Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Suppose that I have a model of random $n \times n$ symmetric, positive definite matrices where the probability distribution is generated by the partition function:

$Z = \int dA \; e^{-V(A)}$

where $V(A)$ is left invariant under orthogonal transformations $A \rightarrow O^T \, A \, O$ and $dA$ is the Haar measure

$dA = \prod_i \, dA_{ii} \, \prod_{i<j} dA_{ij}$.

Because this problem can be reduced to finding the probability distribution for the eigenvalues, it seems intuitive to me that one may be able to show:

$\langle \det A \rangle = \det \langle A \rangle$.

Clearly this wouldn't be true for any random matrix theory as the determinant is a highly non-linear operation, but with the large $O(n)$ symmetry it at least seems possible.

Does anyone know if this is discussed anywhere? Or can someone see how to prove it or show that it is not true?

Thanks

share|improve this question
    
After thinking about this a bit I now feel it is not true. Nevertheless I would like to hear comments from people who actually know about this stuff. –  Kyle May 29 '12 at 6:21
add comment

1 Answer

This cannot be generally true. As you wrote, the problem reduces to a weighted integration over the eigenvalues; we don't have to determine the weighting because $V$ can be chosen arbitrarily to make up for it, so we have

$$Z=\int\mathrm d^n\lambda\,\mathrm e^{-V'(\lambda)}$$

with $\lambda$ the vector of eigenvalues and $V'$ an arbitrary function. If $n$ and $V'$ are even, then $\langle A\rangle=0$ and thus $\det\langle A\rangle=0$, but generally not $\langle\det A\rangle=0$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.