QUANTUM FIELD THEORY II (PHYS7652) LECTURE NOTESpages.physics.cornell.edu/~ajd268/Notes/QFT2-Notes.pdf · 1 QUANTUM FIELD THEORY II (PHYS7652) LECTURE NOTES Lecture notes based on

1

QUANTUM FIELD THEORY II(PHYS7652) LECTURE NOTES

Lecture notes based on a course given by Maxim Perelstein.We begin with discussing the path integral formalism in

Quantum Mechanics and move on to it’s use in Quantum FieldTheory. We then study renormalization and running couplings in

abelian and non-abelian gauge theories in detail.

Presented by: MAXIM PERELSTEIN

LATEX Notes by: JEFF ASAF DROR

2013

CORNELL UNIVERSITY

Contents

1 Preface 41.1 Conventions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

2 Path Integrals in Quantum Mechanics 52.1 Harmonic Oscillator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92.2 Anharmonic Oscillator . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142.3 Correlation Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182.A Jacobian of the Fourier Transformation . . . . . . . . . . . . . . . . . . . 22

3 Quantum Field Theory 243.1 Free Klien Gordan Field . . . . . . . . . . . . . . . . . . . . . . . . . . . 243.2 External Force . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 263.3 Interacting Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 293.4 LSZ Reduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 303.5 Statistical Mechanics and Path Integrals . . . . . . . . . . . . . . . . . . 30

4 Quantization of Electromagnetism 314.1 Classical Electromagnetism . . . . . . . . . . . . . . . . . . . . . . . . . 314.2 Quantization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 344.3 Path Integrals in QED . . . . . . . . . . . . . . . . . . . . . . . . . . . . 354.4 Correlation Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

5 Fermions with Path Integrals 425.1 Grassman Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 425.2 Free Dirac Field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 455.3 QED . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 475.4 Three-point function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 495.5 e+e− → µ+µ− . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 535.6 e−µ− → e−µ− . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 555.7 Compton Scattering . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 575.8 Proof of Ward Identity . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59

2

CONTENTS 3

6 Loop Diagrams and UV Divergences in QED 616.1 Superficial Divergences . . . . . . . . . . . . . . . . . . . . . . . . . . . . 616.2 Electron Self-Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 636.3 Classical Renormalization and Regulators . . . . . . . . . . . . . . . . . . 696.4 Dimensional Regularization . . . . . . . . . . . . . . . . . . . . . . . . . 716.5 Vacuum Polarization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 716.6 Vertex Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76

6.6.1 Physics of F2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78

7 Renormalized Perturbation Theory 807.1 φ4 Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 807.2 QED . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81

8 Running Couplings and Renormalization Group 848.1 “Large Logs” . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 848.2 QED β-function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90

9 Non-Abelian Gauge Theories (“Yang-Mills”) 949.1 Looking Closely at Abelian Gauge Theories . . . . . . . . . . . . . . . . 949.2 SU(2) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 959.3 General Recipe . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 989.4 Universal Couplings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1019.5 Quantum Chromodynamics . . . . . . . . . . . . . . . . . . . . . . . . . 1029.6 Quantization of Yang Mills . . . . . . . . . . . . . . . . . . . . . . . . . . 1039.7 Unitary and Ghosts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1069.8 Renormalized Perturbation Theory . . . . . . . . . . . . . . . . . . . . . 1089.9 RG flows (evolution) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114

10 Non-perturbative results 11610.1 Kallen-Lehmann Representation . . . . . . . . . . . . . . . . . . . . . . . 116

11 Broken Symmetries 11911.1 SSB of Discrete Global Symmetries . . . . . . . . . . . . . . . . . . . . . 11911.2 SSB of Continuous Global Symmetry . . . . . . . . . . . . . . . . . . . . 12211.3 Linear Sigma Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12411.4 More on the Mexican Hat Potential . . . . . . . . . . . . . . . . . . . . . 12511.5 SSB of Gauge Symmetries (“The Higg’s Mechanism”) . . . . . . . . . . . 126

Chapter 1

Preface

This set of notes are based on lectures given by Maxim Perelstein in the Quantum FieldTheory II course at Cornell University during Spring 2013. The course uses both Sred-nicki’s Quantum Field Theory and Peskin and Schroeder’s An Introduction to QuantumField Theory as reference texts. I wrote these notes during lectures and as such may con-tain some small typographical errors and sloppy diagrams. I also make notes for myselfthroughout the text of the form, [Q#:...]. They are just to remind to me to go back andtake a look at something and should be ignored. I’ve attempted to proofread and fix asmany of these problems as I can. I hope to continue to revise and add to these notesuntil I feel they are complete. If you have any corrections feel free to let me know [email protected].

1.1 ConventionsIn these notes we use the conventions used by Peskin and Schroeder. We use a “Westcoast metric”,

gµν =

1 0 0 00 −1 0 00 0 −1 00 0 0 −1

(1.1)

the Weyl basis,

γµ =

(0 σµ

σµ 0

)(1.2)

where σµ ≡ (1, σi) and σ ≡ (1,−σi).When applying Fourier transform (FT) we put the 2π on the momenta integral such

that:f(x) =

∫dp

2πf(p)e−ipx (1.3)

4

mailto:[email protected]

Chapter 2

Path Integrals in Quantum Mechanics

The primary tool of this course will be what’s known as the path integral. In non-relativistic quantum mechanics (NRQM) it is an alternative formulation to using theSchrodinger equation. Consider a particle at xa that evolves with a Hamiltonian

H = H(x, p; t) (2.1)

What we want to compute is the transition amplitude for this particle to go from itsinitial position some other position xb in a time 2T . In QM we denote this amplitudewith a braket:

U = 〈xb, T |xa,−T 〉 (2.2)

To simplify this expression consider inserting in a completeness relation N − 1 complete-ness relations:

=

∫dx1 dx2... dxN−1 〈xb, T |xN−1, T − ∆t〉〈xN−1, T − ∆t|xN−2, T − 2 ∆t〉 ... (2.3)

〈x1, −T + ∆t|xa,−T 〉

Now if we have a braket at two nearby times then you can evolve one side to have anequal time braket1:

〈x, t0|x′, t0 + ∆t〉 = 〈x, t0|e−iH(t1)·(t1−t0)|x′, t0〉 = 〈x|e−iH(t1)∆t|x′〉 (2.4)

where in the last step we rewrote the bra and ket without time dependence as the absolutevalue of time has no significance.

=

∫dx1 dx2... dxN−1 〈xb|e−iH(T−∆t)·∆t|xN−1〉〈xN−1|e−iH(T−2∆t)·∆t|xN−2〉 ...

〈x1|e−iH(−T )·∆t|xa〉 (2.5)

1To do this properly with finite ∆t requires Dyson’s formula

5

6 CHAPTER 2. PATH INTEGRALS IN QUANTUM MECHANICS

Now let t1 = −T, t2 = −T + ∆t, ..., tN = T . Thus we now have

=

∫dx1 dx2... dxN−1 〈xb|e−iH(tN−1)·∆t|xN−1〉〈xN−1|e−iH(tN−2)·∆t|xN−2〉 ...

〈x1|e−iH(t1)·∆t|xa〉 (2.6)

U =

(N−1∏i=1

∫dxi

)(N−1∏i=0

〈xi+1|e−iH(ti)∆t|xi〉

)(2.7)

=

(N−1∏i=1

∫dxi

)(N−1∏i=0

U(xi → xi+1; ∆t)

)(2.8)

Note that technically these equations need two less integrals then the number of xi values.However often people don’t bother to write it like this (since it somewhat ruins the asceticbeauty in the formula). The point is that you write down each of the brakets with everyposition included, but you only integrate over x1, ..., xN−1, not over x0 = xa and xN = xb.Either way they don’t matter much at the end, as long as you know not to integrate overyour end points... Instead this expression is more commonly written as

U =

(N∏i=0

∫dxiU(xi → xi+1; ∆t)

)(2.9)

Pictorially this integral is show below

t

−T12

...

T

xa xbx

This is called a path integral. For simplicity we focus on Hamiltonians are of the form

H =p2

2m+ V (x, t) (2.10)

7

Consider the braket above (if we know the value of this term we almost know everything).We take ∆t to be small.

〈xi+1|e−iH∆t|xi〉 = 〈xi+1|xi〉 − i∆t 〈xi+1|p2

2m+ V (x, ti)|xi〉 (2.11)

= δ(xi+1 − xi)− i∆t 〈xi+1|p2

2m|xi〉 − i∆tδ(xi+1 − xi)V (xi, ti) (2.12)

=

∫dpi2π

e−ipi(xi+1−xi) − i∆t 〈xi+1|p2

2m|xi〉 − i∆tV (xi, ti) (2.13)

×∫dpi2π

e−ipi(xi+1−xi) (no sum)

By insertion of complete set of states (note also that 〈pi|xi〉 = e−xipi) we have,

〈xi+1|p2|xi〉 =

∫dp′dp′′eixi+1p

′e−ixip

′′p′′ 2 〈p′|p′′〉 (2.14)

=

∫dp′

2πeixi+1p

′e−ixip

′p′ 2 (2.15)

〈xi+1|p2|xi〉 =

∫dpi2π

ei(xi+1−xi)pip2i (no sum) (2.16)

Therefore,

〈xi+1|e−iH(ti)∆t|xi〉 =

∫dpi2π

e−ipi(xi+1−xi)(

1− i∆t(p2i

2m+ V (xi, ti)

))(2.17)

=

∫dpi2π

e−ipi(xi+1−xi)e−i∆tH(xi,pi,ti) (2.18)

=

∫dpi2π

e−ipi∆t(xi+1−xi)

∆t e−i∆tH(xi,pi,ti) (2.19)

Note that in the last step we have combined our infinitesimal result into our exponential(in the end we take ∆t → 0). Notice that we no longer have operators; pi and Hi arenumbers!The expression has only numbers, we have moved away from regular quantummechanics. now taking the ∆t→ 0 limit we have

〈xi+1|e−iHi∆t|xi〉 =

∫dpi2π

e−i∆t(pixi−Hi) (2.20)

Thus we have

U =

(N−1∏i=1

∫dxi

)(N−1∏i=0

∫dpi2π

e−i∆t(xipi−Hi)

)(2.21)

=

∫dp0

2π

(N−1∏i=1

∫dxi

∫dpi2π

)e−i

∑N−1i=0 ∆t(xipi−Hi) (2.22)


Taking the limit we finally have

U =

(N−1∏i=1

∫dxi

∫dpi2π

)e−i

∫ T−T dt(xipi−Hi) (2.23)

where now that we have taken the limit we have just dropped the∫dp0 integral and the

value of p0 in the exponential. Since we have an infinite number of integrals, adding onemore shouldn’t effect the final result. Only when we discretize our integrals should sucha distinction be of any importance.

For convenience we define∫Dx = lim

N→∞

N−1∏i=1

∫ ∞−∞

dxi ;

∫Dp = lim

N→∞

N−1∏i=1

∫ ∞−∞

dpi (2.24)

We can now write

U =

∫DxDp e−i

∫ T−T dt(xipi−Hi) (2.25)

Note that the momenta integral has nothing to do with the particular dynamics of theproblems and can be done in general since they are just Gaussian integrals. The integralwe need is (we now go back to the discrete limit for a moment, see equation 2.22)∫

dpi2π

ei∆t(pixi−p2i2m−Vi) =

∫dpi2π

e−i∆t2m

(pi−mxi)2

ei∆t2m

(mxi)2−i∆tVi (2.26)

=

∫dpi2π

e−i∆t2m

(pi)2

ei∆t2m

(mxi)2−i∆tVi (2.27)

= C(∆t,m)ei∆t2m

(mxi)2−i∆tVi (2.28)

= C(∆t,m)ei∆t(m2xi

2−Vi) (2.29)

= C(∆t,m)ei∆tL[xi,ti] (2.30)

where we have denoted the value of the momenta integral by C(∆t,m). Though notobvoius, it’s value will turn out to have no impact on physical quantities. Neverthelesswe include for completeness,

C(∆t,m) =

√2πm

i∆t(2.31)

Thus we have

U = CN−1∏i=1

∫dxi exp

(i∆t

N∑j=1

L(xj, t)

)= C

∫Dx exp

(i

∫ T

−TL[x(t), t]dt

)= C

∫Dx eiS

∣∣∣∣x(−T )=xa,x(T )=xb

(2.32)

2.1. HARMONIC OSCILLATOR 9

where S is the action. This can be thought of as

U =∑paths

eS(path) (2.33)

We worked in natural units. Going back to regular units we have

U =∑paths

eS(path)/~ (2.34)

A classical system is one that has a large action (alternatively one can take the Bohrlimit of ~→ 0). We are dealing with a rapidly oscillating integrand. The trajectory thatends up being the dominant contribution is the one that the smallest phase (S(path)/~).This is the one that has minimum S, as in classical mechanics.

2.1 Harmonic OscillatorWe will now do an example. Consider a simple harmonic oscillator with an external force,f(t) . The Hamiltonian is given by

H =p2

2+ω2x2

2− ω

2︸︷︷︸H0

H′︷︸︸︷−f(t)x (2.35)

(set m = 1). We consider a force that is turned on and then turned off at some timelater:

f(t)

−t0 t0t

Initially the system is in the ground state (|0〉), what’s the probability to remain in |0〉after t0? In standard quantum mechanics we compute probabilities as

P = |〈0, t0|0,−t0〉|2 (2.36)

It’s unclear how this relates to the path integral formulation since so far we’ve onlydiscussed probabilities of a particle moving from some x eigenstate to another x eigen-state.[Q 1: At the end repeat this problem using standard PT.] Consider (we use Dyson’sformula discussed above)

U(xa → xa, 2T ) ≡ 〈xa|e−i∫ T−T Hdt|xa〉 (2.37)


with T > t0. We split the braket into 3 parts:

U(xa → xa, 2T ) = 〈xa|e−2iH0T e−i∫ t0−t0

H′ dt|xa〉 (2.38)

= 〈xa|e−iH0(T−t0)e−i∫ t0−t0

H′ dte−iH0(t0−(−T ))|xa〉 (2.39)

= 〈xa|e−iH0T e−i∫ t0−t0

H dte−iH0T |xa〉 (2.40)

where we have used the fact that H ′ is only applied between −t0 and t0 and in the laststep we were careful to write H ′ in Dysons formula not H0. We now insert two completeset of states:

U(xa → xa, 2T ) =

∫dx+dx− 〈xa|e−iH0T |x+〉〈x+|e−i

∫ t0t0Hdt|x−〉〈x−|eiH0T |xa〉 (2.41)

We now try to simplify the 〈x−|e−iH0T |xa〉 term. Consider,

e−iH0T |xa〉 =∞∑n=0

e−iH0T |n〉〈n|xa〉 (2.42)

=∞∑n=0

ψn(xa)e−iEnT |n〉 (2.43)

where we have inserted the set of eigenstates of H0. Now comes a “dirty trick”. TakeH0 → (1− iε)H (at the end we will take ε→ 0).

limε→0

limT→∞

e−iH0(1−iε)T |xa〉 = limε→0

limT→∞

∞∑n=0

ψn(xa)e−iEnT e−εTEn |n〉 (2.44)

Choose spectrum as En > 0, n > 0 and E0 = 0. If we take T → 0 then the only termthat contributes from the sum is the one that isn’t effected by the damping term, e−εTEn(the |0〉 term).

limε→0

limT→∞

e−iH0(1−iε)T |xa〉 = limε→0

ψ0(xa) |0〉 (2.45)

= ψ0(xa) |0〉 (2.46)

This key step helps us take position eigenstates and relate them to what we really wantwhich is energy eigenstates. Define U as U with H0 → H0(1− iε)

limε→0

limT→∞

U(xa → xa, 2T ) = |ψ0(xa)|2∫dx+dx− 〈0|x+〉〈x+|e−i

∫ t0−t0

Hdt|x−〉〈x−|0〉

(2.47)

= |ψ0(xa)|2 〈0|e−∫ t0−t0

Hdt|0〉 (2.48)

= |ψ0(xa)|2 〈0, t0|0,−t0〉 (2.49)∫ ∞−∞

dxa limε→0

limT→∞

U(xa → xa, 2T ) = 〈0, t0|0,−t0〉 (2.50)


due to the normalization in the wavefunction. We now go back to our formula for thepath integral. Note that don’t go to our “final” form of our path integral equation sinceto simplify the momentum integral we assumed H of the form p2

2m+ V (which is not the

case now that we threw in the iε). Instead we go to our earlier form (see equation 2.25).

limT→∞

U =

∫DxDp exp

(i

∫ ∞−∞

dt(px− (1− iε)H0 + f(t)x)

)(2.51)

=

∫Dx∫ ∏ dp

2πexp

(i∆t

(1− iε)2

− p2 + 2px (1 + iε)− x2 (1 + iε)2

+ x2 (1 + iε)2 + 2 (1 + iε) f(t)x− ω2x2

)(2.52)

where we have used 1+iε ≈ (1− iε)−1, and we ignore the constant term in H0 since it hasno impact on the final result. The term in the exponential can be factored into a constantterm (with respect to dp) and a −(p − x(1 + iε))2 term. We define p′ ≡ p − x(1 + iε).We have the integral∫

dp

2πexp

(−i∆t1− iε

2p2

)=

1

2π

√2π

i∆t(1 + iε)1/2 (2.53)

=

√−i

2π∆t(1 + iε) (2.54)

Finally we have

limT→∞

U = C∫Dx exp

(i

∫ ∞−∞

dtx2

2(1 + iε)− ω2x2

2(1− iε) + f(t)x

)(2.55)

To do this integral we need a way to put x and x in a similar form. We do this using aFourier transform:

x(t) =

∫ ∞−∞

dE

2πe−iEtx(E) (2.56)

x(t) =

∫dE

2π(−iE)e−iEtx(E) (2.57)

f(t) =

∫dE

2πe−iEtf(E) (2.58)

The integration measure can be changed to Dx. This is because every variable in x spacehas a one-to-one correspondence with the variables in k space and the Jacobian to switchspaces is one (this is shown in appendix 2.A). We have∫

dt f(t)x =

∫dE

2πf(E)

∫dE ′

2πx(E ′)2πδ(E + E ′) (2.59)

=

∫dE

2πf(E)x(−E) (2.60)

=1

2

(∫dE

2πf(E)x(−E) + f(−E)x(E)

)(2.61)


Similarly, ∫dt x2 =

∫dE

2πx(E)

∫dE ′

2πx(E ′)2πδ(E + E ′)(−1)EE ′ (2.62)

=

∫dE

2πx(E)x(−E)E2 (2.63)

=1

2

∫dE

2πE2

(x(E)x(−E) + x(−E)x(E)

)(2.64)

limT→∞

U = C∫Dx exp

(i1

2

∫dE

2π

E2−ω2+iε︷︸︸︷(E2(1 + iε)− ω2(1− iε)

)x(E)x(−E)

+ f(E)x(−E) + f(−E)x(E)

)(2.65)

The idea is to complete the square for each dx. The variable that does that is y(E) =

x(E) + f(E)E2−ω2+iε

. The arguement in the exponential turns into,

(...) =i

2

∫dE

2π

(E2 − ω2 + iε

)(y (E)− f(E)

E2 − ω2 + iε

)(y (−E)− f(−E)

E2 − ω2 + iε

)(2.66)

+ f(E)

(y(−E)− f(−E)

E2 − ω2 + iε

)+ f(−E)

(y(E)− f(E)

E2 − ω2 + iε

)(2.67)

=i

2

∫dE

2π

(E2 − ω2 + iε

)y(−E)y(E)− f(E)f(−E)

E2 − ω2 + iε(2.68)

Thus we can write,

limT→∞

U = C∫Dy exp

(i1

2

∫dE

2πy(E)(E2 − ω2 + iε)y(−E)

)exp

(i

2

∫dE

2π

f(E)f(−E)

−E2 + ω2 − iε

)(2.69)

Now consider the special case of f → 0. Then we have no perturbation and the amplitudeto stay in the ground is just 1. So in this case we have (the second exponential is trivial)

1 = C∫Dy exp

(i1

2

∫dE

2πy(E)(E2 − ω2 + iε)y(−E)

)(2.70)

This is a mathematical relation completely independent of the particular system that weconsider. Thus we can insert it into equation 2.69 to get a final succinct expression forthe amplitude,

〈0, t0|0,−t0〉 = limε→0

exp

(i

2

∫dE

2π

f(E)f(−E)

−E2 + ω2 + iε

)(2.71)


The (E2 − ω2)−1 factor is remenicient of the inverse of the simple harmonic oscillatorequation. This hints that this can be rexpressed in terms of the Green’s function forsimple harmonic motion defined by:(

∂2t + ω2

)G(t− t′) = δ(t− t′) (2.72)

Writing both sides our using a Fourier decomposition,∫dE

2π

(−E2 + ω2

)G(E)e−iE(t−t′) =

∫dE

2πe−iE(t−t′) (2.73)

and so,

G =1

E2 − ω2(2.74)

Now taking the inverse Fourier Transform,

G(t′ − t) =

∫dE

2π

e−iE(t−t′)

−E2 + ω2(2.75)

As always there is an ambiguity on the path of integration. We choose the Feynmandecomposition:

G = limε→0

∫dE

2π

e−iE(t−t′)

−E2 + ω2 − iε(2.76)

Using this expression we can rewrite the amplitude in equation 2.71. The argument inthe exponential is

(...) =i

2

∫dE

2π

∫dtdt′f(t)f(t′)

eiEte−iEt′

E2 − ω2 + iε(2.77)

=i

2

dE

2π

∫dtdt′f(t)G(t− t′)f(t′) (2.78)

where we have used the symmetric property of the Green’s functions: G(t−t′) = G(t′−t).Applying this into our equation above we have

〈0, t0|0,−t0〉 = exp

(i

2

∫ t0

−t0dtdt′f(t)G(t− t′)f(t′)

)(2.79)

Alternatively we can express the amplitude using the definition,

W [f(t)] ≡ limT→∞

∫x(T )=x(−T )

Dx exp

[i

∫ T

−Tdt(L0 + fx)

]=

∫Dx exp

[i

∫ ∞−∞

dt(L0 + fx)

](2.80)

where L0 is the free Lagrangian with all the ε inserted. Using Eq. 2.55 we see that wecan write,

〈0, t0|0,−t0〉 = limε→0CW [f ] (2.81)


if f = 0 then

〈0, t0|0,−t0〉 = 1 ⇒ C =1

W [0](2.82)

So in our new notation the amplitude is given by,

〈0, t0|0,−t0〉 =W [f ]

W [0](2.83)

where,

W [f ] = W [0] exp

[i

2

∫dE

2π

f(E)f(−E)

−E2 + ω2 − iε

](2.84)

= W [0] exp

[i

2

∫dtdt′f(t)G(t− t′)f(t′)

](2.85)

since W [0] just cancels away in the calculation of a transition amplitude we don’t needto calculate it explicitly.

2.2 Anharmonic OscillatorUp until now the Hamiltonian of our system was the simply quadratic free Hamiltonianwith a time dependent perturbation. The eigenstates were left unchanged but the f(t)xterm allowed the possibility of a state transitioning into another state. The fact that wewere able to complete the square and exactly integrate the exponential was instrumentalin finding a simple form for the amplitude. This was a consequence of not having anyterms in the exponential that are more then quadratic in x. We now consider modifyingthe Hamiltonian to include such interactions.

Consider the anharmonic oscillator with a driving force:

H0 =p2

2+ω2x2

2(2.86)

Ha =λx4

4!(2.87)

andH = H0 +Ha − f(t)x (2.88)

We have some ground state, |Ω〉. We add a constant to the Hamiltonian such that Eω = 0.Our discussion earlier was rather general and we can reuse some of our earlier results

〈Ω, t0|Ω,−t0〉 =Wλ [f ]

Wλ [0]‘ (2.89)

whereWλ [f ] ≡

∫Dx exp

[i

∫ ∞−∞

dt(L0 − (λx4

4!)(1− iε) + f(t)x)

](2.90)

2.2. ANHARMONIC OSCILLATOR 15

Due to the quartic term we cannot simply do this integral as before. There is not generalway to integrate this Gaussian. What we can do, is do a small λ expansion (perturbationtheory).

The 1 − iε is important in the initial Lagrangian (L0) to pick out the ground stateand relate W to the amplitude of transition from the ground state. However it turns outthat the 1− iε has no effect on the anharmonic term. [Q 2: ellaborate on this.]. Thus wecan take the ε that’s there to zero. We do a small λ expansion (i.e. Taylor expansion):

Wλ [f ] =

∫Dx exp

[i

∫ ∞−∞

dt(L0 − fx)

]exp

[−iλ4!

∫ ∞−∞

dtx4

](2.91)

≈∫Dx exp

[i

∫ ∞−∞

dt(L0 − fx)

]1− iλ

4!

∫dtx4(t)

+λ2

(4!)2

∫dt1 dt2x

4(t1)x4(t2) + ...

(2.92)

We can write this discretely as

Wλ [f ] ≈N−1∏i=1

∫dxi exp

[i∑n

∆t(L0,n − fnxn)

]1− iλ

4!∆t∑j

x4j

+λ2

(4!)2(∆t)2

∑j1,j2

x4j1x4j2

+ ...

(2.93)

or compactly as

Wλ [f ] = W0 [f ]− iλ

4!

∫ ∞−∞

dt

∫Dx x4(t) exp

[i

∫ ∞−∞

dt(L0 + fx)

]+

(−i)2

2!

(λ

4!

)2 ∫dt1 dt2

∫Dx x4(t1)x4(t2) exp

[i

∫dt(L0 + fx)

]+ ...

(2.94)

First consider W0 [f ]. We already know this function since we calculated it last class (wecalled it W [f ] before we had a perturbation):

W0 [f ] =

∫Dx exp

[i

∫ ∞−∞

dt(L0 + fx)

](2.95)

=N∏i=1

∫dxi exp

[iN∑j=1

∆t(L0 j + fjxj)

](2.96)

where fj = f(−T + j∆t) and xj = x(−T + j∆t). We know how to do this integralexplicitly (see Eq. 2.85). We hope to rewrite the other terms in terms of W0 [f ]. To dothis we define something called a “functional derivative”:

δW0

δf(tj′)≡ 1

∆t

∂W0

∂fj′(2.97)


In our case this is given by

δW0

δf(tj′)=

N∏i=1

∫dxiixj′ exp

[iN∑j=1

∆t(L0,j + fjxj)

](2.98)

= i

∫Dx x(tj′) exp

[i

∫dt(L0 + fx)

](2.99)

So at least in this case taking the functional derivative of W0 with respect to f at a timetj′ does the thing you would expect; it pulls out a factor of ix(j′).

With this in mind we can write

Wλ[f ] = W0[f ]− iλ

4!

∫ ∞−∞

dt

(1

i

δ

δf(t)

)4

W0[f ]

+(−i)2

2!

(λ

4!

