Analyze the bouncing ball (Problem 8.5) using the WKB approximation.
(a) Find the allowed energies,${\mathit{E}}_{\mathbf{n}}$ , in terms of , and .
(b) Now put in the particular values given in Problem8.5 (c), and compare the WKB approximation to the first four energies with the "exact" results.
(c) About how large would the quantum number n have to be to give the ball an average height of, say, 1 meter above the ground?

For a free fall under the influence of gravity, the potential energy is V(x)=mgx, whereis the gravitational acceleration and xis the height above the floor.

$$\begin{gathered} \mathrm{Hence}, \\ \mathrm{V}\left(\mathrm{x}\right)=\left\{\begin{array}{ll}\mathrm{mgx}& \mathrm{x}>0\\ \infty & \mathrm{x}<0\end{array}\right. \\ \mathrm{which}\mathrm{corresponds}\mathrm{to}\mathrm{a}\mathrm{potential}\mathrm{well}\mathrm{that}\mathrm{has}\mathrm{one}\mathrm{vertical}\mathrm{wall}(\mathrm{at}\mathrm{x}=0)\mathrm{and}\mathrm{one} \\ \mathrm{sloping}\mathrm{side} \\ {\int }_0^{{\mathrm{x}}_2}\mathrm{p}\left(\mathrm{x}\right)\mathrm{dx}=\left(\mathrm{n}-\frac{1}{4}\right)\mathrm{\pi ħ},\left(\mathrm{n}=1,2,3,...\right)..........\left(1\right) \\ \mathrm{where}{\mathrm{x}}_2\mathrm{is}\mathrm{the}\mathrm{turning}\mathrm{point}\mathrm{satisfying}, \\ \mathrm{E}={\mathrm{mgx}}_2\left(\mathrm{or}\right){\mathrm{x}}_2{\mathrm{x}}_2=\frac{\mathrm{E}}{\mathrm{mg}}. \\ \mathrm{In}\mathrm{this}\mathrm{case}, \\ \mathrm{p}\left(\mathrm{x}\right)=\sqrt{2\mathrm{m}\left(\mathrm{E}-\mathrm{mgx}\right)}=\mathrm{m}\sqrt{2\mathrm{g}}\sqrt{{\mathrm{x}}_2-\mathrm{x}.} \\ \mathrm{Thus}, \end{gathered}$$

$$\begin{gathered} {\int }_0^{x_2}p\left(x\right)dx=m\sqrt{2g}{\int }_0^{x_2}\sqrt{x_2-xdx}=\frac{2}{3}m{x}_2^{3/2}\sqrt{2g}=\frac{2E^{3/2}}{3g}\sqrt{\frac{2}{m},} \\ \mathrm{and}\mathrm{the}\mathrm{quantization}\mathrm{condition}(\mathrm{Equation}(1\left)\right)\mathrm{yields} \\ {\mathrm{E}}_{\mathrm{n}}^{3/2}=\frac{3}{2}\left(\mathrm{n}-\frac{1}{4}\right)\mathrm{\pi ħg}\sqrt{\frac{\mathrm{m}}{2}}, \\ \left(0\mathrm{r}\right) \\ {\mathrm{E}}_{\mathrm{n}}={\left[\frac{9}{8}{\mathrm{\pi }}^2{\mathrm{mg}}^2{\mathrm{ħ}}^2{\left(\mathrm{n}-\frac{1}{4}\right)}^2\right]}^{1/3}. \\ \mathrm{To}\mathrm{draw}\mathrm{the}\mathrm{graph}\mathrm{for}\mathrm{the}\mathrm{potential}\mathrm{of}\mathrm{the}\mathrm{bouncing}\mathrm{ball}, \end{gathered}$$

$$\begin{gathered} \mathrm{using}\mathrm{g}=9.8\mathrm{m}/{\mathrm{s}}^2\mathrm{and}\mathrm{m}=0.100\mathrm{kg},\mathrm{we}\mathrm{have} \\ {\mathrm{E}}_{\mathrm{n}}\approx 1.058\times {10}^{-22}{\left(\mathrm{n}-\frac{1}{4}\right)}^{2/3}\mathrm{J} \\ \mathrm{and}\mathrm{substituting}\mathrm{with}\mathrm{n}=1,2,3,4,\mathrm{the}\mathrm{WKB}\mathrm{approximation}\mathrm{to}\mathrm{first}\mathrm{four}\mathrm{energies} \\ (\mathrm{correct}\mathrm{to}\mathrm{three}\mathrm{significant}\mathrm{digits})\mathrm{are} \\ {E}_1=8.734\times {10}^{-23}\mathrm{J} \\ {\mathrm{E}}_2=1.536\times {10}^{-22}\mathrm{J} \\ {\mathrm{E}}_3=2.077\times {10}^{-22}\mathrm{J} \\ {\mathrm{E}}_4=2.077\times {10}^{-22}\mathrm{J} \end{gathered}$$

we arrived at the formula for the average height of the ball in terms of the energies $E_n$: