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A technical question on the Faddeev-Popov procedure (P&S Chapter 9). P&S introduce the functional integral, which is equal to one and then they choose the gauge-fixing function $G(A)$ to be equal to $$G(A)=\partial_{\mu}A^{\mu}(x)-\omega(x)\tag {9.55}$$ which is totally fine with me. They are allowed to do so, I guess (although some thoughts on their motivation could be useful if there are any).

The resulting expression for the generating functional is proportional to $$Z\sim\int\mathcal{D}Ae^{iS[A]}\delta(\partial_{\mu}A^{\mu}-\omega).$$ Then, they make the claim that the generating functional is independent of the newly introduced scalar function $\omega(x)$ and then multiply by a normalization constant with an integral, which is again understandable, if and only iff the generating functional is indeed independent of $\omega(x)$. I have seen the relevant post asking why are they allowed to do that, but this is not my question.

My question is: can we somewhow show that the generating functional is independent of the scalar function $\omega(x)$? I was thinking something like showing that its functional derivative w.r.t. the latter scalar function is zero, or something like that. Namely, that $$\frac{\delta Z}{\delta \omega}=0.$$ Also, would any form of $G(A)$'s dependence on $\omega(x)$ reproduce a generating functional that is independent of the latter?

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    $\begingroup$ You said yourself that the path integral they introduce equals 1, and 1 does not depend on $\omega$. The fact we can use $\omega$ comes from this equality, which is just a change of coordinate. $\endgroup$
    – Jeanbaptiste Roux
    Jun 23, 2022 at 14:38
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    $\begingroup$ You have to integrate out $\omega(x)$. Otherwise, it depends on $\omega(x)$. To do so, you insert $e^{i\int d^{4}x\omega^{2}}$ into the functional integrand, and then integrate out the $\omega$ field be performing a Gaussian functional integral $\int\mathcal{D}\omega$. $\endgroup$
    – Valac
    Jun 23, 2022 at 16:56
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    $\begingroup$ Would it help you to write $G(A,\omega)$, instead, and note $\partial G/\partial \alpha$ doesn't depend on ω ? $\endgroup$
    – Cosmas Zachos
    Jun 23, 2022 at 19:02
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    $\begingroup$ I think you are getting caught up in notation. Illustrate everything with the trivial case of one point x, so three plain integrals, and some notional gauge invariance leaving S(A) unchanged. It should be straightforward. $\endgroup$
    – Cosmas Zachos
    Jun 23, 2022 at 21:36
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    $\begingroup$ PS an elegant plain-variables toy model relying on rotational invariance instead of shift invariance I proposed above is on pp 190-192 of George Sterman's Introduction to QFT. $\endgroup$
    – Cosmas Zachos
    Jun 23, 2022 at 22:44

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This is a general fact about gauge theories, not just E&M.

  1. In a nutshell the independence of the gauge-fixing function in the path integral/partition function $Z$ is a generalization of the fact that $$ \int_{\Omega}\! d^nx ~\left|\det\frac{\partial f(x)}{\partial x} \right|\delta^n(f(x))~=~1 $$ as long as the pre-image $f^{-1}(\{\vec{0}\})$ of the smooth function $f: \Omega\subseteq\mathbb{R}^n \to \mathbb{R}^n$ is a singleton.

  2. Alternatively, the independence of the gauge-fixing function in $Z$ follows from that the Faddeev-Popov (FP) term plus the gauge-fixing (GF) term $S_{FP}+S_{GF}$ in the gauge-fixed action is BRST-exact. This is e.g. explained in my Phys.SE answer here.

  3. For more general gauge theories this can be proven via the Batalin-Vilkovisky (BV) formalism.

References:

  1. M.E. Peskin & D.V. Schroeder, An Intro to QFT, 1995; eq. (9.55).

  2. M. Srednicki, QFT, 2007; Chapter 71. A prepublication draft PDF file is available here.

  3. G. Sterman, An Intro to QFT, 1993; p. 190-192. (Hat tip: Cosmas Zachos).

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