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  Why is the D0 oscillation so different from the K0 and B0?

+ 3 like - 0 dislike

I have looked for this answer into many articles and books but I am not able to figure out why $D^0\to\bar{D}^0$ is so highly suppressed if compared to the $B^0 \to \bar{B}^0$ and $K^0 \to \bar{K}^0$ diagrams. In principle, I guess that the GIM mechanism acts to the cancellation of diagrams which include vertices with opposite sign CKM factors. However, this effect should be the same for $K^0$,$B^0$ and $D^0$ mesons. I suspect then that this difference could be introduced by the different masses of quarks c,b,s, but I don't understand exactly how. Could anyone clarify me this difference and also cite a reference? Thank you very much.

This post imported from StackExchange Physics at 2015-12-12 20:25 (UTC), posted by SE-user MVal

asked Dec 4, 2015 in Theoretical Physics by MVal (15 points) [ revision history ]
edited Dec 12, 2015 by Dilaton
I suspect it is related to the lifetime: the $D^0$ is the system having the smallest lifetime (an order of magnitude w.r.t $B^0$, and several orders w.r.t $K^0$).

This post imported from StackExchange Physics at 2015-12-12 20:25 (UTC), posted by SE-user Paganini
Good remark. Maybe the two things are related but, as mixing only depends by the mass eigenstates $|D_1>,|D_2>$ which behave different masses and lifetimes, this difference should be dictated by the off-diagonal terms in the weak interaction hamiltonian. Hence, these factors are smaller in the $D^0$ case if compared to $B^0, K^0$. My idea is to try to understand this effect with Feynman diagrams in the context of CKM formalism.

This post imported from StackExchange Physics at 2015-12-12 20:25 (UTC), posted by SE-user MVal

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