# Traceless GUE : Four Centered Fermions

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, {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.

• The link to the image seems to be broken. – jeq Nov 10 '17 at 22:24

Traceless GUE was studied by Tracy and Widom in their paper "On the distributions of the lengths of the longest monotone subsequences in random words", Probab. Theory Relat. Field 119, 350-380 (2001). In Section 4.4 of that paper they basically show (actually for the largest eigenvalues, but the same argument applies to the average eigenvalue density) that $$\rho_k(\lambda) = \sqrt{\frac{k}{\pi}}\int_{-\infty}^{\infty}e^{-k \eta^2}\rho_k^0(\lambda-\eta)d\eta$$ where $\rho_k$ denotes the average eigenvalue density of $k\times k$ GUE and $\rho_k^0$ denotes the average eigenvalue density of the corresponding traceless ensemble. Thus, going from $\rho_k$ to $\rho_k^0$ amounts for solving the backward heat equation for time $1/4k$. This explains the more pronounced humps for small $k$ and will prove the semicircle law for traceless ensembles as $k\to\infty$. For fixed $k$ you can use Fourier transform calculations to get an explicit formula for $\rho_k^0$. That way I have obtained a plot for the $4\times 4$ ensemble:
Folkmar's answer gives much more detailed information, but here's a more quick-and-dirty answer. Assuming I'm reading your normalization right, for you the diagonal entries of $H$ are $N$ i.i.d. standard normal random variables, and it is $N^{-1/2} H$ which has an approximately semicircular eigenvalue distribution. I'll write $H' = H - N^{-1} \operatorname{tr} (H)$. Then, using $\lambda_j$ to denote the $j$th largest eigenvalue, $$\lambda_j(N^{-1/2} H') = \lambda_j(N^{-1/2} H) + N^{-1}Z,$$ where $Z = N^{-1/2}\operatorname{tr}(H)$ is a standard normal random variable. Starting from here it's easy to see that the eigenvalue distributions of $N^{-1/2}H'$ and $N^{-1/2}H$ have the same limit.
• @Folkmar: I don't think so, since that would require understanding the dependence between $\operatorname{tr}(H)$ and the individual eigenvalues. The localization is the main thing I had in mind is saying your answer gives more detailed information. – Mark Meckes Jan 30 '12 at 15:11