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
5,353 answers , 22,786 comments
1,470 users with positive rep
820 active unimported users
More ...

  Why can we simply absorb the positive coefficient of $i\epsilon$ in a propagator?

+ 1 like - 0 dislike
2710 views

As far as I know, absorbing of the positive coefficient of $i\epsilon$ in a propagator seems to be a trivial operation without even the need of justification.

In Peskin page 286, he did this: $$k^0\rightarrow k^0(1+i\epsilon)$$ $$(k^2-m^2)\rightarrow (k^2-m^2+i\epsilon)$$

In M. Srednicki's Quantum Field Theory, page 51,

The factor in large parentheses is equal to $E^2-\omega^2+i(E^2+\omega^2)\epsilon$, and we can absorb the positive coefficient in to $\epsilon$ to get $E^2-\omega^2+i\epsilon$.

Why and does this kind of manipulation affect the final result of calculation?
Although $\frac{1}{k^2-m^2+i\epsilon k^2}-\frac{1}{k^2-m^2+i\epsilon}$ is infinitesimal, but the integration of such terms may lead to divergences, and this is my worry.

Also the presence of $k^0$ in the coefficient of $i\epsilon$ could potentially influence the poles of an integrand and consequently influence the validity of Wick Rotation.

This post imported from StackExchange Physics at 2014-05-04 11:30 (UCT), posted by SE-user LYg
asked Apr 28, 2014 in Theoretical Physics by LYg (15 points) [ no revision ]
retagged May 4, 2014
Looks to me like Srednicki's doing $\epsilon'=(E^2+\omega^2)\epsilon$ and then dropping the prime in the latter equation.

This post imported from StackExchange Physics at 2014-05-04 11:30 (UCT), posted by SE-user Kyle Kanos
@Kyle Kanos, But I don't think he can simply do that and pretend that ϵ′ doesn't depend on E anymore.

This post imported from StackExchange Physics at 2014-05-04 11:30 (UCT), posted by SE-user LYg
Why would $\epsilon'$ not depend on $E$ anymore? If $\epsilon$ is a function of $E$ then $\epsilon'$ is a function of $E$ as well.

This post imported from StackExchange Physics at 2014-05-04 11:30 (UCT), posted by SE-user Kyle Kanos
@Kyle Kanos, But their subsequent calculations all treat $\epsilon$ as independent of $k$, for example in the formalism of Wick Rotation, such dependence could influence the position of the poles in the $k^0$ plane and therefore influence the validity of Wick Rotation method.

This post imported from StackExchange Physics at 2014-05-04 11:30 (UCT), posted by SE-user LYg

1 Answer

+ 3 like - 0 dislike

The size of the parameter $\epsilon$ does not matter, as long as it is infinitesimally small. Rescaling it by that function does not change this. Recall that the whole procedure is just a mathematical trick which allows us to perform a contour integral over the real axis of the complex plane. The shift is really arbitrary, as long as it is small. The precise size is should not affect any results we gain from it.

This post imported from StackExchange Physics at 2014-05-04 11:30 (UCT), posted by SE-user Frederic Brünner
answered Apr 28, 2014 by Frederic Brünner (1,130 points) [ no revision ]

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:
$\varnothing\hbar$ysicsOverflow
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
...