Consider the Hamiltonian

$$H=\mathbf{B}(t) \cdot \mathbf{S}$$

for a particle of spin $\frac{1}{2}$ fixed in space, in a rotating magnetic field, where

$$S_{1}=\frac{\hbar}{2}\left(\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right), \quad S_{2}=\frac{\hbar}{2}\left(\begin{array}{cc} 0 & -i \\ i & 0 \end{array}\right), \quad S_{3}=\frac{\hbar}{2}\left(\begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array}\right)$$

and

$$\mathbf{B}(t)=B(\sin \alpha \cos \omega t, \sin \alpha \sin \omega t, \cos \alpha)$$

with $B, \alpha$ and $\omega$ constant, and $B>0, \omega>0$.

There is an exact solution of the time-dependent Schrödinger equation for this Hamiltonian,

$$\chi(t)=\left(\cos \left(\frac{1}{2} \lambda t\right)-i \frac{B-\omega \cos \alpha}{\lambda} \sin \left(\frac{1}{2} \lambda t\right)\right) e^{-i \omega t / 2} \chi_{+}+i\left(\frac{\omega}{\lambda} \sin \alpha \sin \left(\frac{1}{2} \lambda t\right)\right) e^{i \omega t / 2} \chi_{-}$$

where $\lambda \equiv\left(\omega^{2}-2 \omega B \cos \alpha+B^{2}\right)^{1 / 2}$ and

$$\chi_{+}=\left(\begin{array}{c} \cos \frac{\alpha}{2} \\ e^{i \omega t} \sin \frac{\alpha}{2} \end{array}\right), \quad \chi_{-}=\left(\begin{array}{c} e^{-i \omega t} \sin \frac{\alpha}{2} \\ -\cos \frac{\alpha}{2} \end{array}\right)$$

Show that, for $\omega \ll B$, this exact solution simplifies to a form consistent with the adiabatic approximation. Find the dynamic phase and the geometric phase in the adiabatic regime. What is the Berry phase for one complete cycle of $\mathbf{B}$ ?

The Berry phase can be calculated as an integral of the form

$$\Gamma=i \oint\left\langle\psi \mid \nabla_{\mathbf{R}} \psi\right\rangle \cdot d \mathbf{R}$$

Evaluate $\Gamma$ for the adiabatic evolution described above.