I was asked to show that in the baryon octet, the $\Sigma^0$ baryon is the only particle which decays electromagnetically. Since it is an electromagnetic decay, strangeness should be conserved but I don't really get why can't a neutron (which doesn't have strangeness) also decay ellectromagnetically into a meson plus a photon. Is there any conservation law violated? I don't really know whether the quark content has to be conserved or not...

I also was asked to show why in the baryon octet, the neutron is the only particle which can decay into leptons. In this case, I came up with another decay $\Sigma^-$ $\rightarrow$ n + e$^-$+ $\bar{\nu_{e}}$ which doesn't seem to violate any conservation law.

Is there something that I'm missing? Any help is welcome!

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    $\begingroup$ In the same paragraph, further down, you do intend to conserve baryon number? And the negative member of the isotriplet in your middle paragraph has a BR ~ 1/1000 as here, Bourquin 1983, no? Are you requiring dominant semileptonic mode? $\endgroup$ Jan 19, 2020 at 15:50
  • $\begingroup$ It's actually a introductory course in particle physics, nothing very profound, so I'm asked to look just quantum numbers violation. I totally missed the baryon number conservation in the first paragraph (so the question is solved), but when it comes to the second one, I'm supposed to demonstrate that the neutron is the only particle in the octet baryon which decays into lepton, which if we glanze at the upper decay isn't true? . (I don't have a clue about what dominant semileptonic mode is). Thanks. $\endgroup$ Jan 19, 2020 at 16:02
  • $\begingroup$ dominant means the largest term. semileptonic means that there are also other particles coming from the decay than the lepton.en.wikipedia.org/wiki/Semileptonic_decay . you can use google and dictionaries on the web if you do not understand a word. try semileptonic on google $\endgroup$
    – anna v
    Jan 20, 2020 at 5:23

1 Answer 1


You may be basically asked to understand the PDG. You can't change strangeness, electromagnetically.

So you may only decay by emitting a photon and rearranging your quarks in the case where your baryon charge stays the same (!), $\Sigma ^0 \to \Lambda \gamma$, which makes the neutral Σ nine orders of magnitude shorter-lived than its isopartners. No other options are available, energetically, and this is the only spot in the octet where this (same-charge members) happens. (And, as you appreciated, you must conserve baryon number, so that option is closed.)

Now, in a different world, the neutron would prefer to decay (by rearranging quarks) to a proton and a $\pi^-$ if it could, energetically: but check that the n-p mass difference is too small. So its only option is the usual weak decay to an electron and an antineutrino and a proton (semileptonic: some of the decay products are leptons) and it takes it forever to do that, about a quarter of an hour...

The $\Sigma^-$, however, has that option, since it is so much heavier than the neutron, to decay weakly to $n~\pi^-$, which it does most of the time. But your mode $n~e^- \bar \nu$ is also OK, except subdominant (BR ~ $10^{-3}$), and is, in fact, measured, Bourquin 1983, providing valuable WI information.

NB on comment/good-followup-question of @Marco Villalobos

I don't know. Related, especially the Wu & Rosner 1986 reference on non-leptonic channel enhancements... Looks like it is a just-so accepted mystery on just hyperons. As one moves on to charm and bottom, the enhancement abates. Go figure...

  • $\begingroup$ Okay, and then... why is the pion non-leptonic decay way more dominant? I mean, there is more phase space available (the difference of masses is greater with the semileptonic decay). Why the neutrino is not the dominant? $\endgroup$ May 26, 2022 at 17:41

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