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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 $\bar s$?

References

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

Qmechanic
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Quantum spaghettification
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    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. – dukwon Aug 22 '16 at 17:37

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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.

AccidentalFourierTransform
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Fabian
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