Past exam of the chemistry A course of the University of Cambridge / 2024 / Paper 1 / A1 / b

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.

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.

Solving $-i\hslash u^{\prime }(\varphi )=\lambda u(\varphi )$ gives an exponential. The periodic boundary condition quantizes its phase winding:These are normalized with respect to $\mathrm{d}\varphi$. The corresponding complex Fourier series isUnlike the wrapped sine subset, this orthonormal basis includes the constant $k=0$ state and both sine and cosine combinations. If one retains arc-length normalization instead, replace $1/\sqrt{2\pi }$ by $1/\sqrt{a}$ and integrate with $r\mathrm{d}\varphi$.

For the allowed $n=2k$ states, the normalized standing wave is a quantum superposition of two opposite angular momentum eigenstates:The Born rule therefore predictsThe expectation value is zero, but a single measurement never gives zero for these $k\ge 1$ states. Their uncertainty is $\Delta l_z=k\hslash$. The two components have the same rotor energy ${\hslash }^2{k}^2/(2I)$, so a definite energy does not imply definite angular momentum.