The left-handed up quark and down quark are an $SU(2)$ doublet $(u, d)_{L}$. Why do their right-handed counterparts, right-handed up quark and down quark, not form an $SU(2)$ doublet $(u, d)_{R}$?

  • $\begingroup$ Do you mean, charged under the same SU(2) gauge field as the lefthanded quarks, or under a different SU(2)? $\endgroup$ – Mitchell Porter Aug 6 '18 at 8:24
  • $\begingroup$ Whatever. There is never a $(u, d)_{R}$ doublet. $\endgroup$ – Shen Aug 6 '18 at 16:20
  • $\begingroup$ But there can be, e.g. in the Pati-Salam model. $\endgroup$ – Mitchell Porter Aug 6 '18 at 20:54
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    $\begingroup$ But there are fanciful $(u,d)_R$ doublet models, and have been for several decades, e.g. reviewed in Langacker and Sankar 1989. It's just that experimental evidence has persisted against them, so far. $\endgroup$ – Cosmas Zachos Aug 8 '18 at 13:45

An important feature of the models you need to specify is the symmetry group and the representations under which the fields transform. The absence of $SU(2)_R$ doublets in the Standard Model can be seen as part of the definition of the model itself.

There is, however, a special limit of the Standard Model in which this symmetry is recovered. Indeed, the right-handed quarks do form a $SU(2)_R$ multiplet in the limit of vanishing quark masses. But in the Standard Model this symmetry is explicitly broken by the Yukawa interactions with the Higgs. So, the question can be recasted in the following way: why is the Standard Model working so well?

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  • $\begingroup$ You said the $SU(2)_{R}$ symmetry is explicitly broken by the Yukawa interactions with the Higgs. But why is the $SU(2)_{L}$ symmetry not broken by the same Yukawa interactions with the Higgs? $\endgroup$ – Shen Aug 6 '18 at 16:05
  • $\begingroup$ This is so because the Higgs field is a doublet under $SU(2)_L$. You can make an $SU(2)_L$ invariant interaction by using two doublets and contracting the SU(2) indexes with the epsilon tensor. The Yukawa interaction is of the form $\sim \bar{q}_L H u_R$ and it is perfectly SU(2) invariant. By writing explicitly the SU(2) indexes, you have $\epsilon_{ab}\bar{q}_L^a H^b u_R$. Only when the Higgs takes a vev, then SU(2)_L is spontaneously broken. $\endgroup$ – apt45 Aug 6 '18 at 16:08
  • $\begingroup$ How do we know the Higgs field is a doublet under $SU(2)_{L}$ but not a doublet under $SU(2)_{R}$? $\endgroup$ – Shen Aug 6 '18 at 16:29
  • $\begingroup$ @Shen It's an assumption. For the same reason because the quarks are triplet under SU(3). All the representations of the fields specified in a given model are assumptions which give phenomenological consequences that you should verify "a posteriori". As far as I understand, you are learning Particle physics and the Standard Model. For now, just take those as assumptions to be verified experimentally. $\endgroup$ – apt45 Aug 6 '18 at 16:34
  • $\begingroup$ @Shen BTW, this was the answer to your question $\endgroup$ – apt45 Aug 13 '18 at 13:33

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