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For the Gaussian-unit flux term $e{\Phi }_B\dot{\varphi }/(2\pi c)$, canonical momentum is shifted by $e{\Phi }_B/(2\pi c)$. Periodic wavefunctions then give the displayed spectrum for integer $n$. The flux period is $hc/|e|$ in these units; the corresponding SI expressions omit $c$. A gauge change transfers the flux to a twisted boundary condition rather than removing it from the physical problem. Half-integer $f$ gives a pair of degenerate nearest-momentum levels.

Define a transformed wavefunction byIndeed, writing $\psi =e^{iax/\hslash }\tilde{\psi }$ givesThis is the electromagnetic gauge transformation generated by $\chi (x)=-A_{x}x$: $A_x^{\prime }=A_x+{\partial }_x\chi =0$ and ${\psi }^{\prime }=e^{iq\chi /\hslash }\psi$. The transformed Hamiltonian operator is consequently $p^2/(2m)$, but the original periodic boundary condition becomes the twisted boundary conditionFor the domain of the self-adjoint kinetic Hamiltonian operator, the derivative has the same twist, ${\tilde{\psi }}^{\prime }(x+L)=e^{-iaL/\hslash }{\tilde{\psi }}^{\prime }(x)$. Thus the magnetic flux has moved into the boundary condition. The transformation is a single-valued gauge on periodic representatives only when $aL/\hslash$ is an integer multiple of $2\pi$; otherwise it is understood on the cut interval with the displayed identification of its endpoints.

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.