- #1
lalo_u
Gold Member
- 27
- 0
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?
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?
Last edited: