It is known that the couplings to the Higgs are proportional to the mass for fermions; $$g_{hff}=\frac{M_f}{v}$$ where $v$ is the VEV of the Higgs field. I'm trying to figure out why this is true without explicitly constructing all the interaction terms.
First I can say that, the lagrangian, being massless requires the Higgs field $\vec{\phi}=\left( \begin{array}{c} \phi^1\\ \phi^2\\ \end{array} \right)$ (weak doublet) which adds a symmetry breaking part and a coupling to the fermions.
The coupling is proportional to $$g(\bar{\psi}_L\vec{\phi}\psi_R+\bar{\psi}_R\phi^\dagger\psi_L)$$ where $\psi_L$ is a weak doublet $\psi_R$ a weak singlet.
So, after the symmetry breaking, we can write the Higgs field as $\vec{\phi}=\left( \begin{array}{c} 0\\ v+h(x)\\ \end{array} \right)$ because of the local $SU(2)_W$ symmetry. Therefore the coupling becomes $$gv(\bar{\psi}^{(2)}_L\psi_R+\bar{\psi}_R\psi^{(2)}_L)+gh(\bar{\psi}^{(2)}_L\psi_R+\bar{\psi}_R\psi^{(2)}_L)$$ where the $(2)$ exponent indicates the component in the doublet. Then, we can rewrite the fields as Dirac fields: $\psi=\psi^{(2)}_L+\psi_R$ which gives $$gv\bar{\psi}\psi+gh\bar{\psi}\psi$$
From which we identify the mass of the fermionic field as $M_f=gv$ which implies that the coupling to the higgs is $$g_{hff}=\frac{M_f}{v}$$
Of course I only considered one family of fermions, so my question is how do I generalize the argument when the coupling includes family mixing? $$g(\bar{L}\vec{\phi}\Lambda R+\bar{R}\Lambda^\dagger\vec{\phi}^\dagger L)$$ where $\Lambda$ is a family mixing matrix. Is it straightforward from the diagonalization of the mixing matrix?
