Bound state energy levels \(E<0\) of a two-dimensional hydrogen atom are determined by the Schrödinger equation \[\left( -\frac{\hbar^2}{2 m} (\partial_x^2 + \partial_y^2) - \frac{q^2}{\sqrt{x^2 + y^2}} \right) \psi = E \psi \,,\] where \(m\) and \(q\) are the mass and charge of the electron, respectively, and \(\psi(x,y)\) is the electronic wave function. This equation can be solved in terms of Kummer's confluent hypergeometric function (see this post for details). Here we show how the two-dimensional hydrogen atom can be mapped onto a two-dimensional harmonic oscillator. This mapping has been discussed, e.g., in Quantum Mechanics of H-Atom from Path Integrals.

We begin by making a coordinate transformation from \((x,y)\) to \((u,v)\) defined by \[\begin{align} x &= u^2 - v^2 , \\ y &= 2 u v \,. \end{align}\] From \[\begin{pmatrix} \partial_u \\ \partial_v \end{pmatrix} = J \begin{pmatrix} \partial_x \\ \partial_y \end{pmatrix}\] with \[J = \begin{pmatrix} \partial_u x & \partial_u y \\ \partial_v x & \partial_v y \end{pmatrix} = 2 \begin{pmatrix} u & v \\ -v & u \end{pmatrix} \,,\] it follows that \[\begin{pmatrix} \partial_x \\ \partial_y \end{pmatrix} = J^{-1} \begin{pmatrix} \partial_u \\ \partial_v \end{pmatrix} = \frac{1}{2 (u^2 + v^2)} \begin{pmatrix} u & -v \\ v & u \end{pmatrix} \begin{pmatrix} \partial_u \\ \partial_v \end{pmatrix} \,,\] or \[\begin{align} \partial_x &= \frac{1}{2 (u^2 + v^2)} (u \partial_u - v \partial_v) \,, \\ \partial_y &= \frac{1}{2 (u^2 + v^2)} (v \partial_u + u \partial_v) \,. \end{align}\] It is now straightforward to show that \[\partial_x^2 + \partial_y^2 = \frac{1}{4 (u^2 + v^2)} (\partial_u^2 + \partial_v^2) \,.\] Using the last equality, along with the identity \[\sqrt{x^2 + y^2} = u^2 + v^2 \,,\] we rewrite the Schrödinger equation as \[\frac{1}{4 (u^2 + v^2)} \left( -\frac{\hbar^2}{2 m} (\partial_u^2 + \partial_v^2) - 4 q^2 \right) \psi = E \psi \,.\] Rearranging the last equation, and keeping in mind that \(E < 0\), we arrive at the Schrödinger equation describing a two-dimensional harmonic oscillator: \[\left[ -\frac{\hbar^2}{2 m} (\partial_u^2 + \partial_v^2) + \frac{m \omega^2}{2} (u^2 + v^2) \right] \psi = \mathcal{E} \psi\] with \[\omega = \sqrt{\frac{8 (-E)}{m}} \,, \qquad \mathcal{E} = 4 q^2 \,.\] It is interesting to observe that the oscillator frequency, \(\omega\), is related to the atom energy, \(E\), whereas the oscillator energy, \(\mathcal{E}\), is related to the electron charge, \(q\).

In order to obtain the energy spectrum of the two-dimensional hydrogen atom, let us recall that the allowed energy values of the oscillator are given by \[\mathcal{E} = \hbar \omega (n_1 + n_2 + 1) \,,\] where \(n_1\) and \(n_2\) are non-negative integers. A careful consideration of the \((x,y) \to (u,v)\) coordinate transformation reveals that the \((u,v)\)-coordinate wave function \(\psi\), corresponding to the oscillator energy \(\hbar \omega (n_1 + n_2 + 1)\), has a physically meaningful (i.e. single-valued and normalizable) pre-image in the original \((x,y)\) coordinates only if \(n_1 + n_2\) is an even number, i.e. only if \(n_1 + n_2 = 2 n\) with \(n = 0, 1, 2, \ldots \,.\) This implies that the allowed energy levels of the hydrogen atom are determined by \(\mathcal{E} = \hbar \omega (2 n + 1)\), or \[4 q^2 = \hbar \sqrt{\frac{8 (-E)}{m}} (2 n + 1) \,.\] Solving the last equation for \(E\), we obtain the bound state energies of the two-dimensional hydrogen atom: \[E = -\frac{m q^4 / 2 \hbar^2}{\left( n + \frac{1}{2} \right)^{2}} \qquad (n = 0, 1, 2, \ldots) \,.\]