Past exam of the mathematics course of the University of Cambridge / 2016 / ib / Paper 3 / 16B / a / Solution

Use the physical Coulomb bound state boundary condition: the spherically symmetric wavefunction is regular at the origin and square integrable with measure $4\pi r^{2}dr$. For $E<0$, setAt large $r$, the decaying branch behaves exponentially as $e^{-Kr}$. Substituting $\psi =f{e}^{-Kr}$ into the radial Schrodinger equation givesThe regular solution has a power series $f(r)={\sum }_{j\ge 0}{a}_j{r}^j$, with recurrenceA nonzero regular solution has $a_0\ne 0$. If the series does not terminate, its coefficients eventually have one sign after choosing an overall real phase. For any fixed $0<\delta <1/2$ and sufficiently large $j$, the ratio of consecutive magnitudes is at least $2K(1-\delta )/(j+2)$. The tail therefore grows at least as a positive constant times $e^{2K(1-\delta )r}$ divided by a power of $r$; a finite polynomial of earlier terms cannot cancel it. Multiplication by $e^{-Kr}$ leaves exponential growth, contradicting normalizability.

Thus the series must terminate at degree $N-1$, where $KN=\lambda$ and $N$ is a positive integer. Conversely, termination gives a regular polynomial times $e^{-\lambda r/N}$, which is normalizable. ThereforeThese are the $\ell =0$ members of the hydrogen spectrum. Regularity at the origin is essential: square integrability alone on the punctured radial interval would also admit a singular $1/r$ branch. Such a branch is not a physical eigenfunction of the ordinary three-dimensional Coulomb Hamiltonian; its singularity introduces an inadmissible point-source contribution. In the reduced radial variable $u=r\psi$, the physical condition is $u(0)=0$.