Non-zero probability current for a stationary state? Is there a normalizable stationary state that has a non-zero probability current?
According to Wikipedia, the probability current is
$$
\mathbf{J} = \frac{\hbar}{2im}(\Psi^* \nabla \Psi - \Psi \nabla \Psi^*)
$$
There is a continuity equation due to the conservation of probability
$$
\nabla \cdot \mathbf{J} + \partial_t |\Psi|^2 = 0
$$
But for a stationary state, $\partial_t |\Psi|^2 = 0$, so I am looking incompressible flow. I am interested in the 2D case because, for the 1D case, it is zero as explained in this answer. I have 2 scenarios in mind. Please write an answer if you can show that they're impossible, or otherwise if you have a concrete example.

*

*The probability flows from infinity to infinity. There is a probability source and probability sink at two different infinities.


*The probability is flowing in some kind of circular motion. No need for any sinks or sources.
Any reference is also welcome.
 A: I have comb up with this example. Consider a 2D problem with polar symmetry described by polar coordinates $(s, \varphi)$. This in general is described by a wave function of the kind
\begin{equation}
\psi(s, \varphi) = \chi(s)\exp(i m \varphi)
\end{equation}
We consider a potential $V(s)$ with bound states so that $\chi$ is pure real. Now compute
\begin{align}
\nabla \psi(s, \varphi) &= \left[\mathbf{e}_s \partial_s + \mathbf{e}_\varphi \frac{1}{s}\partial_\varphi\right]\psi(s, \varphi) \\
&= \left[ \frac{\partial \chi}{\partial s} \mathbf{e}_s + im \frac{\chi}{s} \mathbf{e}_{\varphi} \right] \exp(i m \varphi)
\end{align}
Using the fact that $\chi$ is real then you will find for the current density
\begin{equation}
\mathbf{J} = \frac{\hbar m}{M}\frac{\chi^2(s)}{s} \mathbf{e}_\varphi
\end{equation}
I capitalized the mass to distinguish from the quantum number $m$.
Note that $\hbar m $ is the angular momentum, so in a sense this is the current associated with the rotation of the particle.
