Assume a sexaquark contains 3 up and 3 down quarks. What is the difference between this and a deuterium nucleus containing a proton bound to a neutron?

Is there any difference?

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    $\begingroup$ Could be considered as a duplicate of physics.stackexchange.com/q/310820 (and I am partial to my answer there, of course), and even if not it and the links therein are worthy background reading. Also related: physics.stackexchange.com/q/171037 physics.stackexchange.com/q/11487 $\endgroup$ Mar 20 '18 at 19:42
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    $\begingroup$ I would say that a deuteron is the ground state (and only bound state) of this sexaquark system. In this state, there are strong three-body correlations that make it natural for us to describe the system as a composite of two three-quark subcomposite systems. $\endgroup$
    – user4552
    Mar 20 '18 at 20:49
  • $\begingroup$ What's the difference between three helium-4 nuclei and a carbon-12 nucleus? Both contain 6 protons and 6 neutrons. The difference is in their distribution. The same applies here. $\endgroup$ Jul 2 '18 at 22:09
  • $\begingroup$ @BenCrowell if we hit them with the right gamma frequency, can we excite deuterons into becoming sexaquarks? If not, why not? $\endgroup$
    – lurscher
    Jul 3 '18 at 15:39

My advisor posed me the same riddle ages ago, and it drove me stark raving mad.

To expand on Ben Crowell's comment ...

Deuterons have exceptionally weak binding energies, out of line with heavier nuclei, so they may be atypical. Alpha particles seem more representative of differences between nuclear matter and quark matter. A naive shell model says that the 1s shell can accommodate four nucleons or twelve light quarks, so no difference there.

In nuclear matter at normal density, quarks somehow clump in color-singlet groups of three. A naive shell model in the Hartree-Fock tradition (described below) does not predict the observed correlations. You would have to include perturbative admixtures of two- and many-body excited states. Unfortunately, low-energy QCD is so poorly understood that you cannot expect good answers.

(The Hartree-Fock approximation uses a Slater determinant of single-particle wave functions as a trial wavefunction. The actual wavefunction would be a sum of many such determinants.)

At somewhat higher densities, possibly achieved in neutron stars, the nucleons would overlap in space and lose their distinct identities, so you could best picture the nuclear matter as quark matter. QCD forces would get weaker as well, thanks to higher Fermi momenta, so perturbation theory would be more accurate.


There is a difference. The sexaquark (uuddss) is actually possibly a deeply bound state, possibly even stable. I would refer you to Glennys Farrar's paper on the 'Stable Sexaquark' for an in depth explanation.

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    $\begingroup$ Uhm ... you seem to be badly off track. A sexaquark is note required apriori to have any particular valence content, and comparing a state with valence content $uuddss$ to a deterium (which has valence content $uuuddd$) would seem non-sensical from the get-go. $\endgroup$ Jul 3 '18 at 2:19

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