# Dyson Brownian motion

In mathematics, the Dyson Brownian motion is a real-valued continuous-time stochastic process named for Freeman Dyson. Dyson studied this process in the context of random matrix theory.
There are several equivalent definitions:
Definition by stochastic differential equation:$d\lambda _{i}=dB_{i}+\sum _{1\leq j\leq n:j\neq i}{\frac {dt}{\lambda _{i}-\lambda _{j}}}$where $B_{1},...,B_{n}$ are different and independent Wiener processes. Start with a Hermitian matrix with eigenvalues λ$_1$(0), λ$_2$(0), ..., λ$_n$(0){\textstyle \lambda _{1}(0), \lambda _{2}(0), ..., \lambda _{n}(0)}, then let it perform Brownian motion in the space of Hermitian matrices. Its eigenvalues constitute a Dyson Brownian motion. This is defined within the Weyl chamber $W_{n}:=\{(x_{1},\dots ,x_{n})\in \mathbb {R} ^{n}:x_{1}<\dots <x_{n}\}$, as well as any coordinate-permutation of it.
Start with n{\textstyle n} independent Wiener processes started at different locations λ$_1$(0), λ$_2$(0), ..., λ$_n$(0){\textstyle \lambda _{1}(0), \lambda _{2}(0), ..., \lambda _{n}(0)}, then condition on those processes to be non-intersecting for all time. The resulting process is a Dyson Brownian motion starting at the same λ$_1$(0), λ$_2$(0), ..., λ$_n$(0){\textstyle \lambda _{1}(0), \lambda _{2}(0), ..., \lambda _{n}(0)}.

