 Solution

For a stationary scattering state, the time-independent Schrödinger equation is $-{\hslash }^2{\psi }^{\prime \prime }/(2m)+V\psi =E\psi$. The left side contains the incident and reflected waves, while the solid has no incident wave arriving from positive infinity:Multiplying by $e^{-iEt/\hslash }$ gives the time-dependent wavefunctions. The kinetic energy rises inside the lower-potential solid, so $k_2>k_1$ and the transmitted wavelength is shorter. A downward step permits a propagating transmitted wave for every $E>0$, but still generally reflects some incident flux. There is no forbidden region merely because $E<V_0$; the relevant transmitted kinetic energy is $E+V_0$.

Classically every particle crosses a downward step and accelerates, so this partial reflection has no classical counterpart. If the step is instead upward to $U>0$, $E>U$ gives oscillatory transmission with possible reflection, while $E<U$ gives a decaying evanescent wave. A semi-infinite forbidden region carries no transmitted stationary flux; a finite barrier can permit quantum tunnelling to a propagating region beyond it.