 Solution

A smooth quantum rotor with no flux or twist requires bothFor the sine family, ${\Phi }_n(0)={\Phi }_n(2\pi )=0$ holds for every positive integer $n$, butThe derivatives match only for $\boxed{n=2k,k=1,2,3,\dots }$. Equivalently, $\sin [n(\varphi +2\pi )/2]=(-1{)}^n\sin (n\varphi /2)$ must be periodic for every $\varphi$, not merely vanish at the chosen join. Odd $n$ would produce a cusp in a periodic continuation and do not belong to the smooth free-ring Hamiltonian's domain.