## Random matrix theory

In Random Matrix Theory, the Gaussian unitary ensemble is a fundamental ensemble. It is defined as a probability distribution over the space of A∈R$^{n×n}${\textstyle A\in \mathbb {R} ^{n\times n}} Hermitian matrices, with probability density function $\rho (A)\propto e^{-{\frac {1}{2}}tr(A^{2})}$.
Consider a Hermitian matrix A∈R$^{n×n}${\textstyle A\in \mathbb {R} ^{n\times n}}. The space of Hermitian matrices can be mapped to the space of real vectors R$^{n2}${\textstyle \mathbb {R} ^{n^{2}}}: $A\mapsto (A_{11},\dots ,A_{nn},{\sqrt {2}}Re(A_{12}),\dots ,{\sqrt {2}}Re(A_{n-1,n}),{\sqrt {2}}Im(A_{12}),\dots ,{\sqrt {2}}Im(A_{n-1,n}))$This is an isometry, where the matrix norm is Frobenius norm. By reversing this process, a standard Brownian motion in R$^{n2}${\textstyle \mathbb {R} ^{n^{2}}} maps back to a Brownian motion in the space of n×n{\textstyle n\times n} Hermitian matrices:$dA={\begin{bmatrix}dB_{11}&{\frac {1}{\sqrt {2}}}(dB_{12}+idB_{12}')&{\frac {1}{\sqrt {2}}}(dB_{13}+idB_{13}')&\cdots &{\frac {1}{\sqrt {2}}}(dB_{1n}+idB_{1n}')\\{\frac {1}{\sqrt {2}}}(dB_{12}-idB_{12}')&dB_{22}&{\frac {1}{\sqrt {2}}}(dB_{23}+idB_{23}')&\cdots &{\frac {1}{\sqrt {2}}}(dB_{2n}+idB_{2n}')\\{\frac {1}{\sqrt {2}}}(dB_{13}-idB_{13}')&{\frac {1}{\sqrt {2}}}(dB_{23}-idB_{23}')&dB_{33}&\cdots &{\frac {1}{\sqrt {2}}}(dB_{3n}+idB_{3n}')\\\vdots &\vdots &\vdots &\ddots &\vdots \\{\frac {1}{\sqrt {2}}}(dB_{1n}-idB_{1n}')&{\frac {1}{\sqrt {2}}}(dB_{2n}-idB_{2n}')&{\frac {1}{\sqrt {2}}}(dB_{3n}-idB_{3n}')&\cdots &dB_{nn}\end{bmatrix}}$The claim is that the eigenvalues of $A$ evolve according to$d\lambda _{i}=dB_{i}+\sum _{1\leq j\leq n:j\neq i}{\frac {dt}{\lambda _{i}-\lambda _{j}}}$
Since each dB{\textstyle dB} is on the order of O(√(dt)){\textstyle O({\sqrt {dt}})}, we can equivalently write dA=√(dt)G{\textstyle dA={\sqrt {dt}}G}, where G{\textstyle G} is a random Hermitian matrix where each entry is on the order of O(1){\textstyle O(1)}. By construction of the standard Brownian motion, dA{\textstyle dA} is independent of A{\textstyle A}, so G{\textstyle G} is independent of A{\textstyle A}, and can be written as $dA={\begin{bmatrix}X_{11}&{\frac {1}{\sqrt {2}}}(X_{12}+iY_{12})&{\frac {1}{\sqrt {2}}}(X_{13}+iY_{13})&\cdots &{\frac {1}{\sqrt {2}}}(X_{1n}+iY_{1n})\\{\frac {1}{\sqrt {2}}}(X_{12}-iY_{12})&X_{22}&{\frac {1}{\sqrt {2}}}(X_{23}+iY_{23})&\cdots &{\frac {1}{\sqrt {2}}}(X_{2n}+iY_{2n})\\{\frac {1}{\sqrt {2}}}(X_{13}-iY_{13})&{\frac {1}{\sqrt {2}}}(X_{23}-iY_{23})&X_{33}&\cdots &{\frac {1}{\sqrt {2}}}(X_{3n}+iY_{3n})\\\vdots &\vdots &\vdots &\ddots &\vdots \\{\frac {1}{\sqrt {2}}}(X_{1n}-iY_{1n})&{\frac {1}{\sqrt {2}}}(X_{2n}-iY_{2n})&{\frac {1}{\sqrt {2}}}(X_{3n}-iY_{3n})&\cdots &X_{nn}\end{bmatrix}}$ where each random variable X$_{ij}$, Y$_{ij}${\textstyle X_{ij}, Y_{ij}} is standard normal. In other words, G{\textstyle G} is distributed according to the GUE(n).
By the first and second Hadamard variation formulas and Ito’s lemma, we have $\lambda _{i}(t+dt)=\lambda _{i}(t)+{\sqrt {dt}}u_{i}^{*}Gu_{i}+dt\sum _{j\neq i}{\frac {\mathbb {E} [|u_{i}^{*}Gu_{j}|^{2}]}{\lambda _{i}-\lambda _{j}}}$
Since G{\textstyle G} is sampled from GUE(n), it is rotationally symmetric. Also, by assumption, the eigenvector u$_i${\textstyle u_{i}} has norm 1, so √(dt)u$_i$$^∗$Gu$_i${\textstyle {\sqrt {dt}}u_{i}^{*}Gu_{i}} has the same distribution as √(dt)e$_1$$^∗$Ge$_1${\textstyle {\sqrt {dt}}e_{1}^{*}Ge_{1}}, which is distributed as dB{\textstyle dB}.
Similarly, E[|u$_i$$^∗$Gu$_j$|$^2$]=1{\textstyle \mathbb {E} [|u_{i}^{*}Gu_{j}|^{2}]=1}.

