Consider the following $d+id$ mean-field Hamiltonian for a spin-1/2 model on a triangular lattice
$$H=\sum_{<ij>}(\psi_i^\dagger\chi_{ij}\psi_j+H.c.)$$, with $\chi_{ij}=\begin{pmatrix}
0 & \Delta_{ij}\\
\Delta_{ij}^* & 0
\end{pmatrix}$, 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