## (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.

G_{N}=
R

R
x^{N}

N!e^{−}^{1}^{2}^{ax}^{2}^{+P(x)}dx
R

R

e^{−}^{1}^{2}^{ax}^{2}^{+P(x)}dx

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

Z[J]=

∞

X

N=0

J^{N}
N!

Z

R

x^{N}e^{1}^{2}^{ax}^{2}^{+P(x)}dx=ZX

N

J^{N}G_{N}

We can labelGN with 1PI graphs:

GN =X

Γ

λ^{#V(Γ)}J^{E}^{E}^{(Γ)}

#Aut(Γ)a^{#E}^{I}^{(Γ)}

where E_{E}(Γ) denotes the external edges of graph Γ, E_{I} 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^{∞} (R^{D}→R)
The Lagrangian density of a field theory, in general, is

L(φ)= 1

2(∂φ)^{2}−1

2m^{2}φ^{2}−P(φ)

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

S_{L}(φ)=Z

R^{D}

L(φ)d^{D}x

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

i^{S}^{L}^{(φ)}

~

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

hO(φ)i=

R O(φ)e^{iS}^{L}^{(φ)}Dφ
R e^{iS}^{L}^{(φ)}Dφ

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:

G^{L}_{N}(x_{1},...,x_{N})=

R φ(x_{1})...φ(x_{N})e^{iS}^{L}^{(φ)}Dφ
R e^{iS}^{L}^{L(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

e^{iS}^{L}^{(φ)+}^{hJ,φi}Dφ

with the pairing

hJ,φi=Z

R^{D}

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

R^{D}

1

2(∂φ)^{2}+1

2m^{2}φ^{2}+P(φ)

!
d^{D}x

We can derive the Feynman rules from this expression. As a matter of fact, we could already
have derived them from G_{N}: 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

R^{D}

1

2(∂φ)^{2}+1

2m^{2}φ^{2}+J(x)φ(x)

!
d^{D}x

Dφ

Z[J]

Z[0] =

∞

X

N=0

J(x_{1})...J(x_{N})G_{N}(x_{1},...,x_{N}) d^{D}x_{1}...d^{D}x_{N}
We derive the propagator from the Green’s function via Fourier transform:

(∂φ)^{2}+m^{2}φ^{2}→(p^{2}−m^{2})φ^{2}

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

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

4F^{a}_{µν}F^{µνa}

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

F^{a} =∂_{µ}A^{a}−∂νA^{a}+g^{abc}A^{b}A^{c}

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

1

2A^{a}_{µ} g^{µν}−∂^{µ}∂^{ν}
A^{a}_{ν}

but the operator (g^{µν}−∂^{µ}∂^{ν}) is actually not a good one to invert, because it is really a projector.

For the transversal metric, we know

g^{µν}_{tr} g^{νρ}_{tr} =g^{µρ}_{tr}

So we can write for the generating function:

Z[J]=Z

DA^{1}_{µ}DA^{2}_{µ}DA^{3}_{µ} exp
(

i Z

d^{D}x

L+J_{µ}^{a}A^{µa}
)

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

### Μ

### 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 e^{iS(x,y)} =Z

d~r e^{iS(~}^{r)} , ~r=(r,θ)
S(~r)≡S(~r_{ϕ}) , ~r_{ϕ}=(r,θ+ϕ)

⇒ W=2πZ

dr re^{iS(r)}

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

dϕ δ(θ−ϕ). RewriteW:

⇒ W=Z dϕZ

d~r e^{iS(~}^{r)}δ(θ−ϕ)

| {z }

CWϕ

=Z

dϕWϕ

W_{ϕ}=W_{ϕ}^{0} ⇒ W=2πW_{ϕ}

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+ϕ^{0}

|{z}

Cϕ00

)

=Z

dϕ^{00}δ

g(~r_{ϕ}^{00})

Obviously, the functiong(~r_{ϕ}) is invariant under rotation.

LetA^{a}_{µ}be connection fields. They transform under gauge transformation:

A^{a}_{µ} → A^{a}_{µ}^{θ}
A^{a}_{µ}^{θ}τ^{a}

2 =U(θ)

"

A^{a}_{µ}τ^{a}
2 +1

gU^{−1}(θ)∂µU(θ)

#

U^{−1}(θ)
U(θ)=exp

(

−i θ^{a}τ^{a}
2

!)