)2 ∫dt1 dt2

(1

i

δ

δf1

)4(1

i

δ

δf2

)4

W0[f ] + ... (2.100)

We begin by considering O(λ):

− iλ

4!W0 [0]

∫dt

(1

i

δ

δf(t)

)4

exp

[i

2

∫dτdτ ′f(τ)G(τ − τ ′)f(τ ′)

](2.101)

For brevity we write,

exp [...] ≡ exp

[i

2

∫dτdτ ′G(τ − τ ′)f(τ)f(τ ′)

](2.102)

We now take the functional derivatives:

1

i

δ

δf(t)exp [...] =

1

i∆t

∂

∂fj′exp

[i

2

∑i,j

∆t∆tG(τi − τj)fifj

](2.103)

=1

i

1

∆t

i

2∆t2(

∑i

G(τi − tj′)fi +∑j

G(tj′ − τj)fj) exp [...] (2.104)

=1

2∆t(∑i

G(τi − tj′)fi +∑j

G(tj′ − tj)fj) exp [...] (2.105)

=∑j

∆tG(tj′ − τj)fj exp [...] (2.106)

=

∫dτ ′G(t− τ ′)f(τ ′) exp [...] (2.107)

So again the functional derivative does what one would expect. It takes the derivative ofthe arguement of the exponential and uses the product rule on it. Continuing on gives,(

1

i

δ

δf(t)

)2

exp[...]

=1

iG(t−t) exp

[...]+

∫dτ ′dτ ′′G(t−τ ′)G(t−τ ′′)f(τ ′)f(τ ′′) exp

[...]

(2.108)

2.2. ANHARMONIC OSCILLATOR 17(1

i

δ

δf(t)

)3

exp[...]

=

1

iG(t− t)

∫dτ ′G(t− τ ′)f(τ ′) +

2

iG(t− t)

∫dτ ′G(t− τ ′)f(τ ′)

+

∫dτ ′dτ ′′dτ ′′′G(t− τ ′)G(t− τ ′′)G(t− τ ′′′)f(τ ′)f(τ ′′)f(τ ′′′)

exp [...] (2.109)

=

3

iG(t− t)

∫dτ ′G(t− τ ′)f(τ ′)

+

∫dτ ′dτ ′′dτ ′′′G(t− τ ′)G(t− τ ′′)G(t− τ ′′′)f(τ ′)f(τ ′′)f(τ ′′′)

exp [...] (2.110)

(1

i

δ

δf(t)

)4

exp[...]

=

− 3G2(t− t) +

3

iG(t− t)

∫dτ ′dτ ′′G(t− τ ′)G(t− τ ′′)f(τ ′)f(τ ′′)

+3

iG(t− t)

∫dτ ′dτ ′′G(t− τ ′)G(t− τ ′′)f(τ ′)f(τ ′′) (2.111)

+

∫dτ ′dτ ′′dτ ′′′dτ ′′′′G(t− τ ′)G(t− τ ′′)G(t− τ ′′′)G(t− τ ′′′′)

× f(τ ′)f(τ ′′)f(τ ′′′)f(τ ′′′′)

exp [...] (2.112)

=

− 3G2(t− t) +

6

iG(t− t)

∫dτ ′dτ ′′G(t− τ ′)G(t− τ ′′)f(τ ′)f(τ ′′)

+


× f(τ ′)f(τ ′′)f(τ ′′′)f(τ ′′′′)

exp [...] (2.113)

We now consider Wλ [0] (i.e. the above expression with f = 0). This is the normal-ization term for every amplitude.

Wλ [0] = W0 [0] + i3λ

4!W0 [0]

∫ ∞−∞

G2(t− t) dt (2.114)

and in summary we have,

Wλ[f ] = W0[f ]−W0 [f ]iλ

4!

∫ ∞−∞

dt

− 3G2(t− t)

+6

iG(t− t)

∫dτ ′dτ ′′G(t− τ ′)G(t− τ ′′)f(τ ′)f(τ ′′)

+


× f(τ ′)f(τ ′′)f(τ ′′′)f(τ ′′′′)

(2.115)


Recall that the transitions amplitudes are,

〈0,−T |0, T 〉 =Wλ [f ]

Wλ [0](2.116)

So to first order the transition amplitude is,

〈0,−T |0, T 〉 = 1− iλ

4!

∫ ∞−∞

dt

− 3G2(t− t) +

6

iG(t− t)

∫dτ ′dτ ′′G(t− τ ′)G(t− τ ′′)

× f(τ ′)f(τ ′′) +


× f(τ ′)f(τ ′′)f(τ ′′′)f(τ ′′′′)

(2.117)

This calculation was quite tedius and was only to first order. Nevertheless there aretwo main independent ingredients that we can identify. The order of the interaction(which is always accompanied by an integration of t) and the Green’s function. You maythink that the factors f(τ)dτ are also independent but that’s not true. The number ofthese factors and their arguements are competely determined by the form of the Green’sfunctions. Furthemore, as we will see for our purposes we will take these to zero at theend of the calculations. So we can now identify the pieces that will always be part of thetransition amplitude calculations. These will be our “Feynman Rules”:

→ −iλ∫∞−∞ dt

→ G(t1 − t2)

Furthermore, a time will be given to each intersection point in a diagram and a symmetryfactor is also required for each diagram. So for example we can write,

Wλ [0] = W0 [0]

(1 + + + + +...

)

2.3 Correlation Functions

Now that we have an idea of how to generalize our discussion to more complicated po-tentials we quickly go back again to consider the simple harmonic oscillator. Considerthe following derivative with t1 < t2:

1

i2δ2W0 [f ]

δf(t1)δf(t2)

∣∣∣∣f→0

= W0 [0]G(t1 − t2) (2.118)

2.3. CORRELATION FUNCTIONS 19

Alternatively we can rewrite this expression in terms of the Lagrangian,

1

i2δ2W0

δf1δf2

∣∣∣∣f→0

=

∫Dx x(t1)x(t2) exp

(i

∫dtL0

)(2.119)

= limT→∞

∫ T

t2

Dx∫dx2

∫ t2

t1

Dx∫dx1

∫ t1

−TDx x2x1

× exp

[i

(∫ t1

−T+

∫ t2

t1

+

∫ T

t2

)dtL]

(2.120)

= limT→∞

∫ T

t2

Dx∫dx2 x2

∫ t2

t1

Dx∫dx1 x1

∫ t1

−TDx

× exp

[i

(∫ t1

−T+

∫ t2

t1

+

∫ T

t2

)dtL]

(2.121)

=

∫dx1

∫dx2x1x2 〈0, T |x2, t2〉〈x2, t2|x1, t1〉〈x1, t1|0,−T 〉W0 [0] (2.122)

where in the second line we have split the integration interval into the different parts.This is trivial in the exponential. However, on the outside extra spatial integrals overdx1 and dx2 appear due to the way the Dx are defined; they have the end points fixedso to split them up you need to have an integral over each midpoint.

We now switch to the Heisenberg picture. We choose our reference time −T . Bydefinition then we have

|0〉 ≡ |0,−T 〉 (2.123)

likewise we define|x〉 ≡ |x,−T 〉 . (2.124)

With this reference point we can move into the Heisenberg interperpotation. The Heisen-berg form of the position operator is, x(t) = e−iH0(t+T )x(−T )eiH0(t+T ). We want to thinkof x1 as an eigenvalue of this operator. In other words

xi |xi〉 = x(ti) |xi〉 (2.125)

This enables us to get rid of the x integrals:

1

i2δ2W

δf1δf2

∣∣∣∣f→0

= W0 [0] 〈0|x(t2)x(t1)|0〉 (2.126)

This expression holds if t2 > t1. If t1 > t2 everything is the same by symmetry butt1 ↔ t2 and so we have,

1

W0 [0]

1

i2δ2W

δf1δf2

∣∣∣∣f→0

= 〈0|T (x(t2)x(t1)) |0〉 (2.127)

where T denotes the time ordering operator which puts the operator with the earliesttime to the right. So we see that correlation functions (key objects in Quantum Field


Theory!) can be related to W by taking derivatives. In the case of the anharmonicoscillator we have already taken the first derivative of W0 [f ] (see Eq 2.107) we just needto take a second at a different time,

〈0|T (x(t2)x(t1)) |0〉 =1

W0 [0]

1

i

δ

δf1

(∫dτ ′G(t2 − τ ′)f(τ ′) exp [...]

) ∣∣∣∣f→0

(2.128)

=1

iG(t2 − t1) (2.129)

Our discussion did not depend much on the form of the Hamiltonian. If we had Eq2.113 with or without λ = 0 the results would be the same as long as we label our groundstate by |Ω〉 (the ground state of the full system (in our case the anharmonic theory),

〈Ω|T (x(t1)x(t2)) |Ω〉 =1

Wλ [0]

1

i2δ2Wλ [f ]

δf1δf2

∣∣∣∣f→0

(2.130)

=1

Wλ [0]

1

i2δ2

δf1δf2

(W0 −

iλ

4!

∫dt

(1

i

δ

δf

)4

W0 + ...

)∣∣∣∣f→0

(2.131)

We already sketched out Wλ [0]. Calculating

δ2Wλ

δf1δf2

∣∣∣∣f→0

(2.132)

explicitly gives any possible diagram with two external points (two unconnected times, t1and t2). We show this to first order. The expression for Wλ is given in Eq 2.113. Takingits derivatives gives to first order,

1

i2δ2Wλ

δf1δf2

∣∣∣∣f→0

=

1

iG(t1 − t2)− iλ

4!

∫dt 3iG2(0)G(t1 − t2)

+ 12iG(0)G(t− t1)G(t− t2)

W0[0] (2.133)

The zero’th order contribution is a Green’s function starting at the initial point andending at the second (i.e. from t1 to t2). The first order contributions are the a doublebubble diagram (due to the G2(0) factor) with the zero’th order Green’s function and thesecond term is a bubble and Green’s function going in and one going out. Now to find theamplitudes we need to divide by Wλ[0] to get the correlation functions. We calculatedthis quantity earlier on and to first order is given by,

Wλ [0] = W0[0]

(1− 3λi

4!

∫dtG2(0)

)(2.134)


So to first order we have,

〈Ω|T (x(t1)x(t2)) |Ω〉 =1iG(t1 − t2) + λ

4!

∫dt3G2(0)G(t1 − t2) + 12G(0)G(t− t1)G(t− t2)

1− 3iλ4!

∫dtG2(0)

(2.135)

=

(1iG(t1 − t2) + iλ

4!

∫4G(0)G(t− t1)G(t− t2)

) (1− iλ

4!

∫dt3G2(0)

)1− 3iλ

4!

∫dtG2(0)

(2.136)

=1

iG(t1 − t2) +

iλ

4!

∫4G(0)G(t− t1)G(t− t2) (2.137)

To all orders we have,

+ + + + + + ...

+ +1 + ...

+ += + ...

We now briefly summarize what we have done. We have the correlation functions foranharmonic oscillator (x4):

〈Ω|T (x(t1)x(t2)) |Ω〉 =1

Wλ [0]

δ2Wλ [f ]

δf1δf2

∣∣∣∣f→0

=∑

connected Feynman diagrams

(2.138)where Wλ [f ] is the generating functional and fi ≡ f(ti). We found the propagator (tolowest order) is given by,

→ G(t1 − t2) =∫

dE2π

e−iE(t1−t2)

−E2+ω2+iε

and the vertex is given by

→ −iλ∫dt

x → −iλ∫dt. These rules are the space time Feynman rules. By Fourier trans-

forming both sides we have the momentum space rules:

→ 1−E2+ω2+iε

→ −iλδ(∑

iEi)


2.A Jacobian of the Fourier TransformationThe discrete Fourier transform is given by

xk =1√N

∑n

e−2πink/Nxn (2.139)

The Jacobian is then the matrix

Jn,k =∂xk∂xn

=1√Ne−2πink/N (2.140)

We want to show that the absolute value of the determinate of this matrix is equal to 1.Note that the determinate any orthonormal matrix is ±1 since:

1 = det(A−1A

)= detAT detA = (detA)2 (2.141)

thus if we can show that J is orthonormal then |det J | = 1.A matrix is orthonormal if its rows and columns are orthonormal. We need to show

that

〈Jk|Jj〉 =1

N

(e2πik/N , e2πi2k/N , ..., e2πik

)

e−2πij/N

e−2πi2j/N

...e−2πij

(2.142)

=1

N

(e2πi(k−j)/N + e2πi2(k−j)/N + ...e2πi(k−j)) (2.143)

is equal to 0 for all row and column numbers, k − j 6= 0. Let k − j = m ∈ Z. Then wehave,

=1

N

N∑j=1

e2πijm/N (2.144)

=1

N

N/2m∑j=1

e2πijm/N +

2N/2m∑j=N/2m+1

e2πijm/N + ...+N∑

j=(2m−1)N/2m+1

e2πijm/N

(2.145)

=1

N

2m−2∑`=0

(`+1)N/2m∑j=`N/2m+1

e2πijm/N (2.146)

shifting the exponential such that each sum starts at j = 1 gives,

(`+1)N/2m∑j=`N/2m+1

e2πijm/N → e`πiN/2m∑j=1

e2πijm/N (2.147)

2.A. JACOBIAN OF THE FOURIER TRANSFORMATION 23

Thus we finally get that the dot product between an arbitrary row and column is givenby,

=1

N

N/2m∑j=1

e2πijm/N

(2m−2∑`=0

e`πi

)︸︷︷︸

=0

(2.148)

= 0 (2.149)

There is one loophole. If k − j = m = 0 then we can’t divide by m. In this case we get,

1

N

N∑j=1

e2πijm/N =1

N

N∑j=1

= 1 (2.150)

Thus 〈Jj|Jk〉 = δjk and the Jacobian is indeed orthonormal and thus

|det J | = 1 (2.151)

as required.

Chapter 3

Quantum Field Theory

3.1 Free Klien Gordan Field

To keep things simple we consider one non-interacting real scalar field (real Klien Gordantheory). We work in 1 + 1 dimensions.

Since we are working in 1 dimension we have

φ(x, t)discretize

space−−−−−→ φ = φ1(t), φ2(t), ..., φN(t) (3.1)

where φ`(t) = φ(

x`︷︸︸︷`∆ , t)

√∆ and the

√∆ is a normalization factor. The equations of

motion areφ+m2φ = 0 (3.2)

To discretize them we need to find the second derivative in terms of finite differences. Wehave,

∂2φ

∂x2=

∂

∂xlimh→0

φ(x+ h/2)− φ(x− h/2)

h(3.3)

= limh→0

1

h2(φ(x+ h)− 2φ(x) + φ(x− h)) (3.4)

→ 1

∆2(φ`+1 − 2φ` + φ`−1) (3.5)

So the discrete equations of motion are,

φ` +

(2

∆2+m2

)φ` −

1

∆2(φ`+1 + φ`−1) = 0 (3.6)

Since we are working in one spatial dimesion we have L =∫dxL with

L =1

2(∂µφ)(∂µφ)− 1

2m2φ2 (3.7)

24

3.1. FREE KLIEN GORDAN FIELD 25

where our metrix is

gµν =

(1 00 −1

)(3.8)

Note that this Lagrangian is equivalent to having N of Harmonic oscillators (no sourcesor interaction terms) with one at each ` value. Discretizing our Lagrangian gives

L =N∑`=1

1

2

(φ2` −

(φ`+1 − φ`

∆

)2

−m2φ2`

)(3.9)

=N∑`=1

1

2φ2` −

N∑`,m=1

1

2φ`V`mφm (3.10)

where

V ≡

. . . . . . 0 . . . 00 − 2

∆2 −m2 1∆2 0 0

0 0. . . . . . 0

...... 0

. . . . . .0 0 0 0

(3.11)

is a coupling matrix between the fields at different spatial points.Note the following oscillator correspondence:

Klien-Gordan NR oscillatort tx label `φ` x`(t)

φ`(t) x`(t)J (x,t) f`(t)

(for the time being we use hats to denote operators). We work in the Schrodinger picture:

φ =φ1, ..., φN

(3.12)

and limN→∞ φ/√

∆ = φ(x). A definite field state is given by |φ, t〉 :

φ |φ0, t〉 = φ0 |φ0, t〉 (3.13)

we are interested in a transition between two states 〈φf , T |φi,−T 〉. Each state consistsof a direct product of states, one at each point in space. Thus finding the braket betweentwo field configurations is equivalent to a product of brakets. In other words we canwrite,

〈φf , T |φi,−T 〉 = 〈φ′1, T |φ1,−T 〉〈φ′2, T |φ2,−T 〉 ... 〈φ′N , T |φN ,−T 〉 (3.14)

26 CHAPTER 3. QUANTUM FIELD THEORY

with 〈φ′`, T |φ`,−T 〉 being the product of oscillator states at ∆`. But we already knowwhat the value of each of these brakets is for a singlet oscillator. Now we are doing thesame calculation but for many oscillators. It is given by (c.f. equation 2.32)

〈φf , T |φi,−T 〉 = CN−1∏`=1

M−1∏j=1

∫φ`,0=φ`iφ`,M=φ`f

dφ`,j exp

[i

M−1∑n=1

∆tL [φn ]

](3.15)

where the label ` refers to spatial points and the j denotes to time slices. Furthermore,we have introduced the notation dφ`j ≡ dφ`(tj). We now take the limit of N → ∞ andM →∞:

〈φf , T |φi,−T 〉 = C∫Dφ exp

[i

∫ T

T

dt

∫ ∞−∞

dxL [φ(x, t)]

], (3.16)

where ∫Dφ ≡

N−1∏`=1

M−1∏j=1

dφi,j. (3.17)

The interpretation is similar to before. The amplitude of going from one field config-uration to another is equal to the sum over all possible field configurations. Note theextension to multidimensional derivation is completely analogous. For 3 spatial dimen-sions:

〈φf , T |φi,−T 〉 = C∫Dφ exp

[i

∫ T

−Tdt

∫ ∞−∞

d3xL [φ(x, t)]

](3.18)

3.2 External ForceConsider some one dimensional external force, J(x, t) which discretized is given by J`(t)

√∆

(one force for each point and since this is just a field it discretizes in the same way asthe field φ). We take the force to turn on at −t0 and end at t0. We want to compute theprobability for the system to stay in the ground state,|0〉. We take the energy of the stateto be 0 (E0 = 0). Everything we did for a single oscillator can be done for N oscillators.Each of these oscillators has a force which is dependent on time. We denote the force by(J = J1(t), J2(t), ..., JN−1(t)) (here Ji(t) is the f(t) for each oscillator).

〈0, t0|0,−t0〉 = limε→0

∏i

W [Ji]

W0 [0]≡ lim

ε→0

W0 [J(t)]

W0 [0](3.19)

where (c.f. Eq. 2.55)

W0 [J(t)] = limT→∞

N∏j=1

∫φj(−T )=φj(T )

Dφj exp

[i

∫ T

−Tdt

( N∑`=1

1

2φ2`(1 + iε)−

N∑`,n=1

1

2(1− iε)φ`V`nφn

+N∑`=1

φ`J`(t)

)](3.20)

3.2. EXTERNAL FORCE 27

here we have added the 1 + iε and 1 − iε factors in an analagous way to our discussionof Non-relativistic QM. While they are present for notational convenience we will justignore them. More typically we will write

limT→∞

N−1∏j=1

∫φj(−T )=φ`(T )

Dφ` →∫Dφ (3.21)

and dt dx→ d2x such that

W0 [J(t)] =

∫Dφ exp

[i

∫d2x

1

2(∂µφ)(∂µφ)− 1

2m2φ2 + J(xµ)φ(xµ)

](3.22)

As before then if we can do the Gaussian integral then we are set. The way you do it isto diagonalize the argument so you just a product of individual integrals. Since the termwhich gives us trouble connects different space points (the derivative) we can diagonalizeit using a Fourier transform,

W0 [J(t)] = C∫Dφ exp

[i

∫dtdk

2π

1

2

∣∣ ˙φ∣∣2 − 1

2k2∣∣φ∣∣2 − 1

2m2∣∣φ∣∣2 + J∗φ

](3.23)

= C∫Dφ exp

[i

∫dtdk

2π

1

2

∣∣ ˙φ∣∣2 − 1

2ω(k)2

∣∣φ∣∣2 + J∗φ

](3.24)

where

φ(x, t) =

∫dk

2πe−ikxφ(k, t) , J(x) =

∫dk

2πe−ikxJ(k, t)

ω(k)2 ≡ k2 +m2 (3.25)

and we have changed our integration measure using,

N−1∏`=1

∫Dφ` =

N−1∏`=1

M−1∏m=1

∫dφ`m = J

N−1∏j=1

M−1∏m=1

∫dφj,m (3.26)

where J is a Jacobian which we can forget about since it just adds to the overall constant.Thus we can write

N−1∏`=1

∫Dφ` →

N−1∏j=1

∫Dφj (3.27)

Now we have (we don’t need to worry about the factor of 2π as it goes to the constantin front),

W0 [J(t)] = CN∏k=1

∫Dφ exp

[i

∫ ∞−∞

dt

(1

2| ˙φk|2 −

1

2ω2k|φk|2 + φ∗kJk

)](3.28)

= CN∏k=1

W SHO0

[Jk(t), ωk

](3.29)


where W SHO0 is a quantity that we already calculated earlier,

W SHO0 [f, ω] = exp

i

2

∫dt dt′f(t)G(t− t′)f(t′)

(3.30)

= exp

[i

2

∫dE

2π

f(E)f(−E)

−E2 + ω2 − iε

](3.31)

We can finally write

W0 [J] = W0 [0] exp

[i

2

N∑k=1

∫dE

2π

Jk(E)Jk(−E)

−E2 + ω2k − iε

](3.32)

where Jk(E) =∫dtJk(t)e

−iEt =∫dt∑N

j=1 eikjJj(t)e

−iEt. We have xj = ∆j and wedefine the Fourier conjugate variable to x as p and it must be given by k/∆. Taking thecontinuum limit gives

Jk(E) =

∫dt

(∫dx

∆

)eikj

(J(x, t)

√∆)e−iEt (3.33)

=1√∆

∫dt dxJ(x, t)ei∆jpe−iEt (3.34)

=1√∆

∫d2xJ(x, t)e−i(Et−xp) (3.35)

=1√∆

∫d2xJ(x, t)e−ipµx

µ

(3.36)

≡ 1√∆J(p, E) (3.37)

now∑

k → ∆∫

dp2π

and (Note: We got J(−p) by relating J(−p) to J∗(p).

W0 [J] = W0 [0] exp

[i

2

∫dp

2π

∫dE

2π

J(p)J(−p)−E2 + p2 +m2 − iε

](3.38)

In 3 + 1 dimensions we have

W0 [J ] = W0 [0] exp

[− i

2

∫d4p

(2π)4

J(p)J(−p)p2 −m2 − iε

](3.39)

we can equally well write this equation as

W0 [J ] = W0 [0] exp

[− i

2

∫d4x d4x′J(x)G(x− x′)J(x′)

](3.40)

where the Green’s function is defined by (x +m2 − iε)G(x− x′) = iδ4(x− x′)⇒

G(x− x′) =

∫d4p

(2π)4

e−ip·(x−x′)

p2 −m2 + iε(3.41)

3.3. INTERACTING THEORY 29

3.3 Interacting Theory

Suppose we turn on an interaction term H =∫d3x λ

4!φ4(x)→

∑`λ4!φ4` . We obtain

Wλ [J ] =

∫Dφ exp

[i

∫d4x

(L0 −

λφ4

4!+ J(x)φ

)](3.42)

= W0 [J ]− iλ

4!

∫d4x

∫Dφ φ4(x) exp

[i

∫d4x

(L0 + Jφ

)]+ ... (3.43)

= W0 [J ]− iλ

4!

∫d4x

∫Dφ

(1

i

δ

δJ(x)

)4

exp

[i

∫d4x

(L0 + Jφ

)]+ ... (3.44)

= W0 [J ]− iλ

4!

∫d4x

(1

i

δ

δJ(x)

)4

W0 [J ] + ... (3.45)

This equation has the exact same form as for a single harmonic oscillator.We have

Wλ [0] = W0 [0]− iλ

4!

∫d4x− 3G(x− x)2 + ... (3.46)

As before every term is made up of two types of terms. We have

→ −iλ∫dx

→ G(x1 − x2)x1 x2

In Quantum Mechanics we had

〈Ω|T (x(t1)x(t2)) |Ω〉 =1

Wλ [0]

δ2Wλ [f ]

δf(t1)δf(t2)

∣∣∣∣f→0

(3.47)

In field theory we have [Q 3: show]

〈Ω|T (φ(x1)φ(x2))|Ω〉 =1

Wλ [0]

δ2Wλ [J ]

δJ(x1)δJ(x2)

∣∣∣∣J→0

(3.48)

(all we did was replace the quantum mechanics source to the field theory source). Onecan then develop perturbation theory for fields as we did for the harmonic oscillator. Weget

〈Ω|T (φ(x1)φ(x2))|Ω〉 = + + + ...


3.4 LSZ Reduction[Q 4: Maybe insert a proof here instead of just quoting the final result.]

The correlation functions are defined as,

G(n)(x1, ..., x2) ≡ 〈Ω|T(φ(x1)...φ(x2)

)|Ω〉 (3.49)

The Fourier transform is

G(n)(p1, ..., pn) =

∫ ( n∏i=1

d4xieipixi

)G(n)(x1, ..., xn) (3.50)

The LSZ formula says that

G(n)(p1, ..., pn) =n∏j=1

(i√Z

p2j −m2 + iε

)(out 〈p3...pn|p1p2〉in + permutations) (3.51)

+ (terms with < n poles) (3.52)

where the product is the product over the external lines. The bottom line is once youknow that correlators you can then get all the matrix elements. The scattering problemis reduced to a problem of calculating correlators.

3.5 Statistical Mechanics and Path IntegralsSuppose you have a harmonic oscillator that’s in equilibrium with it’s environment. Wewant to treat this oscillator quantum mechanically. If you studied quantum statisticalmechanics then you know that the fundamental object in statistical mechanics is thepartition function which is defined as,

Z = tr[e−βH

]=∑n

〈n|e−βH |n〉 (3.53)

where β = 1/kBT . Let’s define β = it0 (since β is real we know that t0 is imaginary).With this we know that

Z =∑n

〈n|e−iHt0|n〉 (3.54)

=∑n

U (n, t = 0→ n, t = t0) (3.55)

Where U(n, t = 0→ n, t = t0) is the transition amplitude for a state to go to itself aftertime t0. Using our path integral formalism we can write this as∫

x(0)=x(t0)

Dx exp

[i

∫ t0

0

dtL

] ∣∣∣∣t0→− i

kBT

(3.56)

In other words computing the partition function at some temperature is given by a pathintegral with imaginary time.

