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Since by introducing one Higgs Boson we can give a mass to the leptons and gauge bosons of the weak interaction:

Why should we consider more than one Higgs (doublet) once we go beyond the standard model?

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    $\begingroup$ People consider more than one Higgs boson to address some of the problems of the SM. For example, the hierarchy problem is greatly alleviated by SUSY (which can be invoked for totally other reasons too) which basically forces you yo add more Higgses to the game. $\endgroup$ – TwoBs Jun 3 '14 at 9:00
  • $\begingroup$ Maybe its worth noting, that the confidence in SUSY, which requires more than one Higgs, is shrinking rapidly after the first LHC runs, without finding any signs of SUSY. See for example scientificamerican.com/article/… The standard model does not need more than one Higgs and there is even a recent discussion if the problems solved by SUSY, are really problems of the Standard model $\endgroup$ – JakobH Jun 3 '14 at 9:16
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    $\begingroup$ You can probably find some answers to why, and most of what you want to know about how, in this review article: arxiv.org/abs/1106.0034 $\endgroup$ – Robin Ekman Jun 3 '14 at 9:52
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    $\begingroup$ @JakobH LHC is probing only low-energy SUSY, where the SUSY breaking scale is supposed to be around the TeV scale. It says nothing about higher scales SUSY. The SM is a renormalizable theory and therefore it is fine at it is. Except that it does not explain dark matter, matter-antimatter asymmetry, dark energy, and it does not include gravity. So it is almost certainly not the end of the story, and several theories (supersymmetric or not) include extra Higgs bosons. $\endgroup$ – TwoBs Jun 3 '14 at 12:45
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    $\begingroup$ @AnneO'Nyme I am not saying that extra Higgs bosons have to do with all those problems, necessarily. They may or may not arise (and in fact often they do arise), in several theories that try to address those problems. I've mentioned SUSY that can solve the Hierarchy problem and provide for a DM candidate as an example that require an extra Higgs doublet. But for instance, you can think of Higgs doublets in other contexts too. For instance, the so-called inert-doublet model is a candidate for DM. Other Higgses arise sometimes in composite Higgs model to alleviate the little hierarchy problem... $\endgroup$ – TwoBs Jun 3 '14 at 14:11
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After some research on the net, I found the following answer (with the framework of the supersymmetric model of two Higgs doublets.) Please feel free to comment if I'm wrong and give some additional information.


In the Minimal Supersymmetry Standard Model one needs two Higgs doublets to break the symmetry to $SU(2)_L \times U(1)_\text{em}$. There are two reasons:

  1. In the SM the Higgs doublet has hypercharge $Y=1/2$ and one needs to introduce an adjoint doublet $\tilde{H}=i\sigma_2H^*$, which has hypercharge $Y=-1/2$, to give the upper component of the left-handed quark doublets a mass.

However, in a supersymmetric model, the conjugate of a (super)field is not allowed in the Lagrangian. Therefore, one cannot use the $\tilde{H}$-'trick' and we are forced to introduce a second doublet to give the upper-components of the fermion-doublet a mass.

  1. The other reason is related to chiral anomalies related to triangular fermionic loops. For the anomalies to vanish (we don't want the theory to be renormalisable) there turns out to be a simple criterion: $$Tr(Y_f)= Tr(Q_f)=0. $$ (Since $Q=I^3_W+Y$ and the generators of a unitary group are always traceless: $Tr(I_w^3)$.) This is indeed the case in the SM: $$\sum_f Q_f= (-1)+3 \times (2/3)+3\times (-1/3)=0. $$ But in supersymmetry we get an additional charged spin-$1/2$ particle, the higssino, associated to the Higgs and the sum will no longer add up to zero. However, by introducing a second doublet with opposite hypercharge we again get a nice cancelation.
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    $\begingroup$ Is there a typo here we don't want the theory to be renormalisable ? $\endgroup$ – Andre Holzner Sep 26 '14 at 15:48

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