Again, thanks to the SU(2) PSG proposed by prof.Wen, I can answer my question now, THT−1 is in fact SU(2) gauge equivalent to H, and the statement "H is also not SU(2) gauge equivalent to the time-reversal transformed Hamiltonian THT−1" in my question is wrong.
Let's rewrite the Hamiltonian as H(ψi)=∑<ij>(ψ†iχijψj+H.c.), where ψi=(fi↑,f†i↓)T and χij=(tij00−t∗ij). And divide the square lattice into two sublattices(nearest-neighbour sites belong to different sublattices) denoted as A and B. Now it's easy to see that TH(ψi)T−1=H(Giψi),Gi∈SU(2), with
Gi={iσy if i∈A−iσy if i∈B or
Gi={−iσy if i∈Aiσy if i∈B.
Thus, the projected spin-state ψspin indeed has the time-reversal symmetry as well as the translation symmetry.
Remarks:
In fact, as long as the mean-field Hamiltonian H(ψi) on the suqare lattice has the above form(containing only nearest-neighbour terms)
(1)with χij=(tijΔijΔ∗ij−t∗ij), the mean-field Hamiltonian H(ψi) always satisfies the above identity under time-reversal transformation, and thus the projected spin-state always has the time-reversal symmetry.
(2)on the other hand, if χij=(tij00−t∗ij), where tij are parametrized by four complex parameters t1,2,3,4 as shown in the Fig.1. in the paper, as long as t1,2,3,4 have equal magnitudes(no need for equal phase), then one can also show that the projected spin-state has the translation symmetry.
This post imported from StackExchange Physics at 2014-03-09 08:39 (UCT), posted by SE-user K-boy