Past exam of the theoretical physics II course of the University of Cambridge / 2013 / Paper 1 / 1 / c / Solution

Take $\omega >0$ and $T=2\pi /\omega$. A full spin one-half rotation gives $R(T)=-I$, an essential sign. The cyclic phase of a spin in a rotating field has factors and total quantum phasesTo separate the adiabatic dynamical phase, set $B=\sqrt{{\Omega }_0^2+{\Omega }_R^2}>0$ and $\cos \theta ={\Omega }_0/B$. Then $\Omega =B-\omega \cos \theta +O({\omega }^2/B)$. The instantaneous laboratory energies are $\pm \hslash B/2$, so the adiabatic dynamical phases are $\mp BT/2$. Subtracting them from ${\Phi }_{\pm }$ givesThe oriented solid angle $\mathcal{A}$ is swept out by the field direction. These are the Berry phases for the eigenstates parallel and antiparallel to that direction. Retaining the spinor sign makes the quantum phase vanish modulo $2\pi$ when the field does not trace a nontrivial loop. The assumption $\omega \ll B$ suffices for adiabatic following; near degeneracy it must not be silently dropped.^[ref]