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To my understanding the decay of a neutral pion into an electron-positron pair can only happen by the electromagnetic force and the mediation of two virtual photons in a triangle-diagram, so it is loop-suppressed.

What I'm failing to understand is: What is forbidding the direct decay into an electron-positron pair rather than 2 gamma rays? And why is the weak decay forbidden?

I assume I'm missing a conservation law, the question is which one?

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  • $\begingroup$ en.wikipedia.org/wiki/Pion#Neutral_pion_decays $\endgroup$ – fqq Feb 1 '16 at 16:38
  • $\begingroup$ yes, that's what I was summarizing... They say it's electromagnetic, not weak and loop-supressed... The question is why? $\endgroup$ – myname Feb 1 '16 at 16:55
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    $\begingroup$ Just from a SM-theoretical point of view, how would the neutral pion couple to the charged leptons? How do charged leptons communicate with the world (according to the SM)? $\endgroup$ – Your Majesty Feb 2 '16 at 12:54
  • $\begingroup$ Theoretically the pion should be able to interact through the weak or the electromagnetic force with charged leptons. Since the pi0 is neutral that would mean a Z or a photon. Now, naively, I would assume that since the pi+ decays to positron + electron-neutrino through a W+, the pi0 can decay into a positron + electron pair through a Z. I'm not seeing what is violated here. I know the Z does decay into an electron-positron pair, so the problem must be pi0 -> Z? $\endgroup$ – myname Feb 2 '16 at 14:20
  • $\begingroup$ @myname The title LaTeX looked broken - I took the liberty to fix it, but obviously feel free to rollback or re-edit if it does not capture your intent. $\endgroup$ – Emilio Pisanty Feb 2 '16 at 21:41
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By comparing it to the $\pi^- \rightarrow l^- + \bar{\nu_{l} }$ decay, where the lepton and neutrino have the same helicity but is also helicity suppressed since the lepton is massive and contain contributions from left- and right-handed chirality. The antineutrino has right-handed chirality so it participates in the weak exchange.

The process $\pi^0 \rightarrow e^+e^-$ decay is doubly weak-suppressed since both the decay electron and positron have the same right-handed helicity. Only right-handed chiral antiparticles and left-handed chiral particles participate in the weak interaction.

The electron and positron are not massless therefore this right-handed helicity also contains contributions from both left- and right-handed chirality. The neutral pion can decay electromagnetically via loops with a branching ratio of $10^{-9}$.

$\pi^0 \rightarrow e^+e^-$ decay

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In the pion reference frame the two outgoing leptons are very boosted, hence helicity and chirality almost coincide. The angular momentum conservation forces them to have opposite spins, since the pion spin is zero. Therefore, they will have the same helicity, which is highly suppressed in this kinematic regime, because of the vector nature of the QED interactions (see for example Thomson, Modern Particle Physics, chapter 6). Just my two cents.

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    $\begingroup$ Welcome on the Physics SE! Thank you very much your worthy answer. I suggest to change your username to a more remarkable one, you can do it in your profile page. Good luck! $\endgroup$ – peterh Aug 6 '16 at 11:05
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The weak decay is not forbidden. You can have $q\overline q \to ZZ \to e^+e^-$ using a loop like in the EM decay. You need a loop to conserve angular momentum, because a single $Z$ or $\gamma$ has spin $1$ while the $\pi^0$ has spin $0$. If an EM decay is possible, it is so much faster that it will dominate. You can see this in the overall $\pi^0$ decay rate compared to the charged pion decay rate. The neutral pion decays $10^9$ times faster because it is an EM decay. The charged pions have to decay weakly as there is nothing they can decay into electromagnetically.

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  • $\begingroup$ Do you have a source to that? All I have found said that the direct decay is forbidden and you need to go through a loop with two photons to get a decay into $e^{+}e^-$ $\endgroup$ – myname Feb 8 '16 at 16:14
  • $\begingroup$ I have corrected it. I realized the angular momentum problem. It is still true that the EM decays will dominate. $\endgroup$ – Ross Millikan Feb 8 '16 at 17:55
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    $\begingroup$ Sorry for the late reaction. According to this answer about angular momentum for charged pions: [physics.stackexchange.com/questions/89684/… the angular momentum has only to be conserved for in-going and out-going particles. Not for mediating particles. So why does this seem to hold true for charged pions but not for the neutral one? $\endgroup$ – myname Feb 29 '16 at 15:49

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