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  Log-interaction term calculation

+ 3 like - 0 dislike
1621 views

I have a question regarding calculating the following integral with cutoff. dω|ω|cos(ω(τiτj)1)

How should I set up the correct cutoff so that the result of the integral becomes: 2lnτiτjτc

where τc is the short time cutoff~E1F.

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 , abbreviate τiτj by t, differentiate with respect to t to get rid of the denominator, use cosx=(eix+eix)/2, and change eiz to eizϵω to be able to perform the integration. Then set ϵ=0 and integrate the result from t=τc0 to t=τi-τj

answered Apr 13, 2014 by Arnold Neumaier (15,797 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.

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