Chapter 4

Quantization of Electromagnetism

4.1 Classical ElectromagnetismThe electromagnetic (EM) field is

Aµ(x) = (φ(x),A(x)) (4.1)

with E = −∇φ−A and B = ∇×A. The field strength tensor is defined as

Fµν = ∂µAν − ∂νAµ (4.2)

=

0 Ex Ey Ez−Ex 0 Bz −By

−Ey −Bz 0 Bx

−Ez By −Bx 0

(4.3)

The gauge symmetry is the observation that there is some redundancy in our description.The physics is given by the electric and magnetic fields. One can show that the onlytransformation you can do that won’t change the EM fields is

Aµ → Aµ + ∂µξ(x) (4.4)

where ξ(x) is an arbitrary function. Under this gauge symmetry,

Fµν → Fµν (4.5)

We want to get the equations of motion for this system (Maxwell equations) using theLagrangian density.One possibility is to make the Lagrangian out of Fµν . We want tohave a gauge invariant Lagrangian so we use Fµν . To make a Lorentz scalar out of Fµν wetake the product of two Fµν (Alternatively we could take the trace, but Fµν is traceless).This gives us two possible terms

Lgeneral = −1

4FµνF

µν + εµναβFµνFαβ (4.6)

31

32 CHAPTER 4. QUANTIZATION OF ELECTROMAGNETISM

These are the only two non-renormalizable terms. It also turns out that the second termviolates parity. Since we know that electromagnetism does respect parity we don’t havethe second term in EM. Thus

L = −1

4FµνF

µν (4.7)

= −1

4(∂µAν − ∂νAµ) (∂µAν − ∂νAµ) (4.8)

= −1

2∂µAν (∂µAν − ∂νAµ) (4.9)

Note that a m2AµAµ term is not gauge invariant! So we don’t have such a term in our

Lagrangian. The Euler Lagrange equations give

∂µδL

δ∂µAν= 0 (4.10)

Aν − ∂ν (∂µAµ) = 0 (4.11)

Note that we have yet to make a gauge choice. We now fix the gauge to our convenience.Pictorially this is given by

Aµ Aµ + dξµ

gauge-fixing curve

gauge orbits

The gauge orbits are lines which show all the possible values of Aµ along some gaugechoice. All points along the gauge orbit correspond to physically equivalent systems. Allpoints along the gauge fixing curve correspond to the different values of Aµ with a singlegauge. One of the simplest gauge choices is the Lorentz gauge: ∂µAµ = 0.

The solutions of 4.11 in the Lorentz gauge are plane waves,

Aµ =

∫d4k

(2π)4εµe−ik·x (4.12)

with k2 = 0 and εµ is the polarization vector (and also the Fourier transform of the vectorfield). These are just wave solutions with no mass. To make sure that we are still in the

4.1. CLASSICAL ELECTROMAGNETISM 33

Lorentz gauge we require,

∂µAµ = 0 (4.13)∫d4k

(2π)4ikµεµe

−i(k·x) = 0 (4.14)

⇒ k · ε = 0 (4.15)

by uniqueness of the Fourier transform. As an example consider a photon moving the zdirection:

kµ = (E, 0, 0, E) (4.16)

Our ε when dotted into kµ must give zero. A basis of polarization vectors that obeys thecondition above is,

ε(1) =1√2

(0, 1,+i, 0) (4.17)

ε(2) =1√2

(0, 1,−i, 0) (4.18)

ε(3) =1√2

(1, 0, 0, 1) (4.19)

(4.20)

ε(1) and ε(2) correspond to left and right handed circular polarized waves. The thirdpolarization vector corresponds to a longitudinally polarized wave. However we knowthat such a wave does not exist! This extra possible state arose because we didn’t full fixthe gauge of our field. To see this consider some gauge transformed field Aµ. This gaugetransformed field also satisfies the Lorentz gauge if the field connecting the two gaugefields obeys,

Aµ = Aµ + ∂µξ (4.21)⇒ξ = 0 (4.22)

To get rid of this extra freedom consider the component of the Fourier transform of ξthat matches the wave. We have

ξ(x) =

∫d4q

(2π)4ξ(q)e−iq·x (4.23)

and we consider the particular contribution ξ(k) with k2 = 0. We want to know howadding this ξ to Aν changes the solution. We have,

ε′ν(k) = εν(k) + kν ξ(k) (4.24)

By choosing ξ(k) = −ε(3)ν we can get ε′(3) = 0. If one computed the magnetic and electric

fields corresponding to this polarization they would turn out to be zero. This is anunphysical solution. With these two conditions we can fully fix the gauge.


4.2 QuantizationTo zeroth order we treat Aµ as a set of 4 fields. The most naive generalization of ourdiscussion on scalar fields is

W0 [Jµ] =

∫DA exp

(i

(S [A] +

∫Aµ(x)Jµ(x) d4x

))(4.25)

where DA = DA0DA1DA2DA3 and S =∫d4x

(−1

4FµνF

µν). Note however that we

wrote a term Aµ(x)Jµ(x) which does not appear gauge invariant. Applying a gaugetransformation on the source term gives,∫

AµJµ d4x→

∫AµJµ d

4x+

∫∂µξJµ d

4x (4.26)

=

∫AµJµ d

4x−∫ξ∂µJµ d

4x (4.27)

If we restrict our attention to sources with no divergence, i.e. that obey ∂µJµ = 0, thenthe extra term is in fact gauge invariant.

We are going to want to evaluate this path integrals explicitly. At first sight this seemsfeasible since the kinetic term is quadratic and our integral appears Gaussian. Howeverit turns out that there are small technical problems which we will discuss.

We take a short detour into how we treat photons with spin s using canonical quan-tization.

Asµ ∼ εsµe−ip·x (4.28)

with p2 = 0 or equivalently, p0 = ± |p|. We have a restriction in the Lorentz Gauge onour ε given by ε ·p = 0 and ε0 = 0. We have an added the superscript for the polarizationvector for spin, s = 1, 2. We found a particular solution. The general solution is a linearcombination of the solutions we found:

Aµ =

∫d3p

(2π)3

2∑s=1

(As(p)εsµe

−ip·x∣∣∣∣p0=|p|

+ As∗(p)εs∗µ eip·x∣∣∣∣p0=|p|

)(4.29)

This looks identical to the KG field only we now have a spin index s and we havepolarization vectors.

One can go about performing the Canonical Quantization method:

As(p)→ 1√2p0

asp , As∗(p)→ 1√2p0

as†p (4.30)

where a, a† are creation and annihilation operators respectively. We would then definethe vacuum, |0〉 such that

asp |0〉 (4.31)

for all p, s. There is then a one-photon state given by

as†p |0〉 = |p, s〉 (4.32)

4.3. PATH INTEGRALS IN QED 35

4.3 Path Integrals in QED

Recall that

W0 [Jµ] =

∫DA exp

[i

∫d4x(−1

4FµνF

µν︸︷︷︸F term

+AµJµ)

](4.33)

where we restrict our attention to sources such that ∂µJµ = 0. We can write the expo-nential argument as

F term =

∫d4x

(−1

2

)∂µAν (∂µAν − ∂νAµ) (4.34)

=

∫d4x

1

2Aν (gµν − ∂µ∂ν)Aµ (4.35)

where we have integrated by parts to play with the derivatives. Now (here we write theFourier Transform of the vector potential as A(k))

Aµ(x) =

∫d4k

(2π)4eik·xAµ(k) (4.36)

Thus we have

F term =1

2

∫d4x

∫d4k d4k′

(2π)4(2π)4ei(k+k′)·xAν(k

′)(−k2gµν + kµkν)Aµ(k) (4.37)

= −1

2

∫d4k

(2π)4Aν(−k)(k2gµν − kµkν︸︷︷︸

Dµν

)Aµ(k) (4.38)

The second term is given by∫d4xAµJ

µ =

∫d4k

(2π)4Aµ(k)Jµ(−k) (4.39)

Recall that for a scalar field the source term was,∫d4k

(2π)4φ(k)J(k) (4.40)

and the kinetic term was ∫d4k

(2π)4φ(k)(k2 −m2 + iε)φ(−k) (4.41)

and we completed the square by changing variables to

φ′(k) = φ(k) +(k2 −m2 + iε

)−1J(k) (4.42)


With this in mind, we define a new A such that

A′µ = Aµ + (D−1)µν Jν (4.43)

and then our Lagrangian is given by,

L =1

2A′µD

µνA′ν −1

2Jµ(D−1)µν J

ν (4.44)

So we haveW0 [Jν ] = W0 [0] exp

(− i

2

∫Jµ (D−1)µν︸︷︷︸

Green’s function

Jνd4k

(2π)4

)(4.45)

where we have our Gaussian integral included in our definition ofW0 [0]. Everything seemsto have worked flawlessly. The subtlety here is that Dµν is not invertible! To prove thiswe study the eigenvalues of the D matrix. One such eigenvector is kν = (k0,−k) with aneigenvalue of zero: (

k2gµν − kµkν)kν = 0 (4.46)

Since the determinant is the product of the eigenvalues it must be zero, implies that Disn’t invertible.

This problem arose because while we are integrating over the gauge fixed field weare also integrating over gauge inequivalent fields. We haven’t done anthing to set thesecontributions to zero. We need to find a way such that we integrate only about inequiva-lent fields. This corresponds to a single gauge-fixing curve (All the Aµ values for a singlegauge choice):

The procedure is called the Faddeev-Popov procedure. Suppose we have some field A0,µ.We have a gauge transformed field parameterized by α that is given by,

Aα,µ = A0,µ + ∂µα (4.47)

We define a gauge transformation function as G [A]. For the Lorentz gauge we have

G [A] = ∂µAµ = 0 (4.48)


We will also later use the Generalized Lorentz Gauge given by

G [A] = ∂µAµ − ω(x) = 0 (4.49)

In order to ensure we stay on our gauge we introduce a functional delta function (thiscan be thought of as a product of delta functions, one at each spacetime point):∫

Dαδ (G [Aα,µ]) (4.50)

By inserting this into our path integral the path integral will be zero unless G [A] = 0.However, we can’t insert this into the path integral unless it’s equal to 1.

Recall that if f(x) has one zero at x0 then,∫dx

∣∣∣∣df(x)

dx

∣∣∣∣x=x0

δ (f(x)) = 1 (4.51)

We want to generalize this to instead of having f(x) we have, g(a) for vectors of arbitrarysize. To do this consider the Taylor expansion of g around its root (we assume it onlyhas one root, a0):

gi(a) =gi(a0) +

∑j

∂gi∂aj

∣∣∣∣a0

(aj − a0,j) + ... (4.52)

We want to insert this into a delta function, δ(n)(g(a)). This will only be nonzero neara = a0. Thus we have,

δ (g(a)) =∏i

δ (gi(a)) (4.53)

=∏i

δ

(∑j

Jij(aj − a0,j)

)(4.54)

where Jij is the Jacobian matrix defined by Jij ≡ ∂gi∂aj

∣∣a0. We have,

δ (g(a)) = δ

(∑j

J1j(aj − a0,j)

)δ

(∑j

J2j(aj − a0,j)

)... (4.55)

We now use the identiy,

δ(α(a− a0)) =δ(a− a0)

|α|(4.56)

We choose to isolate each delta function in Eq. 4.55 for a different aj:

δ (g(a)) =δ(a1 − a0,1)

|J1,1|δ(a2 − a0,2)

|J2,2|... (4.57)


If we take the Jacobian matrix to be greater then zero then we have the product:

(J1,1J2,2..)−1 =

1

det J(4.58)

where we have used the fact that the determinant of J is independent of a unitarytransformation. So we finally have,(∫ ∏

i

dai

)δ(n) (g(a)) det

(∂gi∂aj

)= 1 (4.59)

where it is understood that the Jacobian matrix is evaluated at the root of g.We write the continuum generalization of this equation as,∫

Dα(x)δ (G(Aα)) det

(δG(Aα)

δα

)= 1 (4.60)

We consider the generalized Lorentz gauge: G(Aα) = ∂µAµ + 1

e∂µα(x)−ω(x). In this

case it is straight forward to calculate the determinate explicitly. The discretized gaugetransformation is (we work here in 1D just for brevity however, its easy to extend theresults to four dimensions.

Gi(αi) =1

∆2e(αi+1 − 2αi + αi−1) + (∂µA

µ0 − ω(x))i︸︷︷︸

Indepdent of αj

(4.61)

Taking the derivative we have,

∂Gi(αi)

∂αjδij =

1

∆2e(δi+1,j − 2δi,j + δi−1,j) δij (4.62)

which doesn’t have any A dependence so it is already evaluated at the root of G. Thusthe determinant has no A or α dependence and we can write:

W0 [Jµ] = det

(δG

δα

)∫Dα

∫DAδ (G [Aα]) exp

(i

(S +

∫d4xAµJ

µ

))(4.63)

However J , DA, and S are gauge invariant thus we can change our equation to the gaugeshifted Aαµ:

W0 [Jµ] = det

(δG

δα

)∫Dα

∫DAαδ (G [Aα]) exp

(i

(S [Aα] +

∫d4xAαµJ

µ

))(4.64)

We now see that the α integral is trivial, it is just the volume of the α space (we includethe determinant in what we call V ol):

W0 [Jµ] = V ol ×∫DAδ (G [A]) exp

(i

(S [A] +

∫d4xAµJ

µ

))(4.65)


where we have redefined our integration variable from Aα → A. We use the GeneralizedLorentz Gauge without α as we redefined, Aα → A. (we ignore the constant in frontsince as we saw for a scalar field, it cancels away when calculating observables):

W0 [Jµ] =

∫DAδ (∂µA

µ − ω(x)) exp

(i

(S [A] +

∫d4xAµJ

µ

))(4.66)

= N (ξ)

∫Dωe−i

∫d4xω

2

2ξ︸︷︷︸1

∫DAδ (∂µA

µ − ω(x)) exp

(i

(S [A] +

∫d4xAµJ

µ

))(4.67)

where ξ is just a constant and we are allowed to through in this integral since we knowour W0 is independent of ω (it is gauge invariant). Thus we have

W0 [Jµ] = N(ξ)

∫DA exp

(−i∫

d4x(∂µA

µ)2

2ξ+ i

∫d4x (L+ AµJ

µ)

)(4.68)

where we have performed our ω integral (which is trivial due to the delta function). Thusour “gauge-fixing procedure” is equivalent to taking L → L− 1

2ξ(∂µA

µ)2. Recall that wehad

S =

∫d4k

(2π)4

1

2Aµ(−k)

(k2gµν − kµkν

)Aν(−k) (4.69)

Adding this extra term we have (note that we can replace 12ξ

(∂µAµ)2 → − 1

2ξ(Aµ∂µ∂νA

ν))

S =

∫d4k

(2π)4

1

2Aµ(−k)

(k2gµν − kµkν

(1− 1

ξ

))︸︷︷︸

Dµνξ

Aν(k) (4.70)

We now have detDµνξ 6= 0. We finally have

W0 [Jµ] = W0 [0] exp

(−1

2

∫d4k

(2π)4Jµ(k)(iD−1

ξ )µν Jν(−k)

)(4.71)

where (D−1ξ

)µν≡ Gµν = − i

k2 + iε

(gµν − (1− ξ)kµkν

k2

)(4.72)

This is known as the photon propagator. To see that this in indeed the inverse of Dµν

just multiply it out directly:(D−1ξ

)µνDξ,νρ =

1

k2


k2

)(k2gνρ − kνkρ

(1− 1

ξ

))(4.73)

=1

k2

(k2δµρ − kµkρ

(1− 1

ξ

)− (1− ξ)kµkρ +

(1− 1

ξ

)(1− ξ) kµkρ

)(4.74)

= δµρ (4.75)


Naively one would say that the propagator result depends on ξ. How can this besince we said that ξ is a mathematical convenience? Being more careful, one noticesthat all the ξ terms look like ξ(k · J)2. However these are zero by our choice of source(∂µJµ = 0⇒ kµJ

µ = 0). Any choice of ξ is valid. There are a few popular choices. Onepopular choice is ξ = 1. This is called the Feynman gauge. Another popular gauge iscalled the Landau gauge (ξ = 0). Alternatively, you can keep ξ as a parameter and watchit’s dependence drop at the end.

There is also something interesting in the sign of the photon propagator. The spatialcomponents of Aµ propagate with the same sign as the scalar field propagator, howeverthe time-like component has a relative negative sign. This is a consequence of the factthat A0 doesn’t actually propagate since the momentum conjugate to A0 is zero:

∂L∂0A0

= −1

4

∂

∂0A0FµνF

µν (4.76)

= 0 (4.77)

4.4 Correlation FunctionsConsider the correlation function:

〈0|T(Aµ(x1)Aν(x2)

)|0〉 =

1

W0 [0]

1

i2δ2W0 [J ]

δJµ(x1)δJν(x2)

∣∣∣∣J→0

(4.78)

=1

iGµν(x1 − x2) (4.79)

However we constructed our Jµ such that ∂µJµ = 0. Having a δJµ implies we can “wiggle”Jµ in any direction we want but in fact we have some restrictions. We now set out tofind what those are

Aµ(x) =

∫d3kAµ(k)e−ik·x (4.80)

and ∫d4xAµJ

µ →∫

d4kAµJµ (4.81)

Now

G(k1, ...) =

∫ (d4x1 d

4x2...) (eik1x1 ...

)〈0|T (Aµ(x1)...) |0〉 (4.82)

∝[

δ

δJµ(k)

δ

δJν(k)...

]W0

[J]∣∣∣∣J→0

(4.83)

where G is the Fourier transform of the correlator. This can be done for any k. We takek2 = 0 with k = (E, 0, 0, E).

Aµ(k) =3∑s=0

εsµ (4.84)


with a basis

ε0µ =1√2

(1, 0, 0,−1)

ε1µ = (0, 1, 0, 0)

ε2µ = (0, 0, 1, 0)

ε3µ =1√2

(1, 0, 0, 1)

NowJµ(k) =

∑Jsεsµ (4.85)

andkµJ

µ = 0⇒ J0 = 0 (4.86)

so

A · J =3∑s=1

AsJsεsµ · εsµ = A1J1 + A2J2 (4.87)

Thus what we can find out is the correlation function between A1 and A2. We can onlyknow

〈0|T(A1(k)...

)|0〉 (4.88)

〈0|T(A2(k)...

)|0〉 (4.89)

With this we can write down the LSZ reduction formula for photons:

Gµ(k, ...) =i√Zγ

k2 + iε

2∑s=1

εsµ 〈(k, s) ...〉︸︷︷︸S matrix

+... (4.90)

Chapter 5

Fermions with Path Integrals

Electrons are fermions, thus you can’t have more then one electron in a single state.That’s why we never thought of electrons as fields until now. On the other hand photonsare bosons and they can all be in the same state. That’s why we had to think of them asfields right from the start. However the concept of path integrals requires one to think offermion fields.

5.1 Grassman NumbersIt turns out the correct way to think about fermion fields is through the concept ofGrassman numbers, θ, η. They are not like ordinary numbers in many ways. Howeverthey still obey rules of addition and multiplication. For Grassman numbers the addition istypical (commutative). However the multiplication of such number is anti commutative.i.e.

η + θ = θ + η (5.1)ηθ = −θη (5.2)

This implies that

θθ = −θθ (5.3)⇒θ2 = 0 (5.4)

Furthermore,f(θ) = a+ bθ = a+ θb (5.5)

where a and b are ordinary numbers. Note that this is analogous to a Taylor expansionhowever we have θ2 = 0 so all higher order terms vanish. These numbers are non-intuitiveand one needs to be careful when working with them.

You can define derivatives of Grassman numbers such as

df

dθ= b (5.6)

42

5.1. GRASSMAN NUMBERS 43

and integration ∫dθf(θ) =

∫dθf(θ + η) (5.7)

The definition of the integral implies that (Exercise: use equation 5.5 to derive thefirst equation. Maxim isn’t sure how to derive the second equation)∫

dθ = 0 (5.8)∫dθθ = 1 (5.9)

This means that we have ∫dθf(θ) = b

df

dθ(5.10)

One can generalize this discussion to complex Grassman Numbers,

θ∗ =1√2

(θ1 + iθ2), θ∗ =1√2

(θ1 − iθ2) (5.11)

and ∫dθθ = 1 =

∫θ∗θ∗ (5.12)

As an example consider integration over a Gaussian∫dθ∗dθe−iθ

∗bθ =

∫dθ∗dθ (1− θ∗bθ) (5.13)

=

∫dθ∗dθ (θθ∗b) (5.14)

= b (5.15)

This should be compared to the result for real numbers,∫dxe−

12bx2

=

√2π

b(5.16)

We now move on to multidimensional integrals. First consider the real variable inte-gral: ∫

dx1 dx2... exp

(−1

2

∑i,j

xiAijxj

)(5.17)

Suppose that A can be diagonalized by some orthogonal transformation,

A = OTADO (5.18)

which implies that,

xTAx = (Ox)TAD(Ox) (5.19)= x′TADx

′ (5.20)

44 CHAPTER 5. FERMIONS WITH PATH INTEGRALS

where we have defined, Ox ≡ x′. The Jacobian for this transformation is,

Jij ≡∂x′i∂xj

= Oij (5.21)

Thus we have,

= detO∫

dx1; dx2; ... exp

(−1

2

∑i

x′iAD,iix′j

)(5.22)

= detO∏∫

dxi exp

(−1

2xiAD,iixi

)(5.23)

= detO(2π)N/2(∏i

AD,ii)−1 (5.24)

= (2π)N/2detOdetA

(5.25)

We now want to find the analogous result for the Grassman integral,∫dnθdnθ∗ exp

(−∑i,j

θ∗iBijθj

)(5.26)

We again want to tranform the integration variable and diagonalize B, however we needto be a bit careful as we are working with Grassman numbers. A set of N Grassmanvariables obey the relations:

θ1θ2...θN =1

N !εi1...iN θi1θi2 ...θiN (5.27)

εi1i2...iN θ1θ2...θN = θi1θi2 ...θiN (5.28)

These relations are a consequence of the anticommutating nature of the Grassman vari-ables and are easy to show for N = 3 though I can’t figure out a general proof yet.

If we consider a unitary transformation on the Grassman variables: θ′i = Uijθj then,∏i

θ′i =1

N !εi1i2...iN θ′i1θ

′i2...θ′iN (5.29)

=1

N !εi1i2...iNUi1,i′1θi′1Ui2,i′2θ

′i′2...UiN ,I′N θi′N (5.30)

=1

N !εi1i2...iNUi1,i′1Ui2,i′2UiN ,I′N εi′1,i′2,...,i′N

(∏j

θj

)(5.31)∏

i

θ′i = detU∏j

θj (5.32)

5.2. FREE DIRAC FIELD 45

Thus when changing variables we pick up a determinate, analogous to the earlier situationwith the real variables. With this we have,∏

i

dθ∗i∏j

dθj = (detU)∗ detU∏i

dθ∗i∏j

dθj (5.33)

=∏i

dθ∗i∏j

dθj (5.34)

and so the integral is,∫dnθdnθ∗ exp

(−∑i,j

θ∗iBijθj

)=

∫dnθ′dnθ′∗ exp

(−∑i,j

θ′∗i BD,ijθ

′j

)(5.35)

=

∫dnθdnθ∗ exp

(−∑i

θ∗iBD,iiθj

)(5.36)

=∏i

BD,ii (5.37)

= detB (5.38)

5.2 Free Dirac Field

The Dirac field is given as

ψ(x) =

ψ1(x)ψ2(x)ψ3(x)ψ4(x)

(5.39)

To have Fermi Dirac statistics we make ψa(x) into a complex Grassman number valuedfield.

The free Dirac field is given by

LD = ψ(i/∂ −m)ψ (5.40)

with ψ = ψ†γ0 and /∂ = ∂µγµ.

We want to produce our generating functional. We need to introduce sources. Wewant something of the form of source multiplied by the Dirac field. We want to add aterm for the source that takes the form of a source multiplied by a scalar. A simplestthing one can imagine is

η =

η1(x)η2(x)η3(x)η4(x)

(5.41)


where this term transforms in the same way as the Dirac spinor. The η source is also acomplex Grassman number valued field. Since we also want a Hermitian Lagrangian oursource takes the form

ηψ + ψη (5.42)

One can treat η and η as independent as well as ψ, ψ as independent. Our generatingfunctional is

W0 [η, η] =

∫DψDψ exp

[i

∫d4x

(LD + ηψ + ψη

)](5.43)

where Dψ ≡∏4

a=1Dψa = ψ1ψ2ψ3ψ4. The procedure is just as before. We now completethe square in an analogous way to what we did earlier.

We have,∫d4x (...) =

∫d4x

[ ∫d4k d4k′

(2π)4(2π)4eikxe−ik

′x

ψ(k) (−iγµ(ikµ)−m)ψ(k)

+ η(k)ψ(k) + h.c.

](5.44)

=

∫d4k

(2π)4ψ(k) (/k −m)ψ(k) + η(k)ψ(k) + h.c. (5.45)

Now we shift the fermion fields:

ψ → ψ − (/k +m)η

k2 −m2 + iε(5.46)

ψ → ψ − η(/k +m)

k2 −m2 + iε(5.47)

With this we have,

ψ(/k −m)ψ + ηψ + h.c.→ ψ(/k −m)ψ +η(/k +m)(/k −m)(/k +m)η

(k2 −m2)2(5.48)

= ψ(/k −m)ψ +η(/k +m)η

(k2 −m2)(5.49)

So our path integral becomes,

W0 [η, η] =

∫DψDψ exp

[i

∫d4kψ(/k −m)ψ +

η(/k +m)η

k2 −m2

](5.50)

= W0[0] exp

[i

∫d4k

η(/k +m)η

k2 −m2

](5.51)

In real space,

W0 [η, η] = W0 [0] exp

[−∫

d4x d4yη(x)S(x− y)η(y)

](5.52)

5.3. QED 47

where S is the Green’s function, given by

(i/∂x −m14×4)S(x− y) = iδ4(x− y)14×4 (5.53)

where 14×4 is the four by four unit matrix. Note that this is really four equations. Solvingthis equation via Fourier transforms we get

S =

∫d4k

(2π)4

i(/k +m)e−ik·(x−y)

k2 −m2 + iε(5.54)

We now move on to functional differentiation. One can show that

〈0|T(ψ(x1)... ˆψ(y1)...

)|0〉 =

1

W0 [0]

(1

i

δ

δη(x1)...−1

i

δ

δη(y1)...

)W0

∣∣∣∣η,η→0

(5.55)

where(δδη...)that we have one functional derivative for each fermionic field being time

ordered. The set of fields can be both Grassman fields and scalar fields. We haveδ

δηaexp

(iψη)

=δ

δηa(1 + i(ψ†)c(γ0)cbη

b)

(5.56)

= −i(ψ†)c(γ0)cbδba (5.57)

= −iψa (5.58)

Further note that 〈0|T (ψ(x1) ˆψ(x2))|0〉 is antisymmetric since the η’s are Grassmannumbers.

