= Solution
For a <wavefunction> $\Psi(x,t)$, the <probability density> and <probability current> are
$$
\rho=|\Psi|^2,
\qquad
j=\frac{\hbar}{2mi}\left(\Psi^*\frac{\partial\Psi}{\partial x}-\Psi\frac{\partial\Psi^*}{\partial x}\right).
$$
Multiply the <Schrödinger equation> by $\Psi^*$, multiply its <complex conjugate> by $\Psi$, and subtract. The real <potential energy> terms cancel, leaving
$$
\frac{\partial|\Psi|^2}{\partial t}
=-\frac{\partial}{\partial x}
\left[\frac{\hbar}{2mi}
\left(\Psi^*\Psi_x-\Psi\Psi_x^*\right)\right],
$$
which is the <probability continuity equation> $\rho_t+j_x=0$.
For a <stationary state>, $\rho_t=0$, so $j_x=0$ and $j$ is constant in space. A <normalizable wavefunction> has vanishing current at spatial infinity, forcing this constant to be zero. By contrast the nonnormalizable <plane wave>
$$
\Psi(x,t)=A e^{i(kx-\omega t)}
$$
has stationary density $|A|^2$ and nonzero current
$$
\boxed{j=\frac{\hbar k}{m}|A|^2}
$$
when $k\ne0$.
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