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 , 22,789 comments
1,470 users with positive rep
820 active unimported users
More ...

  Change of variable for functional integration

+ 4 like - 0 dislike
987 views

First of all, could someone suggest a book on Functional Integration with emphasis on QFT? It's difficult to find a good one and had really only access to Peskin & Schroeder (in which it isn't discussed in much length) and Ramamurti Shankar (which is really good but only covers the basics).

Second, I'm actually having trouble with the change of variable that P&S does in p.285,

\[\cal{D}\phi\left(x\right)=\prod_{i} d\phi\left(x_{i}\right)\]

\[\cal{D}\phi\left(x\right)=\prod _{k^{0}_{n}\gt0}d\Re\phi\left(k_{n}\right)d\Im\phi\left(k_{n}\right)\]

The first line is simply the definition of the functional integration. But in the second one he did a change of variable \(\phi\left(k\right)\rightarrow\{\Re\phi\left(k\right),\Im\phi\left(k\right)\}\)and restricted the time part of the momentum to positive values. I understand why this is needed but I didn't understand formally how to go from one to the other. What I understood is that he doubled the number of differentials so he restricted the time component to obtain the same overall number number of differentials, and, in the exponential (which is the functional integration's integrand), he doubled the number of variables when \(|\phi\left(k\right)|^{2}\rightarrow |\Re\phi\left(k\right)|^{2}+|\Im\phi\left(k\right)|^{2}\).  Any way, I need more information about what is actually happening.

Third, I had an idea on how to proceed in another way to solve the same functional integral but I need to know how to change variables, so, how do I proceed to change the integration variable if I want to let \(\phi\left(x\right)\rightarrow f\left(x\right)+g\left(x\right)\)?

Thanks

asked Nov 18, 2015 in Theoretical Physics by pniau7 (25 points) [ no revision ]
recategorized Nov 19, 2015 by Dilaton

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$\hbar$ysicsOver$\varnothing$low
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
...