For Ornstein-Uhlenbeck (OU) process, \begin{equation}\label{eq:ouprocess} dX_t=-X_tdt+\sqrt{2}dB_t. \end{equation} define \begin{equation} E[f(X_t)|X_0=x]=P_tf(x), \end{equation} for all $f\in C_b(R^n)$.
We know that let $\pi$ be the standard Gaussian measure on $R^n$ (i.e., $\pi(x)=(2\pi)^{-n/2}e^{-\|x\|^2/2}$). The OU process admits $\pi$, where convergence holds in $L^2(\pi)$ for any bounded and smooth function $f:R^n\to R$: $$ \| P_tf-\pi(f)\|_2\to 0 $$ where $\pi(f)=\int fd\pi$.
Question:
I try to get the stationary distribution $\pi$ for the following SDE: \begin{equation}\label{eq:ld} dX_t=-\nabla H(X_t)dt+\sqrt{2}dB_t, \, X_0=x, \end{equation} where $H(t)$ is the potential function on $R^n$. When we take $H(X_t)=X_t\cdot X_t/2$, the above Langevin dynamics are same as the OU process.
How to show that the $L^2(\pi)$ convergence $$ \| P_tf-\pi(f)\|_2 \to 0 $$ where $\pi(dx)=e^{-H(x)}dx/Z$ and $Z=\int e^{-H(x)}dx$ called Gibbs measure.