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Consider a random hermitian matrix $B$ of size $N\times N$ with Gaussian probability measure given by

$$ dx(B) = e^{-\frac{N}{2}Tr(B^2)}\prod_{i=1}^N dB_{ii} \prod_{i<j} d\Re(dB_{ij})d\Im(dB_{ij}) $$ where $B_{ii}, \Re(B_{ij}), \Im(B_{ij})$ are independent Gaussian random variables.

How can we prove the following integral? I am looking for a detailed proof that involves reducing it to the usual Gaussian Matrix integral that we can do, thanks.

$$ \int dx(B) = 2^N \left(\frac{\pi }{N}\right)^{\frac{N^2}{2}} $$

Jeff Faraci
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    Where did you get this equation from? Are you asking about the normalization constant in the GUE? – Mike Hawk Nov 06 '18 at 16:38
  • @MikeHawk I'm not sure why it matters where the equation came from. It's a gaussian integral that needs to be solved. One place it arises is random matrix theory - yes, but thats not the only place. I am looking for a detailed proof. – Jeff Faraci Nov 06 '18 at 16:55
  • i'm asking because I don't think the equation is correct as written, in particular I think 2^n should be 2^{n/2} unless I have misinterpreted something about the setup – Mike Hawk Nov 06 '18 at 17:03

2 Answers2

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We have $Tr(B^2)=\sum_i\sum_j B_{ij}B_{ji}=\sum_i\sum_j \mid B_{ij}\mid^2=\sum_i B_{ii}^2+2\sum_{i<j}Re(B_{ij})^2+Im(B_{ij})^2$. Thus the integrand becomes $\prod_i e^{-nx_{ii}^2/2}dx_{ii}\prod_{i<j}e^{-2nr_{ij}^2/2}dr_{ij} e^{-2nI_{ij}^2/2}dI_{ij}$, by independence of the components (I used a different notation for the dummy variables just for simplicity). Then, each term in the first product is $\int e^{-nx^2/2}dx=(2\pi/n)^{1/2}$, while each term in the second is $((\pi/n)^{1/2})^2$. There are N terms in the first product and $N(N-1)/2$ in the second, so we get $(2\pi/n)^{n/2}(\pi/n)^{n(n-1)/2}=2^{n/2}(\pi/n)^{n^2/2}$

Edit: I also think it is incorrect to say that $dB_{ii}$ is a gaussian measure. If you assume the entries are jointly distributed according to dx (normalized to 1), then it is true they are independent gaussians, but in the equation defining dx, the measure $dB_{ii}$ is just Lebesgue measure on the line (try to replace it with a gaussian measure and you will get a different answer)

Mike Hawk
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  • Can you explain more and format this answer a little nicer? How do you know 'then, each term in the first product is...'? Where does that $n+1$ come from? For such a bounty I'm definitely looking for something a bit cleaner and detailed. – Jeff Faraci Nov 06 '18 at 17:19
  • To evaluate the integral, I just used the formula given here:https://en.wikipedia.org/wiki/Gaussian_integral. the n+1 was a typo, which i fixed – Mike Hawk Nov 06 '18 at 17:21
  • Why are you dividing the integrand by $\sqrt{2\pi}$, $\int e^{-nx^2/2}dx/\sqrt{2\pi}$? But when you do that integral its as if you ignore that $\sqrt{2\pi}$, otherwise you'd just get $1/\sqrt{N}$, not $(2\pi/N)^{1/2}$. – Jeff Faraci Nov 06 '18 at 17:31
  • you're right, that sqrt(2pi) shouldn't be there; corrected – Mike Hawk Nov 06 '18 at 17:42
  • Just to be clear, an $N\times N$ hermitian matrix has $N^2$ real parameters, coming from the diagonal , real part, and imaginary: $N+N(N-1)/2+N(N-1)/2=N^2$. So how come you dont have two factors of $N(N-1)/2$ for the product $\prod_{i – Jeff Faraci Nov 07 '18 at 03:20
  • I lumped both the real and imaginary parts in the product over i – Mike Hawk Nov 07 '18 at 15:26
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Your question involves a fairly standard suite of arguments in random matrices literature so I will provide you references which you can use. I don't know if there's a very short and elementary proof of the result that is not involving the steps i describe below

The first step is to reduce degrees of freedoms: in the Hermitian ensemble, any hermitian matrix $B=UDU^*$ with $U$ a unitary matrix and $D$ the eigenvalue matrix where $U$ and $D$ are independent. The change of variable $B\rightarrow(U,D)$ can be worked out in detail and reads

$$\Pi_{ij} d B_{ij} = VDM(\lambda_1,\ldots,\lambda_n)^2 \Pi_{i=1}^nd\lambda_i d\Omega $$

where $VDM$ is the vandermonde of the eigenvalues $\lambda_i$ and $d\Omega$ is the Haar measure on the unitary group. The proof is not straightforward and details are provided here

Lectures on Random Matrices

Now the particular integral with gaussian measure defines what is called the Gaussian Unitary Ensemble because the integrand is invariant under the unitary transformation so the integral to be evaluated becomes

$$\int \Pi_{i=1}^nd\lambda_i \exp\left(-\frac{N}{2}\sum_i\lambda_i^2\right)\Pi_{i<j}(\lambda_i-\lambda_j)^2$$

The next step to of the computation is to project the vandermonde polynomials onto the basis of orthogonal polynomials with respect to the gaussian measure, aka Hermite polynomials in this case.

This is worked out in details in the following 2 papers where the expected value of the product of 2 vandermonde is explained

Orthogonal polynomial ensembles in probability theory

Dimers and orthogonal polynomials: connections with random matrices

Ezy
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