How to apply the Faddeev-Popov method to a simple integral

Question

Some time ago I was reviewing my knowledge on QFT and I came across the question of Faddeev-Popov ghosts. At the time I was studying thеse matters, I used the book of Faddeev and Slavnov, but the explanation there is not very transparent, specially not for someone like me who was just starting to learn QFT. Therefore, I never understood fully what was meant. To clear my doubts how the method works and what are the gauge orbits I decided to think how the method will work on a simple toy problem.

The local gauge transformation in the non-Abelian case acts non-linearly i.e.

$$ F[\mathscr{A}_{\mu}] = g\mathscr{A}_{\mu}g^{-1} + g_{\mu}g_{\mu}^{-1} $$

Since in the generating functional we integrate over the fields,

$$ Z=\int \mathcal{D}\mathscr{A}_{\mu}e^{iS[\mathscr{A}_{\mu}]} $$

double counting is introduced, due to the integration over many equivalent fields generated by the local gauge transformation. To fix this L.D. Faddeev and V. Popov proposed to introduce the constraint of the gauge transformation in the form:

$$ \Delta_L(\mathscr{A}) \int \delta(F[\mathscr{A}_{\mu}^{\omega}])d\omega=1 $$ There are different methods how to get $\Delta_L(\mathscr{A})$, but I think that the simplest one is to use just the definition of the delta function. Of course using the properties of the Haar measure, the above expression is gauge invariant. Say $U(1)$ with $\omega=e^{i\phi}$ this can be checked by using fixed $U'$

$$ \mathcal{D}\omega\omega' = \mathcal{D}\omega, $$

which in my opinion is just the product rule.

Plugging into the generating functional we get

$$ Z=\iint \mathcal{D}\mathscr{A}_{\mu} d \omega \Delta_L(\mathscr{A}) \delta(F[\mathscr{A}_{\mu}^{\omega}])e^{iS[\mathscr{A}_{\mu}]}, $$

which produces a multiplicative volume factor.

Now comes my question, how do we use this on a toy problem. Suppose we were integrating

$$ I=\iint e^{-(x^2+y^2)}dxdy $$

The integration is redundant and by going to cylindrical coordinates $(r,\phi)$ we can easily factor out the $\int d\phi$ part. Let's do this with the Faddeev-Popov method.

Our integral is rotational invariant and the only real contribution comes from moving in the direction $r \to \infty$. I visualise our gauge transformation as rotation around the origin and I have the feeling that the gauge orbits are concentric circles. Since we would like to use only non-equivalent orbits we fix the $y$ variable. To do so, use the value $y_{\phi} = x\sin\phi+y\cos\phi$

For our unity integral, we have $$ 1=\int d\phi\delta(x\sin\phi+y\cos\phi)|\frac{\partial(x\sin\phi+y\cos\phi)}{\partial \phi}| $$ Since we have a rotational-invariant integral, let's pick $\phi=0$ this gives

$$ I=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} e^{-(x^2+y^2)}\delta(y) |x| dx dy d \phi $$ $$ I=\int_0^{\infty} e^{-x^2} x dx \times \int_0^{2\pi}d\phi = \pi $$

What we have done above is just rotating, so that the integral is taken along the positive real axis $y_{\phi}=0$. This looks like a complicated way of doing changes of variables or introducing constraints.

If the above is correct, what are the gauge orbits in the general case? According to Faddeev himself, his intuition was purely geometrical and the non-Abelian case produces lines that intersect the gauge orbits at different angles.

Coming back to my example instead of circles $F[\mathscr{A}_{\mu}]$ defines a manifold and the Gauge condition $\partial_{\mu}\mathscr{A}^{\mu}$ gives a cut trough this manifold equivalent to the intersection $y_{\phi}=0$.

I would appreciate your critical review of my question.

