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: a discontinuous wavefunction would generate a delta derivative, and integrating the equation over a vanishing interval gives derivative continuity. Matching yields

r = \frac{k - q}{k + q}, t = \frac{2 k}{k + q}, R = (\frac{E - E - V _{0}}{E + E - V _{0}})^{2} (E > V_{0}) .

For

0 < E < V_{0}

, take

q = iκ

and the bounded right-side state

t e^{- κ x}

. It carries no transmitted flux and

R = 1

. The threshold limit also has reflection probability one. Thus at

V_{0} = 1 eV

R (0.9 eV) = 1, R (1.1 eV) ≃ 0.288.

For the two-step profile, set

k_{0} = 2 m E /ℏ

k_{1} = 2 m (E - V_{1}) /ℏ

and

k_{2} = 2 m (E - V_{0}) /ℏ

. With all regions propagating, the net reflection amplitude is

r_{tot} = \frac{r _{01} + r _{12} e ^{2 i k_{1} a}}{1 + r _{01} r _{12} e ^{2 i k_{1} a}}, r_{ij} = \frac{k _{i} - k _{j}}{k _{i} + k _{j}} .

The given quarter-wave condition makes

e^{2 i k_{1} a} = - 1

. Zero reflection then requires

r_{01} = r_{12}

, equivalent to

k_{1}^{2} = k_{0} k_{2}

. Consequently

E - V_{1} = E (E - V_{0}), or E = \frac{V _{1}^{2}}{2 V _{1} - V _{0}} .

For 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.