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In terms of charge, a neutral kaon could consist of a down and anti-strange quark, or an anti-down and a strange quark. However, which of these is considered to be the particle, and which the antiparticle?

Would the down and anti-strange pair be considered the particle due to its strangeness of +1, or would it be the anti-down strange pair, due to the strange quark not being an anti-quark?

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For mesons with non-zero strangeness, charm, beauty or truth, the name indicates the heaviest quark, using the following convention:

$s \to \overline{K}$, $c \to D$, $b \to \overline{B}$, $t \to T$.

http://pdg.lbl.gov/2014/reviews/rpp2014-rev-naming-scheme-hadrons.pdf

The state where the heaviest flavour quantum number is negative is the 'antiparticle'. The sign of the flavour number reflects the sign of the electric charge.

For neutral kaons, specifically $d\overline{s} \to K^0$ and $\overline{d}s \to \overline{K}^0$

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  • $\begingroup$ Perfect answer. $\endgroup$ Commented May 18, 2017 at 14:06

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