# Two puzzles on the Projective Symmetry Group(PSG)?

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Recently I'm studying PSG and I felt very puzzled about two statements appeared in Wen's paper. To present the questions clearly, imagine that we use the Shwinger-fermion $\mathbf{S}_i=\frac{1}{2}f_i^\dagger\mathbf{\sigma}f_i$ mean-field method to study the 2D spin-1/2 system, and get a mean-field Hamiltonian $H(\psi_i)=\sum_{ij}(\psi_i^\dagger u_{ij}\psi_j+\psi_i^T \eta_{ij}\psi_j+H.c.)+\sum_i\psi_i^\dagger h_i\psi_i$, where $\psi_i=(f_{i\uparrow},f_{i\downarrow}^\dagger)^T$, $u_{ij}$ and $\eta_{ij}$ are $2\times2$ complex matrices, and $h_i$ are $2\times2$ Hermitian matrices. And the projection to the spin subspace is implemented by projective operator $P=\prod _i(2\hat{n}_i-\hat{n}_i^2)$(Note here $P\neq \prod _i(1-\hat{n}_{i\uparrow}\hat{n}_{i\downarrow})$). My questions are:

(1)How to arrive at Eq.(15) ? Eq.(15) means that, if $\Psi$ and $\widetilde{\Psi}$ are the mean-field ground states of $H(\psi_i)$ and $H(\widetilde{\psi_i})$, respectively, then $P\widetilde{\Psi}\propto P\Psi$, where $\widetilde{\psi_i}=G_i\psi_i,G_i\in SU(2)$. How to prove this statement?

(2)The statement of translation symmetry above Eq.(16), which can be formulated as follows: Let $D:\psi_i\rightarrow \psi_{i+a}$ be the unitary translation operator($a$ is the lattice vector). If there exists a $SU(2)$ transformation $\psi_i\rightarrow\widetilde{\psi_i}=G_i\psi_i,G_i\in SU(2)$ such that $DH(\psi_i)D^{-1}=H(\widetilde{\psi_i})$, then the projected spin state $P\Psi$ has translation symmetry $D(P\Psi)\propto P\Psi$, where $\Psi$ is the mean-field ground state of $H(\psi_i)$. How to prove this statement?

I have been struggling with the above two puzzles for several days and still can't understand them. I will be very appreciated for your answer, thank you very much.

This post imported from StackExchange Physics at 2014-03-09 08:42 (UCT), posted by SE-user K-boy
retagged Mar 9, 2014
I retagged your post if you don't mind (see the edit summary). By the way, +1 to this question and its answers.

This post imported from StackExchange Physics at 2014-03-09 08:42 (UCT), posted by SE-user Dimensio1n0
@ DImension10 Abhimanyu PS, thank you very much.

This post imported from StackExchange Physics at 2014-03-09 08:42 (UCT), posted by SE-user K-boy

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I just found that I can prove my second puzzle (2) if (1) is true:

Note that in (2), the ground state of $H(\widetilde{\psi_i})$ is $\widetilde{\Psi}=D\Psi$, then according to (1), we have $P\widetilde{\Psi}\propto P\Psi$. And $[P,D]=0$, therefore, $D(P\Psi)=PD\Psi=P\widetilde{\Psi}\propto P\Psi$.

Remark: More generally, the statement (2) can be generalized to any kind of symmetry represented by an unitary(or antiunitary, like time-reversal) operator $A$. But its correctness is based on the fact $[P,A]=0$.

This post imported from StackExchange Physics at 2014-03-09 08:42 (UCT), posted by SE-user K-boy
answered Sep 20, 2013 by (975 points)
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Finally, I can answer the puzzle (1) now:

Let $\mathbf{J}_i=\frac{1}{2}\psi_i^\dagger\mathbf{\sigma}\psi_i$ and $R_i$ be the local $SU(2)$ rotation operators of $\mathbf{J}_i$, then $R_i\psi_iR_i^{-1}=G_i\psi_i,G_i\in SU(2)$. Thus in (1), $H(G_i\psi_i)=RH(\psi_i)R^{-1}$, where $R=\prod _iR_i$. So the ground state of $H(G_i\psi_i)$ is $\widetilde{\Psi}=R\Psi$, therefore, $P\widetilde{\Psi}=PR\Psi=P\Psi$.

Note here we have used the fact $PR=RP=P.$

This post imported from StackExchange Physics at 2014-03-09 08:42 (UCT), posted by SE-user K-boy
answered Sep 21, 2013 by (975 points)

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