= Solution
Applying the <Hamiltonian operator> to $\psi_n$ gives $\psi_n''=-n^2\psi_n$, and therefore
$$
\boxed{E_n=\frac{\hbar^2n^2}{2I}=\frac{\hbar^2n^2}{2mr^2},\qquad n\in\mathbb Z.}
$$
The <ground state> $n=0$ has zero <kinetic energy>. Every positive <energy level> is of <quantum degeneracy> two, because $n$ and $-n$ correspond to opposite <angular momenta> with equal squared magnitude.
The confinement scale suggests a tangential <momentum> scale $\hbar/r$ and thus an excitation <energy> scale $\hbar^2/(mr^2)$, consistent with the <Heisenberg uncertainty principle>. More precisely the <de Broglie wavelength> must fit an integer number of times around the circumference: $2\pi r=|n|\lambda$, so $|p|=h/\lambda=|n|\hbar/r$. This explains both the $r^{-2}$ dependence and the $n^2$ sequence.
One must not infer a nonzero <ground-state energy> by applying $\Delta\phi\Delta L\ge\hbar/2$ naively. Angle is periodic and is not a globally canonical position operator on the circle. The uniform $n=0$ <wavefunction>, with $\Delta L=0$, is a concrete counterexample to that naive inequality. The ordinary Cartesian <Heisenberg uncertainty principle> is not violated by the zero-<angular momentum> state.
Back to article page
