It is my understanding (which may be flawed) that protons and neutrons are stable because the 3 (R, G, and B) quarks form a "white" color singlet. Wouldn't 6 quarks or even 9 quarks create a white singlet? What about RGBGR?

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    $\begingroup$ A free neutron is actually not stable. It is slightly heavier than a free proton and can therefore beta decay into one. Lifetimes become even shorter as the mass increases which at some level explains why it's a big accomplishment every time a new multi-quark particle is observed. But you're right that they are perfectly allowed by confinement. $\endgroup$ Aug 27, 2021 at 2:03
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    $\begingroup$ You can look up Hexaquark on Wikipedia. $\endgroup$
    – Lorenzo B.
    Aug 27, 2021 at 13:35
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    $\begingroup$ Free neutrons aren't just unstable in the "they decay after a million years" way either, the lifetime is about 15 minutes $\endgroup$
    – llama
    Aug 27, 2021 at 15:11
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    $\begingroup$ The "color rule" doesn't convey stability, it merely conveys the ability to (independently) exist at all. $\endgroup$ Aug 27, 2021 at 16:16

2 Answers 2


The deuteron (or $^{2}$H nucleus) is a color-neutral bound state of three $u$ quarks and three $d$ quarks. Because six quarks can combine to make a color-neutral state, this six-quark state is possible. However, the quarks inside are not arranged in anything like a symmetric state. They are tightly bound into two substructures which are already themselves color neutral, one with composition $uud$ and the other $udd$; these are just the proton and neutron constituents of the deuterium nucleus.

There are also nine-quark structures, the nuclei of $^{3}$H and $^{3}$He. The former is unstable, but its decay lifetime is a matter of years, not the yoctoseconds typical of really unstable strong states. The latter is stable outright. As in the deuteron, these states are almost entirely combinations of smaller color-neutral nucleons.

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    $\begingroup$ According to this answer, any nucleus is potentially a multi-quark state as per OPs question. I don’t think OP was asking in the context of multi nucleon states. $\endgroup$
    – joseph h
    Aug 28, 2021 at 0:55

In principle, a hadron with any number of quarks can be formed provided that the overall color is neutral. However, hadrons with more than three quarks (observed in particle accelerators) are unstable$^1$ and decay rapidly.

Tetraquarks and pentaquarks have been observed in high energy collisions, but rapidly decay. A particle of the form $\mid RGBGR\rangle$ does not appear to be viable since the overall color is not neutral.

However, something like $\mid RGBG\bar G\rangle$ where $\bar G$ means anti-green, could be a possible quark "molecule" or pentaquark, since we have $$\mid\underbrace{RGB}_{neutral} \ \underbrace{G\bar G}_{neutral}\rangle$$ or four quarks and one antiquark bound together.

As per the link, a tetraquark, or "meson molecule" (two mesons) the combination $$\mid\underbrace{q\bar q}_{neutral} +\underbrace{Q\bar Q}_{neutral}\rangle$$ is also possible, provided we have color-anti-color, or net color neutral (the bar here means anti-color and not antiparticle).

There has in fact been experimental evidence at CERN for a six-quark state, “dibaryon” or Hexaquark. It would appear that provided we have enough energy, the synthesis of a fleeting hadron with any number of quarks is possible, once again, only if the net color of the combination is neutral.

As for a "nine-quark state", it could possibly be a "tribaryon" or something of the form $$\mid\underbrace{RGB}_{neutral} \ \underbrace{RGB}_{neutral}\ \underbrace{RGB}_{neutral}\rangle$$ but you would need a lot of energy to synthesize this particle, and I do not think anyone has observed a particle with greater quark number than a dibaryon (six).

$^1$ This is because there are more stable lower mass hadron states that they can decay into.


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