Past exam of the mathematics course of the University of Cambridge / 2023 / ib / Paper 4 / 15D / b

For a bound state, $E<0$ and therefore $\gamma >0$. As $r\to \infty$, the terms proportional to $1/r$ are subleading and the Radial Schrodinger equation has dominant balanceIts exponential behaviours are $e^{\gamma r}$ and $e^{-\gamma r}$. The growing solution is not normalizable, so

Put $R=f{e}^{-\gamma r}$. ThenAfter substitution and cancellation of the ${\gamma }^{2}f$ terms, the equation becomesFor the power series $f={\sum }_{n=0}^{\infty }{a}_n{r}^n$, the coefficient of $r^{n-1}$ isThus

If the series does not terminate, then $a_n/a_{n-1}\sim 2\gamma /n$, so $f$ acquires the large-$r$ behaviour $e^{2\gamma r}$ and $R$ grows like $e^{\gamma r}$. Normalizability therefore requires termination, which occurs whenfor some positive integer $N$. HenceThe ground state has $N=1$, so the hydrogen ground-state energy isThis is the series-termination quantization of the Coulomb radial equation.

Take the spherical harmonic $Y_{00}$ to have unit angular norm. The radial normalization condition is thenThereforeEquivalently, if $R$ denotes the entire spherically symmetric wavefunction rather than the radial factor multiplying normalized $Y_{00}$, then $C=({\gamma }^3/\pi {)}^{1/2}$.

The expected radius isFor the ground state, $\gamma =m{q}^2/{\hslash }^2$. The Bohr radius is $a_0={\hslash }^2/(m{q}^2)=1/\gamma$, soThe mean radius is therefore of the Bohr-radius scale and is one and a half times $a_0$. This is the radial normalization of the hydrogen ground state.