= Berry phase of a Bloch-sphere latitude
{c}
{title2=$\gamma=-\pi(1-\cos\theta)=-\Omega/2$}
Choose the <computational basis> $|0\rangle=|\uparrow_z\rangle$, $|1\rangle=|\downarrow_z\rangle$ for a <spin one-half> <qubit>. On the <Bloch sphere>, let $\mathbf n=(\sin\theta\cos\phi,\sin\theta\sin\phi,\cos\theta)$ and use the normalized <quantum state>
$$
|\psi(\theta,\phi)\rangle=\cos\frac{\theta}{2}|0\rangle+e^{i\phi}\sin\frac{\theta}{2}|1\rangle.
$$
This is the $+1$ <eigenstate> of $\mathbf n\cdot\boldsymbol\sigma$, where $\boldsymbol\sigma$ are the <Pauli matrices>. To make it an instantaneous <ground state>, take the model <Hamiltonian operator> $H=-\Delta\,\mathbf n\cdot\boldsymbol\sigma/2$ with constant $\Delta>0$. Its <energy eigenvalues> are $-\Delta/2$ and $+\Delta/2$, so its <spectral gap> is $\Delta$ everywhere. The <spin> points along $\mathbf n$; this explicit convention fixes the sign of the <Berry phase>.
Keep the polar angle $\theta$ fixed and increase the azimuth $\phi$ from $0$ to $2\pi$, slowly enough for the <adiabatic theorem>. The state vector written above is periodic in $\phi$. Directly differentiating gives
$$
\partial_\phi|\psi\rangle=i e^{i\phi}\sin\frac{\theta}{2}|1\rangle,\qquad
\langle\psi|\partial_\phi\psi\rangle=i\sin^2\frac{\theta}{2}.
$$
Thus the azimuthal <Berry connection> is $A_\phi=-\sin^2(\theta/2)$ and its <line integral> is
$$
\gamma=i\int_0^{2\pi}\langle\psi|\partial_\phi\psi\rangle\,d\phi
=-2\pi\sin^2\frac{\theta}{2}=-\pi(1-\cos\theta)=-\frac{\Omega}{2}\pmod{2\pi}.
$$
The north polar cap has oriented <solid angle> $\Omega=\int_0^{2\pi}\int_0^\theta\sin\vartheta\,d\vartheta\,d\phi=2\pi(1-\cos\theta)$, positive for increasing $\phi$. Reversing the loop reverses the <Berry phase>. The opposite <spin> <eigenstate> has the opposite <Berry phase> modulo $2\pi$.
\Image[/berry-phase-bloch-latitude.png]
{title=A 60-degree latitude on the Bloch sphere encloses a north polar cap of solid angle pi}
At $\theta=60^\circ$, the <solid angle> is $\pi$ and the <Berry phase> is $-\pi/2$. At $\theta=90^\circ$, the <solid angle> is $2\pi$ and the <Berry phase> is $-\pi$. At $\theta=0$, the loop collapses and $\gamma=0$. At $\theta=\pi$, the same expression gives $-2\pi\equiv0$: the <Bloch vector> stays at the south pole, while this representative is $e^{i\phi}|1\rangle$. Its winding is only a choice of <global phase>, so the physically relevant phase factor is again one. The displayed formula is a continuous, unwrapped representative, not a unique real-valued <observable>.
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