 Solution

Applying the Hamiltonian operator to ${\psi }_n$ gives ${\psi }_n^{\prime \prime }=-n^2{\psi }_n$, and thereforeThe 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 $\hslash /r$ and thus an excitation energy scale ${\hslash }^2/(m{r}^2)$, 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 $|p|=h/\lambda =|n|\hslash /r$. This explains both the $r^{-2}$ dependence and the $n^2$ sequence.

One must not infer a nonzero ground-state energy by applying $\Delta \varphi \Delta L\ge \hslash /2$ 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.