= Solution
For a finite <potential energy> discontinuity, the <stationary state> $\psi$ and its first <derivative> are <continuous> at $a$. Integrating the <Time-independent Schrödinger equation> across a shrinking interval about $a$ gives continuity of $\psi'$; a jump in $\psi$ would produce a <distributional derivative> incompatible with a finite <potential energy>.
For $E>V_0$, put $k=\sqrt{2mE}/\hbar$, $q=\sqrt{2m(E-V_0)}/\hbar$ and use unit incident amplitude:
$$
\psi(x)=\begin{cases}e^{ik(x-a)}+r e^{-ik(x-a)},&x<a,\\t e^{iq(x-a)},&x>a.\end{cases}
$$
There is no wave incident from the right. The matching equations $1+r=t$, $k(1-r)=qt$ give $r=(k-q)/(k+q)$ and $t=2k/(k+q)$. Comparing reflected and incident <probability currents> gives the <reflection coefficient>
$$
\boxed{R(E)=\left(\frac{\sqrt E-\sqrt{E-V_0}}{\sqrt E+\sqrt{E-V_0}}\right)^2\quad(E>V_0)}.
$$
The transmitted fraction is $T=(q/k)|t|^2=4kq/(k+q)^2$, so $R+T=1$.
For $0<E<V_0$, write $\kappa=\sqrt{2m(V_0-E)}/\hbar$ and retain only the decaying right-hand solution $t e^{-\kappa(x-a)}$. Matching gives $r=(k-i\kappa)/(k+i\kappa)$, which has modulus one. The <evanescent wave> carries no transmitted <probability current>. At $E=V_0$, the bounded right-hand solution is constant and matching again gives $r=1$. Thus
$$
\boxed{R(E)=1\quad(0<E\leq V_0)}.
$$
Classically, a particle is wholly transmitted for $E>V_0$ and wholly reflected for $E<V_0$; there is no classical partial reflection above the step. At the threshold it has zero right-hand speed and no transmitted flux, a marginal case requiring a convention about motion exactly on the discontinuity. The quantum result approaches $R=0$ as $E/V_0\to\infty$.
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