# Why is probability density unchanging for stationary state?

I do understand that the solution obtained from the separation of variables to a system where $$V$$ is constant, is in a stationary state. Mathematically, I do get that the probability density only depends on $$x$$ since the time term of the wave function cancels out to one.

But, conceptually, how is it possible that after a long time, when the wave function has ceased, the probability density to not change? I've always thought that the probability density was a statistical representation of changing quantum states given by the wave function. So, it is confusing to think how the probability density exists same as before when the wave equation bare does.

• This doesn't actually have anything to do with separation of variables, does it? The same ideas would apply to the Schrodinger equation in one dimension. Like Max Stammer, I'm unclear on what you mean by saying "when the wave function has ceased."
– user4552
Nov 24, 2019 at 13:44
• The separation of variables is one way of solving the time-dependent Schrödinger equation. This would also apply for a one-dimensional case, since we separate the temporal from the spatial coordinates. And yes, you are right, the fact that the probability density for a stationary state does not change with time has nothing to do with the separation of variables.
– MST
Nov 24, 2019 at 23:26

If you write down your wavefunction as a simple superposition of two states $$\psi_1(x) e^{-i E_1 t /\hbar }$$ and $$\psi_2(x) e^{-i E_2 t/\hbar }$$, you can readily check that the probability density is time-dependent.