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PHY456H1F: Quantum Mechanics II. Lecture 12 (Taught by Mr. Federico Duque Gomez). WKB Method

Posted by peeterjoot on October 21, 2011

[Click here for a PDF of this post with nicer formatting and figures if the post had any (especially if my latex to wordpress script has left FORMULA DOES NOT PARSE errors.)]


Peeter’s lecture notes from class. May not be entirely coherent.

WKB (Wentzel-Kramers-Brillouin) Method.

This is covered in section 24 in the text [1]. Also section 8 of [2].

We start with the 1D time independent Schr\”{o}dinger equation

\begin{aligned}-\frac{\hbar^2}{2m} \frac{d^2 U}{dx^2} + V(x) U(x) = E U(x)\end{aligned} \hspace{\stretch{1}}(2.1)

which we can write as

\begin{aligned}\frac{d^2 U}{dx^2} + \frac{2m}{\hbar^2} (E - V(x)) U(x) = 0\end{aligned} \hspace{\stretch{1}}(2.2)

Consider a finite well potential as in figure (\ref{fig:qmTwoL13:qmTwoL12fig1})

\caption{Finite well potential}


\begin{aligned}k &= \frac{2m (E - V)}{\hbar},\qquad E > V \\ \kappa &= \frac{2m (V - E)}{\hbar}, \qquad V > E,\end{aligned} \hspace{\stretch{1}}(2.3)

we have for a bound state within the well

\begin{aligned}U \propto e^{\pm i k x}\end{aligned} \hspace{\stretch{1}}(2.5)

and for that state outside the well

\begin{aligned}U \propto e^{\pm \kappa x}\end{aligned} \hspace{\stretch{1}}(2.6)

In general we can hope for something similar. Let’s look for that something, but allow the constants k and \kappa to be functions of position

\begin{aligned}k^2(x) &= \frac{2m (E - V(x))}{\hbar},\qquad E > V \\ \kappa^2(x) &= \frac{2m (V(x) - E)}{\hbar}, \qquad V > E.\end{aligned} \hspace{\stretch{1}}(2.7)

In terms of k Schr\”{o}dinger’s equation is just

\begin{aligned}\frac{d^2 U(x)}{dx^2} + k^2(x) U(x) = 0.\end{aligned} \hspace{\stretch{1}}(2.9)

We use the trial solution

\begin{aligned}U(x) = A e^{i \phi(x)},\end{aligned} \hspace{\stretch{1}}(2.10)

allowing \phi(x) to be complex

\begin{aligned}\phi(x) = \phi_R(x) + i \phi_I(x).\end{aligned} \hspace{\stretch{1}}(2.11)

We need second derivatives

\begin{aligned}(e^{i \phi})'' &=(i \phi' e^{i \phi})'  \\ &=(i \phi')^2 e^{i \phi} + i \phi'' e^{i \phi},\end{aligned}

and plug back into our Schr\”{o}dinger equation to obtain

\begin{aligned}- (\phi'(x))^2 + i \phi''(x) + k^2(x) = 0.\end{aligned} \hspace{\stretch{1}}(2.12)

For the first round of approximation we assume

\begin{aligned}\phi''(x) \approx 0,\end{aligned} \hspace{\stretch{1}}(2.13)

and obtain

\begin{aligned}(\phi'(x))^2 = k^2(x),\end{aligned} \hspace{\stretch{1}}(2.14)


\begin{aligned}\phi'(x) = \pm k(x).\end{aligned} \hspace{\stretch{1}}(2.15)

A second round of approximation we use 2.15 and obtain

\begin{aligned}\phi''(x) = \pm k'(x)\end{aligned} \hspace{\stretch{1}}(2.16)

Plugging back into 2.12 we have

\begin{aligned}-(\phi'(x))^2 \pm i k'(x) + k^2(x) = 0,\end{aligned} \hspace{\stretch{1}}(2.17)


\begin{aligned}\begin{aligned}\phi'(x) &= \pm \sqrt{ \pm i k'(x) + k^2(x) } \\ &= \pm k(x) \sqrt{ 1 \pm i \frac{k'(x)}{k^2(x)} } .\end{aligned}\end{aligned} \hspace{\stretch{1}}(2.18)

If k' is small compared to k^2

\begin{aligned}\frac{k'(x)}{k^2(x)} \ll 1, \end{aligned} \hspace{\stretch{1}}(2.19)

we have

\begin{aligned}\phi'(x) = \pm k(x) \pm i \frac{k'(x)}{2 k(x)}  \end{aligned} \hspace{\stretch{1}}(2.20)


\begin{aligned}\phi(x) &= \pm \int dx k(x) \pm i \int dx \frac{k'(x)}{2 k(x)}  + \text{const} \\ &= \pm \int dx k(x) \pm i \frac{1}{{2}} \ln k(x) + \text{const} \end{aligned}

Going back to our wavefunction, if E > V(x) we have

\begin{aligned}U(x) &\sim A e^{i \phi(x)} \\ &= \exp \left(i\left( \pm \int dx k(x) \pm i \frac{1}{{2}} \ln k(x) + \text{const} \right)\right) \\ &\sim \exp \left(i\left( \pm \int dx k(x) \pm i \frac{1}{{2}} \ln k(x) \right)\right) \\ &= e^{\pm i \int dx k(x)} e^{\mp \frac{1}{{2}} \ln k(x)} \\ \end{aligned}


\begin{aligned}U(x) \propto \frac{1}{{\sqrt{k(x)}}} e^{\pm i \int dx k(x)} \end{aligned} \hspace{\stretch{1}}(2.21)

FIXME: Question: the \pm on the real exponential got absorbed here, but would not U(x) \propto \sqrt{k(x)} e^{\pm i \int dx k(x)} also be a solution? If so, why is that one excluded?

Similarly for the E < V(x) case we can find

\begin{aligned}U(x) \propto \frac{1}{{\sqrt{\kappa(x)}}} e^{\pm i \int dx \kappa(x)}.\end{aligned} \hspace{\stretch{1}}(2.22)

\item V(x) changes very slowly \implies k'(x) small, and k(x) = \sqrt{2 m (E - V(x))}/\hbar.
\item E very far away from the potential {\left\lvert{(E - V(x))/V(x)}\right\rvert} \gg 1.


\caption{Example of a general potential}

\caption{Turning points where WKB won’t work}

\caption{Diagram for patching method discussion}

WKB won’t work at the turning points in this figure since our main assumption was that

\begin{aligned}{\left\lvert{\frac{k'(x)}{k^2(x)}}\right\rvert} \ll 1\end{aligned} \hspace{\stretch{1}}(3.23)

so we get into trouble where k(x) \sim 0. There are some methods for dealing with this. Our text as well as Griffiths give some examples, but they require Bessel functions and more complex mathematics.

The idea is that one finds the WKB solution in the regions of validity, and then looks for a polynomial solution in the patching region where we are closer to the turning point, probably requiring lookup of various special functions.

This power series method is also outlined in [3], where solutions to connect the regions are expressed in terms of Airy functions.


[1] BR Desai. Quantum mechanics with basic field theory. Cambridge University Press, 2009.

[2] D.J. Griffiths. Introduction to quantum mechanics, volume 1. Pearson Prentice Hall, 2005.

[3] Wikipedia. Wkb approximation — wikipedia, the free encyclopedia, 2011. [Online; accessed 19-October-2011].


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