Consider weak interactions of quarks involving the Z0 bosons. Show that there is no d → s + Z0 interaction in the Standard Model.
(d') = (cosθ sinθ) (d)
(s') (-sinθ cosθ) (s)
This is the change of variables matrix.
θ is the Cabibbo angle, with the rotation matrix on the down and strange quark
The Attempt at a Solution
I'm not sure how to prove this, I think it may be something to do with that there is a sinθ x both d and s if you're doing the transformation so they're both suppressed? But that's a wild guess, my knowledge isn't that deep on this matter.
Thank for any help, much appreciated!