Consider the following d+id mean-field Hamiltonian for a spin-1/2 model on a triangular lattice
H=∑<ij>(ψ†iχijψj+H.c.), with χij=(0ΔijΔ∗ij0), fermionic spinons \psi_i=\binom{f_{i\uparrow}}{f_{i\downarrow}^\dagger}, and the mean-field parameters \Delta_{ij}=\Delta_{ji} defined on links have the same magnitudes and their phases differ by \frac{2\pi}{3} with each other referring to the three bond-direction.
My question is, does the projected spin state \Psi=P\phi have the TR symmetry? Where \phi is the mean-field ground state of H, and P removes the unphysical states with empty or doubly occupied sites.
Notice that from the viewpoint of Wilson loop, you can check that the Wilson loops W_l=tr(\chi_{12}\chi_{23}\chi_{31})=0 on each triangle plaquette, thus all the Wilson loops are invariant under the TR transformation W_l\rightarrow W_l^*=W_l. Thus, the TR symmetry should be maintained.
On the other hand, from the viewpoint of SU(2) gauge-transformation, if there exist SU(2) matrices G_i such that \chi_{ij}\rightarrow\chi_{ij}^*=G_i\chi_{ij}G_j^\dagger, then the projected spin state \Psi is TR invariant. But so far, I can not find out those SU(2) matrices G_i. So can anyone work out the explicit form of those SU(2) matrices G_i? Or they do not exist at all?
Thanks in advance.
By the way, I think it would be awkward to explicitly write the form of state \Psi to check the TR symmetry.
This post imported from StackExchange Physics at 2014-03-09 08:38 (UCT), posted by SE-user K-boy