Past exam of the physics A course of the University of Cambridge / 2018 / Paper 1 / Section B / B8

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.

Expand the initial wavefunction in the angular momentum eigenbasis using ${\cos }^2\varphi =1/2+(e^{2i\varphi }+e^{-2i\varphi })/4$. With the normalized $u_m$ from part (a), this givesThe coefficient squares sum to one, also verifying the stated normalization. The Born rule therefore gives angular momentum outcomesFor an energy measurement the two distinct $m=\pm 2$ states belong to the same energy eigenspace. Sum their probabilities, rather than adding their amplitudes, to obtainTheir relative probabilities are $4:1:1$ for the angular momenta and $2:1$ for the energies. In particular, zero mean angular momentum does not imply zero rotational energy.

Each energy eigenstate evolves by its own quantum phase $e^{-i{E}_{m}t/\hslash }$. The two $m=\pm 2$ components have the same energy, while the $m=0$ quantum phase stays one. ThusIn angle representation this isAt $t=0$, $1+\cos 2\varphi =2{\cos }^2\varphi$ recovers the initial state. The fixed coefficient magnitudes keep the angular momentum and energy probabilities constant even though the density in angle changes.

Taking the modulus squared gives the three-mode revival of a planar quantum rotor:The angular density has period $\pi$, while both the state and its density revive after $\boxed{T=2\pi /\Omega =\pi I/\hslash }$. Its integral over $0\le \varphi <2\pi$ is one because $\int \cos 2\varphi d\varphi =0$ and $\int {\cos }^{2}2\varphi d\varphi =\pi$.

At $t=0$, the density is $4{\cos }^4\varphi /(3\pi )$, with maxima at $0,\pi$ and zeros at $\pi /2,3\pi /2$. At $t=T/2$ it becomes $4{\sin }^4\varphi /(3\pi )$, interchanging those maxima and zeros. At $t=T/4$ and $3T/4$, it is $[1+{\cos }^{2}2\varphi ]/(3\pi )$, with four equal maxima along the coordinate axes and positive minima halfway between them. At $t=T$ the original profile returns.

The changing profile is interference between the stationary $m=0$ mode and the degenerate counter-rotating $m=\pm 2$ modes. It does not translate rigidly around the circle; the standing-wave weights alternate between two orientations.