5.3 QEDQuantum Electrodynamics(QED) describes both spin 1

2objects,“matter” (electrons, pro-

tons, muons, ...), and a spin-1 object “light” (photons). The most general Lagrangian isgiven by

L = ψ(i/∂ −m)ψ − 1

4FµνF

µν + Lint (5.59)

The interaction Lagrangian must be Lorentz invariant, and renormalizable. Renormaliz-ability requires that each term must have dimensionality less than or equal to 4. By ourLagrangian we know that [ψ] = 3/2 and [Aµ] = 1. There are very few allowed terms.We cannot have an odd number of ψ terms since then we won’t have a Lorentz invariantLagrangian. The only way to couple them to A′s in a renormalizable way is to have 2 ψand 1 A. There are two potential bilinears we can use:

Aµψγµψ (5.60)

Aµψγµγ5ψ (5.61)

The second term violates parity. Since we know that electromagnetism conserves Paritywe have,

Lint = eAµψγµψ (5.62)


where e is a dimensionless number. It will turn out that e is the electric charge of thequantized ψ fields. One may wonder why the charge has no units. That is a result ofusing ~ = 1. This can be seen since

α ≈ 1

137=

e2

4π~cNatural Units−−−−−−−−−−→ e2

4π(5.63)

and hence e =√

4πα ≈ 0.3.We now derive the same result in a deeper way. The Dirac Lagrangian, LD has a

“global U(1)” symmetry:ψ → eiαψ (5.64)

i.e. LD → LD. There is a Noether current associated with this symmetry given by

jµ = ψγµψ (5.65)

with ∂µjµ = 0. Promoting this global symmetry to a local symmetry, “local U(1)”:

ψ → eiα(x)ψ , ψ → ψe−iα(x) (5.66)

This is no longer a symmetry since

∂µψ → ∂µ(eiαψ) = eiα∂µψ + i∂µαeiαψ (5.67)

andLD → LD + i(∂µα)ψγµψ (5.68)

and our Lagrangian is not invariant. The procedure to “fix” this is to introduce a gaugefield that has the following transformation:

Aµlocal U(1)−−−−−−→ Aµ +

1

e∂µα (5.69)

We now introduce another term known as the covarient derivative:

Dµ = ∂µ − ieAµ (5.70)

Dµψlocal U(1)−−−−−−→ eiαDµψ (5.71)

and we end with

ψDµψlocal U(1)−−−−−−→ ψDµψ (invariant under local U(1)) (5.72)

Our Dirac Lagrangian becomes

L = ψ(i /D −m)ψ = ψ(i/∂ −m)ψ + eAµψγµψ (5.73)

Note that if we had Parity violating U(1) then we would have (notice the barred fielddoesn’t gain a negative in the phase):

ψ → eiγ5α(x)ψ , ψ → ψeiγ5α(x) (5.74)

This transformation would have required a slightly different covarient derivative resultingin a γ5 term in our final Lagrangian.

Our final QED Lagrangian is given by

L = ψ(i/∂ −m)ψ − 1

4FµνF

µν + eAµψγµψ (5.75)

5.4. THREE-POINT FUNCTION 49

5.4 Three-point functionWe get our Feynman rules by looking at different correlation functions. We begin bylooking at a 3-pt function (|Ω〉 is the messy QFT ground state).

〈Ω|T(ψa(x1)ψb(x2)Aµ(x3)

)|Ω〉 =

1

We [0]

1

i

δ

δηa(x1)

1

i

δ

δηb(x2)

1

i

δ

δJµ(x3)We [η, η, Jµ]

∣∣∣∣η→0,J→0

(5.76)

where

We [η, η, Jµ] =

∫DψDψDA exp

[i

∫d4x

(LD +

−14F 2

↑Lg + eAµψγ

µψ + ηψ + ψη + JµAµ

)](5.77)

We almost know how to calculate this. Without the interaction term we could do itexplicitly. We expand in powers of e:

We [η, η, Jµ] = W0 + ie

∫d4x

(1

i

δ

δJµ(x)

)(−1

i

δ

δηc(x)

)(γµ)cd

(1

i

δ

δηd(x)

)W0 +O

(e2)

(5.78)where we define

W0 = exp

[−1

2

∫d4w d4zJµ(w)Gµν(w − z)Jν(z)

]exp

[−∫

d4w d4zη(w)S(w − z)η(z)

](5.79)

We now calculate this explicitly:

1

i

δ

δηd(x)exp [...] = i

∫d4zSdb(x− z)ηb(z) exp [...] (5.80)

−1

i

δ

δηc(x)

1

i

δ

δηd(x)exp [...] = −Sdc(0) exp [...]

−∫d4zSdb(x− z)ηb(z) · (−1)

∫d4wηa(w)Sac(w − x) exp [...] (5.81)

and1

i

δ

δJµ(x)exp [...] = i

∫d4yJµ(y)Gµν(x− y) exp [...] (5.82)

which gives,

We [η, ε, Jµ] =

1− ie

∫d4xd4yJµ(y)Gµν(x− y)

(Sdc(0)

−∫d4zd4wSdb(w − z)ηb(z)ηa(x)Sac(w − x)

)γνcd

W0 (5.83)


To calculate the correlation function to leading order we need to take three derivativesof We. This is straight forward and the answer is [Q 5: Do this calculation!]

〈Ω|T(ψa(x1)ψb(x2)Aµ(x3)

)|Ω〉 = ie

∫d4xGµν(x3 − x)

(Sbc(x2 − x)γνcdSda(x− x1)

+ Sba(x2 − x1)(γν)cdSdc(x− x)

)(5.84)

We develop the position space Feynman rules given by

ψa(x) ψb(y)→ Sab(x− y)

Aµ(x) Aν(y)→ Gµν(x− y)

c

d

→ −ie∫

d4xγµcd

In principle we can now work out every correlation function using these Feynman rules.The momentum space Feynman rules are

a b →i(/p+m)ab

p2 −m2 + iε

µ ν → − i

k2 + iε


k2

)

c

d

→ −ieγµab + momentumconservation

Note that when the momentum flows opposite to the direction in the fermion propagatorwe get a negative sign in front of p.

We can now calculate the correlation function,

Gabµ(p1, p2, p3) =

b, p2

a, p1

p3+unconnected

diagrams

5.4. THREE-POINT FUNCTION 51

we ignore unconnected diagrams as they don’t contribute to the S matrix. This diagramis

Gabµ(p1, p2, p3) =i(/p2

+m)bc

p22 −m2 + iε

(−ie) (γµ)cd

i(/p1

+m)da

p21 −m2 + iε

(−i)(gνµ − (1− ξ)p3,µp3,ν

p23

)p2

3 + iε(5.85)

We know from our experience with scalars that in order to get the S matrix elementsthrough LSZ reduction we need to know the poles of this expression.

Note that if you try to solve the three equations:

p1 + p2 = p3 (5.86)p2

1 = p22 = m2 (5.87)

p23 = 0 (5.88)

you get a contradiction since squaring the first equation gives

p1 · p2 = 0 (5.89)

and if we work in the center of mass frame we have p1 = (m,0), p2 = (E,p) which wouldimply

mE = 0 (5.90)

which is impossible since m > 0. Hence it’s impossible to have a 3 point process thatconserves energy. The first real S matrix process is a four point process:

e

γ

e

γ

e

However in order to see the structure of the S matrix it is sufficient to consider the 3point function.

Consider an incoming fermionp→Fp.

G→ (...) i/p+m

p2 −m2 + iε(5.91)

LSZ reductions for Fermions tells us that

G =i

p2 −m2 + iε

2∑s=1

〈...|...(p, s)〉 us(p) + terms with pole in p2 (5.92)

here the p2 poles are not important. G is of this form because,

|(p, s)〉 = as†p |0〉 (5.93)


and

ψ =

∫d3k

(2π)3√

2Ek

∑s

(asku

s(k)e−ik·x + bs †k vs(k)eik·x

)(5.94)

ψ =

∫d3k

(2π)3√

2Ek

∑s

(as†k u

s(k)eik·x + bskvs(k)e−ik·x

)(5.95)

So we cannot directly get the S matrix from the G, but we can use trace technology tosimplify our result

∑2s=1 us(p)us(p) = /p+m. Using this result we have,

G = (...)i∑2

s=1 us(p)us(p)

p2 −m+ iε(5.96)

Comparing with the LSZ formula we can write,

〈...|...(p, s)〉 = (...)us(p) (5.97)

So for an incoming fermion we see we have a factor of u(p). The derivation for the otheran incoming antifermion as well as for the outgoing fermions is similar.

Now consider the full photon terms:

(−i)(gµν − (1− ξ)pµpν/p2)

p2 + iεN ν (5.98)

where N ν is just the rest of the object which has no µ dependence. Using LSZ reductionwe have for an incoming photon,

Gµ =i

p2 + iε

2∑s=1

〈...|...(p, s)〉 εs∗µ + (no pole at p2 = 0) (5.99)

where we have used,

Aµ =

∫d3k

(2π)3√

2Ek

∑s

asεsµeik·x + as†εs∗µ e

−ik·x (5.100)

Using the Ward identity (we discuss this identity more later) we know that

pµN ν = 0 (5.101)

This means that the (1− ξ)pµpν/p2 term in the incoming and outgoing photon contribu-tions go to zero and we simply have,

−igµνp2 + iε

N ν = − i

p2 + iε

∑s

εs ∗µ (p) 〈...|...(p, s)〉 (5.102)

In the fermion case we rearrange the numerator of the propagator using a spin sum. Wewould like to do the same thing here with gµν and write it in terms of εµ’s. To do this

5.5. e+e− → µ+µ− 53

we use a trick that we will prove later on which says that as long as gµν is going to bemultiplied by an amplitude we can make the replacement,

gµν → −∑s

εsµεs∗ν (5.103)

Since the LSZ formula is true for any s,

N νεsµ(p) = 〈...|...(p, s)〉 (5.104)

Hence we can replace an incoming photon by εsµ.We summarize all the rules for incoming and outgoing particles below

incoming fermionp→Fp = us(p)

incoming antifermionp→p = vs(p)

outgoing fermion p p→F = us(p)

outgoing antifermion p p→ = vs(p)

incoming photon µ p→gp = εsµ(p)

outgoing photon µ p→gp = εs∗µ (p)

5.5 e+e− → µ+µ−

Now that we have all our Feynman rules we can finally do some calculations. As anexample lets consider two electrons turning into two muons. The interaction part of theLagrangian is

Lint = −eAν(ψeγνψe + ψµγνψµ) (5.105)

The leading order connected diagram is

e, p1

e, p2

k = p1 + p2

γ

µ, p3

µ, p4

iMs1,s2,s3,s4 = [vs2e (p2)(−ieγα)us1e (p1)](−i)k2 + iε

(gαβ − (1− ξ)kαkβ

k2

)×[us3µ (p3)(−ieγβ)vs4µ (p4)

](5.106)


(note that here µ, e are not variables just labels so we know what mass each u and vcorresponds to. Consider the term

kαk2

[v(p2)γαu(p1)] = (−ie)v(p2)/ku(p1) (5.107)

= (−ie)v(p2)(/p1+ /p2

)u(p1) (5.108)

= (−ie) (v(p2)meu(p1) + v(p2)(−me)u(p1)) (5.109)= 0 (5.110)

since (/p − m)u = 0, (/p + m)v = 0. Thus the amplitude is indendent of the kαkβcontribution and hence independent of gauge parameter ξ. The amplitude simplifies to

iMs1s2s3s4 = ie2

s[v2γ

αu1] [u3γαu4] (5.111)

where s ≡ (p1 + p2)2. Typical experiments have polarized beams. In this case we canaverage over incoming particle spins, and sum over outgoing particle spins.(

1

2

)2 ∑s1,s2,s3,s4

|Ms1,s2,s3,s4|2 ≡ |M|2 (5.112)

In our case we have

|M|2 =1

4

e4

s2tr[(/p2−me)γ

α(/p1+me)γ

β]× tr

[(/p3

+mµ

)γα(/p4

−mµ)γβ

](5.113)

=1

4

e4

s2(4)(pα2p

β1 + pα1p

β2 − gαβ

(p1 · p2 +m2

e

))(4)

×(pα4p

β3 + pα3p

β4 − gαβ

(p3 · p4 +m2

µ

))(5.114)

In the high energy regime (E me,mµ) we have

|M|2 =8e4

s2((p1 · p4)(p2 · pe) + (p1 · p3)(p2 · p4)) (5.115)

but,p1 + p2 = p3 + p4 (5.116)

and

(p1 − p4)2 = (p3 − p2) (5.117)p1 · p4 ≈ p3 · p2 (5.118)

since we have in the high energy regime. So after switching the Mandlestam variablesthe amplitude squared takes the form,

|M|2 =2e4

s2

(t2 + u2

)(5.119)

If we work in the COM frame:

5.6. e−µ− → e−µ− 55

e eθ

then

p1 = (E, 0, 0, E)

p2 = (E, 0, 0,−E)

p3 = (E,E sin θ, 0, E cos θ)

p4 = (E,−E sin θ, 0,−E cos θ)

where s = 4E2, t = −2E2(1− cos θ), u = −2E2(1 + cos θ). We have the simple result

|M|2 = e4(1 + cos2 θ) (5.120)

anddσ

dΩ=|M|2

64π2s=

e4

32πs(1 + cos2 θ) (5.121)

The functional dependence is as follows,

# of Events

cos θ

5.6 e−µ− → e−µ−

We have found the amplitude and cross sections for e+e− → µ+µ−. We now considerelastic e−µ− scattering but now in the non-relativistic regime,

p2 → p3 →

p1 → p4 →

q = p1 − p4


We can do a similar calculation to last time:

iM =ie2

t(u4γ

µu1)(u3γµu2) (5.122)

In the non-relativistic limit we have

p1 = (me +O(p2),p)

p2 = (mµ,0)

p3 = (mµ +O(p2),q)

p4 = (me +O(p2),p′)

where q = p− p′. Hence we have

t = (p1 − p4)2 = −q2 +O(p4) (5.123)

The Dirac spinor in the non-relativistic limit is

us(p) =√m

(ξs

ξs

)+O(p) (5.124)

where ξ1 =

(10

), ξ2 =

(01

). The scalars are:

u4γ0u1 =

√m(ξs4†ξs1 + ξs4†ξs1) = 2mξs4†ξs1 = 2mδs4s1 (5.125)

u4γiu1 = O(p) (5.126)

Thus to leading order we have

iM≈ −ie2(2me)(2mµ)δs4s1δs3s2

q2(5.127)

The cross-section is given by,

dσ

dΩ=

1

2Ee2mµ

Eepe

1

32π2

2pCMeECM

|M|2 (5.128)

We work in the limit of a stationary muon (though the cross-section is Lorentz invariant).In this case pcme ≈ pe and Ecm ≈ mµ. We have,

dσ

dΩ=

1

64π2m2µ

|M|2 (5.129)

=4α2

q4m2e (5.130)

where the spin averaging is trivial due to the spin delta functions.In non-relativistic quantum mechanics we’d expect the electron to be moving slowly

and the muon to be at rest acting as a source of charge through the potential, V (r) = αr.

5.7. COMPTON SCATTERING 57

pe →

µ

p′e→

In this case the Born approximation tells us that

dσ

dΩ=m2e

4π2V 2(q) (5.131)

where q = p′ − p. Comparing the two expressions we identify,

V 2(q) =α216π2

q4(5.132)

⇒ V (q) =4πα

q2(5.133)

We have 1

V (r) =

∫d3q

(2π)3e−iq·r

4πα

q2(5.134)

=α

π

∫d cos θdqeiqr cos θ (5.135)

= −2α

π

∫dq

1

qsin(qr) (5.136)

= −αr

(5.137)

where the integral was done by inserting a regulator and taking the regulator to 0 atthe end. We see that in the non-relativistic limit QED repreduces classical electromag-netism as expected. Furthermore, this is the first confirmation that the coupilng in theLagrangian, e, really represents an electric charge. Up to now we just assumed the mostnatural interpretation and we finally see that this is indeed the case.

5.7 Compton ScatteringConsider e−γ → e−γ. There are two second order vertices:

+

1Where there is no ambiguity with four-vectors we write, r ≡ |r| and q ≡ |q|.


We have the amplitudeMs = εsµ(p)Mµ (5.138)

To get the cross-section we need to average over spins:∑s

|Ms|2 =(∑

s

εsµεs∗ν

)Mν∗ (5.139)

The claim is that we can make the replacement:∑s

εsµε∗sν → −gµν (5.140)

To see why this is consider the photon with the momenta, p = E(1, 0, 0, 1). The Wardidentity tells us that pµMµ = 0. Hence

M0 =M3 (5.141)

Since we can choose any basis we wish for the polarization lets consider linear polarization,

ε1 = (0, 1, 0, 0)

ε2 = (0, 0, 1, 0)

Explicitly we can now calculate

∑s

εsµεs∗ν =

0 0 0 00 1 0 00 0 1 00 0 0 0

(5.142)

This is not quite

− gµν =

−1 0 0 00 1 0 00 0 1 00 0 0 1

(5.143)

But∑

s εsµεs ∗ν and gµν never appear on their own. Instead they come as as a product of

amplitudes. We have,

− gµνMµMν∗ =−

∣∣M0∣∣2 +

∣∣M1∣∣2 +

∣∣M2∣∣2 +

∣∣M3∣∣2 =

∣∣M1∣∣2 +

∣∣M2∣∣2 (5.144)∑

s

εµεs∗ν MµMν∗ =

∣∣M1∣∣2 +

∣∣M2∣∣2 (5.145)

Hence due to the Ward Identity we can always make the substituion,∑s

εsµεs ∗ν → −gµν (5.146)

5.8. PROOF OF WARD IDENTITY 59

5.8 Proof of Ward IdentityWe finally prove the Ward identity and then discuss the intuition behind it. The definingequation for the vector potential without choosing a gauge is given by

∂ν∂νAµ(x) = 0 (5.147)

This equation is solved by the an satz,

Aµ(x) =

∫d4p

(2π)4εµ(p)e−ip·x (5.148)

The gauge transformation of the vector potential under which physics is invarient is,

A′µ = Aµ + ∂µα (5.149)

Plugging in the potential as well as the Fourier transform of α =∫

d4p(2π)4 e

−ip·xα(p) we get∫d4p

(2π)4

(ε′µ − εµ

)e−ip·x =

∫d4p

(2π)4(−ipµ)e−ip·xα (5.150)

Now multiplying by e−ip′·x and integrating over d4x we have∫d4p

(2π)3

(ε′µ − εµ

)δ(p− p′) = −i

∫d4p

(2π)4pµαδ(p− p′) (5.151)

ε′µ(p) = εµ(p)− iα(p)pµ (5.152)

This is the effect of a gauge transformation on the polarization vector.Now in general any amplitude with N gauge bosons will take the form,

εµ1εµ2 ...εµNMµ1µ2...µN (5.153)

If a theory is gauge invariant then this amplitude must be invariant under any gaugetransformation. Applying equation 5.152 to some polarization vector i we have

εµ1εµ2 ...εµNMµ1µ2...µN → εµ1εµ2 ...εµNMµ1µ2...µN − iα pµiMµ1,...,µN (5.154)

In order for the amplitude to be gauge invariant we must have Ward Identity,

pµiMµ1,...,µN = 0 (5.155)

which is just the Ward Identity.We now try to explore how this relationship holds. Consider the following

p→

p2

p1


where we have p is off-shell since otherwise it would violate energy conservation. Theamplitude is,

M = εν u(p1)γνv(p2)(−ie)︸︷︷︸Mν

(5.156)

Then the sum, pνMν , is zero since,

pνMν = u(p1)/pv(p2)(−ie) (5.157)= u(p1)(/p1

+ /p2)v(p2)(−ie) (5.158)

= (−ie)u1v2(me −me)u1v2 (5.159)= 0 (5.160)

Now if we are considering a more complicated diagram,

×+

×

where the “×” represents some combination of external lines. Then the Ward identitymeans that these contributions must cancel. See Peskin chapter 7.4 for details. [Q6: Expand on this section]

Chapter 6

Loop Diagrams and UV Divergences inQED

6.1 Superficial Divergences

The first thing one thinks about when hearing about loop diagrams is estimating howdivergent diagrams are. We use what’s known an “naive power-counting” to find thesuperficial degree of divergence. For D greater then or equal to zero we expected ∼ ΛD

and we expect logarithmic divergence if D = 0. If D < 0 then our diagrams are notsuperficially divergent. Each loop contributions a factor of

∫d4p

(2π)4 (4 factors of momenta).Fermion propagators have dimension −1 and photon propagators have dimensions −2hence (Pi the number of i propagators)

D = 4L− Pf − 2Pγ (6.1)

One can relate this expression to the number of external lines by (Exercise)

D = 4− 3

2Nf −Nγ (6.2)

61

62 CHAPTER 6. LOOP DIAGRAMS AND UV DIVERGENCES IN QED

In reality the power counting is not precise and diagrams may or may not diverge inde-pendently of the superficial degree of freedom. Consider the D ≥ 0 diagrams in QED:

Diagram D Expected # of CT’s Namep D = 4 vacuum energy → unphysicalpg D = 3 zero!(HW)gpg D = 2 3 counterterms( vacuum polarization

vpguD = 1 zero!(HW)

gpggg D = 0 1 countertermFpF D = 1 2 counterterms electron self-energy

DpgED = 0 1 counterterm vertex correction

Our Lagrangian is

L = ψ(i/∂ −m0)ψ − 1

4FµνF

µν + e0Aµψγµψ (6.3)

We want to introduce renormalized fields ψ =√Z2ψr and Aµ =

√Z3A

rµ. We then have

L = Z2ψr(i/∂ −m0)ψr − Z31

4F rµνF

µνr + e0Z2

√Z3A

rµψrγ

µψr (6.4)

We introduce a physical mass and we have δm = m−m0:

L = Z2ψr(i/∂ −m)ψr − Z31

4F rµνF

µνr + e0Z2

√Z3A

rµψrγ

µψr + δmZ2ψrψr (6.5)

We still need to renormalize the coupling constant. We define the new coupling constantas

e =e0Z2

√Z3

Z1

(6.6)

How we define a coupling is not physical since it’s not a number like mass which has adefinition on it’s own. We just need to require that it won’t be divergent. This formturns out to be convenient. With this we have

L =ψr(i/∂ −m)ψr −1

4F rµνF

µνr + eArµψrγ

µψr+

(Z1 − 1)Arµψrγµψr + (Z2 − 1)ψr(i/∂ −m)ψr − (Z3 − 1)

1

4F rµνF

µνr + δmZ2ψrψr (6.7)

We define

δ1 ≡ Z1 − 1

δ2 ≡ Z2 − 1

δ3 ≡ Z3 − 1 (6.8)δm ≡ (Z2 − 1)(m)− δmZ2

6.2. ELECTRON SELF-ENERGY 63

Our counterterms are

LCT = δ1Arµψrγ

µψr + ψr(iδ2 − δm)ψr −δ3

4F rµνF

µνr (6.9)

Note that from here on we drop the r to denote renormalized fields. Naively we expectthe number of counter terms to be given by the sum of all the divergences→ 3+1+2+1.However we can only possibly write down 4 terms! We know this theory is renormalizableso the superficial degree of divergence must be overestimating the divergence of thistheory. Consider our renormalized propagator

µ

q→gpg

q→ν = Πµν(q) = A(q2)gµν +B(q2)qµqν (6.10)

The Ward identity tells us that qµΠµν = 0⇒

Aqν +Bq2qν = 0 (6.11)

since this must be true for all qµ we have

A = −Bq2e (6.12)

and henceµ

q→gpg

q→ν = (gµνq

2 − qµqν)B(q2) (6.13)

The Ward identity provides a constraint and shows that we only need 1 CT for thepropagator. Similarly one can show in the same way that the four photon diagram is infact finite and does not need a counterterm. So only need the 4 CT’s.

6.2 Electron Self-Energy

Gab(p) =

∫d4xe−ix·p 〈Ω|T

(ψa(x)ψb(0)

)|Ω〉 (6.14)

where in general we have ψb(y) but we just choose y = 0. Using Feynman diagrams wecan write

G =F+FFyFF+FFyFyFF+FFyFFFyFF︸︷︷︸

not 1PI+... (6.15)

=i

/p−m0 + iε+

i

/p−m0 + iε(−iΣ(p))

i

/p−m0 + iε(6.16)

where Σ(p) is a matrix that’s equal to the sum of all the possible diagrams with twoexternal lines and p is the incoming momentum.


We define 1 particle irreducible (1PI) graphs to be diagrams that you cannot cut into2 by cutting a single line. We define

Σ =(sum of 1PI graphs only,without external lines

)=FyF+FyFyF+ ... (6.17)

we have (for brevity we initially suppress the +iε in the denominator)

G =i

/p−m0

+i

/p−m0

(−iΣ

) i

/p−m0

+i

/p−m0

(−iΣ

) i

/p−m0

(−iΣ

) i

/p−m0

+ ...

(6.18)

=i

/p−m0

∞∑n=0

(Σ

1

/p−m0

)n(6.19)

=i

/p−m0

1

1− Σ/p−m0

(6.20)

=i(/p−m0 + iε

)/p−m0 + iε

(1

/p−m0 + iε− Σ(p)

)(6.21)

=i

/p−m0 − Σ(p) + iε(6.22)

where we have used a geometric series for matrices,∞∑k=0

Ak = (1− A)−1 (6.23)

for any matrix A with eigenvalues less than 1.To better see what this means, we switch to the “proper form” of the propagator (this

is straightforward to check by multiplying the /p−m+ iε with its inverse),

(/p−m− Σ(p) + iε)−1 =/p+m+ Σ(p)

p2 − (m+ Σ(p))2 + iε(6.24)

Hence the physical mass is shifted to the point where we still have a pole in the propagator,

p2 = (m0 + Σ(p))2 (6.25)

or m ≡ m0 + Σ(p2 = m2). Thus Σ represents the change in mass from the “bare” masscalculated to all orders in perturbation theory. We can also write from 6.22

G−1 = G−10 −

1

iΣ(p) (6.26)

so the two-point function contains, apart from the inverse bare propagator, only 1PIcontributions.


Now we have

G = 〈Ω|ψr(x)ψr(y)|Ω〉 (6.27)= Z2 〈Ω|ψ(x)ψ(y)|Ω〉 (6.28)

=iZ2

/p−m0 + iε(6.29)

where Z2 is the wave function renormalization factor (this is just the old unrenormalizedpropagator). By comparing this expression with equation 6.22 we can write

Σ = −(/p−m0 + iε

Z2

− /p+m0 − iε)

(6.30)

dΣ

d/p= −Z−1

2 + 1 (6.31)

⇒ Z2 =

(1− dΣ

d/p

)−1

(6.32)

Thus if we know how to calculate Σ we can calculate both mass renormalization andwavefunction renormalization. Note that we have been a little careless above with ournotation. Formally we can write

Σαβ = −(pµγµ −m0 + iε)αβ(Z−1

2 − 1) (6.33)

so∂Σαβ

∂pµ= −(Z−1

2 − 1)(γµ)αβ (6.34)

or1

i

∂Σ

∂pµ= (Z2 − 1)γµ (6.35)

This is what we really mean when we write ∂∂/p.

