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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?

Qmechanic
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schris38
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1 Answers1

3

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).

Qmechanic
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