We saw in a previous post that for a non-interacting theory (i.e. $V(\varphi) = 0$) that the generating functional can be written as

$$ Z[J] = e^{\frac{1}{2} J \cdot K^{-1} \cdot J}. $$

We hinted that it is not always the case that $K$ can be naively inverted. The issue arises when we consider the *Maxwell action* for a $U(1)$ gauge potential $A_\mu$:

$$ S(A) = \int d^4 x \left[ \frac{1}{2} A_\mu \left( \partial^2 g^{\mu \nu} - \partial^\mu \partial^\nu \right) A_\nu + A_\mu J^\mu \right]. $$

Here the operator we are looking to invert is

$$ \begin{equation} Q^{\mu \nu} \equiv \partial^2 g^{\mu \nu} - \partial^\mu \partial^\nu \end{equation} $$

which has zero eigenvalues for vectors of the form $\partial_\mu \theta(x)$:

$$ Q^{\mu \nu} \partial_\mu \theta(x) = \left(\partial^2 g^{\mu \nu} \partial_\nu - \partial^\nu \partial_\nu \partial^\mu \right) \theta(x) = 0 $$

So, $Q^{\mu \nu}$ is not invertible and the issue is gauge redundancy - we need to fix a gauge.

## the Faddeev-Popov procedure

An interpretation of the issue is that the redundancy in the gauge field, where acting with $g \in U(1)$, takes $A_\mu \to A_\mu - \partial_\mu \theta \equiv A_g$; which is not a physical symmetry of the system, but still gets naively integrated over in the generating functional:

$$ \begin{align*} Z &= \int \mathcal{D}A \ e^{iS(A)} \\\ &= \int \mathcal{D}A \int \mathcal{D}g \ e^{iS(A_g)} \\\ \end{align*} $$

Which diverges.

The naive path integral effectively over counts and blows up. What we need to do is insert a constraint on the gauge (e.g. $f(A_g) = 0$), which can be done by inserting

$$ \begin{equation} 1 = \int \mathcal{D}g \ \delta\left( f(A_g) \right) \det \left( \frac{\partial f(A_g)}{\partial{g}} \right) = \Delta(A)\int \mathcal{D}g \ \delta\left( f(A_g) \right), \end{equation} $$

where $\Delta(A)$ is *the Faddeev-Popov determinant* which is gauge invariant. We then get

$$ \begin{aligned} Z &= \int \mathcal{D}A \ e^{iS(A)} \\\ &= \int \mathcal{D}A \ e^{iS(A)} \ \Delta(A) \int \mathcal{D}g \ \delta\left( f(A_g) \right) \\\ &= \int \mathcal{D}g \int \mathcal{D}A \ e^{iS(A)} \ \Delta(A) \ \delta\left( f(A_g) \right) \\\ &= \left( \int \mathcal{D}g \right) \int \mathcal{D}A \ e^{iS(A)} \ \Delta(A) \ \delta\left( f(A) \right) \\\ \end{aligned} $$

Where we arrived at the final result by noting that the path integral measure $\mathcal{A}$ is gauge invariant and shifting $A \to A_{g^{-1}}$ removes the gauge dependence from $\delta(f(A_g))$. The factor outside, $\left( \int \mathcal{D}g \right)$ is an infinite constant that we can drop, as it has no physical effect.

To see how this procedure works with QED check out Zee’s wonderful book *Quantum Field Theory in a Nushell*, which was my primary reference for this post.

## ghosts in non-abelian gauge theories

We will now turn our attention towards *quantum Yang-Mills theory*, where the gauge group being considered is non-abelian, for all intents and purposes is $SU(N)$. Here things get very *spooky*.

In a pure non-abelian gauge theory, each component of the gauge potential $A_\mu$ transforms under the adjoint representation of the gauge group, in such a way that the action remains gauge invariant–i.e.

$$ A_\mu(x) \to U(x)A_\mu(x)U^\dagger(x) + iU(x)\partial_\mu U^\dagger(x) $$

One can note that if $U$ does not vary throughout space, $A$ really does transform in the adjoint representation; the extra term comes from the requirement for gauge invariance in a theory where the gauge group can act differently at each spacetime coordinate.

Now, given a set of generators for the gauge group, ${T^a}$, really a Lie algebra, we can see that under an infinitesimal gauge transformation ($U(x) \simeq 1 + i\theta^a T^a$), and noting we can decompose $A_\mu = A^a_\mu T^a $, we see that

$$ A^a_\mu \to A^a_\mu - f^{abc}\theta^b A^c_\mu + \partial_\mu \theta^a. $$

Where the structure constants $f^{abc}$ are defined by $[T^a, T^b] = f^{abc}T^c$.

