For the “half-harmonic oscillator” (Example 9.3), make a plot comparing the normalized WKB wave function for n = 3 to the exact solution. You’ll have to experiment to determine how wide to make the patching region.

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Example 9.3 Potential well with one vertical wall. Imagine a potential well that has one vertical side (at x = 0)) and one sloping side (Figure 9.11). In this case (0) = 0, so Equation 9.47 says 1/16 20 or S² ["podx = (x - 1) xh. p(x) For instance, consider the "half-harmonic oscillator", л p(x)dx + = nπ, (n = 1, 2, 3, ...). 4 V(x) = In this case where x2 = p(x)=√√2m [E- (1/2)mw²x²] = mw√√ mw²x², (x > 0), En 2E (otherwise). 1 w m is the turning point. So #x2 ΠΕ S • 105 - p(x)dx = mw √√√x²-x²dx = ==mwx² = 20 and the quantization condition (Equation 2.48) yields = ( 2₁ - ²2 ) Fio = (2²/₁ 3 7 11 2' 2' 2 **** ħw. E K (9.48) Figure 9.11: Potential well with one vertical wall. (9.49) In this particular case the WKB approximation actually delivers the exact allowed energies (which are precisely the odd energies of the full harmonic oscillator-see Problem 2.41). V(x) A (9.50)