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\begin{document}

For tunneling problems in WKB, you have 3 regions I (allowed), II
(barrier), III(allowed)

\begin{itemize}

\item \( \Psi _{I}=\frac{A}{\sqrt{k}}e^{i\int _{x}^{x_{1}}k(x)dx} \)\( +\frac{B}{\sqrt{k}}e^{-i\int _{x}^{x_{1}}k(x)dx} \)

\item \( \Psi _{II} \)\( =\frac{C}{\sqrt{K}}e^{-\int _{x_{1}}^{x}K(x)dx} \)\( +\frac{D}{\sqrt{K}}e^{\int _{x_{1}}^{x}K(x)dx} \)

\item \( \Psi _{I}=\frac{F}{\sqrt{k}}e^{i\int _{x_{2}}^{x}k(x)dx} \)

\end{itemize}

Rewriting in terms of cosines and sines, you get: \( \Psi _{I}=(Ae^{-i\frac{\pi }{4}}+Be^{-i\frac{\pi }{4}})\frac{cos(\int _{x}^{x_{1}}kdx-\frac{\pi }{4})}{\sqrt{k}} \)\( +i(-Ae^{-i\frac{\pi }{4}}+Be^{i\frac{\pi }{4}})\frac{sin(\int kdx-\frac{\pi }{4})}{\sqrt{k}} \)

so \( 2C=Ae^{-i\frac{\pi }{4}}+Be^{-i\frac{\pi }{4}} \)\( ,D=i(Ae^{-i\frac{\pi }{4}}-Be^{i\frac{\pi }{4}}) \)

Now match region 2 and region 3.

Define \( r=e^{\int _{x_{1}}^{x_{2}}K(x)dx} \). Then \( \Psi _{II} \)\( =\frac{C}{\sqrt{K}}\frac{1}{r}e^{-\int _{x}^{x_{2}}K(x)dx} \)\( +\frac{Dr}{\sqrt{K}}e^{\int _{x}^{x_{2}}K(x)dx} \)

Matching, you get:\( 2Dr=Fe^{i\frac{\pi }{4}}+Ge^{-i\frac{\pi }{4}} \) and \( \frac{-C}{r}=i(Fe^{i\frac{\pi }{4}}-Ge^{-i\frac{\pi }{4}}) \)

This gives an equation:

\( \begin{array}{c}
C\\
D
\end{array}
=\frac{1}{2}*\begin{array}{cc}
e^{-i\pi /4} & e^{i\pi /4}\\
i2e^{-i\pi /4} & -i2e^{i\pi /4}
\end{array}
*\begin{array}{c}
A\\
B
\end{array}
 \) and a similar equation

This can be used to get

\( \begin{array}{c}
A\\
B
\end{array}
=\frac{1}{2}*\begin{array}{cc}
2r+\frac{1}{2r} & i(2r-\frac{1}{2r})\\
-i(2r-\frac{1}{2r} & 2r+\frac{1}{2r}
\end{array}
*\begin{array}{c}
F\\
G
\end{array}
 \)

This means T=\( |\frac{F}{A}|^{2}=|\frac{1}{r+\frac{1}{4r}}|^{2}\rightarrow r^{-2} \) for a tall, thick barrier. This means \( T=e^{-2\int _{x_{1}}^{x_{2}}\sqrt{\frac{2m}{\hbar ^{2}}(E-V(x)}dx} \)

example:

parabolic potential tunnelling

WKB tunnelling predicts transmission of 1 when E\(>\)V. This is
the classical limit.

\end{document}
