From https://en.wikipedia.org/wiki/Beta_decay

In $β^+$ decay, or positron emission, the weak interaction converts an atomic nucleus into a nucleus with atomic number decreased by one, while emitting a positron ($e^+$) and an electron neutrino ($ν_e$).

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If the positron is emitted, why is its arrow directed towards the $W^+$?


1 Answer 1


The motivation of this convention in Feynman diagrams is that the positron is antimatter, whereas the electron neutrino is matter. By CPT symmetry, an emitted positron is equivalent to an absorbed electron travelling backwards in time. Under time reversal, the rightmost interaction vertex is equivalent to $e^-\to\nu_e+W^-$, and the arrow directions on the "electron" and neutrino are less surprising.

  • $\begingroup$ This answers my question, thanks, but to follow up: if an electron is being absorbed then what happens if a W+ is produced without there being an electron around to absorb? Or is that situation isn't possible then is it reasonable to claim that an absorbed electron (which has to initially exist) is really equivalent to an emitted positron (which does not have to initially exist)? $\endgroup$
    – spraff
    Commented Sep 22, 2021 at 9:37
  • $\begingroup$ @spraff To use CPT we have to wind everything backwards. The outcome is not "a $W^+$ is produced, but we have to hope an electron turns up to absorb it"; it's that $e^-+\bar{\nu}_e\to W^-$ occurred, followed by $\bar{d}+W^-\to\bar{u}$. $\endgroup$
    – J.G.
    Commented Sep 22, 2021 at 9:57

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