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Recently I am studying the projective symmetry group (PSG) and the associated concept of quantum order first proposed by prof.Wen(http://prb.apsproposed by prof.org/abstract/PRB/v65/i16/e165113Wen, http://prl.aps.org/abstract/PRL/v90/i1/e016803 and http://www.sciencedirect.com/science/article/pii/S0375960102008083).

In Wen's paper(http://www.sciencedirect.com/science/article/pii/S0375960102008083paper), see the last line of Eq.(8), the local SU(2) gauge transformation for spinor operators is deﬁned as $$\psi_i\rightarrow G_i\psi_i$$, where $$\psi_i=(\psi_{1i},\psi_{2i})^T$$ are fermionic operators and $$G_i\in SU(2)$$. Why we define it like this?

Since as we know, the Shcwinger fermion representation for spin-1/2 can be written as $$\mathbf{S}_i=\frac{1}{4}tr(\Psi_i^\dagger\mathbf{\sigma}\Psi_i)$$, where $$\Psi_i=\begin{pmatrix} \psi_{1i} & -\psi_{2i}^\dagger \\ \psi_{2i} & \psi_{1i}^\dagger \end{pmatrix}$$, and $$G_i\Psi_i$$ which is the same as the above transformation $$\psi_i\rightarrow G_i\psi_i$$ is in fact a spin rotation of $$\mathbf{S}_i$$, while $$\Psi_iG_i$$ does not change spin $$\mathbf{S}_i$$ at all.

So in Eq.(8), why we define the SU(2) gauge transformation as $$G_i\Psi_i$$ rather than $$\Psi_iG_i$$?

Recently I am studying the projective symmetry group (PSG) and the associated concept of quantum order first proposed by prof.Wen(http://prb.aps.org/abstract/PRB/v65/i16/e165113, http://prl.aps.org/abstract/PRL/v90/i1/e016803 and http://www.sciencedirect.com/science/article/pii/S0375960102008083).

In Wen's paper(http://www.sciencedirect.com/science/article/pii/S0375960102008083), see the last line of Eq.(8), the local SU(2) gauge transformation for spinor operators is deﬁned as $$\psi_i\rightarrow G_i\psi_i$$, where $$\psi_i=(\psi_{1i},\psi_{2i})^T$$ are fermionic operators and $$G_i\in SU(2)$$. Why we define it like this?

Since as we know, the Shcwinger fermion representation for spin-1/2 can be written as $$\mathbf{S}_i=\frac{1}{4}tr(\Psi_i^\dagger\mathbf{\sigma}\Psi_i)$$, where $$\Psi_i=\begin{pmatrix} \psi_{1i} & -\psi_{2i}^\dagger \\ \psi_{2i} & \psi_{1i}^\dagger \end{pmatrix}$$, and $$G_i\Psi_i$$ which is the same as the above transformation $$\psi_i\rightarrow G_i\psi_i$$ is in fact a spin rotation of $$\mathbf{S}_i$$, while $$\Psi_iG_i$$ does not change spin $$\mathbf{S}_i$$ at all.

So in Eq.(8), why we define the SU(2) gauge transformation as $$G_i\Psi_i$$ rather than $$\Psi_iG_i$$?

Recently I am studying the projective symmetry group (PSG) and the associated concept of quantum order first proposed by prof.Wen.

In Wen's paper, see the last line of Eq.(8), the local SU(2) gauge transformation for spinor operators is deﬁned as $$\psi_i\rightarrow G_i\psi_i$$, where $$\psi_i=(\psi_{1i},\psi_{2i})^T$$ are fermionic operators and $$G_i\in SU(2)$$. Why we define it like this?

Since as we know, the Shcwinger fermion representation for spin-1/2 can be written as $$\mathbf{S}_i=\frac{1}{4}tr(\Psi_i^\dagger\mathbf{\sigma}\Psi_i)$$, where $$\Psi_i=\begin{pmatrix} \psi_{1i} & -\psi_{2i}^\dagger \\ \psi_{2i} & \psi_{1i}^\dagger \end{pmatrix}$$, and $$G_i\Psi_i$$ which is the same as the above transformation $$\psi_i\rightarrow G_i\psi_i$$ is in fact a spin rotation of $$\mathbf{S}_i$$, while $$\Psi_iG_i$$ does not change spin $$\mathbf{S}_i$$ at all.

So in Eq.(8), why we define the SU(2) gauge transformation as $$G_i\Psi_i$$ rather than $$\Psi_iG_i$$?

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# A simple question on SU(2) gauge transformationtransformations in Wen's papers on PSG?

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