Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

PO is now at the Physics Department of Bielefeld University!

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

205 submissions , 163 unreviewed
5,082 questions , 2,232 unanswered
5,354 answers , 22,792 comments
1,470 users with positive rep
820 active unimported users
More ...

  Why does this condition ensure that the residue of the propagator is 1?

+ 3 like - 0 dislike
1591 views

The corrected propagator is given by $$\Delta'(q)=\frac{1}{q^2+m^2-\Pi^*(q^2)-i\epsilon}$$ ($\Pi^*$ is the sum of all irreducible one-particle amplitudes) I get that the residue of the original propagator around the pole $q^2=-m^2$ is $$\frac{1}{2\pi i}\oint_{\text{around }q^2=-m^2} \frac{dq^2}{q^2+m^2-i\epsilon}=\lim_{q^2\rightarrow -m^2}\frac{q^2+m^2}{q^2+m^2}=1$$ and that the corrected propagator must have the same residue $$\frac{1}{2\pi i}\oint \Delta'(q)dq^2=1$$ So how does the condition $$\left[\frac{d\Pi^*(q^2)}{dq^2}\right]_{q^2=-m^2}=0$$ ensure the second integral above?

EDIT: Devouring complex analysis literature. Have already edited some things that weren't quite right. For anyone interested, I'm using Weinberg Vol 1 and this in section 10.3, ~p. 430.

This post imported from StackExchange Physics at 2014-09-16 10:47 (UCT), posted by SE-user 0celo7
asked Sep 15, 2014 in Theoretical Physics by 0celo7 (50 points) [ no revision ]
You know that the denominator vanishes at the pole mass. Just Taylor expand the denominator around that point, and then you get your condition for the residues by taking the limit as in the unperturbed propagator.

This post imported from StackExchange Physics at 2014-09-16 10:47 (UCT), posted by SE-user TwoBs
Does this also imply that all higher derivatives of $\Pi^*$ vanish at the mass shell as well? The first term in the denominator's Taylor series is obviously 0. The second term is $[1-d\Pi^*/dq^2]_{-m^2}(q^2+m^2)$. Integrating this obviously leads to the desired condition, but how can I be sure that the series terminates after the first-order $q^2$ derivative?

This post imported from StackExchange Physics at 2014-09-16 10:47 (UCT), posted by SE-user 0celo7
The higher order terms in the expansions vanish faster than the $q^2+m^2$ term, and so they do not contribute to the limit. Just take the limit.

This post imported from StackExchange Physics at 2014-09-16 10:47 (UCT), posted by SE-user TwoBs
Yeah thanks. It's essentially a case of L'Hospital's rule.

This post imported from StackExchange Physics at 2014-09-16 10:47 (UCT), posted by SE-user 0celo7

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
p$\hbar$ysicsOverflo$\varnothing$
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
Please complete the anti-spam verification




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...