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I calculated the Compton amplitude of three diagrams but how can I verify it under gauge invariance structure? $$ {\cal A} = 2 e^2 \left[ \frac{ p_3 \cdot \epsilon_1 p_2 \cdot \epsilon_4^* }{ p_2 \cdot p_4 } - \frac{ p_2 \cdot \epsilon_1 p_3 \cdot \epsilon_4^*}{ p_2 \cdot p_1 } + \epsilon_1 \cdot \epsilon_4^* \right] $$

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  • $\begingroup$ On this site all math is expected to be in MathJax. Scans of handwriting are not considered acceptable. Here is a MathJax tutorial. $\endgroup$
    – G. Smith
    Jul 24, 2020 at 0:05
  • $\begingroup$ Thank you so much 😊 $\endgroup$ Jul 24, 2020 at 0:06
  • $\begingroup$ Writing four 4-vectors next to each other doesn’t mean anything. $\endgroup$
    – G. Smith
    Jul 24, 2020 at 0:07
  • $\begingroup$ Oh no the p1 and p4 are the initial and final photon momenta and p2 and p4 is for electron and of course the epsilon is the polarization for photons. $\endgroup$ Jul 24, 2020 at 0:26
  • $\begingroup$ I was referring to your omission of the scalar products, which Prahar added when he did the MathJax for you. You mean $p_2$ and $p_3$ for the electron. $\endgroup$
    – G. Smith
    Jul 26, 2020 at 0:07

1 Answer 1

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Gauge invariance requires that the amplitude vanishes if you replace the polarization with its corresponding momentum. If we replace $\epsilon_4 \to p_4$, we get \begin{align} {\cal A} &\to 2 e^2 \left[ \frac{ p_3 \cdot \epsilon_1 p_2 \cdot p_4 }{ p_2 \cdot p_4 } - \frac{ p_2 \cdot \epsilon_1 p_3 \cdot p_4}{ p_2 \cdot p_1 } + \epsilon_1 \cdot p_4 \right]\\ &= 2 e^2 ( p_3 + p_4 - p_2 ) \cdot \epsilon_1 \\ &= 2 e^2 p_1 \cdot \epsilon_1 \\ &= 0 \end{align} In the second line, I have used the fact that $p_3 \cdot p_4 = p_1 \cdot p_2$ which is implied by momentum conservation. In the third, we have again used momentum conservation, $p_1 + p_2 = p_3 + p_4$.

You can try to verify gauge invariance by also replacing $\epsilon_1 \to p_1$.

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