Dyson Brownian motion

Since each ${\textstyle dB}$ is on the order of ${\textstyle O({\sqrt {dt}})}$, we can equivalently write ${\textstyle dA={\sqrt {dt}}G}$, where ${\textstyle G}$ is a random Hermitian matrix where each entry is on the order of ${\textstyle O(1)}$. By construction of the standard Brownian motion, ${\textstyle dA}$ is independent of ${\textstyle A}$, so ${\textstyle G}$ is independent of ${\textstyle A}$, and can be written as $${\displaystyle dA={\begin{bmatrix}X_{11}&{\frac {1}{\sqrt {2}}}(X_{12}+iY_{12})&{\frac {1}{\sqrt {2}}}(X_{13}+iY_{13})&\cdots &{\frac {1}{\sqrt {2}}}(X_{1n}+iY_{1n})\\{\frac {1}{\sqrt {2}}}(X_{12}-iY_{12})&X_{22}&{\frac {1}{\sqrt {2}}}(X_{23}+iY_{23})&\cdots &{\frac {1}{\sqrt {2}}}(X_{2n}+iY_{2n})\\{\frac {1}{\sqrt {2}}}(X_{13}-iY_{13})&{\frac {1}{\sqrt {2}}}(X_{23}-iY_{23})&X_{33}&\cdots &{\frac {1}{\sqrt {2}}}(X_{3n}+iY_{3n})\\\vdots &\vdots &\vdots &\ddots &\vdots \\{\frac {1}{\sqrt {2}}}(X_{1n}-iY_{1n})&{\frac {1}{\sqrt {2}}}(X_{2n}-iY_{2n})&{\frac {1}{\sqrt {2}}}(X_{3n}-iY_{3n})&\cdots &X_{nn}\end{bmatrix}}}$$ where each random variable ${\textstyle X_{ij},Y_{ij}}$ is standard normal. In other words, ${\textstyle G}$ is distributed according to the GUE(n).

By the first and second Hadamard variation formulas and Ito’s lemma, we have $${\displaystyle \lambda _{i}(t+dt)=\lambda _{i}(t)+{\sqrt {dt}}u_{i}^{*}Gu_{i}+dt\sum _{j\neq i}{\frac {\mathbb {E} [|u_{i}^{*}Gu_{j}|^{2}]}{\lambda _{i}-\lambda _{j}}}}$$

Since ${\textstyle G}$ is sampled from GUE(n), it is rotationally symmetric. Also, by assumption, the eigenvector ${\textstyle u_{i}}$ has norm 1, so ${\textstyle {\sqrt {dt}}u_{i}^{*}Gu_{i}}$ has the same distribution as ${\textstyle {\sqrt {dt}}e_{1}^{*}Ge_{1}}$, which is distributed as ${\textstyle dB}$.

Similarly, ${\textstyle \mathbb {E} [|u_{i}^{*}Gu_{j}|^{2}]=1}$.

Let the GUE(n) probability distribution over ${\textstyle V_{n}}$ be defined as ${\textstyle \rho (M)dM}$, where ${\textstyle dM=dM_{11}\dots dM_{nn}d(Re(M_{12}))\dots d(Im(M_{n,n-1}))}$, and ${\textstyle \rho (M)=C_{n}e^{-{\frac {1}{2}}tr(M^{2})}}$ and ${\textstyle C_{n}}$ is a constant. Similarly, the eigenvalue distribution for the GUE(n) is $${\displaystyle \rho (\lambda )={\frac {1}{(2\pi )^{n/2}c_{n}}}\Delta _{n}(\lambda )^{2}e^{-\|\lambda \|_{2}^{2}/2}d\lambda }$$ where ${\textstyle d\lambda =d\lambda _{1}\dots d\lambda _{n}}$, and ${\textstyle c_{n}}$ is another constant., and ${\textstyle \Delta _{n}=\prod _{1\leq i<j\leq n}\left(\lambda _{i}-\lambda _{j}\right)}$ is the Vandermonde determinant.

If ${\textstyle f:V_{n}\to \mathbb {C} }$ is unitarily invariant, with sufficient regularity and decay, then it can be decomposed as ${\textstyle f(M_{n})=f(\lambda (M_{n}))}$. By Riesz representation theorem, there exists some function ${\textstyle w}$ such that ${\textstyle \int f(M)dM=\int f(\lambda )w(\lambda )d\lambda }$, which by the above argument equals $${\displaystyle w(\lambda )=\rho (\lambda )/\rho (M)={\frac {\Delta _{n}(\lambda )^{2}}{c_{n}C_{n}(2\pi )^{n/2}}}}$$

Given two such unitarily invariant functions ${\textstyle f,g}$ with sufficient regularity and decay, then consider their heat kernel convolution $${\displaystyle X=\iint dAdB\;f(A){\frac {C_{n}}{t^{n^{2}/2}}}e^{-{\frac {tr(A-B)^{2}}{2t}}}g(B)}$$

We compute ${\textstyle X}$ in one way.

