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?

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    $\begingroup$ This is one of the main results of BRST theory. Give Weinberg V.II a read. $\endgroup$ Aug 11, 2017 at 11:18

1 Answer 1


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.

Finally let us mention that via the BRST formulation, or more generally the Batalin-Vilkovisky (BV) formalism, much more general gauge-fixing choices are available.


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