= Solution
Take the threshold from the above-step side used in the preceding parts, $E\to V_0^+$. Then $k_2\to0$ while $k_1\to\sqrt{2mV_0}/\hbar$ 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 therefore
$$
\psi(x)\longrightarrow\begin{cases}
A(e^{ik_1x}+e^{-ik_1x})=2A\cos(k_1x),&x\leq0,\\
2A,&x>0.
\end{cases}
$$
Its <probability density> is
$$
\boxed{|\psi(x)|^2\longrightarrow\begin{cases}
4|A|^2\cos^2(k_1x),&x\leq0,\\
4|A|^2,&x>0.
\end{cases}}
$$
\Image[/past-exam-of-the-physics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2019-ia-physics-q8-threshold-density.png]
{title=Probability density at the positive-step threshold, normalized to the incident amplitude squared}
The left-hand <standing wave> has zero-density nodes at $k_1x=-\pi/2,-3\pi/2,\ldots$ 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: $j_{\rm tr}=\hbar k_2|C|^2/m\to0$, so $R\to1$ and $T\to0$. The exact threshold solution with no incoming component from the right is the limiting constant solution of $\psi''=0$ in the raised region. These stationary scattering states are not normalized over the whole real line; finite-interval density ratios are nevertheless well defined.
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