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I have a question about residues in QFT. I am calculating a fermionic loop and the expression I obtained for the loop has the form $\not{p}\Sigma_1(p^2)+\Sigma_2(p^2)$. Here, $\Sigma_1$ and $\Sigma_2$ both have real and imaginary parts. The propagator then becomes

$$\frac{Z(\not{p}+M)}{p^2-M^2}$$

where $Z=(1-\Sigma_1)^{-1}$ and $M=(M_0+\Sigma_2)Z$, $M_0$ being the bare mass. I was under the impression that $Z(\not{p}+M)$ represents the residue of the pole. However, my advisor told me that we want the real part of the residue to be equal to unity and that therefore the residue is not $Z=(1-\Sigma_1)^{-1}$.

So, is the residue just $Z$? And should I set the real part of $$Z=\frac{1}{1-\operatorname{Re}\Sigma_1(p^2)-i\operatorname{Im}\Sigma_1(p^2)}$$ equal to unity? Sorry for the many questions and thanks in advance.

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  • $\begingroup$ Good question! You want the residue with respect to $z\equiv \not\! p$, not with respect to $z=p^2$. The details are spelled out in this answer of mine. Hope you find it useful. $\endgroup$
    – AccidentalFourierTransform
    Commented May 5, 2018 at 0:01
  • $\begingroup$ Thank you for your reply. Basically, what you obtained is the same as mine, with my Sigmas 1 and 2 replaced by a and b. However, I am still confused as to how you obtained the line after Eq.(1). Are you expanding alpha in powers of m? $\endgroup$
    – christianwos
    Commented May 6, 2018 at 1:53
  • $\begingroup$ I'm expanding both $a(p^2)$ and $b(p^2)$ around $p^2=m^2$. $\endgroup$
    – AccidentalFourierTransform
    Commented May 6, 2018 at 2:18
  • $\begingroup$ @AccidentalFourierTransform I figured that out. Thanks. So Eq.(2) is the condition for the real part of the residue being unity? And is there another way of obtaining this condition on the residue without expanding around $m^2$? $\endgroup$
    – christianwos
    Commented May 6, 2018 at 3:08
  • $\begingroup$ 1) Eq. (2) is the condition for the residue being unity, not its real part. But in general the residue is real anyway; only beyond the threshold of pair-production you get an imaginary part. 2) What other way could there be? A residue is, by definition, what you get when you expand around the singularity. How do you expect to get a condition about the residue without expanding around $m^2$? $\endgroup$
    – AccidentalFourierTransform
    Commented May 6, 2018 at 3:27

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