= Solution
Taking the modulus squared gives the <three-mode revival of a planar quantum rotor>:
$$
\boxed{\rho(\phi,t)=\frac1{3\pi}[1+\cos^22\phi+2\cos2\phi\cos(\Omega t)],\qquad \Omega=\frac{2\hbar}{I}.}
$$
The angular density has period $\pi$, while both the state and its density revive after $\boxed{T=2\pi/\Omega=\pi I/\hbar}$. Its <integral> over $0\leq\phi<2\pi$ is one because $\int\cos2\phi\,d\phi=0$ and $\int\cos^22\phi\,d\phi=\pi$.
At $t=0$, the density is $4\cos^4\phi/(3\pi)$, with maxima at $0,\pi$ and zeros at $\pi/2,3\pi/2$. At $t=T/2$ it becomes $4\sin^4\phi/(3\pi)$, interchanging those maxima and zeros. At $t=T/4$ and $3T/4$, it is $[1+\cos^22\phi]/(3\pi)$, with four equal maxima along the coordinate axes and positive minima halfway between them. At $t=T$ the original profile returns.
\Image[/past-exam-of-the-physics-a-course-of-the-university-of-cambridge/2018/paper-1-rotor-revival.png]
{title=Angle-time probability density and snapshots over one planar-rotor revival}
The changing profile is <interference> between the stationary $m=0$ mode and the degenerate counter-rotating $m=\pm2$ modes. It does not translate rigidly around the circle; the standing-wave weights alternate between two orientations.
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