# Weak isospin 1/2 vector bosons

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.

• 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. Dec 15, 2017 at 20:54
• @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? Dec 15, 2017 at 21:23
• @MoziburUllah yes, the adjoint of SU(2) is a 3-dimensional representation. Dec 15, 2017 at 21:28
• @MoziburUllah yes, precisely. There are 8 gluons because the adjoint of SU(3) color is 8-dimensional. Dec 15, 2017 at 21:38
• @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 $SU(3)_{L+R}$ . Dec 15, 2017 at 21:49