Here,τ^{a}are the generators of the Lie group. For the connection fields, we choose a condition
fa(A^{1}_{µ},A^{2}_{µ},A^{3}_{µ})=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 f_{a}and that we can invert it. In the Lie algebra, Taylor expand
the transformation to get a trasformation law for infinitesimal anglesθ:

U(θ)=1+iθ^{a}τ^{a}

2 +O(θ^{2})
with the measure

[dθ]=

3

Y

a=1

dθ_{a}=Dθ
dθ=dθ^{0}

From 1= ∆g(~r)R dϕ δ

g(~rφ)

, it follows that
1= ∆f(A^{a}_{µ})

Z δ

f_{a}(A^{1}_{µ},A^{2}_{µ},A^{3}_{µ})
Dθ

∆f(A^{a}_{µ})=detMf

(M_{f})_{ab} = δf_{a}
δθ_{b}

A^{a}_{µ}^{θ}=A^{a}_{µ}+^{abc}θ^{b}A^{c}_{µ}−1
g∂_{µ}θ^{a}

⇒ fa

A^{a}_{µ}^{θ}(x)

= fa

A_{µ}(x)
+

Z
d^{4}y

M_{f}(x,y)

abθ_{b}(y)+...

⇒ Zf[J]= Z

D Y3

a=1

A^{a}_{µ}(detMf)δ
fa(A^{)}_{µ}

e^{i}

Rd^{4}x

L(x)+J^{µa}A^{a}_{µ}

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 DeterminantdetM_{f}

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

detM_{f} =expn
Tr

ln (M_{f})o

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

expn Tr

ln (Mf)o

→exp (

TrL+1

2TrL^{2}+· · ·+1
nTrL^{n}

)

→exp (Z

d^{4}xL_{aa}(x,x)+1
2

Z

d^{4}xd^{4}yL_{ab}(x,y)L_{ba}(y,x)+...

)

detM_{f} ∼
Z Y

a

Dc^{a}Y

b

Dc^{b}^{†}exp

i

Z

d^{4}xd^{4}yX

a,b

c^{†}_{a}M_{f}(x,y)c_{b}(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 detM_{f}.

1.1.2 Exponentiating theδFunctionδ

f_{a}(A^{a}_{µ})
δ

f_{a}(A~_{µ})

connesponds to the constraint f_{a}(A~_{µ})=0. For the derivation, assume that f_{a},0, but
some arbitrary field B.~

f_{a}(A~_{µ})=B_{a}(x)
Z

Dθ∆f(A~_{µ})δ

f_{a}(A~_{µ})−B_{a}(x)

=1 Z[J]=

Z

DA~_{µ}DB~(detMf)δ(fa−Ba) exp
(

i Z

d^{4}x L+J~_{µ}A~^{µ}− 1
2ξ~B·~B

!)

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+L_{g f} +L_{FPGh}
L_{g f} =− 1

2ξ Z

d^{4}x f_{a}(A~_{µ})^{2}
L_{FPGh}=Z

d^{4}xd^{4}y X

a,b

c^{†}_{a}(x)M_{f}(x,y)_{ab}c_{b}(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,a^{p}µ

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)

• p_{i}: external momenta

• µ_{i}: 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
p_{i}. Furthermore, it should transform covariantly. It must also carry the three group indices. The
kinetic term would be

−1

4F^{a}_{µν}F^{µνa} with F^{a}_{µν}=∂_{µ}A^{a}_{ν}−∂_{ν}A^{a}_{µ}+ fabcA^{b}_{µ}A^{c}_{ν}

where the SU(2) structure constant f_{abc}, is actually _{abc}, the total antisymmetric tensor. In
general, the fa_{1}a_{2}a_{3} is the group structure constant.

Moreover, the vertex should depend on

(p_{1}−p2)_{µ}_{3}g_{µ}_{1}_{µ}_{2} and cyclic.

To sum up,

f_{a}_{1}_{a}_{2}_{a}_{3}n

(p_{1}−p_{2})_{µ}_{3}g_{µ}_{1}_{µ}_{2}+(p_{2}−p_{3})_{µ}_{1}g_{µ}_{2}_{µ}_{3}+(p_{3}−p_{1})_{µ}_{2}g_{µ}_{3}_{µ}_{1}o
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.