European Physical Journal C: Particles and Fields (Aug 2019)

Flavor changing neutral current decays $$t\rightarrow c X$$ t→cX ($$X=\gamma ,\,g,\, Z,\, H$$ X=γ,g,Z,H ) and $$t\rightarrow c{{\bar{\ell }}}\ell $$ t→cℓ¯ℓ ($$\ell =\mu ,\,\tau $$ ℓ=μ,τ ) via scalar leptoquarks

  • A. Bolaños,
  • R. Sánchez-Vélez,
  • G. Tavares-Velasco

DOI
https://doi.org/10.1140/epjc/s10052-019-7211-8
Journal volume & issue
Vol. 79, no. 8
pp. 1 – 21

Abstract

Read online

Abstract The flavor changing neutral current decays $$t\rightarrow c X$$ t→cX ($$X=\gamma ,\,g,\, Z,\, H$$ X=γ,g,Z,H ) and $$t\rightarrow c{{\bar{\ell }}}\ell $$ t→cℓ¯ℓ ($$\ell =\mu ,\,\tau $$ ℓ=μ,τ ) are studied in a renormalizable scalar leptoquark (LQ) model with no proton decay, where a scalar SU(2) doublet with hypercharge $$Y=7/6$$ Y=7/6 is added to the standard model, yielding a non-chiral LQ $$\varOmega _{5/3}$$ Ω5/3 . Analytical results for the one-loop (tree-level) contributions of a scalar LQ to the $$f_i\rightarrow f_j X$$ fi→fjX ($$f_i\rightarrow f_j {\bar{f}}_m f_l$$ fi→fjf¯mfl ) decays, with $$f_a=q_a, \ell _a$$ fa=qa,ℓa , are presented. We consider the scenario where $$\varOmega _{5/3}$$ Ω5/3 couples to the fermions of the second and third families, with its right- and left-handed couplings obeying $$\lambda _R^{\ell u_i}/\lambda _L^{\ell u_i}=O(\epsilon )$$ λRℓui/λLℓui=O(ϵ) , where $$\epsilon $$ ϵ parametrizes the relative size between these couplings. The allowed parameter space is then found via the current constraints on the muon $$(g-2)$$ (g-2) , the $$\tau \rightarrow \mu \gamma $$ τ→μγ decay, the LHC Higgs boson data, and the direct LQ searches at the LHC. For $$m_{\varOmega _{5/3}}=1$$ mΩ5/3=1 TeV and $$\epsilon =10^{-3}$$ ϵ=10-3 , we find that the $$t\rightarrow c X$$ t→cX branching ratios are of similar size and can be as large as $$10^{-8}$$ 10-8 in a tiny area of the parameter space, whereas $${\mathrm{Br}}(t\rightarrow c{{\bar{\tau }}}\tau )$$ Br(t→cτ¯τ) [$${\mathrm{Br}}(t\rightarrow c{{\bar{\mu }}}\mu )$$ Br(t→cμ¯μ) ] can be up to $$10^{-6}$$ 10-6 ($$10^{-7}$$ 10-7 ).