 Solution

For an incident particle from the left, take $E>0$ and $E>V_0$, and defineThe time-independent Schrödinger equation gives an incident plus reflected plane wave on the left and an outgoing transmitted plane wave on the right:Continuity of the wavefunction and its derivative at a finite potential jump yields $1+r=t$ and $k(1-r)=qt$. Hence $r=(k-q)/(k+q)$ and $t=2k/(k+q)$. The incident and reflected probability currents have equal wave-number magnitude, soAs a check, the transmitted current fraction is $(q/k)|t{|}^2=4kq/(k+q{)}^2$, and the two probabilities sum to one.

For a positive step, $E\downarrow V_0$ makes $q\downarrow 0$, so $P\to 1$. For a negative step with fixed positive incident energy, $V_0\to -\infty$ makes $q\to \infty$, so again $P\to 1$. The latter is quantum reflection from an increasingly large wave-number mismatch, even though the step is downward.