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

$$E_n = - \frac 1 2 \cdot \frac{\mu e^4}{\hbar^2 (n + \ell+1)^2}.\tag{7}$$

  • 1
    $\begingroup$ To get from (1) to (2) try to find the equation for $u(r)$ when it something like $R(r) \equiv \frac{u(r)}{r}$. $\endgroup$
    – gatsu
    Commented Feb 8, 2014 at 19:27

3 Answers 3


The hydrogen potential

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.

  • $\begingroup$ Megaelectronvolts?! $\endgroup$
    – Ruslan
    Commented Mar 1, 2014 at 15:21

How do I transform equation (1) to equation (2)

plug $R(r)=u(r)/r$ into (1), you'll get (2) immediately, where $k(x)$ would the expression before $R(r)$ in the second term of (2)

and what do I use for the bounds of integration in equation (4) to get the energy eigenvalue?



You are deal with action variables so the integrations extrem are always give by the classical trajectory. So for the hidrogen atom in bound state they are 2 times the distance between R1 and R2.


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