 514] with G given above as G = 1

M2Pl

.


1.1.2 Geometrical Units

In classical General Relativity the constants c and G occur most often andgeometrical units are used with c and G set equal to unity. Recall that innatural units everything was expressed in terms of GeV . In geometricalunits everything is expressed in terms of cm.

Example Evaluate G when c 1.

Solution

G = 6:67 1011m3kg1sec2= 6:67 108cm3g1sec2

and when c 1 we have sec = 3 1010cm giving

G = 6:67 108cm3g1(3 1010cm)2= 7:4 1029cm g1

Now imposing G 1 gives the geometrical units

sec = 3 1010cmg = 7:4 1029cm

It is important to realize that geometrical and natural units are not com-patible. In natural units c = h = 1 and we deduce that G = 1

M2Pl

asin a previous Example. In geometrical units c = G = 1 we deduce thath = 2:6 1066cm2. (see Problems) Note that in these units h = L2Pl whereLPl 1:6 1033cm. In particle physics, gravity becomes important whenenergies (or masses) approach the Planck mass MPl. In gravitation (GeneralRelativity), quantum efiects become important at length scales approachingLPl.

1.2. COVARIANT AND CONTRAVARIANT VECTORS 11

1.2 Covariant and Contravariant vectors

The subject of covariant and contravariant vectors is discussed in [Jackson],which students should consult for a thorough introduction. In this sectionwe summarize the basic results.

The metric tensor that is used in this book is

= =

0BBB@1 0 0 00 1 0 00 0 1 00 0 0 1

1CCCAContravariant vectors are written in 4-dimensional form as

A = (Ao; Ai) = (Ao; ~A)

Covariant vectors are formed by \lowering" the indices with the metric ten-sor as in

A = gA = (Ao; Ai) = (Ao;Ai) = (Ao; ~A)

noting that

Ao = Ao

Thus

A = (Ao; ~A) A = (Ao; ~A)

Now we discuss derivative operators, denoted by the covariant symbol@ and deflned via

@

@x @ = (@o; @i) = ( @

@xo;@

@xi) = (

@

@t; ~5)

The contravariant operator @ is given by

@

@x @ = (@o; @i) = g@ = ( @

@xo;@

@xi) = (

@

@xo; @

@xi) = (

@

@t;~5)

Thus

@ = ( @@t ; ~5) @ = ( @@t ;~5)


The length squared of our 4-vectors is

A2 AA = AA = A2o ~A2

and

@2 @@ 22 = @2

@t252 (1.1)

Finally, note that with our 4-vector notation, the usual quantum mechanicalreplacements

pi ! ih@i ih~5and

po ! ih@o = ih @@t

can be succintly written as

p ! ih@

giving (with h = 1)

p2 ! 22

1.3 Classical point particle mechanics

1.3.1 Euler-Lagrange equation

Newtons second law of motion is

~F =d~p

dt

or in component form (for each component Fi)

Fi =dpidt

where pi = m _qi (with qi being the generalized position coordinate) so thatdpidt = _m _qi + mqi. (Here and throughout this book we use the notation_x dxdt .) If _m = 0 then Fi = mqi = mai. For conservative forces ~F = ~5Uwhere U is the scalar potential. Rewriting Newtons law we have

dUdqi

=d

dt(m _qi)

1.3. CLASSICAL POINT PARTICLE MECHANICS 13

Let us deflne the Lagrangian L(qi; _qi) T U where T is the kinetic energy.In freshman physics T = T ( _qi) = 12m _q

2i and U = U(qi) such as the harmonic

oscillator U(qi) = 12kq2i . That is in freshman physics T is a function only

of velocity _qi and U is a function only of position qi. Thus L(qi; _qi) =T ( _qi) U(qi). It follows that @L@qi = dUdqi and @L@ _qi = dTd _qi = m _qi = pi. ThusNewtons law is

Fi =dpidt

@L

@qi=

d

dt(@L

@ _qi)

with the canonical momentum deflned as

pi @L@ _qi

The next to previous equation is known as the Euler-Lagrange equation ofmotion and serves as an alternative formulation of mechanics [Goldstein]. Itis usually written

d

dt(@L

@ _qi) @L

@qi= 0

or just

_pi =@L

@qi

We have obtained the Euler-Lagrange equations using simple arguments.A more rigorous derivation is based on the calculus of variations [Ho-Kim47,Huang54,Goldstein37, Bergstrom284] as follows.

In classical point particle mechanics the action is

S Z t2t1L(qi; _qi; t)dt

where the Lagrangian L is a function of generalized coordinates qi, general-ized velocities _qi and time t.

According to Hamiltons principle, the action has a stationary value forthe correct path of the motion [Goldstein36], i.e. S = 0 for the correct path.To see the consequences of this, consider a variation of the path [Schwabl262,BjRQF6]

qi(t)! q0i(t) qi(t) + qi(t)


subject to the constraint qi(t1) = qi(t2) = 0. The subsequent variation inthe action is (assuming that L is not an explicit function of t)

S =Z t2t1

(@L

@qiqi +

@L

@ _qi _qi)dt = 0

with _qi = ddtqi and integrating the second term by parts yieldsZ@L

@ _qi _qidt =

Z@L

@ _qid(qi) (1.2)

=@L

@ _qiqijt2t1

Zqid(

@L

@ _qi)

= 0Zqi

d

dt(@L

@ _qi)dt

where the boundary term has vanished because qi(t1) = qi(t2) = 0. Weare left with

S =Z t2t1

@L

@qi ddt

(@L

@ _qi)qidt = 0

which is true for an arbitrary variation qi indicating that the integral mustbe zero, which yields the Euler-Lagrange equations.

1.3.2 Hamiltons equations

We now introduce the Hamiltonian H deflned as a function of p and q as

H(pi; qi) pi _qi L(qi; _qi) (1.3)

For the simple case T = 12m _q2i and U 6= U( _qi) we have pi @L@ _qi = m _qi so that

T = p2i

2m and pi _qi =p2im so that H(pi; qi) =

p2i2m + U(qi) = T + U which is the

total energy. Hamiltons equations of motion immediately follow as

@H

@pi= _qi

Now L 6= L(pi) and @H@qi = @L@qi so that our original deflnition of the canon-ical momentum above gives

@H@qi

= _pi

1.4. CLASSICAL FIELD THEORY 15

1.4 Classical Field Theory

Scalar flelds are important in cosmology as they are thought to drive in-ation. Such a fleld is called an inaton, an example of which may be theHiggs boson. Thus the fleld ` considered below can be thought of as aninaton, a Higgs boson or any other scalar boson.

In both special and general relativity we always seek covariant equationsin which space and time are given equal status. The Euler-Lagrange equa-tions above are clearly not covariant because special emphasis is placed ontime via the _qi and ddt(

@L@ _qi

) terms.Let us replace the qi by a fleld ` `(x) where x (t;x). The generalized

coordiante q has been replaced by the fleld variable ` and the discrete indexi has been replaced by a continuously varying index x. In the next sectionwe shall show how to derive the Euler-Lagrange equations from the actiondeflned as

S ZLdt

which again is clearly not covariant. A covariant form of the action wouldinvolve a Lagrangian density L via

S ZLd4x =

ZLd3xdt

where L = L(`; @`) and with L R Ld3x. The term @L@qi in the Euler-

Lagrange equation gets replaced by the covariant term @L@`(x) . Any timederivative ddt should be replaced with @ @@x which contains space as wellas time derivatives. Thus one can guess that the covariant generalization ofthe point particle Euler-Lagrange equation is

@@L

@(@`) @L@`

= 0

which is the covariant Euler-Lagrange equation for a fleld `. If there is morethan one scalar fleld ì then the Euler-Lagrange equations are

@@L

@(@ì) @L@ì = 0

To derive the Euler-Lagrange equations for a scalar fleld [Ho-Kim48, Gold-stein548], consider an arbitrary variation of the fleld [Schwabl 263; Ryder83; Mandl & Shaw 30,35,39; BjRQF13]

`(x)! `0(x) `(x) + `(x)


again with ` = 0 at the end points. The variation of the action is (assumingthat L is not an explicit function of x)

S =Z X2X1

"@L@`

`+@L

@(@`)(@`)

#d4x = 0

where X1 and X2 are the 4-surfaces over which the integration is performed.We need the result

(@`) = @` =@

@x`

which comes about because `(x) = `0(x) `(x) giving

@`(x) = @`0(x) @`(x) = @`(x)

showing that commutes with difierentiation @. Integration by parts onthe second term is a bit more complicated than before for the point particlecase, but the flnal result is (see Problems)

S =Z X2X1

"@L@` @ @L

@(@`)

#` d4x = 0

which holds for arbitrary `, implying that the integrand must be zero,yielding the Euler-Lagrange equations.

In analogy with the canonical momentum in point particle mechanics,we deflne the covariant momentum density

@L@(@`)

so that the Euler-Lagrange equations become

@ =@L@`

The canonical momentum is deflned as

0 = @L@ _`

The energy momentum tensor is (analagous to the deflnition of the pointparticle Hamiltonian)

T @` gL


with the Hamiltonian density

H ZHd3x

H T00 = _` LIn order to illustrate the foregoing theory we shall use the example of theclassical, massive Klein-Gordon fleld.

Example The massive Klein-Gordon Lagrangian density is

LKG = 12(@`@`m2`2)

=12

[ _`2 (5`)2 m2`2]

A) Derive expressions for the covariant momentum density andthe canonical momentum.B) Derive the equation of motion in position space and momen-tum space.C) Derive expressions for the energy-momentum tensor and theHamiltonian density.

Solution A) The covariant momentum density is more easilyevaluated by re-writing LKG = 12(g@`@` m2`2). Thus = @L@(@`) =

12g(fi@` + @`

fi ) =

12(

fi@

` + @`fi ) =12(@

fi`+ @fi`) = @fi`. Thus for the Klein-Gordon fleld we have

fi = @fi`

giving the canonical momentum = 0 = @0` = @0` = _`,

= _`

B) Evaluating @L@` = m2`, the Euler-Lagrange equations givethe fleld equation as @@`+m2` or

(22 +m2)` = 0`52`+m2` = 0

which is the Klein-Gordon equation for a free, massive scalarfleld. In momentum space p2 = 22, thus

(p2 m2)` = 0


(Note that some authors [Muirhead] deflne 22 52 @2@t2

dif-ferent from (1.1), so that they write the Klein-Gordon equationas (22 m2)` = 0 or (p2 +m2)` = 0.)C) The energy momentum tensor is

T @` gL= @`@` gL

= @`@` 12g(@fi`@fi`m2`2)

Therefore the Hamiltonian density isH T00 = _`2 12(@fi`@fi`m2`2) which becomes [Leon]

H = 12

_`2 +12

(5`)2 + 12m2`2

=12

[2 + (5`)2 +m2`2]

1.5 Noethers Theorem

Noethers theorem provides a general and powerful method for discussingsymmetries of the action and Lagrangian and directly relating these sym-metries to conservation laws.

