0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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." $\endgroup$ – user4552 Nov 24 '19 at 13:44
  • $\begingroup$ 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. $\endgroup$ – Max Stammer Nov 24 '19 at 23:26
2
$\begingroup$

From my point of view it seems that you are mixing two different cases together:

It is correct that your probability density is time-independent if you just choose one solution of the time-dependent Schrödinger equation. But in this case the wavefunction does not cease (I am still not completely sure what you mean by this), since we have just figured out that it does not change with time.

The second, and maybe more interesting case, is when your wavefunction is a superposition of two or more solutions of the time-dependent Schrödinger equation. Then, when calculating the probability density, you will have these nice interference terms between the different states and your probability density is no longer time-independent. Such a wavepacket can spread over time, and I assume that this is what you mean by a ceasing wavepacket.

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.