Non-zero probability current for a stationary state?

Question

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.

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

2. The probability is flowing in some kind of circular motion. No need for any sinks or sources.

Any reference is also welcome.

Answer 1

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.