X(3872) tetraquarks in B and Bs decays

Luciano Maiani,Veronica Riquer,Antonio D Polosa

doi:10.1103/physrevd.102.034017

Luciano Maiani, Veronica Riquer + Show 1 more

Open Access

https://doi.org/10.1103/physrevd.102.034017

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Abstract

We discuss how the latest data on X(3872) in B and B_s decays speak about its tetraquark nature. The established decay pattern, including the up to date observations by CMS, are explained by the mixing of two quasi-degenerate, unresolvable, neutral states. The same mechanism also explains isospin violations in X decays and strongly suggests that the lurking charged partners are required to have very small branching fractions in J/psi rho^pm, well below the current experimental limits. In addition, a new prediction on the decay into J/psi omega final states is attained. The newest experimental observations are found to give thrust to the simplest tetraquark picture and call for a definitive, in-depth study of final states with charged rho mesons.

Highlights

The discussion on the nature of Xð3872Þ has been going on, with conflicting conclusions, for about two decades since its first observation at Belle [1]
The proximity of the Xð3872Þ to the D0D 0Ã threshold, isospin violations, and the lack of evidence so far of a complete multiplet of charged and neutral states has convinced a large part of the community working on this problem that the Xð3872Þ should be a sort of deuteron made of neutral D mesons, namely a D0D 0Ã molecule, with a very small binding energy, which is still unknown because of the uncertainties in the determination of the Xð3872Þ mass value
One might suppose that only color forces determine the structure of the Xð3872Þ, which is often referred as to the compact tetraquark interpretation

Summary

Introduction

The discussion on the nature of Xð3872Þ has been going on, with conflicting conclusions, for about two decades since its first observation at Belle [1]. Despite its decay modes involving the J=Ψ, Xð3872Þ cannot be interpreted as a pure charmonium state. One of the simplest reasons for this is the fact that it decays in J=ψρ and J=ψω with similar rates, violating isospin.

Results

Conclusion