Many books [Kaku] discuss Noethers theorem in a piecemeal fashion,for example by treating internal and spacetime symmetries separately. Itis better to develop the formalism for all types of symmetries and then toextract out the spacetime and internal symmetries as special cases. Thebest discussion of this approach is in [Goldstein, Section 12-7, pg. 588]and [Greiner FQ, section 2.4, pg. 39]. Another excellent discussion of thisgeneral approach is presented in [Schwabl, section 12.4.2, pg. 268]. Howevernote that the discussion presented by [Schwabl] concerns itself only with thesymmetries of the Lagrangian, although the general spacetime and internalsymmetries are properly treated together. The discussions by [Goldstein]and [Greiner] treat the symmetries of both the Lagrangian and the actionas well.

In what follows we rely on the methods presented by [Goldstein].The theory below closely follows [Greiner FQ 40]. We prefer to use the

notation of [Goldstein] for flelds, namely r(x) or r(x) rather than using

1.5. NOETHERS THEOREM 19

`r(x) or r(x) because the latter notations might make us think of scalaror spinor flelds. The notation r(x) is completely general and can refer toscalar, spinor or vector fleld components.

We wish to consider how the Lagrangian and action change under acoordinate transformation

x ! x0 x + xLet the corresponding change in the fleld (total variation) be [Ryder83,Schwabl263]

0r(x0) r(x) + r(x)

and the corresponding change in the Lagrangian

L0(x0) L(x) + L(x)

withL(x) L((x); @(x); x)

where @(x) @(x)@x and 1

L0(x0) L(0(x0); @00(x0); x0)

(Note: no prime on L on right hand side) where @00(x0) @0(x0)@x0Notice that the variations deflned above involve two transformations,

namely the change in coordinates from x to x0 and also the change in theshape of the function from to 0.

However there are other transformations (such as internal symmetries orgauge symmetries) that change the shape of the wave function at a singlepoint. Thus the local variation is deflned as (same as before)

0r(x) r(x) + r(x)1This follows from the assumption of form invariance [Goldstein 589]. In general the

Lagrangian gets changed to

L(r(x); @r(x); x)! L0(0r(x0); @00r(x0); x0)with @0 @@x0The assumption of form invariance [Goldstein 589] says that the Lagrangian has the samefunctional form in terms of the transformed quantities as it does in the original quantities,namely

L0(0r(x0); @00r(x0); x0) = L(0r(x0); @00r(x0); x0)


The local and total variations are related via

r(x) = 0r(x) r(x)= 0r(x) 0r(x0) + 0r(x0) r(x)= [0r(x0) 0r(x)] + r(x)

Recall the Taylor series expansion

f(x) = f(a) + (x a)f 0(a) + :::= f(a) + (x a)@f(x)

@xjx=a + :::

or

f(x) f(a) (x a)@f@a

which gives

(x0) (x) (x0 x)@(x0)

@x0jx0=x

(x0 x)@@x

= x@

@x

Thus

r(x) = r(x) @0r

@xx

To lowest order 0r r. We do this because the second term is second orderinvolving both @0 and x. Thus flnally we have the relation between thetotal and local variations as (to flrst order)

r(x) = r(x) @r@x

x

Now we ask whether the variations and commute with difierentia-tion. (It turns out does commute but does not.) From the deflnition(x) 0(x) (x) it is obvious that (see before)

@

@x(x) =

@(x)@x


showing that \commutes" with @ @@x . However does not commute,but has an additional term, as in (see Problems) [Greiner FQ41]

@

@x(x) =

@(x)@x

+@(x)@x

@x

@x

Let us now study invariance of the action [Goldstein 589, Greiner FQ41]. The assumption of scale invariance [Goldtein 589] says that the actionis invariant under the transformation 2 (i.e. transformation of an ignorableor cyclic coordinate)

S0 Z

0d4x0 L0(0r(x0); @00r(x0); x0)

=Z

d4x L(r(x); @r(x); x) S

Demanding that the action is invariant, we have (in shorthand notation)

S Z

0d4x0 L0(x0)

Zd4xL(x) 0

Note that this S is deflned difierently to the S that we used in the deriva-tion of the Euler-Lagrange equations. Using L0(x0) L(x) + L(x) gives

S Z

0d4x0L(x) +

Z0d4x0 L(x)

Zd4xL(x) = 0

We transform the volume element with the Jacobian

d4x0 =flflflfl@x0 @x

flflflfl d4xusing x0 = x + x which gives [Greiner FQ 41]

2Combining both form invariance and scale invariance gives [Goldstein 589]

S S0 S =Z

0d4x0 L(0r(x0); @00r(x0); x0)

Z

d4xL(r(x); @r(x); x) = 0

In the flrst integral x0 is just a dummy variable so thatR0 d

4x L(0r(x); @0r(x); x)R

d4x L(r(x); @r(x); x) = 0

which [Goldstein] uses to derive current conservation.


d4x0 =flflflfl@x0 @x

flflflfl d4x =flflflflflflflflfl

1 + @x0

@x0@x0

@x1: : :

@x1

@x01 + @x

1

@x1...

... : : : 1 + @x3

@x3

flflflflflflflflfl d4x

= (1 +@x

@x)d4x

to flrst order only. Thus the variation in the action becomes

S =Z

(1 +

@x

@x)d4x L(x) +

Z

(1 +@x

@x)d4x L(x)

Zd4x L(x)

=Z

d4x L(x) +

Zd4x L(x)@x

@x

to flrst order. The second order term @x

@x L(x) has been discarded. Usingthe relation between local and total variations gives

S =Z

d4x (L(x) + @L

@xx) +

Zd4x L(x)@x

@x

=Z

d4x fL(x) + @

@x[L(x)x]g

Recall that L(x) L(r(x); @r(x)). Now express the local variation L interms of total variations of the fleld as

L = @L@r

r +@L

@(@r)(@r)

= " +@L

@(@r)@ r

because \commutes" with @ @@x . Now add zero,

L = @L@r

r @

@L@(@r)

r +

@

@L@(@r)

r +

@L@(@r)

@r

=@L@r @ @L

@(@r)

r + @

@L

@(@r)r

Note: the summation convention is being used for both and r. This ex-pression for L is substituted back into S = 0, but because the region of


integration is abritrary, the integrand itself has to vanish. Thus the inte-grand is

@L@r @ @L

@(@r)

r + @

@L

@(@r)r + Lx

= 0

The flrst term is just the Euler-Lagrage equation which vanishes. For ruse the relation between local and total variations, so that the second termbecomes

@

@L@(@r)

r @r

@xx

+ Lx

= 0

which is the continuity equation

@ j = 0

with [Schwabl 270]

j @L@(@r)

r

@L@(@r)

@r gLx

@L@(@r)

r Tx

with the energy-momentum tensor deflned as [Scwabl 270]

T @L@(@r)

@r gL

The corresponding conserved charge is (See Problems)

Q Zd3x j0(x)

such thatdQ

dt= 0

Thus we have j0(x) is just the charge density

j0(x) (x)This leads us to the statement,

Noethers Theorem: Each continuous symmetry transformationthat leaves the Lagrangian invariant is associated with a con-served current. The spatial integral over this currents zero com-ponent yields a conserved charge. [Mosel 16]


1.6 Spacetime Symmetries

The symmetries we will consider are spacetime symmetries and internal sym-metries. Super symmetries relate both of these. The simplest spacetimesymmetry is 4-dimensional translation invariance, involving space transla-tion and time translation.

1.6.1 Invariance under Translation

[GreinerFQ 43] Consider translation by a constant factor ,

x0 = x +

and comparing with x0 = x + x gives x = . The shape of the flelddoes not change, so that r = 0 (which is properly justifled in Schwabl270) giving the current as

j =

@L@(@r)

@r@x gL

with @j = 0. Dropping ofi the constant factor lets us write down amodifled current (called the energy-momentum tensor)

T @L@(@r)

@r gL

with@T = 0

In general j has a conserved charge Q Rd3x jo(x). Thus T will have

4 conserved charges corresponding to T00; T01; T02; T03 which are justthe energy E and momentum ~P of the fleld. In 4-dimensional notation[GreinerFQ 43]

P = (E; ~P ) =Zd3x T 0 = constant:

withdP

dt= 0

The above expression for T is the same result we obtained before wherewe wrote

T = @` gL (1.4)with

=@L

@(@`)(1.5)

1.6. SPACETIME SYMMETRIES 25

1.6.2 Angular Momentum and Lorentz Transformations

NNN: below is old Kaku notes. Need to revise; Schwabl and Greiner arebest (they do J=L+S)

Instead of a simple translation xi = ai now consider a rotation xi =aijxj . Lorentz transformations are a generalisation of this rotation, namelyx = x . Before for spacetime translations we had x = a andtherefore ` = @`@x x

= x@` = a@`. Copying this, the Lorentztransformation is

x = x

` = x@`@` = x@@`

Now repeat same step as before, and we get the conserved current

M = T x T x

with@ M = 0

and the conserved charge

M =Zd3x M0

withd

dtM = 0

For rotations in 3-d space, the ddtMij = 0 corresponds to conservation of

angular momentum.


