= Solution
For a one-dimensional <wavefunction>,
$$
\boxed{\rho(x,t)=|\psi(x,t)|^2},
\qquad
\boxed{j(x,t)=\frac{\hbar}{2mi}
\left(\psi^*\frac{\partial\psi}{\partial x}
-\psi\frac{\partial\psi^*}{\partial x}\right)}
=\frac{\hbar}{m}\operatorname{Im}\left(\psi^*\partial_x\psi\right).
$$
The <probability continuity equation> is
$$
\partial_t\rho+\partial_xj=0.
$$
For a <stationary state> $\psi(x,t)=\phi(x)e^{-iEt/\hbar}$, the <probability density> is independent of time, so $\partial_xj=0$ and the <probability current> is constant in space.
For the <plane wave>
$$
\psi=Ae^{ikx-iEt/\hbar},
$$
one obtains
$$
\boxed{\rho=|A|^2},
\qquad
\boxed{j=\frac{\hbar k}{m}|A|^2}.
$$
This is a <momentum eigenstate> with momentum $p=\hbar k$ and, when it satisfies the free <Schrödinger equation>, energy $E=\hbar^2k^2/(2m)$. Its constant density and current describe a spatially uniform beam carrying probability in the sign of $k$. It is not a <normalizable wavefunction>, so it represents an idealized state rather than a localized particle.
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