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We know color-neutral bound systems of quarks exist in the form of hadrons, we suspect color-neutral bound systems of gluons exist in the form of glueballs, we have a candidate particle which may be an hybrid meson bound to a gluon. My question is if there exists any configuration consisting of a single quark and an arbitrary number of gluons that results in a color-neutral system that could exist on its own, even if it would be unstable or would mix with other particle states. If the answer is no, why?

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  • $\begingroup$ I specified non-virtual gluons to clarify that I mean the real thing that forms glueballs and such as opposed to the cloud of virtual gluons every hadron has mediating the strong force and holding quarks together. And unfortunately I don't know what a Wilson line is so that doesn't really help my understanding of the situation. $\endgroup$ Commented Aug 10, 2021 at 6:32
  • $\begingroup$ or basically a "sticky quark" :) (i.e. sticky because it's covered in glue) $\endgroup$ Commented Aug 10, 2021 at 6:43
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    $\begingroup$ @The_Sympathizer: Q: What's the definition of a tachyon? A: A gluon that's not completely dry. $\endgroup$ Commented Aug 10, 2021 at 14:09

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It boils down to whether one can make a color neutral particle with a quark and some colored gluons.

The rationale for the concept of color can be highlighted with the case of the omega-minus, a baryon composed of three strange quarks. Since quarks are fermions with spin 1/2, they must obey the Pauli exclusion principle and cannot exist in identical states. So with three strange quarks, the property which distinguishes them must be capable of at least three distinct values.

omega-

See also the evidence for three colors.

Now the gluons do not carry a single color

gluon

Gluon interactions are often represented by a Feynman diagram. Note that the gluon generates a color change for the quarks. The gluons are in fact considered to be bi-colored, carrying a unit of color and a unit of anti-color as suggested in the diagram

I do not think that the algebra of the existing model can model what you want. If it is a red quark it might be neutralized by a gluon carrying antired, but there would be left over the other color carried by the gluon. This would be true for combinations of gluons too, there would be left over color or anticolor, so color neutrality cannot be attained within QCD.

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    $\begingroup$ @JDługosz the meson would be color neutral, the gluon cannot be $\endgroup$
    – anna v
    Commented Aug 10, 2021 at 14:59
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    $\begingroup$ @JDługosz The Qantiblue is color, not neutral, which cannot be free, it has to be within a nucleon/hadron. The question is about a binding of a quark to a gluon, not about how mesons can be formed. $\endgroup$
    – anna v
    Commented Aug 10, 2021 at 15:45
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    $\begingroup$ @annav see my original comment. $\endgroup$
    – JDługosz
    Commented Aug 10, 2021 at 16:17
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    $\begingroup$ Gluons are not color-anti-color but more complex fractions. If you insist they are I will ask you to enumerate all 8 of them and you will get into trouble. I got the real table from somewhere and the math forms were too much for me, so I can't eliminate the possibility of piling up enough gluons to cancel out the color. $\endgroup$
    – Joshua
    Commented Aug 10, 2021 at 21:43
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    $\begingroup$ @annav Let me start over... you ended your post by saying you have a single color or anticolor left over. I pointed out that the obvious solution to that of adding another quark to cancel that remaining color would then produce a meson. $\endgroup$
    – JDługosz
    Commented Aug 10, 2021 at 21:52
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You can prove that this is impossible using Young tableaux. The trivial (color-neutral) representation 1 must have 3 boxes in each vertical column. (You can prove this to yourself using the hook-length formula.) A single quark in 3 contributes one box. A single gluon in 8 adds 3 boxes. Thus, no matter how you arrange the boxes from additional gluons, you can never form the 1 representation.

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    $\begingroup$ Hey! An explanation even I can understand! (+1) And $3\otimes 8\otimes 8$ still has no $1$-component etc. No matter how many $8$:s you tensor it with, all the weights stay in the wrong coset of the root lattice inside the weight lattice (translating this into the language used by algebraists). $\endgroup$ Commented Aug 11, 2021 at 12:59
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No. In general, in any (exotic) hadron made of valence quarks, antiquarks and gluons, the number of quarks minus the number of antiquarks in absolute value must be divisible by 3. Then, add any number of gluons. That's the recipe to form possible color singlets. See the introduction to:

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