1.7 Internal Symmetries

[Guidry 91-92]One of the important theorems in Lie groups is the following :

Theorem: Compact Lie groups can always be represented byflnite-dimensional unitary operators. [Tung p.173,190]

Thus using the notation U(fi1; fi2; :::fiN ) for an element of an N-parameterLie group (fii are the group parameters), we can write any group element as

U(fi1; fi2; :::fiN ) = eifiiXi

1 + iiXi + :::

where the latter approximation is for inflnitessimal group elements fii = i.The Xi are linearly independent Hermitian operators (there are N of them)which satisfy the Lie algebra

[Xi; Xj ] = i fijkXk

where fijk are the structure constants of the group.These group elements act on wave functions, as in [Schwabl 272]

(x)! 0(x) = eifiiXi(x) (1 + iiXi)(x)

giving(x) = 0(x) (x) = iiXi(x)

Consider the Dirac equation (i /@ m) = 0 with 4-current, j = .This is derived from the Lagrangian L = (i /@ m) where y0.Now, for fii = constant, the Dirac Lagrangian L is invariant under thetransformation ! 0 = eifiiXi. This is the signiflcance of group theoryin quantum mechanics. Noethers theorem now tells us tht we can flnd acorresponding conserved charge and conserved current.

The Noether current, with x = 0 and therefore (x) = (x) [Schw-abl 272], is

j =@L

@(@r)r

=@L

@(@r)ii Xi r

1.7. INTERNAL SYMMETRIES 27

and again dropping ofi the constant factor i deflne a new curent (and throwin a minus sign so that we get a positive current in the example below)

ji = @L

@(@r)i Xi r

which obeys a continuity equation

@ ji = 0

Example Calculate ji for the isospin transformation eifiiXi for the Dirac

Lagrangian L = (i /@ m) (i@ m)Solution

ji = @L

@(@)iXi

and@L

@(@)= i

givingji = i iXi

orji =

Xi

where Xi is the group generator. (Compare this to the ordinary Diracprobabilty current j = (;~j ) = ).


From the previous example we can readily display conservation of charge.Write the U(1) group elements as

U() = eiq

where the charge q is the generator. Thus the conserved current is

j = q

where J = (;~j ) = is just the probability current for the Diracequation. The conserved charge is [Mosel 17,34]

Q =Zd3x j0 = q

Zd3x 0

= qZd3x

(Note that = 0 = y00 = y = 4i=1jij2 where i is eachcomponent of . Thus is positive deflnite.)

See also [Mosel 17,34; BjRQM 9]. Note that if is normalized so that[Strange 123] Z

d3x =Zd3xy = 1

then we have Q = q as required. This is explained very clearly in [Gross122-124]. However often difierent normalizations are used for the Diracwave functions [Muirhead 72, Halzen & Martin 110]. For example [Halzen& Martin 110] haveZ

d3x =Zd3xy = uyu = 2E

See also [Griths 223].

1.8. SUMMARY 29

1.8 Summary

1.8.1 Covariant and contravariant vectors

Contravariant and covariant vectors and operators are

A = (Ao; ~A) A = (Ao; ~A)and

@ = ( @@t ; ~5) @ = ( @@t ;~5)

1.8.2 Classical point particle mechanics

The point particle canonical momentum is

pi =@L

@ _qi

and the EL equations are

d

dt(@L

@ _qi) @L

@qi= 0

The point particle Hamiltonian is

H(pi; qi) pi _qi L(qi; _qi) (1.6)giving Hamiltons equations

@H

@pi= _qi @H

@qi= _pi

1.8.3 Classical fleld theory

For classical flelds ì, the EL equations are

@@L

@(@ì) @L@ì = 0

The covariant momentum density is

@L@(@`)

and the canonical momentum is

0 = @L@ _`


The energy momentum tensor is (analagous to point particle Hamiltonian)

T @` gL

with the Hamiltonian density

H T00 = _` L

1.8.4 Noethers theorem

The conserved (@j = 0) Noether current is

j @L@(@r)

r Tx

with

T @L@(@r)

@r gL

The conserved (dQdt = 0) charge is

Q Zd3x j0(x)

which is just the charge density,

j0(x) (x)

If we consider the spacetime symmetry involving invariance under transla-tion then we can derive T from j. The result for T agrees with thatgiven above. For the Klein-Gordon Lagrangian this becomes

T = @`@` gL (1.7)

The momentum is (with E H)

P = (H; ~P ) =Zd3x T 0 = constant (1.8)

where dP

dt = 0.For internal symmetries the group elements can be written

U(fi1; fi2; :::fiN ) = eifiiXi 1 + iiXi + :::

1.8. SUMMARY 31

where[Xi; Xj ] = i fijkXk

The group elements act on wave functions

! 0 = eifiiXi (1 + iiXi)

giving = 0 = iiXi

The Noether current, for an internal symmetry (x = 0 and therefore = ) becomes

ji = @L

@(@r)i Xi r

For the Dirac Lagrangian, invariant under eifiiXi , with fii = constant,this becomes

ji = Xi


1.9 References and Notes

General references for units (Section 1.1) are [Aitchison and Hey, pg.526-531; Halzen and Martin, pg. 12-13; Guidry, pg.511-514; Mandland Shaw, pg. 96-97; Griths, pg. 345; Jackson, pg. 811-821; Misner,Thorne and Wheeler, pg. 35-36]. References for Natural units (Section1.1.1) are [Guidry, pg. 511-514; Mandl and Shaw, pg. 96-97; Griths, pg.345; Aitchison and Hey, pg. 526-531; Halzen and Martin, pg. 12-13] andreferences for Geometrical units (Section 1.1.2) are [Guidry, pg. 514; MTW,pg. 35-36].

For a complete introduction to covariant and contravariant vectors see[Jackson]

The most commonly used metric is

=

0BBB@1 0 0 00 1 0 00 0 1 00 0 0 1

1CCCAThis metric is used throughout the present book and is also used by thefollowing authors: [Greiner, Aitchison & Hey, Kaku, Peskin & Schroeder,Ryder, Bjorken and Drell, Mandl & Shaw, Itzykson & Zuber, Sterman,Chang, Guidry, Griths, Halzen & Martin and Gross].

Another less commonly used metric is

=

0BBB@1 0 0 00 1 0 00 0 1 00 0 0 1

1CCCAThis metric is not used in the present book, but it is used by [Weinberg andMuirhead].

The best references for the classical fleld ELE and Noethers theoremare [Schwabl, Ryder]

Chapter 2

Symmetries & Group theory

2.1 Elements of Group Theory

SUSY nontrivially combines both spacetime and internal symmetries.

2.2 SO(2)

In SO(2) the invariant is x2 + y2. We writex0

y0

!=

cos sin sin cos

!xy

!

or

xi0

= Oijxj

For small angles this is reduced to

x = y and y = x

or

xi = ijxj

where ij is antisymmetric and 12 = 21 = 1

Read Kaku p. 36-38

33

34 CHAPTER 2. SYMMETRIES & GROUP THEORY

2.2.1 Transformation Properties of Fields

References: [Kaku 38; Greiner FQ 95, 96; Elbaz 192; Ho-Kim 28; Mosel 21]

Consider a transformation U which transforms a quantum state [Elbazl92] flflfi0 > U flflfi >

< fi0 j=< fijU yTo conserve the norm, we impose

< fi0flflfl0 >=< fiflflfl >

which means that U is unitary, i.e. UU y = 1 implying U y = U1.Now consider transformation of an operator O, with expectation value

< O >< fijOjfi >=< fi0jO0jfi0 >This gives

< O >=< fijOjfi >=< fijU yO0U jfi >giving the transformation rule for the operator as

O = U yO0U

orO0 = UOU y

To summarize, if a quantum state transforms as [Elbaz 192]

0 = U

orjfi0 >= U jfi >

then an operator transforms as

O0 = UOU y

Fields can be grouped into difierent categories [Mosel 21] according to theirbehavior under general Lorentz transformations, which include spatial rota-tion, Lorentz boost transformations and also the discrete transformations ofspace reection, time reversal and space-time reection. The general Lorentztransformation is written [Mosel 21, Kaku 50]

x0 = x (2.1)

2.3. REPRESENTATIONS OF SO(2) AND U(1) 35

but lets write it more generally (in case we consider other transformations)

x0 = ax (2.2)

or a for rotation, boost, space inversion is very nicely discussed in[Ho-Kim 19-22]. A scalar fleld transforms under (2.1) or (2.2) as

`0(x0) = `(x)

[Mosel 21; Ho-Kim 26; Greiner 95, 96]. Now if `(x) is an operator then itstransformation is also written as [Greiner 96, Kaku 38]

`0(x0) = U`(x0)U y

Thus a scalar fleld transforms under (2.2) as

`0(x0) = U`(x0)U y = `(x)

A vector fleld transforms as [Ho-Kim 30; Kaku 39]

`0(x0) = a`(x) U`(x0)U y

i.e. it just transforms in the same way as an ordinary 4-vector.

2.3 Representations of SO(2) and U(1)

Read Kaku 39-42, 741-748

SO(2) can be deflned as the set of transformations that leave x2 + y2 invari-ant. There is a homomorphism between SO(2) and U(1). A U(1) transfor-mation can be written 0 = ei = U: This will leave the inner product` invariant. Thus the group U(1) can be deflned as the set of transfor-mations that leave ` invariant [Kaku 742, Peskin 496]

2.4 Representations of SO(3) and SU(1)

SO(3) leaves x2 + x2 + z2 invariant.

Read Kaku 42-45.

36 CHAPTER 2. SYMMETRIES & GROUP THEORY

2.5 Representations of SO(N)

Students should read rest of chapter in Kaku.

NNN NOW do a.m.

Chapter 3

Free Klein-Gordon Field

NNN write general introduction

3.1 Klein-Gordon Equation

Relativistic Quantum Mechanics (RQM) is the subject of studying relativis-tic wave equations to replace the non-relativistic Schrodinger equation. Thetwo prime relativistic wave equations are the Klein-Gordon equation (KGE)and the Dirac equation. (However these are only valid for 1-particle prob-lems whereas the Schrodinger equation can be written for many particles.)

