Consider, for small $\epsilon$, the equation

$$\epsilon^{2} \frac{d^{2} \psi}{d x^{2}}-q(x) \psi=0$$

Assume that $(*)$ has bounded solutions with two turning points $a, b$ where $b>a, q^{\prime}(b)>0$ and $q^{\prime}(a)<0$.

(a) Use the WKB approximation to derive the relationship

$$\frac{1}{\epsilon} \int_{a}^{b}|q(\xi)|^{1 / 2} d \xi=\left(n+\frac{1}{2}\right) \pi \text { with } n=0,1,2, \cdots$$

[You may quote without proof any standard results or formulae from WKB theory.]

(b) In suitable units, the radial Schrödinger equation for a spherically symmetric potential given by $V(r)=-V_{0} / r$, for constant $V_{0}$, can be recast in the standard form $(*)$ as:

$$\frac{\hbar^{2}}{2 m} \frac{d^{2} \psi}{d x^{2}}+e^{2 x}\left[\lambda-V\left(e^{x}\right)-\frac{\hbar^{2}}{2 m}\left(l+\frac{1}{2}\right)^{2} e^{-2 x}\right] \psi=0$$

where $r=e^{x}$ and $\epsilon=\hbar / \sqrt{2 m}$ is a small parameter.

Use result $(* *)$ to show that the energies of the bound states (i.e $\lambda=-|\lambda|<0)$ are approximated by the expression:

$$E=-|\lambda|=-\frac{m}{2 \hbar^{2}} \frac{V_{0}^{2}}{(n+l+1)^{2}}$$

[You may use the result

$$\left.\int_{a}^{b} \frac{1}{r} \sqrt{(r-a)(b-r)} d r=(\pi / 2)[\sqrt{b}-\sqrt{a}]^{2} .\right]$$