We now explicitly calculate Σ to second order in e.

Σ(p) =p→Fy`→F p→ (6.36)

= (−ie)2

∫d4`

(2π)4γµ

i(/+m0)

`2 −m20 + iε

γν−igµν

(`− p)2 + iε(6.37)

= (−ie)2

∫d4`

(2π)4γµ

(`αγα +m0)

`2 −m20 + iε

γνgµν

(`− p)2 + iε(6.38)

Now using (see for example [Griffith(2008)]) γµγαγµ = −2γα and γµγµ = 4 it’s easy tosee that

Σ(p) = −e2

∫d4`

(2π)4

i(−2/+ 4m0)

(`2 −m20 + iε) ((`− p)2 + iε)

(6.39)


Note that this expression has a superficial degree of divergence of order 1. The truedivergence turns out to be of zeroth order as we will see later due to some terms beingodd. To solve this integral and incorporate the divergence we introduce an unphysicalmass µ that will regulate the integral. Note that this is not a unique way to regulate.We could have used for example dim-reg.

Σ(p)→ −e2

∫d4`

(2π)4

(−2i/+ 4m0)

(`2 −m20 + iε) ((`− p)2 − µ2 + iε)

(6.40)

The steps are analogous to our discussion in the Fall of scalar loops diagrams. Weintroduce Feynman parameters (where A = ((`− p)2 − µ2 + iε) and B = `2 −m2

0 + iε)

1

AB=

∫ 1

0

dx

(Ax+B(1− x))2(6.41)

Our expression becomes

Σ(p) = −e2

∫dx

∫d4`

(2π)4

(−2i/+ 4m0)

(((`− p)2 − µ2 + iε)x+ (`2 −m20 + iε)(1− x))

2 (6.42)

= −e2

∫dx

∫d4`

(2π)4

(−2i/+ 4m0)

(`2 − 2`px+ p2x2 − p2x2 + (p2 − µ2)x+ (−m20)(1− x) + iε)

2

(6.43)

= −e2

∫dx

∫d4`

(2π)4

(−2i/+ 4m0)

((`− px)2 −∆ + iε)2 (6.44)

(6.45)

where ∆ ≡ −p2x(1− x)− (µ2 −m20)x+ µ2. Now define k = `− px:

Σ = −e2

∫dx

∫d4k

(2π)4

(−2i(/k + /px) + 4m0)

(k2 −∆ + iε)2 (6.46)

(6.47)

Applying a Wick rotation such that k0 → ikE0 and k→ kE. We now get (dk0 = idkE0 andk0 → ikE0 )

Σ = −e2

∫dx

∫d4kE(2π)4

(2(i/kE − /px) + 4m0)

(k2E −∆ + iε)

2 (6.48)

The denominator is even. Thus we can get rid of the top contribution from /kE1:

Σ = −e2

∫dx

∫d4kE(2π)4

−2(/px) + 4m0

(k2E −∆ + iε)

2 (6.49)

(6.50)

1this is where the reduction in the divergence of the diagram occurs


First since our integral is Wick rotated we are no longer near a pole and we can drop theiε. Now taking the integral is a 4D spherical space we have (S3 = 2π2)

Σ = −e2

∫dx

∫dkEk

3E

8π2

−2(/px) + 4m0

(k2E −∆)

2 (6.51)

=−e2

4π2

∫dx((−/px) + 2m0

) ∫dkE

k3E

(k2E −∆)

2 (6.52)

=−e2

4π2

∫dx((−/px) + 2m0

) 1

2

(−Λ2

Λ2 −∆2+ log

Λ2 −∆

∆

)(6.53)

where in the last step we introduced a cutoff. The integral is easy enough to do inMathematica. For simplicity we now assume that Λ2 ∆ then we get the somewhatsimpler result,

Σ = −απ

∫dx(/px− 2m0

)log

(Λ2

∆

)(6.54)

Consider the limit that µ→ 0. In this case ∆→ m2(1− x) and we have to integratethe logarithm from 0 to 1. This is divergent as logs aren’t defined at 0. This causesthe electron normalization to diverge. However, the electron normalization itself isn’tphysical. No observables will depend on µ, however we introduce it for convenience so allthe terms in our calculations will be well defined. The physical mass is shifted by

m = m0 + Σ(m) (6.55)

We can use m or m0 in evaluating the Σ since the correction will be of order α2. To seethis consider the structure of Σ:

Σ(/p) = /pf(/p2) + g(/p

2) (6.56)

Setting /p = m and expanding f, p,m we have:

Σ(/p = m) =(m0 + δm(1) + ...

) (f (0) + αf (1) + ...

)+(g(0) + αg(1) + ...

)= 0 (6.57)

The zero’th order result is just the propagator evaluated at /p = m0:

f (0) =1

p2 −m20 + iε

, g(0) = − m0

p2 −m20 + iε

(6.58)

and trivially satisfies the condition. To first order in α we have,

δm(1)f (0) +m0f(1) + g(1) = 0 (6.59)

Inside f (1) and g(1) there are contributions proportional to m and m2. However, when tothis order we drop any additional factors of α which arise from inserting m0 + αδm(1).Thus we can just use m ≈ m0 to first order.


We then have the rather simple result,

Σ ≈ α

πm0

∫dx (2− x) log

Λ2

−m20x

2(6.60)

This integral can be evaluated in Mathematica to be

Σ ≈ 3α

4πlog

Λ2

m20

(6.61)

So we finally have the mass shift,

m = m0 + Σ(m0) = m0

(1 +

3α

4πlog

Λ2

m20

)(6.62)

There are two approaches towards renormalization. One interpretation is that it isjust mathematical trickery that is used in order to get finite answers. This point of viewis fine and it certainly works however there is a more modern interpretation which saysthat there is some physics in renormalization.

Suppose we are evaluating some loop diagram. Our Feynman rules say to integrateover some loop momentum. If we work in position space we know the relation betweenmomenta and wavelength. Large momenta correspond to short wavelengths. At veryhigh loop momenta it corresponds to fluctuations of the field with short wavelengths.

With every theory there is some scale in which we know the theory works. For examplefor all we know the SM works well until about the energies that we have gotten to so far.However when we do loop integrals we are told to integrate all the way to infinity (allpossible virtual particle momenta are allowed). That means that we are extrapolating toenergy scales further then we know that the theory works. With this point of view, thereprobably means that this extrapolation has to fail at some energy. This is analogous toblackbody radiation problem in quantum mechanics. Physicists tried to extrapolate theirunderstanding of blackbody radiation to regimes where they didn’t know it held.

The suspicion about these divergences in QFT is that they are there because we aremissing some essential pieces of physics that occur at some shorter length scale that weare unable to probe. Putting in a cutoff scale, Λ, is just some crude model for whathappens at some short length scale. Another way to model how a theory behaves atsmall length scales is to switch from field theory to string theory. Depending on theparticle high energy theory you must use a different regulator. Different regulators areessentially different toy models for how to model the effects at higher energy scale. Thefreedom of choice of regulator is a representation of our ignorance of high energy physics.

In particular since we know that gravity cannot be explained by QFTs we believe thatthe highest Λ can be is at the Planck scale (though it is likely much earlier). Hence Λ issome scale that is of the Planck scale or less. Renormalizability tells us that the details ofthe high energy physics don’t effect the effects at the low energy regime. By definition itmeans that the dependence of the cutoff can be absorbed into the CT’s and won’t effectphysics quantities. The cutoff scale is taken is some sense to be a real physical energyscale. The interpretation of a formula such as 6.62 is an observer at a low energy will seem ≈ m0 however at a large energy there are going to be corrections.

6.3. CLASSICAL RENORMALIZATION AND REGULATORS 69

6.3 Classical Renormalization and Regulators

Consider an electron. It has some electric field

e

Einstein told us that energy is equivalent to mass hence the electric field must changethe mass of the particle. This correction is what we calculated using equation 6.62.Classically one can go ahead to calculate this same result.

δm =

∫d3r|E|2

8π(6.63)

=4π

8π

∫ ∞0

r2dr( er2

)2

(6.64)

∝ e2

∫ ∞0

dr

r2(6.65)

→∞ (6.66)

Classically this energy doesn’t converge!The problem occurred since we extrapolated clas-sical EM to a length scale in which it was no longer valid. Classical EM is a low energyeffective theory. While it does a good job at describing the physics at low energies (largelength scales) it fails at high energies. Alternatively one can introduce a length cutoff(analogous to the energy cutoff, Λ that we are used to in QFT:

δm =∝ e2

∫ ∞r0

dr

r2(6.67)

∝ e2

r0

(6.68)

This quantity is no longer infinite. One could ask why this length scale was requiredsince to our current understanding the electron truly is a point charge. The problem isQM. In QFT the vacuum is not static, but electron positron pairs pop in and out of thevacuum. The vacuum of QED is filled with these pairs:


+ -

+ -

+ -

+ -

+ -

+ -

These pairs can appear as long as they disappear in a timescale as determined by theuncertainty principle. Normally we don’t care about them, since they average to zero(equally likely to be oriented in all direction). However if you have a real electron thenthey are more likely to orient themselves along the electric field and screen the field.As you approach the electron of order the inverse electron mass these effects becomeimportant. The wavelength that such effects will occur is the Compton wavelength

λ =1

me

≈ rc (6.69)

is known as the Compton wavelength. We identify r0 with the Compton wavelength andwe get

δm ∝ e2me (6.70)

This is starting to look very similar to our equation that we derived with QFT exceptwith the Logarithmic divergence. Knowing the higher energy theory of QED told us theexact form of the correction. In essence our QED calculation is just an estimate that willbe fixed when more physics is added. One can go ahead and renormalize classical EM ifone wanted as well.

Consider our correction term in equation 6.62. α = 1137

, m0 ≈ 511keV , Λ ≤MPlanck ∼1019GeV . This makes the correction

3α

4πlog

Λ2

m20

≈ 0.2 (6.71)

So even with having this huge cutoff scale the correction is still not large. Note suchan argument doesn’t work using the φ4 theory. In the end we don’t really care since werenormalize our theories anyways.

When picking a regulator it’s important to have it respect the symmetry of yourproblem, since its possible that the high energy theory does respect the theory (of courseit is also possible that it won’t break the symmetry but you don’t want to remove theother possibility). As an example recall our derivation of the photon terms. We were notallowed to write down a mass term:

1

2m2AµA

µ (6.72)

since we wanted gauge invariance. When finding the photon propagator:

µ

q→gp q→gν∝(gµνq

2 − qµqν)

Π(q2) (6.73)

6.4. DIMENSIONAL REGULARIZATION 71

we would in general renormalize the mass and create a mass. However experimentallywe know the photon is massless to a high precision. We want to somehow ensure thatthe photon will remain massless at this high energy scale. We want to introduce aregulator that respects gauge invariance. The cutoff is not a gauge invariant cutoff (thisis not obvious at this point). The preferred regulator in QED is through dimensionalregularization or “dim reg”.

6.4 Dimensional RegularizationUsing dim reg we now consider the integral we found earlier with a hard cutoff

Σ(p) = e2

∫dd`

(2π)dγµ(/+m0)γµ

(`2 −m20 + iε) ((`− p)2 −m2

0 + iε)(6.74)

Note that we have changed the dimension of each object in this expression from 4 → d.It’s not clear what the γ matrices even mean in more then 4 dimensions. A discussion inPeskin and Schroeder tells us that γµγµ = d, γµγνγν = −(d− 2)γν . These expression areanalytic continuations of the expressions of integer d.

Σ(p) = 2e2

∫ 1

0

dx

∫ddkEd(2π)d

−(d2− 1)x/p+ d

2m0

(k2E + ∆)2

(6.75)

=2e2

(4π)d/2Γ(2− d/2)

∫ 1

0

dx

(d

2m0 −

(d

2− 1

)x/p

)(1

∆

)2−d/2

(6.76)

In 4 dimensions this expression is infinite because of the Γ term. However in 4 − εdimensions we have

Σ(p) =e2

8π2

(∫ 1

0

dx(x/p− 2m0

)(2

ε− γ + log 4π − log ∆

)+

∫ 1

0

dx(x/p−m0

)+O(ε)

)(6.77)

where γ is the Euler constant and equal to 0.566....

6.5 Vacuum PolarizationConsider

Gµν(q) =q→gp q→g (6.78)

=g+gcg+gcgcg+ ... (6.79)

= − i

q2

(gµν −

qµqνq2

)+

(− i

q2

(gµα −

qµqαq2

))(iΠαβ)

(− i

q2

(gβν −

qβqνq2

))+ ... (6.80)


where we use the Landau gauge for the propagator. We define

Π =c+cgc+ ... (6.81)

andΠ =

∑1PI graphs (6.82)

Note that this is completely analogous to Σ and Σ from earlier. Then

Gµν = − i

q2

(gµν −

qµqνq2

)+

(− i

q2

(gµα −

qµqαq2

))(iΠαβ

)(− i

q2

(gβν −

qβqνq2

))+ (...)µα

(iΠ)αβ

(...)βσ

(iΠ)σρ

(...)ρν + ... (6.83)

The Ward identity says that (see equation 6.13)

Παβ = q2

(gαβ −

qαqβq2

)Π(q2) (6.84)

which implies(gµα −

qµqαq2

)Παβ = q2

(gµα −

qµqαq2

)(gαβ − qαqβ

q2

)Π(q2) (6.85)

= q2

(gβµ −

1

q2

(qµq

β + qµqβ)

+qµq

β

q2

)Π(q2) (6.86)

= q2

(gβµ −

qµqβ

q2

)Π(q2) (6.87)

and hence

Gµν = − i

q2

(gµν −

qµqνq2

)+

(gβµ −

qµqβ

q2

)(gβν −

qβqνq2

)((−i)(i)Π(q2)

)+ (...) (...) (...)

(−iΠ(q2)

)2+ ...

(6.88)

We define Aαβ ≡ gαβ − qαqβq2 . With this we can write as

Gµν =−iq2

Aµν + A β

µ AβνΠ(q2) + AµαAαβAβν

(Π(q2)

)2+ ...

(6.89)

but (gβµ −

qµqβ

q2

)(gβν −

qβqνq2

)= gµν −

qµqνq2

(6.90)

6.5. VACUUM POLARIZATION 73

or equivalently A βµ Aβν = Aµν . This gives

Gµν =−iq2

Aµν + Aµν

Π(q2)

q2+ Aµν

(Π(q2)

)2+ ...

(6.91)

=−iAµνq2

[1− Π(q2))

]−1 (6.92)

=−iq2

(gµν −

qµqνq2

)1

1− Π(q2)(6.93)

(6.94)

So we have our old propagator with an extra factor, since the photon propagator is

−iq2

(gµν −

qµqνq2

)(6.95)

Note that here the Π doesn’t come additive to the q2 value as it did for the massive fields.The pole is still at q2 = 0 ⇒ mγ = 0. One can trace the reason that this happened tothe Ward identity (and hence gauge invariance). The only renormalization comes aboutin the wavefunction renormalization:

Gµν = − i

q2

(gµν −

qµqνq2

)Z3 (6.96)

or equivalently in Feynman gauge:

Gµν = −igµνq2

Z3 (6.97)

The amplitude we want is,

−iΠµν =q → q →

← `

`+ q →

− iΠµν(q) = (−1) (−ie)2

∫d4`

(2π)4

Tr[γµ(−/+m

)γν(/+ /q +m

)](`2 −m2)

((`+ q)2 −m2

) (6.98)

The most divergent part is (using a hard cutoff)∫d4` (a`2gµν + b`µ`ν)

`4∼ gµνΛ

2 (6.99)


However this object doesn’t obey qµΠµν = 0. Hence this is not true at O(Λ2)! To solvethis integral in a gauge invariant way we use dim reg. We can greatly simplify the amountof work that we need to do by identifying that the final answer must be in the form,

Πµν = (q2gµν − qµqν)Π(q2) (6.100)

Thus we just need to calculate either the qµqν or the gµν part and just infer the otherpart. As we will see it will be easier to calculate the qµqν contribution. The denomenatoris:

1

`2 −m2 + iε

1

(`+ q)2 −m2 + iε=

∫dx

1[(`+ qx)2 −∆ + iε

]2 (6.101)

where∆ ≡ −q2x (1− x) +m2 (6.102)

We now shift the integration variable and conside the trace:

Tr [...]→ Tr[(−/q + /+ /qx+m

)γν(−/− /qx+m

)γµ]

(6.103)= −Tr

[(/+ /q (x− 1)

)γν(/+ /qx

)γµ]

+m2Tr [γµγν ] (6.104)= −Tr

[/γν /γµ

]− Tr

[/q (x− 1) γν/qxγµ

]+m2Tr [γµγν ] + linear in ` (6.105)

= 4(2`µ`ν − `2gµν

)+ 4x (1− x)

(2qµqν − q2gµν

)(6.106)

∆ only depends on q2 so there is only one term that has a qµqν contribution. We workout this term and infer the value of the rest of the terms. We need the integral,∫

d4`

(2π)4

1

[`2 −∆]2=

i

(4π)2(1 + ε log 4π)

(2

ε− γ)

(1− ε log ∆) (6.107)

=i

16π2

(2

ε− log

∆

4π− γ)

(6.108)

Thus

Πµν = −qµqν(− e2

4π2

∫dx2x (1− x)

(2

ε− log

∆

4π− γ))

+ (...) gµν (6.109)

⇒ Π(q2) = −2α

π

∫dx (1− x)

(2

ε− log

∆

4π− γ)

(6.110)

Recall that earlier we showed that for an electron scattering off a muon in the non-relativistic limit given by

the amplitude went as iM∼ e2

q2 . The Born approximation tells us that

〈p|iT |p〉 ∼ V (p− p′) (6.111)

6.5. VACUUM POLARIZATION 75

We had

V =e2

r→ V ∼ e2

|q|2(6.112)

what we really need to do is

(we need to have the renormalized propagator for the photon). This results in iM∼ e20Z3

|q|2

using instead of the bare charge we have e = e0

√Z3. At higher energies we need to

include the following correction as well:

iM∼ e20Z3

|q|2

(1− Π(0)

1− Π(−q2)

)(6.113)

Going through the calculation one can show

V (r) =α

r

(1 +

α

4√π

e−2mr

(mr)3/2+ ...

)(6.114)

The interaction gets stronger then the Coulomb expectation as the charges get close.This goes back to why this effect is called vacuum polarization. Imagine the muon:

+ -

+ -+

-+ -

+ -+

-

+ -

+ -

Hence at low energy what you see is the screened potential. This is the effect of havinge2 → e2

0Z3. As you start throwing more energetic electrons toward the muon, then youget the electron very close the muon. You can get rid of some of the screening effect andthe stronger the field will be. You’re interaction with the muon will not be determinedby e2 but more by something that’s more close to the bare charge. Eventually when youare very close the center of the muon you will be able to measure Z3. This is the conceptbehind the running coupling constant. This effect is observed and well tested.


6.6 Vertex Function

Consider the vertex function which represents the following diagram

Gµ(p, p′) =

p

p′

q ≡ p′ − pµ

in the limit of p2 → m2e, p′2 → m2

e but q has to stay off shell by energy momentumconservation. Now recall that

i

/p′ −m+ iε=

i(/p′ +m)

p′2 −m2 + iε=i∑2

s=1 us(p′)us(p′)

p′2 −m2 + iε(6.115)

so we have a

Gµ =

∑us(p′)us(p′)

p′2 −m2 + iεΓν(p, p′)

∑us′(p)us

′(p)

p2 −m2 + iεDµν(q) (6.116)

The interesting object is Γν . To leading order we have just a single vertex:

→ Γν = −ieγν

(in the terminology we will introduce shortly, F1 = 1, F2 = 0). Next to leading order wehave the diagram:

p→

``− p ↓

`− qq →

q −p→

Γν = (−ie)3

∫dd`

(2π)dγρ

i(/+ /q +m)

(`+ q)2 −m2 + iεγν

i(/+m

)`2 −m2 + iε

γσ(−igρσ)

(`− p)2 + iε(6.117)

6.6. VERTEX FUNCTION 77

We won’t solve this explicitly however we consider the Lorentz structure. The onlynontrivial scalar in this problem is q2. From the fact it’s a four vector we already knowthat we will have something of the form,

Γν = γνA(q2) + (p+ p′)νB(q2) + (p− p′)νC(q2) (6.118)

In general B and C can be matrices as well however since mu(p) = /pu(p) we can writethe expressions in terms of ordinary numbers. From the Ward identity we know that

qµGµ = 0 (6.119)

and so,

u(p′)qνΓνu(p) = 0 (6.120)

u(p′)

(/q︸︷︷︸

=u(p)(/p−/p′)u(p′)=0

A(q2) +

=p2−p′2=0︷︸︸︷q · (p+ p′)B(q2) + q · (p− p′)︸︷︷︸

q2 6=0

C(q2)

)u(p) = 0 (6.121)

where we have used the fact that q ≡ p′ − p and hence we have C = 0. Now we use theGordon identity:

u(p′)γµu(p) = u(p′)

((p′ + p)µ

2m+iσµνqν

2m

)u(p) (6.122)

where σµν = i2

[γµ, γν ].We quickly prove this identity below. Consider

−pµ + iσµνpν = −pµ − 1

2(γµγν − γνγµ) pν (6.123)

= −pµ − 1

2

(2γµ/p− 2gµνpν

)(6.124)

= −pµ −(γµ/p− pµ

)(6.125)

= −γµ/p (6.126)

Furthermore,

p′µ + iσµνp′ν = p′µ +1

2p′ν (γνγµ − γµγν) (6.127)

= p′µ +1

2

(2/p′γµ − 2p′µ

)(6.128)

= /p′γµ (6.129)

Subtracting these two equations gives

(p′ + p)µ

+ iσµν(p′ − p)ν = /p′γµ + γµ/p (6.130)

up′

(p′ + p)µ

+ iσµν(p′ − p)νup = up′mγ

µup + up′γµmup (6.131)

u(p′)γµu(p) = −up′(

(p′ + p)µ

2m+ iσµν

(p′ − p)µ

2m

)up (6.132)


Using this identity we see that

Γν(p, p′) = (−ie)[γνF1(q2) +

iσµνqµ2m

F2(q2)

](6.133)

where the Fi(q2) are known as form factors. To lowest order we have no loop and F1(q2) =1 and F2(q2) = 0. To second order they are given by [Q 7: Show this]

F1 =e2

(4π)d/2

∫ 1

0

dxdydzδ(x+ y + z − 1)

Γ(2− d/2)

∆2−d/2(2− ε)2

2+

Γ(3− d/2)

∆3−d/2

(q2 [2(1− x)(1− y)− εxy] +m2

[2(1− 4z + z2)− ε(1− z)2

])(6.134)

where ∆ = (1 + z)2m2 − xyq2 + zµ2 and ε = 4− d as well as

F2 =α

2π

∫ 1

0

dxdydzδ(x+ y + z − 1)2m2z(1− z)

∆

∣∣∣∣µ2→0

(6.135)

where F2(q2 → 0) = α2π. Note that the UV divergences are completely hidden inside F1

[Q 8: Is this true to all orders?]

6.6.1 Physics of F2

Consider the semi classical picture of a fermion in an EM field

p→p′ →

EM field

Lint = eAµψγµψ (6.136)

To leading order we have

〈p′, s′|iT |p, s〉 = us′(p′)(−ieγµ)us(p′)Aµ(p− p′) (6.137)

and the Feynman rule is,

×q → −ieγµAµ(q)

6.6. VERTEX FUNCTION 79

In non relativistic electron scattering we have (for a scalar potential, Aµ = (φ((x)),0)⇒

Aµ = (φ(q),0)). Recall that in the Weyl basis us =√

2m

(ξs

ξs

)〈p′, s′|iT |p, s〉 = −ie u(p′)γ0u(p)φ(q) = −ie(2m)δss′φ(q) (6.138)

If we have a vector potential Aµ = (0,A) with B = ∇×A then

〈p′, s′|iT |p, s〉 = −ieus′(p′)γius(p)Ai(q) (6.139)

= −i(2m)e

m

(ξs′†σiξs

)Bi(q) (6.140)

[Q 9: I don’t see how we got from A to B.] (recall that γi =

(0 σi

σi 0

)).

Now in the quantum mechanics Born approxiamtation we have,

〈p′, s′|iT |p, s〉 ∼ V (q) (6.141)

This equation shows that the potential is given by V = emS · B = µ · B. From non-

relativistic QM we know that µ = g e2m

S, where g is the Landé g factor. Hence to leadingorder it was found that g = 2. See Peskin and Schroeder pg 187-188 for more details.

Now consider the next to leading order expression

×

= −ieu(p′)

(γµF1(0) +

iσµνqν2m

F2(0)

)u(p)Aµ(q)

The scalar field is V = eF1(0) ·φ(x). With the vector field one can show that you get thesame structure we found before, V ∝ S ·B with a different g factor

g = 2 + 2F2(0) = 2 +α

π(6.142)

This was the first radiative correction that was predicted and confirmed by experiment.

Chapter 7

Renormalized Perturbation Theory

7.1 φ4 Theory

Recall φ4 theory:

L =1

2(∂µφ)2 − 1

2m2

0φ2 − 1

4!λ0φ

4 (7.1)

where

G(p) =iZ

p2 −m2 + iε+ (no poles) (7.2)

Define new renormalized fields as φr = Z−1/2φ which implies that Gr(p) = ip2−m2+iε

+ ...and we have

L =Z

2(∂µφr)

2 − Z

2m2

0φ2r −

Z2

4!λ0φ

4r (7.3)

Experimentally we want to measure the mass, m and the coupling, λ. λ can be measuredfrom φφ→ φφ at p→ 0. More precisely we can measure

iM4(s = 4m2, u = t = 0) = λ (7.4)

Notice that when we do this measurement we don’t know about what’s a loop diagram,what’s tree level. This term includes all diagrams. We want to rewrite our Lagrangianin terms of these physical parameters:

L =1

2(∂µφr)

2 − 1

2m2φ2

r − 4!λφ4

r +1

2δz(∂µφ2)2 − 1

2δµφ

2r −

1

4!δλφ

4r (7.5)

where δz = Z − 1, δm = Zm20 −m2, δλ = Z2λ0 − λ. Our new Feynman rules are

80

7.2. QED 81

=i

p2 −m2 + iε

= −iλ

× = i(p2δz − δm)

× = −iδλ

we impose our renormalization conditions:(G2r(p

2))−1

∣∣∣∣p2=m2

= 0 (7.6)

d

dp2

(G2r(p

2))−1

∣∣∣∣p2=m2

= −i (7.7)

At first order these conditions are trivially satisfied. At higher orders these conditionsgive us our values of δz, δm.

