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?

  • $\begingroup$ Hi Barefeg - it seems like you're not really asking a question here, as you have the derivation of the couplings pretty much completely written out. The only actual question in the post is "What do you think?" which is not the sort of thing this site is for. Could you edit your post to ask specifically about the issue that is confusing you, instead of just asking for general feedback? Once you do that, ping me in a comment or flag this for mod attention and I'll be happy to reopen it. (Also, you might get a better response if you shorten the question a bit.) $\endgroup$
    – David Z
    Apr 13, 2013 at 21:22
  • 1
    $\begingroup$ @DavidZaslavsky, I have a concern in the fermion part when including family mixing, as I mentioned. Is it OK if I remove the last sentence and remove the part for bosons? $\endgroup$
    – Prastt
    Apr 13, 2013 at 21:46
  • $\begingroup$ Oh, sorry, I missed that! (despite reading the question twice, somehow) It definitely would help if you take the boson part out and remove the last sentence. I'd also suggest phrasing the thing you are actually asking as a question (people often focus on the question mark), and making a more specific title, perhaps something along the lines of "Can you diagonalize the Higgs coupling term with a mixing matrix?" I'll reopen the question now, with the expectation that you'll make the appropriate changes soon. $\endgroup$
    – David Z
    Apr 13, 2013 at 21:56

1 Answer 1


The fermions are mixed by off-diagonal elements in the Yukawa matrices, which result in off-diagonal elements in the up- and down-type quark mass matrices.

Diagonalising these mass matrices results in the mass eigenstates identified with the observed quarks. Because the Higgs-quark couplings are proportional to the Yukawas, they are diagonlised by these same mass eigenstates.

The mismatch, however, between the up- and down-type Yukawa results in the CKM matrix in the charged-weak interaction - the source of CP violation. NB there is no equivalent in the lepton sector, because there are no right-handed neutrinos in the Standard Model.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.