Path integral quantization of the EM field in Peskin and schroeder

Question

I'm studying path integral quantization of the electromagnetic field using Peskin and Schroeder secdtion 9.4. We want to compute the functional integral

$$\tag{9.50} \int \mathcal{D}A\,e^{iS[A]}.$$

We use the method by Faddeev and Popov, let $$G(A)=\partial^\mu A_\mu(x)-\omega(x)$$ be the function that we wish to set to zero, use the following equation

$$\tag{9.53} 1=\int\mathcal{D}\alpha(x)\,\delta(G(A^\alpha))\text{det}\bigg(\frac{\delta G(A^\alpha)}{\delta\alpha}\bigg).$$

We stick 9.53 into 9.50, then integrate over $w(x)$ with respect to the Gaussian weight $\exp[-i\int d^4x \frac{\omega^2}{2\xi}]$, this shows that (9.50) is given by

$$\tag{9.56} N(\xi)\det\bigg(\frac{1}{e}\partial^2\bigg)\bigg(\int\mathcal{D}\alpha\bigg)\int\mathcal{D}A e^{iS[A]}\exp\bigg[-i\int d^4x\frac{1}{2\xi}(\partial^\mu A_\mu)^2\bigg].$$

Peskin and Schroeder then claims that we have worked out the denominator of

$$\langle \Omega|T\mathcal{O}(A)|\Omega\rangle=\lim_{T\rightarrow\infty}\frac{\int \mathcal{D}A\mathcal{O}(A)\exp[i\int_{-T}^Td^4x \mathcal{L}]}{\int \mathcal{D}A\exp[i\int_{-T}^Td^4x \mathcal{L}]}$$

We can write a similar expression for the numerator, then Peskin and Schroeder claims the "awkward constant factors in (9.56) cancel" and we find for its correlation function

$$\tag{9.57}\langle \Omega|T\mathcal{O}(A)|\Omega\rangle=\lim_{T\rightarrow\infty}\frac{\int \mathcal{D}A\mathcal{O}(A)\exp[i\int_{-T}^Td^4x (\mathcal{L}-\frac{1}{2\xi}(\partial^\mu A_\mu)^2)}{\int \mathcal{D}A\exp[i\int_{-T}^Td^4x (\mathcal{L}-\frac{1}{2\xi}(\partial^\mu A_\mu)^2)]}.$$

My questions are:

1. In (9.57), by "awkward constant factors have canceled", do we mean $N(\xi)\det(\frac{1}{e}\partial^2)(\int\mathcal{D}\alpha)$? That is are we treating $\int\mathcal{D}\alpha$ as a constant factor?

2. Peskin and Schroeder says we need $\mathcal{O}(A)$ to be gauge invariant, what is an example of a gauge invariant $\mathcal{O}(A)$? I don't think expressions like $A(x_1)A(x_2)$ which we use for scalar fields work here.

3. Peskin and Schroeder then claims the method by Faddeev and Popov shows that correlation function is independent the choice of $\xi$. But how? In (9.57) we clearly still have $\xi$ in it.

Answer 1

1. Yes and yes.

2. E.g. $\mathcal{O}(A)=F_{\mu\nu}$ is gauge invariant.

3. Independence of the gauge-fixing choice is e.g. discussed in this related Phys.SE post.