 Solution

On periodic ring states the angular-momentum index $m$ is an integer. At the stated resonance,so $U_T$ is the identity on every orbital state. After $N$ steps from the normalized ground state,Its phase varies in space but its position probability density remains uniform. Periodicity permits integration by parts, givingConsequentlyEquivalently, the angular-momentum probabilities are $P_m=J_m(N\eta {)}^2$, with ${\sum }_m{m}^2{J}_m(z{)}^2=z^2/2$. The mean angular momentum remains zero while its variance grows as $N^2$. This quantum resonance of the kicked rotor gives coherent ballistic energy growth. The Bessel second-moment identity connects the momentum spreading to that quadratic energy law; an incoherent random walk would instead give linear energy growth.