= Solution
{c}
For a real potential, the <Time-dependent Schrödinger equation> is
$$
i\hbar\partial_t\psi=-\frac{\hbar^2}{2m}\nabla^2\psi+V\psi.
$$
Multiply this equation by $\psi^*$, multiply its conjugate by $\psi$, and subtract. The potential terms cancel and the product rule converts the remaining terms into a divergence. Consequently
$$
\boxed{\partial_t|\psi|^2+\nabla\cdot\mathbf j=0,\qquad
\mathbf j=\frac{\hbar}{2mi}(\psi^*\nabla\psi-\psi\nabla\psi^*)
=\frac\hbar m\operatorname{Im}(\psi^*\nabla\psi).}
$$
This is the <probability current> and its <continuity equation>.
Put $R(r)=\sin(kr)/r$. The supplied radial <Laplacian> formula gives $\nabla^2R=-k^2R$ for $r>0$. Also $i\hbar\partial_t\psi=E\psi$. Therefore the free-particle equation is satisfied when
$$
\boxed{k=\sqrt{2mE}/\hbar.}
$$
The apparent singularity at zero is removable: $R(r)=k-k^3r^2/6+\cdots$ extends smoothly there and satisfies the same equation. Since $R$ is real and the time dependence is only a common phase, $\psi^*\nabla\psi=R\nabla R$ is real, so $\mathbf j=0$. The <density> $R^2$ is time-independent, proving the continuity equation explicitly. This is a standing spherical wave with zero net radial current.
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