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

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

145 submissions , 122 unreviewed
3,930 questions , 1,398 unanswered
4,852 answers , 20,624 comments
1,470 users with positive rep
501 active unimported users
More ...

Log-interaction term calculation

+ 3 like - 0 dislike
98 views

I have a question regarding calculating the following integral with cutoff. $$ \int_{-\infty}^{\infty} \frac{d\omega}{|\omega|} \cos(\omega(\tau_i-\tau_j)-1)$$

How should I set up the correct cutoff so that the result of the integral becomes: $$-2\ln \frac{\tau_i-\tau_j}{\tau_c}$$ where $\tau_c$ is the short time cutoff~$E_F^{-1}$.

This calculation is from certain resonant tunneling problem of Luttinger liquid, for which you can obtain renormalization group flow equation.

This post imported from StackExchange Physics at 2014-04-13 11:24 (UCT), posted by SE-user huyichen
asked Apr 13, 2014 in Theoretical Physics by huyichen (40 points) [ no revision ]
retagged Apr 13, 2014 by huyichen

1 Answer

+ 3 like - 0 dislike

rewrite the cosine in terms of exponenials, and add a tiny real part to the exponents.

In more detail, rewrite the integral as twice the integral from 0 to $\infty$, abbreviate $\tau_i-\tau_j$ by $t$, differentiate with respect to $t$ to get rid of the denominator, use $cos x = (e^{ix}+e^{-ix})/2$, and change $e^{iz}$ to $e^{iz-\epsilon\omega}$ to be able to perform the integration. Then set $\epsilon=0$ and integrate the result from $t=\tau_c \approx 0$ to $t=\tau_i$-$\tau_j$. 

answered Apr 13, 2014 by Arnold Neumaier (12,355 points) [ revision history ]
edited Apr 15, 2014 by Arnold Neumaier

Could you explain a little bit more in detail?

I added some details.

So how do you handle the 0 for integral of (1/t), wouldn't that produce -infinity there?

In the final integration w.r. to $t$, one needs the cutoff mentioned.

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$\hbar$ysic$\varnothing$Overflow
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).
To avoid this verification in future, please log in or register.




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

Your rights
...