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

  1. Jul 12, 2013 #1

    fmj

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    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?
     
  2. jcsd
  3. Aug 13, 2013 #2
    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,

     
  4. Aug 21, 2013 #3

    DrDu

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    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:

    [itex]\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 [/itex]


    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?
     
  5. Aug 21, 2013 #4

    DrDu

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    Science Advisor

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