Editing WKB approximation (section)

=== Quantum bouncing ball ===
Consider the following potential a bouncing ball is subjected to:

<math>V(x) = \begin{cases}
mgx & \text{if }  x \geq 0\\
  \infty & \text{if } x < 0 \\
 \end{cases}</math>

The wavefunction solutions of the above can be solved using the WKB method by considering only odd parity solutions of the alternative potential <math>V(x) = mg|x|</math>. The classical turning points are identified <math display="inline">x_1 = - {E \over mg}   </math> and <math display="inline">x_2 = {E \over mg}    </math>. Thus applying the quantization condition obtained in WKB:

<math display="block">\int_{x_1}^{x_2} \sqrt{2m \left( E-V(x)\right)}\,dx = (n_{\text{odd}}+1/2)\pi \hbar</math>

Letting <math display="inline">n_{\text{odd}}=2n-1 </math> where <math display="inline">n = 1,2,3,\cdots      </math>, solving for <math display="inline">E </math> with given <math>V(x) = mg|x|</math>, we get the quantum mechanical energy of a bouncing ball:<ref>{{Cite book |last1=Sakurai |first1=Jun John |title=Modern quantum mechanics |last2=Napolitano |first2=Jim |date=2021 |publisher=Cambridge University Press |isbn=978-1-108-47322-4 |edition=3rd |location=Cambridge}}</ref>

<math display="block">E = {\left(3\left(n-\frac 1 4\right)\pi\right)^{\frac 2 3} \over 2}(mg^2\hbar^2)^{\frac 1 3}. </math>

This result is also consistent with the use of equation from bound state of one rigid wall without needing to consider an alternative potential.