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Show that there is no d → s + Z^0

  • Thread starter Poirot
  • Start date
  • #1
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Homework Statement


Consider weak interactions of quarks involving the Z0 bosons. Show that there is no d → s + Z0 interaction in the Standard Model.

Homework Equations


(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!
 

Answers and Replies

  • #2
34,055
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Which course is this?

The nonexistence of this interaction (at tree-level) is an experimental result. Sure, you can look at equations and point out that it does not happen according to them (the standard model Lagrangian is probably the best point), but those equations were made to reflect our measurements, so that's kind of recursive reasoning.
 
  • #3
94
2
Which course is this?

The nonexistence of this interaction (at tree-level) is an experimental result. Sure, you can look at equations and point out that it does not happen according to them (the standard model Lagrangian is probably the best point), but those equations were made to reflect our measurements, so that's kind of recursive reasoning.
It's a particle physics module in a 2nd year physics degree, I had the exam today and it went as well as it could have done, so thank you for all your help, with this and with other questions, it's greatly appreciated!
 

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