Science Talk

Lou Pagnucco · Guest

I have a question on Berry phase for this very simple example:

Define D as the 4X4 diagonal 0-1 matrix

[1000]
[0000]
[0010]
[0000]

Define M as the symmetric 4X4 0-1 matrix

[0001]
[0011]
[0100]
[1100]

Define the parameterized Hamiltonian

H(x,y) = (x^2+y^2)*M + D (x,y real)

The maximum eigenvalue of H(x,y) has only one degeneracy at
(x,y) = (0,0)

Let P(t) be the circular path (sin(t),cos(t)) [0<=t<=2*pi].

The maximum eigenvalue of H on P(t) is always 2 and has an
eigenvector = |1,1,1,1>/2 independent of t.
The Hamiltonian on the path H(sin(t),cos(t)) is constant.

Since P(t) encircles the single degeneracy, shouldn't
adiabatically transporting the max eigenvector around P(t)
accumulate a Berry phase of pi, i.e., change the polarity
of the max eigenvector?

However this seems not to occur since the eigenvector
(and Hamiltonian) is constant.

What is my mistake?

Thanks,
Lou Pagnucco