The standard model has 12 massive leptons and 2 massive bosons other than the Higgs. My understanding of the Higgs mechanism is at about the level of this article, which goes as follows. Start with the massless Klein-Gordon equation for two fields, H and Z: $\nabla^2 H=0$ and $\nabla^2 Z=0$. Then modify the two equations to let H (the Higgs) have a v.e.v., and by adding an interaction term to the Z field's wave equation, $\nabla^2 Z=kH^2Z$. Since $H$ has a v.e.v., this looks exactly like the Klein-Gordon equation for a particle of mass $m$, with $m^2=kH^2$.

So extrapolating naively, I'd guess we have 14 coupling constants, $k_1$ through $k_{14}$, that have to be put in arbitrarily by hand and that determine the 14 masses that we observe in the low-energy limit for the standard model's 14 massive, fundamental particles.

Is this naive extrapolation accurate, or does $m_j$ actually depend not just on $k_j$ but also on the $k_i$ with $i\ne j$? If there is an interdependence, is it something that can be expressed linearly? Does chiral symmetry breaking (which I know nothing about) change the whole picture qualitatively? If there is an interdependence, do we get any prospect of a natural explanation for the features of the mass spectrum we see (e.g., why the masses cover such a huge range, from meV to TeV)?

Regardless of whether there is interdependence, is there any naturalness argument to explain why we see real-valued masses, as opposed to, say, imaginary masses, which would seem perfectly natural if there were no reason to prefer $k>0$?


1 Answer 1


You ask good questions.

  1. The massive gauge bosons, the $W$ and the $Z$, obtain their masses from electroweak gauge interactions with the Higgs field $$ \sim g^2 H^2 A^2 $$ Upon $H\to v+ h$, the boson acquires a mass. Such terms were already present, and thus we require no extra couplings.
  2. The massive fermions, leptons and quarks (but in the Standard Model neutrinos are massless), obtain their masses from Yukawa interactions $$ YH\psi\psi $$ Upon $H\to v+ h$, the fermion acquires a mass. The Yukawa couplings, Y, are without loss of generality $3\times 3$ complex matrices: one for up-type quarks $(u,c,t)$, down-type quarks $(d,s,b)$ and leptons $(e,\mu,\tau)$. We do not need a fourth matrix, because neutrinos are massless. In princple, it looks like we have $2\times3\times3\times3=54$ new real parameters.

    Now, we can be clever and rotate the fields $\psi$ so that the Yukawa matrices are real and diagonal (and hence the masses are real). For the leptons, its easy, we can remove all complex phases and off-diagonal elements leaving 3 diagonal elements - the three real lepton masses.

    For the quarks, its more difficult, because we want to simultaneously diagonalize the up-type quarks $(u,c,t)$ and down-type quarks $(d,s,b)$ matrices. In fact, it's impossible, because of the structure of the electroweak interaction. We rotate the up-type fields so that their matrix is real and diagonal, with three real masses. We then see what we can do with the down-type matrix. It turns out, that along with its three masses, we are left with $4$ angles.

So it turns out we have no new parameters for the gauge bosons, 3 for the three lepton masses, 6 for the six for the quark masses plus 4 angles that we couldn't get rid of by field rotations, making 13 in total.

Note that of the $4$ angles that we couldn't get rid of, $3$ are just rotation angles in a $3\times3$ matrix, but the final one is a complex phase that is the only source of CP-violation in the Standard Model.

  • $\begingroup$ +1, helpful answer. How about the naturalness issues raised in the question? Is it natural for the mass spectrum to span so many orders of magnitude? From your answer, I wasn't clear on whether, e.g., the lepton masses have to come out real, or whether you're just saying that we can choose Yukawa coupling matrices such that they come out real. $\endgroup$
    – user4552
    Commented Jun 24, 2013 at 22:35
  • 1
    $\begingroup$ Well, by naturalness one might expect the dimensionless Yukawas to be order 1. Only the top Yukawa is. The Yukawas span 1 to $10^{-5}$. Somewhat unnatural, but usually thought of as a problem. $\endgroup$
    – innisfree
    Commented Jun 25, 2013 at 5:54
  • $\begingroup$ Since neutrinos have experimentally confirmed mass, how does this alter the accepted answer? $\endgroup$
    – JPattarini
    Commented Jul 19, 2020 at 23:35

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