= Solution
For a stationary scattering state, the time-independent <Schrödinger equation> is $-\hbar^2\psi''/(2m)+V\psi=E\psi$. The left side contains the incident and reflected waves, while the solid has no incident wave arriving from positive infinity:
$$
\psi_<(x)=Ae^{ik_1x}+Be^{-ik_1x},\qquad
\psi_>(x)=Ce^{ik_2x},
$$
$$
k_1=\frac{\sqrt{2mE}}{\hbar},\qquad
k_2=\frac{\sqrt{2m(E+V_0)}}{\hbar}.
$$
Multiplying by $e^{-iEt/\hbar}$ 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. \b[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.
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