Fadeev-Popov ghosts

This post has been migrated from my old blog, the math-physics learning seminar.

Today I want to review the Fadeev-Popov procedure, with a view toward BRST and eventually BV.

Gauge-Invariance and Gauge-Fixing

First we’ll review the Fedeev-Popov method, to motivate the introduction of ghosts. Suppose we have a gauge theory involving a (G)-connection (A) and some field (\phi) charged under (G). Under a gauge transformation (g(x)), (\phi) transforms as

[ \phi \mapsto g \cdot \phi. ]

We would like that the covariant derivative transforms in the same way, i.e.

[ \nabla \phi \mapsto g \nabla \phi. ]

In terms of the connection 1-form (A), the covariant derivative is

[ \nabla = d + A. ]

Let (\nabla') denote the gauge-transformed covariant derivative, and (\phi') the gauge-transformed field. Then we want

[ \nabla' \phi' = g \nabla \phi. ]

We compute

\begin{align}

\nabla' \phi' &= (d + A')(g \phi) \

&= dg \phi + g d\phi + A'(g\phi) \

&= gg^{-1}dg \phi + gd\phi + g g^{-1} A' g\phi \

&= g(d\phi + g^{-1} A' g \phi + g^{-1} dg \phi \

&= g(d\phi + A\phi)

\end{align}

Comparing terms, we see that

[ A = g^{-1} A' g + g^{-1} dg, ]

so upon re-arranging we have

[ A' = g A g^{-1} - dg g^{-1} ]

This causes a problem: at critical points of the action, the Hessian of the action is degenerate in directions tangent to the gauge orbits. This means that the propagator is not well-defined, and there is no obvious way to derive the Feynman rules for perturbation theory. The solution is to take the quotient by gauge-transformations. To do this, we pick some gauge-fixing function (G(A)) which ought to be transverse to the orbits. Then we can restrict to the space (G(A) = 0), on which the Hessian of the action is non-degenerate, leading to a well-defined propagator. Formally, the path integral is

[ Z = \int_{{G(A) = 0}} \exp{\frac{i}{\hbar} S[A, \phi]} \mathcal{D}A \mathcal{D}\phi ]

Formally, this suggests that the path integral should be something like

[ Z = \int \delta(G(A)) \exp{\frac{i}{\hbar} S[A, \phi]} \mathcal{D}A \mathcal{D}\phi, ]

but this is not quite right! To understand the source of the problem, we’ll first study the finite-dimensional case and then use this to solve the problem in infinite-dimensions.

The Fadeev-Popov Determinant

Suppose we are on (\mathbb{R}^n), and we would like to integrate a function (f(x)) over a submanifold (M) defined by (M = G^{-1}(0)) for some smooth function (G: \mathbb{R}^n \to \mathbb{R}^k). Naively, we might expect that the answer is

[ \int_M f(x) \stackrel{?}{=} \int f(x) \delta(G(x)) dx. ]

To see why this is not correct, write the delta function as

[ \delta(G(x)) = \frac{1}{(2\pi)^k}\int e^{ip\cdot G(x)} d^k p. ]

We can regularize this by taking the limit as (\epsilon \to 0) of

[ \frac{1}{(2\pi)^k}\int \exp\left{ip\cdot G(x) -\frac{\epsilon}{2} |p|^2\right} d^k p ]

This integral is Gaussian, so we obtain explicitly

[ \left(\frac{2\pi}{\epsilon} \right)^{\frac{k}{2}} \exp\left{-\frac{1}{2\epsilon} |G(x)|^2 \right}. ]

So our original guess becomes

[ \left(\frac{1}{2\pi \epsilon}\right)^\frac{k}{2} \int f(x) \exp\left{ -\frac{1}{2\epsilon} |G(x)|^2 \right} dx .]

As (\epsilon \to 0), this integral localizes on the locus ({G(x) = 0}), as desired, but does not give the right answer! To see this, let (u) be a coordinate on (M = G^{-1}(0)) and (v) coordinates normal to (M). Then we have

[ G(x) = G(u,v) = v^T H(u) v + o(|v|^3) ]

where (H(x)) is the Hessian of (|G|^2) at the point (x = (u, 0)). So the integral becomes (as (\epsilon \to 0))

\begin{align}

I_\epsilon &= \left(\frac{1}{2\pi \epsilon}\right)^\frac{k}{2} \int\int

f(u, v) \exp\left{ -\frac{1}{2\epsilon} v^T H(u) v \right} du dv \

&= \int_M \frac{f(u)}{\sqrt{\det H(u)}} du.

\end{align}

This is not correct. We have to account for the determinant of the Hessian. Now, the Hessian is given by

\begin{align} H_{ij} &= \frac{1}{2} \frac{\partial^2 |G|^2}{\partial v^i \partial v^j} \

&= \frac{\partial}{\partial v^i} \left( G^a \partial_j G^a \right) \

&= \left(\partial_i G^a \partial_j G^a + G^a \partial_{ij} G^a \right) \

&= \partial_i G^a \partial_j G^a

\end{align}

where we have used the fact that (G = 0) on (x = (u, 0)). Hence we see that

[ \det H = (\det A)^2 ]

where (A) is the (k \times k) matrix with entries (\partial_i G^a). Hence

[ \sqrt{\det H} = \det A. ]

Now there is a straightforward way to eliminate the determinant. We introduce Fermionic coordinates (\eta^i, \theta^i), (i = 1, \ldots, k). Then by Berezin integration, we have

[ \int e^{\eta^i G^i(0, \theta^j)} d\theta d\eta = \det A. ]

So in the end, we find

[ \int_M f(x) d\mu = \int_{\mathbb{R}^n}

f(x) \delta(G(x)) \exp\left(\eta \cdot G(x+ \theta) \right)

dx d\theta d\eta. ]