If the three quarks $u, d, s$ had the same mass, they would have an $SU(3)$ flavor symmetry ($u, d, s$). This symmetry is broken because these three quarks have acquired different masses through interactions with the Higgs field (Yukawa interactions). However, in the Standard Model, Yukawa interactions are between the Higgs field and the doublet ($u, d$). What about the triplet ($u, d, s$)? How does this triplet interact with the Higgs field so that these quarks acquire their different masses?

  • $\begingroup$ The symmetry is broken because the quarks have different masses. However, the up and down quacks have similar masses. $\endgroup$
    – knzhou
    Aug 10, 2018 at 20:05
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    $\begingroup$ If you have more questions about the quark model, the SM, etc. I recommend an intro particle physics text like Griffiths. It covers everything you’ve asked in your past 10 questions very nicely! $\endgroup$
    – knzhou
    Aug 10, 2018 at 20:06
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    $\begingroup$ If that’s what you really wanted to know (specifically, how to write down the quark masses within the SM), you should edit your question to reflect that. $\endgroup$
    – knzhou
    Aug 10, 2018 at 20:11
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    $\begingroup$ The mass of the electron, as opposed to the neutrino, includes the energy of the electromagnetic field created by the electron's charge. This is why the electron is heavier than the neutrino, as other than the charge they are the same particle. The same idea applies to the quarks of the same generation, such as u and d. The Higgs field (if it exists) makes the particle massive, but does not necessarily give it all its mass. (Personally I believe the quantum gravity will get rid of the Higgs field, but this is a different topic.) I hope SM experts here will post an answer you are looking for. $\endgroup$
    – safesphere
    Aug 10, 2018 at 21:07
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    $\begingroup$ "How does the triplet (u,d,s) interact with the Higgs field?" It does not interact as a triplet. As Cosmas Zachos states, there are interactions between right-handed-quark SU(2) singlets, left-handed-quark SU(2) doublets, and the Higgs field, whose structure is determined by the quantum numbers of each field. $\endgroup$ Aug 11, 2018 at 4:28

1 Answer 1


In the SM, all six quarks, d,u,s,c,b,t, (and leptons) get their varied masses through gauge-invariant Yukawa interactions; their strong or generation symmetries are completely irrelevant, and the size or systematics or such masses is not part of the SM to explain. They are six arbitrary parameters (Yukawa couplings) completely unconstrained by SM symmetries; but, of course, beyond the SM model-building seeks to predict them, somehow.

Typically, e.g., the weak-gauge-invariant couplings responsible for the mass of the d are $$ -y_d \overline{ \begin{pmatrix} u_{L} \\ d_L \end{pmatrix} } \cdot \Phi ~ ~d_R +\hbox{h.c.}, $$ where the v.e.v. of the Higgs amounts to $$ \langle \Phi \rangle = \frac{v}{\sqrt{2}} \begin{pmatrix} 0 \\ 1 \end{pmatrix},$$ for v ~ 0.25 TeV . You then see $m_d=y_d v/\sqrt{2}$.

The mass of the u in the weak doublet knows nothing about that coupling, and arises out of a completely independent Yukawa, $$ -y_u \overline{ \begin{pmatrix} u_{L} \\ d_L \end{pmatrix} } \cdot \tilde{\Phi} ~ ~u_R +\hbox{h.c.}, $$ where, of course, $$ \langle \tilde{\Phi} \rangle =\langle i\tau_2 \Phi^* \rangle = \frac{v}{\sqrt{2}} \begin{pmatrix} 1 \\ 0 \end{pmatrix}. $$

You write two such terms of each kind for the other four quarks, and you are done.

The sizes of the Yukawas, and so the masses are experimental inputs: the structure of the SM accommodates them all, and gives model-builders something to do in inferring them out as something beyond the SM. Thus, there never could be an issue of them acquiring different masses:

  • There never was a good reason for any quark masses, or any fermion masses, to not be as different as they please. Expectations of the contrary in the SM rises to the level of metaphysical falsehood.

Corrections of these masses due to electromagnetism or chiral symmetry breaking effects of QCD are implicit in the SM basic interactions, but messier to estimate.

  • small practical complication in "real life": Actually, for the 3 generations of the real world, there are more yukawas, cross generational, yielding more elaborate, non-diagonal mass matrices. Diagonalization of such ends up producing the CKM mixing matrix.
  • $\begingroup$ The final observation in your answer: "small practical complication in "real life": Actually, for the 3 generations of the real world, there are more yukawas, cross generational, yielding more elaborate, non-diagonal mass matrices. Diagonalization of such ends up producing the CKM mixing matrix." is not a consensus interpretation. In the SM, the four parameters involved in the CKM matrix are arbitrary experimentally measured inputs as well and not necessarily a function of the mass matrices at all. $\endgroup$
    – ohwilleke
    Aug 11, 2018 at 2:06
  • $\begingroup$ @Cosmas Zachos - I can see such explanation in the textbook. What I am concerned with is: How did the supposed $SU(3)$ flavor symmetry ($u, d, s$) break down into a doublet ($u, d$) and a separate $s$? I know that $s$ and $c$ form a doublet ($c, s$) and they acquire their masses through interactions with the Higgs field similar to the doublet ($u, d$). But what is the role of the flavor triplet ($u, d, s$)? How did it break down? $\endgroup$
    – Shen
    Aug 11, 2018 at 8:59
  • $\begingroup$ It was born broken, by virtue of the different quark masses, effected by the different Yukawas. It is actually the other way round: you are talking about this triplet because the masses of these quarks are similar, and lower than the QCD scale that determines masses of hadrons after chiral symmetry breaking. Note that neither you nor anyone else is talking about flavor sextets, (u,d,s,c,b,t)! $\endgroup$ Aug 11, 2018 at 11:00

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