Past exam of the physics A course of the University of Cambridge / 2011 / Paper 2 / Section B / B9

Put $\kappa =e^2/(4\pi {ϵ}_0)$. Substitution of $\psi =R(r)\Phi (\varphi )$ into the time-independent Schrödinger equation, followed by division by $R\Phi$ and multiplication by $r^2$, separates the angular dependence:Thus $\Phi =e^{im\varphi }/\sqrt{2\pi }$ andThe single-valued wavefunction condition, $\Phi (\varphi +2\pi )=\Phi (\varphi )$, requires $m\in \mathbb{Z}$. The regular boundary condition at the origin selects $R\sim r^{|m|}$. Accordingly use $s=|m|$ in the radial ansatz; the printed $x^m$ notation describes the $m\ge 0$ representatives.

For the two-dimensional Coulomb bound state, substitution of $R=x^s{e}^{-x/2}G$ with the stipulated radial and energy scales givesThe coefficient of $x^n$ in the power series isThe regular solution otherwise has an exponentially growing contribution at large $x$, incompatible with being normalizable. Termination at degree $n_r\ge 0$ therefore requiresThe resulting $G$ is proportional to a Generalized Laguerre polynomial.

Define $R_{\mathrm{y}}=\mu {\kappa }^2/(2{\hslash }^2)$, the hydrogenic Rydberg energy for the stipulated mass. The first three distinct two-dimensional energies are $-4R_{\mathrm{y}},-4R_{\mathrm{y}}/9,-4R_{\mathrm{y}}/25$, whereas in three dimensions they are $-R_{\mathrm{y}},-R_{\mathrm{y}}/4,-R_{\mathrm{y}}/9$. Both series approach zero from below, but the planar ground state is four times more tightly bound. Without spin the respective quantum degeneracies are $1,3,5$ and $1,4,9$; if individual degenerate states are counted, the second and third planar states already share the first excited energy. This comparison retains the supplied $1/r$ interaction; it does not replace it by the logarithmic electrostatic potential of a genuinely two-dimensional electromagnetic theory.