Let’s now choose a gauge fixing condition to be $f(A) = \partial^\mu A_\mu - \sigma$, with $\sigma$ arbitrary. By (2) we then have

$$ \begin{align*} \Delta(A) &= \left\{\int \mathcal{D}\theta \ \delta\left[ f(A) \right] \right\}^{-1} \\\ &= \left\{\int \mathcal{D}\theta \ \delta\left[ \partial^\mu A^a_\mu - \sigma^a \right] \right\}^{-1} \\\ &= \left\{\int \mathcal{D}\theta \ \delta\left[ \partial^\mu A^a_\mu - \sigma^a - \partial^\mu \left( f^{abc}\theta^b A^c_\mu + \partial_\mu \theta^a \right)\right] \right\}^{-1} \\\ ``&= \text{’’} \ \left\{\int \mathcal{D}\theta \ \delta\left[ \partial^\mu \left( f^{abc}\theta^b A^c_\mu + \partial_\mu \theta^a \right)\right] \right\}^{-1} \\\ \end{align*} $$

Where there are quotes around the equality of the last line because we have yet to muliply by the gauge constraint $\delta[f(A)]$.

Now lets formally define an operator $K^{ab}(x,y)$ by

$$ \partial^\mu \left( f^{abc}\theta^b A^c_\mu + \partial_\mu \theta^a \right) = \int d^4y \ K^{ab}(x, y) \theta^b(y). $$

that is

$$ K^{ab}(x,y) \equiv \partial^\mu \left( f^{abc} A^c_\mu - \partial_\mu \delta^{ab} \right) \delta^{(4)}(x - y) $$

A generalization of the delta function identity, $\int d\theta \ \delta(K\theta) = 1/K$, to a vector $\theta$ and matrix $K$, gives us $\int d\theta \ \delta(K\theta) = 1 / \det K$. We also have that anti-commuting Grassmann variables allow us to write $\det K = \int d\eta d\eta^\dagger \exp\left[-\eta^\dagger \cdot K \cdot \eta\right]$. Pulling all of this together we see that

$$ \Delta(A) = \det K^{ab}(x,y) = \int \mathcal{D} c \mathcal{D} c^\dagger e^{i S_{\text{ghost}}(c^\dagger, c)} $$

where

$$ \begin{aligned} S_{\text{ghost}}(c^\dagger, c) &= \int d^4x d^4y \ c_a^\dagger(x)K^{ab}(x, y)c_b(y) \\\ &= \int d^4x \ c_a^\dagger(x) \left[ \partial^\mu f^{abc} A^c_\mu(x) c_b(x) - \partial^2 c_b(x) \right] \\\ &= \int d^4x \ \left[ \partial^\mu c_a^\dagger(x) \partial_\mu c_b(x) - \partial^\mu c^\dagger_a f^{abc} A^c_\mu(x) c_b(x) \right] \\\ &= \int d^4x \ \partial^\mu c_a^\dagger(x) \left[ \partial_\mu c_b(x) - f^{abc} A^c_\mu(x) c_b(x) \right] \\\ &= \int d^4x \ \partial^\mu c_a^\dagger(x) D_\mu c_a(x) \\\ \end{aligned} $$

Here $D_\mu$ is the covariant derivative of the adjoint representation; $c_a, c_a^\dagger,$ and $A_\mu^a$ all transform in this way.

$c_a$ and $c_a^\dagger$ are *ghost fields* and seemingly violate the *spin-statistics* connection, because they are grassmann anti-commuting variables, but they are not physical, just a convenient representation of $\Delta(A)$.

Zooming out, we now have

$$ Z = \int \mathcal{D}A\mathcal{D}c\mathcal{D}c^\dagger \ e^{iS(A) + iS_{\text{ghost}}(c^\dagger, c)} \ \delta(\partial A - \sigma) $$

which after integrating over $\sigma$ with a Gaussian weight, $e^{ -(i/2\xi) \int d^4x \sigma^a(x)^2}$, we arrive at a final gauge fixed (with gauge parameter $\xi$) generating functional for a pure non-abelian Yang-Mills theory:

$$ \boxed{ Z(J) = \int \mathcal{D}A\mathcal{D}c\mathcal{D}c^\dagger \ \exp \left\{iS(A) + iS_{\text{ghost}}(c^\dagger, c) - \frac{i}{2\xi} \int d^4x (\partial^\mu A_\mu)^2 + \int d^4x J^\mu A_\mu\right\} } $$