Our quantum wave equation (both relativistic and non-relativistic) iswritten

H^ = E^

withH^ T^ + U^

andE^ = ih

@

@t

Non-relativistically we have T^ = p^2

2m with p^ = ih~r giving h

2

2mr2 + U

! = ih

@

@t

The time-independent Schrodinger equation simply has Eb instead of E^where Eb is the binding energy, i.e.

h2

2mr2 + U

! = Eb

37

38 CHAPTER 3. FREE KLEIN-GORDON FIELD

However in special relativity we have (with c = 1)

E = T +m

andE2 = p2 +m2

giving

T =qp2 +m2 m

With the replacement ~p! ih~r, the relativistic version of the free particle(U = 0) Schrodinger equation would be (T^ = Eb)q

h2r2 +m2 m = Eb

In the non-relativistic case E = Eb, but relativistically E = Eb +m, givingqh2r2 +m2 = E

which is called the Spinless Salpeter equation. There are two problems withthis equation; flrstly the operator

ph2r2 +m2 is non-local [Landau 221]

making it very dicult to work with in coordinate space (but actually itseasy in momentum space) and secondly the equation is not manifestly co-variant [Gross, pg. 92]. Squaring the Spinless Salpeter operator gives theKlein-Gordon equation (KGE)

(h2r2 +m2)` = E2`= h2 @

2

@t2`

or (with h = c = 1, see Halzen & Martin, pg. 12, 13; Aitchison & Hey, pg.526-528)

`r2`+m2` = 0Recall the wave equation (with y00 @2y=@x2, y @2y=@t2)

y00 1c2y = 0

where c is the wave velocity. Thus the KGE (c 6= 1) is (with r2 = @2=@x2)

`00 1c2

` = m2`

3.2. PROBABILITY AND CURRENT 39

which is like a massive (inhomogeneous) wave equation. The KGE is writtenin manifestly covariant form as

(22 +m2)` = 0

which in momentum space is (using p2 ! 22)

(p2 m2)` = 0

A quick route to the KGE is with the relativistic formula p2 pp = m2(6= ~p2) giving p2 m2 = 0 and (p2 m2)` = 0 and p2 ! 22 giving(22+m2)` = 0. The KGE can be written in terms of 4-vectors, (p2m2)` =0 and is therefore manifestly covariant. Finally, note that the KGE is a 1-particle equation!

(Note that some authors [Muirhead] use g =

0BBB@1

11

1

1CCCA andso have 22 r2 @2

@t2and (22 m2)` = 0 or (p2 + m2)` = 0 for the

Klein-Gordon equation.)

3.2 Probability and Current

The Klein-Gordon equation was historically rejected because it predicted anegative probability density. In order to see this lets flrst review probabilityand current for the Schrodinger equation. Then the KG example will beeasier to understand.

3.2.1 Schrodinger equation

The free particle Schrodinger equation (SE) is

h2

2mr2 = ih@

@t

The complex conjugate equation is (SE)

h2

2mr2 = ih@

@t

Multiply SE by and SE by

h2

2mr2 = ih@

@t


h2

2mr2 = ih@

@t

and subtract these equations to give

h2

2m(r2 r2) = ih

@

@t+

@

@t

= ih

@

@t()

= h2

2m~r [~r (~r)]

or@

@t() +

h2mi

~r [~r (~r)] = 0

which is just the continuity equation @@t + ~r ~j = 0 if

~j h2mi

[~r (~r)]

which are the probability density and current for the Schrodinger equation.

3.2.2 Klein-Gordon Equation

The free particle KGE is (22 +m2)` = 0 or (using 22 = @2

@t2r2)

@2`

@t2r2`+m2` = 0

and the complex conjugate equaiton is (KGE)

@2`

@t2r2` +m2` = 0

Multiplying KGE by ` and KGE by ` gives

`@2`

@t2 `r2`+m2`` = 0

`@2`

@t2 `r2` +m2`` = 0

and subtract these equations to give

`@2`

@t2 `@

2`

@t2 `r2`+ `r2` = 0


=@

@t

`@`

@t `@`

@t

~r

h`~r` `~r`

iwhich is the continuity equation @=@t+ ~r ~j = 0 if

`@`@t `@`

@t

~j `~r` `~r`or j = `@` `@`

but to get this to match the SE wave function we should deflne ~j in thesame way, i.e.

~j h2mi

h`~r` `~r`

i=ih2m

`~r` `~r`

which is h2mi times ~j above. Thus for

@@t + ~r ~j = 0 to hold we must have

ih2m

`@`

@t `@`

@t

The problem with (in both expressions above) is that it is not positivedeflnite and therefore cannot be interpreted as a probability density. Thisis one reason why the KGE was discarded. (Note: because can be eitherpositive or negative it can be interpreted as a charge density. See [Landau,pg. 227]

Notice how this problematic comes about because the KGE is 2nd orderin time. We have @@t and itself constains

@@t ; this does not happen with

the SE or DE.By the wave, note that we can form a 4-vector KG current

j =ih2m

(`@` `@`)

3.3 Classical Field Theory

Reference: [Schwabl, Chapter 13; Kaku, Chapter 3]Some of the key results for the free and real Klein-Gordon fleld were

worked out in an Example in Chapter 1. Lets remind outselves of theseresults.


The massive Klein-Gordon Lagrangian was

LKG = 12(@`@`m2`2)

which gave the equation of motion in position and momentum space as

(22 +m2)` = 0(p2 m2)` = 0

The covariant momentum density was

= @`

giving the canonical momentum

o = _`(x)

The Hamiltonian density was (with H R d3xH)H = 1

2[2 + (~r`)2 +m2`2]

Finally, the momentum operator of the Klein-Gordon fleld is (see Problems)

~P = Zd3x _`(x)~r`(x)

3.4 Fourier Expansion & Momentum Space

[See Greiner FQ 76 , Jose and Saletin 589]As with the non-relativistic case we expand plane wave states as

`(~x; t) =Zd~k a(~k; t)ei~k~x

but now withd~k = Nkd3k

where Nk is a normalization constant, to be determined later.Substitute `(~r; t) into the KGE

22 +m2` = 0

@2t r2 +m2` = 0

3.4. FOURIER EXPANSION & MOMENTUM SPACE 43

giving Zd~ka+ k2a+m2a

ei~k~x = 0

Deflning

! !(~k) = pk2 +m2and requiring the integrand to be zero gives

a+ !2a = 0

which is a 2nd order difierential equation, with Auxilliary equation

r2 + !2 = 0

orr = i!

The two solutions are crucial ! They can be interpreted as positive andnegative energy, or as particle and antiparticle. In the non-relativistic (NR)case we only got one solution. Thus [Teller, pg. 67]

a(~k; t) = c(~k)ei!t + a(~k)ei!t

giving our original expansion as

`(~x; t) =Zd~kha(~k)ei(~k~x!t) + c(~k)ei(~k~x+!t)

i(Remember that ! !(~k) = pk2 +m2) Now the Schrodinger wave function(~x; t) is complex but the KG wave function `(~x; t) is real. (The Schrodingerequation has an i in it, but the KGE does not.) Thus [Greiner, pg. 77]

`(~x; t) = `(~x; t) = `y(~x; t)

giving Zd~kha(~k)ei(~k~x!t) + c(~k)ei(~k~x+!t)

i=Zd~khay(~k)ei(~k~x!t) + cy(~k)ei(~k~x+!t)

iRe-write this as Z

d~kha(~k)ei(~k~x!t) + c(~k)ei(~k~x!t)

i=Zd~khay(~k)ei(~k~x!t) + cy(~k)ei(~k~x!t)

i


where we have made the substitution ~k ! ~k in two terms.(This does not mean k ! k; it cannot. k is the magnitude of ~k. We

are simply reversing the direction (magnitude) of ~k. This does not afiect thevolume element d3k.)

Obviously thenc(~k) = ay(~k)

cy(~k) = a(~k)Re-writing our expansion as

`(~x; t) =Zd~kha(~k)ei(~k~x!t) + c(~k)ei(~k~x!t)

igives [same conventions as Kaku 152, Mandl and Shaw 44, Schwabl 278]

`(~x; t) `+(x) + `(x)

=Zd~kha(~k)ei(~k~x!t) + ay(~k)ei(~k~x!t)

i

=Zd~kha(~k)eikx + ay(~k)eikx

i

=X~k

1p2!V

ha~keikx + ay~ke

ikxi

(3.1)

[see Greiner FQ 79 for box vs. continuum normalization] where k x kx = koxo ~k ~x = !t ~k ~x.

The conjugate momentum is

(~x; t) = _`(~x; t) = iZd~k !

ha(~k)ei(~k~x!t) ay(~k)ei(~k~x!t)

i

= iZd~k !

ha(~k)eikx ay(~k)eikx

i +(x) + (x)

(3.2)

Note that we are still studying classical fleld theory when we Fourier expandthese classical flelds [Goldstein 568, Jose & Saletan 588].

3.5. KLEIN-GORDON QFT 45

3.5 Klein-Gordon QFT

Recall the commutation relations from non-relativistic quantum mechanics,namely

[x; p] = ih

and[a; ay] = 1

To develop the QFT we impose similar relations, but this time for flelds.The equal time commutation relations

[`(~x; t);(~x0; t)] = i(~x ~x0) (3.3)and

[`(~x; t); `(~x0; t)] = [(~x; t);(~x0; t)] = 0 (3.4)

where we had

= _`

We now need to flnd the commutation relations for a(~k) and ay(~k). Weexpect the usual results

[a(~k); ay(~k0)] = (~k ~k0) (3.5)and

[a(~k); a(~k0)] = [ay(~k); ay(~k0)] = 0 (3.6)

3.5.1 Indirect Derivation of a; ay Commutators

[Greiner, pg. 77]To check whether our expection above is correct evaluate

[`(~x; t);(~x0; t)]= [`+(~x; t) + `(~x; t);+(~x0; t) + (~x0; t)]= [`+(~x; t);+(~x0; t)] + [`+(~x; t);(~x0; t)]

+[`(~x; t);+(~x0; t)] + [`(~x; t);(~x0; t)]

= iZd~kZd~k0 !0

n[a(~k); a(~k0)]ei(kx+k

0x0) [a(~k); ay(~k0)]ei(kxk0x0)

+ [ay(~k); a(~k0)]ei(kxk0x0) [ay(~k); ay(~k0)]ei(kx+k0x0)]

owhere k x = !t ~k ~x and t0 t and !0 !(~k0) =

q~k0 2 +m2.


Inserting (3.5) and (3.6) gives

[`(~x; t);(~x0; t)] = iZd~kZd~k0 !0

h(~k ~k0)ei(kxk0x0)

(~k0 ~k)ei(kxk0x0)i

= iZd3k N2k !

hei~k(~x~x0) + ei~k(~x~x

0)i

Now if [Greiner, pg. 77]

Nk =1p

2!(2)3

then we obtain

[`(~x; t);(~x0; t)] = iZd3k

12!(2)3

!hei~k(~x~x0) + ei~k(~x~x

0)i

= i(~x ~x0)

as required. Here we have used the result,

1(2)3

Zd3k ei~k(~x~x

0) = (~x ~x0)

With the above normalization we have

d~k d3kp

2!(2)3

(which is difierent to IZ114, but same as Kaku). Using the result [IZ114,Kaku 64-65]

d3k

2!= d4k (k2 m2) (k0)

gives

d~k =

s2!

