### Two-dimensional hydrogen atom

Consider a particle of mass $$M$$ moving in the $$xy$$ plane in the presence of the Coulomb potential $V = -\frac{\alpha}{\sqrt{x^2 + y^2}} \qquad (\alpha > 0) \,.$ Bound states $$\psi$$ and energy lelvels $$E < 0$$ are determined by the time-independent Schrödinger equation $-\frac{\hbar^2}{2 M} \nabla^2 \psi + V \psi = E \psi \,.$ In polar coordinates ($$x = r \cos \theta$$, $$y = r \sin \theta$$), the equation reads $-\frac{\hbar^2}{2 M} \left( \frac{\partial^2}{\partial r^2} + \frac{1}{r} \frac{\partial}{\partial r} + \frac{1}{r^2} \frac{\partial^2}{\partial \theta^2} \right) \psi - \frac{\alpha}{r} \psi = E \psi \,.$ Using separation of variables, along with the condition that $$\psi$$ is single-valued, we get $\psi(r,\theta) = R(r) e^{i m \theta} \qquad (m \in \mathbb{Z}) \,,$ where the radial wave function $$R(r)$$ satisfies $\frac{d^2 R}{d r^2} + \frac{1}{r} \frac{d R}{d r} + \left( \frac{2 M E}{\hbar^2} + \frac{2 M \alpha}{\hbar^2 r} - \frac{m^2}{r^2} \right) R = 0 \,.$ Introducing dimensionless (positive) energy $$\mathcal{E}$$ and radius $$\rho$$ according to $E = -\frac{M \alpha^2}{2 \hbar^2} \mathcal{E} \qquad (\mathcal{E} > 0)\,,$ $r = \frac{\hbar^2}{2 M \alpha} \rho \,,$ we obtain $\frac{d^2 R}{d \rho^2} + \frac{1}{\rho} \frac{d R}{d \rho} + \left( -\frac{\mathcal{E}}{4} + \frac{1}{\rho} - \frac{m^2}{\rho^2} \right) R = 0 \,.$

Let us first consider the asymptotic behavior of $$R$$ at small and large $$\rho$$:

• Limit $$\rho \to 0$$. Suppose $$R \sim \rho^{\nu}$$ with $$\nu \ge 0$$. Then, the most singular terms in the equation balance each other out if $$\nu$$ is such that $$\nu (\nu-1) + \nu - m^2 = 0$$, or $$\nu = |m|$$. Hence, $R \sim \rho^{|m|} \,.$

• Limit $$\rho \to \infty$$. Suppose $$R \sim e^{-\lambda \rho}$$ with $$\lambda > 0$$. Since the left-hand side of the equation is dominated by $$R'' - (\mathcal{E}/4) R$$, we get $$\lambda^2 - \mathcal{E}/4 = 0$$, or $$\lambda = \sqrt{\mathcal{E}}/2$$. Hence, $R \sim e^{-\sqrt{\mathcal{E}} \rho / 2} \,.$

In view of the asymptotic behavior of $$R$$, we make the following substitution: $R = \rho^{|m|} e^{-\sqrt{\mathcal{E}} \rho / 2} S(\rho) \,,$ where $$S(\rho)$$ is a yet unknown function. A straightforward calculation shows that $$S$$ must satisfy $\rho \frac{d^2 S}{d \rho^2} + \Big( 2 |m| + 1 - \sqrt{\mathcal{E}} \rho \Big) \frac{d S}{d \rho} - \left( \frac{2 |m| + 1}{2} \sqrt{\mathcal{E}} - 1 \right) S = 0 \,.$ Introducing $z = \sqrt{\mathcal{E}} \rho \,,$ we find that $$S$$ is described by Kummer's equation $z \frac{d^2 S}{d z^2} + (b - z) \frac{d S}{d z} - a S = 0$ with $a = |m| + \frac{1}{2} - \frac{1}{\sqrt{\mathcal{E}}} \,,$ $b = 2 |m| + 1 \,.$ The solution (nonsingular at the origin) is given by Kummer's confluent hypergeometric function $M(a, b, z) = \sum_{k=0}^{\infty} \frac{a^{(k)}}{b^{(k)} k!} z^k = 1 + \frac{a}{b} z + \frac{a (a+1)}{b (b+1) 2!} z^2 + \ldots \,,$ where $$a^{(k)}$$ and $$b^{(k)}$$ are the rising factorials defined as $r^{(0)} = 1 \,, \qquad r^{(1)} = r \,, \qquad r^{(2)} = r (r + 1) \,, \qquad \ldots$ $r^{(k)} = r (r + 1) \ldots (r + k - 1) = \frac{\Gamma(r+k)}{\Gamma(r)} \,.$ Except when $$a = 0, -1, -2, \ldots \,,$$ $M(a,b,z) \sim \frac{z^{a-b} e^z}{\Gamma(a)} \qquad \text{as} \qquad z \to \infty \,,$ implying $$R \to \infty$$, which is physically impossible. When $$a = -n$$, with $$n = 0, 1, 2, \ldots \,,$$ function $$M(-n,b,z)$$ is a polynomial in $$z$$ of degree not exceeding $$n$$. In this case, $$R$$ decays to zero at infinity sufficiently fast and is normalizable. This yields the following quantization codition: $|m| + \frac{1}{2} - \frac{1}{\sqrt{\mathcal{E}}} = -n \,,$ or $\mathcal{E} = \frac{1}{\left( n + |m| + \frac{1}{2} \right)^2} \qquad (n = 0, 1, 2, \ldots; \; m = 0, \pm 1, \pm 2, \ldots) \,.$ The corresponding (unnormalized) radial eigenfunction is given by $R = \rho^{|m|} e^{-\sqrt{\mathcal{E}} \rho / 2} M(-n, \, 2 |m| + 1, \, \sqrt{\mathcal{E}} \rho) \,,$ or $R = \rho^{|m|} \exp \left[ -\frac{\rho}{2 n + 2 |m| + 1} \right] M \left( -n, \, 2 |m| + 1, \, \frac{2 \rho}{2 n + 2 |m| + 1} \right) \,.$

An alternative (and more conventional) way to label the bound states is to use quantum numbers $N \equiv n + |m| + 1 = 1, 2, 3, \ldots$ and $m = -(N-1), \, \ldots \,, N-1 \,.$ Then, $\mathcal{E} = \frac{1}{\left( N - \frac{1}{2} \right)^2} \qquad (N = 1, 2, 3, \ldots)$ and $R = \rho^{|m|} \exp \left[ -\frac{\rho}{2 N - 1} \right] M \left( -N + |m| + 1, \, 2 |m| + 1, \, \frac{2 \rho}{2 N - 1} \right) \,.$