= 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 phases>
$$
\boxed{e^{i\Phi_\pm}=-e^{\mp i\pi\Omega/\omega},\qquad
\Phi_\pm=\pi\mp\frac{\pi\Omega}{\omega}\pmod{2\pi}}.
$$
To 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 $\pm\hbar B/2$, so the adiabatic <dynamical phases> are $\mp BT/2$. Subtracting them from $\Phi_\pm$ gives
$$
\boxed{\gamma_\pm=\mp\pi(1-\cos\theta)=\mp\frac{\mathcal A}{2}\pmod{2\pi},\qquad
\mathcal A=2\pi(1-\cos\theta)}.
$$
The 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.<Quantal phase factors accompanying adiabatic changes by Michael Victor Berry>[Section 4, spinning particles in slowly changing magnetic fields]{ref}
\Image[/past-exam-of-the-theoretical-physics-ii-course-of-the-university-of-cambridge/2013/paper-1-rotating-spin-phase.png]
{title=Original rotating-field cone and convergence of the exact cyclic phase to the spin-one-half geometric phase}
Back to article page
