Past exam of the mathematics course of the University of Cambridge / 2007 / ii / Paper 3 / 33A / Solution

Let $\epsilon =\omega /B\ll 1$. The exact frequency has expansion $\lambda =B-\omega \cos \alpha +O({\omega }^2/B)$. The ratio $(B-\omega \cos \alpha )/\lambda =1+O({\epsilon }^2)$, while the coefficient of the orthogonal state is uniformly $O(\epsilon )$. Hence the supplied solution reduces towith phase error $O({\omega }^{2}t/B)$, which is $O(\epsilon )$ over one cycle. This is the adiabatic approximation: the system remains in the instantaneous positive-energy eigenstate, with energy $E_{+}=\hslash B/2$.

Thus the dynamic phase is $-Bt/2$ and the geometric phase is $-\omega t(1-\cos \alpha )/2$, in the gauge for ${\chi }_{+}$ printed in the PDF. For one cycle $T=2\pi /\omega$, the Berry phase isminus half the solid angle swept by the field direction.

To calculate the requested integral, use $|\psi (\alpha ,\phi )⟩=(\cos (\alpha /2),e^{i\phi }\sin (\alpha /2){)}^T$. It has $i⟨\psi |{\partial }_{\phi }\psi ⟩=-{\sin }^2(\alpha /2)$; its radial and polar-angle connection components vanish in this gauge. Along the circle in parameter space, $\phi$ runs from zero to $2\pi$, so $\Gamma =-{\int }_0^{2\pi }{\sin }^2(\alpha /2)d\phi =-\pi (1-\cos \alpha )$, agreeing with the exact-solution limit.