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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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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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