The Tracy–Widom distributions admit many interpretations.

One of them is related to quantum mechanics: If we consider $N$ non-interacting fermions confined by the potential $V(x) = x^2$, then in the ground state, the position of the rightmost fermion (approximately) has probability density $P=\sqrt{2N} + \frac{N^{-1/6}}{\sqrt 2} F_2$ where $F_2$ is the PDF of the unitary Tracy-Widom distribution.

In the ground state, the wave-function $\Phi(x_1, \dots , x_N ) = \frac{1}{\sqrt{N!}} \det[\phi_i(x_j)]$ where $\phi _i$ runs over the first $N$ Hermitian polynomials and $j$ runs over $1 \dots N$. The probability density of observing the particles, $\Phi^2$, can be written as a determinant $\frac{1}{N!}\det[K_N(x_i,x_j)]$ where $K_N$ is a kernel depending on $N$. When $N\rightarrow \infty$, the kernel $K_N$ converges to the Airy kernel $K_A(u,v) = \int_0^{+\infty} Ai(u+x) Ai(v+x)dx$ at the edge location $\sqrt{2N}$ after scaling. This is why we have the Fredholm determinant representation $F_2=\det(1-K_A)_{L^2([s,+\infty])}$.

When the temperature is above zero, there is no determinantal structure for $\Phi^2$ when $N$ is fixed. However, if we consider the grand canonical potential, then there is a similar edge kernel $K_b(u,v)=\int_{-\infty}^{+\infty}\frac{ Ai(u+x) Ai(v+x)}{1+e^{-b x}}dx$ where $b$ is related to the temperature. Thus we call the Fredholm determinant $F_{2,b}=\det(1-K_b)_{L^2([s,+\infty])}$ the **Finite-temperature Tracy-Widom distribution** function.

**Question:** How can we find a similar distribution that degenerates to the orthogonal/symplectic Tracy-Widom distribution when $T\rightarrow 0$?

`\Ai`

is undefined. $\endgroup$