= Solution
Take $0<E<V_0$ and equal electron <mass> in all regions. Define $k=\sqrt{2mE}/\hbar$ and $\kappa=\sqrt{2m(V_0-E)}/\hbar$. Then
$$
\begin{aligned}
\mathrm I:&\quad-\frac{\hbar^2}{2m}\psi''=E\psi,\qquad\psi_I=Ae^{ikx}+Be^{-ikx},\\
\mathrm {II}:&\quad-\frac{\hbar^2}{2m}\psi''+V_0\psi=E\psi,\qquad\psi_{II}=Ce^{\kappa x}+De^{-\kappa x},\\
\mathrm {III}:&\quad-\frac{\hbar^2}{2m}\psi''=E\psi,\qquad\psi_{III}=Fe^{ikx}.
\end{aligned}
$$
There is no wave incident from the far right. Region I contains oscillatory incident and reflected waves; region II is \b[evanescent], with real exponentials rather than propagating <plane waves>. Both exponential terms are allowed in a finite barrier; deleting the growing term before boundary matching is generally incorrect. A complex superposition in the barrier can carry a nonzero conserved tunnelling current.
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