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.


1 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.

  • $\begingroup$ If we take $\mathcal{O}(A)=F_{\mu\nu}$, what does the time ordering $\mathcal{T}$ mean? Since $F_{\mu\nu}$ has derivatives of $A^\mu$ in it. $\endgroup$ Commented Sep 5, 2022 at 18:02
  • $\begingroup$ The covariant time ordering $T_{\rm cov}$. $\endgroup$
    – Qmechanic
    Commented Sep 5, 2022 at 18:05

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