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 $\slashed{p}\Sigma_1(p^2)+\Sigma_2(p^2)$. Here, $\Sigma_1$ and $\Sigma_2$ both have real and imaginary parts. The propagator then becomes

$\frac{Z(\slashed{p}+M)}{p^2-M^2}$

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 $Z(\slashed{p}+M)$ 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 $Z=\frac{1}{1-Re\Sigma_1(p^2)-iIm\Sigma_1(p^2)}$ equal to unity? Sorry for the many questions and thanks in advance.

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.