Particle In A Circle

## Particle in a circular well: Quantum Mechanical description

Consider a particle of mass $m \neq 0$ confined to a circular well with

(1)\begin{align} V(r = R, \phi) = \infty \end{align}

and

(2)\begin{align} V(r, \phi) = 0 \end{align}

inside (i.e for $r < R$). The Schrodinger equation written for $r < R$ reads

(3)\begin{align} \hat{H}\psi = E\psi \end{align}

with

(4)\begin{align} \hat{H} = -\frac{\hbar^2}{2m}\nabla^{2} + V(r,\phi) \end{align}

The TISE reduces to

(5)\begin{align} \left(\frac{\partial^{2}}{\partial r^2} + \frac{1}{r}\frac{\partial}{\partial r} + \frac{1}{r^2}\frac{\partial^{2}}{\partial \phi^{2}}\right)\psi(r,\phi) = -\frac{2mE}{\hbar^2}\psi(r,\phi) \end{align}

We attempt a solution by separating variables:

(6)\begin{align} \psi(r,\phi) = \psi_{r}(r)\psi_{\phi}(\phi) \end{align}

This gives

(7)\begin{align} \left(\psi_{\phi}(\phi)\frac{\partial^{2}\psi_{r}(r)}{\partial r^2} + \frac{\psi_{\phi}(\phi)}{r}\frac{\partial \psi_{r}(r)}{\partial r} + \frac{\psi_{r}(r)}{r^2}\frac{\partial^{2}\psi_{\phi}(\phi)}{\partial \phi^{2}}\right) = -\frac{2mE}{\hbar^2}\psi_{r}(r)\psi_{\phi}(\phi) \end{align}

Dividing both sides by $\psi(r,\phi) = \psi_{r}(r)\psi_{\phi}(\phi)$ we get

(8)\begin{align} \frac{1}{\psi_{r}(r)}\frac{d^{2}\psi_{r}(r)}{dr^2} + \frac{1}{r\psi_{r}(r)}\frac{d\psi_{r}(r)}{dr} + \frac{1}{r^2\psi_{\phi}(\phi)}\frac{d^{2}\psi_{\phi}(\phi)}{d \phi^{2}} = -\frac{2mE}{\hbar^2} \end{align}

**Under construction**

page revision: 5, last edited: 03 Aug 2007 17:43