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I Berry phase without a magnetic field

  1. Nov 22, 2016 #1
    It is common to calculate Berry phases for quantum systems in, for example, a magnetic field. In this case we compute the Berry phase ##\gamma## using

    $$\gamma[C] = i\oint_C \! \langle n,t| \left( \vec{\nabla}_R |n,t\rangle \right)\,\cdot{d\vec{R}} \,$$

    where ##R## parametrizes the cyclic adiabatic process, in this case, the magnetic field.

    I was wondering what the Berry phase is for a system that has no external fields. Say, you have a particle in a box and the box rotates in three-dimensional box about some point. How do you compute the Berry phase for this system?
     
  2. jcsd
  3. Nov 22, 2016 #2

    PeterDonis

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    Taken at face value, your equation says that the Berry phase is zero if the magnetic field is zero. Do you have some reason to think that should not be the case?
     
  4. Nov 22, 2016 #3

    radium

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    The Berry phase is the phase gained by a wavefunction moving around a closed path. In a solid it would be from the variance of the wavefunction within the unit cell. You can formulate it as the loop integral of the Berry curvature, which is like a gauge potential in momentum space. This is just a general definition.

    The Berry phase is often used as a diagnostic of nontrivial topology in the wavefunction without any physical field. For example, in a Z2 topological insulator, the Berry phase corresponding to the topological invariant is the integral of half the Brillouin zone. The spin orbit interaction acts as a magnetic field for spin up and spin down electrons but there is no external magnetic field.
     
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