Possible explanation why τB±∼τB⁰ but τD±∼2τD⁰

Abstract

Data show that τB±∼τB0, but τD±∼2τD0. The naive interpretation which attributes τD±∼2τD0 to a destructive interference between two quark diagrams for D± decays definitely fails in the B case. We investigate the suggestion of Close and Lipkin that the phases for producing radially excited states ψ2s in the decay products of B mesons can possess an opposite sign to the integrals for ψ1s decay products. Their contributions can partially compensate each other to result in τB±∼τB0. Since D mesons are much lighter than B mesons, such possibilities do not exist in D decays.