= Solution
{c}
Since $P\psi_n=n\hbar\psi_n$, applying the <Hamiltonian operator> gives
$$
H\psi_n=\frac{P^2}{2m}\psi_n=\frac{n^2\hbar^2}{2m}\psi_n.
$$
Thus the <energy eigenvalues> and corresponding full <stationary states> are
$$
\boxed{E_n=\frac{n^2\hbar^2}{2m},\qquad \Psi_n(x,t)=e^{-in^2\hbar t/(2m)}\psi_n(x).}
$$
The <ground state> is $\psi_0$ with <energy> zero and <quantum degeneracy> one. For each $n=1,2,\ldots$, the positive <energy level> $n^2\hbar^2/(2m)$ has <quantum degeneracy> two: its <energy eigenspace> is $\operatorname{span}\{\psi_n,\psi_{-n}\}$. This is the <free particle spectrum on a circle>.
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