7.2 QEDThe QED Lagrangian in terms of the renormalized parameters is,

L = ψr(i/∂ −m)ψr − eψrγµψrAr,µ −1

4(Fr,µν)

2

+ ψr(iδ2/∂ − δm

)ψr − eδ1ψ1γ

µψrAr,µ −1

4δ3 (Fr,µν)

2 (7.8)

The original theory was gauge invariant so we better maintain that gauge invariance. Thecounter-terms must also be gauge invariant. The only way that we happen is if δ1 = δ2.To see this consider part of QED Lagrangian (the rest is trivialy gauge invariant on itsown),

∆L = δ2ψi/∂ψ + ieδ1ψ1γµψAµ (7.9)

Now consider a gauge shift Aµ → Aµ − 1e∂µα(x), ψ → e−iα:

δL = −δ2ψiγµi∂µαψ − iδ1ψ1γ

µψ∂µα (7.10)= (δ2 − δ1)ψγµψ∂µα (7.11)

82 CHAPTER 7. RENORMALIZED PERTURBATION THEORY

In order for this to be gauge invariant we require δ1 = δ2. This highly non-trivial resultimplies that the renormalization of the QED coupling is equal to the renormalization ofthe fermion wavefunction renormalization! This implies that Z1 = Z2 and the renormal-ization couplings is simplify given by (see Eq. 6.6),

e = e0

√Z3 (7.12)

We have the Feynman rules

→ i/p−m+iε

→ −igµνq2+iε

× → i(/pδ2 − δm)

× −i(gµνq2 − qµqν)δ3

−ieγµ

The renormalization conditions are( )−1∣∣∣∣/p=m

= 0

dd/p

( )−1∣∣∣∣/p=m

= −i

ddq2

( )−1∣∣∣∣=0

= −igµν

= −ieγµ

The first two conditions provide us with δz, δm, the third conditions gives δ3 and the lastcondition gives δ1.

We now consider the two point function at 1-loop. We have already derived thisexpression (the amputated version):

gcg = i(gµνq2 − qµqν

)Π(q2) (7.13)

7.2. QED 83

We have

gpg =g+gcg+gxg+ ... = − igµνq2 + iε

1

1− ˆΠ(q2)(7.14)

where Π =∑

1PI diagrams. Our condition is that Π(q2 = 0) = 0. We have (theamputated version)

gcg+gxg = i(gµνq2 − qµqν

) (Πloop(q

2) + δ3

)(7.15)

and hence we know thatδ3 = −Πloop(q

2 = 0) (7.16)

(calculated last lecture)

Chapter 8

Running Couplings andRenormalization Group

8.1 “Large Logs”Consider λφ4 theory. The phenomena happens in almost all QFT’s however it is conve-nient and easier to work with φ4 theory. We consider 2→ 2 scattering. We will work ata COM energy of

√s = µ m. We will basically be computing Green’s function like

this:

= + + ++ + + ...

s channel t channel u channel

The first diagram is trivial and given by λ. Based on dimensional analysis we know thatthe second diagram is

Γ = cλ2

16π2

(log

(Λ√s

)+ non-log

Λ-ind terms

)+ δλ (8.1)

δλ is given when Γ = 0. When s = 4m2 ⇒ δλ = −c λ2

16π2

(log Λ

m+ ...

). Hence Γ is given by

Γ = cλ2

16π2

(log

(Λ√s− log

Λ

m

)+ non-log

Λ-ind terms

)(8.2)

= cλ2

16π2

(log

(m√s

)+ non-log

Λ-ind terms

)(8.3)

(8.4)

Going to higher order terms we have higher order powers of these logs:

Γ = cλ2

16π2

(log

m√s

+ non-logΛ-ind terms

)+ c′

λ3

(16π2)2

(log

m√s

)2

+ ... (8.5)

84

8.1. “LARGE LOGS” 85

Our perturbation theory is not a series in λ but it’s a series in λ log√sm. This is an issue

since you lose control of your calculation at sufficiently high energies. This issue can beavoided by organizing your perturbation theory in a better way.

We define

λ ≡ i

∣∣∣∣∣√s=µ

We impose a different renormalization condition given by δλ : Γ = 0 when s = µ2 ⇒ δλ =

−c λ2

16π2

(log Λ

µ

)and we have

Γ = cλ2

16π2

(log

µ√s

+ non-log Λ-ind terms)

(8.6)

Instead of on shell renormalization condition you can impose your renormalization con-dition at any scale you want. In the end your results do not depend on the scale that yourenormalize however that’s only true for the sum of the series. Since we always truncateour series, the error made by that truncation depends on the scale that you chose foryour renormalization scheme. Our λ(µ) is called the running coupling.

Type of PT Coupling NotesBare PT λ0 → “bare cou-

pling”doesn’t include screening (scat-tering at

√s ≈ Λ)

RenormalizedPT (on-shell)

λ → “physicalcoupling”

includes screening (scattering at√s ≈ 2m )

Off-shell Renor-malized PT

λ(µ)→ “runningcoupling”

partial screening 2m <√s < Λ

o

screening

The “Renormalization Group evolution curve” is shown below

86 CHAPTER 8. RUNNING COUPLINGS AND RENORMALIZATION GROUP

λ(µ)

µ

λ

λ0

Λ

Now consider the Lagrangian:

L =1

2(∂µφ)2 − 1

2m(µ)2φ2 − 1

4!λ(µ)φ4 +

1

2δZ(∂µφ)2 − 1

2δmφ

2 − 1

4!δλφ

4 (8.7)

where as before we have

δZ = Z − 1, δm = Zm20 − m2, δλ = Z2λ0 − λ

Before we had a certain scale and that was m. now our scale is µ instead. Before we goahead and start calculating. We try to think what our δ ’s will depend on :

δi = δi

(m, λ,Λ, µ

)(8.8)

We can invert our formulas for δi to get

λ = λ (m0, λ0, µ,Λ) (8.9)m = m (m0, λ0, µ,Λ) (8.10)

Z = Z(m0, λ0, µ,Λ) (8.11)

We fix m0, λ0,Λ (UV physics). We define the Beta functions as

β ≡ µ∂λ

∂µ

∣∣∣∣m0,λ0,Λ fixed

, βm ≡ µ∂m

∂µ, γ ≡ 1

2Zµ∂Z

∂µ

∣∣∣∣m0,λ0,Λ fixed

where for historical reasons the last expression has a different name (the γ function).Note that if you use a different regulator you would keep that fixed instead of Λ.

To 1 loop in φ4 theory one can show that (assuming µ m)

β(λ) ≡ µ∂λ

d log µ=

3λ2

16π2(8.12)


This is called the RG equation (RGE). λ(Λ) ≡ λ0 ⇒∫ λ0

λ(µ)

dλ

λ2=

3

16π2

∫ Λ

µ=µ

d log µ (8.13)

1

λ(µ)− 1

λ0

=3

16π2log

Λ

µ(8.14)

1

λ(µ)=

1 + 3λ0

16π2 log Λµ

λ0

(8.15)

λ(µ) =λ0

1 + 3λ0

16π2 log Λµ

(8.16)

(8.17)

This is the plot we showed earlier.We don’t need to use λ0 we could use the non-relativistic coupling to estimate the

coupling at a different scale: λ(µ) = λNR ⇒ λ(µ) = λNR

1+3λNR16π2 log µ

µ0

. The large log will no

longer be in the PT. Note as µ→ 0, λ→ 0. Hence at arbitrarily low energies the particleis free. These kinds of theories are called IR−free theories. It turns out that in practicethe running coupling does not run to zero. Instead when the coupling gets close to m, itbegins to level off. We will not show this. This turns out to be due to particles having anonzero mass.

Note further that the coupling blows up at

logΛ

µ= −16π2

3λ0

(8.18)

ΛL = µ exp

(− 16π2

3λ(λ0)e

)(8.19)

This is called the “Landau pole”. The coupling becomes large at some scale. At this pointperturbation theory no longer makes sense. The theory is strongly coupled.

We now develop the formalism required to calculate these β functions. Consider thefollowing

〈Ω|T (φr(x1)...φr(xn)) |Ω〉 = G(n)r (x1, ..., xn; λ, m,

via CT︷︸︸︷Λ, µ) (8.20)

we also have

〈Ω|T (φ(x1)...φ(xn)) |Ω〉 = G(n)b (x1, ..., xn;λ0,m0,Λ) (8.21)

φr = Z−1/2φ , G(n)b = Zn/2G

(n)r . But G(n)

b is clearly independent of µ (the unrenormalized


propagator doesn’t depend on what renormalization scale you use). Hence we know that

µd

dµG

(n)b = 0 (8.22)

µd

dµ

(Zn/2G(n)

r

)= 0 (8.23)

µ

(Zn/2∂λ

∂µ

∂

∂λ+ Zn/2∂m

∂µ

∂

∂m+ Zn/2 ∂

∂µ+

n

2ZZn/2∂Z

∂µ

)G(n)r = 0 (8.24)(

βλ∂

∂λ+ βm

∂

∂m+

∂

∂µ+ nγ

)G(n)r = 0 (8.25)

(8.26)

where in the last step we used the definitions of β and γ functions as well as divided byZn/2.

This is called the Callan-Symanzik equation. It applies very generally and not just toλφ4. One can show that to one loop one doesn’t need to worry about the βm. Conceptuallyit is the same as the other functions however it is more convenient to ignore this term.In other words we take m→ 0.

We consider 1-loop calculations: G(2)r and G(4)

r .

G(2)r =

p→f + cp→fffp→ +p→fδZxfp→︸︷︷︸

0, by construction

+ O(λ2) (8.27)

Our one loop diagram gives the CT. To first order we have,

Π(p2) = λ

∫d4`

(2π)4

1

`2 + iε= λNΛ2 (8.28)

where N is a number. Note that this integral is independent of p. Our renormalizationcondition is Π(µ2) + δZ = 0, which implies that δZ = −Π(µ2). To first order we have

G(2)r =

i

p2 + iε(8.29)

which is independent of λ, m, an µ. Thus the Callan-Symanzik equation says that in λφ4

the Gamma function is zero to second order. i.e. γ = 0 + O(λ2). Recall that our fourpoint function was given by

G(4)r =

= + + ++ + + ...

s channel t channel u channel

(8.30)

=

(4∏i=1

i

p2i + iε

)(−iλ+

(−iλ

)2

(V (s) + V (t) + V (u))− iδZ)

+O(λ3) (8.31)

Consider the second loop:


p1

p2

p3

p4

It contribution to the vertex is

V (s) =1

2

∫dd`

(2π)d1

`2

1

(`+ p)2

∣∣∣∣p=p1+p2

(8.32)

=1

2

∫dd`

(2π)d

∫dx

1

(x(2 + 2p`+ p2) + (1−x)`2)2(8.33)

=1

2

∫dd`

(2π)d

∫dx

1

(`2 + 2`px+ p2x2 + ∆)2(8.34)

where ∆ ≡ −p2x2 + p2x.

=1

2

∫dd`

(2π)d

∫dx

1

((`+ px)2 + ∆)2(8.35)

=1

2

∫dx

∫dd`

(2π)d1

(`2 + ∆)2(8.36)

Using the known integral we have

=1

2

∫ 1

0

dx iΓ(2− d/2)

(4π)d/2Γ(2)(−∆)d/2−2 (8.37)

We now take d = 4− ε:

=1

2i

∫dx

Γ(ε)

(4π)2Γ(2)(−∆)−ε (8.38)

=i

32π2

∫dx

(1

ε− γE +

1

12(6γ2

E + π2)ε

)(1− ε log

(p2(x2 − x)

))(8.39)

=i

32π2

∫dx

(1

ε− log(p2(x2 − x))− γE

)(8.40)

=i

32π2

(1

ε+ 2− log(p2)− γE

)(8.41)

or using the Mandelstam variable s ≡ p1 + p2 = p we can finally write

iM =i

32π2

(1

ε− log s+ 2− γE

)(8.42)

or (the two constant are equivalent...)

= − i

32π2

(1

ε− log s+ γE log 4π

)(8.43)


We set the renormalization condition given by s = µ2 ⇒ (∵ we have massless particles inthe CM frame) t = µ2

2(1− cos θ), u = µ2

2(1 + cos θ). Our renormalization condition gives

us (using crossing symmetry)

δZ = −iλ2 (V (s) + V (t) + V (u)) (8.44)

=λ2

32π2

(3

ε− 3 log µ2 + 3γE log 4π

)(8.45)

We have (µ∂

∂µ+ β

∂

∂λ+ nγ

)G(4)r = 0 (8.46)

At order λ we have 0 = 0. At O(λ2) we have µ dependence:

µ∂

∂µG(4)r = µ

∂

∂µ(−iδλ) =

3iλ2

16π2(8.47)

and we have

β∂G

(4)r

∂λ= −iβ +O(λ) (8.48)

To order λ2 the gamma term still doesn’t contribute since O(γ) = λ,O(Gr) = λ2. Usingour equation relating the β functions we have

β =3λ2

16π2(8.49)

8.2 QED β-function

We begin by considering the two photon Green’s function

G(2γ)µν (p) =

∫d4xe−ix·p 〈Ω|T (Aµ(x)Aν(0))|Ω〉 (8.50)

=g+gFFg+gxg+O(e4) (8.51)

= i

(gµν −

pµpνp2

)1

p2 + iε

(1 + e2f(p, ε)

convention︷︸︸︷− δ3(e, ε, µ)

)(8.52)

Here we also set me = 0 (or equivalently µ me). Thus we have βm = 0. The Callan-Symanzik (CS) equation is given by (this is exactly the same as before with renamingthe variables) [

µ∂

∂µ+ β

∂

∂e+ 2γA

]G(2γ)µν = 0 (8.53)

8.2. QED β-FUNCTION 91

We need to compute δ3 in order to know our µ dependence.

q→gpg =i(gµν − pµpν

p2

)q2

1

1− Π(q2)(8.54)

Our on-shell renormalization condition was Π(q2 = 0) = 0. We now abandon this condi-tion and have the condition Π(q2 = µ2) = 0. At one loop we have

q→FFq→+ x = Πloop(q

2)− δ3 (8.55)

and hence we have δ3 = Πloop(µ2). We found earlier that

Πloop(q2) =

−8e2

(4π)d/2

∫ 1

0

dx(1− x)x︸︷︷︸1/6

Γ(2− d/2)

∆2−d/2 (8.56)

where ∆ =

0︷︸︸︷m2e −x(1− x)q2. Expanding the denominator we have

Πloop(q2) = −4

3

e2

16π2

(1

ε− log µ2 +

(µ-independent

constants

))(8.57)

where we have used∆ε = elog ∆ε

= eε log ∆ = 1 + ε log ∆ + ... (8.58)

Now β = O(e3)(we’ll see why soon). If we are consistent to O(e3) we have the CSequation gives:

µ∂

∂µ(−Pµνδ3) + 2γAPµν = 0 (8.59)

where we have defined Pµν = i(gµν − pµpν

p2

)1

p2+iε. Hence we have

γA =1

2

∂

∂µδ3 =

e2

12π2(8.60)

Hence〈Ω|ψ(x)ψ(0)|Ω〉 ⇒ γψ =

1

2µ∂

∂µδ2 =

e2

16π2(8.61)

This function is somewhat abstract and doesn’t have a physical meaning. However wenow go on to find the β function which is physical:

〈Ω|T(ψ(x)ψ(y)Aµ(0)

)|Ω〉 =

x

x

x x

+ + + +

++++(8.62)


The CS equation is given by[µ∂

∂µ+ β

∂

∂e+ γA + 2γψ

]G(3) = 0 (8.63)

To lowest order we have

G(3) = ED=

i

/p′(−ieγν) i

/pDµν(q) ≡ (−ie)Fµ(p, p′) (8.64)

The counterterms are

x

x

x x

+++= −ieFµ +

1

/p(i/pδ2)Fµ + ... (8.65)

= −ie(δ1 − 2δ2 − δ3)Fµ (8.66)

so we have

0 = (−ie)µ ∂

∂µ(δ1 − 2δ2 − δ3)Fµ + β(−i)Fµ + (2γψ + γA)(−ieFµ) (8.67)

β = e

[−µ ∂

∂µ(δ1 − 2δ2 − δ3)− 2γψ − γA

](8.68)

β = eµ∂

∂µ

(−δ1 + δ2︸︷︷︸

=0

+1

2δ3

)(8.69)

where δ1 = δ2 is zero by gauge invariance.Thus we have

β =eµ

2δ∂

∂µδ3 =

e3

12π2(8.70)

Recall thatα(µ) =

α(µ0)

1− 2α(µ0)3π

log µµ0

(8.71)

The coupling goes as

mL

( )

1/137

e

8.2. QED β-FUNCTION 93

where ΛL = me exp(

3π2αNR

) mPlanck

Chapter 9

Non-Abelian Gauge Theories(“Yang-Mills”)

9.1 Looking Closely at Abelian Gauge TheoriesQED is an “Abelian Gauge Theory”. You start with a Dirac Lagrangian:

L = iψγµ∂µψ −mψψ (9.1)

This is the only thing you can write down that is renormalizable, Lorentz invariant, andonly has a fermion field. Once you impose these requirements you see an accidental“global U(1)” symmetry:

ψ → eiαψ, ψ → e−iαψ

and hence L → L. What if we promote the global symmetry to a local symmetry? Thenwe have “local U(1)” (this is also known as “gauge U(1)”).

ψ → eiα(x)ψ, ψ → e−iα(x)ψ

⇒ψµψ → eiαψ + i∂µαe

iα∂µψ (9.2)

and hence L′ 6= L. We can fix this by letting ∂µ → Dµ where the covarient derivativeobeys

Dµψ → eiαDµψ (9.3)

This holds ifDµ = ∂µ − ieAµ (9.4)

where Aµ → Aµ + 1e∂µα. This procedure is unique. To turn this into a real field we need

to add kinetic terms for A. These terms should be quadratic in ∂µAµ and require gaugeinvariance. The only way this can be done is by taking FµνF µν or εµναβF µνFαβ. Thesecond term violates parity so we are left with the first term. We get our Lagrangian andadding in a normalization we have

L = iψγµDµψ −mψψ −1

4FµνF

µν (9.5)

94

9.2. SU(2) 95

Here it was easy to find Fµν . However in different gauge theories it would be harder toconstruct. We now consider a useful way to build such terms. Note that phase factorscommute with the covarient derivative (by construction). In other words

Dµ(eiα...) = eiαDµ(...) (9.6)

with that in mind consider the commutator:

[Dµ, Dν ]ψ → eiα [Dµ, Dν ]ψ (9.7)(∂µ − ieAµ)(∂ν − ieAν)− (∂ν − ieAν)(∂µ − ieAµ)ψ → eiα [Dµ, Dν ]ψ (9.8)

−ie(∂µAν − ∂νAµ)ψ → eiα [Dµ, Dν ]ψ (9.9)−ieFµνψ → eiα [Dµ, Dν ]ψ (9.10)

(9.11)

but Fµν doesn’t transform under gauge symmetry. So we know that Fµνψ → eiαFµνψ.Thus we have

−eiαieFµνψ = eiα [Dµ, Dν ]ψ (9.12)

Thus we can finally write[Dµ, Dν ]ψ = −ieFµνψ (9.13)

9.2 SU(2)

We now generalize U(1)→ SU(2). We put two fields with the same mass in a “doublet”:

Ψ(x) ≡(ψ1(x)ψ2(x)

)(9.14)

L =2∑

a=1

(iψaγ

µ∂µψa −mψaψ)

(9.15)

= iΨγµ∂µΨ−mΨΨ (9.16)

where Ψ ≡(ψ1

ψ2

). This Lagrangian has an accidental global symmetry given by rotation

between the two doublets,Ψ→ VΨ (9.17)

where V is a 2×2 unitary matrix. The possible V matrices form a group known as U(2).We can write V as

V = exp

(iα0 + i

3∑k=1

αkσk2

)(9.18)

96 CHAPTER 9. NON-ABELIAN GAUGE THEORIES (“YANG-MILLS”)

where

1, σk2

are the generators of U(2) algebra. We can split this symmetry into two

parts: U(2) = U(1)× SU(2). SU(2) is defined in terms of its algebra:[σi2,σj2

]= iεijk

σk2

(9.19)

where εijk are the structure constants of SU(2).We first consider the first identity generator. If we demand a local symmetry out of

this generator we would get back QED for two different fields. Now consider the caseof detV = 1 (an SU(2) symmetry). The identity operator does not obey this conditionsince it is not traceless. Thus we only have the Pauli matrices as our generators and wehave

V = exp

(i

3∑k=1

αkσk2

)(9.20)

We promote this symmetry to a local (gauge) SU(2) symmetry.

V (x) = exp(iαk(x)

σk2

)(9.21)

whereΨ→ VΨ , Ψ→ V †Ψ

but∂µΨ→ V ∂µΨ + (∂µV )Ψ (9.22)

and hence L is not invariant under this transformation. We need to introduce a covarientderivative. We define

DµΨ→ V DµΨ (9.23)

We have12×2∂µ − igAµ (9.24)

where Aµ is some matrix.

(∂µ − igAµ)Ψ→ V ∂µΨ + (∂µV )Ψ− igA′µVΨ (9.25)

To find our covarient derivative. We enforce the requirement in equation 9.23. In otherwords we set this equation to V (∂µ − igAµ)Ψ:

A′V = V A− i

g(∂µV ) (9.26)

A′ = V AV † − i

g(∂µV )V † (9.27)

For linear order in α we can write V = 1 + i∑3

k=1 αkσk2. It is sufficient for us to restrict

ourselves to Hermitian, traceless 2× 2 matrices. In this case we can write

Aµ =3∑

a=1

Aaµσa2

(9.28)

9.2. SU(2) 97

With this one can show that our requirement for the covarient derivatives finally says

Aaµ → Aa′µ = Aaµ +1

g∂µα

a + εabcAbµαc (9.29)

Note that our vector boson transformation looks identical to the U(1) transformationwith an extra part. All of this can be applied equally well to U(1) however since thegroup is so simple it is fairly trivial.

We write our Lagrangian as

L = iΨγµDµΨ−mΨΨ (9.30)

= LD [ψ1] + LD [ψ2] + g

3∑a=1

Aaµ

(2∑i,j

ψi

(σa2

)ijγµψj

)(9.31)

where LD [ψ] = iψ(/∂ − m)ψ. We can try to guess our Feynman rules. This is a bitpremature since we haven’t quantized our gauge fields. However let’s give it a shot. Wehave the interactions

Ejµ,agi

→ −igγµ(σa

2

)ij

For example for σ1 =

(0 11 0

)we have different fermion interaction.

We have yet to make these gauge fields dynamical. We need to add a kinetic term forthese fields. To do this we use the trick we learned from QED:

[Dµ, Dν ] Ψ =

(∂µ − igAaµ

σa

2

)(∂ν − igAaν

σa

2

)−(∂ν − igAaν

σa

2

)(∂µ − igAaµ

σa

2

)Ψ

(9.32)

= −ig(∂µA

aν − ∂νAaµ − igAaµAbν

[σa

2,σb

2

])Ψ (9.33)

= −ig(∂µA

aν − ∂νAaµ + gεabcgAbµA

cν

)︸︷︷︸Faµν

σa

2Ψ (9.34)

(9.35)

We call this F aµν our field strength. This object is antisymmetric, F a

µν = −F aνµ. We define

−igFµνΨ ≡ [Dµ, Dν ] Ψ or Fµν = F kµν

σk

2. We have

− igFµνΨ→ −igFµνVΨ (9.36)

andFµν → V FµνV

† (9.37)


We have two options of terms that conserve parity to add into our Lagrangian FµνFµν

or FµµF νν . However Fµν = 0. We’re not done since Fµν is still a matrix. The simplestthing we can do is take the trace of this (we can take the determinate but that’s a mess):

tr(FµνF

µν)

(9.38)

The trace is gauge invariant:

tr(F F)→ tr

(V FV †V FV †

)(9.39)

= tr(V †V F F

)(9.40)

= tr(F F)

(9.41)

Lastly we need to normalize this term. In terms of the non-matrix fields we have

L = −1

2tr(FµνF

µν)

(9.42)

= −1

2

∑k,j

F kµνF

j,µν

12δk,j︷︸︸︷

tr(σk2

σi

2

)(9.43)

= −1

4

3∑k=1

F kµνF

k,µν (9.44)

so this normalization gives the same normalization we had in QED. This is the kineticterm of 3 “photons” + cubic and quartic terms in A (due to the extra gεijkAjµAkν addition).These are self interactions of the gauge field!In QED these vertices did not exist at treelevel. However this is a property of non-Abelian gauge fields. We have diagrams such as

vgu andvuuv even without any other particles. What’s important to note is that

the gεijkAjµAkν where not optional, they were required by gauge invariance.As an example, consider ψψ → V V where V is a gauge boson. This an analogue to

e+e− → γγ. This corresponds to( fgbfg+ photon

crossing

)+ d uge v

Note that the last diagram is fundamental to non-Abelian gauge theories and is requiredin order for the Ward Identity to work.

9.3 General RecipeWe now discuss gauge theories more generally. To construct the theory you need to

9.3. GENERAL RECIPE 99

• Choose a symmetry group (any Lie group, G). In our example this was SU(2).

• Choose a representation r(G). In our example we chose the fundamental represen-tation (∼ , in Young Tableaux notation).

• Introduce n ≡ dim (r(G)) fields - “n-tuplet”. In our example we had n = 2ψ fields.These fields are called “matter fields” (as opposed to gauge fields). These n fieldsmust be in the same representation of the Lorentz group. For Yang-Mills theoriesthese have to be spin 0 or spin 1/2. In our example we have spin 1/2.

• Consider a global symmetryψ → V ψ (9.45)

where V ∈ r(G). We build the most general Lagrangian that is invariant under thistransformation:

L = iψγµ∂µψ − ψψ (9.46)

• Promote the global symmetry to local. In other words we rotate our ψ such thatψ → V (x)ψ with

V (x) = ei∑Ngi αi(x)Ti (9.47)

where Ng is the number of generators and the Ti are generators of the algebraG in the representation r(G). The number of generators is independent of therepresentation chosen. The number, Ng is equal to the dimension of the ad-jointrepresentation (dim(Adj(G))). Where for the ad-joint representation we have

(T i)jk = fijk (9.48)

The generators are given in terms of the structure constants by [Ti, Tj] = fijkTk,where fijk are the structure constants of G. These Ti ’s are n× n matrices.

To promote to a local symmetry we need to require

∂µ → Dµ = ∂µ − igAµ (9.49)

where Aµ always has the same form and given by

Aµ =

Ng∑i=1

Aiµ(x)T i (9.50)

and the T i are the generators in the representation r(G). We introduce one newnumber that is our gauge coupling. Note that in order to enforce gauge invarianceyou are forced to have the same coupling for each Aµ. This is the case for simplegroups (groups that do not have any generators that commute with all other gen-erators). For non-simple groups we are allowed to have such coupling differences.

