0
$\begingroup$

We know that the current symmetry of GSW is $SU(2)_{fermions, left}*U(1)$, and the correct representation of the $SU(2)_{fermions, left}$ is the $2+2$ representation. I want to know what is the reason we don't consider the symmetry group to be $SU(2)_{lepton, left}*SU(2)_{quark, left}*U(1)$? Considering quarks and leptons are both represented in the fundamental representations.

I guess that it can lead to a fine-tuning problem like why the couplings to both sectors (leptons and quarks) are the same? (Surely if the couplings are the same then according to the unique Higgs vev the masses of bosons in both sectors turn out to be the same). Am I right?! Then to ease this problem doesn't one have to introduce another symmetry?!!

But is there any deeper reason that can really falsify this symmetry group as an alternative to the current one?

Improve this question
$\endgroup$

2 Answers 2

Reset to default
3
$\begingroup$

Because it is already falsified, by any scattering experiment that produces leptons from weak decay of hadrons, or the other way around. Your model forbids the $W$ bosons emitted by quarks to decay into leptons, and the other way around. We observe such decays.

Improve this answer
$\endgroup$
4
  • $\begingroup$ And even if I include an inner product of the two W fields in the lagrangian to add a vertex for the W particle to change the first W to the second one, it seems that I lose gauge invariance explicitly. True? $\endgroup$
    – Bastam Tajik
    Commented May 24, 2020 at 23:57
  • $\begingroup$ @BastamTajik in your construction, the two $SU(2)$ from a Cartesian product group. The corresponding Lie algebra generators commute with each other. Therefore, a Yang-Mills theory built around your gauge groups will have the two $W/Z$ bosons which don’t interact with each other. You could always try writing down something other than Yang-Mills, but I’m not aware of any such successful approach. $\endgroup$
    – Prof. Legolasov
    Commented May 25, 2020 at 0:00
  • $\begingroup$ Well first of all, we're to all extent still within the paradigm of Yang-Mills theory and gauge invariance is the golden rule. I was considering to write down a new gauge invariant interaction term between the two sectors. And it's nothing but the renormalizable operator of the trace of the multiplication of the two W bosons field strengths. $Tr(F_{1}*F_{2})$ Idk if this term can modify the perturbation theory significantly or not. $\endgroup$
    – Bastam Tajik
    Commented May 25, 2020 at 14:04
  • $\begingroup$ Also by imposing gauge invariance on the four point function of the new interactive term, the coupling should be the same! $\endgroup$
    – Bastam Tajik
    Commented May 25, 2020 at 14:13
0
$\begingroup$

I think the main problem with this symmetry is that we have 6 gauge bosons while after spontaneous symmetry breaking the number of Goldstone's bosons is 3, and consequently only 3 of the total 6 bosons can become massive! Since we know the weak force is a short-range force then the non-existence of long-range weak interaction by a massless copy of the 3 weak interaction bosons can falsify this new symmetry.

Improve this answer
$\endgroup$

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.