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I understand that if the Hamiltonian does not depend on the time, the Schrödinger Equation becomes separable, so you get

$$ H \psi(x) = E \psi(x) $$

and

$$ \Psi(x,t) = \psi(x)\exp\left(-\frac{\imath E}{\hbar}t\right). $$

But $-\frac{\imath E}{\hbar}t$ is a purely imaginary number, so $$ \left|\exp\left(-\frac{\imath E}{\hbar}t\right)\right| = 1 $$

If that is correct, then how can there be any probability density flow in time? The $exp$ term is only changing the phase of $\psi$, but does not contribute anything to its absolute value.

What did I understand wrong?

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What you've written is only true if $\psi(x)$ is an eigenstate of $H$.

For some general $\psi(x)$ that is not an eigenstate (i.e. $H \psi(x) \ne E \psi(x)$), then $\Psi(x, t)$ will be more complicated than just the time independent wavefunction multiplied by a phase factor.

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The separable solutions are exactly the eigenstates of the Hamiltonian which are exactly the ones where the probability density does not change.

However you can have a flux or flow of probability even if the probability doesn't change. This is like electromagnetism where you can have current floe through a wire even if the wire has no change of charge density.

For instance if you had a circular wire with now resistance it is possible to have a steady current flow in a circle even without and change in charge density.

Similarly you can have a probability current even if the probability density doesn't change.

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