= Solution
For $E\gg V_0$, the two <wavenumbers> nearly coincide:
$$
k_2=k_1\sqrt{1-\frac{V_0}{E}}
\simeq k_1\left(1-\frac{V_0}{2E}\right).
$$
Consequently $B/A\simeq V_0/(4E)\to0$ and $C/A\to1$. To leading order the complete <wavefunction> is
$$
\boxed{\Psi(x,t)\simeq A\exp\!\left[i\left(k_1x-\frac{Et}{\hbar}\right)\right].}
$$
The 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 $k_2x$ on the transmitted side; even a small wavenumber difference accumulates over a sufficiently long distance.
Back to article page
