Past exam of the options I course of the University of Oxford / 2010 / J / J1 / Solution

For a real potential, the time-dependent Schrödinger equation isMultiply it by ${\psi }^{\ast }$, combine with its conjugate multiplied by $\psi$, and note that the real-potential terms cancel. This giveswhere the probability current isThe product rule gives the divergence identity because the gradient cross terms cancel.

For $F(r)=\sin (kr)/r$, the supplied radial identity gives $\Delta F=-k^{2}F$ for $r>0$. The time factor gives $i\hslash {\psi }_t=E\psi$, so the free equation holds forThe apparent origin singularity is removable: $F(0)=k$ and $F(r)=k-k^3{r}^2/6+\cdots$, so the equation extends smoothly there as well. Since the spatial factor is real and its time dependence is a common phase, ${\psi }^{\ast }\nabla \psi =F\nabla F$ 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.