I generate an \( n \times n \) random matrix in the GUE with the following code

RandomMatrixGUE[n_] := Module[{A, B, H},

A = RandomVariate[NormalDistribution[], {n, n}];

B = RandomVariate[NormalDistribution[], {n, n}];

H = 1/2 (A + Transpose[A]) + I / 2 (B - Transpose[B]);

H]

Given a matrix \( H \), I calculate the sum \( \sum_{ i \neq j} \frac{1}{ \left( \lambda_i - \lambda_j \right)^2} \) withf[H_] := Module[{lambda = Eigenvalues[H], i, j, n = Length[H], result = 0},

For[i = 1, i <= n, i++,

For[j = 1, j < i, j++, result += 1/ (lambda[[i]] - lambda[[j]])^2]];

` 2*result] `

I then use

`randomsamples[n_, NSamples_] := `

Table[A = RandomMatrixGUE[n]; f[A], {k, NSamples}]and

`SeedRandom[1]`

For [n = 2, n <= 5, n++,

NSamples = 20000;

result = randomsamples[n, NSamples];

vev = Mean[result];

vev = NumberForm[vev, {3, 2}];

std = StandardDeviation[result]/Sqrt[NSamples];

std = NumberForm[std, {3, 2}];

exact = 1/2 n (n - 1);

Print["\n n = ", n, ": exact = ", exact, ", Monte Carlo gives ",

vev, " with std ", std];

difference = Abs[vev - exact];

If[difference <= 3 std,

Print["the difference is smaller than 3 std"],

Print["The Monte Carlo result is different from exact result, \

namely ", difference/std, " standard deviations"]];]

The last code produces the following output:

n = 2: exact = 1, Monte Carlo gives 0.94 with std 0.02

n = 3: exact = 3, Monte Carlo gives 3.20 with std 0.31

n = 4: exact = 6, Monte Carlo gives 6.06 with std 0.14

n = 5: exact = 10, Monte Carlo gives 10.30 with std 0.48

The Monte Carlo results above lie within 3 standard deviations of the exact value for \( n = 2,3,4,5 \). This is thus a good test on \eqref{eq:20151201a}.

__Remarks__:

- The proof of \eqref{eq:20151201a} can be found in the chapter on Brownian motion in Mehta's book. The proof is based on setting up an Ornstein-Uhlenbeck process and then analyzing the related heat equation. The argument is directly taken from a paper by Dyson. I do not know if there is a more direct way of calculating \eqref{eq:20151201a}.
- Illustrations of the Dyson Ornstein-Uhlenbeck process can be found in a previous post.

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