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

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.