Do precision electroweak constraints guarantee $e^+e^-$ collider discovery of at least one Higgs boson of a two-Higgs-doublet model?

Details Do precision electroweak constraints guarantee $e^+e^-$ collider discovery of at least one Higgs boson of a two-Higgs-doublet model? Do precision electroweak constraints guarantee $e^+e^-$ collider discovery of at least one Higgs boson of a two-Higgs-doublet model?

2000-09-22

arxiv.org/abs/hep-ph/0009271v2

Phys.Lett. B496 (2000) 195-205

Physics

High Energy Physics

High Energy Physics - Phenomenology

14 pages, 4 figures. 1st two figures have been reformatted to improve readability

Scientific paper

10.1016/S0370-2693(00)01293-4

We consider a CP-conserving two-Higgs-doublet type II model with a light scalar or pseudoscalar neutral Higgs boson ($\h=\hl$ or $\h=\ha$) that has no $ZZ/WW$ coupling and, thus, cannot be detected in $\epem\to Z\h$ (Higgs-strahlung) or $\nu\anti\nu \h$ (via $WW$ fusion). Despite sum rules which ensure that the light $\h$ must have significant $t\anti t$ or $b\anti b$ coupling, for a wedge of moderate $\tanb$, that becomes increasingly large as $\mh$ increases, the $\h$ can also escape discovery in both $b\anti b \h$ and $t\anti t \h$ production at a $\rts=500-800\gev$ $\epem$ collider (for expected luminosities). If the other Higgs bosons happen to be too heavy to be produced, then no Higgs boson would be detected. We demonstrate that, despite such high masses for the other Higgs bosons, only the low-$\tanb$ portion of the no-discovery wedges in $[\mh,\tanb]$ parameter space can be excluded due to failure to fit precision electroweak observables. In the $\tanb\gsim 1$ regions of the no-discovery wedges, we find that the 2HDM fit to precision electroweak observables has small $\Delta\chi^2$ relative to the best minimal one-doublet SM fit.

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