Let ${\textstyle f_{t}(A):=f(A)e^{-{\frac {tr(A^{2})}{2t}}},g_{t}(B):=g(B)e^{-{\frac {tr(B^{2})}{2t}}}}$, then the quantity is $${\displaystyle {\begin{aligned}X&={\frac {C_{n}}{t^{n^{2}/2}}}\int _{V_{n}}\int _{V_{n}}f_{t}(A)e^{-{\frac {tr(AB)}{t}}}g_{t}(B)dAdB\\&={\frac {C_{n}}{t^{n^{2}/2}}}\int _{V_{n}}\int _{V_{n}}\int _{U(n)}f_{t}(A)e^{\frac {tr(AUBU^{*})}{t}}g_{t}(B)dAdBdU\\&={\frac {C_{n}}{t^{n^{2}/2}}}\int _{V_{n}}\int _{V_{n}}f_{t}(A)K_{t}(A,B)g_{t}(B)dAdB\\\end{aligned}}}$$ where we integrate over the Haar measure of the unitary group ${\textstyle U(n)}$, and use the fact that ${\textstyle g_{t}(B)}$ is unitarily invariant, and we define the kernel
$${\displaystyle K_{t}(A,B):=\int _{U(n)}e^{\frac {tr(AUBU^{*})}{t}}dU}$$

Since ${\textstyle K_{t},f_{t},g_{t}}$ are all unitarily invariant, we have $${\displaystyle X={\frac {1}{C_{n}(2\pi )^{n}c_{n}^{2}t^{n^{2}/2}}}\int _{W_{n}}\int _{W_{n}}f_{t}(\lambda )g_{t}(\nu )K_{t}(\lambda ,\nu )\Delta _{n}(\lambda )^{2}\Delta _{n}(\nu )^{2}d\lambda d\nu }$$

We compute ${\textstyle X}$ in another way.

Fix ${\textstyle B}$, then set ${\textstyle G=(A-B)/{\sqrt {t}}}$, then we have $${\displaystyle X=\int _{V_{n}}\int _{V_{n}}f(B+{\sqrt {t}}G)C_{n}e^{-{\frac {1}{2}}tr(G^{2})}dGdB}$$

Apply the Johansson formula, and convert the domain of integral to the Weyl chamber: $${\displaystyle X={\frac {1}{C_{n}c_{n}(2\pi )^{n}t^{n/2}}}\iint f(\lambda )g(\nu )\Delta _{n}(\lambda )\Delta _{n}(\nu )\det \left[e^{-{\frac {(\lambda _{i}-\nu _{j})^{2}}{2t}}}\right]_{i,j}d\lambda d\nu }$$

Equate the two results, and simplify, we obtain the equality.

The GUE is constructed as the ${\textstyle t=1}$ distribution when starting with ${\textstyle A_{0}}$. So we take ${\textstyle t=1}$ and ${\textstyle \nu \to 0}$ in the Johansson formula.

$${\displaystyle \rho (\lambda )={\frac {\Delta _{n}(\lambda )}{(2\pi )^{n/2}}}\lim _{\nu \to 0}{\frac {1}{\Delta _{n}(\nu )}}\operatorname {det} \left(e^{-{\frac {1}{2}}\left(\lambda _{i}-\nu _{j}\right)^{2}}\right)_{1\leq i,j\leq n}}$$

Since ${\textstyle e^{-\left(\lambda _{i}-\nu _{j}\right)^{2}/2}=e^{-\lambda _{i}^{2}/2}e^{-\nu _{j}^{2}/2}e^{\lambda _{i}\nu _{j}}}$, we have $${\displaystyle \operatorname {det} \left(e^{-\left(\lambda _{i}-\nu _{j}\right)^{2}/2}\right)_{1\leq i,j\leq n}=e^{-|\lambda |^{2}/2}e^{-|\nu |^{2}/2}\operatorname {det} \left(e^{\lambda _{i}\nu _{j}}\right)_{1\leq i,j\leq n}}$$

Now by a property of Vandermonde matrix, at the ${\textstyle \nu \to 0}$ limit,

$${\displaystyle \operatorname {det} \left(e^{\lambda _{i}\nu _{j}}\right)_{1\leq i,j\leq n}={\frac {1}{1!\ldots n!}}\Delta _{n}(\lambda )\Delta _{n}(\nu )+o\left(\Delta _{n}(\nu )\right)}$$ uniformly in ${\textstyle \nu }$.