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

\[ \begin{aligned} \nabla' \phi' &= (d + A')(g \phi) \cr &= dg \phi + g d\phi + A'(g\phi) \cr &= gg^{-1}dg \phi + gd\phi + g g^{-1} A' g\phi \cr &= g(d\phi + g^{-1} A' g \phi + g^{-1} dg \phi \cr &= g(d\phi + A\phi) \end{aligned} \]

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 \(\epsilon \to 0\))

\[ \begin{aligned} I_\epsilon &= \left(\frac{1}{2\pi \epsilon}\right)^\frac{k}{2} \int\int f(u, v) \exp\left\{ -\frac{1}{2\epsilon} v^T H(u) v \right\} du dv \cr &= \int_M \frac{f(u)}{\sqrt{\det H(u)}} du. \end{aligned} \]

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

\[ \begin{aligned} H_{ij} &= \frac{1}{2} \frac{\partial^2 |G|^2}{\partial v^i \partial v^j} \cr &= \frac{\partial}{\partial v^i} \left( G^a \partial_j G^a \right) \cr &= \left(\partial_i G^a \partial_j G^a + G^a \partial_{ij} G^a \right) \cr &= \partial_i G^a \partial_j G^a \end{aligned} \]

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

\[ \int_M f(x) d\mu = \int_{\mathbb{R}^n} f(x) \delta(G(x)) \exp\left(\eta \cdot G(x+ \theta) \right) dx d\theta d\eta. \]