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Take fixed $E>0$ and define $k=\sqrt{2mE}/\hslash$. For sufficiently large $V_0$, the barrier region has $\kappa =\sqrt{2m(V_0-E)}/\hslash$ and solves ${\psi }^{\prime \prime }={\kappa }^2\psi$. Solving this constant-coefficient equation gives the exact transfer relationIn the specified limit, $a=U/V_0$ and $\kappa a\to 0$, whileThe transfer matrix therefore tends to $\left(\begin{array}{cc}1& 0\\ \eta & 1\end{array}\right)$. As the two boundaries collapse to the origin, this proves continuity and the derivative jump for a delta potential:The vanishing exterior phase shift $e^{ika}\to 1$ introduces no further factor.

Write the incoming and reflected waves as $e^{ikx}+r{e}^{-ikx}$ on the left and the transmitted wave as $t{e}^{ikx}$ on the right. Matching givesThe denominator is nonzero for real $\eta$ and $k>0$, so the limiting matching system is well defined. The probability currents on either side have the same factor $\hslash k/m$, making the transmission probability $|t{|}^2$. Thus scattering by a delta potential givesThe limit retains a finite reflection probability despite the barrier's vanishing width because its integrated strength remains finite.

For a wave incident from the left, define the reflection and transmission amplitudes byReflection invariance gives the same amplitudes for incidence from the right. If the incoming amplitudes from the left and right are ordered first, and the outgoing amplitudes toward the left and right second, the One-dimensional S-matrix is

For $V(x)=V_0\delta (x)$, the wavefunction is continuous and its derivative has the jumpThe scattering ansatz givesWithwe obtain the scattering by a delta potential amplitudesThey satisfyThereforeso the scattering matrix is unitary.