Probability current density of a stationary state

[Thomas Rigby]

TL;DR

How do I calculate the probability current density for a solution of the time-independent Schrodinger equation?

I have written a finite difference program to solve 1D time-independent Schrödinger equation. It seems to work correctly for harmonic oscillator, particle in a box, etc. But I can't figure out how to calculate the probability current density. It should be constant, but what is it? The program returns a real function, so I can't use the usual formula for current density. And I don't know how to get the velocity so I can't use density times velocity.

I will post this question as is. If it needs more clarification, then I will add it.

 

[vanhees71]

Why do you think, there's something wrong? In a stationary state the probability density $\rho=\psi^* \psi$ is time-independent for an energy eigenstate $\psi$ (that's why the energy eigenstates are the stationary states of the system). Thus the only constraint from unitarity, $\partial_t \rho + \vec{\nabla} \cdot \vec{j}=\vec{\nabla} \cdot \vec{j}=0$, i.e., $\vec{j}=0$ is not a priori wrong :-).

 

[Thomas Rigby]

I am not interested in the trivial cases where j=0.

 

[DrClaude]

I am not interested in the trivial cases where j=0.

How could a stationary state have $j \neq 0$?

 

[Thomas Rigby]

How could a stationary state have $j \neq 0$?

Plane wave in free space.

 

[Demystifier]

The program returns a real function, so I can't use the usual formula for current density.

You can use it and the result is 0. Many stationary states have j=0. For instance, the ground state often has j=0.

If you want to get nonzero j, you should modify your program to allow for more general solutions.

 

[Wanuke]

How do you show that the particle current density J vanishes for a stationary states.