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Consider a hamiltonian with degeneracies of energy in parameter space $(R)$.Now the Geometric phase(Wilczek Zee Potential) will acquire a non abelian nature. To prove the non abelian nature of Wilczek Zee potential one generally take an unitary transformation of wave function $\psi(t)$ to $\Omega(t)\psi(t)$ and show the potential has a non abelian gauge transformation,Is there any other way to show the non abelian nature of the Wilczek Zee potential? ($\psi$ is the wave function of the hamiltonian $\psi_{a}=U_{ab}(t)\eta_{b}(t)$ where $\eta(t)$ is the basis vector at each point of degeneracy. )

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This question is clearly incomplete, but for what's here, what did you have in mind exactly? The point is simply to generalize Berry's geometric phase calculation to hamiltonians with degeneracies. You don't assume the transformation of the wavefunction is of that form -- it's a standard bit of linear algebra. – wsc Feb 5 '12 at 20:02
I corrected it. – WInterfell Feb 5 '12 at 22:39
This arXiv preprint sounds related: – Qmechanic Feb 5 '12 at 23:06
I read the paper.Although I didn't know quantum computation but it seems to me is that, in the end they also showed the non abelianity of the potential by changing the basis by unitary transformation.All I am asking is whether there any different method to show the non ablianity of the potential. – WInterfell Feb 6 '12 at 20:28

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