III Advanced Quantum Field Theory - Non-abelian gauge theory

6Non-abelian gauge theory

III Advanced Quantum Field Theory

6.4 Faddeev–Popov ghosts

To understand how to do this, we will first consider a particular finite-dimensional

example. Suppose we have a field (

x, y

) :

{pt} → R

on a zero-dimensional

universe and an action

[

x, y

]. For simplicity, we will often ignore the existence

of the origin in R

The partition function we are interested in is

dx dy e

−S[x,y]

Suppose the action is rotationally invariant. Then we can write the integral as

dx dy e

−S[x,y]

2π

dθ

∞

r dr e

−S[r]

= 2π

∞

r dr e

−S[r]

We can try to formulate this result in more abstract terms. Our space

fields has an action of the group

(2) by rotation. The quotient/orbit space of

this action is

\ {0}

SO(2)

∼

Then what we have done is that we have replaced the integral over the whole of

, namely the (

x, y

) integral, with an integral over the orbit space

, namely

the r integral. There are two particularly important things to notice:

–

The measure on

is not the “obvious” measure d

. In general, we have

to do some work to figure out what the correct measure is.

–

We had a factor of 2

sticking out at the front, which corresponds to the

“volume” of the group SO(2).

In general, how do we figure out what the quotient space

is, and how do we

find the right measure to integrate against? The idea, as we have always done,

is to “pick a gauge”.

We do so via a gauge fixing function. We specify a function

→ R

, and

then our gauge condition will be

(

) = 0. In other words, the “space of gauge

orbits” will be

C = {x ∈ R

: f(x) = 0}

f(x) = 0

For this to work out well, we need the following two conditions:

(i) For each x ∈ R

, there exists some R ∈ SO(2) such that f(Rx) = 0.

(ii) f

is non-degenerate. Technically, we require that for any

such that

f(x) = 0, we have

∆

(x) =

∂

∂θ

f(R

(x))



θ=0

6= 0,

where R

is rotation by θ.

The first condition is an obvious requirement — our function

does pick out a

representative for each gauge orbit. The second condition is technical, but also

crucial. We want the curve to pick out a unique point in each gauge orbit. This

prevents a gauge orbit from looking like

where the dashed circle intersects the curved line three times. For any curve that

looks like this, we would have a vanishing ∆

(

) at the turning points of the

curve. The non-degeneracy condition forces the curve to always move radially

outwards, so that we pick out a good gauge representative.

It is important to note that the non-degeneracy condition in general does not

guarantee that each gauge orbit has a unique representative. In fact, it forces

each gauge orbit to have two representatives instead. Indeed, if we consider a

simple gauge fixing function

(

x, y

) =

, then this is non-degenerate, but the

zero set looks like

It is an easy exercise with the intermediate value theorem to show that there

must be at least two representatives in each gauge orbit (one will need to use

the non-degeneracy condition).

This is not a particularly huge problem, since we are just double counting

each gauge orbit, and we know how to divide by 2. Let’s stick with this and

move on.

To integrate over the gauge orbit, it is natural to try the integral.

dx dy δ(f(x))e

−S(x,y)

Then the δ-function restricts us to the curve C, known as the gauge slice.

However, this has a problem. We would want our result to not depend on

how we choose our gauge. But not only does this integral depend on

. It in

fact depends on f as well!

To see this, we can simply replace f by cf for some constant c ∈ R. Then

δ(f(x)) 7→ δ(cf(x)) =

|c|

δ(f(x)).

So our integral changes. It turns out the trick is to include the factor of ∆

previously defined. Consider the new integral

dx dy δ(f(x))|∆

(x)|e

−S(x)

. (∗)

To analyze how this depends on

, we pretend that the zero set

actually

looks like this:

rather than

Of course, we know the former is impossible, and the zero set must look like the

latter. However, the value of the integral (

∗

) depends only on how

behaves

locally near the zero set, and so we may analyze each “branch” separately,

pretending it looks like the former. This will make it much easier to say what

we want to say.

Theorem. The integral (∗) is independent of the choice of f and C.

Proof.

We first note that if we replace

(

) by

(

)

(

) for some

c >

0, then we

have

δ(cf) =

|c|

δ(f), |∆

(x)| = c(r)|∆

and so the integral doesn’t change.

Next, suppose we replace

with some

, but they have the same zero set.

Now notice that

(

) and

∆

depend only on the first-order behaviour of

. In particular, it depends only on

∂f

∂θ

. So for all practical purposes,

changing

is equivalent to multiplying

by the ratio of their derivatives.

So changing the function

while keeping

fixed doesn’t affect the value of (

∗

Finally, suppose we have two arbitrary

and

, with potentially different

zero sets. Now for each value of r, we pick a rotation R

θ(r)

∈ SO(2) such that

f(x) ∝ f(R

θ(r)

x).

By the previous part, we can rescale

, and assume we in fact have equality.

We let

θ(r)

. Now since the action only depends on the radius, it

in particular is invariant under the action of

θ(r)

. The measure d

is also

invariant, which is particularly clear if we write it as d

θ r

instead. Then we

have

dx dy δ(f(x))|∆

(x)|e

−S(x)

δ(f(x

))|∆

)|e

−S(x

)

δ(

f(x))|∆

(x)|e

−S(x

)

dx dy δ(

f(x))|∆

(x)|e

−S(x)

Example.

We choose

to be the

-axis with

(

) =

. Then under a rotation

f(x) = y 7→ y sin θ − x sin θ,

we have

∆

(x) = −x.

