Ring wave guide

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In quantum mechanics, the ring wave guide starts from the one dimensional, time independent Schrödinger equation:

$-\frac{\hbar^2}{2m}\nabla^2 \psi = E\psi$

Using polar coordinates on the 1 dimensional ring, the wave function depends only on the angular coordinate, and so

$\nabla^2 = \frac{1}{r^2} \frac{\partial^2}{\partial \theta^2}$

Requiring that the wave function be periodic in $\ \theta$ with a period $2 \ \pi$ (from the demand that the wave functions be single-valued functions on the circle), and that they be normalized leads to the conditions

$\int_{0}^{2 \pi} \left| \psi ( \theta ) \right|^2 \, d\theta = 1\$ ,

and

$\ \psi (\theta) = \ \psi ( \theta + 2\ \pi)$

Under these conditions, the solution to the Schrodinger equation is given by

$\psi(\theta) = \frac{1}{\sqrt{2 \pi}}\, e^{\pm i \frac{r}{\hbar} \sqrt{2 m E} \theta }$

The energy eigenvalues $E$ are quantized because of the periodic boundary conditions, and they are required to satisfy

$e^{\pm i \frac{r}{\hbar} \sqrt{2 m E} \theta } = e^{\pm i \frac{r}{\hbar} \sqrt{2 m E} (\theta +2 \pi)}$ , or

$e^{\pm i 2 \pi \frac{r}{\hbar} \sqrt{2 m E} } = 1 = e^{i 2 \pi n}$

This leads to the energy eigenvalues

$E = \frac{n^2 \hbar^2}{2 m r^2}$ where $n = 0,1,2,3, \ldots$

The full wave functions are, therefore

$\psi(\theta) = \frac{1}{\sqrt{2 \pi}} \, e^{\pm i n \theta }$

Quantum states found:

$n = 0$ :

ψ

is a constant function, and

E = 0

. This represents a stationary particle (no angular momentum spinning around the ring).

$n = 1$ :

$E = \frac{\hbar^2}{2 m r^2}$

and

$\psi(\theta) = \frac{1}{\sqrt{2 \pi}} \, e^{\pm i \theta }$

This produces two independent states that have the same energy level (degeneracy) and can be linearly combined arbitrarily; instead of $\exp(\pm\cdots)$ one can choose the sine and cosine functions. These two states represent particles spinning around the ring in clockwise and counterclockwise directions. The angular momentum is $\pm\hbar$ .

$n = 2$ (and higher):

the energy level is proportional to

n 2

, the angular momentum to n. There are always two (degenerate) quantum states.

Except for the case $n = 0$ , there are two quantum states for every value of $n$ (corresponding to $\ e^{\pm i n \theta}$ ). Therefore there are 2n+1 states with energies less than an energy indexed by the number n.

[edit] Application

In organic chemistry, aromatic compounds contain atomic rings, such as benzene rings (the Kekulé structure) consisting of five or six, usually carbon, atoms. So does the surface of "buckyballs" (buckminsterfullerene). These molecules are exceptionally stable.

The above explains why the ring behaves like a circular wave guide. The excess (valency) electrons spin around in both directions.

To fill all energy levels up to n requires $2\times(2n+1)$ electrons, as electrons have additionally two possible orientations of their spins.