4.2 The Potential Barrier
Consider a potential barrier:
$$
U=\begin{cases}
0 & x<0\ \text{or}\ x>L\\
U_0 & 0<x<L.
\end{cases}
$$
As discussed in section 4.1, we can see that the wavefunction propagates in both direction in regions where $x<0$, and $0L$.
(i) $E>U_0$
Let the $\psi_I(x)$ be the wavefunction to the left of the barrier, $\psi_{II}$ be the wavefunction inside the barrier, and $\psi_{III}$ be the wavefunction to the right of the barrier. Using the result from section 4.1 directly, we have
$$
\begin{cases}
\psi_I(x)=A\mathrm{e}^{ikx}+B\mathrm{e}^{-ikx},\\
\psi_{II}(x)=C\mathrm{e}^{ik'x}+D\mathrm{e}^{-ik'x},\\
\psi_{III}(x)=F\mathrm{e}^{ikx},
\end{cases}
$$
where
$$
k=\sqrt{\frac{2mE}{\hbar^2}},\ \text{and}\ k'=\sqrt{\frac{2m(E-U_0)}{\hbar^2}}.
$$
Applying boundary conditions
$$
\begin{cases}
\psi_I(0)=\psi_{II}(0),\\
\psi_I'|_{x=0}=\psi_{II}'|_{x=0},\\
\psi_{II}(L)=\psi_{III}(L),\\
\psi_{II}'|_{x=L}=\psi_{III}'|_{x=L},
\end{cases}
$$
gives
$$
\begin{cases}
A+B=C+D,\\
k(A-B)=k'(C-D),\\
C\mathrm{e}^{ik'L}+D\mathrm{e}^{-ik'L}=F\mathrm{e}^{ikL},\\
ik'(C\mathrm{e}^{ik'L}+D\mathrm{e}^{-ik'L})=ikF\mathrm{e}^{ikL}.
\end{cases}
$$
Similar to section 4.1, we can find the reflection and transmission probabilities as following
$$
R=\frac{\sin^2(k'L)}{\sin^2(k'L)+4{k'^2k^2}/{(k^2-k'^2)^2}},
$$
$$
T=\frac{4{k'^2k^2}/{(k^2-k'^2)^2}}{\sin^2(k'L)+4{k'^2k^2}/{(k^2-k'^2)^2}}.
$$
Or in terms of $E$ and $U_0$
$$
\bbox[silver]{R=\frac{\sin^2[\sqrt{2m(E-U_0)}L/\hbar]}{\sin^2[\sqrt{2m(E-U_0)}L/\hbar]+4{(E/U_0)}/{[(E/U_0)-1]}},}
$$
$$
\bbox[silver]{T=\frac{4{(E/U_0)}/{[(E/U_0)-1]}}{\sin^2[\sqrt{2m(E-U_0)}L/\hbar]+4{(E/U_0)}/{[(E/U_0)-1]}}.}
$$
Notice the reflection probability can be 0, when
$$
\frac{\sqrt{2m(E-U_0)}}{\hbar}L=n\pi.
$$
This occurs when the waves that are reflected at $x=0$ and at $x=L$ destructively interfere in the region where $x<0$. A classical analogy can be thin-film interference.
(ii) $E<U_0$
In side the barrier, the wavefunction now becomes
$$
\psi_{II}(x)=C\mathrm{e}^{\alpha x}+D\mathrm{e}^{-\alpha x},
$$
where $\alpha=\sqrt{2m(U_0-E)/\hbar^2}$.
With procedures similar as case (i), the reflection and transmission probabilities in this case are
$$
\bbox[silver]{R=\frac{\sinh^2[\sqrt{2m(E-U_0)}L/\hbar]}{\sinh^2[\sqrt{2m(E-U_0)}L/\hbar]+4{(E/U_0)}/{[(E/U_0)-1]}},}
$$
$$
\bbox[silver]{T=\frac{4{(E/U_0)}/{[(E/U_0)-1]}}{\sinh^2[\sqrt{2m(E-U_0)}L/\hbar]+4{(E/U_0)}/{[(E/U_0)-1]}}.}
$$
Notice that even when particle have insufficient energy to overcome the potential barrier, it still has probability to transmit through. This is referred to as tunneling.