Berry phase ($\gamma$)

A Berry phase is the geometric phase acquired when a nondegenerate energy eigenstate follows a closed cycle of parameters of a Hamiltonian operator in the ideal limit of the adiabatic theorem. The energy eigenstate must stay separated from other energy eigenstates by a nonzero spectral gap, and the parameters must change sufficiently slowly. Returning the Hamiltonian operator to its starting value returns the state to the same ray, but its quantum phase can remember the journey. The total quantum phase contains both the dynamical phase and the Berry phase.

The beauty of the Berry phase is that, within the adiabatic theorem's limit, it depends on the oriented path of the parameters rather than how long the journey takes. For a spin one-half qubit, the Berry phase of a Bloch-sphere latitude turns that memory into half a solid angle. Berry-phase interference with a reference path converts this geometry into a detection probability.

Michael Berry's 1984 paper formulated the geometric phase for cycles governed by the adiabatic theorem and derived the solid angle result for spin. Its bibliographic details are Proceedings of the Royal Society of London A, volume 392, pages 45–57, DOI 10.1098/rspa.1984.0023. ^[ref]; ^[ref]. The wider concept of geometric phase does not require every cycle to obey the adiabatic theorem: Aharonov and Anandan formulated a geometric phase for general cyclic quantum evolution in 1987. ^[ref].

For a normalized, smoothly chosen instantaneous energy eigenstate $|\psi (\mathbf{R})⟩$, the Berry connection has components $A_j=i⟨\psi |{\partial }_{R_j}\psi ⟩$. Its line integral around a closed parameter path $C$ gives the Berry phase in a periodic choice of state vector:Normalization makes the inner product $⟨\psi |{\partial }_{R_j}\psi ⟩$ purely imaginary, so the Berry connection is real. Changing the representative to $e^{i\chi (\mathbf{R})}|\psi ⟩$ changes the Berry connection to $A_j-{\partial }_{R_j}\chi$. A single-valued phase factor along the closed path changes the line integral by an integer multiple of $2\pi$, leaving $e^{i\gamma }$ unchanged. If the chosen vector is not periodic, include the endpoint phase:Here $\psi$ denotes the instantaneous energy eigenstate representative, with its dynamical phase removed. This is freedom to choose a representative of a state ray. ^[ref].

Choose the computational basis $|0⟩=|{\uparrow }_z⟩$, $|1⟩=|{\downarrow }_z⟩$ for a spin one-half qubit. On the Bloch sphere, let $\mathbf{n}=(\sin \theta \cos \varphi ,\sin \theta \sin \varphi ,\cos \theta )$ and use the normalized quantum stateThis is the $+1$ eigenstate of $\mathbf{n}\cdot \mathbf{\sigma }$, where $\mathbf{\sigma }$ are the Pauli matrices. To make it an instantaneous ground state, take the model Hamiltonian operator $H=-\Delta \mathbf{n}\cdot \mathbf{\sigma }/2$ 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 $\varphi$ from $0$ to $2\pi$, slowly enough for the adiabatic theorem. The state vector written above is periodic in $\varphi$. Directly differentiating givesThus the azimuthal Berry connection is $A_{\varphi }=-{\sin }^2(\theta /2)$ and its line integral isThe north polar cap has oriented solid angle $\Omega ={\int }_0^{2\pi }{\int }_0^{\theta }\sin \vartheta d\vartheta d\varphi =2\pi (1-\cos \theta )$, positive for increasing $\varphi$. Reversing the loop reverses the Berry phase. The opposite spin eigenstate has the opposite Berry phase modulo $2\pi$.

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 $-2\pi \equiv 0$: the Bloch vector stays at the south pole, while this representative is $e^{i\varphi }|1⟩$. 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.

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 $\Delta T/(2\hslash )$, so their relative dynamical phase vanishes in the ideal limit of the adiabatic theorem.

The joint quantum state before recombination then factors asThe Berry phase is now a relative quantum phase between the two paths. The calibrated plus output projects onto $|+⟩=(|\mathrm{ref}⟩+|\mathrm{loop}⟩)/\sqrt{2}$. The Born rule givesFor 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.