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Is it possible to have a non-zero probability current in a stationary state? In a stationary state, the probability density $|\Psi|^2$ doesn't depend on time. For this to be true, I'd expect no flow of probability so that $|\Psi|^2$ can maintain a constant shape. However from my work below, it seems like probability can flow even in a stationary state. Is this correct?

1-dimensional probability current is given by: $$ J(x,t) = \frac{i\hbar}{2m}\Big(\Psi \frac{\partial \Psi^*}{\partial x} - \Psi^*\frac{\partial\Psi}{\partial x}\Big)$$

For a stationary state $\Psi(x,t) = Af(x)e^{-iwt}$ where $A$ is a complex constant, the probability current is:

$$J(x,t) = \frac{i\hbar}{2m}\Big(|A|^2f(x)\frac{\partial f^*(x)}{\partial x}- |A|^2 f^*(x)\frac{\partial f(x)}{\partial x}\Big) $$

If $f(x)$ is real-valued, then $J = 0$. However assuming the more general case that $f(x)$ is complex-valued $f(x) = u(x) + iv(x)$, I get that

$$ J(x,t) = \frac{i\hbar}{2m}|A|^2\big( \;2i(u'v - v'u)\;\big) \neq 0$$

Can $f(x)$ even be complex-valued?

DWade64
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  • You can derive the conservation equation for the probability, the divergence of the probability current plus the time derivative of the probability density equals zero. So, it's completely analogous to e.g. fluid flow, you can have a stationary solution where there is fluid flow, if you have a sink and a source. In a 1D problem you can put these at minus and plus infinity, in a 3D scattering problem you can put a source at the origin and let the current flow out to infinity, so there is then a sink at infinity. – Count Iblis Aug 28 '17 at 01:27
  • In 1-d, you can always choose your stationary states to be real. Or better put, any stationary state is equivalent to a real state, up to a phase difference, so your bracketed term would indeed be zero. – CDCM Aug 28 '17 at 01:34

1 Answers1

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In 1d one can always choose steady-state solutions to be real. This follows because 1d states are not degenerate (see this post) and the usual argument that, if $\phi(x)$ is solution with energy $E$, so if $\phi^*(x)$ if $V$ is real, and thus so is $\phi(x)+\phi^*(x)$, which is also real.

ZeroTheHero
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