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Let $M$ come from an ensemble of $N\times N$ matrices. The Wigner surmise is density function $p^W_0(s)=\frac{\pi}{2}se^{-\pi s^2/4}$. From a random matrix point of view, we can write $\rho^W_0(s)=\frac{d^2}{ds^2}E((0,s))$, where $E(I)$ is the eigenvalue gap probability: $M$ has no eigenvalues in interval $I$. $E(I)$ and it's derivatives are intimately related to the correlations between nearest neighbors.

Question 1: What are known random matrix ensembles which have their eigenvalue gap probability (in limit) exactly equal to the Wigner surmise? To be specific: either all $N\times N$ matrices have Wigner surmise OR the limiting eigenvalue distribution is exactly Wigner surmise.

Question 2: What are known interacting-particle systems which have their particle gap probability (in limit) exactly equal to the Wigner surmise?To be specific: either all $N\times N$ particle systems exhibit Wigner surmise OR the limiting particle distribution is exactly Wigner surmise.

One example that I've seen is the real Ginibre ensemble which takes a random Gaussian matrix and focuses only on real eigenvalues. Then the probability of there being an even number of eigenvalues in $[0,s]$ matches the Wigner surmise. This is equivalent to certain statistics of creation/annihilation processes on the line. In addition to this, there are some statistical physics spin systems which seem to give an exact surmise as well. Unfortunately I'm not an expert in the latter area.

Some more background:

It's a well known fact that many random matrix ensembles exhibit a (limiting) density function of the form:

$$p_0(s)=\frac{2u(\pi^2 s^2/4)}{s}\exp\left(-\int_0^{\pi^2 s^2/4}\frac{u(t)}{t}dt\right),$$

where $u$ satisfies a Painleve equation and of which $\rho_0^W(s)$ is a special case. So in short, $\rho^W_0(s)$ is usually an approximation, not an exact answer. One can certainly derive some conditions on $u$ and the resulting Painleve equation to get an exact Wigner surmise but this doesn't answer from which random matrix ensembles it comes from.

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  • $\begingroup$ one answer to question 1 is the Gaussian ensemble of 2x2 matrices; this is how Wigner arrived at his surmise... $\endgroup$ – Carlo Beenakker Oct 3 '14 at 20:55
  • $\begingroup$ @CarloBeenakker: That's cute! I completely forgot about that example. I've added emphasis that the matrices don't need to have exact Wigner surmise, but their limiting distribution should be exact. $\endgroup$ – Alex R. Oct 3 '14 at 21:08

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