# Particle on a circular wire

Required math: calculus

Required physics: Schrödinger equation

References: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 2.46.

Suppose we have a particle that slides on a frictionless circular loop of circumference ${L}$. Since there is no friction, ${V=0}$, but unlike the free particle, we have a periodic constraint that ${\psi(x+L)=\psi(x)}$.

Since ${V=0}$, the solution is the same as that for the free particle, with the added condition that ${0:

$\displaystyle \Psi_{k}(x,t)=Ae^{i(kx-\hbar k^{2}t/2m)} \ \ \ \ \ (1)$

Normalizing this by integrating the square modulus over ${[0,L]}$ gives ${A=1/\sqrt{L}}$.

The constraint here is that ${\Psi(x)=\Psi(x+L)}$, where ${L}$ is the circumference of the wire. This leads to ${e^{i(kx+kL)}=e^{ikx}}$, so the condition on ${k}$ becomes ${k=2n\pi/L}$. Since the solution above assumes that ${k}$ can be positive or negative, ${n}$ takes on all positive and negative integer values. From this we can get the allowed energies:

$\displaystyle E_{\pm n}=\frac{2n^{2}\pi^{2}\hbar^{2}}{mL^{2}} \ \ \ \ \ (2)$

Note that the energy is the same for ${\pm n}$, so in this case there are two wave functions for each energy meaning that we have a one-dimensional system with degenerate states. However, in the proof that such states cannot exist, we assumed that the wave function was defined over all ${x}$, not just a restricted range. Thus the condition that the wave function goes to zero at infinity is not true here, so the conditions for non-degeneracy specified in the proof do not hold here.