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.

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    $\begingroup$ All of this is absolutely correct and is basically how FP gauge fixing is explained in modern field theory/string theory books. Good going if you figured out this picture on your own! $\endgroup$ Dec 2, 2015 at 17:39
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    $\begingroup$ I'm not sure which book does this in more detail than what you have here. But just as some basic references, there are a couple of places in the "Quantum fields and strings" volumes where this is explained: one in Fadeev's own lecture and one in D'hoker's lecture. There's also a lecture by Davide Gaiotto on BRST quantization where he explains this in his string theory course from the 2014/2015 Perimeter Scholars program. $\endgroup$ Dec 2, 2015 at 18:08
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    $\begingroup$ What you really want is the slightly smoother discussion of this very situation on pages 190-192 of George Sterman's An Introduction to Quantum Field Theory ISBN-13: 978-0521311328. $\endgroup$ Apr 5, 2016 at 19:32
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    $\begingroup$ Notice that the gauge fixing is not necessary because there is a double counting. If this was the case, there would also be a gauge fixing necessary for the fermionic path integal $\int\mathcal{D}\Psi\,\mathcal{D}\overline{\Psi}$, in which one integrates over $\Psi(x)$ and $e^{\mathrm{i}\,\alpha(x)}\Psi(x)$. Rather, gauge fixing is necessary because the operator $k^\mu k^\nu-k^2 \eta_{\mu\nu}$ is not invertible, i.e. it has a nontrivial kernel. Gauge fixing allows one to fix this problem. See e.g. Pokorski's book "Gauge field theories" for a clear explanation. $\endgroup$
    – user178876
    Dec 16, 2017 at 21:13
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    $\begingroup$ @Nogueira I am not disputing that it is necessary to fix the gauge, but it is really for the reason I mentioned. Please have a look at e.g. Pokorski's book. (I know that in several books it is suggested that one needs to fix the gauge because one otherwise overcounts configurations, but this is not entirely correct.) $\endgroup$
    – user178876
    Apr 30, 2018 at 23:37