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From reading the answer in Difference between the CKM and the PMNS matrix , I gather that the transition $W\to ub$ where $u$ and $b$ mean flavour eigenstates is not possible, but it is possible where $u$ and $b$ means mass eigenstates. Is this understanding correct?

If this is so, then starting with a $B+$, which is in a flavour eigenstate of $u\bar b$, not a mass eigenstate (again is this correct?) then is the following Feynman diagram possible? I believed it was, but if the $b$ that comes from the $W$ is not a flavour eigenstate, then surely the annihilation vertex at the top of the diagram $b\bar b \to y$ is not possible, as electromagnetism does not change flavour and hence requires that both the $b$ and $\bar b$ are flavour eigenstates?

Thanks for any help.enter image description here

Jack
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The flavor eigenstates for the quarks are defined to be the same as the mass eigenstates. What you're talking about is a basis that diagonalizes the weak interaction matrix. There are many bases you could choose in principle, but the standard one is $u, c, t, d', s', b'$ where the primed particles are related to $d, s, b$ by the CKM matrix. $W\to\bar ub'$ (note the bar) can't happen because the amplitude is proportional to an off-diagonal entry of the diagonal interaction matrix.

Your Feynman diagram is allowed, but it would be disallowed if you replaced either or both of the $b$s by $b'$s.

benrg
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