A simple question on $SU(2)$ gauge transformations in Wen's papers on projective symmetry group (PSG)?

Question

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 $\Psi_iG_i$?

Answer 1

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.

Answer 2

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!