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I am following Peskin & Schroeder to read a Feynman diagram. But in this given image, they used the non-relativistic limit to write the incoming fermion gamma mu product. How did he derive the bottom formula? Any explanations in detail?

enter image description here

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  • $\begingroup$ What is your question explicitly? $\endgroup$
    – aQuestion
    Mar 27, 2015 at 11:10
  • $\begingroup$ That seems pretty clear to me. The OP is asking how to derive the nonrelativistic limit form of the quoted relation (or equality) below. Simply enough. I don't say I have an answer though, but I can spend some time thinking about it. $\endgroup$ Dec 22, 2019 at 5:48

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How did he get it?

How did he get what? Which part of the calculation it is that's confusing you. You have listed multiple equalities, which one is at issue?

Eq 3.55 of the pdf online version of Peskin and Schroder shows that: $$ u^\dagger u=2E_p\xi^\dagger \xi $$ and since $$ \gamma_0^2=1 $$ we know that $$ \bar u\gamma_0u=u^\dagger u $$ and, in the non-relativistic limit (i.e. $p<<mc$) $$ E_p\to m\;. $$ (N.b., $c=1$)

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