New physics contributions to $\bar{B}_s \rightarrow \pi^0(\rho^0 )\,\eta^{(')} $ decays

TL;DR

This paper explores how a new $U(1)'$ gauge symmetry could significantly enhance certain rare $ar{B}_s$ decay rates beyond Standard Model predictions, with potential observable effects in specific decay channels.

Contribution

It introduces a model with an additional $U(1)'$ gauge symmetry and analyzes its impact on $ar{B}_s$ decay branching ratios, highlighting scenarios with substantial enhancements.

Findings

Maximum enhancement of $ar B^0_s o \, ho^0\,\eta$ branching ratio up to ten times SM prediction.

Enhancement of $ar B^0_s o \, ho^0\,\eta$ occurs under fine-tuned parameter conditions.

Significant increase in $ar B_s ightarrow ho^0 \, \eta$ branching ratio in certain $Z'$ coupling scenarios.

Abstract

The decay modes $\bar{B}_s \rightarrow \pi^0(\rho^0 )\,\eta^{(')} $ are dominated by electroweak penguins that are small in the standard model. In this work we investigate the contributions to these penguins from a model with an additional $U(1)'$ gauge symmetry and show there effects on the branching ratios of $\bar{B}_s \rightarrow \pi^0(\rho^0 )\,\eta^{(')} $. In a scenario of the model, where $Z^\prime$ couplings to the left-handed quarks vanish, we show that the maximum enhancement occurs in the branching ratio of $\bar B^0_s\to \,\pi^0\,\eta'$ where it can reach $6$ times the SM prediction. On the other hand, in a scenario of the model where $Z^\prime$ couplings to both left-handed and right-handed quarks do not vanish, we find that $Z^\prime$ contributions can enhance the branching ratio of $B^0_s\to\,\rho^0\,\eta$ up to one order of magnitude comparing to the SM prediction for several sets of the parameter space where both $ \Delta M_{B_s}$ and $S_{\psi\phi}$ constraints are satisfied. This kind of enhancement occurs for a rather fine-tuned point where $ \Delta M_{B_s}$ constraint on $\mid S_{SM} (B_s) + S_{Z'} (B_s)\mid $ is fulfilled by overcompensating the SM via $S_{Z'} (B_s) \simeq -2 S_{SM} (B_s)$.