(2)3d4k(k2 m2) (ko)

(which is difierent from IZ 114, but same as Kaku)

Note: Some authors [IZ114] use the normalization Nk = 1(2)32! in which

case [a(~k); ay(~k0)] = (2)32!k3(~k ~k0) [Greiner, pg. 77, footnote].


3.5.2 Direct Derivation of a; ay Commutators

The best way to obtain the commutators directly is to invert the Fourierexpansions to obtain a and ay in terms of ` and . The commutators arethen derived directly.

Inverting (3.1) and (3.2) gives [see Problems]

)

a~k =1p

2!(2)3

Zd3x eikx[!`(x) + i(x)]

ay~k =1p

2!(2)3

Zd3x eikx[!`y(x) iy(x)]

where a~k a(~k) and `(x) `(~x; t) and k x !t ~k ~x and by directevaluation of the commutators we arrive at (3.5) and (3.6) [do Problem 5.7]

But remember for KGE ` is real and therefore ` = `y and = y giving

ay~k =1p

2!(2)3

Zd3x eikx[!`(x) i(x)]

3.5.3 Klein-Gordon QFT Hamiltonian

The 2nd quantized fleld Hamiltonian is (see Problems)

H =Zd3k

!

2(ay~k a~k + a~k a

y~k)

and using the commutator

[ a~k ; ay~k0

] = (~k ~k0)

gives

H =Zd3k

N~k +

12

!

orX~k

N~k +

12

!

withN~k ay~ka~k

(see Problems)


We can also calculate the momentum ~P as [Kaku 66, 67; Schwabl 279]

~P =Zd3k

N~k +

12

~k

orX~k

N~k +

12

~k

3.5.4 Normal order

References [Kaku68, Mosel 27, Schwabl 280]The previously derived Hamiltonian is actually inflnite! This is beacuse theterm Z

d3k12! =

12

Zd3k

q~k 2 +m2 =1

This is one of the flrst (of many) places where QFT gives inflnite answers.Now because only energy difierences are observable, we are free to simplythrow away the inflnite piece, and re-write the Hamiltonian as

H =Zd3k N~k !:

A formal way to always get rid of these inflnite terms (there is also onein the previous expression for the momentum ~P ), is to introduce the idea ofnormal order.

In a normal ordered product all annihilation operators are placedto the right hand side of all creation operators.

Two colons :: are used to denote a normal ordered product. For example[Schwabl 280]

: a~k1a~k2ay~k3

: = ay~k3a~k1a~k2

: ay~ka~k + a~kay~k

: = 2ay~ka~k

In calculating the Hamiltonian (see Problems) we arrived at

H =Zd3k(ay~ka~k + a~ka

y~k)!

2

and using the commutator

[ a~k ; ay~k0

] = (~k ~k0)


gave

H =Zd3k(ay~ka~k + a

y~ka~k + 1)

!

2

=Zd3k(ay~ka~k + 1=2)!

which is inflnite. If we deflne H to be normal ordered then

: H : =Zd3k : (ay~ka~k + a~ka

y~k) :

!

2

=Zd3k(ay~ka~k + a

y~ka~k)

!

2

=Zd3kay~ka~k!

which is flnite. Note that normal ordering is equivalent to treating the bosonoperators as if they had vanishing commutator. [Schwabl 280]. (nnn Are weback to a classical theory?)


3.5.5 Wave Function

[Kaku 68-69]Now that we have expanded the KG fleld and Hamiltonian in terms of

creation and destruction operators, we need something for them to operateon. These are just the many-body states introduced earlier [Bergstrom &Goobar 289]

j n~ki n~kj i =Yi

jn~kii (3.7)

withjn~kii =

1qn~k!

ay~kn~k j0i

where the vacuum state is deflned via

a~kj0i = 0

The physical interpretation [Bergstrom & Goobar 290] is provided by

Hj n~ki n~kj i =X~k

nk (~k)j n~ki n~kj i

where (~k) = h!(~k). Equation (3.7) is interpreted as a many particle statewhere n~k1 have momentum

~k1, n~k2 have momentum~k2 etc.

Thus a 1-particle state is written

j1~ki j~ki = ay~kj0i

orh1~kj h~kj = h0ja~k

The states are normalized as

h~kj~k0i = (~k ~k0)

givingh0ja~kay~k0 j0i = (~k ~k

0):

3.6. PROPAGATOR THEORY 51

3.6 Propagator Theory

[Halzen and Martin 145-150, Kaku, Bj RQM Chapter 6]The Klein-Gordon equation for a free particle is

(22 +m2)` = 0

or with the replacement 22 ! p2 in momentum space

(p2 m2)` = 0

Lets write the non-free KGE as

(22 +m2)` = J(x)

where J(x) is referred to as a source term. This is solved with the Greenfunction method by deflning a propagator F (x y) as

(22 +m2)F (x y) 4(x y)

so that the solution is

`(x) = `0(x)Zd4yF (x y)J(y)

where `0(x) is the solution with J = 0. (see Problems) Deflne the Fouriertransform

F (x y) Z

d4k

(2)4eik(xy)F (k)

Our usual method of solution for propagators or Green functions is to solvefor F (k) and then do a contour integral to get F (x y) rather thansolving for F (x y) directly. The Problems show that

F (k) =1

k2 m2


Example Derive the momentum space Green function for theSchrodinger equation.

Solution The free particle Schrodinger equation is h

2

2mr2 ih @

@t

! = 0

and with the inclusion of a source this is h

2

2mr2 ih @

@t

! = J(x) U(~x)(~x; t)

or +

h2

2mr2 + ih @

@t

! = J(x) = U(~x)(x)

Deflne the Green functionh2

2mr2 + ih @

@t

!G0(x x0) = 4(x x0)

and deflne the Fourier transform

G0(x x0) =Z

d4p

(2)4eip(xx

0)G0(p)

Now operate on this to give, and using p (!; ~p), with h = 1h2

2mr2 + ih @

@t

!G0(x x0) =

Zd4p

(2)4

1

2m(i~p)2 + i(i!)] eip(xx0)G0(p)

= 4(x x0) =Z

d4p

(2)4

" ~p

2

2m+ !

#eip(xx

0)G0(p)

but recall that

4(x y) =Z

d4k

(2)4eik(xy)

which implies

G0(p) =1

! ~p 2=2m


Example Derive the position space Green function leaving thepole on the Real axis.

Solution Substituting for G0(p) into G0(xx0) we need to eval-uate (h = 1)

G0(x x0) =Z

d4p

(2)41

! ~p 2=2meip(xx0)

=Z

d3p

(2)3ei~p(~x~x

0)Z 11

d!

2ei!(tt0)

! ~p 2=2mand we see that the integrand is singular at ! = ~p 2=2m. Thissimple pole is shown in the flgure. We need to decide whether tointegrate in the upper half plane (UHP) or the lower half plane(LHP). This is dictated by the boundary conditions as follows.Write ! Re ! + i Im !, so that

ei!(tt0) = eiRe!(tt

0)e+ Im!(tt0)

For t t0 > 0, the term eIm!(tt0) will blow up for Im ! > 0but will go to zero for Im ! < 0. Thus the boundary conditiontt0 > 0 dictates we integrate in the LHP. Similarly for tt0 < 0we use the UHP. This is shown in the flgure. [See also Halzenand Martin 148]

XRe w

Im w

t < t'

XRe w

Im w

t > t'

w =w 0=p2/2m

C1C

Now apply the Cauchy Residue Theorem, and remember coun-terclockwise integration is a positive sign. In the left flgure the


coutour does not enclose any poles so that

0 =Z !01

+ZC1

+Z 1!0+

+ZC

and withRC

= 0 (Jordans lemma) and lim!0

we have

Z 11

= ZC1

Thus for t > t0

G0(x x0)t>t0 = + iZ

d3p

(2)3ei~p(~x~x

0) 12ei!0(tt

0)

= i2

Zd3p

(2)3ei~p(~x~x

0)ei!0(tt0)

with !0 ~p 2=2m. Similarly for t < t0 we obtain

G0(x x0)t t0 but certainly not at anearlier time t < t0.

Because of the singularity, the integral is not well deflned until we specifythe limiting process. Lets now try a difierent method for evaluating theintegral that is consistent with the boundary condition, i.e. we wantG0(x x0)t


Example Evaluate the position space Green function by shiftingthe pole ofi the real axis by a small amount , and take lim

!0.

Solution We can shift the pole either above or below the realaxis. We choose to shift it below because then the upper contourwill enclose no poles and will give a zero integral consistent withour boundary conditions. This is shown in the flgure.

Ct>t

t t0 we have1R1

+RC

= 2i Residues and withRC

= 0 we

get

G0(x x0)t>t0 = 2iZ

d3p

(2)3ei~p(~x~x

0)Z 11

d!

2ei!(tt0)

! ~p 2=2m+ i= i

Zd3p

(2)3ei~p(~x~x

0)ei!0(tt0)

with lim!0 e

(tt0) = 0, which is exactly double our previous answer.For t < t0 we get

G0(x x0)t


An excellent discussion of all of these issues can be found in [Arfken 4thed. 417, 427; Landau 84-88; Bj RQM 85; Cushing 315; Greiner QED 27;Halzen and Martin 145-150]

The key integral that we have been considering is of the form

I =Z 11

d!ei!t

! !0and with the pole shifted below the real axis this is

I() =Z 11

d!ei!t

! !0 + iand lets summarize our results as

It>0 = i ei!0tIt0 = 2 i ei!0tI()t


and using [Merzbacher 3rd ed., p. 631]

(x) =1

lim!0+

x2 + 2

we have1

! i =!

!2 + 2 i (!)

The flrst term on the right hand side becomes 1! as ! 0 except if ! = 0.If f(!) is a well behaved function we have [Merzbacher, old ed., p. 85]

lim!0

Z 11

f(!)!

!2 + 2d! = lim

!0

Z 1

f(!)d!

!+Z 1

f(!)d!

!

+Z f(!)

!d!

