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

  1. Jul 9, 2009 #1
    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)
    [tex]J^{-}_{\mu}=\bar{\nu}_{e}\gamma_{\mu}(1-\gamma_5)e[/tex]

    [tex]J^{+}_{\mu}=(J^{-}_{\mu})^\dagger [/tex]

    [tex]J^{em}_{\mu}=-\bar{e}\gamma_{\mu}e[/tex]
    From these one may hope to build a SU(2) group using as generators the three charges
    [tex]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[/tex]

    [tex]T_-(t)=T^\dagger_+(t)[/tex]

    [tex]Q(t)=\int d^3 x J^em_0(x)=-\int d^3 e^\dagger e[/tex]
    At this point one should show that the generators [tex] T_+, \, T_-, \, Q[/tex] do not form a close algebra by computing the commutator [tex]\[T_+,T_-\][/tex] and showing that it is not equal to [tex]Q[/tex] (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. jcsd
  3. Jul 9, 2009 #2

    Avodyne

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    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

    [tex]\bar{\nu}_{e}\gamma_{\mu}(1-\gamma_5){\nu}_{e} - {\bar e}\gamma_{\mu}(1-\gamma_5)e[/tex]
     
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