The proof of the Wigner Semicircle Law comes from studying the GUE Kernel \[ K_N(\mu, \nu)=e^{-\frac{1}{2}(\mu^2+\nu^2)} \cdot \frac{1}{\sqrt{\pi}} \sum_{j=0}^{N-1}\frac{H_j(\lambda)H_j(\mu)}{2^j j!} \] The eigenvalue density comes from setting $\mu = \nu$. The Wigner semicircle identity is a Hermite polynomial identity \[ \rho(\lambda)=e^{-\mu^2} \cdot \frac{1}{\sqrt{\pi}} \sum_{j=0}^{N-1}\frac{H_j(\lambda)^2}{2^j j!} \approx \left\{\begin{array}{cc} \frac{\sqrt{2N}}{\pi} \sqrt{1 - \lambda^2/2N} & \text{if }|\lambda|< 2\sqrt{N} \\\\ 0 & \text{if }|\lambda| > 2 \sqrt{N} \end{array} \right. \] The asymptotics come from calculus identities like Christoffel-Darboux formula.

For finite size matrices the eigenvalue distribution is a semicircle yet.

Plotting the eigenvalues of a random $4 \times 4$ matrix, the deviations from semicircle law are noticeable with 100,000 trials and 0.05 bin size. GUE is in brown, GUE|trace=0 is in orange.

Axes not scaled, sorry!

Mathematica Code:

num[] := RandomReal[NormalDistribution[0, 1]] herm[N_] := (h = Table[(num[] + I num[])/Sqrt[2], {i, 1, N}, {j, 1, N}]; (h + Conjugate[Transpose[h]])/2) n = 4; trials = 100000; eigen = {}; Do[eigen = Join[(mat = herm[n]; mat = mat - Tr[mat] IdentityMatrix[n]/n ; Re[Eigenvalues[mat]]), eigen], {k, 1, trials}]; Histogram[eigen, {-5, 5, 0.05}] BinCounts[eigen, {-5, 5, 0.05}]; a = ListPlot[%, Joined -> True, PlotStyle -> Orange] eigen = {}; Do[eigen = Join[(mat = herm[n]; mat = mat; Re[Eigenvalues[mat]]), eigen], {k, 1, trials}]; Histogram[eigen, {-5, 5, 0.05}] BinCounts[eigen, {-5, 5, 0.05}]; b = ListPlot[%, Joined -> True, PlotStyle -> Brown] Show[a, b]

My friend asks if

**traceless**GUE ensemble $H - \frac{1}{N} \mathrm{tr}(H)$ can be analyzed. The charts suggest we should still get a semicircle in the large $N$ limit. For finite $N$, the oscillations (relative to semicircle) are very large. Maybe has something to do with the related harmonic oscillator eigenstates.

The trace is the average eigenvalue & The eigenvalues are being "recentered". We could imagine 4 perfectly centered fermions - they will repel each other. Joint distribution is: \[ e^{-\lambda_1^2 -\lambda_2^2 - \lambda_3^2 - \lambda_4^2} \prod_{1 \leq i,j \leq 4} |\lambda_i - \lambda_j|^2 \] On average, the fermions will sit where the humps are. Their locations should be more pronounced now that their "center of mass" is fixed.