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By applying a discrete Z2 symmetry to the theory of Two Higgs Doublet Model it is ensured that fermions of one type couples to only one doublet. But how FCNC is removed by doing so? Because if all the leptons for example couple to the 1st doublet then there is a chance of coupling of different flavours of leptons to the same neutral mediator. How will it differentiate among different flavours?

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  • $\begingroup$ The statement is that certain discrete symmetries kill tree-level FCNCs. You need to make sure that there is not one (or more) Higgs that couples both to the $u$- and $d$-type quarks at the same time. For instance, in the MSSM you have one Higgs that couples to the $u$-type and another one that couples to the $d$-type quarks, such that this model does not have tree-level FCNCs. $\endgroup$ – user178876 Jun 19 '19 at 0:04
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It is not true that any $\mathbb{Z}_2$ symmetry eliminates the flavor-changing neutral currents (FCNCs). A more accurate statement is that some $\mathbb{Z}_2$ symmetries eliminate the tree-level FCNCs. By assigning the quarks and leptons appropriate $\mathbb{Z}_2$ charges, one can avoid that two Higgs doublets couple both to $u$- and $d$-type quarks. The following table, which I found in this link, surveys the possibilities:

enter image description here

Scheme (i) is like the standard model plus a Higgs that does not couple to the fermions at all, so it is fine. The dangerous cross terms are also avoided in schemes (ii) and (iii) but not (iv).

As for "How will it differentiate among different flavours?": the simple traditional models don't, they only distinguish the $u$- and $d$-type quarks and leptons in their $\mathbb{Z}_2$ charges. Settings (i)-(iii) work fine, i.e. do not require additional ad hoc assumptions. On the other hand, models in which the couplings are flavor-dependent are usually less simple.

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