 Solution

For a real potential, the Time-dependent Schrödinger equation isMultiply this equation by ${\psi }^{\ast }$, multiply its conjugate by $\psi$, and subtract. The potential terms cancel and the product rule converts the remaining terms into a divergence. ConsequentlyThis is the probability current and its continuity equation.

Put $R(r)=\sin (kr)/r$. The supplied radial Laplacian formula gives ${\nabla }^{2}R=-k^{2}R$ for $r>0$. Also $i\hslash {\partial }_t\psi =E\psi$. Therefore the free-particle equation is satisfied whenThe apparent singularity at zero is removable: $R(r)=k-k^3{r}^2/6+\cdots$ extends smoothly there and satisfies the same equation. Since $R$ is real and the time dependence is only a common phase, ${\psi }^{\ast }\nabla \psi =R\nabla R$ 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.