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Showing posts with the label Two-dimensional systems

Two-dimensional hydrogen atom as a harmonic oscillator

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

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}