### Infinitesimal generator

Define the adjoint Dyson operator:$D^{*}F:={\frac {1}{2}}\sum _{i=1}^{n}\partial _{\lambda _{i}}^{2}F+\sum _{1\leq i,j\leq n:i\neq j}{\frac {\partial _{\lambda _{i}}F}{\lambda _{i}-\lambda _{j}}}.$For any smooth function $F:\mathbb {R} ^{n}\to \mathbb {R}$ with bounded derivatives, by direct differentiation, we have the Kolmogorov backward equation $\partial _{t}\mathbb {E} [F]=\mathbb {E} [D^{*}F]$. Therefore, the Kolmogorov forward equation for the eigenspectrum is $\partial \rho =D\rho$, where $D$ is the Dyson operator by$D\rho :={\frac {1}{2}}\sum _{i=1}^{n}\partial _{\lambda _{i}}^{2}\rho -\sum _{1\leq i,j\leq n:i\neq j}\partial _{\lambda _{i}}\left({\frac {\rho }{\lambda _{i}-\lambda _{j}}}\right)$Let $\rho (t,\lambda )=\Delta _{n}(\lambda )u(t,\lambda )$, where $\Delta _{n}:=\prod _{i<j}(\lambda _{i}-\lambda _{j})$ is the Vandermonde determinant, then the time-evolution of eigenspectrum is equivalent to the time-evolution of $u$, which happens to satisfy the heat equation $\partial _{t}u={\frac {1}{2}}\sum _{i}\partial _{i}^{2}u$,
This can be proven by starting with the identity $\partial _{\lambda _{i}}\Delta _{n}=\Delta _{n}\sum _{1\leq j\leq n:i\neq j}{\frac {1}{\lambda _{i}-\lambda _{j}}}$, then apply the fact that the Vandermonde determinant is harmonic: $\sum _{i}\partial _{i}^{2}\Delta _{n}=0$.

### Johansson formula

Each Hermitian matrix with exactly two eigenvalues equal is stabilized by $U(2)\times U(1)^{n-2}$, so its orbit under the action of $U(n)$ has $\mathrm {dim} (U(n))-\mathrm {dim} (U(2)\times U(1)^{n-2})=n^{2}-n-2$ dimensions. Since the space of $n-1$ different eigenvalues is $(n-1)$-dimensional, the space of Hermitian matrix with exactly two eigenvalues equal has $n^{2}-3$ dimensions.
By a dimension-counting argument, $\rho$ vanishes at sufficiently high order on the border of the Weyl chamber, such that $u$ can be extended to all of $\mathbb {R} ^{n}$ by antisymmetry, and this extension still satisfies the heat equation.
Now, suppose the random matrix walk begins at some deterministic $A(0)$. Let its eigenspectrum be $\nu =\lambda (A(0))$, then we have $u(0,\lambda )={\frac {1}{\Delta _{n}(\nu )}}\sum _{\sigma \in S_{n}}(-1)^{|\sigma |}\delta (\lambda -\sigma (\nu ))$, so by the solution to the heat equation, and the Leibniz formula for determinants, we have
Johansson formula — Let A$_0${\textstyle A_{0}} be a Hermitian matrix with simple spectrum ν=(ν$_1$, …, ν$_n$){\textstyle \nu =\left(\nu _{1}, \ldots , \nu _{n}\right)}, let t>0{\textstyle t>0}, and let A$_t$=A$_0$+t$^{1/2}$G{\textstyle A_{t}=A_{0}+t^{1/2}G} where G{\textstyle G} is drawn from GUE. Then the spectrum λ=(λ$_1$, …, λ$_n$){\textstyle \lambda =\left(\lambda _{1}, \ldots , \lambda _{n}\right)} of A$_t${\textstyle A_{t}} has probability density function
$\rho (t,\lambda )={\frac {1}{(2\pi t)^{n/2}}}{\frac {\Delta _{n}(\lambda )}{\Delta _{n}(\nu )}}\operatorname {det} \left(e^{-\left(\lambda _{i}-\nu _{j}\right)^{2}/2t}\right)_{1\leq i,j\leq n}$
on the Weyl chamber.

