In Euclidean space, $\mathbb R^d$, the Langevin diffusion $${\rm d}X_t=b(X_t){\rm d}t+\sigma(X_t){\rm d}W_t\tag1,$$ where $\sigma:\mathbb R^d\to\mathbb R^{d\times k}$, $$b:=\frac{\Sigma+U}2\nabla\ln p+\frac12\nabla\cdot\Sigma,$$ $\Sigma:=\sigma\sigma^\ast$ and $(W_t)_{t\ge0}$, $U\in\mathbb R^{d\times d}$ is any anti-symmetric matrix is a $\mathbb R^k$-valued standard Wiener process, has an unique invariant measure with density $p$ with respect to the Lebesgue measure on $\mathbb R^d$.

Now, in my application, I'm working on $[0,1)^d$ instead. So I have a given density $p:[0,1)^d\to[0,\infty)$. In my application, it works nicely to consider $[0,1)^d$ with toroidal boundary and the standard MCMC sampling approach is Metropolis-Hastings with proposal kernel given by the wrapped normal distribution.

What I would like to do now is defining a (continuous-time) Markov process, with values in $[0,1)^d$, with unique invariant measure with density $p$ with respect to the restriction of the Lebesgue measure on $\mathbb R^d$ to $[0,1)^d$. Is this possible?

One might think that we can simply use $(1)$ with toroidal wrapping, but as this paper shows, the resulting process is not even Markov anymore.

Another naive idea is to simply consider $p$ on all of $\mathbb R^d$ by simply setting it to $0$ outside $[0,1)^d$. However, then the $\nabla\ln p$ is not even well-defined everywhere. But I have to say that my target density is not strictly positive on $[0,1)^d$ anyways, so I have to deal with this problem in any case.