!2 + 2

= P

Z 11

f(!)d!

!+ f(0) lim

!0

Z

!d!

!2 + 2

= PZ 11

f(!)d!

!+ 0

where the last integral vanishes because the integrand is an odd function of!. Thus we can write [Merzbacher, old ed., p. 85]

lim!0

1! i = P

1! i (!)

and we can also write [Cushing 315] (with the pole at ! = !0 instead of! = 0)

P

Zf(!)d!! !0 =

Zf(!)d!

! !0 i i f(!0)


Example Verify the above formula for our previous integralI =

Rd! e

i!t!!0 .

Solution For t > 0 we had

P

Zei!td!! !0 = i e

i!0t

and Zei!td!

! !0 + i = 2i ei!0t

Now f(!0) = ei!0t, so thatZei!td!

! !0 + i + i f(!0) = 2i ei!0t + i ei!0t

= i ei!0t

= PZei!td!! !0

in agreement with the formula. For t < 0 we had

P

Zei!td!! !0 = +i e

i!0t

and Zei!td!

! !0 + i = 0

so that Zei!td!

! !0 + i + i f(!0) = 0 + i ei!0t

= PZei!td!! !0

also in agreement with the formula.


Example Show that the function deflned as

(t t0) =(

1 if t > t0

0 if t < t0

can be written as

(t t0) = lim!012i

Z 11

d!ei!(tt0)

! + i

Solution We showed previously that with I() 1R1

d! ei!t

!!0+iwe get

I()t>0 = 2i ei!0tI()tt0 = 2iIt


Example Evaluate the position space Green function, consistentwith the causality boundary condition, in terms of (t t0). Alsowrite the answer in terms of the plane wave states

`p(x) eipx

(2)3=2

Solution To satisfy causality we use the i prescription to eval-uate the integral which gave

G0(x x0)t>t0 = iZ

d3p

(2)3ei~p(~x~x

0)ei!0(tt0)

andG0(x x0)t


Klein-Gordon Propagator

We originally wrote the Klein-Gordon propagator F (k) = 1k2m2 butthen digressed on a lengthy discussion of the Schrodinger propagator in orderto illustrate the integration techniques. We now return to the Klein-Gordonpropagator. Consider the following i prescription

F (k) =1

k2 m2 + iWrite this as

1k2 m2 + i =

1

k20 ~k2 m2 + i=

1k20 E2 + i

with E +q~k2 +m2

We can show (see Problems)

1k2 m2 + i =

1k20 E2 + i

=1

2k0

1

k0 E + i +1

k0 + E i

Recall that the momentum space propagator F (k) was deflned in terms ofthe Fourier transform

F (x y) =Z

d4k

(2)4eik:(xy)F (k)

Performing the integration as before we get, with d~~k d3k(2)32!k

(see Prob-lems)

F (x x0) = i(t t0)Zd~~k eik(xx

0) i(t0 t)Zd~~k eik(xx

0)

= i(t t0)Zd~~k `k(x)`k(x

0) i(t0 t)Zd~~k `k(x)`k(x

0)


Example Show that iF (x x0) = h0jT`(x)`(x0)j0i(Note this is also equal to

= h0jT`(x)`y(x0)j0i

for Hermitian flelds [BjRQF42].)

Solution The time ordered product is

TA(t1)B(t2) (A(t1)B(t2) if t1 > t2B(t2)A(t1) if t2 > t1

(t1 t2)A(t1)B(t2) + (t2 t1)B(t2)A(t1)

Thus for the flelds

T`(x)`(x0) = (t t0)`(x)`(x0) + (t0 t)`(x0)`(x)

or

h0jT`(x)`(x0)j0i = (t t0)h0j`(x)`(x0)j0i+ (t0 t)h0j`(x0)`(x)j0i

Thus we want to evaluate h0j`(x)`(x0)j0i and h0j`(x0)`(x)j0i. Now

`(x) =Zd~k (a~ke

ikx + ay~keikx)

with d~k d3kp(2)32!

and ! q~k2 +m2 and !0

q~k02 +m2. Evaluate

h0j`(x)`(x0)j0i=h0j

(2)3

Zd3kp

2!d3k0p

2!0(a~ke

ikx + ay~keikx)(a~k0e

ik0x0 + ay~k0eik0x0)j0i

=Zd~k d~k0 h0ja~ka~k0ei(kx+k

0x0) + a~kay~k0ei(kxk

0x0)

+ay~ka~k0ei(kxk0x0) + ay~ka

y~k0ei(kx+k

0x0)j0i

However a~kj0i = 0 and h0jayk = 0. Thus the 1st, 3rd and 4th terms are zero.We are only left with

h0ja~kay~k0 j0i = (~k ~k0)


Continuing

h0j`(x)`(x0)j0i = 1(2)3

Zd3kp

2!d3k0p

2!0(~k ~k0)ei(!t~k~x!0t0+~k0~x0)

=1

(2)3

Zd3k

2!ei~k:(~x~x0)i!(tt0)

=1

(2)3

Zd3k

2!eik(xx

0)

Obviously we also have

h0j`(x0)`(x)j0i = 1(2)3

Zd3k

2!eik(x

0x)

=1

(2)3

Zd3k

2!eik(xx

0)

Thus

h0jT`(x)`(x0)j0i = (t t0)h0j`(x)`(x0)j0i+ (t0 t)h0j`(x0)`(x)j0i= (t t0)

Zd3k

(2)32!eik(xx

0) + (t0 t)Z

d3k

(2)32!eik(xx

0)

= iF (x x0)


Other Propagators

Recall that

`(x) =Zd~k[a~ke

ikx + ay~keikx]

`+(x) + `(x) [Kaku 152; Mandl & Shaw 44]

From this we can deflne other propagators (see Problems)

i(x y) [`(x); `(y)] = Z

d3k

(2)32!~keik(xy)

where !~k +q~k2 +m2 giving (see Problems)

i(x y) [`(x); `(y)] = i+(x y) + i(x y)

or just (x) = +(x) + (x). Thus one has [Mandl & Shaw 51]

(x) = +(x) + (x) =1

(2)3

Zd3k

!~ksin k x

=i

(2)3

Zd4k (k2 m2)(k0)eikx

with (k0) = k0jk0j =(

+1 for k0 > 01 for k0 < 0

Note also that (x) = +(x)These propagators are related to the Feynman propagator by [Mandl

and Shaw 54]F (x) = (t)+(x) (t)(x)

or [Schwabl 283]

F (x x0) = (t t0)+(x x0) (t0 t)(x x0)

which can be written [Schwabl 283, Mandl and Shaw 54]

F (x) = (x) for t >


Example Show that

F (x x0) = (t t0)+(x x0) (t0 t)(x x0)where

iF (x x0) = h0jT`(x)`(x0)j0iand

i(x y) [`(x); `(y)]:

Solution

iF (x x0) = h0jT`(x)`(x0)j0i= (t t0)h0j`(x)`(x0)j0i+ (t0 t)h0j`(x0)`(x)j0i

Thus to do this problem we really need to show that

h0j`(x)`(x0)j0i = [`+(x); `(x0)]and

h0j`(x0)`(x)j0i = [`(x); `+(x0)]Lets only do the flrst of these. The question is, how do these relationscome about ? The answer is easy. Commutators are just c-numbers, i.e.classical functions or numbers. Thus h0jcj0i = ch0j0i = c for any c-number.Therefore

[`+(x); `(x0)] = h0j[`+(x); `(x0)]j0i= h0j`+(x)`(x0)j0i h0j`(x0)`+(x)j0i= h0j`+(x)`(x0)j0i

because `+j0i = 0.Now ` = `+ + `. Thus

`j0i = `+j0i+ `j0i = 0 + `j0i = `j0iand similarly

h0j` = h0j`+giving

[`+(x); `(x0)] = h0j`(x)`(x0)j0ias required. (Proof of the other relation follows similarly).


3.7 Complex Klein-Gordon Field

[Schwabl 285; Kaku 69; Huang 24-25][W. Pauli and V. F. Weisskopf, Helv. Phys. Acta, 7, 709 (1934)]Charged particles cannot be described with a real scalar fleld because it

only has one component, whereas a minimum of two components is neededto describe charge. Deflne

` 1p2

(`1 + i`2) and `y 1p2

(`1 i`2)

where `1 and `2 are real flelds. The normalization factor 1p2 is chosen sothat ì has the same renormalization as the real scalar fleld discussed before.A suitable classical Lagrangian is (with j`j2 = ``y) [Huang 24]

L = @`y @`m2j`j2

=12

2Xi=1

(@ì @ì m2`2i )

which is just the sum of the Lagrangian for the real flelds.The Euler-Lagrange equations give the equations of motion [Schwabl

285](22 +m2)` = 0 and (22 +m2)`y = 0

with the conjugate momenta being [Kaku 70; Schwabl 285]

= _`y and y = _`

with the equal time commutation relations [Kaku 70; Schwabl 285]

[`(~x; t);(~x0; t)] = [`(~x; t); _`y(~x0; t)] = i(~x ~x0)[`y(~x; t);y(~x0; t)] = [`y(~x; t); _`(~x0; t)] = i(~x ~x0)

and [Schwabl 285]

[`(~x; t); `(~x0; t)] = [(~x; t);(~x0; t)] = 0[`y(~x; t); `y(~x0; t)] = [y(~x; t);y(~x0; t)] = 0

We Fourier expand each component ì(x) exactly as before [Kaku 70; Huang24]

ì(x) =Zd~k (a

i~keikx + ay

i~keikx)

3.7. COMPLEX KLEIN-GORDON FIELD 67

where the commutation relations are [Kaku 70; Huang 25]

[ai~k; ayj~k0

] = (~k ~k0)ijHowever rather than treating the i flelds separately, we can combine themas we did with the wave functions `(x). Deflne [Kaku 70; Huang 25]

a~k 1p2

(a1~k

+ ia2~k

) and b~k 1p2

(a1~k ia

2~k)

so that the Fourier expansion now becomes [Huang 25; Schwabl 285]

`(x) =Zd~k (a~ke

ikx + by~keikx)

(Exercise: Prove this result) and for the other fleld, obviously it is

`y(x) =Zd~k(ay~ke

ikx + b~keikx)

The new commutation relations are [Kaku 70; Schwabl 286; Huang 25]

