Where does the Berry phase of $\pi$ come from in a topological insulat

[fmj]

The Berry connection and the Berry phase should be related. Now for a topological insulator (TI) (or to be more precise, for a quantum spin hall state, but I think the Chern parities are calculated in the same fashion for a 3D TI). I can follow the argument up to defining the Chern parity $\nu$, where the Berry connection $\textbf{A}$ is explicitly within this equation where it gets integrated over half the Brillouin zone:

$\nu = \frac{1}{2\pi} \left( \oint\limits_{\partial(A + B)} d\textbf{k} \cdot \textbf{A} - \iint\limits_{A + B} dk_x dk_y \nabla \times \textbf{A} \right) \text{mod } 2 $


Is there a simple way to show that the surface states must have a $\pi$ Berry phase? Maybe
by invoking time-reversal symmetry on this equation?

For the Integer Quantum Hall Effect the conductance was used to show how the Berry phase is connected to the Berry connection. But
I would like to avoid using charge polarization, I just want to see the direct relation between the Berry connection on the BZ and the resulting Berry phase of the wavefunctions that is exactly equal to $\pi$, e.g. a fermion must undergo two complete rotations to acquire a phase of $2\pi$. Is there a simple argument to relate this equation with the time reversal contraint to the Berry phase?

 

[Jolb]

I saw this very nice question a week or two ago and I was hoping a real expert would weigh in because I am studying topological states of matter and have some questions myself. Anyway, I just found a simple explanation of the π Berry phase that hopefully answers your question. From http://arxiv.org/abs/1002.3895,

Unlike an ordinary metal, which has up and down spins at every point on the Fermi surface, the surface states are not spin degenerate. Since T symmetry requires that states at momenta k and -k have opposite spin, the spin must rotate with k around the Fermi surface, as shown in Fig. 7(b). This leads to a nontrivial Berry phase acquired by an electron going around the Fermi circle. T symmetry requires that this phase be 0 or π. When an electron circles a Dirac point, its spin rotates by 2π, which leads to a π Berry phase.

 

[DrDu]

Just nicifying the original question (OP, use two hashes instead of the dollar symbol to enter latex):

The Berry connection and the Berry phase should be related. Now for a topological insulator (TI) (or to be more precise, for a quantum spin hall state, but I think the Chern parities are calculated in the same fashion for a 3D TI). I can follow the argument up to defining the Chern parity $\nu$, where the Berry connection $\textbf{A}$ is explicitly within this equation where it gets integrated over half the Brillouin zone:

$\nu = \frac{1}{2\pi} \left( \oint\limits_{\partial(A + B)} d\textbf{k} \cdot \textbf{A} - \iint\limits_{A + B} dk_x dk_y \nabla \times \textbf{A} \right) \text{mod } 2$Is there a simple way to show that the surface states must have a $\pi$ Berry phase? Maybe
by invoking time-reversal symmetry on this equation?

For the Integer Quantum Hall Effect the conductance was used to show how the Berry phase is connected to the Berry connection. But
I would like to avoid using charge polarization, I just want to see the direct relation between the Berry connection on the BZ and the resulting Berry phase of the wavefunctions that is exactly equal to $\pi$, e.g. a fermion must undergo two complete rotations to acquire a phase of $2\pi$. Is there a simple argument to relate this equation with the time reversal contraint to the Berry phase?

 

[DrDu]

Are you referring to a particular article? How is A defined exactly in your case? What are the regions A and B?