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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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schris38
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    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. – Jeanbaptiste Roux Jun 23 '22 at 14:38
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    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 $\int\mathcal{D}\omega$. – Valac Jun 23 '22 at 16:56
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    Would it help you to write $G(A,\omega)$, instead, and note $\partial G/\partial \alpha$ doesn't depend on ω ? – Cosmas Zachos Jun 23 '22 at 19:02
  • Hi all and thanks for the reply. Okay, I guess that by reverse engineering the derivation, one can show that the generating functional is indeed independent of $\omega(x)$, but is there a way of attacking the problem from the final expression (like demonstrating that the functional derivative wrt $\omega(x)$ vanishes?)? – schris38 Jun 23 '22 at 20:12
  • @LibertarianFeudalistBot I think that even before integrating out $\omega(x)$ the generating functional is independent on $\omega$, no?? – schris38 Jun 23 '22 at 20:13
  • @CosmasZachos maybe, but there still exists an $\omega(x)$ term in the delta functional and I would like to see/understand/explain why this term (in $\delta$) does not yield any dependence of the generating functional on it! – schris38 Jun 23 '22 at 20:15
  • @CosmasZachos oh I guess you mean to define $G$ s.t. it does not exhibit $\omega$ dependence and only choose such functions by definition... If so, yes, this is considerably better, but still, what of the $\delta$ functional argument $\omega(x)$... How does $Z$ not depend on this $\omega$? – schris38 Jun 23 '22 at 20:18
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    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. – Cosmas Zachos Jun 23 '22 at 21:36
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    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. – Cosmas Zachos Jun 23 '22 at 22:44
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    I also have a trouble here, why in the paragraph below (9.55), the book said "The functional determinant is the same as in Lorentz gauge, $\operatorname{det}\left(\delta G\left(A^\alpha\right) / \delta \alpha\right)=\operatorname{det}\left(\partial^2 / e\right)$". Why the determinant in this general case equal to that with Lorentz gauge? Could you please explain for me? – Daren Sep 28 '22 at 13:14
  • Hi @Daren. Do you mean that you have difficulty in convincing yourself that $\text{det}(\delta G(A^a)\delta \alpha)=\text{det}(\partial^2/e)$?? – schris38 Sep 30 '22 at 07:07
  • @schris38 Hello! I can convince myself that for Lorentz gauge, $\text{det}(\delta G(A^a)\delta \alpha)=\text{det}(\partial^2/e) $. I just thinking if this determinant is gauge invariant, since the book said for a general $\omega(x)$, still satisfy this. However, today I checked Srednicki's book, and basically understand that. Previously I may lost on Peskin's logic. If you have any comment, I appreciate it. – Daren Sep 30 '22 at 08:52

1 Answers1

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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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