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Looking at the Particle Data Group tables of the $\Lambda$ baryon, I find that the rate of the hadronic decay $\Lambda \to p \pi^-$ is 64% while the semileptonic decay $\Lambda \to p e^-\nu_e$ has the rate of $8 \times 10^{-4}$.

I cannot explain what is the reason for such large suppression of the semileptonic decay of the $\Lambda$. Both $\Lambda \to p \pi^-$ and $\Lambda \to p e^-\nu_e$ decays originate from the same Feynman diagram, $s \to u W^-$ at the quark level, with $W^-$ going either to $\overline{u}d$ or $e^-\nu_e$. So, I would expect the factor 3 difference between these two diagrams, due to the color factor; which could then be diminished by the smaller phase-space in the hadronic decay compared to the semileptonic one. However, the actual difference in rate of $\Lambda \to p \pi^-$ and $\Lambda \to p e^-\nu_e$ is factor 800, so my estimation is wrong.

The assumption that a three-body decay would somehow be more suppressed than a two-body decay, also does not work, as the charged kaon, $K^-$, decays (via the same Feynman diagrams as the decays discussed above, modulo one spectator quark) to $\pi^0\pi^-$ in about 21% of times, and to $\pi^0 e^-\nu_e$ in about 5% of times (so, factor 4 difference instead of 800). There should be some suppression affecting only baryonic decays but not mesonic... but I do not see where it might come from.

In a related question, a suggestion has been made that semileptonic decays might be affected by the helicity suppression. However, under that assumption, I would expect the $\Lambda \to p \mu^-\nu_\mu$ decay to dominate over $\Lambda \to p e^-\nu_e$ (like in a pion decay), which is not the case.

What am I missing?

(This question is motivated by my previous question on hyperon lifetimes.)

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  • $\begingroup$ The comparison $\Gamma_e/\Gamma_\mu\approx 5 $ is not too far from the customary 5th power of available momenta, $(163/131)^5\approx 3 $... It's the cheapest dimensional guess for widths that are inversely proportional to the inverse fourth power of $M_W$... $\endgroup$
    – Cosmas Zachos
    Jun 5, 2020 at 22:39
  • $\begingroup$ So far, I'm stumped. They all do this, $\Lambda, \Sigma^{\pm}, \Xi, \Omega$. The standard review refers to it as an experimental fact. But BR factors of 800 do not arise from a few wave function peculiarities and extra diagrams... Good question. $\endgroup$
    – Cosmas Zachos
    Jun 6, 2020 at 16:23
  • $\begingroup$ But only hyperons. The charmed baryons or the $\Lambda_b$ have a few % rate of their heavy-quark semileptonic decays. (It would be interesting to look at strangeness-changing decays of $\Xi_c$ or $\Xi_b$ where phase-space is as small as in hyperon decays, but only $\Xi_b^- \to \Lambda_b \pi^-$ has been measured out of them.) $\endgroup$
    – Martino
    Jun 6, 2020 at 22:07
  • $\begingroup$ Yes, $\Lambda_c^+$ has 6% for β-decay... Looks like as the lost quark gets heavier, the situation "normalizes". All this, points, then, to a freak wavefunction enhancement of the hadronic modes for the lighter quarks... still intriguing. $\endgroup$
    – Cosmas Zachos
    Jun 7, 2020 at 0:15
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    $\begingroup$ OK, I checked with real experts that it is a murky long-distance / hadronization business, involving P waves, penguin diagrams, etc... cf Wu & Rosner 1986. The takeaway is that it is a recondite "just so" situation, and there is no quick-and-dirty "of course!" explanation... the strong interactions conspire... But your e-print discussion above is unsettling! $\endgroup$
    – Cosmas Zachos
    Jun 8, 2020 at 20:52

1 Answer 1

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The hadronic decay can also go through the exchange of a W. (The gluon can come from anywhere)

enter image description here

Apologies for low-quality diagram...

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  • $\begingroup$ there are two weak vertices there. But it could explain how for the higher lamdas ( c ) equillibrioum is found. see comment in my "so called" answer. $\endgroup$
    – anna v
    Jun 6, 2020 at 13:17
  • $\begingroup$ There are two weak vertices in the alternative with a W producing $\overline u d$ or $e \nu$. $\endgroup$
    – RogerJBarlow
    Jun 6, 2020 at 14:00
  • $\begingroup$ If you read my answer, you will see that since W couples to electromagnetic also, a higer order quark antiquark pair from virtual photons is much more probable than with weak vertices. $\endgroup$
    – anna v
    Jun 6, 2020 at 14:03
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    $\begingroup$ I don't see how you can draw the diagram. Btw the PDG gives the W branching ratio to hadrons as 67%. (Top of page 6 in the reference you kindly link to). This is what you expect from 2 generations * 3 colours of quark modes versus 3 lepton modes, ignoring phase space and the b quark. $\endgroup$
    – RogerJBarlow
    Jun 6, 2020 at 15:24
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    $\begingroup$ The $Ω^−$ has the same dominance of hadronic modes, without extra diagrams, though... An extra diagram and the interferences it enables is hard to parlay to a factor of 800 in the rate, no? – $\endgroup$
    – Cosmas Zachos
    Jun 6, 2020 at 17:00

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