 Solution

Keep the electromagnetic units of the supplied Lagrangian, including its factor $c$, and regard $e$ as the signed coupling used there. Set $a_B=e{\Phi }_B/(2\pi c)$. The canonical momentum is $p_{\varphi }=I\dot{\varphi }+a_B$, so the Legendre transform givesSingle-valued periodic wavefunctions have normalized eigenstates $e^{in\varphi }/\sqrt{2\pi }$ with integer $n$. Defining dimensionless flux $f=e{\Phi }_B/(hc)$ gives $E_n={\hslash }^2(n-f{)}^2/(2I)$, and the trace is thereforeThe flux-threaded quantum rotor spectrum is periodic under $f\mapsto f+1$ after relabelling $n$, corresponding to flux period $hc/|e|$ in these units. A local gauge change can remove the magnetic vector potential from the differential operator but then imposes a twisted boundary condition with phase $e^{-2\pi if}$; it cannot remove the flux dependence from both the Hamiltonian and boundary condition simultaneously. In SI conventions the flux coupling has no $c$, so one must not mix the two unit systems.