 Solution

A stationary state has $\Psi (t,x)=e^{-iEt/\hslash }\psi (x)$. Substitution into the Schrödinger equation gives the energy eigenvalue equationThe trial wavefunction $C{e}^{ikx}$ satisfies it with $E={\hslash }^2{k}^2/(2m)$. The periodic boundary condition requires $e^{2\pi ikR}=1$, hence $k=n/R$, $n\in \mathbb{Z}$. The wavefunction normalization is $1=2\pi R|C{|}^2$, so a convenient common global phase gives the orthonormal basisIndeed, ${\int }_0^{2\pi R}{\varphi }_n^{\ast }{\varphi }_{q}dx={\delta }_{nq}$. This is the free particle spectrum on a circle. The ground state ${\varphi }_0$ has energy zero and is nondegenerate. Each positive energy level ${\hslash }^2{n}^2/(2m{R}^2)$, $n=1,2,\dots$, has energy eigenspace $\mathrm{span}\{{\varphi }_n,{\varphi }_{-n}\}$ and quantum degeneracy two. Real sine and cosine combinations give another basis of the same energy eigenspace.