# (Approach to) Feynman Rules for Non-Abelian Gauge Theory

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## (Approach to) Feynman Rules for Non-Abelian Gauge Theory

06.06.2012

### 1 Towards the Path Integral

Consider the Green’s function of a zero dimensional field theory.

GN= R

R xN

N!e12ax2+P(x)dx R

R

e12ax2+P(x)dx

P(x) is a polynomial describing the interaction. Cut through the polynomials with a generating function:

Z[J]=

X

N=0

JN N!

Z

R

xNe12ax2+P(x)dx=ZX

N

JNGN

We can labelGN with 1PI graphs:

GN =X

Γ

λ#V(Γ)JEE(Γ)

#Aut(Γ)a#EI(Γ)

where EE(Γ) denotes the external edges of graph Γ, EI the internal edges, V the vertices, and

#(·) generally denotes the number of (argument).

The whole idea of the path integral is to mimick this structure, but to appoint xto a field, as a function of spacetime points. The measure is transformed:

Z

dx → Z

Dφ φ∈C (RD→R) The Lagrangian density of a field theory, in general, is

L(φ)= 1

2(∂φ)2−1

2m2φ2−P(φ)

whereP(φ) denotes the interacting polynomial, as above. The action is defined as the spacetime integral over the Lagrangian density,

SL(φ)=Z

RD

L(φ)dDx

Quantum physics comes from a probability amplitude, which is somewhat associated with the action, which is a function of the fields. The amplitude is given by exp

iSL(φ)

~

. To compute a vacuum expectation value of an observableO, we write

hO(φ)i=

R O(φ)eiSL(φ)Dφ R eiSL(φ)

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The vacuum expectation value is normalized to one, by definition. Up until this point, we have not actually defined what the new measureDactually is.

The Green’s function depends on the Lagrangian:

GLN(x1,...,xN)=

R φ(x1)...φ(xN)eiSL(φ)Dφ R eiSLL(φ)Dφ This can be considered a formal definition of the Green’s function.

In principal, the (original) generating function is a Taylor expansion.

Z

eiSL(φ)+hJ,φi

with the pairing

hJ,φi=Z

RD

J(x)φ(x)dx

If we Wick rotate, the action acquires a relative sign change before the second and third term in the integrand,

S(φ)=Z

RD

1

2(∂φ)2+1

2m2φ2+P(φ)

! dDx

We can derive the Feynman rules from this expression. As a matter of fact, we could already have derived them from GN: For each line, we get a propagator, which is the inverse of the coefficient of the quadratic term, and we get aλfor each vertex).

Continue with the generating function:

Z[J]=Z exp









− Z

RD

1

2(∂φ)2+1

2m2φ2+J(x)φ(x)

! dDx









 Dφ

Z[J]

Z[0] =

X

N=0

J(x1)...J(xN)GN(x1,...,xN) dDx1...dDxN We derive the propagator from the Green’s function via Fourier transform:

(∂φ)2+m2φ2→(p2−m22

### 1.1 An Abelian Gauge Theory: SU (2)

We know that the interaction Lagrangian inSU(2) is given by L=−1

4FaµνFµνa

Here,µandνare the Lorentz indices, which are contracted when they appear up and down. a, on the other hand, is a group index, it is contracted whenever it appears twice. Write the field tensor as

Fa =∂µAa−∂νAa+gabcAbAc

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As far as the kinetic part is concerned, nothing has changed.

1

2Aaµ gµν−∂µν Aaν

but the operator (gµν−∂µν) is actually not a good one to invert, because it is really a projector.

For the transversal metric, we know

gµνtr gνρtr =gµρtr

So we can write for the generating function:

Z[J]=Z

DA1µDA2µDA3µ exp (

i Z

dDx

L+JµaAµa )

The Lagrangian is gauge invariant under local phase transformation. But the measure might not be? We find a symmetry:

### SU(2) fibre

Figure 1: TheSU(2) fibre, with group action etc., over a manifoldM, with a connection field between theSU(2) fibre and a fibre in the neighborhood.

Let us try and isolate the volume of the gauge group.

W=Z

dxdy eiS(x,y) =Z

d~r eiS(~r) , ~r=(r,θ) S(~r)≡S(~rϕ) , ~rϕ=(r,θ+ϕ)

⇒ W=2πZ

dr reiS(r)

2πis the volume of the gauge group. Introduce: 1=R

dϕ δ(θ−ϕ). RewriteW:

⇒ W=Z dϕZ

d~r eiS(~r)δ(θ−ϕ)

| {z }

CWϕ

=Z

dϕWϕ

Wϕ=Wϕ0 ⇒ W=2πWϕ

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θ=ϕ δ(θ−ϕ)

Figure 2: If a functiong(~r) vanishes, theng(~rϕ) vanishes as well.

Write down the identity:

1= ∆g(~r) Z

dϕ δ g(~rφ)

g(~r)= ∂g(~r)

∂θ g=0 Z

dϕ δ

g(~rϕ+ϕ0

|{z}

Cϕ00

)

=Z

00δ

g(~rϕ00)

Obviously, the functiong(~rϕ) is invariant under rotation.

