I List of unstable and unknown hexaquarks

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One hexaquark is stable and well known.
All the others must decay weakly.
But of all the other possible hexaquarks, how many are bound, how many are unbound to strong interaction but decay by elastic collision, how many undergo exothermic quark exchange and how many are unknown?
 
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... the deuteron?
Yes.
Why? Why not strongly?
I expressly considered two strong decay mechanisms. Unbound to elastic collision (confirmed to be the case with dineutron and diproton) and unstable to exothermic quark exchange.
What I was saying is that a hexaquark other than deuteron that is stable to strong decay should still decay weakly, because unlike neutron, the other baryons (total of, I think, 24 baryons that only decay weakly, inclusive of neutron) have so large weak decay energies that they are hard to stabilize by strong binding.
 

ohwilleke

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... the deuteron?
A "true" hexaquark is a hadron with no substructure with all quarks bound directly to each other via gluons. This is theoretically much easier to realize in forms that according to theory should be extremely unstable than in a stable form.

A deuteron, in contrast, is not a "true" hexaquark, it is a "hadron molecule" made up of two hadrons (a proton and a neutron), with the quarks within the proton and neutron, respectively, bound directly to each other via gluons, and the nucleons themselves, bound to each other through nuclear binding forces mediated by mesons, mostly pions and additionally far fewer kaons and other mesons (often virtual), in a "spillover" of the strong force between quarks in a hadron. (The term "hadron molecule" is something of a "false friend" in terms of terminology, as it is more analogous to an atomic nucleus than what chemists call a molecule.)

We can observe in a deuteron interactions indicative of clustering within its nucleus of a proton and a neutron that are separate structures, rather than of one unitary composite particle made of six quarks.

One of the challenges of QCD experimenters is two distinguish between "true" tetraquarks, pentaquarks, and hexaquarks that are truly quarks bound directly to each other by gluons, and "exotic" hadron "molecules" that are bound to each other either in the manner of protons and neutrons being bound to each other in a nucleus, or via electromagnetism in the manner of the forces that bind atoms to each other.

Distinguishing the three different ways that system of more than three quarks can be bound to each other is subtle and tricky because both "true" tetra-, penta-, hexa-quarks, and hadron molecules bound by mesons to each other with the same valence quark content, are both (1) very short lived, (2) few in number because they have to be created artificially in collider type contexts, and (3) have similar decay products but with different frequencies, polarities and kinematics.

It also doesn't help that the math necessary to predict what the decay products of each scenario looks like is cumbersome and never accurate to much more than the percent level since the underlying physical constants like the strong force coupling constant isn't know very precisely.

But of all the other possible hexaquarks, how many are bound, how many are unbound to strong interaction but decay by elastic collision, how many undergo exothermic quark exchange and how many are unknown?
This is an area of active investigation that is in its infancy. Almost all possible cases are unknown. We are still struggling to determine the underlying structure and nature of lots of X, Y, Z resonances, with most such resonances having multiple plausible explanations. Some of them may be hexaquarks, most of them are probably not. Lots of hexaquark resonances that are naively possible given what we known about QCD at a qualitative level, have never been observed.

We are very confident that in high energy physics interactions simple two valence quark mesons and three valence quark baryons are vastly more likely to be produced than hadrons with more valence quarks than three, although honestly, we're not entirely sure why that is the case beyond some vague heuristic arguments.

We are less confident, but still quite confident, that there are no "true" tetraquarks, pentaquarks or hexaquarks that are stable in an unbound state. The reasoning for this is a mix of the state of what we have studied and observed and our theoretical understanding of the Standard Model. An outline of the reasons for this runs roughly as follows.

If there were, there would be lots of them (because stable particles accumulate). And, we haven't observed them. So, unless they only interact very weakly with everything else (in which case they could make up dark matter), they probably don't exist.

There is certainly no way would could miss large numbers of charged stable "true" hexaquarks given experiments that have been conducted to date, which are highly sensitive to even tiny electrically charged particles, so they would have to be electromagnetically neutral.

This mass would have to show up as missing traverse momentum in collider experiment, since that is what stable exotic, neutral hadrons with few interactions with other matter should look like in a collider because they wouldn't leave decay products and would leave electromagnetic traces within distance from the collision to the collider's instruments which are located by design to catch only quickly decaying hadrons and leptons. And, we haven't seen that in any statistically significant frequency yet.

This mass range isn't one particularly favored for dark matter or for being unobservable in dark matter detection experiments.

A dineutron made up to two neutrons bound by nuclear forces mediated by mesons was first definitively observed in 2012 (see also here) and, as predicted, is was much less stable than a free neutron, decaying in about 10^-22 seconds rather than about 881 seconds for a free neutron. It seems unlikely that a "true" hexaquark with the same valence quarks would be more stable, although one would have to do the calculation (or prove the hypothesis by definitively observing an unstable one of that type, or definitely disprove that hypothesis by definitively observing one was that was stable or metastable) to be sure.

Also, there are no known substances in which strange, charm, or bottom quarks hadrons are stable (due to weak force decay), and there are no cases observed in which top quarks even hadronize (they decay via the weak force faster than the strong force can form a hadron made of them in the vast majority of cases, even though it is theoretically possible in an extreme rare case for an extremely brief time).

So, for a stable hadron with more than three valence quarks, we'd need one make up entirely of up quarks and down quarks as valence quarks that are, in sum, electromagnetically neutral, which confines them to a fairy narrow mass range of approximately 2 GeV/c^2 plus or minus a fraction of a GeV, and that is a mass range that existing collider experiments have been able to study quite exhaustively.

