At the end of a Feynman diagram I have (p and p' are two incoming momenta): $$Tr[\not p' \gamma^\mu \not p \gamma^\nu(\frac{1+\gamma^5}{2})]=Tr[\gamma^\mu p_\mu' \gamma^\mu \gamma^\nu p_\nu \gamma^\nu(\frac{1+\gamma^5}{2})]$$ I think that the term with $\gamma^5$ doesn't contribute since there would be an odd number of gamma matrices inside the trace so the remaining is: $$\frac{1}{2}Tr[\gamma^\mu p_\mu' \gamma^\mu \gamma^\nu p_\nu \gamma^\nu]$$ And from this point forward I don't know how to find the result which is: $$2(p'^\mu p^\nu + p'^\nu p^\mu - g^{\nu\mu}p\cdot p'-i\epsilon^{\alpha \mu \beta \nu}p'_\alpha p_\beta)$$ Can someone explain this to me or point to a reference in which is well explained. (I'm currently using Peskin-Schroeder and I'm not getting too much out of it)

  • $\begingroup$ en.wikipedia.org/wiki/Gamma_matrices#Trace_identities $\endgroup$ May 8, 2018 at 16:25
  • $\begingroup$ Also the term with the $\gamma^5$ does not vanish. You have an even number of matrices. $\endgroup$
    – FrodCube
    May 8, 2018 at 16:38
  • $\begingroup$ Another thing to note is that you have too many of the same indices which leads to ambiguities in the summation, you should introduce different indices for the summations on the momenta, which when you write them as four vectors can be taken outside the trace while you are performing it. Then just use the identities earlier comments indicate $\endgroup$
    – Triatticus
    May 8, 2018 at 16:43
  • $\begingroup$ Thank you all for the help; the wikipedia page helped clarify a lot of things. By using the identities and changing the name to the index I arrived at a final result similar to the one provided by P-S but with a + instead of a - for the $\epsilon$ term while the other sign are right; any idea of why this is the case? $\endgroup$
    – Ringo_00
    May 8, 2018 at 17:02


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