# Charge operator applied to matrix multiplets

#### lalo_u

Gold Member
In the context of SM ($SU(3)_C\otimes SU(2)_L\otimes U(1)_Y$) the charge operator is $Q_{SM} = T_3 + \frac{Y}{2}\mathbb{I}_2$ and gives us the fermions charges. Here $T_3=\frac{1}{2}\sigma_3$ is the third $SU(2)$ generator.
For example, assuming $Y=-1$ for the left lepton doublet, $Q_{SM}\Psi=\begin{pmatrix}0&0\\0&-1\end{pmatrix}\begin{pmatrix}\nu_L \\ e_L\end{pmatrix}=\begin{pmatrix}0 \nu_L \\ -1 e_L\end{pmatrix}$ and the charge components are obtained.
This process works fine for the scalar doublet too, $Q_{SM}\begin{pmatrix}\phi^+\\\phi^0\end{pmatrix}=\begin{pmatrix}+1 \phi^+\\0 \phi^0\end{pmatrix}$, with $Y=+1$.
On the other hand, the same calculation can be applied for $SU(3)_C\otimes SU(3)_L\otimes U(1)_X$ extension in which the charge operator is $Q_{331} = T_3-\sqrt{3}T_8+X\mathbb{I}_3$ where $T_3=\frac{1}{2}\lambda_3$ and $T_8=\frac{1}{2}\lambda_8$ are the $SU(3)$ diagonal generators.
For example, $Q_{331}\Psi_1=\begin{pmatrix}0&0&0\\0&-1&0\\0&0&1\end{pmatrix}\begin{pmatrix}\nu_L \\ e_L \\e_L^c\end{pmatrix}=\begin{pmatrix}0 \nu_L \\ -1 e_L \\ +1 e_L^c\end{pmatrix}$ and the charge components are obtained again with $X=0$.
In the same way as before, we can do this with the scalar multiplets. For example with one of the triplets called $\eta$, $Q_{331}\eta=\begin{pmatrix}1&0&0\\0&0&0\\0&0&2\end{pmatrix}\begin{pmatrix}\eta^+ \\ \eta^0 \\ \eta^{++}\end{pmatrix}=\begin{pmatrix}+1 \eta^+ \\ 0 \eta^0 \\ +2 \eta^{++}\end{pmatrix}$ and the charge components are obtained again with $X=+1$.
The question is, ¿what about the matrix scalar multiplets?, for example a 331 sextet is defined as $S=\begin{pmatrix} \sigma_1^0&h_2^-&h_1^+ \\ h_2^- & H_1^{--} & \sigma_2^0 \\ h_1^+ & \sigma_2^0 & H_2^{++}\end{pmatrix},$ with hypercharge $X=0$.
When you try to do the same process, but with $Q_{331}^{\dagger}SQ_{331}$ for being $S$ a matrix and not a vector, i can't obtain the charges. Here $Q_{331}$ is the same as for $\Psi_1$.
¿Any idea?

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