1

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.

Gold
  • 35,872
christianwos
  • 23
  • 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. – AccidentalFourierTransform May 05 '18 at 00:01
  • 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? – christianwos May 06 '18 at 01:53
  • I'm expanding both $a(p^2)$ and $b(p^2)$ around $p^2=m^2$. – AccidentalFourierTransform May 06 '18 at 02:18
  • @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 $m^2$? – christianwos May 06 '18 at 03:08
  • 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 $m^2$?
    • – AccidentalFourierTransform May 06 '18 at 03:27
  • @AccidentalFourierTransform I see. I thought there was some particular way of finding the residue in QFT, but it makes sense that we use the standard definition of a residue from complex analysis. Thanks again. – christianwos May 06 '18 at 12:34
  • @AccidentalFourierTransform One more question. How does the residue condition, $\Sigma'(m)=0$ change the form of the propagator? For the pole I set $\Sigma_i=\Sigma_i(p^2)-\Sigma_i(m^2)$, $i=1,2$, which ensures that $\Sigma$ vanished when $p^2=m^2$, but I fail to see how setting the condition for the residue changes the form of the propagator. – christianwos May 06 '18 at 13:17
  • $\Sigma_i(\not p)\to\Sigma_i(\not p)-\Sigma_i(m)-(\not p-m)\Sigma'_i(m)$. – AccidentalFourierTransform May 07 '18 at 16:59
  • @AccidentalFourierTransform You are using the on-shell condition. However, there is a more general way of imposing conditions on the two functions, I believe. Asking for the pole to be at $M^2$ and for the residue to be unity, gives $M\Sigma_1(M^2)+\Sigma_2(M^2)=0$ and $M\Sigma'_1(M^2)+\Sigma'_2(M^2)=-\frac{\Sigma_1(M^2)}{2M}=\frac{\Sigma_2(M^2)}{2M^2}$. This is more useful to me, since I want to plot the renormalized propagator and I wouldn't know what to do with a $\not{p}$ in the expression. However, I am having a hard time imposing these two conditions satisfactorily. Thanks again. – christianwos May 15 '18 at 13:15