The other way that you might get a stable "true" hexaquark is in a bound structure that is stable in the same way that a bound neutron is stable, by being bound into a larger structure. This would probably take the interior of a neutron star to create, in which gravitational attraction and Fermi contact force pressure would make up for not quite sufficient standard model binding forces. But, of course, we can't directly observe the interiors of neutron stars and can't externally observe them with sufficient precision to notice the difference yet (although we are improving on that front). Dineutrons (but not "true" hexaquarks) have also been theoretically predicted to exist in at least one isotype of Terbium atoms, although this has not been observed experimentally.

Also, since anything that contains a quark and anti-quark of the same type is prone to annihilation in a short time frame (all forms of quarkonium have short mean lifetimes), that combination of quarks shouldn't be possible in a stable hadron with more than four valence quarks.

Considering all of those limitations, the spectrum of stable "true" hexaquarks, which are stable in an unbound state, if there are any, should be confined to structures with exactly four up quarks and exactly two down quarks (or with exactly four anti-up quarks and exactly two anti-down quarks). This isn't exactly one possibility and it anti-particle, because there are different combinations of spin states that are possible (there are fifteen possible distinguishable spin states for these hexaquarks matched by fifteen possible anti-hexaquark states by my calculation in the ground state, plus theoretically infinitely many unstable excited states). But, the possibilities are still very limited and the mass range of the possible stable "true" hexaquarks is very narrow making it a target that it pretty easy to rule out experimentally.

It is also possible (and indeed likely) that if we run the numbers that this qualitatively possible set of stable "true" hexaquarks aren't actually stable in fact once one does the QCD calculations necessary to determine its properties.

Why is it likely that an allowed stable "true" hexaquark by other measures would not be stable?

A free neutron, which has proportionately the same mix of quarks, is not stable (and is only stable when bound in an atomic nucleus dynamically), although it comes close, and generally speaking, adding more valence quarks to the mix appears to make a hadron less stable rather than more stable.

In particular, a neutron doesn't have to have both of its down quarks in the same spin state, while the hypothetical hexaquark that might be stable would have to have at least some down quarks in the same spin state (two, three or four of them); the up quarks could have mixed spin states which would theoretically be more stable, or the same spin states, which would theoretically be less stable, and hadrons with identical quarks in the same spin state tend to be less stable than those that do not have that property.

Simple omega baryons with three valence strange quarks some of which are in same spin state, are an exception to this rule and are more stable than some other particles with mixed spin states with a mean lifetime of about 8*10^-11 seconds, because they can only decay via the weak force which happens more slowly than decays via the strong force, but they are still unstable. The delta +2 baryon and delta -1 baryon made up of three up quarks and three down quarks respectively, are also extremely unstable with mean lifetimes on the order of 10^-24 seconds.

Of the fifteen possible spin states of a hypothetical four down quark and two up quark spin states, the most stable ought to be the one with two down quarks in each possible spin state and one quark quark in each possible spin state, while the other fourteen possible configurations ought to be at least somewhat less stable by comparison. And anti-hexaquarks ought to be just as stable as their ordinary matter counterparts, so you'd only need to examine one possible maximally stable undetected hexaquark state (which should also be the lightest of the fifteen possible hexaquark states) to see if a stable "true" hexaquark that has not yet been detected was possible theoretically within the range of precision in QCD calculations at this point in time.

Another theoretical consideration is that there is no obvious reason that a hexaquark, of the only kind that might be stable considering the factors addressed above, couldn't decay into a pair of neutrons, or into a proton and a delta -1 baryon (made of three valence down quarks). So, the hexaquark would have to be lighter than the possible decay products with the same quark content to be stable. But, generally, the more "stuff" a hadron had to overcome to hold itself together, the more massive it is relative to its quark content. So, it seems unlikely that this structure would be lighter than its possible decay products, and hence it is more likely that it would be unstable.

Furthermore, this hexaquark would have to have some means of balancing out the tendency of down quarks to decay to up quarks that is shown in free neutrons, as in the case of a bound neutron, something that is not obviously present in this structure.

In sum, there are very good reasons to expect that there are no "true" hexaquarks that are stable in an unbound state, although there might be extraordinary bound states, such as those arising in the cores of neutron stars, where they might be able to exist in stable form.
 
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Checking just the combinations that have weirdness but no charm nor beauty...
Weirdless 0 - just the deuteron, charge +1.
Weirdness -1... consider the options.
Charge +3 would have to include Δ++. So strong decay.
Charge +2 could include Δ++ or be p+Σ+.
Charge +1 could be p+Λ or n+Σ+. But here exothermic quark exchange is possible: n+Σ+ would be 939+1189=2128 MeV, while p+Λ is just 938+1115=2053 MeV.
Charge 0 could be p+Σ- or n+Λ. Again exothermic exchange is possible: p+Σ- is 938+1197=2135 MeV, while n+Λ is just 939+1115=2054 MeV.
Charge -1 could be n+Σ-
Charge -2 would have to include Δ-, so strong decay.
The combinations for weirdness -1 are thus:
p+Σ+; p+Λ; n+Λ; n+Σ-.
Are any of those bound?
 

ohwilleke

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Checking just the combinations that have weirdness but no charm nor beauty... Weirdless 0 - just the deuteron, charge +1.
Weirdness -1... consider the options.
As a slight pedantic point, the customary term is "strangeness" (parallel to the "strange quark") and not "weirdness", a term that I have never seen used to refer to a quantum number of a hadron. "Weirdness" is when your physics professor shows up to class in a Halloween costume, or delivers a lecture in Sumerian or pig Latin, or forgets to wear pants and has unicorn underwear on.

Perhaps you are just joking, and if so, I apologize for not realizing that the term is a joke. It is hard to figure out context and tone in short written communications on the Internet.
 

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