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
5,355 answers , 22,793 comments
1,470 users with positive rep
820 active unimported users
More ...

  Matrix integral identity

+ 7 like - 0 dislike
1594 views

1) How to prove that $N\times N$ matrix integral over complex matrices $Z$ $$ \int d Z d Z^\dagger e^{-Tr Z Z^\dagger} \frac{x_1\det e^Z -x_2 \det e^{AZ^\dagger}}{\det(1-x_1e^Z)\det(1-x_2e^{AZ^\dagger})} $$ does not depend on the external Hermitian matrix $A$? $x_1$ and $x_2$ are numbers. The statement is trivial for $1\times1$ case.

2)The same for

$$ \int d Z d Z^\dagger e^{-Tr Z Z^\dagger} \frac{x_1\det e^Z -x_2 \det e^{AZ^\dagger}}{\det(1-x_1e^Zg)\det(1-x_2e^{AZ^\dagger}g)} $$

where g - arbitrary $GL(N)$ matrix.

This post imported from StackExchange MathOverflow at 2014-07-29 11:48 (UCT), posted by SE-user Sasha
asked Nov 29, 2010 in Mathematics by Sasha (110 points) [ no revision ]
retagged Jul 29, 2014
I understand that $dZ$ is the Lebesgue measure on $N\times N$ complex matrices, that is $dZ=\prod_{i,j}d\Re z_{ij}d\Im z_{ij}$, but what $dZ^\dagger$ stands for ?

This post imported from StackExchange MathOverflow at 2014-07-29 11:48 (UCT), posted by SE-user Adrien Hardy
This is just a notation sometimes used in the mathphys literature to show that you integrate over $2N^2$ real variables contrary to $N^2$ for the Hermitian model.

This post imported from StackExchange MathOverflow at 2014-07-29 11:48 (UCT), posted by SE-user Sasha

How exactly is the integral defined? Isn't the denominator always going to hit 0? For instance in the 1 by 1 case, the denominator contains $1-x_1 e^z$, where $z$ is a complex number, and then at the points $z=\log (x_1)^{-1}$ the integrand diverges.

If it were true, you would just do a unitary transformation (which leaves the measure invariant) to diagonalize A, and then rescale the coordinates to get rid of the dependence on the eignevalues. But this integral is A dependent in the 1 by 1 case, which is not trivial.

Did you try to prove it for $N=1$? If it does not hold in that case I do not think it is valid for $N>1$.

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$\varnothing$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
...