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In a temporal winding sector of a quantum rotor, decompose the lifted angle into a straight winding path and a fluctuation vanishing at both endpoints. The cross term in the kinetic action vanishes, so all sectors have the same Gaussian path integral. With inverse-energy imaginary time its value is the displayed diagonal free-particle kernel on the line. Integrating the initial angle separately supplies $2\pi$. Thus this fixed-endpoint factor is an amplitude density, not an additional rotor partition function or an uncounted angular zero mode.

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.

Set $x=r\varphi$ and $a=2\pi r$. The wrapped spatial form isHence the differential expression for the quantum rotor acts asThe box expression becomes $E_n=n^2{\hslash }^2/(8m{r}^2)$, so consistency givesthe moment of inertia of a point mass a distance $r$ from the axis. Equivalently, ${\partial }_x=r^{-1}{\partial }_{\varphi }$ transforms the kinetic operator directly. Being an eigenfunction of this differential expression is not yet sufficient: the ring boundary conditions must also hold.

For a wavefunction normalized with the angular measure $\mathrm{d}\varphi$, use ${\Phi }_n(\varphi )=\sqrt{r}{\Psi }_n(r\varphi )={\pi }^{-1/2}\sin (n\varphi /2)$. This change of normalization convention does not change the energy eigenvalue.

For the quantum rotor, the transition kernel is the path integralThe integral includes all paths on the circle, not just paths whose chosen real lifts have the same endpoint difference. Operator evolution and the stated energy-state normalization giveThe factor $1/(2\pi )$ comes from the completeness relation; it makes the zero-time limit the periodic Dirac delta distribution. For real time the oscillatory sum is understood through $T\to T-i0$, or as a distributional limit.

Single-valued rotor wavefunctions are periodic in $\varphi$. The normalized eigenfunctions are $⟨\varphi |n⟩=e^{in\varphi }/\sqrt{2\pi }$ with integer $n$, angular momentum $\hslash n$ and energy $E_n={\hslash }^2{n}^2/(2I)$. The quantum rotor consequently hasThe nonzero levels have the $n\leftrightarrow -n$ degeneracy; the ground state is $n=0$. No twisted boundary condition or applied flux has been introduced.

The quantum rotor has orientation angle $\varphi$ of period $2\pi$, so physical wavefunctions satisfy $\psi (\varphi +2\pi )=\psi (\varphi )$. The exponential solutions of $Hu=Eu$ are thereforeThey are normalized and orthogonal on $[0,2\pi )$. The corresponding angular momentum operator is $L_z=-i\hslash {\partial }_{\varphi }$, with eigenvalue $m\hslash$. The $m=0$ ground level is nondegenerate, and every $|m|>0$ energy level has the twofold degeneracy $m\leftrightarrow -m$. This uses ordinary periodic boundary conditions, with no flux or twisted quantum phase.

For the untwisted quantum rotor, periodicity on the angle circle requires $\psi (\varphi +2\pi )=\psi (\varphi )$. The normalized energy eigenstates and their angular momentum areTaking the trace in this orthonormal basis therefore givesThe n=0 state is nondegenerate and each nonzero $|n|$ has two opposite-angular-momentum states. No flux or twisted angular boundary condition is present.

The quantum rotor with angle period $2\pi$ has both the spectral kernel $(2\pi {)}^{-1}{\sum }_n{e}^{in\Delta -i\hslash n^{2}T/(2I)}$ and the displayed winding representation. The Poisson summation formula converts one into the other by a Gaussian integral. Each winding sector has a constant-velocity classical path and action $I(\Delta +2\pi m{)}^2/(2T)$. Quadratic fluctuations give the same prefactor in every sector, so this is exact. Real time is understood by $T\to T-i0$, which fixes the square-root branch and convergence prescription.

For the quantum rotor, low temperature makes the angular-momentum sum converge fastest: n=0 gives $Z\to 1$. High temperature makes the temporal winding sum converge fastest: m=0 gives $Z\sim \sqrt{2\pi I/(\beta {\hslash }^2)}$. At low temperature many winding sectors contribute and cancel the vanishing fixed-endpoint prefactor. A zero-winding approximation valid at high temperature cannot be extended to low temperature merely because winding paths look geometrically longer.

For a quantum rotor with angle period $2\pi$, let $u_m=e^{im\varphi }/\sqrt{2\pi }$ and prepare $\psi (\varphi ,0)=\sqrt{4/(3\pi )}{\cos }^2\varphi$. Its only nonzero angular-momentum amplitudes are $c_0=\sqrt{2/3}$ and $c_{\pm 2}=1/\sqrt{6}$. Their energies are $0$ and $2{\hslash }^2/I$, respectively, soThe state revives after $T=\pi I/\hslash$ and its density at $T/2$ is the initial density rotated by $\pi /2$. At $T/4$ there are four equal angular maxima. This is interference between a stationary component and a degenerate pair of counter-rotating components; it is not a uniformly rotating rigid profile. Angular-momentum and energy probabilities stay constant.