How can we prove that Compton scattering has two equivalent terms in the $S$-matrix expansion?

Question

Consider the Compton scattering $$e^{-}(p,s)+\gamma(k,\lambda)\rightarrow \gamma(k',\lambda')+e^{-}(p',s')$$ To calculate the process' amplitude one has to compute the matrix element $$S_{fi}=<f|\hat{S}^{(2)}|i>=<0|\hat{b}_{p',s'}\hat{a}_{k',\lambda'}\hat{S}^{(2)}\hat{b}_{p,s}\hat{a}_{k,\lambda}|i>$$ where the physical relevant contributions of the scattering matrix are the two following terms: $$\hat{S}^{(2)}=\frac{(-ie)^{2}}{2}∫ d^{4}x_{1}d^{4}x_{2}<0|T[\hat{\psi}_{1}\hat{\bar{\psi}}_{2}]|0>:\hat{\bar{\psi}}_{1}\gamma_{\mu}\gamma_{\nu}\hat{\psi}_{2}A_{1}^{\mu}A_{2}^{\nu}:$$

$$+\frac{(-ie)^{2}}{2}∫ d^{4}x_{1}d^{4}x_{2}<0|T[\hat{\bar{\psi}}_{1}\hat{\psi}_{2}]|0>:\gamma_{\mu}\hat{\psi}_{1}\hat{\bar{\psi}}_{2}\gamma_{\nu}A_{1}^{\mu}A_{2}^{\nu}:$$

It's then said that those two terms are the same up to exchange of $x_{1}\leftrightarrow x_{2}$. My question is, how can I show that? I'm searching for a mathematical proof, since this is always explained with the diagrams. I know that since we're integrating we can exchange them, but then I still have to prove that

$$\hat{\bar{\psi}}_{2}\gamma_{\mu}\gamma_{\nu}\hat{\psi}_{1}=\gamma_{\mu}\hat{\psi}_{1}\hat{\bar{\psi}}_{2}\gamma_{\nu}$$

when integrated. I'm not even sure about the equivalence between the Wick contracted terms. I've tried using simple spinor and Clifford algebra but I can't find the right path.

Answer 1

First, when you write the relevant matrix term, you have to keep the order of the gamma matrixes (also included in the electron propagator) so the first term is the following:

$$\hat{S}^{(2)}=\frac{(-ie)^{2}}{2}∫ d^{4}x_{1}d^{4}x_{2}\hat{\bar{\psi}}_{1}\gamma_{\mu}A_{1}^{\mu}<0|T[\hat{\psi}_{1}\hat{\bar{\psi}}_{2}]|0>\gamma_{\nu}A_{2}^{\nu}\hat{\psi}_{2}$$

$$+\frac{(-ie)^{2}}{2}∫ d^{4}x_{1}d^{4}x_{2} \hat{\psi}_{2}\gamma_{\mu}A_{2}^{\mu}\ <0|T[\hat{\psi}_{2}\hat{\bar{\psi}}_{1}]|0>\gamma_{\nu}A_{1}^{\nu}\hat{\psi}_{1}$$

The second term is obtained because $\hat{\psi}_{2}\hat{\psi}_{1} = -\hat{\psi}_{1}\hat{\psi}_{2}$ and you exchange two fermionic operators four times and $A_2^\nu \hat{\psi}_{2} = \hat{\psi}_{2}A_2^\nu$. After placing the operators of the second term in the same position as the operators of the first term, you can exchange $x_1$ and $x_2$ as they are integration variables over the same space.