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<x<L}:

\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.

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