2
$\begingroup$

Does 2HDM predict two higgs boson? Why two doublets are needed? I need a simplified answer as i am new to BSM

$\endgroup$
  • $\begingroup$ It predicts 5 higgs bosons including the usual one. Usually called h (light scalar), H (heavy scalar), A (pseudo-scalar), H+ (charged positive Higgs boson) and H- (charged negative Higgs boson). I'd answer but don't have time to look up a citation. Not entirely obvious in 2HD models if the SM Higgs is h or H. $\endgroup$ – ohwilleke Feb 9 at 20:49
2
$\begingroup$

You may guess it by counting the number of degrees of freedom. One doublet corresponds to 4 real degrees of freedom. However the Higgs breaks $U(1)\times SU(2)$ gauge symmetry to $U(1)$. This requires the appearance of three Goldstone modes - one neutral and two charged ones conjugated to each other. Those 3 Goldstones are not physical and instead neutral $Z$ and charged $W^\pm$ bosons get longitudinal polarizations. This leaves us with only one neutral Higgs boson in the physical sector.

Two Higgs doublets give us 8 real degrees of freedom. But then 3 of them again correspond to unphysical Goldstone modes. Thus we are left with 5 physical modes (not two!) - 3 neutral bosons and charged boson and its antiparticle.

2HDM models are one of the simplest extensions of the Standard Model. You also have to introduce at least two Higgs doublets into the supersymmetric extensions like MSSM where you can't give masses to both up and down quarks with only one Higgs doublets. Another motivation comes from the axion models where you can't construct the nontrivial Peccei-Quinn symmetric potential for the one Higgs doublet.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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