Analysis of $W^\pm+4\gamma$ in the 2HDM Type-I at the LHC

TL;DR

This paper investigates a specific light charged Higgs boson scenario in the 2HDM Type-I, proposing a promising LHC detection channel with a background-free signature involving a W boson and four photons.

Contribution

It introduces a novel LHC search strategy for a light charged Higgs in the 2HDM Type-I through a background-free $W^ ext{±} + 4 ext{γ}$ signal, supported by detailed Monte Carlo analysis.

Findings

The $H^ ext{±} o W^{ ext{±}*} h$ decay dominates in the considered parameter space.

The neutral Higgs $h$ predominantly decays into two photons.

The $pp o H^ ext{±} h o W^{ ext{±}*} h h o l u + 4 ext{γ}$ process is background free and detectable at the LHC.

Abstract

We analyse a light charged Higgs boson in the 2-Higgs Doublet Model (2HDM) Type-I, when its mass satisfies the condition $M_{H^{\pm}} < M_{t}+M_{b}$ and the parameter space is consistent with theoretical requirements of self-consistency as well as the latest experimental constraints from Large Hadron Collider (LHC) and other data. Over such a parameter space, wherein the Standard Model (SM)-like state discovered at the LHC in 2012 is the heaviest CP-even state of the 2HDM, it is found that the decay modes of the charged Higgs boson are dominated by $H^{\pm} \rightarrow W^{\pm *} h$. Furthermore, the light neutral Higgs boson $h$ dominantly decays into two photons. Under these conditions, we find that the production and decay process $ p p \to H^\pm h \to {W^\pm}^{(*)} h h \to l \nu_{l} + 4 \gamma$ ($l=e,\mu$) is essentially background free. However, since the $W^{\pm(*)}$ could be largely off-shell and the $h$ state is very light, so that both the lepton coming from the former and the photons coming from the latter could be rather soft, we perform here a full Monte Carlo (MC) analysis at the detector level demonstrating that such a $W^{\pm} + 4\gamma$ signal is very promising, as it would be yielding significant excesses at the LHC with an integrated luminosity of $L=$ 300 $fb^{-1}$ at both $\sqrt{s}= 13$ and $14 ~\text{TeV}$.