How to make sense of a kaon decay $$K^+ \to \pi^0 + e^+ + \nu_e$$ if we take into account the quark structure of the kaon $K^+ = u \bar{s}$ and pion $\pi^0 = (u \bar{u} - d \bar{d})/\sqrt{2}$? Obviously, we have the weak decay of the $\bar{s}$ quark which produces $\bar{u}$. But, how do we get the extra $d \bar{d}$ quarks?
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The quarks of the K must go into a W+. The leptons come out of a W +. So consider π emission. – Cosmas Zachos Dec 07 '16 at 16:43
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The notation $\pi^0 = (u\bar{u} - d\bar{d})/\sqrt{2}$ means that the $\pi^0$ has equal probability of being either $u\bar{u}$ or $d\bar{d}$. It definitely doesn't mean they're tetraquarks!
They can be formed from a $u\bar{u}$ pair without having to consider the possibility of $d\bar{d}$, and vice-versa.
dukwon
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So what? I don't see the relevance of that. Have a look at http://physics.stackexchange.com/q/11039/73214 – dukwon Dec 07 '16 at 16:58
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The pi0 made out of u quarks should be less massive than the one made of d quarks. But the pi0 has a well defined mass – Dec 07 '16 at 17:04
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The $\pi^0$ indeed has a well-defined mass. The $u\bar{u}$ and $d\bar{d}$ states do not have well-defined masses, though. You can't really say that one is heavier than the other. – dukwon Dec 07 '16 at 19:46