Magister said:
But is the CKM matrix in the the [itex]B^0 bar-B^0[/itex] system the same as in the [itex]k^0 bar-k^0[/itex] system?
Yes!
I'll try to explain why your question doesn't make so much sense :)
When you consider an elementary interaction involving a quark q, another quark q', and a W, in the calculation of the amplitude you will have to multiply for g (the weak interaction coupling, which is *universal*, i.e. does not depend on the process), and a factor which depends on the quarks, let's call it Vqq'. This factor is the element in the q-th row and the q'-th column of the CKM matrix.
The [itex]K^0[/itex] is composed of a strange and a down quark (one is a quark and the other an antiquark, and which is which depends on whether it is [itex]K^0[/itex] or [itex]\bar K^0[/itex]). So the coefficients of CKM involved will be Vus, Vcs,Vts and Vud, Vcd,Vtd (because the weak interaction can couple the "low" quarks only to "high" quarks, I mean that d,s,b can only couple to u,c,t and not to other d,s,b).
In the [itex]B^0[/itex] the quark composition is one bottom and one down quark, in the [itex]B^0_s[/itex] the composition is one bottom and one strange. So the coefficients will be different coefficients in general (Vub, Vcb, etc.) but the matrix is always the same good CKM matrix.