LetAaµbe connection fields. They transform under gauge transformation:

Aaµ → Aaµθ Aaµθτa

2 =U(θ)

"

Aaµτa 2 +1

gU−1(θ)∂µU(θ)

#

U−1(θ) U(θ)=exp

(

−i θaτa 2

!)

Here,τaare the generators of the Lie group. For the connection fields, we choose a condition fa(A1µ,A2µ,A3µ)=0

to act as constraints. As a consequence,gcan be any line form, not just straight lines, as long as there are no double values. Every radius can only be hit once.

Assume that we have a constraint faand that we can invert it. In the Lie algebra, Taylor expand the transformation to get a trasformation law for infinitesimal anglesθ:

U(θ)=1+iθaτa

2 +O(θ2) with the measure

[dθ]=

3

Y

a=1

a=Dθ dθ=dθ0

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From 1= ∆g(~r)R dϕ δ

g(~rφ)

, it follows that 1= ∆f(Aaµ)

Z δ

fa(A1µ,A2µ,A3µ) Dθ

f(Aaµ)=detMf

(Mf)ab = δfa δθb

Aaµθ=Aaµ+abcθbAcµ−1 g∂µθa

⇒ fa

Aaµθ(x)

= fa

Aµ(x) +

Z d4y

Mf(x,y)

abθb(y)+...

⇒ Zf[J]= Z

D Y3

a=1

Aaµ(detMf)δ fa(A)µ

ei

Rd4x

L(x)+JµaAaµ

In the second line, we have introduced the Faddeev-Popov determinant. In the next step, we will exponentiate the determinant and the δ function in order to write everything in just one exponential function, so to speak as one single Lagrangian density.

1.1.1 Exponentiating the DeterminantdetMf

At first, we rewrite the determinant in the following way:

detMf =expn Tr

ln (Mf)o

And we expand Mf =1+L. Then we can express the logarithm as a series in terms ofL:

expn Tr

ln (Mf)o

→exp (

TrL+1

2TrL2+· · ·+1 nTrLn

)

→exp (Z

d4xLaa(x,x)+1 2

Z

d4xd4yLab(x,y)Lba(y,x)+...

)

detMf ∼ Z Y

a

DcaY

b

Dcbexp







 i

Z

d4xd4yX

a,b

caMf(x,y)cb(y)









The c and c are called Grassmann variables. They only show up in loops, not externally, because they were introduced for the sole reason to exponentiate detMf.

1.1.2 Exponentiating theδFunctionδ

fa(Aaµ) δ

fa(A~µ)

connesponds to the constraint fa(A~µ)=0. For the derivation, assume that fa,0, but some arbitrary field B.~

fa(A~µ)=Ba(x) Z

Dθ∆f(A~µ

fa(A~µ)−Ba(x)

=1 Z[J]=

Z

DA~µDB~(detMf)δ(fa−Ba) exp (

i Z

d4x L+J~µA~µ− 1 2ξ~B·~B

!)

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Formally, this is a Gaussian, but with the constraint fa

=! Ba.

With these two functions exponentiated, the Lagrangian density acquires two more terms: one for gauge fixing, and one for the Faddeev Popov ghost particle.

L → L+Lg f +LFPGh Lg f =− 1

2ξ Z

d4x fa(A~µ)2 LFPGh=Z

d4xd4y X

a,b

ca(x)Mf(x,y)abcb(y) By choosing a gauge, we fix how the determinant has to be computed.

### 2 Remark on Feynman Rules

From foregoing discussions, we know that there must certainly be Feynman rules for

, , , ,

The three-gluon vertex,apµ

p a µ

p a µ 1

1 1

2

3 2 2

3 3

, should involve terms inL(the kinetic Lagrangian with- out subscript):

• 1,2,3: indeces ofSU(2)

• pi: external momenta

• µi: Lorentz indices

The three-gluon vertex must be completely symmetric unter interchange of indices. Since the cubic interaction has one derivative, its Fourier transformation must be linear in the momenta pi. Furthermore, it should transform covariantly. It must also carry the three group indices. The kinetic term would be

−1

4FaµνFµνa with Faµν=∂µAaν−∂νAaµ+ fabcAbµAcν

where the SU(2) structure constant fabc, is actually abc, the total antisymmetric tensor. In general, the fa1a2a3 is the group structure constant.

Moreover, the vertex should depend on

(p1−p2)µ3gµ1µ2 and cyclic.

To sum up,

fa1a2a3n

(p1−p2)µ3gµ1µ2+(p2−p3)µ1gµ2µ3+(p3−p1)µ2gµ3µ1o is basically the only candidate one could write down for the three-gluon vertex.

We have written down the Feynmal rule for the three-gluon vertex just by pure thought. A similar argument is possible for the four-gluon vertex, but then, one would eventually run into the problem of radiative corrections.

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