The Hydrogen (Coulomb potential) radial equation is solved by finding the behavior at large \bgroup\color{black}$r$\egroup, then finding the behavior at small \bgroup\color{black}$r$\egroup, then using a power series solution to get

\begin{displaymath}\bgroup\color{black}R(\rho)=\rho^\ell\sum\limits_{k=0}^\infty a_k\rho^k e^{-\rho/2}\egroup\end{displaymath}

with \bgroup\color{black}$\rho=\sqrt{-8\mu E\over\hbar^2}r$\egroup. To keep the wavefunction normalizable the power series must terminate, giving us our energy eigenvalue condition.

\begin{displaymath}\bgroup\color{black}E_n=-{Z^2\alpha^2mc^2\over 2n^2}\egroup\end{displaymath}

Here \bgroup\color{black}$n$\egroup is called the principle quantum number and it is given by


where \bgroup\color{black}$n_r$\egroup is the number of nodes in the radial wavefunction. It is an odd feature of Hydrogen that a radial excitation and an angular excitation have the same energy.

So a Hydrogen energy eigenstate \bgroup\color{black}$\psi_{n\ell m}(\vec{x})=R_{n\ell}(r)Y_{\ell m}(\theta,\phi)$\egroup is described by three integer quantum numbers with the requirements that \bgroup\color{black}$n\geq 1$\egroup, \bgroup\color{black}$\ell<n$\egroup and also an integer, and \bgroup\color{black}$-l\leq m\leq\ell$\egroup. The ground state of Hydrogen is \bgroup\color{black}$\psi_{100}$\egroup and has energy of -13.6 eV. We compute several of the lowest energy eigenstates.

The diagram below shows the lowest energy bound states of Hydrogen and their typical decays.


Jim Branson 2013-04-22