The WKB approximation (see Challenge 40.64) can be used to

Chapter 40, Problem 40.65

(choose chapter or problem)

The WKB approximation (see Challenge Problem 40.64) can be used to calculate the energy levels for a harmonic oscillator. In this approximation, the energy levels are the solutions to the equation

                                              \(\int_a^b \sqrt{2 m[E-U(x)]} d x=\frac{n h}{2} \quad n=1,2,3, \ldots\)

Here E is the energy, U(x) is the potential-energy function, and x = a and x = b are the classical turning points (the points at which E is equal to the potential energy, so the Newtonian kinetic energy would be zero). (a) Determine the classical turning points for a harmonic oscillator with energy E and force constant \(k^{\prime}\). (b) Carry out the integral in the WKB approximation and show that the energy levels in this approximation are \(E_n=\hbar \omega\), where \(\omega=\sqrt{k^{\prime} / m}\) and \(n=1,2,3, \ldots\). (Hint: Recall that \(\hbar=h / 2 \pi\). A useful standard integral is

                                                     \(\int \sqrt{A^2-x^2} d x=\frac{1}{2}\left[x \sqrt{A^2-x^2}+A^2 \arcsin \left(\frac{x}{|A|}\right)\right]\)

where arcsin denotes the inverse sine function. Note that the integrand is even, so the integral from -x to x is equal to twice the integral from 0 to x.) (c) How do the approximate energy levels found in part (b) compare with the true energy levels given by Eq. (40.46)? Does the WKB approximation give an underestimate or an overestimate of the energy levels?