Penguin operators in nonresonantB−→MM¯π−

Abstract

We investigate the contributions coming from the penguin operators in the nonresonant B − → M M π − (M = π − , K − , K 0 ) decays. The effective Wilson coefficients of the the strong penguin operators O 4,6 are found to be relatively larger. We calculate the contributions arising from the O 4 and O 6 operators in the nonresonant decays B − → M M π − (M = π − , K − , K0 ) using a model combining heavy quark symmetry and the chiral symmetry, developed previously. We find that the forbidden nonresonant B − → K 0 K0 π − decay occurs through the strong penguin operators. These penguin contributions affect the branching ratios for B − → M M π − (M = π − , K − ) by only a few percent. The branching ratio for B − → K 0 K0 π − is estimated to be of the order 10 −6 .

Nonresonant three-body decays ofDandBmesons

“…Nevertheless, its study is important for several reasons. First, the interference between resonant and nonresonant decay amplitudes in B decays may provide information on the CP-violating phase angles [1][2][3][4][5][6]. For example, the interference between B − → (π + π − π − ) NR and B − → χ c0 π − could lead to a measurable CP asymmetry characterized by the phase angle γ [1], while the Dalitz plot analysis of B → ρπ → πππ allows one to measure the angle α.…”

Search for the decayB+→Ks0Ks0π+

“…A model based on the factorization approximation, which makes use of heavyquark and chiral symmetries, predicts a nonresonant branching fraction for B þ ! K 0 " K 0 þ of order 10 À6 [9]. Decays via intermediate resonant states can also lead to the K 0 S K 0 S þ final state.…”

CPviolation in the partial width asymmetries forB−→π+π−Fajfer1, Pham2, Prapotnik3 2004Phys. Rev. DSelf Cite360270View full textAdd to dashboardCite

“…The approach used in the calculation of these matrix elements was already explained in [5,23,24,25]. Here, we follow the same method, but add the contributions of the B * 0 scalar meson resonances.…”

“…The B − → K − K + K − amplitude can be factorized in the same way by replacing π ± with K ± in (8)- (10). However, in this case, the contribution coming from…”