## Harish-Chandra-Itzykson-Zuber integral formula

Dyson Brownian motion allows a short proof of the Harish-Chandra-Itzykson-Zuber integral formula.
Harish-Chandra-Itzykson-Zuber integral formula — If A, B{\textstyle A, B} have no repeated eigenvalues, and t{\textstyle t} is a nonzero complex number, then - $\int _{U(n)}\exp \left(t\operatorname {tr} \left(AUBU^{*}\right)\right)dU=c_{n}{\frac {\det \left[\exp \left(t\lambda _{i}(A)\lambda _{j}(B)\right)\right]_{1\leq i,j\leq n}}{t^{\left(n^{2}-n\right)/2}\Delta _{n}(\lambda (A))\Delta _{n}(\lambda (B))}}$
where U{\textstyle U} is integrated over the Haar probability measure of the unitary group U(n){\textstyle U(n)}, and c$_n$=∏$_{i=1}$$^n$i!{\textstyle c_{n}=\prod _{i=1}^{n}i!}.
Let the GUE(n) probability distribution over V$_n${\textstyle V_{n}} be defined as ρ(M)dM{\textstyle \rho (M)dM}, where dM=dM$_{11}$…dM$_{nn}$d(Re(M$_{12}$))…d(Im(M$_{n, n−1}$)){\textstyle dM=dM_{11}\dots dM_{nn}d(Re(M_{12}))\dots d(Im(M_{n, n-1}))}, and ρ(M)=C$_n$e$^{−1/2tr(M2)}${\textstyle \rho (M)=C_{n}e^{-{\frac {1}{2}}tr(M^{2})}} and C$_n${\textstyle C_{n}} is a constant. Similarly, the eigenvalue distribution for the GUE(n) is $\rho (\lambda )={\frac {1}{(2\pi )^{n/2}c_{n}}}\Delta _{n}(\lambda )^{2}e^{-\|\lambda \|_{2}^{2}/2}d\lambda$ where dλ=dλ$_1$…dλ$_n${\textstyle d\lambda =d\lambda _{1}\dots d\lambda _{n}}, and c$_n${\textstyle c_{n}} is another constant., and Δ$_n$=∏$_{1≤i<j≤n}$(λ$_i$−λ$_j$){\textstyle \Delta _{n}=\prod _{1\leq i<j\leq n}\left(\lambda _{i}-\lambda _{j}\right)} is the Vandermonde determinant.
If f:V$_n$→C{\textstyle f:V_{n}\to \mathbb {C} } is unitarily invariant, with sufficient regularity and decay, then it can be decomposed as f(M$_n$)=f(λ(M$_n$)){\textstyle f(M_{n})=f(\lambda (M_{n}))}. By Riesz representation theorem, there exists some function w{\textstyle w} such that ∫f(M)dM=∫f(λ)w(λ)dλ{\textstyle \int f(M)dM=\int f(\lambda )w(\lambda )d\lambda }, which by the above argument equals $w(\lambda )=\rho (\lambda )/\rho (M)={\frac {\Delta _{n}(\lambda )^{2}}{c_{n}C_{n}(2\pi )^{n/2}}}$
Given two such unitarily invariant functions f, g{\textstyle f, g} with sufficient regularity and decay, then consider their heat kernel convolution $X=\iint dAdB\;f(A){\frac {C_{n}}{t^{n^{2}/2}}}e^{-{\frac {tr(A-B)^{2}}{2t}}}g(B)$
We compute X{\textstyle X} in one way.
Let f$_t$(A):=f(A)e$^{−tr(A2)/2t}$, g$_t$(B):=g(B)e$^{−tr(B2)/2t}${\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 ${\begin{aligned}X&={\frac {C_{n}}{t^{n^{2}/2}}}\int _{V_{n}}\int _{V_{n}}f_{t}(A)e^{-{\frac {tr(AB)}{t}}}g_{t}(B)dAdB\\&={\frac {C_{n}}{t^{n^{2}/2}}}\int _{V_{n}}\int _{V_{n}}\int _{U(n)}f_{t}(A)e^{\frac {tr(AUBU^{*})}{t}}g_{t}(B)dAdBdU\\&={\frac {C_{n}}{t^{n^{2}/2}}}\int _{V_{n}}\int _{V_{n}}f_{t}(A)K_{t}(A,B)g_{t}(B)dAdB\\\end{aligned}}$ where we integrate over the Haar measure of the unitary group U(n){\textstyle U(n)}, and use the fact that g$_t$(B){\textstyle g_{t}(B)} is unitarily invariant, and we define the kernel
$K_{t}(A,B):=\int _{U(n)}e^{\frac {tr(AUBU^{*})}{t}}dU$
Since K$_t$, f$_t$, g$_t${\textstyle K_{t}, f_{t}, g_{t}} are all unitarily invariant, we have $X={\frac {1}{C_{n}(2\pi )^{n}c_{n}^{2}t^{n^{2}/2}}}\int _{W_{n}}\int _{W_{n}}f_{t}(\lambda )g_{t}(\nu )K_{t}(\lambda ,\nu )\Delta _{n}(\lambda )^{2}\Delta _{n}(\nu )^{2}d\lambda d\nu$
We compute X{\textstyle X} in another way.
Fix B{\textstyle B}, then set G=(A−B)/√t{\textstyle G=(A-B)/{\sqrt {t}}}, then we have $X=\int _{V_{n}}\int _{V_{n}}f(B+{\sqrt {t}}G)C_{n}e^{-{\frac {1}{2}}tr(G^{2})}dGdB$
Apply the Johansson formula, and convert the domain of integral to the Weyl chamber: $X={\frac {1}{C_{n}c_{n}(2\pi )^{n}t^{n/2}}}\iint f(\lambda )g(\nu )\Delta _{n}(\lambda )\Delta _{n}(\nu )\det \left[e^{-{\frac {(\lambda _{i}-\nu _{j})^{2}}{2t}}}\right]_{i,j}d\lambda d\nu$
Equate the two results, and simplify, we obtain the equality.

