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.

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    $\begingroup$ This state has time reversal symmetry, but if hopping is turned on in the mean field Hamiltonian the resulting state will no longer have time reversal symmetry. You may be interested in the supplementary information to a recent paper I was involved with, at arxiv.org/abs/1307.0829 $\endgroup$ Jan 25, 2014 at 22:48
  • $\begingroup$ @Jim Garrison Yes, I agree with you. If the nearest neighbor hopping $t$ is turned on, then the triangle Wilson loop $W_l$ will take a nonzero imaginary value $\propto it\Delta^2$ and $W_l$ is changed to $-W_l$ under TR operation, thus TR symmetry would be broken. $\endgroup$
    – Kai Li
    Jan 26, 2014 at 14:59
  • $\begingroup$ @Jim Garrison But I want to know that whether the $SU(2)$ matrices mentioned in my question exist? And from which viewpoint(Wilson loop or $SU(2)$ matrices) you infer that the projected spin state has TR symmetry? $\endgroup$
    – Kai Li
    Jan 26, 2014 at 15:03
  • $\begingroup$ See the discussion of TR symmetry and Wilson loops in arXiv:1409.7820 $\endgroup$
    – Kai Li
    Apr 25, 2016 at 7:45

1 Answer 1


I just found that the solution of $SU(2)$ matrices is really simple.

When there is no hopping term, the projected spin state of the above $d+id$ mean-field Hamiltonian indeed has TR symmetry. Because there exist global $SU(2)$ matrices $G_i$ which implement the TR transformation, say $G_i=i\tau ^x$.


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