# 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. – TwoBs Dec 15 '17 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? – Mozibur Ullah Dec 15 '17 at 21:23
• @MoziburUllah yes, the adjoint of SU(2) is a 3-dimensional representation. – TwoBs Dec 15 '17 at 21:28
• @MoziburUllah yes, precisely. There are 8 gluons because the adjoint of SU(3) color is 8-dimensional. – TwoBs Dec 15 '17 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 $SU(3)_{L+R}$ . – TwoBs Dec 15 '17 at 21:49