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Berry Phase & Topological Phases of Matter

Geometric phase from adiabatic transport, and how it gives rise to topological invariants.

Consider a Hamiltonian $\hat H(\mathbf R)$ depending on parameters $\mathbf R$. Slowly varying $\mathbf R$ along a closed loop, an instantaneous eigenstate $|n(\mathbf R)\rangle$ acquires not only a dynamical phase but also a Berry phase

$$\gamma_n = i \oint \langle n(\mathbf R) | \nabla_{\mathbf R} | n(\mathbf R)\rangle \cdot d\mathbf R = \oint \mathbf A_n \cdot d\mathbf R,$$

where $\mathbf A_n = i \langle n | \nabla_{\mathbf R} | n\rangle$ is the Berry connection. By Stokes,

$$\gamma_n = \int_S \mathbf F_n \cdot d\mathbf S, \quad \mathbf F_n = \nabla \times \mathbf A_n \quad (\text{Berry curvature}).$$

For a spin-$\tfrac{1}{2}$ in a slowly rotating magnetic field, the Berry phase is half the solid angle swept: $\gamma = \Omega/2$. The Berry curvature acts as a magnetic field in parameter space.

Topological band insulators have nontrivial integrated Berry curvature (Chern number $C \in \mathbb Z$) over the Brillouin zone. The integer quantum Hall conductivity is $\sigma_{xy} = (e^2/h)\, C$. Edge states protected by topology cannot be removed by smooth deformations that preserve the gap.

Interactive: Berry phase for a spin in a rotating field

The path traces a cone of half-angle $\theta$ on the parameter sphere; the Berry phase equals half the enclosed solid angle.

Quiz

1. Berry phase is:
2. For a spin-1/2 in a slowly rotating magnetic field tracing solid angle $\Omega$:
3. The Chern number of a band is:
4. Quantum Hall conductivity in units of $e^2/h$ is:
5. Topological edge states are:
6. The Berry connection $\mathbf A$ under gauge transformations $|n\rangle \to e^{i\chi(\mathbf R)}|n\rangle$: