Scrutinizing $$B^0$$ B 0 -meson flavor changing neutral current decay into scalar $$K_0^*(1430)$$ K 0 ∗ ( 1430 ) meson with $$b→ s \ell ^+\ell ^-(ν \bar{ν })$$ b → s ℓ + ℓ - ( ν ν ¯ ) transition - Yin-Long Yang; Ya-Xiong Wang; Hai-Bing Fu; Tao Zhong; Ya-Lin Song (European Physical Journal C: Particles and Fields)
Abstract In this paper, we investigate the rare decay $$B^0→ K_0^*(1430)\ell ^+\ell ^-$$ B 0 → K 0 ∗ ( 1430 ) ℓ + ℓ - with $$\ell =(e,μ ,τ )$$ ℓ = ( e , μ , τ ) and $$B^0→ K_0^*(1430)ν \bar{ν }$$ B 0 → K 0 ∗ ( 1430 ) ν ν ¯ induced by the flavor changing neutral current transition of $$b→ s\ell ^+\ell ^-(ν \bar{ν })$$ b → s ℓ + ℓ - ( ν ν ¯ ) . Firstly, the $$B^0→ K_0^*(1430)$$ B 0 → K 0 ∗ ( 1430 ) transition form factors (TFFs) are calculated by using the QCD light-cone sum rule approach up to next-to-leading order accuracy. In which the $$K_0^*(1430)$$ K 0 ∗ ( 1430 ) -meson twist-2 and twist-3 LCDAs have been calculated both from the SVZ sum rule in the background field theory framework and light-cone harmonic oscillator model. Then, we obtain the three TFFs at large recoil point, i.e., $$f_+^{B^0→ K_0^*}(0)= 0.470_{-0.101}^{+0.086}$$ f + B 0 → K 0 ∗ ( 0 ) = 0 . 470 - 0.101 + 0.086 , $$f_-^{B^0→ K_0^*}(0)= -0.340_{-0.068}^{+0.068}$$ f - B 0 → K 0 ∗ ( 0 ) = - 0 . 340 - 0.068 + 0.068 , and $$f_T^{B^0→ K_0^*}(0)= 0.537^{+0.112}_{-0.115}$$ f T B 0 → K 0 ∗ ( 0 ) = 0 . 537 - 0.115 + 0.112 . Meanwhile, we extrapolate TFFs to the whole physical $$q^2$$ q 2 -region by using the simplified $$z(q^2)$$ z ( q 2 ) -series expansion. Furthermore, we calculate the $$B^0→ K_0^*(1430)\ell ^+\ell ^-(ν \bar{ν })$$ B 0 → K 0 ∗ ( 1430 ) ℓ + ℓ - ( ν ν ¯ ) decay widths, branching fractions, and longitudinal lepton polarization asymmetries of $$B^0→ K_0^*(1430)\ell ^+\ell ^-$$ B 0 → K 0 ∗ ( 1430 ) ℓ + ℓ - , which lead to $$\mathcal{B}(B^0→ K_0^*(1430)e^+e^-) = (6.65^{+2.52}_{-2.42})× 10^{-7}$$ B ( B 0 → K 0 ∗ ( 1430 ) e + e - ) = ( 6 . 65 - 2.42 + 2.52 ) × 10 - 7 , $$\mathcal{B}(B^0→ K_0^*(1430)μ ^+μ ^-)=(6.62^{+2.51}_{-2.41})× 10^{-7}$$ B ( B 0 → K 0 ∗ ( 1430 ) μ + μ - ) = ( 6 . 62 - 2.41 + 2.51 ) × 10 - 7 , $$\mathcal{B}(B^0→ K_0^*(1430)τ ^+τ ^-)=(1.88^{+1.10}_{-0.97})× 10^{-8}$$ B ( B 0 → K 0 ∗ ( 1430 ) τ + τ - ) = ( 1 . 88 - 0.97 + 1.10 ) × 10 - 8 , $$\mathcal{B}(B^0→ K_0^*(1430)ν \bar{ν })= 3.85^{+1.55}_{-1.48}× 10^{-6}$$ B ( B 0 → K 0 ∗ ( 1430 ) ν ν ¯ ) = 3 . 85 - 1.48 + 1.55 × 10 - 6 and the integrated longitudinal lepton polarization asymmetries $$\langle A_{P_L} \rangle = (-0.99, -0.96, -0.03)$$ ⟨ A P L ⟩ = ( - 0.99 , - 0.96 , - 0.03 ) for the cases $$\ell =(e, μ , τ )$$ ℓ = ( e , μ , τ ) respectively.