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 $H(\psi_i)=\sum_{<ij>}(\psi_i^\dagger\chi_{ij}\psi_j+H.c.)$, where $\psi_i=(f_{i\uparrow},f_{i\downarrow}^\dagger)^T$ and $\chi_{ij}=\begin{pmatrix}
t_{ij} & 0\\
0 & -t_{ij}^*
\end{pmatrix}$. 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 $$TH(\psi_i)T^{-1}=H(G_i\psi_i),G_i\in SU(2)$$, with
$G_i=\begin{cases}
i\sigma_y& \text{ if } i\in A \\
-i\sigma_y& \text{ if } i\in B
\end{cases}$ or
$G_i=\begin{cases}
-i\sigma_y& \text{ if } i\in A \\
i\sigma_y& \text{ if } i\in B
\end{cases} .$
Thus, the projected spin-state $\psi_{spin}$ indeed has the time-reversal symmetry as well as the translation symmetry.

**Remarks:**
In fact, as long as the mean-field Hamiltonian $H(\psi_i)$ on the suqare lattice has the above form(containing only **nearest-neighbour** terms)

(1)with $\chi_{ij}=\begin{pmatrix}
t_{ij} & \Delta_{ij}\\
\Delta_{ij}^* & -t_{ij}^*
\end{pmatrix}$, the mean-field Hamiltonian $H(\psi_i)$ 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 $\chi_{ij}=\begin{pmatrix}
t_{ij} & 0\\
0 & -t_{ij}^*
\end{pmatrix}$, where $t_{ij}$ are parametrized by four complex parameters $t_{1,2,3,4}$ as shown in the Fig.1. in the paper, as long as $t_{1,2,3,4}$ 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