Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Geometric phase of a parallel transport over the surface of a sphere

  1. Nov 1, 2013 #1
    I have this question on the calculation of the geometric phase (Berry phase) of a parallel transporting vector over the surface of a sphere, illustrated by Prof. Berry for example in the attached file starting on page 2.
    The vector performing parallel transport is defined as ψ=(e+ie')/√2,
    satisfying the parallel transport law, Imψ*=0.
    Then another local basis was defined, n(r)=(u(r)+iv(r))/√2,
    and ψ=n(r)exp(-iα).
    Together the geometric phase (or so called anholonomy) is given as

    I can't see the difference between n and ψ here, except for a phase factor α. I think both of them performing the same parallel transport with α being constant. But why
    Imψ*=0 while Imn*dn≠0, even with the latter being a gauge of the geometric phase?

    Thanks in advance.

    Attached Files:

  2. jcsd
  3. Nov 2, 2013 #2


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    The phase is local, expressed as ##\alpha=\alpha(\mathbf{t})##, where ##t## is a parameter on the path. You should be able to compute that ## \mathrm{Im}\mathbf{n}^* \cdot d\mathbf{n} = d\alpha##.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook