Chapter 8: Q7P (page 334)

Use the WKB approximation to find the allowed energies of the harmonic oscillator.

Short Answer

Expert verified

The allowed energies of the harmonic oscillator are $(\frac{1}{2}, \frac{3}{2} \frac{, 5}{2}, \dots) ℏ ω .$

Step by step solution

To find the energies of harmonic oscillator.

Consider a harmonic oscillator with mass and frequency $ω$ . The associated potential is

$V (x) = \frac{1}{2} m ω^{2} x^{2} w h i c h c o r r e s p o n d s t o a p o t e n t i a l w e l l w i t h n o v e r t i c a l w a l l s . T h u s, t h e E q u a t i o n s a y s \int_{x_{1}}^{x_{2}} p (x) d x = (n - \frac{1}{2}) πħ, (n = 1, 2, 3, . . .) \dots \dots \dots . (1) where x_{1} and x_{2} are the turning points satisfying E = \frac{1}{2} {mω}^{2} {x_{1}}^{2} = \frac{1}{2} {mω}^{2} {x_{2}}^{2}, (or) x_{1} = x_{2} = \frac{1}{ω} \sqrt{\frac{2 E}{m}} . In this case, p (x) = \sqrt{2 m (E - \frac{1}{2} {mω}^{2} x^{2})} p (x) = mω \sqrt{{x_{2}}^{2} - x^{2}} Thus, \int_{x_{1}}^{x_{2}} p (x) dx = 2 mω \int_{0}^{x_{2}} \sqrt{{x_{2}}^{2} - x^{2}} dx This integral is done using the substitution x \equiv x_{2} \sin θ, so \int_{x_{1}}^{x_{2}} p (x) dx = \frac{π}{2} mω {x_{2}}^{2} = \frac{πE}{ω}, and the quantization condition (Equation (1)) yields E_{n} = (n - \frac{1}{2}) ħω = (\frac{1}{2}, \frac{3}{2}, \frac{5}{2}, . . .) ħω . In this particular case the WKB approximation actually delivers the exact allowed energies,$