Register to reply

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

Share this thread:
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)
Phys.Org News Partner Physics news on Phys.org
New complex oxides could advance memory devices
'Squid skin' metamaterials project yields vivid color display
Scientists control surface tension to manipulate liquid metals (w/ Video)
fzero
#2
Nov2-13, 11:09 AM
Sci Advisor
HW Helper
PF Gold
P: 2,602
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##.


Register to reply

Related Discussions
Parallel Transport on a sphere Calculus & Beyond Homework 0
Parallel transport on sphere Special & General Relativity 15
Geometric understanding of integration / surface area of sphere Calculus 2
Parallel transport on the sphere Calculus & Beyond Homework 1
Lie transport vs. parallel transport Differential Geometry 10