= Solution
For a real <potential energy> and a <stationary state>, the <probability current> is
$$
j(x)=\frac{\hbar}{2mi}\bigl(\psi^*\psi'-\psi\psi'^*\bigr)=\frac\hbar m\operatorname{Im}(\psi^*\psi').
$$
The time-independent <Schrödinger equation> gives $\psi''=2m(V-E)\psi/\hbar^2$, so $j'=\hbar\operatorname{Im}(\psi^*\psi'')/m=0$. In an exterior <plane wave> superposition, interference terms contribute no current and
$$
j_{\rm left}=\frac{k}{m}(|A|^2-|B|^2),\qquad j_{\rm right}=\frac{k}{m}(|C|^2-|D|^2).
$$
Consequently $|A|^2+|D|^2=|B|^2+|C|^2$ for every choice of incoming amplitudes. If $u=(A,D)^T$, this says $u^\dagger(I-S^\dagger S)u=0$ for every complex <vector> $u$. Since $I-S^\dagger S$ is <Hermitian>, vanishing of its <quadratic form> implies that the <matrix> itself vanishes. Therefore
$$
\boxed{S^\dagger S=I,\qquad R+T=1.}
$$
The second identity follows by choosing $u=(1,0)^T$. Since $S$ is a square <matrix>, the first also implies $SS^\dagger=I$, 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.
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