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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.

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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.

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  • $\begingroup$ Mm still not sure how you switched the fermonic operators with the gamma matrices tbh. I mean those operators are built with spinors, so shouldn't commute right? $\endgroup$
    – Filippo
    Commented Jul 20, 2023 at 14:09
  • $\begingroup$ I only moved the operators: the fermionic operators anticommute with each other while a bosonic and a fermionic operators commute with each other. More details are given in many textbooks, such as Quantum Field Theory of Mandl & Shaw (Chapter 7 of the 2nd edition) $\endgroup$
    – nomeruk
    Commented Jul 25, 2023 at 10:45
  • $\begingroup$ I don't understand why u kept that order of gamma matrices. The Wick contraction are just c-numbers so they can be put out of the normal ordering, so doing that what you are left is what I've written in the question... can you convince me that is not the case? $\endgroup$
    – Filippo
    Commented Oct 4, 2023 at 9:33

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