Graduate Covariant derivative in Standard Model

[Shen712]

The covariant derivative in standard model is given byD^μ = ∂^μ + ig_s G_a^μ L_a + ig W_b^μ T_b + ig'B^μYwhere G_a^μ are the eight gluon fields, W_b^μ the three weak interaction bosons and B^μ the single hypercharge boson. The L_a's are SU(3)_C generators (the 3×3 Gell-Mann matrices ½ λ_a for triplets, 0 for singlets), the T_b's are the SU(2)_L generators (the 2×2 Pauli matrices ½ τ_b for doublets, 0 for singlets), and the Y's are the U(1)_Y charges.

It looks like the second term is a 3×3 matrix, the third term is a 2×2 matrix, and the last term is a single term. How can they add up? Is D^μ a 3×3 matrix or something else?

 

[Orodruin]

They all act on the representation space of whatever fermion representation you are considering. You have to use the correct representation for whatever fermion field you are operating on. For example, the left-handed quark fields are in a doublet representation of SU(2) and a triplet representation of SU(3) and so they are in the representation space $2 \otimes 3$ (where the first factor is the SU(2) representation and the second the SU(3) representation), which is a 6-dimensional space (I am suppressing the phase factor from the U(1) here). The representation of the SU(3) generators on this space is of the form $1 \otimes (\lambda_a/2)$ and that of the SU(2) generators $(\tau_b/2)\otimes 1$. These are all 6x6 matrices. For the right-handed quark fields, the representation space is 3-dimensional ($1\otimes 3$) and for the left-handed lepton fields it is 2-dimensional ($2\otimes 1$). For the right-handed charged lepton fields, the representation space is the trivial 1-dimensional representation ($1\otimes 1$) - only their U(1)-representation, i.e., hypercharge, is non-trivial.