Relation of Higgs couplings to masses of fundamental particles 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 $k>0$?
 A: You ask good questions.


*

*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.

*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. 
