Given a potential function $U: \mathbb{R}^n \to \mathbb{R}$, Langevin diffusion is gradient descent plus a Brownian motion term: $X' = -\nabla U(X) + \sqrt{2} \text{ }dW$.

It happens that the stationary distribution of Langevin diffusion is very nice: it is proportional to $\exp(-U)$. I can verify this by plugging $\exp(-U)$ into a PDE and checking that some second-order terms miraculously cancel. But seeing as $\exp(-U)$ is such a simple solution, I expect an elegant reason for it... or at the very least, some intuition for why the stationary distribution at $x$ should depend only on $U(x)$. Is there any intuition to be found here?