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I am trying to get a better understanding of why positronium decays while a hydrigen atom is stable. In the case of positronium, I can write an elementary process were the leptons annihilate into two photons. But for the case of an hydrogen atom, I can not write a similar simple process where quark and electron would annihilate, due to charge conservation. Is it indeed impossible to write such a process at any order? More speculatively, if we lived in a different world were quarks had integer charge, would a quark/electron bound state be stable?

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  • $\begingroup$ Can you better explain the "elementary process"? I'm not sure how two anti-matter particles would annihilate, or how a quark and an electron could annihilate. $\endgroup$ – HDE 226868 Aug 6 '14 at 14:21
  • $\begingroup$ The hydrogen atom is neutral, (electric) charge conservation has nothing to do with this. $\endgroup$ – fqq Aug 6 '14 at 14:26
  • $\begingroup$ Perhaps I should have said 'simple' process rather : tree-level or one loop order. Positronium annihilation is straightforward to write using simple Feynman diagrams, But I was not able to write one for quark and electron, and I was wondering if one reason was the lack of interaction vertex between up quark (charge +2/3) and electron (-1). $\endgroup$ – Whelp Aug 7 '14 at 7:06
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The symmetries that you're missing are conservation of baryon number $B$ and lepton number $L$.

We strongly suspect that baryon number is not an exact symmetry, because the universe appears to contain very many baryons and very few antibaryons. Actually, a better metric for the baryon asymmetry of the universe is to compare the baryon density to the density of cosmic microwave photons, since that's what the gas of baryons and antibaryons at the Big Bang turned into; by that standard, the universe has a baryon asymmetry of about 10-9. But no process that violates baryon number — starting with a proton, $B=1$, and ending up only with pions, $B=0$, or that sort of thing — has ever been observed in any laboratory.

We also have no evidence for any processes that change lepton number. Weak decays, like $\beta$ decay, always create leptons and antileptons in pairs. I'm not sure whether lepton number violation is required by cosmology in the same way that baryon number violation is: it's possible that the matter electrons are balanced by an equal number of antimatter neutrinos that haven't interacted with anything since the Big Bang. Although the theoretical/phenomenological arguments for $L$-violation are actually stronger than for $B$-violation, since it's possible that the neutrino and antineutrino are actually the same particle.

If you get deep into the literature you'll find discussions of theories that break $B$ and $L$ but conserve the difference $B-L$, predicting, for instance, that the proton should undergo $p\to\pi^0+e^+$. (This was actually the original purpose the giant water tank at Super Kamiokande in Japan: to put lots and lots of hydrogen in view of lots and lots of detectors and to wait for a proton to decay. The mean time before decay for that particular mode is somewhere above 1033 years.) Your hydrogen atom disappearance would have the same Feynman diagram, $e^-+p\to\pi^0$. There are also comparable limits on interactions where $B$ changes by two units, such as $pp\to\pi\pi$ in the confines of a heavy nucleus.

Postironium decay, however, has lepton number zero in the initial state (with one lepton and one antilepton) and lepton number zero in the final state (with zero leptons and two photons), and so is not forbidden by these conservation laws.

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The electron is the lightest lepton and the proton is the lightest baryon, so it's hard to see what reaction could occur without violating lepton number or baryon number.

I suppose if proton decay (to a pion and positron) occurs then there could be a reaction to give a pion and two photons.

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  • $\begingroup$ $p^{+} + e^{-} \rightarrow n + \mu_{e}$, but that reaction is endothermic $\endgroup$ – Jerry Schirmer Aug 6 '14 at 18:10

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