# Symmetry, gauge, and projective symmetry group (PSG)?

My following questions come from the understanding of the relations between the PSGs for two gauge-equivalent mean-field (MF) Hamiltonians (or MF ansatz). Considering the Schwinger-fermion ($f_{\sigma}$) MF approach to spin-liquid phases of spin-1/2 system. Let $H$ and $H'$ be two $SU(2)$ gauge-equivalent MF Hamiltonians, i.e., $H'=RHR^{-1}$, where the unitary operator $R$ represents an $SU(2)$ gauge rotation generated by $J=\frac{1}{2}\psi^\dagger \mathbf{\tau} \psi$ with $\psi=(f_\uparrow,f_\downarrow^\dagger)^T$.

Now, if $G_UU\in PSG(H)$, then simple calculation shows that $RG_UUR^{-1}\in PSG(H')$, where $U$ represents a symmetry operator and $G_U$ is the gauge operator associated with $U$. But we are used to the combined form of a symmetry operator followed by a gauge operator for an element in PSG, thus there are several ways to rewrite the expression $RG_UUR^{-1}$ as: (1) $(RG_UUR^{-1}U^{-1})U$; (2) $(RG_U)U'$ with $U'=UR^{-1}$; or (3) $(RG_UR^{-1})U''$ with $U''=RUR^{-1}$. So how to understand these expressions?

As for (1): The question is whether $RG_UUR^{-1}U^{-1}$ is an $SU(2)$ gauge operator? More specifically, it seems generally impossible to write $UR^{-1}U^{-1}$ as an gauge operator generated by $J=\frac{1}{2}\psi^\dagger \mathbf{\tau} \psi$. However, if we generalize the definition of $SU(2)$ gauge operators $R$ to those satisfying 3 properties A: Unitary; B: $R\psi_iR^{-1}=W_i\psi_i, W_i\in SU(2)$ matrices, which implies that physical spins should be gauge invariant (e.g., $R\mathbf{S}_iR^{-1}=\mathbf{S}_i$); and C: $RP=PR=P$ with projection operator $P$, which implies that physical spin-space should be gauge invariant. Then one can show that $UR^{-1}U^{-1}$ indeed fulfills the above 3 properties A,B,C where $U$ is time reversal, $SU(2)$ spin rotation, or lattice symmetries. (Furthermore, $R$ respects A,B,C $\Rightarrow R^{-1}$ respects A,B,C; $R_1,R_2$ both respect A,B,C $\Rightarrow R_1R_2$ respects A,B,C.) Therefore, the expression $RG_UUR^{-1}U^{-1}$ in (1) is an $SU(2)$ gauge operator in the sense A,B,C.

As for (2) or (3): We may ask: If $U$ represents some symmetry (e.g., time reversal, $SU(2)$ spin rotation, or lattice symmetries), then does $UR$ or $RU$ still represent the same physical symmetry? Where $R$ is an $SU(2)$ gauge operator (in the sense A,B,C mentioned above). One can show that $U'=UR$ or $RU$ represents the same physical symmetry as $U$ in the following sense: $U'\mathbf{S}_iU'^{-1}=U\mathbf{S}_iU^{-1}$ and $U'\phi=U\phi$, where $\phi=P\phi\in$ physical spin space. (Note that $U'\phi=U\phi$ is still a physical spin state due to $[P,U]=[P,U']=0$.) Therefore, the $U'$ and $U''$ in expressions (2) and (3) indeed represent the same physical symmetry as $U$.

Are my understandings correct? Thanks in advance.

A useful formula: Let $G_iU_i\in PSG(H), i=1,2,...,n$, then the $SU(2)$ gauge operator $G_U$ associated with the combined symmetry $U=U_1U_2\cdots U_n$ has the following form $$G_U=G_1U_1G_2U_2\cdots G_{n-1}U_{n-1}G_nU_{n-1}^{-1}\cdots U_2^{-1}U_1^{-1}$$ such that $G_UU\in PSG(H)$.

• Another simple formula: If $G_UU\in PSG(H)$, then $G_{U^{-1}}=U^{-1}G_U^{-1}U$ such that $G_{U^{-1}}U^{-1}\in PSG(H)$. – Kai Li Nov 5 '14 at 15:13