For example you can split the couplings for U(2) through U(2) =

g︷︸︸︷U(1)×SU(2)︸︷︷︸

g′

.


We have the gauge field transformation

Aµ → V AµV† − i

g(∂µV )V † (9.51)

It’s a simple exercise to show that

Aiµ → Aiµ +1

g∂µα

i + f ijkAjµαk (9.52)

Note that there is no dependence on r(G). Further note that you can have anymultiple of matter fields. However there are a single set of gauge fields that arefixed by G.

One last comment is often people say “gauge fields belong in the adjoint repre-sentation, Adj(G)” (The more correct way to say it is transforms as the adjointrepresentation). For global transformations (α = const) we have

Aiµ = Aiµ + f ijkAjµαk (9.53)

= Aiµ + (T i)jkAjµαk (9.54)

so Aiµ transforms through adjoint representation. However this is only for a globaltransformation and is strictly wrong in general due to the ∂ααi term.

• The Field Strength Tensor is defined as

igFµνψ [Dµ, Dν ]ψ (9.55)

where Dµ are n×n matrices where n is the dimension of the representation. Further

Fµν =

Ng∑i=1

F iµνT

i (9.56)

where F iµν = ∂µA

iν − ∂νAiµ + gf ijkAjµA

kν . This is independent of r(G). The trans-

formation law is given byFµν → V FµνV

† (9.57)

• The Gauge Field Lagrangian is given by

L = −1

2tr(FµνF

µν)

(9.58)

assuming that the generators are normalized as tr (T iT j) = 12δij.

9.4. UNIVERSAL COUPLINGS 101

• The general Feynman rules are as follows given below (the first result will be derivedsoon):

µ,ag k→gg

ν,b=−igµνδabq2 + iε

+(gauge depend.

term

)if p→Ff

j=

iδij

/p−m+ iε

jDgµ,ai = −igγµ (T a)ij

µ,akv ←qg

ρ,cpuν,b = gfabc (gµν(k − p)ρ + gνρ(p− q)µ + gρµ(q − k)ν)

vuuv = −ig2(fabef cde (gµρgνσ − gµσgνρ + (b↔ c, ν ↔ ρ) + (b↔ d, ν ↔ σ))

)

where i, j = 1, ..., dim(r(G)) and a, b = 1, ..., Ng. We can motivate the cubic andquartic vertices by looking at the kinetic term for the gauge bosons:Ng∑a=1

F aµνF

a,µν =∑a

(∂µA

aν − ∂νAaµ + gfabcAbµA

cν

) (∂µAa,ν − ∂νAa,µ + gfadeAd,µAe,ν

)(9.59)

9.4 Universal CouplingsAn important property of gauge theories is that every particle charge under that thegauge group will have the same coupling constant unless the group is Abelian. We nowprove this statement below.

Suppose you have some Lie group with abstract generators τa, then the generatorsobey,

[τa, τb] = ifabcτc (9.60)

and a representation of the group with generators Ta obeys the same algebra,

[Ta, Tb] = ifabcTc (9.61)

Now if the theory has two fields ψA, ψB (we consider fermions though one could just aseasily discuss bosons) in the same reprsentation of a gauge group then by the definitionof a gauge group they must transform the same way:

ψA → eiTaθa

ψA (9.62)ψB → eiTaθ

a

ψB (9.63)


Now the Lagrangian of these fields is of the form,

L = ψADAµ γψA + ψBiD

Bµ γ

µψ + ... (9.64)

where the covariant derivatives are,

DAµ = ∂µ − gAW a

µTa (9.65)

DBµ = ∂µ − gBW a

µTa (9.66)

where here we assume different gauge couplings. But the first derivative implies that theW aµ field transforms as,

W aµ → Wµ + fabcWµ,bθc +

1

gA∂µθ

a (9.67)

and the second derivative implies that

W aµ → Wµ + fabcWµ,bθc +

1

gB∂µθ

a (9.68)

However W aµ are the same fields in either case. Thus we must have gA = gB.

In the above discussion we assumed that the fields were in the same representationof the gauge group. This discussion fails if the fields are in different representations.In an Abelian group since there are no commutation relations, every field can be in adifferent representation of the Lie group. Thus every Abelian field can have differentgauge couplings.

9.5 Quantum ChromodynamicsQuantum Chromodynamics (QCD) is Yang-Mills theory with an SU(3) gauge group. ForSU(N) the number of generators is Ng = N2 − 1. In this case since N is 3 we have 8generators which in turn gives 8 gauge fields Aaµ (a = 1, 2, ..., 8). We call these fieldsgluons. There is one coupling constant g. They are given by structure constants fabc.

We want to write the matter fields that are charged under this gauge group. Thefundamental matter fields are the quarks. We only consider QED so we call these spin -12Dirac fermions (in reality due to the weak interaction we should split these into Weyl

fermions). The quarks are qf = (u, d, s, c, b, t) (f = 1, ..., 6) we call this the “flavor” ofthe quarks. The spin of the particles are the same. The electric charges split into 3 setsof doublets. Their could be a symmetry in between these sets. However it would notexact since they have different masses. We take one of the quarks for simplicity (theyall have the same interactions). These transform as the fundamental representation ofSU(3). The transformations act on vectors given by U . dim() = N (Young Tableauxnotation).

U =

U1

U2

U3

(9.69)

9.6. QUANTIZATION OF YANG MILLS 103

where 1, 2, 3→ r, g, b (“colour”). The Lagrangian is given by

L =∑f

iqfγµDµqf −mqfqf −

1

4

9∑a=1

(F aµνF

a,µν)

(9.70)

where Dµ = ∂µ − ig∑8

a=1AaµT

a and T a are the generators of SU(3) in the fundamentalrepresentation, the “Gell-Mann matrices”.

9.6 Quantization of Yang MillsWe have

W0 =

∫DA exp

[i

∫d4x

(−1

4

Ng∑a=1

F aµνF

a,µν

)](9.71)

where DA =∏Ng

a=1

∏3µ=0

∏x dA

a,µx . We tried this in QED and we ran into a problem. The

procedure was to go to Fourier space and we got a Green’s function which we needed toinvert and it was not invertible. We traced that problem back to the fact that this integralis redundant (we were dealing with all the gauge fields). We required the Faddeev-Popovprocedure. The idea is the same here. The gauge freedom gives

Aaµα(x)−−→ Aaµ +

1

g∂µα

a + fabcAbµαc (9.72)

There is a huge amount of redundancy in this integral which we need to eliminate.The procedure is as follows. You choose some functional of the gauge field and you

choose this to be zero, G [A] = 0. Geometrically:

Int space G[A]=0

Orbits

A

We have a mathematical identity:

1 =

∫Dαδ (G [Aα]) det

(δG [Aα]

δα

)(9.73)

We have

W0 =

∫DAαDα exp

[i

∫d4x (L [Aα])

]δ (G [Aα]) det

(δG [Aα]

δα

)(9.74)


where we shifted our integration measure to Aα. Furthermore, the Lagrangian is bydefinition gauge invariant so we have L [Aα] = L [A]. We now relabel our integrationvariable.

W0 =

∫DADα exp

[i

∫d4x (L [A])

]δ (G [A]) det

(δG [Aα]

δα

) ∣∣∣∣Aα→A

(9.75)

Now it’s clear that∫Dα = N is just a constant. So we can forget about it

W0 =

∫DA exp

[i

∫d4x (L [A])

]δ (G [A]) det

(δG [Aα]

δα

) ∣∣∣∣Aα→A

(9.76)

For U(1) we had G [A] = ∂µAµ − ω(x) and G [Aα] = ∂µ

(Aµ + 1

e∂µα

)− ω(x). Clearly

δGδα

= const (A independent).For non-Abelian gauge theories we have

G [A] =

Ng∏a=1

(∂µAa,µ − ωa(x)) (9.77)

andG [Aα] =

∏a

(∂µ

(Aa,µ +

1

g∂µαa + fabcAb,µαc

)− ωa(x)

)(9.78)

So we have

δG

δαc=δacg∂µ∂

µ + fabc∂µAb,µ (9.79)

The definition of the first term will become clear soon. In order to make sense of thisexpression we go back to the discrete version of our integrals. x → xi, α(x) → αi,Aaµ(x)→ Aaµ,i.

Ga [Aα(x)] = ∂µ

(Aa,µ +

1

g∂ααa + fabcAb,µαc

)− ωa(x) (9.80)

→Aa,µi+1 − A

a,µi

∆+

1

g

1

∆2

(αi+1 + αai−1 − 2αai

)+ fabc

Ab,µi+1 − Aµi

∆αc

+ fabcAb,µαci+1 − αci

∆− ωai (9.81)

So we have

δGa

δαc→ dGa

i

dαcj(9.82)

=1

g∆2(δac )(δi+1,j + δi−1,j − 2δi,j) + ... (9.83)

≡Ma,ic,j (9.84)

9.6. QUANTIZATION OF YANG MILLS 105

The next step was to convert everything in the path integral into a term in the Lagrangian.The delta function becomes a gauge fixing term. In QED we had an extra∫

Dωe−∫d4xω2/2ξ (9.85)

and we have

Lgauge fix. =1

2ξ

∑a

(∂µAa,µ)2 (9.86)

which gives

W =

∫DA exp

[i

∫d4x

(−1

4F aµνF

a,µν +1

2ξ(∂µA

a,µ)2

)](9.87)

The last question is how to convert the determinant into a term into the Lagrangian.This is done by something which are known as ghosts. These in some sense behave likeordinary fields. They take part into the Feynman rules but just cannot propagate.

Ghosts are essentially just way to exponentiate the determinate term so we will be able

to perform a Taylor expansion. We want to rewrite the determinate: det

M i,aj,b︷︸︸︷[

∂µ∂µδab − gfabc∂µAc,µ

]in an exponential form. Faddeev and Popov showed that you can write

det[∂µ∂

µδab − gfabc∂µAc,µ]

=

∫DcDc exp

(i

∫d4xca

(∂µ∂

µδab − gfabc∂µAc,µ)cb)

(9.88)

The new ghost fields, c, c are Grassman variables. Recall that for two Grassman variables:∫dθdθeθAθ = A (9.89)

Earlier we showed that, ∫ (∏i

dθidθi

)e∑i,j θiAi,jθj = detA (9.90)

So any determinate you wish can be written as a product of integrals over Grassmannumbers. The c(x) are “ghost fields”. They are anti-commuting however unlike Diracfields they don’t have an extra index and are spin-0 particles. Anti-commuting spin-0fields violate spin-statistics!On the other hand spin-statistics is only for real asymptoticparticles. So far we have not introduced a source for these ghost fields and addingthis source will violate spin-statistics theorem. In other words these cannot be externalparticles (there are no S matrix elements for these particles).


Adding in these ghost fields we have

W =

∫DADcDc exp

[i

∫d4x

(−1

4F aµνF

a,µν +1

2ξ(∂µA

a,µ)2 + ca∂µ (Dµ)ab cb

)](9.91)

=

∫DADcDc exp

[i

∫d4x

(−1

4F aµνF

a,µν +1

2ξ(∂µA

a,µ)2 − (∂µca) (Dµ)ab cb

)](9.92)

where we have just integrated by parts to which the derivative term. Furthermore Dµ isjust the just regular covarient derivative since we have

∂µ∂µδab − gfabc∂µAc,µ = ∂µ

(∂µδab − gfabcAc,µ

)(9.93)

= ∂µ

(∂µδab − g

(T cAdj

)abAc,µ

)(9.94)

The Feynman rules for these ghosts are,

µ, a ν, b

k →= δab

(−i

k2 + iε

)(gµν − (1− ξ)kµkν

k2

)

µ, a ν, b

k →= δab

(−i

k2 + iε

)

b

a

c, µ = gfabcpµ

←p

Notice that ghosts can only show up in pairs. With this in mind and the fact that theycannot show up in external states, they only appear in loop diagrams. Thus in tree levelyou can just forget about ghosts.

9.7 Unitary and Ghosts

The fact that ghosts can only show up in loops is not obviously consistent with theunitarity of the S matrix. This topic is a bit subtle. We will not do many calculationshowever we will go over the logic and series of statements that describe the situation.

The statement of unitarity is the statement that (this is the same as saying thatprobability is correctly normalized)

S†S = 1 (9.95)

9.7. UNITARY AND GHOSTS 107

gn where S is the S matrix. We split S into the part that’s associated with no collisionand the interested part which is due to the interactions, S = 1 + iT (see for example[Peskin and Schroeder(1995)] pg. 104).

S = 1 + iT (9.96)

Plugging this into 9.95 we have

(1 + iT )(1− iT †

)= 1

1 + i(T − T †

)+ T †T = 1

⇒ −i(T − T †

)= T †T (9.97)

This is further discussed in [Peskin and Schroeder(1995)]. What this says is that (for φ4)

+ ... = ( )*

d3p/E1 d3p/E2

This is known as the optical theorem. Unitarity becomes more intricate for non-Abeliantheories. We denote V as the following amplitude: (ψiψj → VaVb, two fermions into twogauge bosons). The square of this corresponds to ψiψ → ψkψ`.

+ +

= + +

+

Im

This is equivalent to∫d3k1

E1

d3k2

E2

(Vs1,s2 (1, 2→ k1k2)V ∗s1,s2 (k1, k2 → 3, 4)

)=∑s1s2

Mµνεs1µ (k1)εs2ν (k2)εs1∗ρ εs2∗σ M∗ρσ

(9.98)

Recall that in QED we had∑2

s=1 εsµ(k)εs∗ν (k) = −gµν + Akµkν where we relied on the

Ward identity to simplify our results.This was all relatively simple however things get complicated in non-Abelian gauge

theories. It turns out that for 2 or more external gauge bosons this replacement is notallowed (we don’t have (((((

((((∑εµε

νν → −gµν). This is because when you have two gauge

bosons you can make one boson unphysical. If you make one boson longitudinal thenWard identity for the other boson will not work. What always holds is

3∑s=0

εsµε∗,sν = −gµν (9.99)


however this requires summing over all the polarizations. You can work this out and we’llbasically be working this out on the homework.

This now sounds like a problem. Since on the left hand side of the unitarity relationwe have ε ’s (we have external gauge bosons) and on the right hand side we don’t haveany such terms. The term with ghosts fixes this problem. This is quite neat. The ghostsare in a sense there to preserve unitarity. The formal proof of this is known as the BRSTtheorem.

We now go back to tree level. We just said that at tree level we don’t worry aboutghosts. However consider ψiψj → VaVb. We have

M =Mµνεs1µ (p1)εs2ν (9.100)

In QED we would have

∑s1,s2

|M|2QED︷︸︸︷= gµαgνβMµνM∗,αβ (9.101)

However in non-Abelian theories we can’t do that. We have two options

1. We can have explicit ε ’s. If p = (E, 0, 0, E) then ε = 1√2(0, 1,±i, 0). You can just

compute∑

s1,s2|Ms1,s2|2 without using the spin sum formula. We will do this on

the homework. This always works and you never need to mention ghosts.

2. There is a second way to deal with this that is more convenient but a little bitweird. This is to remember that we still have

∑3s=0 ε

sµε∗,sν = −gµν (notice our

summing limits). What you can do is use∑εµε∗ν → −gµν but then you always need

to include an extra diagram:

∣∣∣∣2∣∣∣∣

In other words our tactic is to add an extra diagram with the external gauge bosonsare replaced with external ghosts.

9.8 Renormalized Perturbation TheoryThis just means that we have counter terms. We start with the Yang-Mills Lagrangian.We assume that the fermion is massless. This makes the calculations for the β functionsa lot easier but this is not necessary.

L = ψii /Di,jψj − 1

4F aµνF

a,µν +1

2ξ(∂µA

a,µ)2 − (∂µca)(Dµ)abcb + LCT (9.102)

9.8. RENORMALIZED PERTURBATION THEORY 109

where

LCT = δ2ψi/∂ψ − δ1gψγµAaµT

aψ − 1

4δ3(∂µA

aν − ∂νAaµ)2 + (4 more CT’s) (9.103)

In the case of QED we did not have the last 4 terms since ghosts didn’t exist. In thiscase gauge invariance no longer implies that δ1 = δ2. This is because now for example(∂µA

aν−∂νAaµ) is not even gauge invariant since the gauge invariance is only saved by the

ghost CT’s.What we are really interested in is the β function. To do this we need to use the CS

equations. We need to compute

〈AA〉 gave−−→ γA⟨ψψ⟩ gave−−→ γψ⟨

ψψA⟩ gave−−→ β = M

∂e(M)

∂M

The formula we had in QED for the β function was

β = eM∂

∂M

(−δ1 + δ2 +

1

2δ3

)(9.104)

If you look back at that procedure. Nothing in it used the Feynman rules of the theoryin anyway. We just used the structure of counterterms. In non-Abelian Yang Mills wecan do the exact same thing. Just like in QED the three point function will be

+ + + + +× ×

××

i j

+ +

This calculation is historically significant and lead to a Nobel prize. The countertermshave the following conditions

δ2 : Σ2(/k = M) = 0 (9.105)

We evaluate the following


k → k →+ ×

The first diagram gives,

iM = (ig)2

∫dd`

(2π)dγµ(T a)ij

iδjk/+ /k + iε

γν(T b)k`

Feynman gauge︷︸︸︷(−igµν)k2 + iε

δab (9.106)

= (ig)2∑a

(T aT a)i`︸︷︷︸C2(rf )δi`

∫dd`

(2π)dγµ

1

/+ /kγµ

1

k2︸︷︷︸same as QED

(9.107)

=ig2

(4π)d/2Γ(2− d/2)

(k2)2−d/2 /kC2(rf )δi` (9.108)

This gives (using our renormalization condition for Σ2)

δ2 = − g2

(4π)d/2Γ(2− d/2)

(M2)2−d/2 C2(rf ) (9.109)

In this calculation we got lucky. We were able to factor our answer into a QED contri-bution and a group theory contribution. This will not happen again. We now calculateδ1. For this we use

→ igδ1γµ(T a)ij

Our renormalization condition is to set F1(q2 = M2) = 0. We need to calculate

i j

= (T bT aT b)ijΓµQED(q, ...)

we have the matrix coefficient:

T bT bT a + T b[T a, T b

]= C2(rf )T

a + T bfabcT c (9.110)


Structure constants are antisymmetric so we can write this as

T bT bT a + T b[T a, T b

]= C2(rf )T

a +1

2fabc

(T bT c + T bT c

)(9.111)

= C2(rf )Ta +

1

2fabc

(T bT c − T bT c

)(9.112)

= C2(rf )Ta +

1

2fabc

[T b, T c

](9.113)

= C2(rf )Ta +

1

2fabcf bceT e (9.114)

= C2(rf )Ta − 1

2f bacf bce︸︷︷︸

(T bAdj)ac(TbAdj)ce

T e (9.115)

= C2(rf )Ta − 1

2(T bAdjT

bAdj)aeT

e (9.116)

= C2(rf )Ta − 1

2C2(Adj)δaeT

e (9.117)

(9.118)

We now discuss the group structure of another diagram contribution:

i

a

j

This has the group theory contribution of

(T bT c)ijfabc =

1

2fabc

[T b, T c

]=

1

2fabcf bce︸︷︷︸−C2(Adj)δae

T e (9.119)

The full calculation gives

δ1 =−g2

(4π)d/2Γ(2− d/2)

(M2)2−d/2 (C2(rf ) + C2(Adj)) (9.120)

There is one more set of diagrams to calculate which is

+×

+ +


We now calculate the group factors. The fermionic loop has a group theory factor of,

(T a)ij(Tb)ji = tr

(T aT b

)= C(rf )δ

ab (9.121)

while for the gluon loop we have,

facdf bcd = C2(Ajd)δab (9.122)

A full calculation of the last counterterm gives

δ3 =g2

(4π)d/2Γ(2− d/2)

(M2)2−d/2

5

3C2(Adj)− 4

3

∑rf

C(rf )

(9.123)

We have (M2)ε = eε logM2= 1 + ε logM2 + ... (and a similar expansion for 4π). Using

these expressions we have

1

(4π)d/2Γ(2− d/2)

(M2)2−d/2

∣∣∣∣d=4−ε

=1

16π2

(1

ε+ γE − logM2 − log(4π) +O(ε)

)(9.124)

Using M ∂∂M

logM2 = 2, Our expression for the β function is

β(g) = − g3

16π2

[11

3C2(Adj)− 4

3

∑f

C(rf )

](9.125)

We now interpret this result. The first interesting thing to notice is that this is nonzero if there are no fermions at all. What’s even more interesting is that β < 0. Thisis completely different from the φ4 case and the QED case where the β function waspositive. This is interesting since we know that β controls how the coupling changes withscale that we are studying the system at.

We had the graph in QED of

g

M


The reason this worked in this way was because the β function was positive. However inthis case we will have the opposite effect. Instead we have

g

M

1

This is called an asymptotically free theorem. However at low energies this theory be-comes non-perturbative. You cannot take a zero momentum scattering (or equivalentlytake the quantum mechanics limit). QCD of course has fermions however it doesn’t haveenough fermions to change the sign of the β function. In QCD the energy that we havenon perturbative results is about 1GeV or below.

In QED we had screening:

+

- +

- +

- +-+

-+

-+

-+

In non-Abelian Yang Mills we have “anti-screening” due to gauge fields. To understandwhere this anti-screening comes from we consider a SU(2) gauge theory. We have

L = −1

4F aµνF

a,µν + gAaµja,µ + Lmatter (9.126)

where ja,µ is the current (it’s easy to show that our earlier form of the Lagrangian isequivalent to this with the appropriate conserved current) and F a

µν = ∂µAaν − ∂νA

aµ +

gεabcAbµAcν .

We consider the equations of motion for the gauge field,

δLδAaµ

= ∂νδL

δ∂νAaµ(9.127)


Recall that in QED we had Ei = F0i. We can extend this idea to a non-Abelian gaugetheories and write, Ea

i = F a0i. This is often called the “chromo-electric field”. Note this

is a bit of a misnomer in this case since we are dealing with SU(2) but people often usethis term for all non-Abelian gauge theories. The equations of motion are

∂iEai = gρa − gεabcAbiEc

i (9.128)

where ρa ≡ ja,0 is called (in analogy with QED) the charge density. This is the analogueto Gauss’s law.

Consider a point-like charge, ρa = δ3(x)δa,1. Note that we needed to specify a directionin the group space for the charge. Notice that Gauss’s law is a non-linear equation. TheQED classical solution is E1 = g

4πr2 r,E2 = E3 = 0. You can choose a gauge such that

Aa = 0 and Aa0 6= 0. If we do this then (since εaab = 0) the difficult term disappears. Sowe can recycle the QED solution.

This was a classical computation. However the β function arose from a quantumsolution (it came from loop corrections which necessarily imply the use of quantum me-chanics). These non-linear terms inevitably come in at the quantum level. In quantummechanics we can’t just set A = 0. In other words, δA 6= 0. We have

+1 +3

A2->

E1->

+1 -1

anti-screening!

Consider δA2. The sources are δE3. So we have

∇ · E3 = gδA2 · E⊥ (9.129)

So we have∇ · δE1 = −igδA2 · δE3 (9.130)

This is demonstrated above.

9.9 RG flows (evolution)So far to display our running couplings we have shown plots of g as a function of therenormalization scale. There is an alternative way to represent this.

9.9. RG FLOWS (EVOLUTION) 115

g( ) gIR free

Such theories are called IR free. These include λφ4, QED , Yang-Mills when β > 0.Alternatively we can have UV or asymptotically free theories:

g( ) g

Asymptotically Free

This includes theories such as Yang-Mills with no/little matter (β < 0). The alternative,which we didn’t see in our calculation but is possible is called “fixed-point”. This occurswhen β = 0, but λ 6= 0. You have an interacting theory but the coupling doesn’t run.For example

* *

IR-stable fixed point

The theory is “attracted” to some stable coupling. We have

β ∼ aλ+ bλ2 + ... (9.131)

Another thing that’s possible is so called Conformal Field Theories. These have β = 0for any λ. For this to be the case you require some symmetry to make it this way. Thisis also known as scale invariance.

Chapter 10

Non-perturbative results

10.1 Kallen-Lehmann RepresentationThe pole structure present in correlation functions has a very particular structure. Thiscan already by seen in the two-point function where we can make powerful non-perturbativeobservations. To this end consider the 2-point function:

〈0|φ(x)φ†(y)|0〉 (10.1)

Recall that for a free-field this is given by,

∆+(x− y;µ2) ≡ 〈φ(x)φ(y)†〉 =

∫d4p

(2π)4e−ip·(x−y)θ(p0)2πδ(p2 − µ2) (10.2)

More generally we can insert in a complete set of states and use the fact that the vacuumis translationally invariant to write:

〈φ(x)φ(y)†〉 =∑n

〈0|φ(x)|n〉〈n|φ(y)†|0〉 (10.3)

=∑n

e−ipn·(x−y) |〈0|φ(0)|n〉|2 (10.4)

where the sum here runs over all continuous and discrete indices (in the simplest case ofφ4 theory where internal states cannot carry any other indices this is just and integralover momenta of the possible internal states. More generally this can be spin, flavor, etc.of all the possible internal states.).

We can pretty this up a bit by inserting in a delta function,

〈φ(x)φ(y)†〉 =∑n

[∫d4pδ(4)(p− pn)

]e−ipn·(x−y) |〈0|φ(0)|n〉|2 (10.5)

Swapping around the sums and enforcing the delta function relation onto the pn we find,

〈φ(x)φ(y)†〉 =

∫d4pe−ip·(x−y)

∑n

δ(4)(p− pn) |〈0|φ(0)|n〉|2 (10.6)

116

10.1. KALLEN-LEHMANN REPRESENTATION 117

The sum over n is a scalar function and, in terms of the external states, can only dependon the four-vector, pµ. Furthermore, if the external state has negative energy this shouldbe equal to zero. Thus we can write,

〈φ(x)φ(y)†〉 =

∫d4pe−ip·(x−y) 1

(2π)3θ(p0)ρ(p2) (10.7)

where the added factor of (2π)3 is conventional. ρ is known as the spectral function. Sinceρ is equal to a sum over positive definite quantities, it is real and positive. Furthermore,introducing an additional delta function we can write,

〈φ(x)φ(y)†〉 =

∫dµ2ρ(µ2)

∫d4p

(2π)42πδ(p2 − µ2)e−ip·(x−y)θ(p0) (10.8)

This is exactly the free-field two-point function!Finally we have,

〈φ(x)φ(y)†〉 =

∫dµ2ρ(µ2)∆+(x− y, µ2) (10.9)

Now consider,

〈φ(y)†φ(x)〉 =∑n

e−ipn·(y−x)∣∣〈0|φ(0)†|n〉

∣∣2 (10.10)

=

∫d4pe−pn·(y−x)

∑n

δ(4)(p− pn)∣∣〈0|φ(0)†|n〉

∣∣2 (10.11)

=

∫d4p

(2π)3e−pn·(y−x)ρ(p2)θ(p0) (10.12)

where ρ is the same as ρ with φ→ φ†. We conclude that we get the same result as abovewith x↔ y and ρ→ ρ:

〈φ(y)†φ(x)〉 =

∫dµ2ρ(µ2)∆+(y − x, µ2) (10.13)

To understand the relation between ρ and ρ consider the expectation value of the com-mutator,⟨[

φ(x), φ(y)†]⟩

=

∫ ∞0

dµ2(ρ(µ2)∆+(x− y;µ2)− ρ(µ2)∆+(y − x;µ2)

)(10.14)

Now assume that x−y is space-like. In this case ∆+(z;µ2) ∝ 1√z2K1(µ

√z2) = ∆+(−z;µ2).

