= Solution
For a real potential, the time-dependent <Schrödinger equation> is
$$
\boxed{i\hbar\partial_t\psi=-\frac{\hbar^2}{2m}\Delta\psi+V\psi.}
$$
Multiply it by $\psi^*$, combine with its conjugate multiplied by $\psi$, and note that the real-potential terms cancel. This gives
$$
\partial_t|\psi|^2=\frac{i\hbar}{2m}(\psi^*\Delta\psi-\psi\Delta\psi^*)
=-\nabla\cdot\mathbf j,
$$
where the <probability current> is
$$
\boxed{\mathbf j=\frac{\hbar}{2mi}(\psi^*\nabla\psi-\psi\nabla\psi^*)
=\frac\hbar m\operatorname{Im}(\psi^*\nabla\psi).}
$$
The product rule gives the divergence identity because the gradient cross terms cancel.
For $F(r)=\sin(kr)/r$, the supplied radial identity gives $\Delta F=-k^2F$ for $r>0$. The time factor gives $i\hbar\psi_t=E\psi$, so the free equation holds for
$$
\boxed{k=\frac{\sqrt{2mE}}\hbar.}
$$
The apparent origin singularity is removable: $F(0)=k$ and $F(r)=k-k^3r^2/6+\cdots$, so the equation extends smoothly there as well. Since the spatial factor is real and its time dependence is a common phase, $\psi^*\nabla\psi=F\nabla F$ is real. Thus $\mathbf j=0$, while $\rho=F^2$ is independent of time. Both terms in the <continuity equation> vanish separately. This is a standing <spherical wave>; the real spatial factor carries no net radial <probability current>.
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