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$K^{0}$ meson consists of a $d$ quark and an $\bar{s}$ antiquark. Its antiparticle $\bar{K^{0}}$ consists of an $s$ quark and a $d$ antidown quark.

$$|K^{0}\rangle=|d\bar{s}\rangle$$ $$|\bar{K}^{0}\rangle=|\bar{d}s\rangle$$

There are two versions of $K^{0}$ meson. One of them is the K-zero-long:

$$K_{L}^{0}=\frac{1}{\sqrt{2}}\left[|K^{0}\rangle-|\bar{K}^{0}\rangle\right]$$

My question is how can both decays be possible, as in my opinion the decay to electron violates conservation of lepton number. And is this interaction only possible via Z-Boson in the std. model?

$$K_{L}^{0} \xrightarrow{weak} \pi^{±} e^{\mp} \nu_e.$$

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    $\begingroup$ That should be electron+antineutrino or positron+neutrino. Lepton number is conserved, and the process runs via W, not Z. $\endgroup$ – Bert Barrois Jun 17 '18 at 10:35
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    $\begingroup$ lepton number is conserved by the appropriate neutrino for positron, antineutrino for electron $\endgroup$ – anna v Jun 17 '18 at 10:35
  • $\begingroup$ so for example in the wikipedia article for the kaon (table) it is implicitly assumed with $\nu_e$ that the appropriate choice for the neutrino has to be made? $\endgroup$ – 0xkev Jun 17 '18 at 10:38
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    $\begingroup$ yes, in order not to write explicitly two lines $\endgroup$ – anna v Jun 18 '18 at 3:03

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