 Solution

For a real potential energy and a stationary state, the probability current isThe time-independent Schrödinger equation gives ${\psi }^{\prime \prime }=2m(V-E)\psi /{\hslash }^2$, so $j^{\prime }=\hslash \mathrm{Im}({\psi }^{\ast }{\psi }^{\prime \prime })/m=0$. In an exterior plane wave superposition, interference terms contribute no current andConsequently $|A{|}^2+|D{|}^2=|B{|}^2+|C{|}^2$ for every choice of incoming amplitudes. If $u=(A,D{)}^T$, this says $u^{†}(I-S^{†}S)u=0$ for every complex vector $u$. Since $I-S^{†}S$ is Hermitian, vanishing of its quadratic form implies that the matrix itself vanishes. ThereforeThe second identity follows by choosing $u=(1,0{)}^T$. Since $S$ is a square matrix, the first also implies $S{S}^{†}=I$, proving unitarity. A real delta potential preserves this proof: continuity of $\psi$ and the real derivative jump make the current continuous across the interaction point.