Update The plane wave state for a free particle given by $\psi(x,t)=A\exp[i(kx-\omega t)]$ is completely delocalized in space and time. Therefore, the wavefunction is present everywhere in space at all times. We can, however, 'normalize' this wavefunction, pretending that it is confined in a box of length $L$. In this case, we find, $$\psi(x,t)=\frac{1}{\sqrt{L}}\exp[i(k_nx-\omega t)]$$ with quantized wavenumber $k\rightarrow k_n$. The position probability density has a spatial profile $$\rho(x)=\psi^*\psi=1/L$$ which is independent of time (like any other stationary state). Moreover, the probability of finding the particle is same everywhere at all times. But if we calculate the probability current (flow) it is nonzero: $$j(x)=-\frac{\hbar k}{m}\frac{1}{L}\neq 0.$$ For example, $J_\phi$, the azimuthal component of the current is nonzero in Hydrogen atom for the stationary states.

Therefore, although the spatial profile of position probability density does not change with time, there is a propagation!

How do we interpret this? Should it be interpreted that at any point $x$ the influx of probability is equal to the outflux such that the probability amplitude at any point remains constant in time?

  • 1
    $\begingroup$ Related: physics.stackexchange.com/q/12611/2451 and links therein. $\endgroup$
    – Qmechanic
    Nov 22, 2016 at 18:11
  • 2
    $\begingroup$ "Moreover, the probability of finding the particle is same everywhere." That's inaccurate. The actual probability of finding the particle can only be calculated in a spatial region, here $P=\int_{\Delta x}\psi^*\psi dx$. In a point, $\Delta x=0\implies P=0$. $\endgroup$
    – Gert
    Nov 22, 2016 at 18:21
  • 1
    $\begingroup$ "Therefore, although the spatial profile of position probability density does not change with time, there is a propagation. Aren't these two pictures contradictory?" Why would they be contradictory? The wave function can evolve in time without the time entering the modulus square (that happens whenever the states are eigenstates of the Hamiltonian). $\endgroup$
    – gented
    Nov 22, 2016 at 19:12
  • 2
    $\begingroup$ The charge density in a wire is constant, yet there is current. $\endgroup$
    – garyp
    Nov 22, 2016 at 19:22
  • $\begingroup$ That the state evolves in time doesn't necessarily mean that the probability must depend on time: the state evolves so that the probability still remains the same in every interval. If you want to perform a measure of the position then the state collapses into an eigenstate of the position operator, which is in general a combination of energy eigenstates with different energies and which evolve differently in time (hence the time dependence does eventually appear in the modulus square). $\endgroup$
    – gented
    Nov 23, 2016 at 10:06

1 Answer 1


The solution for a box is a periodic solution. You can imagine that the particle exits the box, say, through the right wall of the box (I did not check the sign) and immediately enters through the left wall of the box. You may say that this picture is not reasonable, but these solutions are simple and useful. If you want more reasonable solutions for a box, you need to set such boundary conditions at the walls that the current at the walls vanishes. You will then have standing wave solutions, and the probability density will change in time, in accordance with the intuition.

As for the non-vanishing azimuthal component of the current in the hydrogen atom, it corresponds to orbital motion of electrons (along closed trajectories, if we use the correspondence principle).


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.