# Why can't SU(2) be the gauge group of electroweak theory?

1. Jul 9, 2009

### CaptainKidd

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.

2. Jul 9, 2009

### Avodyne

Well, without actually doing the calculation, it's pretty obvious that it can't possibly work. For one thing, only the left-handed electron field (and its hermitian conjugate) appears in J+ and J-, but J_em also includes the right-handed electron field; this can't be generated out of nothing from the commutator. Also, the neutrino field does not appear in J_em, yet it appears symmetrically with the electron field in J+ and J-, which means that it must appear in the commutator as well. The current you would get from the commutator would be something like

$$\bar{\nu}_{e}\gamma_{\mu}(1-\gamma_5){\nu}_{e} - {\bar e}\gamma_{\mu}(1-\gamma_5)e$$