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Use a stationary wavefunction $\Psi (x,t)=\psi (x)e^{-iEt/\hslash }$ solving the Time-independent Schrödinger equation. For energy above the positive step potential, the positive wavenumbers areIncidence only from the left permits a right-moving incident plane wave and a left-moving reflected plane wave for $x\le 0$, but only a right-moving transmitted plane wave for $x>0$:The omitted right-incident component would describe a different scattering experiment. Wavefunction matching at a finite potential step requires continuity of both $\psi$ and $d\psi /dx$ for a finite jump and the same mass on either side:Eliminate $B=C-A$ from the second condition:The first ratio is a transmission amplitude; a transmitted probability current also depends on its different propagation velocity.

For a downward step potential, $V_0=-|V_0|$, the region on the right is classically allowed at every positive incident energy. The same matching calculation applies withAs $E\to 0^{+}$, $k_1\to 0$ and $k_2\to k_{\ast }=\sqrt{2m|V_0|}/\hslash >0$. Henceand the flux coefficients satisfyThis quantum reflection at a downward step occurs even though a classical particle would simply accelerate into the lower-potential region. The abrupt change in wavenumber causes strong reflected-wave interference.

Both this limit and part (d) give total reflection, but their probability amplitudes and spatial densities differ. At the upward-step threshold, $B/A\to +1$ and $C/A\to 2$: an antinode meets the origin and the right-hand density stays finite while its propagation velocity vanishes. Here $B/A\to -1$ and $C/A\to 0$: the incident and reflected waves nearly cancel at the origin, and the transmitted density vanishes while the right-hand propagation velocity remains finite. In both descriptions the incident probability amplitude is held fixed for comparison.

The coefficients above are limiting flux ratios for $E>0$. At exactly $E=0$ the incident probability current vanishes, so directly dividing zero-energy fluxes would not define a scattering probability.

For a positive step potential and left incidence, take $E\to V_0^{+}$ at fixed incident probability amplitude $A$. The transmitted wavenumber tends to zero while the incident wavenumber tends to $k=\sqrt{2m{V}_0}/\hslash$. Wavefunction matching at a finite potential step gives reflected amplitude $A$ and transmitted amplitude $2A$, producing the spatial wavefunction in the title. The left probability density is $4|A{|}^2{\cos }^{2}kx$ and the right density $4|A{|}^2$. The transmitted probability current still tends to zero because its velocity tends to zero. Over adjacent intervals of length $2\pi /k$, the right integrated density is twice the left. These ratios use finite local integrals of a non-normalizable stationary scattering state.