# Tag Info

But they do decay into the channels available from conservation laws. Annihilation happens when all the quantum numbers cancel. In the pio the charge and baryon number of the quark antiquark cancel each other, and the decay particles add up to zero quantum numbers. The $\pi^0$ goes into two photons as soon (electromagnetic interaction rates) as it is ...
It seems to me that there's a confusion about how the flavour is acting. In your example $R_1 \sim \bar{\bf{2}}$ and $R_2 \sim {\bf{2}}$. Therefore, by multiplying these representations we expect a singlet state ($\ell =0$) and a vector ($\ell =1$), i.e., $R_1 \otimes R_2 = \bf{1} \oplus \bf{3}$. Now, if we identify the fundamental doublet as: $R_2 = ... 25 Mesons are not elementary, they are composed of quarks. So take a look at their quark content. The charmed eta meson consists of a charm and an anti-charm quark, denoted$c\overline{c}$. An anti charmed eta meson would therefore be an anti-charm and an anti-anti-charm (which is just a charm) quark, i.e.$\overline{c}c$, which is obviously the same as ... 2 Is there any possible way to extract all the mass of a quark?probably under extreme Gs or heat or pressure? Not with classical means, like Gs and heat and pressure, because the quarks are elementary particles held together within hadrons because of QCD . Quarks can give up all their energy when quark meets antiquark, as happens in proton antiproton ... 3 A similar question is the following. How can$\pi^0$and$\eta$in the$SU(3)_F$meson octet both have the same$SU(3)_F$flavor content? One could answer that this is because$\pi^0$is part of an isospin triplet of pions with$I=1$, while$\eta$is an isospin singlet with$I=0$. Or one may point out that their explicit ket linear combinations of ... 0 I cannot give you an advice on how to do this computation by hand in the most efficient way, but if you need a way to cross-check what you have obtained so far and you have access to Mathematica, you could use FeynCalc for that. The calculation of the$gg \to q \bar{q}$matrix element squared (unpolarized) is shipped with FeynCalc as an example (see here) ... 7 The isospin is different.$I=0$for the$\Lambda^0$and$I=1$for the$\Sigma^{0}$. This makes the$\Lambda^0$an isospin singlet state but the$\Sigma^0$is part of an isospin triplet. There are quite few other examples e.g. compare a proton (uud with$I=1/2$) with a$\Delta^{+}$(uud with$I=3/2\$).