I need help to find the energy eigen values of Hydrogen atom using WKB approach. So far I know, the radial equation is given by
$$\frac{1}{r^2} \frac{\partial }{\partial r} \left( r^2 \frac{\partial R(r)}{\partial r}\right) + \frac{2 \mu}{\hbar^2} \left( E – V(r) – \frac{\ell(\ell+1)\hbar^2}{2 \mu r^2}\right)R(r) = 0 \tag{1}$$
also I know the relation, for equation of the form
$$u''(x) + k(x)^2 u(x) = 0 \tag{2}$$
the solution is of the form (for $E<V$)
$$u(x) = \frac{C_1}{\sqrt{k(x)}} \sin \left( \int^x k(x) dx\right) + \frac{C_2}{\sqrt{k(x)}} \cos \left( \int^x k(x) dx\right)\tag{3}$$
as well as the relation
$$\oint \hbar k(x) dx = \left( n + \frac 1 2 \right) \pi \hbar. \tag{4}$$
I have it's solution on note written in class but it's hardly understandable. How do I transform equation $(1)$ to equation $(2)$ and what do I use for the bounds of integration in equation $(4)$ to get the energy eigenvalue?
I have done similar question for harmonic oscillator where the bounds of integration in equation $(4)$ is $\pm$ turning points (solution of $E(x) = V(x)$) but not sure about this one.
ADDED:: Changing $R(r) = u(r)/r$ changes into
$$ \frac{d^2u(r)}{dr^2} + \frac{2 \mu}{\hbar^2} \left( E – V(r) – \frac{\ell(\ell+1)\hbar^2}{2 \mu r^2}\right)u(r) = 0. \tag{5}$$
Changing $V = -e^2/r$ gives
$$\int_{R_{min}}^{R_{max}} \sqrt{2 \mu \left( E + \frac{e^2}{r}- \frac{\ell(\ell+1)\hbar^2}{2 \mu r^2}\right)}dr = \left( n + \frac 1 2\right) \hbar \pi.\tag{6}$$
Now what do I choose my bounds for $r$? The final answer is given as
$$E_n = – \frac 1 2 \cdot \frac{\mu e^4}{\hbar^2 (n + \ell+1)^2}.\tag{7}$$
Best Answer
The bounds for r should still be the classical turning points, as you mentioned for the harmonic oscillator. Presumably you're in a bound state of Hydrogen, i.e. have an energy of the form $\frac{-13.6 eV}{n^2}$ for some integer n. The problem then reduces to finding the zeros of the equation $$\frac{-13.6 eV}{n^2} = -\frac{e}{r^2} - \frac{l(l+1) \hbar^2}{2 \mu r^2}$$ as a function of r.
EDIT: Changed gs energy from -13.6 MeV to -13.6 eV. Thanks to Ruslan for pointing out the error.