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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.

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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2 Answers

a) For a rotation of axis $\vec n$ and angle $\theta$, the transformation for a spin $1/2$ representation is :

$$\psi \to e^{i \frac{\theta}{2} \vec\sigma. \vec n}\psi = G(\vec n,\theta)\psi\tag{1}$$ where $\psi$ is a 2-component row complex spinor. Here $G(\vec n,\theta)$ is a member of $SU(2)$. The action of a matrix on a 2-component row object is always at the left of the object.

b) Beginning with : $\vec S=\frac{1}{4}tr(\Psi^\dagger \vec \sigma\Psi)$, suppose we have a transformation $\Psi \to \Psi G$. As you noticed, it would not change the value of $\vec S$, because :

$$tr(G^\dagger\Psi^\dagger \vec \sigma\Psi G)=tr(\Psi^\dagger \vec \sigma\Psi GG^\dagger) = tr(\Psi^\dagger \vec \sigma\Psi) \tag{2}$$

But $\vec S$ is a vector (it is in the fundamental or vectorial representation of $SO(3)$, or, if you prefer, in the adjoint representation of $SU(2)$), so it has to change under a rotation. So the transformation $\Psi \to \Psi G$ is non-valid and irrelevant.

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@ Trimok, thanks for your answer. But I am still unclear that why we define the SU(2) gauge transformation as $G_i \Psi_i$ rather than $\Psi_i G_i$? Maybe I'm not familiar with gauge theory. –  K-boy Sep 3 '13 at 7:27
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The transformation law, for the spinor representation, correspond to equation (1), so the transformation does not act on a matrix, but on a 2-row spinor. So, it is mandatory, in equation (1), that $G$ is at the left. Now, using $\Psi$, is a "trick" to present the spinor as a pseudo "matrix" to be used in a trace expression, it is not a fundamental presentation for the spinor representation. Now, you may consider the matrix $\Psi$ as $2 *2-$row spinors, so you may imagine that you apply $G$ on the first-column spinor, then on the second-column spinor, so $G$ must stay at the left, too. –  Trimok Sep 3 '13 at 8:25
    
@ Trimok, ok. And in the gauge transformation $\psi_i \rightarrow G_i \psi_i$, why we restrict $G_i$ belong to $SU(2)$ rather than $U(2)$? Thanks. –  K-boy Sep 4 '13 at 7:38
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@K-boy : Spinor representations are some special representations of orthogonal groups $SO(p,q)$ which cannot be constructed with vectors or tensors (see Clifford algebra). If one is interesting with spinor representations of $SO(3)$, we have to look at the double cover of $SO(3)$, which is $SU(2)$ (this means that $S0(3) = SU(2)/Z_2$). So, we have to look at the representations of $SU(2)$. Each representation of $SU(2)$ is labelled by an integer or a half-integer. The representation corresponding to $1/2$ corresponds to a spinor representation. You may go further and considering the Lorentz –  Trimok Sep 4 '13 at 7:52
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@K-boy : .....group $SO(3,1)$, wich has a double cover $SL(2,C)$. The representations of $SL(2,C)$ are labelled by $2$ numbers $(p,q)$ which can be integers or semi-integers. A left-handed spinor corresponds to the representation $(1/2,0)$, a right-handed spinor correspond to the representation $(0,1/2)$. They transform in the same way under rotations, but not with boosts. –  Trimok Sep 4 '13 at 7:53
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God, I just found that I made a very foolish misunderstanding of prof.Wen's paper. In fact, the operator $\psi_i=(\psi_{1i},\psi_{2i})^T=(f_{1i},f_{2i}^\dagger)^T$ and $\mathbf{S_i}=\frac{1}{2}f_i^\dagger\mathbf{\sigma}f_i$, but I misunderstood $\mathbf{S_i}=\frac{1}{2}\psi_i^\dagger\mathbf{\sigma}\psi_i$ before(Please see Eq.(8) in this paper http://prb.aps.org/abstract/PRB/v65/i16/e165113 at the very beginning).

Now everything goes well. In fact, the $SU(2)$ gauge transformations $\psi_i \rightarrow G_i\psi_i$ are totally equivalent to $\bigl(\begin{smallmatrix} f_{1i} & -f_{2i}^\dagger\\ f_{2i} & f_{1i}^\dagger \end{smallmatrix}\bigr)\rightarrow \bigl(\begin{smallmatrix} f_{1i} & -f_{2i}^\dagger\\ f_{2i} & f_{1i}^\dagger \end{smallmatrix}\bigr)G_i$. And the $SU(2)$ matrices being on the left or right depends on the notation of spinon operartors that you define, which is not the key point here.

Note: The operators $\psi_i$ that appear in all of Wen's papers on PSG in 2002 are not the direct annihilation operators for Schwinger-fermions $f_i$, so please be very careful!

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