Past exam of the physics A course of the University of Cambridge / 2018 / Paper 1 / Section B / B6 / d / Solution

From the second-edge amplitude in part (c),This uses ${\psi }_{\mathrm{R}}=F{e}^{i{k}_2(x-a)}$. If the transmitted wave is instead written $T{e}^{i{k}_{2}x}$, its coefficient is $\boxed{T/A=i\sqrt{k_0/k_2}{e}^{-i{k}_{2}a}}$; the extra factor only changes the quantum phase convention.

Using the probability current of a plane wave,Thus the flux transmission probability is one. The larger transmitted density $|F{|}^2>|A{|}^2$ compensates for the smaller transmitted velocity $\hslash k_2/m$; equal currents need not have equal wave amplitudes.