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

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lld212
#1
Nov1-13, 06:37 AM
P: 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
α(C)=Im∫Cn*dn.

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
File Type: pdf 1988-M.V.Berry-The Quantum Phase-Five years after.pdf (594.9 KB, 1 views)
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fzero
#2
Nov2-13, 11:09 AM
Sci Advisor
HW Helper
PF Gold
P: 2,606
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##.


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