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How is this Yukawa coupling term invariant?

Physics Asked by CauchySchwarzMan on June 14, 2021

I have the seen the following term used to describe Yukawa coupling for the lepton and Higgs field:

$$ epsilon_{ij}phi^{i}bar{e}_{R}f_{L}^{j} $$

Under $SU(2) otimes U(1)$ and so expected the following transformations:
$$phi rightarrow mathrm{e}^{ialpha cdot tau/2}mathrm{e}^{ibeta/2}phi$$
$$ e_{R} rightarrow mathrm{e}^{-ibeta}e_{R}$$
$$f_{L} rightarrow mathrm{e}^{ialpha cdot tau/2}mathrm{e}^{-ibeta/2}f_{L}$$

but this does not give me an invariant term? Is there something obvious that I don’t understand?

One Answer

The transformation laws you wrote are correct, with $Y=2(Q-T_3)$ being 1 for $phi$; -2 for $e_R$; and -1 for $f_L$. $$ f_L=begin{pmatrix} nu e end{pmatrix}, qquad phi= begin{pmatrix} phi^+ v+h + iphi^0 end{pmatrix}. $$

But the Yukawa you wrote is not invariant in mainstream (original) conventions, say, Peskin & Schroeder, Li & Cheng, Donoghue & Holstein, WP, and, and moreover, etc., and thus nonexistent!

First, recall the actual SM Yukawa term giving the electron its mass, $$ bar e_R ~phi^dagger cdot f_L = bar e_R ~phi ^- nu_L +bar e_R e_L ~(-iphi^0 +v + h) , $$ +h.c.. The mass term comes from the v.e.v., and you should be able to tell it is invariant under the correct transformations of your question. Note how the positron component ensures this term has net hypercharge 0, whence net charge 0, much unlike what you wrote.

What you most probably meant to write is the alternate invariant Yukawa term which would give neutrinos a Dirac mass (analogous to how up-like quarks get their masses), $$ -bar nu_R ~phi isigma_2 cdot f_L= bar nu_R ~(iphi^0+v+h)nu_L -bar nu_R ~phi^+ e_L , $$ which is also a hypercharge (and hence charge) singlet, given the null hypercharge of the sterile right-handed neutrino. Linked.


  • NB It became apparent from your comments that you are mixing mainstream conventions (transformation laws, except you redefined hypercharges by halving them all, the "modern", revisionist way) with Srednicki ones. He uses, by dint of his perverse notations, a Higgs doublet with the v.e.v. on the upper component, his (87.13), unlike the rest of the world. He is thus using the adjoint of the conjugate Higgs of the last paragraph above, calling it the Higgs field; which he can, but I did not have the patience to find where he discusses his left turn. Caveat lector. In any case, when in doubt in such checks, always run to the charges. The expressions I wrote conserve charge, and his only do provided you adopt the perverse choice (87.13). The advantages/disadvantages of his choices might be a good separate question.

Answered by Cosmas Zachos on June 14, 2021

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