 Solution

The probability current of the incident wave is $j_{\mathrm{in}}=\hslash k_1|A{|}^2/m$, and that of the reflected wave is $j_{\mathrm{ref}}=-\hslash k_1|B{|}^2/m$. The quantum reflection probability is the positive ratio of reflected to incident flux magnitudes:For comparison, the transmitted probability current is $j_{\mathrm{tr}}=\hslash k_2|C{|}^2/m$, so the transmission probability isThese satisfy $R+T=1$, as required by conservation of probability current for a real, nonabsorbing potential energy. In particular, $|C/A{|}^2$ alone is not the transmitted flux fraction when $k_2\ne k_1$.