The Faddeev-Popov Determinant | Yucheng (Luca) Jin

Source: Perimeter Scholars International, Quantum Field Theory II, Tutorial 9; 10 worked parts.

1 The Faddeev--Popov Determinant

Question prompts paraphrased from the official Perimeter PSI Online sheet; the equations and notation follow the source.

1.1 The Faddeev--Popov Determinant

As a finite-dimensional model of an unfixed gauge zero mode, take \[D=\begin{pmatrix}1&0\\0&0\end{pmatrix}, \qquad A=\begin{pmatrix}a\\b\end{pmatrix}.\tag{1.1}\] Then \[I=\int_{-\infty}^{\infty}dA\,e^{-A^TDA} =\int_{-\infty}^{\infty}da\,db\,e^{-a^2}\tag{1.2}\] is ill-defined because one factor is the volume of the real line.

For the general gauge-invariant integral \[I=\int\mathcal{D}A\,e^{iS(A)},\tag{1.3}\] assume that $A\mapsto A_g$ belongs to a gauge group $\mathcal{G}$ and satisfies \[\mathcal{D}A=\mathcal{D}A_g, \qquad S(A)=S(A_g).\tag{1.4}\] Define the Faddeev--Popov determinant by \[1=\Delta(A)\int\mathcal{D}g\, \delta\!\bigl[f(A_g)\bigr], \qquad \int\mathcal{D}g=\operatorname{Vol}(\mathcal{G}).\tag{1.5}\]

Prove that $\Delta(A)$ is gauge invariant.

Under a gauge transformation \[A \mapsto A_h, \quad (A_h)_g = A_{gh}, \quad \Delta(A) \mapsto\Delta(A_h),\tag{1.6}\] we have \[1 = \Delta(A_h)\int \mathcal{D}g\delta[f(A_{gh})].\tag{1.7}\] Formally, the functional integral over the gauge group is defined as \[\mathcal{D}g = \prod_xd\mu_H(g(x)),\tag{1.8}\] where $d\mu_H$ is Haar measure. It is the group theoretical analogue to the normal translational invariant measure, $dx = d(x+a)$. In group theory, the abelian operation $+$ is replaced by actions like right or left actions. In our present case, we take the Haar measure to be right invariant, namely \[\mathcal{D}g = \mathcal{D}(gh).\tag{1.9}\] Therefore, we have \[1 = \Delta(A_h)\int \mathcal{D}(gh)\delta[f(A_{gh})] = \Delta(A_h)\int \mathcal{D}g'\delta[f(A_{g'})] = \Delta(A)\int \mathcal{D}g\delta[f(A_{g})] \implies \Delta(A) =\Delta(A_h).\tag{1.10}\]
Insert the identity above and show that \[I=\operatorname{Vol}(\mathcal{G}) \int\mathcal{D}A\,e^{iS(A)}\Delta(A)\delta[f(A)].\tag{1.11}\]

Inserting the identity into the path integral, we find \[I = \int \mathcal{D}A e^{iS[A]}\Delta(A)\int\mathcal{D}g\delta[f(A_g)].\tag{1.12}\] Using gauge invariant of the action and the integral measure, $\mathcal{D}A = \mathcal{D}A_g$, we find \[I = \left(\int \mathcal{D}g\right)\int\mathcal{D}A_g\,e^{iS(A_g)}\Delta(A_g)\delta[f(A_g)] = \operatorname{Vol}(\mathcal{G}) \int\mathcal{D}A\,e^{iS(A)}\Delta(A)\delta[f(A)].\tag{1.13}\]

Next use the zero-dimensional toy theory \[Z=\int_{-\infty}^{\infty}dx\,dy\,e^{-S(x,y)}, \qquad S(x,y)=(x-y)^2.\tag{1.14}\]

Identify the global “gauge” symmetry of this zero-dimensional action.

The toy theory is clearly invariant under translation, $x \to x+a, y \to y+a$.
Describe the full configuration space $\mathcal{A}$ and its gauge orbits $\mathcal{G}$.

\[\mathcal{A} = \{x,y\in \mathbb{R}\} = \mathbb{R}^2, \quad \mathcal{G} = (\mathbb{R},+) \implies \mathcal{A}/\mathcal{G} \cong \mathbb{R}.\tag{1.15}\]

The divergent partition function can be factorized as \[Z=\left(\int da\right)X,\tag{1.16}\] where $a$ parametrizes the gauge group and $X$ is independent of $a$.

Find new variables that make this factorization explicit. As an additional challenge, suggest a modified model whose gauge-group volume is finite.

Perform the change of variable, \[a = \frac{x-y}{\sqrt{2}}, \quad b = \frac{x+y}{\sqrt{2} },\tag{1.17}\] which gives \[Z = \left(\int da\right)\int db e^{-b^2}.\tag{1.18}\]
Choose a gauge slice for computing $X$ and give a condition that characterizes it; the choice need not be unique. Denote the condition by $f(x,y)=0$.

It is easiest if we choose the gauge slice $f(x,y) = y = 0$.
Show that \[Z=\int_{-\infty}^{\infty}da \int_{-\infty}^{\infty}dx\,dy\, J\,\delta\!\bigl(f(x,y)\bigr)e^{-(x-y)^2},\tag{1.19}\] where the Jacobian $J$ is independent of $a$.

We have \[1 = \int df(x,y) \delta(f(x,y)) = \int df(x+a,y+a) \delta(f(x+a,y+a)) = \int da \, J\delta(f(x+a,y+a)).\tag{1.20}\] For our particular gauge choice, this is \[1 = \int da \delta(y+a).\tag{1.21}\] Inserting into the partition function, we find \[Z=\int_{-\infty}^{\infty}da \int_{-\infty}^{\infty}dx\,dy\, J\,\delta\!\bigl(f(x,y)\bigr)e^{-(x-y)^2}\tag{1.22}\] by "gauge" invariance of the action and the measure.

Replace the gauge slice $f=0$ by $f=c$, with constant $c\neq0$, so that \[Z=\int da\int dx\,dy\, J\,\delta\!\bigl(f(x,y)-c\bigr)e^{-(x-y)^2}.\tag{1.23}\]

Although the last expression displays $c$ explicitly, prove that $Z$ is independent of $c$.

Since \[1 = \int df(x,y) \delta(f(x,y)) = 1 = \int df(x,y) \delta(f(x,y)-c),\tag{1.24}\] the rest follows straightforwardly from above.
Average over $c$ with the Gaussian weight \[\int_{-\infty}^{\infty}dc\,e^{-c^2}\tag{1.25}\] and derive \[Z=\frac{1}{\sqrt{\pi}} \int da\int dx\,dy\, J\,e^{-f^2}e^{-(x-y)^2}.\tag{1.26}\] Explain how the factor $e^{-f^2}$ supplies a gauge-fixing term.

\[\begin{aligned} Z=\frac{1}{\sqrt{\pi}}\int dc\, e^{-c^2}\int da\int dx\,dy\, J\,\delta\!\bigl(f(x,y)-c\bigr)e^{-(x-y)^2} = \frac{1}{\sqrt{\pi}} \int da\int dx\,dy\, J\,e^{-f^2}e^{-(x-y)^2}. \end{aligned}\tag{1.27}\] This factor is largest when $f=0$ and exponentially suppresses configurations for which $f\neq0$. It therefore imposes the same gauge condition softly.
Exponentiate the remaining Jacobian by representing it as an integral over Grassmann “ghost” variables.

\[J = \int d\bar cdc \, e^{-\bar c J c}.\tag{1.28}\]