= Solution
Replacing $L$ by the <angular momentum operator> gives the <Hamiltonian operator>
$$
\hat H=\frac{\hat L^2}{2I}=-\frac{\hbar^2}{2I}\frac{\partial^2}{\partial\phi^2}.
$$
The time-dependent <Schrödinger equation> is therefore
$$
\boxed{i\hbar\frac{\partial\Psi}{\partial t}=-\frac{\hbar^2}{2mr^2}\frac{\partial^2\Psi}{\partial\phi^2}.}
$$
For a <stationary state>, separate variables as $\Psi(\phi,t)=\psi(\phi)T(t)$. Dividing by the product separates a function of $t$ from one of $\phi$:
$$
i\hbar\frac{\dot T}{T}=-\frac{\hbar^2}{2I}\frac{\psi''}{\psi}=E.
$$
Thus $T=e^{-iEt/\hbar}$ up to a constant, and the time-independent <Schrödinger equation> is $-\hbar^2\psi''/(2I)=E\psi$.
Diagonalize the commuting <angular momentum operator> at the same time: $-i\hbar\psi'=\ell\psi$ has solutions $Ae^{i\ell\phi/\hbar}$. A <single-valued wavefunction> requires $e^{2\pi i\ell/\hbar}=1$, hence $\ell=n\hbar$ with $n\in\mathbb Z$. Normalizing with the angular probability measure $d\phi$ gives
$$
\int_0^{2\pi}|A|^2\,d\phi=1,\qquad
\boxed{\psi_n(\phi)=\frac{e^{in\phi}}{\sqrt{2\pi}},\qquad
\Psi_n(\phi,t)=\frac{e^{in\phi-iE_nt/\hbar}}{\sqrt{2\pi}},\quad n\in\mathbb Z.}
$$
These form an <orthonormal basis> under $\int_0^{2\pi}\psi_m^*\psi_n\,d\phi=\delta_{mn}$. The <periodic boundary conditions> apply to both the <wavefunction> and its first <derivative>, ensuring the physical self-adjoint <Hamiltonian operator>. For probability per <arc length> $ds=r\,d\phi$, the normalization instead reads $\psi_n=e^{in\phi}/\sqrt{2\pi r}$; the two conventions describe the same probabilities. General states are <superpositions> of these <energy eigenstates>.
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