Berry phase in degenerare case

  • Context: Graduate 
  • Thread starter Thread starter wdlang
  • Start date Start date
  • Tags Tags
    Berry phase Phase
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
2 replies · 3K views
wdlang
Messages
306
Reaction score
0
Is there anyone familiar with berry phase?

Wilczek and Zee have a classic paper PRL 52_2111

I can not understand their equation (6)

I can not see why the first equality should hold

\eta_a and \eta_b should be orthogonal to each other, but why should \eta_b be orthogonal to the time derivative of \eta_a ?
 
Physics news on Phys.org
hmmm... yeah. compare this to the usual berry's phase case where U_{ab} is just a number e^{i\gamma} (e.g., in Sakurai "Modern Quantum Mechanics" p. 465). in that case we have
[tex] \frac{d}{dt} (e^{i\gamma(t)} n(t))=0[/tex]
where the time dependence in the n(t) (analogous to the \psi_a(t) in the Zee paper) is due to the changing parameter so that
[tex] 0=i\frac{d \gamma}{dt} e^{i\gamma} n + e^{i\gamma} \frac{d n}{dt}<br /> =i\frac{d\gamma}{dt} e^{i\gamma} n + e^{i\gamma} \frac{d \vec \lambda}{dt} \cdot \nabla_{(\lambda)} n[/tex]
so that (now using normalizaiton of n)
[tex] i\frac{d\gamma}{dt} + \frac{d \vec\lambda}{dt}\cdot (n,\nabla n)=0[/tex]

I guess, the Zee paper is just generalizing this to the case where instead of
[tex] \eta_a = e^{i\gamma}\psi_a[/tex]
with just a number for proportionality there is a matrix
[tex] \eta_a=U_{ab}\psi_b\;.[/tex]

I'm sure this isn't all too helpful, but maybe some other people will have more to say. cheers.
 
olgranpappy said:
hmmm... yeah. compare this to the usual berry's phase case where U_{ab} is just a number e^{i\gamma} (e.g., in Sakurai "Modern Quantum Mechanics" p. 465). in that case we have
[tex] \frac{d}{dt} (e^{i\gamma(t)} n(t))=0[/tex]
where the time dependence in the n(t) (analogous to the \psi_a(t) in the Zee paper) is due to the changing parameter so that
[tex] 0=i\frac{d \gamma}{dt} e^{i\gamma} n + e^{i\gamma} \frac{d n}{dt}<br /> =i\frac{d\gamma}{dt} e^{i\gamma} n + e^{i\gamma} \frac{d \vec \lambda}{dt} \cdot \nabla_{(\lambda)} n[/tex]
so that (now using normalizaiton of n)
[tex] i\frac{d\gamma}{dt} + \frac{d \vec\lambda}{dt}\cdot (n,\nabla n)=0[/tex]

I guess, the Zee paper is just generalizing this to the case where instead of
[tex] \eta_a = e^{i\gamma}\psi_a[/tex]
with just a number for proportionality there is a matrix
[tex] \eta_a=U_{ab}\psi_b\;.[/tex]

I'm sure this isn't all too helpful, but maybe some other people will have more to say. cheers.

Thanks a lot!

I read Simon's classic paper (PRL 51,2167), and i guess the equation in question reflects the parallel transport

similar eqaution also appears in Simon's paper (on the second page, left upper part)