Why does the $\pi$-flux state have time-reversal symmetry?

Question

It's known that the $\pi$-flux state of the antiferromagnetic Heisenberg model on the square lattice is an important concept. The $\pi$-flux state is described by the (simplified) mean-field Hamiltonian $$H=t_1f_{1\sigma}^\dagger f_{2\sigma}+t_2f_{2\sigma}^\dagger f_{3\sigma}+t_3f_{3\sigma}^\dagger f_{4\sigma}+t_4f_{4\sigma}^\dagger f_{1\sigma}+H.c.$$, where $t_i=\left | t \right |e^{i\frac{\pi}{4}}(i=1,2,3,4)$, and the spin-1/2 operator is $\mathbf{S}_i=\frac{1}{2}f_i^\dagger\mathbf{\sigma}f_i$.

It's obvious that the mean-field Hamiltonian $H$ is not invariant under time-reversal operation( $T$ ), say $H\neq THT^{-1}$, and $H$ is also not $SU(2)$ gauge equivalent to the time-reversal transformed Hamiltonian $THT^{-1}$. So due to what reason, the projected spin-state $\psi_{spin}=\hat{P}\psi_{MF}$ is time-reversal invariant? Where $\psi_{MF}$ is the ground state of the mean-field Hamiltonian $H$ and $\hat{P}=\prod (2\hat{n}_i-\hat{n}_i^2)$ is the projection to the spin subspace.

Remarks: Here the effect of translation(with one lattice spacing along the $\hat{x}$ or $\hat{y}$ direction) on the Hamiltonian $H$ is the same as the effect of time-reversal $T$. Thus, if the spin-state $\psi_{spin}$ has $T$ symmetry, it must also have the translation symmetry.

Thanks in advance.

Answer 1

I don't know the article you refer to, but I believe the Hamiltonian you discuss should get a $\pi$-phase shift after one turn around a (2D) lattice cell. So I guess it should read $H=F^{\dagger}\cdot H_{\pi}\cdot F$ with

$$H_{\pi}=t\left(\begin{array}{cccc} 0 & e^{\mathbf{i}\pi/4} & 0 & e^{-\mathbf{i}\pi/4}\\ e^{-\mathbf{i}\pi/4} & 0 & e^{\mathbf{i}\pi/4} & 0\\ 0 & e^{-\mathbf{i}\pi/4} & 0 & e^{\mathbf{i}\pi/4}\\ e^{\mathbf{i}\pi/4} & 0 & e^{-\mathbf{i}\pi/4} & 0 \end{array}\right)$$

and $F^{\dagger}=\left(\begin{array}{cccc} f_{1}^{\dagger} & f_{2}^{\dagger} & f_{3}^{\dagger} & f_{4}^{\dagger}\end{array}\right)$. Then, one has

$$H_{\pi}=\dfrac{t}{\sqrt{2}}\left[\left(1+\tau_{x}\right)\otimes\eta_{x}-\left(1-\tau_{x}\right)\otimes\eta_{y}\right]$$

where the $\eta$ and $\tau$ are the usual Pauli matrices.

Time reversal symmetry operator -- when it exists -- is defined as an anti-unitary operator which commutes with the Hamiltonian. Such an operator can be defined as $T=\mathscr{K}\tau_{z}\otimes\mathbf{i}\eta_{y}$ and thus $H$ is time reversal symmetric. $\mathscr{K}$ is the anti-unitary operator $\mathscr{K}\left[\mathbf{i}\right]=-\mathbf{i}$ and thus $\mathscr{K}\left[\eta_{y}\right]=-\eta_{y}$. One verifies that $\left[H_{\pi},T\right]=0$ as it must.

Please tell me if I started with the wrong Hamiltonian.

A few words about the definition (as follow from the comment below): The time-reversal operator is defined as I did, i.e. one applies it to the Hamiltonian $H_{\pi}$, (call it the Hamiltonian density if you wish, since in my way of writing $H=F^{\dagger}\cdot H_{\pi}\cdot F$, the dots should include summation(s) over phase-space-time [delete as appropriate]). You could prefer to define the action of an operator as transforming the operators (or the wave-function). But you should not use both definitions at the same time. It is clear that you can not do both, since otherwise you transform $H=F^{\dagger}\cdot H_{\pi}\cdot F \rightarrow F^{\dagger}\cdot U^{\dagger}\cdot \left(U \cdot H_{\pi} \cdot U^{\dagger}\right) \cdot U\cdot F = H$ trivially, whatever (anti-)unitary transformation $U$ you choose. It is clear that what your are looking for is something like $H=F^{\dagger}\cdot H_{\pi}\cdot F \rightarrow F^{\dagger}\cdot U^{\dagger}\cdot H_{\pi} \cdot U\cdot F \sim H$ and you see what I just said: apply the transformation to the Hamiltonian (density) or to the fields, but not both. In condensed matter we usually choose the convention I gave to you: we transform the Hamiltonian. One of the reasons is that the operators (especially the fermionic creation/annihilation ones) are seen as encoding the statistics of the fields, whereas the Hamiltonian encodes the dynamics, and it is simple imagination to change the dynamics.

Answer 2

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.