Furthermore, causility demands that the commutator vanishes. Hence we require,

0 =

∫ ∞0

dµ2∆+(x− y;µ2)(ρ(µ2)− ρ(µ2)

)(10.15)

This can only be satisfied if ρ = ρ. We could also have observed this directly from CPTinvariance.

118 CHAPTER 10. NON-PERTURBATIVE RESULTS

We are finally in a position to compute the 2-point function we are actually interestedin, which corresponds to a particle travelling from x to y,⟨T φ(x)φ(y)†

⟩= θ(x0 − y0)

⟨φ(x)φ(y)†

⟩+ θ(y0 − x0)

⟨φ(y)†φ(x)

⟩(10.16)

=

∫dµ2ρ(µ2)

(∆+(x− y;µ2)θ(x0 − y0) + ∆+(y − x;µ2)θ(y0 − x0)

)(10.17)

≡∫dµ2ρ(µ2)∆F (x− y;µ2) (10.18)

where we have introduced the usual Feynman propagator for a free field. In momentumspace this relation takes the form,∫

d4xe−ip·(x−y)⟨T φ(x)φ(y)†

⟩=

∫dµ2ρ(µ2)

i

p2 − µ2 + iε(10.19)

Now lets consider the case where φ is a conventionally normalized (not renormalized)field. In this case we have the equal-time commutation relations,[

∂tφ(x, t), φ†(y, t)]

= iδ(3)(x− y) (10.20)[φ(x, t), ∂tφ

†(y, t)]

= −iδ(3)(x− y) (10.21)

Taking the derivative of equation 10.13 and using the relation,

∂∆+(x− y)

∂x0

∣∣∣∣x0→y0

= iδ(3)(x− y) (10.22)

gives a unitarity condition on the spectral density,∫ ∞0

ρ(µ2)dµ2 = 1 (10.23)

We now make some general remarks,

• The full propagator must have a pole at p2 = m2. This requires ρ(µ2) to take theform,

ρ(µ2) = Zδ(µ2 −m2) + σ(µ2) (10.24)where Z is independent of µ and σ is an known function without a pole at µ2 = m2.The value of Z is the wavefunction renormalization of the field.

• Positivity of ρ requires that σ(µ2) ≥ 0. Furthermore, unitarity of σ gives thecondition:

1 = Z +

∫ ∞0

σ(µ2)dµ2 (10.25)

Since σ ≥ 0, we conclude that Z ≤ 1.

• Above we see that the propagator can’t vanish faster than 1/p2 for |p2| → ∞.Indeed adding in higher order derivatives into the Lagrangian could induce a fasterfall-off the propagator but this general argument shows that such a theory mustviolate either positivity of the states or causility.

Chapter 11

Broken Symmetries

11.1 SSB of Discrete Global Symmetries

Classical Analysis

The symmetries we have discussed so far have been exact. There are two types of sym-metry breaking. Explicit symmetry breaking is fairly trivial and less interesting. Spon-taneous symmetry breaking (SSB) plays a key role and nature and can be very subtle.

A field transformation is φ→ Λφ (where φ is a scalar or a fermion, Λ ∈ Rep(G), andG can be a Lie or discrete group. Λ is x-independent, global or a function of x and local.It is a symmetry if

L → L+ ∂µJ µ (11.1)

As an example consider a local discrete symmetry. We take the Lagrangian to be givenby

L =1

2(∂µφ)2 − V (φ) (11.2)

with V (φ) = µ2

2φ2 + λ

4φ4. This theory is invariant under φ → φ′ = −φ. This is a Z2

symmetry. This theory cannot produce an odd number of external legs. This is cleardiagrammatically:

An odd number of external legs is forbidden due to the Z2 symmetry. This symmetrycan be broken in two different ways.

119

120 CHAPTER 11. BROKEN SYMMETRIES

1. This can be done explicitly by adding the term ∆V = εφ3 to the Lagrangian. Youcan then do perturbation theory in ε. To order ε0 Z2 is a good symmetry.

2. Alternatively this can be broken spontaneously. We denoted the quadratic termas µ2

2φ2. However we don’t really know the sign of µ2. The way to tell the sign

of terms is by the requirement that the Hamiltonian ( 12(φ2 + (∇φ)2) + V (φ)) is

bounded from below. To be bounded from below we must have λ > 0. However wecan have µ2 > 0 or µ2 < 0. In either case we have

V2>0 V

2<0

v-v

The first case is what we’ve done before. Consider µ2 < 0. Classically our systemwill settle down in the lowest energy state. δV

δφ= 0 ⇒ v ≡ −µ2

λ. You don’t know

which point the system will fall to. If we lived in a universe where this was presentand the field had enough time to evolve and settle at a lowest energy state. Bymeasuring this field we could find whether out field will be in either v or −v. Soeven though the Lagrangian is symmetric about this symmetry, to an observer itwill appear the system is not symmetric. Once you specify which configuration yoursystem is in you have spontaneously broken the symmetry.

Quantization

In the normal case we quantize the field by finding the classical equations of motion:

φ+ µ2φ = −λφ3 (11.3)

We first solve the system in the case of λ = 0:

φ ∼ αke−ik·x + α∗ke

ik·x (11.4)

where k2 = µ2. We then replace αk → ak, α∗k = a†k. The vacuum is given by ak |0〉 = 0for all k. Hence we have

〈0|φ|0〉 = 0 (11.5)

It turns out that this approach doesn’t really work when µ2 < 0. The reason is verysimple. If you try to just solve this classically with µ2 < 0 then formally the solution stillworks but now k2 < 0 which means that at least something in k must be imaginary. Inparticular it means that k0 is imaginary, k0 = ik. Hence the solutions aren’t oscillatorby grow in time. i.e.. φ ∝ ekt. Physically this is because we are looking at an unstablesystem (when λ = 0) since the potential goes to negative infinity. We can’t just throw

11.1. SSB OF DISCRETE GLOBAL SYMMETRIES 121

away λ because it’s what stabilizes the potential. We cannot take it to be zero. Thelesson is that we cannot just ignore this λ in the beginning.

The obvious thing to do is instead of ignoring λ which was basically equivalent toexpansions around small perturbations around zero, what we need to do is expand aboutthe actual minimum of the theory. This is the same thing you would do in classicalmechanics. If you want to find the oscillations of a ball in such a potential well then youshould expand about the minima.

Around the minima we have (from the Euler Lagrange equations)

φ+δV

δφ= 0 (11.6)

φ0 = +v or φ0 = −v. We have

φ+δV

δφ

∣∣∣∣φ0

+δ2V

δφ2

∣∣∣∣φ0︸︷︷︸

≡m2

(φ− φ0) +1

2!

δ3V

δφ3

∣∣∣∣φ0

(φ− φ0)2 + ...︸︷︷︸interactions

= 0 (11.7)

It’s easy to show that all the interactions are proportional to λ. We define a new fieldξ(x) = φ(x)− φ0. So we have

ξ +m2ξξ = 0 (11.8)

up to O(λ). Our new field is ξ and it has solutions of

ξ ∼ αe−ik·x + α∗eik·x (11.9)

with k2 = m2ξ wherem2

ξ = −2µ2. The quantization is trivial. We do the same replacementwe had before

ξ ∼ ake−ik·x + a†ke

ik·x (11.10)

We have the vacuum given by ak |0〉 = 0 for all k and then 〈0|x|0〉 = 0. However what isthe condition based on our initial field? Recall that φ = φ0 + ξ and thus

〈0|φ|0〉 = φ0 (11.11)

This is called the vacuum expectation value (VEV) of the system. This is a huge differencein the physics!If for example the electric field (and hence Aµ) had a VEV there would bea background field everywhere which would mean if you had a charged particle it wouldbe accelerated in some direction in space.

We can now create particlesa†k |0〉 = |k〉 (11.12)

These are massive particles that obey E2 = k2 +m2ξ .


You can insert the interactions in two different ways. One way is to rewrite thepotential in terms of ξ:

V (φ) =µ2

2φ2 +

λ

4φ4

=1

2m2ξξ

2 + λ3︸︷︷︸∝λv

ξ3 + λ4︸︷︷︸λ

ξ4 (11.13)

Note that we have gained a cubic interaction out of a Lagrangian invariant under Z2.

11.2 SSB of Continuous Global Symmetry

Classical Analysis

Consider V (Φ) = µ2 |Φ|2 + λ |Φ|4 where Φ = φ1+iφ2√2

. The symmetry of this Lagrangian isΦ→ eiαΦ where α is a constant. If µ2 > 0 then the VEV is at Φ = 0. If µ2 < 0 then wehave

1

2

V

Vacuum Manifold

and δVδΦ

= δVδΦ∗

= 0 which gives |Φ|2 = −µ2

λ. Note that before we had two minima but now

we have an infinite number of vacua. All the different vacua are known as the vacuummanifold. We can get all the vacuums by applying the transformations on the VEV:

Φα = eiα1√2

√−µ2

λ= eiαΦ0 (11.14)

This is a very general property of spontaneous breaking. We have spontaneous breakingof U(1).

Quantization

Our equation of motion are

φ1 +δV

δφ1

= 0 (11.15)

φ2 +δV

δφ2

= 0 (11.16)

11.2. SSB OF CONTINUOUS GLOBAL SYMMETRY 123

Due to the symmetry the physical results will not depend on the vacuum we choose. Wechoose to expand about φ1 = v and φ2 = 0. The equations of motion give

φ1 +

δV

δφ1

∣∣∣∣(φ1,φ2)=(v,0)

+δ2V

δφ21

∣∣∣∣(v,0)

(φ1 −v√2

) +δ2V

δφ1δφ2

∣∣∣∣(v,0)

+

(interaction

terms)

= 0 (11.17)

ξ +δ2V

δφ21

∣∣∣∣(v,0)

ξ +δ2V

δφ1δφ2

∣∣∣∣(v,0)

φ2 = 0 (11.18)

in terms of ξ ≡ φ1 − v√2. We also have a second equation:

φ2 +δ2V

δφ2δφ1

∣∣∣∣(v,0)

ξ +δ2V

δφ22

∣∣∣∣(v,0)

φ2 = 0 (11.19)

We define a mass matrix: m2ij ≡ δ2V

δφiδφj

∣∣∣∣(v,0)

. We can put our equations together:

(ξφ2

)+M

(ξφ2

)= 0 (11.20)

For our example we have V = µ2

2(φ2

1 + φ22) + λ

4(φ2

1 + φ22)

2. We can take the derivatives:

δ2V

δφ21

= µ2 + 3λv2 = −2µ2 (11.21)

δ2V

δφ1δφ2

= 2λφ1φ2(φ1,φ2)=(v,0)−−−−−−−→ 0 (11.22)

Lastly

δ2V

δφ22

= µ2 +λ

2+ λφ2

1 + 3λφ22

(φ1,φ2)=(v,0)−−−−−−−→ 0 (11.23)

Hence not only is the mass matrix diagonal but one of the eigenvalues is zero. In thelanguage of quantum field theory the δ2V

φibecome masses. We have two particles with

masses m2ξ = −2µ2 and mφ2 = 0. The massless particles are known as Goldstone bosons.

This mass also turns out to be included in the quantum corrections.The existence of the particle is easy to understand classically. Consider a mechanical

system at the point near (v, 0) (see diagram of potential above). It’s clear that there isone direction along φ1 that there is a restoring force. However we can also move alongφ2 along the vacuum manifold and not change the energy and hence there is no restoringforce. In the language of oscillator that means you have an oscillator with zero frequency.The fact that the potential is flat was a consequence of the theory. This just follows fromthe symmetry of the theory. Since the symmetry holds to all orders quantum correctionsdo not change the mass of the Goldstone boson.


11.3 Linear Sigma ModelNow that we know how this works in the U(1) case we will generalize these ideas in what’sknown as the linear sigma model. We start with N real fields, φi:

L =N∑i=1

1

2(∂µφi)

2 + V (φi) (11.24)

We consider a potential that is invariant under rotation of the fields about themselves:φi = Ri,jφi. Where R is orthogonal: RT = R−1. The only way this can occur is if thepotential only depends on the magnitude of the vector

φ =

φ1

φ2...φN

We take

V (φ) = µ2φ2 + λφ4 (11.25)

= µ2∑i

φ2i + λ

(∑i

φ2i

)2

(11.26)

As before we take µ2 < 0. The vacuum manifold is given by φ2 = −µ2

λ. The manifold is

one dimension lower then the number of degrees of freedom in the system SN−1. We picka vacuum

φ =

0...0v

We define σ(x) = φN(x)− v and we get

V (φ) =1

2

(−2µ2

)σ2 +

(cubic andquarticterms

)(11.27)

The key point is that we don’t have any mass terms except for σ so we have N − 1Goldstone bosons. We could have predicted how many Goldstone bosons will appear byjust thinking about the symmetry of the theory. We can express the symmetry of theLagrangian as R = eiαiTi where Ti are the generators of O(N). The dimensionality of theO(N) group is N(N−1)

2. This comes from the condition that Ti ’s need to be antisymmetric.

Consider

Rφ0 = (1 + iαiTi + ...)φ0 (11.28)

11.4. MORE ON THE MEXICAN HAT POTENTIAL 125

We have two possibilities,

Tiφ0 = 0 Unbroken gen.Tiφ0 6= 0 Broken gen.

The Ti matrices take the form

Ti =

0

. . .. . . 1−1 0

(11.29)

The number of broken generators is equal to the number of Goldstone bosons. This iscalled Goldstone theorem. We did two examples but you can do any group you want.In nature we don’t think anything like this is happening exactly. But in QCD thereare a lot of combinations of quarks. The pions for example are much lighter then theother particles. You can discuss the physics of pions as Goldstone bosons. You canadd explicit breaking to give them some mass to correct that they are not massless. Inmany beyond the SM theories people speculate that different particles are approximateGoldstone bosons.

11.4 More on the Mexican Hat Potential

We now go back to the original Mexican Hat potential, V (Φ) = µ2 |Φ|2 + λ |Φ|4, withµ2 = −λv2 < 0 and Φ ≡ φ1+iφ2√

2. If this were a mechanics problem then it would be clear

that it is more convenient to use fields that obey the symmetry of the problem. In thiscase this would be a degree of freedom that represents the angle. There is nothing thatstops you from doing this in Quantum Field Theory.

With this in mind we introduce a non-linear field redefinition

Φ = eiπ(x)/v 1√2

(v + σ(x)) (11.30)

Our potential becomes

V (Φ) =µ2

2(v + σ)2 +

λ

4(v + σ)4

=1

2m2σσ

2 + λ3σ3 + λ4σ

4 (11.31)

where we drop the unimportant constant term and the linear term disappeared since


δVδφ

= 0. The kinetic term is given by

|∂µΦ|2 =1

2

∣∣∣i∂µπ (1 +σ

v

)+ ∂µσ

∣∣∣2=

1

2(∂µσ)2 +

1

2(∂µπ)2︸︷︷︸

can. normalization

+1

v(∂µπ)2σ +

1

2v2(∂µπ)2 σ2︸︷︷︸

interactions

Notice that the interaction terms have couplings with dimensions of inverse mass!Theseinteractions appear non-renormalizable. However that is very naive and power counting isnot sufficient to test Renormalizability. This theory is still renormalizable (it is completelyequivalent to the original φ4 theory).

Further notice that π only enters the Lagrangian in terms of it’s derivative. Our U(1)symmetry acts on π and σ in a very simple way:

π → π + α “shift” (11.32)σ → σ (11.33)

To have an invariant Lagrangian we can’t have any dependence on π, since π is notinvariant. For example if we had a aπ2 term we could apply a shift and get a(π + α)2

which differs from the first Lagrangian. Hence we can only have (as we see above),L [σ, ∂µπ].

In the original theory renormalizability was clear and here the symmetry is clear.Further we have

〈Ω|T (π(p)...)|Ω〉 ∝ pµ · (...) (11.34)

11.5 SSB of Gauge Symmetries (“The Higg’s Mecha-nism”)

Abelian Case

Today we will consider the Abelian example of SSB. Later we will the more intricate SMSU(2) × U(1) → U(1) mechanism. We have the same potential as before only now weconsider a gauge symmetry:

L = |Dµφ|2 − V (Φ)− 1

4FµνF

µν (11.35)

where V (Φ) = −µ2 |Φ|2 + λ |Φ|4 (we make the minus sign on µ2 explicit such that nowµ2 > 0) and Dµ = ∂µ + ieAµ. Our VEV is

v2 =µ2

λ(11.36)

and our vacuum is Φα = eiα v√2and Aµ = 0.

11.5. SSB OF GAUGE SYMMETRIES (“THE HIGG’S MECHANISM”) 127

We’ll start by solving this problem heuristically and follow up with a more rigorousderivation. We fix Φ to it’s vacuum value, Φ0 = v√

2. This gives

LΦ kin = |DµΦ|2 =

∣∣∣∣ieAµ v√2

∣∣∣∣2 =e2v2

2AµA

µ︸︷︷︸mass term

(11.37)

We got a mass term for the gauge field,mA = ev. When we first began talking about QEDwe rejected these terms because they were not gauge invariant (Aµ → Aµ − 1

e∂µα). We

started with a theory that’s completely gauge invariant by shifting the Φ field, Φ→ eiα(x)Φspontaneously broke gauge invariance. Picking the vacuum.

We now attack this problem more carefully. We begin by shifting our field:

Φ = eiπ/v1√2

(v + σ(x)) (11.38)

This shift gives

|Dµφ|2 =1

2

∣∣∣∂µ + π(

1 +σ

v

)+ ∂µσ + ieAµ(v + σ)

∣∣∣2=

1

2(∂µσ)2 +

1

2(∂µπ + evAµ)2

(1 +

σ

v

)2

(11.39)

This is the same kinetic term as before except for the extra gauge field terms:

ev∂µπAµ +

1

2e2v2AµAµ (11.40)

Note that 11.39 is still manifestly gauge invariant since σ → σ, π → π + α and Aµ →Aµ − 1

e∂α.

The second mass term we knew would be there based on our earlier heuristic argu-ments. However there is a new interesting term, ev∂µπAµ. This term is very strange.This leads to propagators starting as one field and turning into another. This means thatwe are probably not using a good basis. There is probably more then one way to solvethis problem and understand what the physics is. We will use the standard solution. Theidea is instead of trying to redefine fields, we have another approach. Recall that in QEDLgauge fix = − 1

2ξ(∂µA

µ)2. This allowed us to compute a propagator. This is a remnantof the choice we made that said that G [A] = ∂µA

µ = 0. This was a good choice but itwas not unique.

Having said that, with theories with spontaneously broken symmetries we use what’sknown as Rξ gauge. Our gauge fixing Lagrangian is

Lgauge fix = − 1

2ξ(∂µA

µ − ivξπ)2

= − 1

2ξ(∂µA

µ)2 + evπ(∂µAµ)− e2v2ξ

2π2

= − 1

2ξ(∂µA

µ)2 − ev(∂µπ)Aµ − e2v2ξ

2π2 (11.41)


where we have integrated by parts. The middle term cancels the term we didn’t likeev∂µπA

µ in equation 11.40. However there is a new strange mass term for the π field.Based on physical arguments we there should not be any mass for π.

We have three propagators:

µg k→ggν=−i

k2 −m2A

(gµν −

kµkνk2 − ξmA

(1− ξ))

πh p→hh =1

p2 −m2A

σh p→hh =1

p2 −m2σ

, m2σ = 2µ2

Note that we have two poles in the gauge boson propagator. One makes sense since itcorresponds to that we have a massive gauge boson. However we have a second gaugedependent pole. This gauge dependent pole is cancelled by the π field. The propagatorsalways come together:

d eAge d+d eπhe d→ gauge independent

π cannot be real fields. We cannot have diagrams with external π fields.To understand our system better we introduce Unitary gauge, ξ → ∞. In this limit

π drops out of the theory completely. The photon propagator becomes

µg k→ggν→ −ik2 −m2

A

(gµν −

kµkνm2A

)(11.42)

To understand the particle we consider the case of the particle on shell:

µg k→ggν −→ k2 → m2A

−ik2 −m2

A

3∑s=1

εsµ(p)ε∗sν (p) (11.43)

We started with the theory of the photon that has two degrees of freedom and acomplex scalar field with two degrees of freedom. The overall number of degrees offreedom was 2 + 2 = 4. After the symmetry is broken the picture changes, the gaugeboson gets a mass and has three degrees of freedom while the σ is a real scalar field andhas 1 degree of freedom. So we still have 3 + 1 = 4 degrees of freedom. The gauge boson“ate” the π field and became massive.

Non-Abelian Case

Consider a scalar field, Φ, which is charged under some non-Abelian gauge symmetry, G.In other words that

Φ→ Φ′ = ΛΦ (11.44)


where Λ†Λ = 1 and Λ = eiαa(x)Ta (a = 1...Ng) with Ng = dim (ajd(G)). Our Lagrangianbecomes

L = −1

4F aµνF

a,µν + |DµΦ|2 − V (Φ) (11.45)

where Dµ = ∂µ − igAaµT a. The vacuum state (Φ0) is where

δV

δΦi

∣∣∣∣Φ0

= 0 (11.46)

If Φ0 is a solution then Φα = eiαaTaΦ0 is also a solution. There are two possibilities. Either

we have T bΦ0 6= 0 or T aΦ0 = 0. The first case corresponds to the broken generators whilethe second corresponds to unbroken generators. We Nb such that b = 1, ..., Nb then thevacuum manifold is an Nb-dimensional space. Each broken generator gives a Goldstoneboson.

Just as we did last time parametrize our field non-linearly as

Φ = eiΠbT b/vΦ0 [(σ − fields)] (11.47)

The gauge transformation gives πb(x) → πb + αb(x), and σ → σ. Expanding about thevacuum, Φ0 gives a term

|DµΦ|2 → g2

Nb∑a,b=1

AaµAb,µ(T aΦ0)†(T bΦ0) (11.48)

Note that the sum is over broken generators since the unbroken ones gives 0. We nowexamine this term more closely. Claim: T aΦ0 (where a runs over unbroken generators)

are eigenvectors of the mass matrixM2i,j ≡ δ2V

δΦiδΦj

∣∣∣∣Φ0

with zero eigenvalue. Proof: Suppose

that Φ→ eiαaTaΦ is a symmetry. Then if we consider infinitesimal α such that

Φ→ Φ + δΦ

δΦi = αa (T aΦ)

Since V must be invariant under this transformation (by assumption) then

δV =δV

δΦi

δΦi = 0

which implies that

αa (T aΦ)iδV

δΦi

= 0

Taking the derivative of both sides gives:

αaδ

δΦj

(T aΦ)iδV

δΦi

+ αa (T aΦ)iδ2V

δΦiδΦj

= 0


Plugging in Φ = Φ0δVδΦi

∣∣∣∣Φ0

gives

(T aΦ0)iδ2V

δΦiδΦj

∣∣∣∣Φ0

(11.49)

as required.The Mass matrix,M i,j is Hermitian, so eigenvectors can be chosen to be orthonormal:

(T aΦ0)†(T bΦ0

)∝ v2δab (11.50)

which implies that g2v2∑

bAbµA

a,µ is a mass term for the gauge bosons. The number ofmassive gauge bosons is equal to the number of broken generators. We don’t have anyGoldstone bosons left (all have been “eaten”). The masses are m2 ∝ |T aΦ0|2.

As an example we consider SU(2). We define Φ is a fundamental “spin-1/2” rep-

resentation, Φ =

(Φ1

Φ2

). The potential is given by V = −µ2Φ†Φ + λ

(Φ†Φ

)2 where

Φ†Φ = |Φ1|2 + |Φ2|2. The Lagrangian is invariant under Φ → ΛΦ where Λα = e−iα·σ/2.The vacuum is given by

Φα = Λα

(0

v/√

2

)(11.51)

where v2 = µ2

λ.

We haveσaΦα 6= 0 (11.52)

for a = 1, 2, 3 so we have three broken generators and hence 3 Goldstone bosons. Ourthree gauge bosons are ∣∣∣∣σa2

(0

v/√

2

)∣∣∣∣2 =v2

8(11.53)

for all a = 1, 2, 3 which implies that m2Ag2v2

4.

This is very similar to the situation in the Standard Model, except in that case wehave an extra U(1) symmetry. In the Standard Model we see weak interactions whichgive masses of mW ∼ 80GeV,mZ = 90GeV and only left handed fermions feel the weakinteraction. The Lagrangian is given by

L = iψ†L(∂µ − igW a

µ ta)σµψL + iψ†R∂µσ

µψR (11.54)

where ψL =

(ψL,1ψL,2

).

The naive guess for the mass terms (the Dirac mass terms) is m(ψ†L,aψR + ψ†RψL,a

).

However this is not gauge invariant!We need the Higg’s mechanism to fix this problem.

We have H =

(H∗

H0

). The vacuum is given by

〈H0〉 =v√2

(11.55)


which implies that mW = gv√2. The Yukawa couplings are given by

LY uk = λ(ψ†LHψR + ψ†RH

†ψL

)=

λv√2︸︷︷︸

mF

(ψ†L,1ψR + ψ†RψL,1 + ...

)(11.56)

so but adding the Yukawa couplings to the Higgs field and give it a VEV we immediatelyget fermion masses. Note that to derive this formula we require the transformation

H = eiπata/v

(0

v√2

+ σ

)where σ is the physical Higg’s boson.

Bibliography

[Griffith(2008)] D.J. Griffith. Introduction to Elementary Particles. Wiley-VCH, 2ndedition, 2008.

[Peskin and Schroeder(1995)] M. Peskin and D. Schroeder. An Introduction to QuantumField Theory. West View Press, 1995.

132

Documents

QUANTUM FIELD THEORY II (PHYS7652) LECTURE NOTESpages.physics.cornell.edu/~ajd268/Notes/QFT2-Notes.pdf · 1 QUANTUM FIELD THEORY II (PHYS7652) LECTURE NOTES Lecture notes based on