Radial Schrodinger equation for the hydrogen atom

For zero orbital angular momentum and Coulomb potential $-q^2/r$, the radial equation is

For a normalized maximal-angular-momentum hydrogen state, the probability inside a small radius $r_p$ is the integral of $|R{|}^2{r}^2$. Since $R$ behaves as $r^l$ near zero, it equals $2^{2l+3}(r_p/a{)}^{2l+3}/[(2l+3)!(l+1{)}^{2l+3}]$ times $1+O(r_p/a)$. Increasing angular momentum strongly suppresses penetration toward the Coulomb centre.

A hydrogen atom state with principal quantum number $n=l+1$ has no radial nodes and radial factor $R(r)=A{r}^l{e}^{-r/[a(l+1)]}$, where $a$ is the Bohr radius. Its energy is $-{\hslash }^2/[2m{a}^2(l+1{)}^2]$. With unit-normalized spherical harmonics, the radial normalization gives $|A{|}^2=2^{2l+3}/[a^{2l+3}(l+1{)}^{2l+3}(2l+2)!]$.

After the logarithmic radial coordinate $r=e^x$, the Langer replacement $l(l+1)\mapsto (l+1/2{)}^2$ turns the radial Coulomb problem into a two-turning-point WKB approximation problem. Its action integral giveswhich in this Coulomb case agrees with the exact dependence on the principal quantum number.

For a radial solution regular at the origin, writing $R=e^{-\gamma r}{\sum }_{n\ge 0}{a}_n{r}^n$ givesNormalizability at infinity forces the series to terminate, so $2\gamma N=\beta$ for a positive integer $N$ and $E_N=-m{q}^4/(2{\hslash }^2{N}^2)$.

The lowest Coulomb bound-state energy is