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This article on wikipedia on alpha decay states:

One curiosity is why alpha particles, helium nuclei, should be preferentially emitted as opposed to other particles like a single proton or neutron or other atomic nuclei.[note 1] Part of the answer comes from conservation of wave function symmetry, which prevents a particle from spontaneously changing from exhibiting Bose–Einstein statistics (if it had an even number of nucleons) to Fermi–Dirac statistics (if it had an odd number of nucleons) or vice versa. Single proton emission, or the emission of any particle with an odd number of nucleons would violate this conservation law.

Is this incorrect or misleading somehow? Single proton emission actually exists.

(Note added later: the text quoted above has been deleted from the wikipedia article, apparently by a reader of this post.)

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  • $\begingroup$ Remember that in beta decay the electron is not the only particle emitted. $\endgroup$ – Jon Custer Sep 24 '19 at 13:28
  • $\begingroup$ Oops. Good point. I have removed the reference to beta decay, since it's of secondary importance to the main question. $\endgroup$ – John Sep 24 '19 at 14:19
  • $\begingroup$ OK, now, thinking in simple terms: you have an initial nuclei with some (bound) wave function A. It decays into a free particle and some other (bound) wave function B. If A and B have different symmetry, the overlap integral of A&B is zero and you should not be able to decay from A to B. Now you know by now that various symmetries get violated often, but usually that comes at a high cost in probability of the transition happening ("allowed" vs "disallowed" transitions). $\endgroup$ – Jon Custer Sep 24 '19 at 14:28
  • $\begingroup$ I've never heard of a violation of the rule about wave function symmetry. Seems like a strict rule to me. $\endgroup$ – John Sep 24 '19 at 15:07
  • $\begingroup$ In atomic electronic transitions you often have transitions from one angular momentum state to another separated by more than 1 - they are just less common. While they are 'prohibited', they are only strictly prohibited for a one-electron atom - the presence of all the other electrons in a multi-electron atom, and their far more complex interactions, means they can happen. The same thing in a multi-nucleon nucleus. $\endgroup$ – Jon Custer Sep 24 '19 at 15:54
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The article is total nonsense. As you point out, proton decay exists. The system as a whole doesn't change symmetry, but the final state of the system includes both of the decay products. So for example, if the original nucleus was a fermion, and it proton decayed, the final system will also be fermionic: it will consist of a bosonic nucleus plus a fermionic proton.

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