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

180 submissions , 140 unreviewed
4,534 questions , 1,819 unanswered
5,158 answers , 21,954 comments
1,470 users with positive rep
720 active unimported users
More ...

  Calculating exactly the divergent part of amplitudes at all loop order

+ 0 like - 0 dislike
129 views

Suppose we have the $L$-loop amplitude of the form 

$$\mathcal{I}_L=\int \prod_{i=1}^L \frac{d^D q_i}{(2 \pi)^D} \frac{1}{q_i^2} \frac{1}{(p-\sum_{i=1}^L q_i)^2}.$$

Introducing Feynman parameters to merge the denominators as usual, we may write the amplitude in the form 

$$\mathcal{I}_L = (N-1)! \int_0^1 \prod_{j=1}^{L+1} dx_j \delta \left( \sum_{i=1}^{L+1} x_i-1\right) \int \prod_{i=1}^L \frac{d^D q_i}{(2 \pi)^D} \left[ q_i q_j M_{ij}-2 q_j K_j+J \right]^{-(L+1)}$$

where $M$ is a symmetric matrix. We can evaluate the $q$ integrals using the formula 

$$\int \frac{d^D l}{(2 \pi)^D}\frac{1}{(l^2-\Delta)^N} = \frac{(-1)^Ni}{(4 \pi)^{D/2}}\frac{\Gamma(N-D/2)}{\Gamma(N)}\left( \frac{1}{ \Delta}\right)^{N-D/2},$$

It can be shown that we can write $\mathcal{I}_L$ as 

$$\mathcal{I}_L=\frac{(-1)^L \Gamma(L+1-D)}{(4 \pi)^D} \int_0^1 \prod_{i=1}^{L+1} dx_i \delta \left( \sum_{i=1}^{L+1} x_i-1\right) \frac{\mathcal{U}^{L+1-3D/2}}{\mathcal{F}^{L+1-D}},$$

where we have defined  $\mathcal{U}=\det M$ and $ \mathcal{F}=\det M \left(  K_i M^{-1}_{ij} K_j -J \right).$

By setting $D=4-2 \epsilon$, my goal is to calculate the divergent part 

$$\mathcal{I}_L = \frac{c_L}{\epsilon} + \mathcal{O}(\epsilon^0)$$

for any $L$. With that aim in mind, let's first look at $L=2$. Doing the above computations and calculating $\mathcal{U}$ and $\mathcal{F}$ explicitly, amounts to 

$$c_L \propto \int dx_1 dx_2 dx_3 \delta \left( x_1+x_2+x_3-1\right) \frac{x_1 x_2 x_3}{(x_1 x_2 +x_2 x_3 + x_1 x_3)^3}.$$

The form for larger $L$ follows a similar pattern (product of $x_i$ in the numerator to some power, and sum of products of $x_i$ with one missing in each term, just like above), so evaluating the integral for $L=2$ would probably lead the way for a more general result. However, how does one treat integrals such as this? Any ideas about how one might try to evaluate it completely?

asked May 9 in Theoretical Physics by Valentina Pavlichenko (0 points) [ revision history ]

The sweeping under the carpet of divergencies is one of the reasons I never delved deeply into QFT, and therefore I restrict myself to a comment instead of an answer.

The \(\delta\)-function as such is pernicious evil, obscuring often what is meant. If the above is indeed a volume integral, then the result is 0 (taking the \(\delta\)-function as what it is usually described to be, infinity at one point and zero everywhere else, and using Lebesgue-integrals). If you take it as a distribution then things should be expressed differently, or one could employ a limiting procedure (which would point a way to a solution, as it makes clearer what is meant).

What probably is meant, however, is that there is a contribution to the integral only if \(x_1+x_2+x_3-1=0\), which then suggests that the integral is not intended as a volume integral but as a surface integral. Why not substitute \(x_3=1-x_1-x_2\) and integrate over \(x_1\)and \(x_2\) only?
 

Thanks for the comment! Indeed, I mean the latter. Of course, substituting for $x_3$ is the first thing I tried, but it is still not very tractable. I left it as is because it has a symmetric form that is similar to a more general expression for all $L$. But yes, it is essentially a surface integral. 

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$\hbar$ysics$\varnothing$verflow
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
...