D.46 Derivation of the WKB approximation

The purpose in this note is to derive an approximate solution to the Hamiltonian eigenvalue problem

\begin{displaymath}
\frac{{\rm d}^2\psi}{{\rm d}x^2} = - \frac{p_{\rm {c}}^2}{\hbar^2}\psi
\end{displaymath}

where the classical momentum $p_{\rm {c}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\sqrt{2m(E-V)}$ is a known function for given energy. The approximation is to be valid when the values of $p_{\rm {c}}$$\raisebox{.5pt}{$/$}$$\hbar$ are large. In quantum terms, you can think of that as due to an energy that is macroscopically large. But to do the mathematics, it is easier to take a macroscopic point of view; in macroscopic terms, $p_{\rm {c}}$$\raisebox{.5pt}{$/$}$$\hbar$ is large because Planck’s constant $\hbar$ is so small.

Since either way $p_{\rm {c}}$$\raisebox{.5pt}{$/$}$$\hbar$ is a large quantity, for the left hand side of the Hamiltonian eigenvalue problem above to balance the right hand side, the wave function must vary rapidly with position. Something that varies rapidly and nontrivially with position tends to be hard to analyze, so it turns out to be a good idea to write the wave function as an exponential,

\begin{displaymath}
\psi = e^{{\rm i}\tilde\theta}
\end{displaymath}

and then approximate the argument $\tilde\theta$ of that exponential.

To do so, first the equation for $\tilde\theta$ will be needed. Taking derivatives of $\psi$ using the chain rule gives in terms of $\tilde\theta$

\begin{displaymath}
\frac{{\rm d}\psi}{{\rm d}x} = e^{{\rm i}\tilde\theta}\,{\...
...tilde\theta}\,{\rm i}\frac{{\rm d}^2\tilde\theta}{{\rm d}x^2}
\end{displaymath}

Then plugging $\psi$ and its second derivative above into the Hamiltonian eigenvalue problem and cleaning up gives:
\begin{displaymath}
\left(\frac{{\rm d}\tilde\theta}{{\rm d}x}\right)^2
= \f...
...hbar^2}
+ {\rm i}\frac{{\rm d}^2\tilde\theta}{{\rm d}x^2} %
\end{displaymath} (D.30)

For a given energy, $\tilde\theta$ will depend on both what $x$ is and what $\hbar$ is. Now, since $\hbar$ is small, mathematically it simplifies things if you expand $\tilde\theta$ in a power series with respect to $\hbar$:

\begin{displaymath}
\tilde\theta =
\frac{1}{\hbar}\left(f_0 + \hbar f_1 + {\textstyle\frac{1}{2}} \hbar^2 f_2 + \ldots \right)
\end{displaymath}

You can think of this as writing $\hbar\theta$ as a Taylor series in $\hbar$. The coefficients $f_0,f_1,f_2,\ldots$ will depend on $x$. Since $\hbar$ is small, the contribution of $f_2$ and further terms to $\psi$ is small and can be ignored; only $f_0$ and $f_1$ will need to be figured out.

Plugging the power series into the equation for $\tilde\theta$ produces

\begin{displaymath}
\frac{1}{\hbar^2}f_0'^2 + \frac{1}{\hbar} 2 f_0' f_1' + \l...
...\hbar^2}p_{\rm {c}}^2 + \frac{1}{\hbar} {\rm i}f_0'' + \ldots
\end{displaymath}

where primes denote $x$-​derivatives and the dots stand for powers of $\hbar$ greater than $\hbar^{-1}$ that will not be needed. Now for two power series to be equal, the coefficients of each individual power must be equal. In particular, the coefficients of 1$\raisebox{.5pt}{$/$}$$\hbar^2$ must be equal, $f_0'^2$ $\vphantom0\raisebox{1.5pt}{$=$}$ $p_{\rm {c}}^2$, so there are two possible solutions

\begin{displaymath}
f_0' = \pm p_{\rm {c}}
\end{displaymath}

For the coefficients of 1$\raisebox{.5pt}{$/$}$$\hbar$ to be equal, $2f_0'f_1'$ $\vphantom0\raisebox{1.5pt}{$=$}$ ${{\rm i}}f_0''$, or plugging in the solution for $f_0'$,

\begin{displaymath}
f_1'= {\rm i}\frac{p_{\rm {c}}'}{2p_{\rm {c}}}
\end{displaymath}

It follows that the $x$-​derivative of $\tilde\theta$ is given by

\begin{displaymath}
\tilde\theta' =
\frac{1}{\hbar}
\left(
\pm p_{\rm {c...
...r {\rm i}\frac{p_{\rm {c}}'}{2p_{\rm {c}}} + \ldots
\right)
\end{displaymath}

and integrating gives $\tilde\theta$ as

\begin{displaymath}
\tilde\theta =
\pm \frac{1}{\hbar}
\int p_{\rm {c}}{\,...
...m i}{\textstyle\frac{1}{2}} \ln p_{\rm {c}} + \tilde C \ldots
\end{displaymath}

where $\tilde{C}$ is an integration constant. Finally, $e^{{\rm i}\tilde\theta}$ now gives the two terms in the WKB solution, one for each possible sign, with $e^{{\rm i}\tilde{C}}$ equal to the constant $C_{\rm {f}}$ or $C_{\rm {b}}$.