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I'am trying to figure out the evaluation of the expectation value of the Wilson loop for QED. (Its actually the problem 15.3 in Peskin and Shroeder)

Lets say the Wilson loop is $W(x) = exp(-ie\oint_P A_{\mu}dx^{\mu})$

The expectation value is the path integral \begin{equation} \langle W(P) \rangle = \int DA_{\mu} exp\Big[iS[A_{\mu}] -ie \oint_P dx^{\mu}A_{\mu}\Big] \end{equation} where, \begin{equation} S[A_{\mu}] = \int d^4 x [-F_{\mu\nu}F^{\mu\nu} - \frac{1}{\zeta}(\partial^{\mu}A_{\mu})^2] \end{equation} Peskin and Shroeder have taken the term $\zeta$ as an arbitrary gauge parameter, with $\zeta=0$ being the Landau Gauge and $\zeta=1$ as the Feynman Gauge.

So far so good. I am actually lost in evaluating the integral $langle W(P)\rangle$. I looked at some online material giving the solution, and I came across one. It says " $\langle W(P)\rangle$ is simply a Gaussian integral, and can be worked out directly as (after taking the liming $\zeta \rightarrow 0$)" \begin{equation} \langle W(P) \rangle = exp\Big[ -\frac{1}{2}(-ie\oint_Pdx^{\mu})(-ie\oint_Pdy^{\nu})\int\frac{d^4k}{(2\pi)^4}\frac{-ig_{\mu\nu}}{k^2 + i\epsilon}e^{-k(x-y)} \Big] \end{equation}

Whoa, now I am not able to figure out how the last step was arrived upon from the previous step. Can someone help here, or if someone has a better way of proceeding from the previous step?

Basically, how do I go about solving the expectation value of the Wilson loop? Any clues/hints?

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    $\begingroup$ To start with, try to evaluate the integral $$\int d^dx exp\left (- x_i A_{ij} x_j+x_i J_i\right) $$ You should find something proportional to $exp(-1/4 J_i A^{-1}_{ij} J_j) $. Next, rewrite the kinetic terms in a gaussian form $$- 1/4 F_{\mu \nu} ^2 - \frac{1}{2\xi}(\partial_\mu A^\mu)^2\sim \int d^4 k A(k) _\mu\left(k^2 \eta_{\mu \nu} - k_\mu k_\nu (1-1/\xi)\right) A(k) _\nu$$ You can then evaluate your expectation value in the same way as the first integral (with appropriate substitutions for A and J) $\endgroup$ – Lunaron Jul 16 at 15:56

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