- #1
CaptainKidd
- 1
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I know that there are many reasons why SU(2) can't be the electroweak gauge group, but I want to have some clarifications about the following one, that disergads neutral currents:
in this case the currents are (considering only the lepton sector of the first generation)
J^{-}_{\mu}=\bar{\nu}_{e}\gamma_{\mu}(1-\gamma_5)e
J^{+}_{\mu}=(J^{-}_{\mu})^\dagger
J^{em}_{\mu}=-\bar{e}\gamma_{\mu}e
From these one may hope to build a SU(2) group using as generators the three charges
T_{+}(t)=\frac{1}{2}\int d^3 x J^-_0(x)=\frac{1}{2}\int d^3 \nu_e^\dagger(1-\gamma_5)e
T_-(t)=T^\dagger_+(t)
Q(t)=\int d^3 x J^em_0(x)=-\int d^3 e^\dagger e
At this point one should show that the generators T_+, \, T_-, \, Q do not form a close algebra by computing the commutator \[T_+,T_-\] and showing that it is not equal to Q (and this proves that SU(2) is not the right gauge group).
I just want to know how to compute this commutator explicitly. In Cheng and Li's book it is said that it can be computed using the canonical fermion anticommutation relations but I was not able to do it.
Thanks.
in this case the currents are (considering only the lepton sector of the first generation)
J^{-}_{\mu}=\bar{\nu}_{e}\gamma_{\mu}(1-\gamma_5)e
J^{+}_{\mu}=(J^{-}_{\mu})^\dagger
J^{em}_{\mu}=-\bar{e}\gamma_{\mu}e
From these one may hope to build a SU(2) group using as generators the three charges
T_{+}(t)=\frac{1}{2}\int d^3 x J^-_0(x)=\frac{1}{2}\int d^3 \nu_e^\dagger(1-\gamma_5)e
T_-(t)=T^\dagger_+(t)
Q(t)=\int d^3 x J^em_0(x)=-\int d^3 e^\dagger e
At this point one should show that the generators T_+, \, T_-, \, Q do not form a close algebra by computing the commutator \[T_+,T_-\] and showing that it is not equal to Q (and this proves that SU(2) is not the right gauge group).
I just want to know how to compute this commutator explicitly. In Cheng and Li's book it is said that it can be computed using the canonical fermion anticommutation relations but I was not able to do it.
Thanks.