The generator an OU process is given by


$$A = \operatorname{tr}(QD^2)+\langle Bx,D\rangle.$$


This one possesses an invariant measure given by
$$d\mu(x) = b(x) \ dx \text{ with } b(x) = \frac{1}{(4\pi)^{N/2} \vert Q\_{\infty} \vert^{1/2}} e^{-\langle Q\_{\infty}^{-1}x,x \rangle /4},$$
where $Q\_{\infty}= \int\_0^{\infty} e^{sB} Q e^{sB^\*} \ ds.$


I read that $A$ is self-adjoint on $L^2(\mathbb R^n, d\mu)$ if and only if $QA^\*=AQ,$ but I have never seen an argument for this.


In particular, what I am mainly interested in, is there anywhere an explicit representation of the adjoint operator of $A$?



Adjoint operator of OU generator
The argument is a bit tricky. Assume for simplicity that $Q=I$, then $BQ\_\infty+Q\_\infty B^\*=-I$. Next decompose $A$ as a self-adjoint part plus a remainder, namely introduce the form $a(u,v)=\int\_{\mathbb R^n} \nabla u \nabla v \, d\mu$ which corresponds to $A\_1=\Delta-\frac 12 Q\_\infty ^{-1} x \cdot \nabla$ and write $A=A\_1+C$ with $C=(Bx+\frac 12 Q\_\infty^{-1} x) \cdot \nabla$.


A computation shows that $C$ is skew-adjoint in $L^2\_\mu$, so that $A$ is self-adjoint iff $C=0$ or $B=-\frac 12 Q\_\infty^{-1}$, in particular $B=B^\*$. On the other hand, if $B=B^\*$, we can diagonalize $B$ and check that $B=-\frac12 Q\_\infty^{-1}$ so that $A=A\_1$ is self-adjoint. This answers to the first question. Concerning the second, $A^\*=A\_1-C$, but I dooubt that one can compute it explicitely without knowing $Q\_\infty$.


