# Can we choose other than Gaussian integral for Faddeev-Popov gauge fixing?

for $U(1)$ field $A_\mu$ and its longitudinal gauge component $\partial_\mu \alpha(x)$, Faddeev-Popov gauge fixing written in Peskin (eq.9.56) is: $$N(\xi)\int \mathcal{D}\omega\hspace{0.1cm}\text{exp}\left[-i\int d^4x \frac{\omega^2}{2\xi}\right]\text{det}\left(\frac{1}{e}\partial^2\right)\left(\int \mathcal{D}\alpha\right)\int\mathcal{D}Ae^{iS[A]}\delta(\partial^\mu A_\mu -\omega(x))\\ =N(\xi)\text{det}\left(\frac{1}{e} \partial^2\right)\left(\int\mathcal{D}\alpha\right)\int\mathcal(A)e^{iS[A]}\text{exp}\left[-i\int d^4x\frac{1}{2\xi}(\partial^\mu A_\mu)^2\right]. \tag{9.56}$$

This equation follows from $$\int\mathcal{D}A e^{iS[A]}=\text{det}\left(\frac{1}{e}\partial^2\right)\left(\int\mathcal{D}\alpha\right)\int\mathcal{D}A e^{iS[A]}\delta(\partial^\mu A_\mu-\omega(x)) \tag{9.55b}$$ where $N(\xi)$ is normalization factor.

I think that $$N(\xi)\int \mathcal{D}\omega\hspace{0.1cm}\text{exp}\left[-i\int d^4x \frac{\omega^2}{2\xi}\right]=1$$ justifies equivalence of 2nd equation and eq.9.56 (first line in eq.9.56).

1. If it is true, Can we pick up any functional integral (e.g.$N(\xi)\int \mathcal{D}\omega\hspace{0.1cm} f[\omega]$), where $f[\omega]$ is bounded and path integrable) instead of Gaussian (i.e. $\int \mathcal{D}\omega\hspace{0.1cm}\text{exp}\left[-i\int d^4x \frac{\omega^2}{2\xi}\right]$) and divide it by its value to normalize (like $N(\xi)$ in Gaussian integral used above)?

2. Is there any particular reason to choose $f[\omega]=\text{exp}\left[-i\int d^4x \frac{\omega^2}{2\xi}\right]$?

3. And if I take $f[\omega]$ different from $\exp\left[-i\int d^4x \frac{\omega^2}{2\xi}\right]$, then Gauge fixing term in 2nd line of eq.9.56 (i.e. $\exp\left[-i\int d^4x \frac{(\partial^\mu A_\mu)^2}{2\xi}\right]$ will be changed to a different form($f[\partial^\mu A_\mu])$ and will give different propagators. In this case, even though I have a propagator with a different form, will my final answer of S-matrix element which should be independent of $\xi$ be the same as Gaussian integration case?

• This is one of the main results of BRST theory. Give Weinberg V.II a read. – AccidentalFourierTransform Aug 11 '17 at 11:18

Any integrable function $f$ will in principle do. But the calculations may become more cumbersome.
It should be obvious why we normally choose the function $f$ to be Gaussian, because it is exponentially decaying (after Wick rotation), and the math involved is simple and can be done analytically.