# Why are there five, instead of four fermion fields in the electroweak kinetic part of the Lagrangian?

One can read it this Wikipedia article, about the electroweak interaction, that before symmetry breaking the kinetic part of the Lagrangian is:

$$\mathcal L_f=\bar Q_ji\not DQ_j+\bar u_ji\not Du_j+\bar d_ji\not Dd_j+\bar L_ji\not DL_j+\bar e_ji\not De_j$$

This part represents the interaction of the fermions and the gauge bosons (which are represented by the covariant derivatives). It seems as if there are five fermion fields (well, in fact, fifteen, as a summation over j is implied, where j goes from 1 to 3, representing the three generations of fermions). But why aren't there four (or twelve) fermion fields, representing the four different fermions in each generation?

Or, taking spin into account, why aren't there seven (21, when summed over j) fields, representing all left- and right-handed leptons and quarks except the right-handed neutrinos?

The electroweak force is maximally parity-violating and hence treats left-handed and right-handed solutions differently. All the fermions in the Standard Model are Dirac fermions and so can be decomposed into their left-handed and right-handed components (Weyl spinors) under the projection operators $$\frac12(1\pm\gamma^5)$$ respectively. In the context of the $$\mathrm{SU}(2)$$ gauge group, the right-handed components of all the fermions transform in the trivial representation, while certain pairs of left-handed fermions form doublets and transform in the fundamental representation. Naturally, these terms are accounted for separately in the Lagrangian.

This is what each of the components of the EW Lagrangian signifies:

Term Meaning Components
$$L_j$$ Left-handed lepton doublet $$(e_L,\nu_{e,L})^T$$, $$(\mu_L,\nu_{\mu,L})^T$$, $$(\tau_L,\nu_{\tau,L})^T$$
$$Q_j$$ Left-handed quark doublet $$(u_L, d_L)^T$$, $$(c_L, s_L)^T$$, $$(t_L, b_L)^T$$
$$u_j$$ Right-handed "upper" quark singlet $$u_R$$, $$c_R$$, $$t_R$$
$$d_j$$ Right-handed "lower" quark singlet $$d_R$$, $$s_R$$, $$b_R$$
$$e_j$$ Right-handed lepton singlet $$e_R$$, $$\mu_R$$, $$\tau_R$$

So in total there are 3 lepton doublets, 3 quark doublets, 6 quark singlets, and 3 lepton singlets (rather than 6, since this model discounts the existence of right-handed neutrinos).

It is clear that the Lagrangian must take this form on account of its gauge invariance, with no "mixing" of inequivalent representations: $$\mathcal L_f=\bar Q_ji\not DQ_j+\bar u_ji\not Du_j+\bar d_ji\not Dd_j+\bar L_ji\not DL_j+\bar e_ji\not De_j$$

Finally, note that the gauge covariant derivative $$D_\mu$$ is not the same for each of the terms, varying according to the weak hypercharge, weak isospin and handedness of the family.

• Ah yes, of course. I didn't take the handedness of particles into consideration (in which case there are 12). When you take handedness into consideration there are 21 (so not 24) particles. The Lagrangian contains indeed 21 particles. Apr 29, 2021 at 12:06
• The L, Q, u, d, and e (five, by which I was misled) do indeed contain 21 particles. Apr 29, 2021 at 12:22