So we have

dx dy δ(f)∆

(x)e

−S(x,y)

dx dy δ(y)|x|e

−S(x,y)

∞

−∞

dx |x|e

−S(x,0)

= 2

∞

d|x| |x|e

−S(|x|)

= 2

∞

r dr e

−S(r)

So this gives us back the original integral

∞

r dr e

−S(r)

of along the gauge orbit we wanted, except for the factor of 2. As we mentioned

previously, this is because our gauge fixing condition actually specifies two points

on each gauge orbit, instead of one. This is known as the Gribov ambiguity.

When we do perturbation theory later on, we will not be sensitive to this

global factor, because in perturbation theory, we only try to understand the

curve locally, and the choice of gauge is locally unique.

The advantage of looking at the integral

dx dy δ(f)∆

−S(x,y)

is that it only refers to functions and measures on the full space

, which we

understand well.

More generally, suppose we have a (well-understood) space

with a measure

. We then have a Lie group

acting on

. Suppose locally (near the identity),

we can parametrize elements of

by parameters

for

= 1

, ··· , dim G

. We

write

for the corresponding element of

(technically, we are passing on to

the Lie algebra level).

To do gauge fixing, we now need many gauge fixing functions, say

, again

with a = 1, ··· , dim G. We then let

∆

= det



∂f

∂θ



θ=0



This is known as the Fadeev–Popov determinant.

Then if we have a function

−S[x]

that is invariant under the action of

then to integrate over the gauge orbits, we integrate

dµ |∆

dim G

a=1

δ(f

(x))e

−S[x]

Now in Yang–Mills, our spaces and groups are no longer finite-dimensional, and

nothing makes sense. Well, we can manipulate expressions formally. Suppose we

have some gauge fixing condition f . Then the expression we want is

Z =

A/G

Dµ e

−S

Y M

DA δ[f]|∆

(A)|e

−S

Y M

[A]

Suppose the gauge group is

, with Lie algebra

. We will assume the gauge

fixing condition is pointwise, i.e. we have functions

g → g

, and the gauge

fixing condition is

f(A(x)) = 0 for all x ∈ M.

Then writing n = dim g, we can write

δ[f] =

x∈M

(n)

(f(A(x))).

We don’t really know what to do with these formal expressions. Our strategy is

to write this integral in terms of more “usual” path integrals, and then we can

use our usual tools of perturbation theory to evaluate the integral.

We first take care of the

[

] term. We introduce a new bosonic field

h ∈ Ω

(g), and then we can write

δ[f] =

Dh exp



−

(x)f

(A(x)) d



This

acts as a “Lagrange multiplier” that enforces the condition

(

)) = 0,

and we can justify this by comparing to the familiar result that

ip·x

dp = δ(x).

To take care of the determinant term, we wanted to have

∆

= det

δf

(x)]

δλ

(y)

where λ

(y) are our gauge parameters, and A

is the gauge-transformed field.

Now recall that for a finite-dimensional n × n matrix M , we have

det(M) =

c d

¯c e

¯cMc

where

c, ¯c

are

-dimensional fermion fields. Thus, in our case, we can write the

Fadeev—Popov determinant as the path integral

∆

Dc D¯c exp



M×M

x d

y ¯c

(x)

δf

(x))

δλ

(y)



where

c, ¯c

are fermionic scalars, again valued in

under the adjoint action. Since

we assumed that

is local, i.e.

(

)(

) is a function of

(

) and its derivative

at x only, we can simply write this as

∆

Dc d¯c exp



x ¯c

(x)

δf

)

δλ

(x)c

(x)



The fermionic fields

and

¯c

are known as ghosts and anti-ghosts respectively.

We might find these a bit strange, since they are spin 0 fermionic fields, which

violates the spin statistic theorem. So, if we do canonical quantization with this,

then we find that we don’t get a Hilbert space, as we get states with negative

norm! Fortunately, there is a subspace of gauge invariant states that do not

involve the ghosts, and the inner product is positive definite in this subspace.

When we focus on these states, and on operators that do not involve ghosts,

then we are left with a good, unitary theory. These

and

¯c

aren’t “genuine”

fields, and they are just there to get rid of the extra fields we don’t want. The

“physically meaningful” theory is supposed to happen in A/G, where no ghosts

exist.

Will all factors included, the full action is given by

S[A, ¯c, c, h] =



Y M

µν

a,µν

+ ih

(A) − ¯c

δf

)

δλ



and the path integral is given by

Z =

DA Dc D¯c Dh exp



−S[A, ¯c, c, h]



Example.

We often pick Lorenz gauge

(

) =

∂

. Under a gauge transfor-

mation, we have A 7→ A

= A + ∇λ. More explicitly, we have

)

= A

+ ∂

+ f

So the matrix appearing in the Fadeev–Popov determinant is

δf

)

δλ

= ∂

∇

Thus, the full Yang–Mills action is given by

S[A, ¯c, c, h] =



Y M

µν

a,µν

∂

− ¯c

∂

∇



Why do have to do this? Why didn’t we have to bother when we did

electrodynamics?

If we did this for an abelian gauge theory, then the structure constants were

zero. Consequently, all

terms do not exist. So the ghost kinetic operator

does not involve the gauge field. So the path integral over the ghosts would be

completely independent of the gauge field, and so as long as we worked in a

gauge, then we can ignore the ghosts. However, in a non-abelian gauge theory,

this is not true. We cannot just impose a gauge and get away with it. We need

to put back the Jacobian to take into account the change of variables, and this

is where the ghosts come in.

The benefit of doing all this work is that it now looks very familiar. It seems

like something we can tackle using Feynman rules with perturbation theory.