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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.

Replacing $L$ by the angular momentum operator gives the Hamiltonian operatorThe time-dependent Schrödinger equation is thereforeFor a stationary state, separate variables as $\Psi (\varphi ,t)=\psi (\varphi )T(t)$. Dividing by the product separates a function of $t$ from one of $\varphi$:Thus $T=e^{-iEt/\hslash }$ up to a constant, and the time-independent Schrödinger equation is $-{\hslash }^2{\psi }^{\prime \prime }/(2I)=E\psi$.

Diagonalize the commuting angular momentum operator at the same time: $-i\hslash {\psi }^{\prime }=\ell \psi$ has solutions $A{e}^{i\ell \varphi /\hslash }$. A single-valued wavefunction requires $e^{2\pi i\ell /\hslash }=1$, hence $\ell =n\hslash$ with $n\in \mathbb{Z}$. Normalizing with the angular probability measure $d\varphi$ givesThese form an orthonormal basis under ${\int }_0^{2\pi }{\psi }_m^{\ast }{\psi }_{n}d\varphi ={\delta }_{mn}$. The periodic boundary conditions apply to both the wavefunction and its first derivative, ensuring the physical self-adjoint Hamiltonian operator. For probability per arc length $ds=rd\varphi$, the normalization instead reads ${\psi }_n=e^{in\varphi }/\sqrt{2\pi r}$; the two conventions describe the same probabilities. General states are superpositions of these energy eigenstates.