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The weak vector bosons are spacetime vectors (spin 1) and also incidentally weak isospin vector components (-1, 0, +1). I understand why that is required from nucleon beta decay and other weak interaction decays. Experimentally there are no weak interactions that I'm aware of that require a weak boson field that is a spacetime vector but a weak isospin spinor (weak isospins of +-1/2). Is there a sound theoretical reason to exclude such a field from the interaction Lagrangian density, e.g. no suitable term can be made? I have not seen it discussed anywhere, but perhaps just haven't found the right reference yet.

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    $\begingroup$ massless spin-1 particles must be gauge bosons, and the gauge bosons must transform in the adjoint representation of the group, hence the gauge bosons of SU(2) are necesserily triplets (and gluons as 8 of SU(3)...). Non-massless spin-1 particles could in principle transform in other representations, but then their mass would be most likely at around the cutoff of the theory that would not be renormalizable. $\endgroup$ – TwoBs Dec 15 '17 at 20:54
  • $\begingroup$ @TwoBS: if I'm reading your comment correctly does that mean that the representing space of the adjoint rep of SU(2) is 3d in order to get a triplet? $\endgroup$ – Mozibur Ullah Dec 15 '17 at 21:23
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    $\begingroup$ @MoziburUllah yes, the adjoint of SU(2) is a 3-dimensional representation. $\endgroup$ – TwoBs Dec 15 '17 at 21:28
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    $\begingroup$ @MoziburUllah yes, precisely. There are 8 gluons because the adjoint of SU(3) color is 8-dimensional. $\endgroup$ – TwoBs Dec 15 '17 at 21:38
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    $\begingroup$ @MoziburUllah in fact, no, I am sorry: the eightfold way refers to another 8. It refers to chiral symmetry breaking in the strong interactions: $SU(3)_L \times SU(3)_R$ is broken spontaneously down to the diagonal $SU(3)_{L+R}$, to be identified with the strong isospin. By the goldstone theorem there arise 8+8-8=8 goldstone bosons which trasform in the adjoint of the strong isospin group $SU(3)_{L+R}$ . $\endgroup$ – TwoBs Dec 15 '17 at 21:49
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Thanks for the advice. @TwoB, re the initial reply I believe that the requirement for the gauge bosons to transform as the adjoint rep comes from the need to end up with a scalar lagrangian density, and the fact that normally there is a product of the field with the su(2) (lie alg) generators to cause this. So I guess the question is in those terms whether any other reasonable way to create a scalar term from SU(2) or su(2) and the field reps exists, although, as we all agree, the current one has been pretty well verified by experiment. Possibly something considered by Pauli, Yang and Mills and the others when the theory was forming but discarded by them. Apparently not considered in print so far as I've found. And yes, as someone noticed the motivation is to see if some valid possibility has been overlooked that might help get us out of the tight corner we've painted ourselves into with the SM.

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  • $\begingroup$ Waitwaitwait: what "tight corner we've painted ourselves into with the SM"? Many physicists just can't get enough of the unremitting success of the SM. As indicated, purely formally, a vector isodoublet field can saturate a spinor isodoublet and a spinor isosinglet to produce a Lorentz invariant coupling term invariant under SU(2) as well. It just won't be part of a renormalizable or realistic theory. The vector field won't be a gauge field, unless you switch your group to the 1d affine group, with just 1 nontrivial commutation relation involving the 2 sole generators. $\endgroup$ – Cosmas Zachos Dec 17 '17 at 0:30
  • $\begingroup$ Thanks, Cosmas, that helps. The "...corner." comment of mine was out of line and takes things into the realm of discussion which is not the purpose here. Sorry. $\endgroup$ – Jim Eshelman Dec 18 '17 at 2:09

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