## Ginibre formula

Ginibre formula (Tao 2012, page 251) — $\rho (\lambda )={\frac {1}{(2\pi )^{n/2}1!\ldots n!}}e^{-|\lambda |^{2}/2}\left|\Delta _{n}(\lambda )\right|^{2}$ on the Weyl chamber.
The GUE is constructed as the t=1{\textstyle t=1} distribution when starting with A$_0${\textstyle A_{0}}. So we take t=1{\textstyle t=1} and ν→0{\textstyle \nu \to 0} in the Johansson formula.
$\rho (\lambda )={\frac {\Delta _{n}(\lambda )}{(2\pi )^{n/2}}}\lim _{\nu \to 0}{\frac {1}{\Delta _{n}(\nu )}}\operatorname {det} \left(e^{-{\frac {1}{2}}\left(\lambda _{i}-\nu _{j}\right)^{2}}\right)_{1\leq i,j\leq n}$
Since e$^{−(λi−νj)2/2}$=e$^{−λi2/2}$e$^{−νj2/2}$e$^{λiνj}${\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 $\operatorname {det} \left(e^{-\left(\lambda _{i}-\nu _{j}\right)^{2}/2}\right)_{1\leq i,j\leq n}=e^{-|\lambda |^{2}/2}e^{-|\nu |^{2}/2}\operatorname {det} \left(e^{\lambda _{i}\nu _{j}}\right)_{1\leq i,j\leq n}$
Now by a property of Vandermonde matrix, at the ν→0{\textstyle \nu \to 0} limit,
$\operatorname {det} \left(e^{\lambda _{i}\nu _{j}}\right)_{1\leq i,j\leq n}={\frac {1}{1!\ldots n!}}\Delta _{n}(\lambda )\Delta _{n}(\nu )+o\left(\Delta _{n}(\nu )\right)$ uniformly in ν{\textstyle \nu }.