A common "cartoon model" of a nucleus is that there are a set of bound energy levels for the protons and a similar set of bound energy levels for the neutrons. The existence of these energy levels helps explain why bound neutrons are stable, even though free neutrons decay. If the decay would require the new proton to occupy an energy level that's greater than the original proton's energy level by more than 0.78 MeV, then it's energetically favorable for the neutron to stay a neutron.

But there are other baryons out there. For example, consider $\Lambda^0$ baryon, which usually decays via $$ \Lambda^0 \to p^+ + \pi^- \qquad \text{or} \qquad \Lambda^0 \to n + \pi^0 $$ which would normally release about 37 MeV of energy or 42 MeV of energy, respectively. So if a $\Lambda^0$ found itself inside a heavy nucleus, such that the lowest unoccupied proton & neutron levels were over 40 MeV above the lowest unoccupied $\Lambda^0$ level, would it be stable? If not, why not? As an order of magnitude, 40 MeV isn't all that large compared to the binding energies in heavy nuclei.

I feel like there's something I'm missing here, and my grasp of nuclear physics is not especially deep, so feel free to challenge my frame on this question.

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    $\begingroup$ The binding energy per nucleon in nuclei is below 9MeV, not over 1116-938~178MeV, the substitution energy of a p by a Λ... But what you are speculating about has been speculated about in "strangelet" neutron-star systems and other critical nuclear matter contexts... $\endgroup$ Oct 29, 2020 at 21:25
  • $\begingroup$ I guess my point is that a $\Lambda$ in its ground state in a nucleus would need to "jump up" to the highest unoccupied proton energy level if it decayed into a proton. In other words, it's not the binding energy per nucleon that matters; it's the difference in binding energy of a $\Lambda$ (given no other $\Lambda$s already in the nucleus) vs. the binding energy of a proton (given $Z$ other protons already in the nucleus). That wouldn't be quite the same thing as the average binding energy per nucleon. $\endgroup$ Oct 29, 2020 at 22:48
  • $\begingroup$ Maybe this partly answers cds.cern.ch/record/2726997/files/20-07Jul-31_DanielJohnson.pdf Exotic baryon & meson-like states at LHCb $\endgroup$
    – anna v
    Oct 30, 2020 at 4:55
  • $\begingroup$ Also please read this link en.wikipedia.org/wiki/Exotic_baryon and its links $\endgroup$
    – anna v
    Oct 30, 2020 at 5:26

1 Answer 1


Not an answer, really, but I'd take the opportunity to juxtapose the right numbers, as I believe the numbers provided in the question might be misleading.

The basic, rock-bottom, mass difference to be grappling with is the current mass difference between strange and non strange quarks, $$ m_s-m_d\approx 90MeV, $$ where I have chosen the d for having the higher mass among light quarks and the same electric charge. Note I have ignored the constituent quark mass difference, larger, but involving the strong interactions, and the actual mass differences of the relevant baryons in the same octet, again with the same charge, $$ m_\Lambda-m_n\approx 1116-938\approx 178MeV. $$ The point is you wish to make as wide an allowance for mysterious felicitous strong forces in the nuclear medium that would alter the mass of the bound "Λ", etc... favoring binding. But there is nothing that can possibly affect the mass of the strange quark, set by the "gods" of the SM Yukawa coupling. Ultimately, this s quark must me stabilized against weak decay to u and somehow its >90MeV advantage neutralized by your nuclear medium.

Trading your speculated strange baryon at the bottom of the nuclear potential for a non-strange one at the top, would still require an ineluctable/insurmountable injection of 90MeV of energy in a landslide in the nucleus preserving its stability and integrity. But... considering only the excess kinetic energy in hadronic weak Λ decays and ignoring the pion, with a lot of energy, yields a completely unrealistic energy disadvantage/penalty! The bogey to grapple with in 90MeV.

In effect, you are asking whether a bound p at a high level of the potential can change its identity (inverse-decay) to a Λ fitting at the bottom of the potential, the level difference providing more than 90MeV (plus the small energy of the $e\bar\nu$ released to the world). I am weak in nuclear physics, so I would not opine on how problematic this is... My original comment involved how unlikely it would be for a "Λ" to be captured by a nucleus, the way a neutron would be captured, given the crude small binding energy per nucleon, smaller than 9MeV. Conceivably, elaborate nuclear physics, beyond my depth, could enable capture, but experts should be convinced first.

Of course, people have speculated their heads off on strangelets, chunks of "strange-neutron stars", hypothetical nuclear media where the difference between the s and d relieves Pauli exclusion pressure significantly enough to discourage weak decay and lead to stable binding (cf. Freedman and McLerran, "Quark star phenomenology", PhysRev D17 (1978) 1109); but novel nuclear media or hadrons do not appear to be what you are after here...

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    $\begingroup$ ... I was told by somebody knowledgeable that the "nuclear potential well is only about 40 MeV deep" ... so a Λ decaying to a nucleon would eject it outside the well with great force. $\endgroup$ Oct 30, 2020 at 22:24

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