Let's consider the Aharanov-Bohm effect. Following Girvin & Yang, an infinitely long, very thin flux tube running along the $\hat z$ axis is surrounded by a strong potential barrier preventing charged particles from entering the region containing the field. A particle which is adiabatically taken around this flux tube $n$ times will acquire a phase of $n \Phi$, where $\Phi$ is the amount of flux in the tube. Thus, closed paths have quantized Berry phase.

Now, again following Girvin & Yang, let us consider a Spin-1/2 system. with the Hamiltonian $\vec h \cdot \vec \sigma$ for Pauli matrices $\sigma_i$ and 3-vector corresponding to the parameter space $\vec h$. Now W.L.O.G we take $\left|h\right|=1$, and we adiabatically move the parameters $h$ through a closed contour $C$ on the unit sphere, the system will acquire a Berry phase $\frac 1 2 \Omega_C$ where $\Omega_C$ is the solid angle subtended by $C$. Why is the Berry phase in this example not quantized, but the Berry phase in the previous example quantized? When is the Berry phase quantized?


1 Answer 1


Berry phase is equal to the surface integration of Berry Curvature. In the first case, Berry curvatures are located in the tube, so you move your particle around the tube will collect all the Berry phase and it get quantization result. By the way - if you move the particle into the tube, you also get non-quantized Berry phase.

In the spin model, the source of Berry phase placed at the point $h=0$ spreads curvature into all spaces, and you can no longer collect all Berry phase, which leads to non-quantized and route-dependent results.


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