Answer 1

Faddeev-Popov Toy Model and General Gauge Orbit Review

Review of the Faddeev--Popov Toy Model:

We begin with the rotationally invariant integral $$ I = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} e^{-(x^2+y^2)}\,dx\,dy, $$ which, in polar coordinates, naturally separates into a radial and an angular part. The redundancy due to rotational invariance is analogous to the gauge redundancy in field theory.

1. Insertion of the Unity

In the Faddeev--Popov method the idea is to insert the identity $$ 1 = \int d\phi\, \delta\bigl( x\sin\phi + y\cos\phi \bigr) \left|\frac{d}{d\phi}\Bigl(x\sin\phi+y\cos\phi\Bigr)\right| $$ into the integral. Here, the gauge transformation is a rotation by the angle $\phi$, and the condition $$ y_\phi \equiv x\sin\phi + y\cos\phi = 0 $$ serves as our gauge-fixing condition.

2. Evaluation at a Specific Gauge

Due to the overall rotational invariance, we may choose $\phi=0$ without loss of generality. At $\phi=0$ the gauge condition becomes $y=0$, and the determinant factor is evaluated as $$ \left|\frac{d}{d\phi}(x\sin\phi+y\cos\phi)\right|_{\phi=0} = |x|. $$

Thus, the original integral is rewritten as $$ I = \int d\phi \int_{-\infty}^{\infty}dx\,e^{-x^2}\,|x|\int_{-\infty}^{\infty}dy\,e^{-y^2}\,\delta(y). $$ The $y$-integral simplifies immediately: $$ \int_{-\infty}^{\infty} \delta(y)e^{-y^2}\,dy = 1, $$ and the remaining integrals become $$ I = \left(\int_{0}^{\infty}e^{-x^2}\,x\,dx\right)\left(\int_{0}^{2\pi}d\phi\right). $$

Since $$ \int_{0}^{\infty} x\,e^{-x^2}\,dx = \frac{1}{2}, $$ and the angular integral yields $2\pi$, we obtain $$ I = \frac{1}{2}\times 2\pi = \pi. $$

General Case: Non-Abelian Gauge Theories

In a non-Abelian gauge theory, the fields $A_\mu$ transform as $$ A_\mu \to A_\mu^\omega = g(x)A_\mu g^{-1}(x) + g(x)\partial_\mu g^{-1}(x), $$ with $g(x)$ an element of the gauge group. The generating functional is $$ Z = \int DA_\mu\, e^{iS[A_\mu]}, $$ which suffers from overcounting because the integration includes entire \emph{gauge orbits} (sets of fields related by a gauge transformation).

To remove the redundancy, one inserts the identity $$ 1 = \Delta[A]\int D\omega\, \delta\bigl(F[A^\omega]\bigr), $$ where $F[A]=0$ is the gauge condition and $\Delta[A]$ is the Faddeev--Popov determinant. This determinant accounts for the Jacobian of the transformation when changing variables from $A_\mu$ to the gauge-fixed variables.

Geometrical Interpretation:

In our toy model the gauge orbits are the concentric circles (orbits of the rotation group in the $(x,y)$ plane) and the gauge fixing $y_\phi=0$ selects a unique representative on each orbit. In the non-Abelian case, however, the gauge orbits form curved submanifolds in the infinite-dimensional space of field configurations. The gauge-fixing surface (for example, $\partial_\mu A^\mu = 0$) typically intersects these orbits at non-orthogonal angles. This non-orthogonality is at the heart of the complications in non-Abelian gauge theories, such as the appearance of Gribov copies (multiple intersections of the gauge-fixing surface with a single gauge orbit).

Conclusion:

The toy example demonstrates how the Faddeev--Popov procedure factors out redundant integrations over gauge orbits by introducing a delta function constraint and an associated determinant. While the method is straightforward in finite dimensions with a clear geometrical picture (orbits being circles), the general case is more subtle due to the complicated geometry of the gauge orbits in the space of fields.