= Berry-phase interference with a reference path
{c}
{title2=$P_+=(1+\cos\gamma)/2$}
An isolated <qubit> that acquires only a <global phase> has unchanged measurement <probabilities>. To reveal the <Berry phase>, use two coherent paths of a <Mach-Zehnder interferometer>, represented by a path <qubit>. In one arm the internal <spin one-half> <qubit> follows the loop; in the reference arm it stays in its starting <energy eigenstate>. Assume equal arm amplitudes, identical final internal states, cancelled relative <dynamical phases>, and calibrated fixed apparatus phases. In the model above, hold the reference <Hamiltonian operator> at its initial value for the same duration. Both arms then have <energy eigenvalue> $-\Delta/2$ and acquire the same <dynamical phase> $\Delta T/(2\hbar)$, so their relative <dynamical phase> vanishes in the ideal limit of the <adiabatic theorem>.
The joint <quantum state> before recombination then factors as
$$
\frac{|\mathrm{ref}\rangle+e^{i\gamma}|\mathrm{loop}\rangle}{\sqrt2}\otimes|\psi(\theta,0)\rangle.
$$
The <Berry phase> is now a relative <quantum phase> between the two paths. The calibrated plus output projects onto $|+\rangle=(|\mathrm{ref}\rangle+|\mathrm{loop}\rangle)/\sqrt2$. The <Born rule> gives
$$
P_+=\left|\frac{1+e^{i\gamma}}{2}\right|^2=\frac{1+\cos\gamma}{2},\qquad P_-=1-P_+.
$$
For the <Berry phase of a Bloch-sphere latitude>, the plus-port <probabilities> at $\theta=0,60^\circ,90^\circ,180^\circ$ are respectively $1,1/2,0,1$. This ideal <Mach-Zehnder interferometer> measurement cannot distinguish $\gamma$ from $-\gamma$ or $\gamma+2\pi$ by itself; scanning a known reference phase can resolve the sign. An uncancelled relative <dynamical phase> would shift the fringe as well.
\Image[/berry-phase-versus-loop-angle.png]
{title=Unwrapped Berry phase and ideal reference-path interference probability for latitude loops}
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