 Quantum quarter-wave matching layer

For equal-mass one-dimensional scattering between regions with real wavenumbers $k_0>k_2>0$, insert a layer of width $a$ and wavenumber $k_1$. For a purely right-moving incident wave of amplitude $A$ and no reflection, continuity at the first edge gives interior amplitudes $C=A(1+k_0/k_1)/2$ and $D=A(1-k_0/k_1)/2$. If $k_{1}a=\pi /2$, continuity at the second edge gives a transmitted edge amplitude $F=i(k_0/k_1)A$ and the matching condition $k_1^2=k_0{k}_2$. Thus $F/A=i\sqrt{k_0/k_2}$ and the transmitted probability current equals the incident current, despite their unequal amplitudes.

For potentials $0,V_1,V_2$ with $E>V_2>V_1>0$, the condition is $(E-V_1{)}^2=E(E-V_2)$, or $E=V_1^2/(2V_1-V_2)$, requiring $V_2/2<V_1<V_2$. It is the quantum-wave counterpart of a quarter-wave impedance transformer: the intermediate wavenumber is the geometric mean and the two reflected amplitudes cancel.