 Solution

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.