[a~k; ay~k0

] = [b~k; by~k0

] = ~k~k0

and[a~k; a~k0 ] = [b~k; b~k0 ] = [a~k; b~k0 ] = [a~k; b

y~k0

] = 0

There are now two occupation-number operators, for particles a and forparticles b [Schwabl 286]

Na~k ay~ka~k and Nb~k b

y~kb~k

where ay~k and a~k create and annihilate a particles and by~k

and b~k create andannihilate b particles. The vacuum state is deflned by [Schwabl 286]

a~kj0i = b~kj0i = 0The four momentum is [Schwabl 286; Mandl and Shaw 49; Greiner FQ 93]

P = (H; ~P ) =X~k

k(Na~k

+Nb~k

)

givingH =

X~k

!~k(Na~k +Nb~k)

and~P =

X~k

~k(Na~k

+Nb~k

)


3.7.1 Charge and Complex Scalar Field

Recall the real scalar fleld Lagrangian L = 12(@`@` m2`2) and thecomplex scalar fleld Lagrangian

L = @`y@`m2j`j2 = 122Xi=1

(@`i@`i m2`2i )

with j`j2 ``y. The complex scalar fleld Lagrangian (and action) is invari-ant under the transformation

`(x)! eiq`(x) `y(x)! eiq`y(x)which generates a U(1) symmetry [Kaku 70]. Recall that before we had (forsmall fii = i)

r(x)! 0r(x) = eifiiXir(x) (1 + iiXi)r(x)giving

r(x) = 0r(x) r(x) = iiXir(x)Our two flelds are 1 ` and 2 `y giving (with small = )

`(x) = iq`(x) and `y(x) = iq`y(x)Recall the Noether current

j @L@(@r)

r Tx

with

T @L@(@r)

@r gL

Also recall the relation between local and total variations

r(x) = r(x) @r@x

x

For x = 0, we have (x) = (x) [Schwabl 272] and therefore

j =@L

@(@r)r

=@L

@(@`)`+

@L@(@`y)

`y

=@L

@(@`)iq`+

@L@(@`y)

(i)q`y

3.7. COMPLEX KLEIN-GORDON FIELD 69

and with L = @`y@`m2``y = g@`y@`m2``y gives@L

@(@`)= g@`y = @`y

and@L

@(@`y)= @`

Thus the Noether current is

j = (@`y)iq` (@`)iq`y

Once again we drop the constant factor and insert a minus sign todeflne a new current [Kaku 71, Mosel 19, Schwabl 286, Huang 25]

j = iq[(@`)`y (@`y)`]

which agrees exactly with the 4-current derived previously!The conserved charge (dQdt = 0) is therefore [Kaku 71, Mosel 19]

Q =Zd3x jo = iq

Zd3x( _``y _`y`)

= qZd3k(ay~ka~k b

y~kb~k)

= qZd3k(N

a~kN

b~k)

(Exercise: Prove this result). Here jo matches exactly our expression for derived previously! [Kaku 71] gives an excellent discussion of the physicalinterpretation of this charge. Also note that \charge" is used in a genericsense [Huang 26] since electromagnetic coupling has not yet been turned on.

Compare this to our result from Chapter 1 which was (with probabilitycurrent density j = (;~j ) = )

Q =Zd3x j0 = q

Zd3x 0

= qZd3x

where the probability density was = 0 = y00 = y. This was anexpression in terms of the classical Dirac fleld . Above we now have Q interms of the quantized fleld operators a; ay; b; by for the complex scalar fleld.


3.8 Summary

Two useful integrals are:

1(2)3

Zd3xei(~k~k

0):~x = (~k ~k0)

4(x y) =Z

d4k

(2)4eik(xy)

3.8.1 KG classical fleld

The massive Klein-Gordon Lagrangian is

LKG = 12(@`@`m2`2)

giving the KGE (using p2 ! 22)(22 +m2)` = 0(p2 m2)` = 0

The covariant momentum density is

= @`

giving the canonical momentum

o = _`(x)and Hamiltonian density

H = 12

[2 + (~r`)2 +m2`2]

and momentum~P =

Zd3x _`(x)~r`(x)

The Fourier expansion of the KG fleld is

`(~x; t) `+(x) + `(x) =Zd~kha~keikx + ay~ke

ikxiThe conjugate momentum is

(~x; t) = _`(~x; t) +(x) + (x)= i

Zd~k !

ha~keikx ay~ke

ikxi

3.8. SUMMARY 71

In the above equations (with k0 ! = +q~k2 +m2)

d~k d3kp

2!(2)3=

s2!

(2)3d4k(k2 m2) (ko)

The integrals are cast into discrete form with the replacementZd3k ! ~k:

3.8.2 Klein-Gordon Quantum fleld

To develop the QFT we impose the equal time commutation relations

[`(~x; t);(~x0; t)] = i(~x ~x0)and

[`(~x; t); `(~x0; t)] = [(~x; t);(~x0; t)] = 0which imply

[a(~k); ay(~k0)] = (~k ~k0)and

[a(~k); a(~k0)] = [ay(~k); ay(~k0)] = 0Deflning

N~k ay~ka~kthe Hamiltonian and momentum can be re-written as

H =Zd3k

N~k +

12

!

~P =Zd3k

N~k +

12

~k

The vacuum state is deflned via

a~kj0i = 0and the many body states are

j n~ki n~kj i =Yi

jn~kii

withjn~kii =

1qn~k!

ay~kn~k j0i

Hj n~ki n~kj i =X~k

nk (~k)j n~ki n~kj i

where (~k) = h!(~k).


3.8.3 Propagator Theory

The non-free KGE is(22 +m2)` = J(x)

The Feynman propagator in position space is deflned as

(22 +m2)F (x y) 4(x y)so that the solution to the non-free KGE is

`(x) = `0(x)Zd4yF (x y)J(y)

where `0(x) is the solution with J = 0. The momentum space Feynmanpropagator is deflned through the Fourier transform

F (x y) Z

d4k

(2)4eik(xy)F (k)

The usual method of solution for propagators or Green functions is to solvefor F (k) and then do a contour integral to get F (x y) rather thansolving for F (x y) directly. We get

F (k) =1

k2 m2which is the inverse of the momentum operator in the free KGE. This flnallyleads to

F (x x0) = i(t t0)Zd~k eik(xx

0) i(t0 t)Zd~k eik(xx

0)

One can also show that

iF (x x0) = h0jT`(x)`(x0)j0i= h0jT`(x)`y(x0)j0i

where the second line is true for Hermitian flelds [BjRQF42].Other useful propagators are

i(x y) [`(x); `(y)] = Zd~~keik(xy)

giving

i(x y) [`(x); `(y)] = i+(x y) + i(x y)

3.9. REFERENCES AND NOTES 73

with

(x) = +(x) + (x) = Z

d3k

(2)3!~ksin k x

These other propagators are related to the Feynman propagator by

F (x) = (t)+(x) (t)(x)

orF (x) = (x) for t >

Chapter 4

Dirac Field

We have seen that the KGE gives rise to negative energies and non-positivedeflnite probabilities and for these reasons was discarded as a fundamentalquantum equation. These problems arise because the KGE is non-linear in@@t , unlike the SE. Dirac thus sought a relativistic quantum equation linearin @@t , like the SE. We shall see that therefore Dirac was forced to invent amatrix equation.

The derivation of the DE presented here follows [Griths, pg. 215].Dirac was searching for an equation linear in E or @@t . Instead of startingfrom

p2 m2 = 0his strategy was to factor this relation, which is easy if we only have p0 (i.e.~p = 0), namely (with p0 = p0)

(p0 m)(p0 +m) = 0and obtain two flrst order equations

p0 m = 0 or p0 +m = 0However its more dicult if ~p is included. Then we are looking for some-thing of the form p2 m2 = pp m2 = (flp +m)(p m) where fland are 8 coecients to be determined. The RHS is

flpp +m( fl)p m2

For equality with LHS we dont want terms linear in p , thus = fl ,leaving

p2 = pp = pp

75

76 CHAPTER 4. DIRAC FIELD

i.e. (with (p1)2 = (p1)2)

(p0)2 (p1)2 (p2)2 (p3)2 = (0)2(p0)2 + (1)2(p1)2 + (2)2(p2)2+(3)2(p3)2 + (01 + 10)p0p1+(02 + 20)p0p2 + (03 + 30)p0p3+(12 + 21)p1p2 + (13 + 31)p1p3+(23 + 32)p2p3

Heres the problem; we could pick 0 = 1 and 1 = 2 = 3 = i butwe cant get rid of the cross terms. At this point Dirac had a brilliantinspiration; what if the s are matrices instead of numbers? Since matricesdont commute we might be able to flnd a set such that

(0)2 = 1 (1)2 = (2)2 = (3)2 = 1 + = 0 for 6=

Or, more succinctly, f; g = 2g which is called a Clifiord Algebrawhere the curly brackets denote the anti-commutator fA;Bg AB + BA.It turns out that this can be done, but the smallest set of matrices are 44.These are

0 fl =I 00 I

~ fl~fi

0 ~~ 0

(fl; fl~fi)

Lets introduce everything else for completeness

5 i0123 =

0 II 0

1 =

0 11 0

2 =

0 ii 0

3 =

1 00 1

~fi

0 ~~ 0

fl

1 00 1

y = 00i:e: 0

y= 0; i

y= i

fl1 = fl and 5 = 5These conventions are used by the following authors: [Griths, Kaku,Halzen & Martin].

Thus the Dirac equation is (pm) = 0 with A A . In coordi-nate space (using p ! i@) it is (i@ m) = 0 . In non-covariant notationit is

H = ih@

@twith H ~fi ~p+ flm

4.1. PROBABILITY & CURRENT 77

The DE can be written in terms of 4-vectors (/pm) = 0 and is thereforemanifestly covariant.

4.1 Probability & Current

For the SE and KGE we used SE and KGE to derive the continuity equa-tion. For matrices the generalization of complex conjugate () is Hermitianconjugate (y) which is the transpose of the complex conjugate. The DE is

(i/@ m) = 0 = (i@ m)

and DEy is (using (AB)y = ByAy)

y(i/@ m)y = 0= y(iy@ m) = 0 where @y = @= y(i00@ m) = 0 using y = 00

We want to introduce the Dirac adjoint ( is a column matrix)

y0 ( is a row matrix!)

