A.28.2 So­lu­tion wkb-b

Ques­tion:

Use the equa­tion

\begin{displaymath}
\frac{1}{\hbar} \int_{{\underline x}=x_1}^{x_2} p_{\rm {c}}(...
...e x}) { \rm d}{\underline x}= (n-{\textstyle\frac{1}{2}}) \pi
\end{displaymath}

to find the WKB ap­prox­i­ma­tion for the en­ergy lev­els of the har­monic os­cil­la­tor. The po­ten­tial en­ergy is ${\textstyle\frac{1}{2}}m\omega{x}^2$ where the con­stant $\omega$ is the clas­si­cal nat­ural fre­quency. So the to­tal en­ergy, ex­pressed in terms of the turn­ing points $x_2$ $\vphantom0\raisebox{1.5pt}{$=$}$ $-x_1$ at which $E$ $\vphantom0\raisebox{1.5pt}{$=$}$ $V$, is $E$ $\vphantom0\raisebox{1.5pt}{$=$}$ ${\textstyle\frac{1}{2}}m\omega{x_2}^2$.

In this case too, the WKB ap­prox­i­ma­tion pro­duces the ex­act en­ergy eigen­val­ues. That, how­ever, is just a co­in­ci­dence; the clas­si­cal WKB wave func­tions are cer­tainly not ex­act; they be­come in­fi­nite at the turn­ing points. As the ex­am­ple $h_{50}$ above shows, the true wave func­tions most def­i­nitely do not.

An­swer:

Sub­sti­tut­ing in $p_{\rm {c}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\sqrt{2m(E-V)}$, with $V$ $\vphantom0\raisebox{1.5pt}{$=$}$ ${\textstyle\frac{1}{2}}m\omega^2{x}^2$ and $E$ $\vphantom0\raisebox{1.5pt}{$=$}$ ${\textstyle\frac{1}{2}}m\omega^2{x_2}^2$, pro­duces

\begin{displaymath}
\frac{1}{\hbar} \int_{{\underline x}=-x_2}^{x_2} m \omega\sqrt{x_2^2-x^2}{ \rm d}x = (n-{\textstyle\frac{1}{2}}) \pi
\end{displaymath}

The in­te­gral can be done by mak­ing the sub­sti­tu­tion $x$ $\vphantom0\raisebox{1.5pt}{$=$}$ $x_2\sin(\alpha)$:

\begin{displaymath}
\frac{1}{\hbar} m \omega x_2^2 \int_{\alpha =-\pi /2}^{\pi /2} \cos^2\alpha{ \rm d}\alpha = (n-{\textstyle\frac{1}{2}}) \pi
\end{displaymath}

and the re­main­ing in­te­gral is ${\textstyle\frac{1}{2}}\pi$:

\begin{displaymath}
\frac{1}{\hbar} m \omega x_2^2 {\textstyle\frac{1}{2}} \pi = (n-{\textstyle\frac{1}{2}}) \pi
\end{displaymath}

So, since $E$ $\vphantom0\raisebox{1.5pt}{$=$}$ ${\textstyle\frac{1}{2}}m\omega^2{x_2}^2$, the en­ergy lev­els are found to be $E_n$ $\vphantom0\raisebox{1.5pt}{$=$}$ $(n-{\textstyle\frac{1}{2}})\hbar\omega$. That is ex­act; the fact that in this case the val­ues of $n$ are counted from one in­stead of zero is just a mat­ter of no­ta­tions. De­spite the im­per­fect wave func­tions, it sure is a lot sim­pler than the ex­act de­riva­tion of chap­ter 4.1 as found in its note.