Possible Explanation Why τBτB0 But τD τD0
Abstract
Data show that τBτB0, 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 Close and Lipkin's suggestion that the phases for producing radially excited states 2s in the decay products of B-mesons can possess an opposite sign to the integrals for 1s decay products. Their contributions can partially compensate each other to result in τBτB0. Since D-mesons are much lighter than B-mesons, such possibilities do not exist in D-decays.
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