Past exam of the physics course of the year 1 of the natural sciences course of the University of Cambridge / 2019 / Section B / 8

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.

The probability current of the incident wave is $j_{\mathrm{in}}=\hslash k_1|A{|}^2/m$, and that of the reflected wave is $j_{\mathrm{ref}}=-\hslash k_1|B{|}^2/m$. The quantum reflection probability is the positive ratio of reflected to incident flux magnitudes:For comparison, the transmitted probability current is $j_{\mathrm{tr}}=\hslash k_2|C{|}^2/m$, so the transmission probability isThese satisfy $R+T=1$, as required by conservation of probability current for a real, nonabsorbing potential energy. In particular, $|C/A{|}^2$ alone is not the transmitted flux fraction when $k_2\ne k_1$.

For $E\gg V_0$, the two wavenumbers nearly coincide:Consequently $B/A\simeq V_0/(4E)\to 0$ and $C/A\to 1$. To leading order the complete wavefunction isThe small fractional change in kinetic energy scarcely changes the propagation wavenumber, so reflection vanishes and the particle behaves approximately as a free right-moving plane wave. A more accurate phase at large positive $x$ retains $k_{2}x$ on the transmitted side; even a small wavenumber difference accumulates over a sufficiently long distance.

Take the threshold from the above-step side used in the preceding parts, $E\to V_0^{+}$. Then $k_2\to 0$ while $k_1\to \sqrt{2m{V}_0}/\hslash$ remains finite. The reflected probability amplitude approaches $B=A$, and the transmitted amplitude approaches $C=2A$. At each fixed position, the potential-step threshold wavefunction is thereforeIts probability density is

The left-hand standing wave has zero-density nodes at $k_{1}x=-\pi /2,-3\pi /2,\dots$ and maxima of height $4|A{|}^2$. Its density period is $\pi /k_1$, half the incident de Broglie wavelength. At the origin the density is a maximum, continuous with the constant value on the right; its first derivative is also zero on either side there.

The nonzero right-hand density does not imply transmitted flux: $j_{\mathrm{tr}}=\hslash k_2|C{|}^2/m\to 0$, so $R\to 1$ and $T\to 0$. The exact threshold solution with no incoming component from the right is the limiting constant solution of ${\psi }^{\prime \prime }=0$ in the raised region. These stationary scattering states are not normalized over the whole real line; finite-interval density ratios are nevertheless well defined.

Use adjacent segments $[-\ell ,0]$ and $[0,\ell ]$, with $\ell =2\pi /k_1$. Integrate the threshold probability density from part (d). The left segment contains two complete density periods, so its average ${\cos }^2$ is $1/2$:The right-hand density is constant:The common stationary-state normalization cancels, yieldingThe larger local probability weight on the right is compatible with zero transmitted probability current at threshold: density and flux measure different quantities.

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.