Wigner’s surmise

Consider a 2 x 2 GOE matrix $H_s=\left(\begin{array}{cc}{x}_1& x_3\\ x_3& x_2\end{array}\right)$ with $x_1,x_2\sim \mathcal{N}(0,1)$ and $x_3\sim \mathcal{N}(0,1/2)$. What is the pdf ( \rho(s) ) of the spacing $s={\lambda }_2-{\lambda }_1$ between its two eigenvalues (${\lambda }_2>{\lambda }_1$)?

The two eigenvalues are random variables, given in terms of the entries by the roots of the characteristic polynomial

${\lambda }^2-\text{Tr}(H_s)\lambda +\text{det}(H_s),$

Therefore ${\lambda }_{1,2}=(x_1+x_2\pm \sqrt{(x_1-x_2{)}^2+4x_3^2})/2$ and $s=\sqrt{(x_1-x_2{)}^2+4x_3^2}$.

By definition, we have

$$\rho (s)={\int }_{-\infty }^{\infty }{\int }_{-\infty }^{\infty }{\int }_{-\infty }^{\infty }\frac{e^{-\frac{1}{2}{x}_1^2}{e}^{-\frac{1}{2}{x}_2^2}{e}^{-x_3^2}}{\sqrt{2\pi }\sqrt{2\pi }\sqrt{\pi }}\delta \left(s-\sqrt{(x_1-x_2{)}^2+4x_3^2}\right)d{x}_{1}d{x}_{2}d{x}_3.$$

Changing variables as

$$\{\begin{array}{l}{x}_1-x_2=r\cos \theta \\ 2x_3=r\sin \theta \\ x_1+x_2=\psi \end{array}\Rightarrow \{\begin{array}{l}{x}_1=\frac{r\cos \theta +\psi }{2}\\ x_2=\frac{\psi -r\cos \theta }{2}\\ x_3=\frac{r\sin \theta }{2}\end{array}$$

and computing the corresponding Jacobian

$$J=\text{det}\left(\begin{array}{ccc}\frac{\partial x_1}{\partial r}& \frac{\partial x_1}{\partial \theta }& \frac{\partial x_1}{\partial \psi }\\ \frac{\partial x_2}{\partial r}& \frac{\partial x_2}{\partial \theta }& \frac{\partial x_2}{\partial \psi }\\ \frac{\partial x_3}{\partial r}& \frac{\partial x_3}{\partial \theta }& \frac{\partial x_3}{\partial \psi }\end{array}\right)=\text{det}\left(\begin{array}{ccc}\cos \frac{\theta }{2}& -\frac{r\sin \frac{\theta }{2}}{2}& \frac{1}{2}\\ -\cos \frac{\theta }{2}& \frac{r\sin \frac{\theta }{2}}{2}& \frac{1}{2}\\ \sin \frac{\theta }{2}& \frac{r\cos \frac{\theta }{2}}{2}& 0\end{array}\right)=-\frac{r}{4},$$

One obtains

$$\begin{gathered} p(s)=\frac{1}{8{\pi }^{3/2}}{\int }_0^{\infty }drr\delta (s-r){\int }_0^{2\pi }d\theta {\int }_{-\infty }^{\infty }d\psi e^{-\frac{1}{2}\left[{\left(\frac{r\cos \theta +\psi }{2}\right)}^2+{\left(\frac{\psi -r\cos \theta }{2}\right)}^2+\frac{r^2{\sin }^2\theta }{2}\right]} \\ =\frac{\sqrt{4\pi s}}{8{\pi }^{3/2}}{\int }_0^{2\pi }d\theta e^{-\frac{s^2}{2}\left[\frac{{\cos }^2\theta }{2}+\frac{{\sin }^2\theta }{2}\right]}=\frac{s}{2}{e}^{-\frac{s^2}{4}} \end{gathered}$$