Neutral particle oscillations / quantum number conservation

Garlic
Gold Member
Messages
181
Reaction score
72
Hello everyone,
Why don't neutral particle oscillations have to obey conservation of (quark) flavor quantum numbers, with the example of neutral Kaon oscillations?
 
Physics news on Phys.org
Because weak interactions are not flavour diagonal. In general, you would expect mixing among particles whenever they have the same gauge quantum numbers unless there is a symmetry to forbid it.
 
  • Like
Likes vanhees71
Why should they? There is no conservation law for quark flavors, the weak interaction can change them. The oscillations happen via the weak interaction.

Edit: Too slow.
 
  • Like
Likes vanhees71
Toponium is a hadron which is the bound state of a valance top quark and a valance antitop quark. Oversimplified presentations often state that top quarks don't form hadrons, because they decay to bottom quarks extremely rapidly after they are created, leaving no time to form a hadron. And, the vast majority of the time, this is true. But, the lifetime of a top quark is only an average lifetime. Sometimes it decays faster and sometimes it decays slower. In the highly improbable case that...
I'm following this paper by Kitaev on SL(2,R) representations and I'm having a problem in the normalization of the continuous eigenfunctions (eqs. (67)-(70)), which satisfy \langle f_s | f_{s'} \rangle = \int_{0}^{1} \frac{2}{(1-u)^2} f_s(u)^* f_{s'}(u) \, du. \tag{67} The singular contribution of the integral arises at the endpoint u=1 of the integral, and in the limit u \to 1, the function f_s(u) takes on the form f_s(u) \approx a_s (1-u)^{1/2 + i s} + a_s^* (1-u)^{1/2 - i s}. \tag{70}...

Similar threads

Back
Top