The strange quark can decay via weak charged currents, following the rule that if $|\Delta S|=1$ then $\Delta S=\Delta Q_H$ where $Q_H$ is the hadron charge. The source below states that an $s$-quark can only decay into 'up-type quarks' why is this so? and can a similar thing be said about $\bar s$?


(1) Pal, P.B., 2014. An introductory course of particle physics. Taylor & Francis. (p502, link to Google books)

  • 2
    $\begingroup$ Your source doesn't quite say the same thing as what you wrote. The strange quark is a down-type quark, so when it decays through a charged-current transition, it must become an up-type quark, because electric charge is a conserved quantity. There are flavour-changing neutral currents, but they are forbidden at tree-level in the Standard Model, so their leading-order diagrams must contain at least one loop, making them suppressed. $\endgroup$ – dukwon Aug 22 '16 at 17:37

If you consider a weak decay, a $W^\pm$ is always involved (flavour changing neutral currents have not been observed and are suppressed in the SM). This means that the strange quark "radiates" a $W^\pm$ boson and hence it decays into an up quark. Due to energy considerations it cannot decay into a charm or a top. The argument is the same for strange and anti strange.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.