Using 00 = 1 we get

y(i00@ m00) = 0

) (i0@ +m0) = 0and cancelling out 0 gives, the Dirac adjoint equation,

(i/@ +m) = 0 , (i/@ +m) = 0

Some sloppy authors write this as (i/@+m) = 0 but this cannot be because is a row matrix! The notation

/@ however means that /@ operates on to

the left, i.e. /@ (@ ) .

Its very important not to get confused with this. The DE and DEy areexplicitly

(i/@ m) = 0 , i@ m = 0(i/@ +m) = 0 , i(@ ) +m = 0


Now lets derive the continuity equation. Multiply DE from the left by and DEy from the right by . Now one could get very confused writing

(i/@ m) = 0(i/@ +m) = 0

whereas what is really meant is [Halzen & Martin, pg. 103]

(i@ m) = 0(i(@ ) +m ) = 0

Adding these gives

@ + (@ ) = 0 = @( )

givingj = ( ) (;~j)

or

= 0 = y =4Xi=1

jij2

which is now positive deflnite! The 3-current is

~j = ~

[see also Mosel 17, 34]

4.2 Bilinear Covariants

can be written

0BBB@1234

1CCCA and one can try to construct a scalar, such as

y = (1 2 3 4)

0BBB@1234

1CCCA = j1j2 + j2j2 + j3j2 + j4j2 but this is nota Lorentz scalar (its got all + signs).

Rather, deflne the Dirac adjoint

y0 = (1 2 3 4)

4.3. NEGATIVE ENERGY AND ANTIPARTICLES 79

and = j1j2 + j2j2 j3j2 j4j2 is a Lorentz scalar (see Bjorken andDrell]. One can prove that the following quantities transform as indicated:

scalar (1 component)5 pseudoscalar (1 component) vector (4 component)5 pseudovector (4 component) antisymmetric 2nd rank tensor (6 component)

where i2( ).

4.3 Negative Energy and Antiparticles

4.3.1 Schrodinger Equation

The free particle SE is

h2

2mr2 = ih@

@t

which has solution

(x; t) = (C cos kx+D sin kx) eihEt E

=Aeikx +Beikx

eihEt in 1-dimension.

Substituting gives

hk2

2m(x; t) = E(x; t) or

E +

h2k2

2m

! = 0

yielding

E = h2k2

2m

However it also has solution (x; t) = (Aeikx +Beikx)eihEt E

Substituting gives

h2k2

2m(x; t) = +E(x; t) or

E h

2k2

2m

! = 0

yielding

E = +h2k2

2m


E and E are difierent solutions. The flrst one E corresponds to positiveenergy and the second one E to negative energy. We are free to toss awayone solution as unphysical and only keep E .

However for KGE and DE the same solution gives both positive andnegative energy. Of course we are not free to toss away one energy becauseits in the solution. Whereas in the SE we got two difierent solutions forpositive and negative energy. If we want to get rid of negative energy in theSE we toss away one solution. However in KGE and DE we always get bothpositive and negative energy for all solutions. The only way to toss awaynegative energy is to toss away all solutions; i.e. toss out the whole equation!

?Also reason why Einstein did not reject SR; E = p~p2 +m2 wereseparate solutions; not part of same solution.

4.3.2 Klein-Gordon Equation

The free particle KGE is(22 +m2)` = 0

with

22 = @2

@t2+r2

A solution is` = Neipx = Nei(Et~p~x)

Substituting gives(E2 ~p 2 +m2)` = 0

which implies) E2 = ~p 2 +m2

orE =

q~p 2 +m2

Thus the single solution ` = Neipx has both positive and negative energysolutions. Another solution is

` = Neipx = Nei(Et~p~x)

Substituting also gives the same as above, namely (E2 ~p 2 +m2)` = 0 or

E = q~p 2 +m2

4.3. NEGATIVE ENERGY AND ANTIPARTICLES 81

and again the single solution ` = Neipx has both positive and negativeenergy solutions.

The interpretation of these states is as follows. For ~p = 0 (particle at rest)then E = m. For ~p 6= 0, there will be a continuum of states above andbelow E = m [Landau, pg. 225], with bound states appearing in between.In QM there would be transitions to the negative energy continuum to infl-nite negative energy. This also happens with the Dirac equation. However,the DE describes, automatically, particles with spin. Diracs way out of thenegative energy catastrophe was to postulate that the negative energy seawas fllled with fermions and so the ???

E < 0

E > 0

E

- mc

0

p = 0 2

mc 2 p = 0

However for the KGE the negative energies are a catastrophe.

Also we can now clearly see the problem with as calculated with theKGE. Recall = ih2m

` @`@t `@`

@t

. For ` = eipx we get = hm``E

which gives negative for positive or negative E. (Halzen & Martin, pg. 74]


4.3.3 Dirac Equation

Lets look for plane wave solutions of the form (x) = w(~p)eipx . Substi-

tuting into (i/@ m) = 0 gives the momentum space DE (/pm)w = 0 .Actually, lets flrst look at rest frame (~p = 0) solutions (RF). Writing theDE as

H = ih@

@twith H = ~fi ~p+ flm

independent DE is H = E, which in the RF is

Hw = flmw =

mI 00 mI

!w = Ew;

The eigenvalues are E = m;m;m;m with eigenvectors [Halzen & Martin,pg. 104] 0BB@

1000

1CCA0BB@

0100

1CCA0BB@

0010

1CCA0BB@

0001

1CCAThus the DE has positive (E = +m) and negative (E = m) energy solu-tions! (We shall look at ~p 6= 0 solutions in a moment)

Lets summarize so far (Aitchison & Hey, pg. 71)

Probability () Energies for same solution SE + +KGE +, DE + +,

Thus both the KGE and DE have negative energies.

The DE describes fermions (see next section). Diracs idea was thatthe negative energy sea was fllled with fermions and via PEP prevented thenegative energy cascade of positive energy particles. Dirac realized if givenparticles in sea energy of 2mc2 create holes. Dirac Sea not taken seriouslyuntil positron discovered!! (1932 Anderson)

4.4. FREE PARTICLE SOLUTIONS OF DIRAC EQUATION 83

4.4 Free Particle Solutions of Dirac Equation

Before proceeding recall the following results:

x =

0 11 0

y =

0 ii 0

z =

1 00 1

~ ~p =Pz pp+ pz

pz px ipypx + ipy pz

Lets now look at ~p 6= 0 solutions (i.e. not in rest frame). As before, we lookfor solutions of the form (x) = w(~p)eipx Substituting into (i/@ m) = 0gives the momentum space DE (/p m)w = 0. Using /p =

E ~ ~p~ ~p E

gives, with w

wAwB

(/pm)w =E m ~ ~p~ ~p E m

wAwB

= 0 =

"(E m) wA ~o ~p wB~ ~p wA (E +m) wB

#

giving

wA =~ ~pE m wB and wB =

~ ~pE +m

wA

Combining yields

wA =(~ ~p )2E2 m2wA =

~p 2

E2 m2wA

because (~ ~p)2 = ~p 2. Thus

E2 = ~p 2 +m2 or E = q~p 2 +m2

Thus again we see the negative energy solutions ! (this time for the ~p 6= 0DE)

Our flnal solutions are

w =wAwB

=

wA~~pE+mwA

or

~~pEmwBwB

with wA and wB left unspecifled, which means we are free to choose them[Aitchison & Hey 69] as

wA; wB =

10

or

01


We have the following possibilities

Pick wA =

10

) wB = ~o ~p

E +m

10

=

1E +m

pzp+

(1) p px ipy

Pick wA =

01

) wB = ~o ~p

E +m

01

=

1E +m

ppz

(2)

Pick wB =

10

) wA = ~o ~p

E m

10

=

1E m

pzp+

(3)

Pick wB =

01

) wA = ~o ~p

E m

01

=

1E m

ppz

(4)

But the question is, do we use E = +p~p 2 +m2 or E = p~p 2 +m2? Well,

for (1) and (2) we must use E = +p otherwise 1E+m blows up for ~p = 0.For (3) and (4) we must use E = p otherwise 1Em blows up for ~p = 0.The term E = +p is called the particle solution. The term E = p iscalled the antiparticle solution.

Rewrite as, with (1)

10

and (2)

01

w(1)(~p) = N

0BB@10pz

E+mp+E+m

1CCA = N 1~~pE+m

(1)

w(2)(~p) = N

0BBB@01pE+mpzE+m

1CCCA = N

1~o~pE+m

(2)

with E +q~p 2 +m2

and also

w(3)(~p) = N

0BB@pz

Emp+Em

10

1CCA = N ~~pEm1(1)

w(4)(~p) = N

0BBB@pEmpzEm

01

1CCCA = N ~~pEm

1

(2)

with E q~p 2 +m2

4.4. FREE PARTICLE SOLUTIONS OF DIRAC EQUATION 85

or

w(s)(~p) = N

1~o~pE+m

(s) with E +p~p 2 +m2

w(s+2)(~p) = N ~~pEm

1

(s) with E p~p 2 +m2

9>>>>=>>>>; s = 1; 2

[Halzen & Martin, pg. 105]But all free particles carry positive energy! Thus re-interpret w(3) and w(4)

as positive energy antiparticle states

w(s+2)(~p) = N

~~pp~p 2+m2m

1

!(s)

w(s+2)(~p) = N ~~pp~p 2+m2m

1

!(s) = N

~~pp~p 2+m2+m

1

!(s)

::: Deflne u(p; s) = u(s)(p) w(s)(~p) u(1;2)(p) = w(1;2)(~p)v(p; s) = v(s)(p) w(s+2)(~p) v(1;2)(p) = w(3;4)(~p)

(Me & Kaku)everyone else: v(1;2)(p) = w(4;3)(~p)

Note: Bj writes u(p; s) and w(~p). Gross writes u(~p; s)

u(1) = N

0BB@10pz

E+mp+E+m

1CCA u(2) = N0BBB@

01pE+mpzE+m

1CCCA

v(2) = N

0BB@pz

E+mp+E+m

10

1CCA v(1) = N0BBB@

pE+mpzE+m

01

1CCCA

9>>>>>>>>>>>>>>=>>>>>>>>>>>>>>;

all with E +p~p 2 +m2

= v(1) (Kaku) = v(2) (Kaku) Kaku difierentfrom eve

norbury. quantum field theory (wisconsin lecture notes, 2000)(107s)

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