Functional derivative in Faddeev Popov method (Lorenz Gauge) When applying Faddeev and Popov method (am using Peskin and Schroeder as reference), we use the identity:
$$1=\int \mathcal{D}\alpha(x)\delta(G(A^\alpha)) \det\left(\frac{\delta G(A^\alpha)}{\delta\alpha}\right) \tag{9.53}$$
to write
$$ \int \mathcal{D}Ae^{iS[A]}=\det\left(\frac{\delta G(A^\alpha)}{\delta\alpha}\right)\int\mathcal{D}\alpha(x)\int\mathcal{D}Ae^{iS[A]}\delta (G(A^\alpha))\tag{9.54}$$
When we use the Lorenz gauge, we obtain:
$$G(A)=\partial^\mu A_\mu+\frac{1}{e}\partial^2\alpha(x)$$
My questions is: how do we obtain the following:
$$ \det\left(\frac{\delta G(A^\alpha)}{\delta\alpha}\right) = \det\left(\frac{1}{e}\partial^2\right) $$
I am also confused about how can we obtain an operator by taking a functional derivative, if someone could give an intuitive explanation of this it would be highly appreciated!
 A: $$\begin{align}{\rm Det} \left(\frac{\delta G}{\delta \alpha}\right)
~=~&\int {\cal D}c{\cal D}\bar{c}\exp\left(\int \!d^4x \int \!d^4y ~\bar{c}(x)\frac{\delta G(x)}{\delta \alpha(y)}c(y) \right) \cr
~=~&\int {\cal D}c{\cal D}\bar{c}\exp\left(\int \!d^4x \int \!d^4y ~\bar{c}(x) \frac{1}{e}\partial_x^2\delta(x-y) c(y) \right)\cr
~=~& {\rm Det} \left(\frac{1}{e}\partial^2\right).\end{align}$$
A: This should be roughly familiar from obtaining the usual Feynman rules from the path integral formulation, though perhaps it usually isn't phrased this way.
The long story short is that the functional determinant of an operator is defined by the trace log formula,
$$
\det A=e^{\text{tr}\log A}
$$
which can be proven by thinking about eigenvalues. For these operations to make sense on a differential operator, however, we usually move to the Fourier basis where all the derivatives are just momenta.
In fact, this is what we do when writing down the Feynman rules. Remember that computing the saddle point approximation of the path integral (with source $J\phi$) results in
$$
Z[J]\sim e^{J^T\Delta J},
$$
appropriate integrations implied. From here we employ the usual tricks of taking $J$ derivatives to obtain correlators, this $\Delta$ being the Feynman propagator.
The way this approximation works is by writing the action Taylor expanded to second order around the saddle point (which is usually taken to be $\phi=0$ in Peskin and Schroeder, but need not be and indeed this fact has important implications related to instantons) so we write
$$
S[\phi]=S_0+\frac{1}{2}\phi \frac{\delta^2 S}{\delta \phi\delta\phi}\phi+J\phi + \mathcal{O}(\phi^3),
$$
integrations again implied as necessary. In the typical case of a scalar field,
$$
\frac{\delta^2 S}{\delta \phi\delta\phi}=\partial^2-m^2
$$
up to signs.
This is where it all comes together, we of course know that the Feynman propagator should be the inverse of this operator $\partial^2-m^2$, but the trick we usually employ is moving to Fourier basis where this operator is $-k^2-m^2$ so its inverse is given by
$$
\frac{-1}{k^2+m^2},
$$
which is the propagator up to a factor of $i$.
So asking about the determinant of $\partial^2$ shouldn't be so terrible in light of the fact that we invert the operator $\partial^2-m^2$ all the time (though admittedly it is objectively worse to calculate).
