 Solution

Substitution of a plane wave in the Schrödinger equation gives $\hslash \omega ={\hslash }^2{k}^2/(2m)+V_0$. For a stationary step-scattering state, remove the common phase $e^{-iEt/\hslash }$ and write ${\psi }_L=e^{ikx}+r{e}^{-ikx}$, $k=\sqrt{2mE}/\hslash$. If $E>V_0$, the right-side outgoing wave is ${\psi }_R=t{e}^{iqx}$, $q=\sqrt{2m(E-V_0)}/\hslash$. A finite step with no delta-function potential in the Schrödinger operator requires continuous $\psi$ and ${\psi }^{\prime }$: a discontinuous wavefunction would generate a delta derivative, and integrating the equation over a vanishing interval gives derivative continuity. Matching yieldsFor $0<E<V_0$, take $q=i\kappa$ and the bounded right-side state $t{e}^{-\kappa x}$. It carries no transmitted flux and $\mathcal{R}=1$. The threshold limit also has reflection probability one. Thus at $V_0=1\mathrm{eV}$,For the two-step profile, set $k_0=\sqrt{2mE}/\hslash$, $k_1=\sqrt{2m(E-V_1)}/\hslash$ and $k_2=\sqrt{2m(E-V_0)}/\hslash$. With all regions propagating, the net reflection amplitude isThe given quarter-wave condition makes $e^{2i{k}_{1}a}=-1$. Zero reflection then requires $r_{01}=r_{12}$, equivalent to $k_1^2=k_0{k}_2$. ConsequentlyFor an increasing intermediate step this needs $E>V_0$ and $V_0/2<V_1<V_0$, together with the stipulated thickness condition. In this quarter-wave matching layer for a quantum potential step, two reflected amplitudes cancel while the transmitted flux equals the incident flux.