Consider a 2 x 2 GOE matrix H s = ( x 1 x 3 x 3 x 2 ) H_s = \left( \begin{array}{cc}
x_1 & x_3 \\
x_3 & x_2 \\
\end{array}\right) H s = ( x 1 x 3 x 3 x 2 ) x 1 , x 2 ∼ N ( 0 , 1 ) x_1, x_2 \sim \mathcal{N}(0, 1) x 1 , x 2 ∼ N ( 0 , 1 ) x 3 ∼ N ( 0 , 1 / 2 ) x_3 \sim \mathcal{N}(0, 1/2) x 3 ∼ N ( 0 , 1/2 ) ρ ( s ) \rho(s) ρ ( s ) s = λ 2 − λ 1 s = \lambda_2 - \lambda_1 s = λ 2 − λ 1 λ 2 > λ 1 \lambda_2 > \lambda_1 λ 2 > λ 1
The two eigenvalues are random variables, given in terms of the entries by the roots of the characteristic polynomial
λ 2 − Tr ( H s ) λ + det ( H s ) , \lambda^2 - \text{Tr}(H_s)\lambda + \text{det}(H_s), λ 2 − Tr ( H s ) λ + det ( H s ) ,
Therefore λ 1 , 2 = ( x 1 + x 2 ± ( x 1 − x 2 ) 2 + 4 x 3 2 ) / 2 \lambda_{1,2} = (x_1 + x_2 \pm \sqrt{(x_1 - x_2)^2 + 4x_3^2})/2 λ 1 , 2 = ( x 1 + x 2 ± ( x 1 − x 2 ) 2 + 4 x 3 2 ) /2 s = ( x 1 − x 2 ) 2 + 4 x 3 2 s = \sqrt{(x_1 - x_2)^2 + 4x_3^2} s = ( x 1 − x 2 ) 2 + 4 x 3 2
By definition, we have
ρ ( s ) = ∫ − ∞ ∞ ∫ − ∞ ∞ ∫ − ∞ ∞ e − 1 2 x 1 2 e − 1 2 x 2 2 e − x 3 2 2 π 2 π π δ ( s − ( x 1 − x 2 ) 2 + 4 x 3 2 ) d x 1 d x 2 d x 3 . \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) dx_1 dx_2 dx_3. ρ ( s ) = ∫ − ∞ ∞ ∫ − ∞ ∞ ∫ − ∞ ∞ 2 π 2 π π e − 2 1 x 1 2 e − 2 1 x 2 2 e − x 3 2 δ ( s − ( x 1 − x 2 ) 2 + 4 x 3 2 ) d x 1 d x 2 d x 3 .
Changing variables as
{ x 1 − x 2 = r cos θ 2 x 3 = r sin θ x 1 + x 2 = ψ ⇒ { x 1 = r cos θ + ψ 2 x 2 = ψ − r cos θ 2 x 3 = r sin θ 2 \left\{ \begin{array}{l}
x_1 - x_2 = r \cos \theta \\
2x_3 = r \sin \theta \\
x_1 + x_2 = \psi \\
\end{array} \right.
\quad \Rightarrow \quad
\left\{ \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} \right. ⎩ ⎨ ⎧ x 1 − x 2 = r cos θ 2 x 3 = r sin θ x 1 + x 2 = ψ ⇒ ⎩ ⎨ ⎧ x 1 = 2 r c o s θ + ψ x 2 = 2 ψ − r c o s θ x 3 = 2 r s i n θ
and computing the corresponding Jacobian
J = det ( ∂ x 1 ∂ r ∂ x 1 ∂ θ ∂ x 1 ∂ ψ ∂ x 2 ∂ r ∂ x 2 ∂ θ ∂ x 2 ∂ ψ ∂ x 3 ∂ r ∂ x 3 ∂ θ ∂ x 3 ∂ ψ ) = det ( cos θ 2 − r sin θ 2 2 1 2 − cos θ 2 r sin θ 2 2 1 2 sin θ 2 r cos θ 2 2 0 ) = − r 4 , 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}, J = det ∂ r ∂ x 1 ∂ r ∂ x 2 ∂ r ∂ x 3 ∂ θ ∂ x 1 ∂ θ ∂ x 2 ∂ θ ∂ x 3 ∂ ψ ∂ x 1 ∂ ψ ∂ x 2 ∂ ψ ∂ x 3 = det cos 2 θ − cos 2 θ sin 2 θ − 2 r s i n 2 θ 2 r s i n 2 θ 2 r c o s 2 θ 2 1 2 1 0 = − 4 r ,
One obtains
p ( s ) = 1 8 π 3 / 2 ∫ 0 ∞ d r r δ ( s − r ) ∫ 0 2 π d θ ∫ − ∞ ∞ d ψ e − 1 2 [ ( r cos θ + ψ 2 ) 2 + ( ψ − r cos θ 2 ) 2 + r 2 sin 2 θ 2 ] = 4 π s 8 π 3 / 2 ∫ 0 2 π d θ e − s 2 2 [ cos 2 θ 2 + sin 2 θ 2 ] = s 2 e − s 2 4 p(s) = \frac{1}{8\pi^{3/2}} \int_0^\infty dr r \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}} p ( s ) = 8 π 3/2 1 ∫ 0 ∞ d r r δ ( s − r ) ∫ 0 2 π d θ ∫ − ∞ ∞ d ψ e − 2 1 [ ( 2 r c o s θ + ψ ) 2 + ( 2 ψ − r c o s θ ) 2 + 2 r 2 s i n 2 θ ] = 8 π 3/2 4 π s ∫ 0 2 π d θ e − 2 s 2 [ 2 c o s 2 θ + 2 s i n 2 θ ] = 2 s e − 4 s 2