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On Peskin and Schroeder's QFT book, page 296, the book give the functional integral formula after inserting Faddeev and Popov's trick of identity. $$ \int \mathcal{D} A e^{i S[A]}=\operatorname{det}\left(\frac{1}{e} \partial^2\right)\left(\int \mathcal{D} \alpha\right) \int \mathcal{D} A e^{i S[A]} \delta\left(\partial^\mu A_\mu-\omega(x)\right) .$$ where $\omega(x)$ is defined on (9.55) as a gauge-fixing function: $$G(A)=\partial^\mu A_\mu(x)-\omega(x).\tag{9.55} $$ I am troubled for following statement:

"This equality (my first equation) holds for any $\omega(x)$, so it will also hold if we replace the right-hand side (of my first equation) with any properly normalized linear combination involving different functions $\omega(x)$." $$N(\xi) \int \mathcal{D} \omega \exp \left[-i \int d^4 x \frac{\omega^2}{2 \xi}\right] \operatorname{det}\left(\frac{1}{e} \partial^2\right)\left(\int \mathcal{D} \alpha\right) \int \mathcal{D} A e^{i S[A]} \delta\left(\partial^\mu A_\mu-\omega(x)\right)\tag{9.56} $$ where $N(\xi)$ is an unimportant normalization constant.

So how to understand this linear combination?

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    $\begingroup$ In the first equation, the functional integral has the same value for each such $\omega$, so averaging that over $\omega$ still gives that value. If $f$ is a PDF, $\int cfdx=c$. $\endgroup$
    – J.G.
    Commented Sep 28, 2022 at 14:29
  • $\begingroup$ Thank you very much. In your last formula, does $c$ be the expectation value? Thanks. $\endgroup$
    – Daren
    Commented Sep 29, 2022 at 10:55
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    $\begingroup$ It's a constant, so is equal to its own expectation, yes. $\endgroup$
    – J.G.
    Commented Sep 29, 2022 at 11:16
  • $\begingroup$ @J.G. Thank you very much! Can I ask you a related quick question? In peskin's book page 295, they give Lorentz gauge $G\left(A^\alpha\right)=\partial^\mu A_\mu+(1 / e) \partial^2 \alpha$. And they said (9.55) in my post is a generalization of Lorentz gauge. I am puzzled for their statement below (9.55), "The functional determinant is the same as in Lorentz gauge, det$\left(\delta G\left(A^\alpha\right) / \delta \alpha\right)=\operatorname{det}\left(\partial^2 / e\right)$" Why (9.55) this general case's determinant equal to Lorentz case? Thank you take time to help, I am really appreciate it. $\endgroup$
    – Daren
    Commented Sep 29, 2022 at 12:17
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    $\begingroup$ What's constant is not $\omega$, but the first expression for $\int\mathcal{D}Ae^{iS[A]}$ in which it appears. As for why the determinant is the same, note $\delta G/\delta\alpha=\partial^2/e$. $\endgroup$
    – J.G.
    Commented Sep 29, 2022 at 12:39

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