- #1
jdstokes
- 523
- 1
I've read that the choice of Gauge group [itex]\mathrm{SU}(2)_\mathrm{L}\times \mathrm{U}(1)_\mathrm{Y}[/itex] can be justified by the fact that the electromagnetic charge and the flavour-changing weak charge do not form a closed SU(2) current algebra. The solution is to tack on an additional U(1) group and to interpret the photon as a superposition of the U(1) and the neutral SU(2) gauge fields.
I saw this explained somewhere by the fact that since the charge operator is
[itex]Q = \left(
\begin{matrix}
0 & 0 \\
0 & -1
\end{matrix}
\right)[/itex],
it does not commute with any of the Pauli matrices. Can anyone explain where this definition of the charge operator comes from? I'm failing to understand even what the charge operator is.
I saw this explained somewhere by the fact that since the charge operator is
[itex]Q = \left(
\begin{matrix}
0 & 0 \\
0 & -1
\end{matrix}
\right)[/itex],
it does not commute with any of the Pauli matrices. Can anyone explain where this definition of the charge operator comes from? I'm failing to understand even what the charge operator is.