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What makes the two 'color-neutral' gluons $(r\bar r−b\bar b)/\sqrt2$ and $(r\bar r+b\bar b −2g\bar g )/\sqrt6$ different from the pure $r\bar r +b\bar b +g\bar g $ ?

Why don't they result in long range (photon-like) interactions?

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    $\begingroup$ I am not sure I fully understand the question. Any QCD bound state has to be color neutral. And it has a non vanishing mass set by $\Lambda_{QCD}$ (except for pions that are protected by the Goldstone theorem and their mass is smaller as controlled also by the masses of the light quarks). I mean, QCD has a mass gap. $\endgroup$
    – TwoBs
    Jun 4 '14 at 17:27
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    $\begingroup$ @Melquíades Since both questions are from the same user, I suspect that user believes they are not equivalent. $\endgroup$
    – rob
    Jun 4 '14 at 21:44
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    $\begingroup$ @PeterShor The above assignments are riding on one individual gluon. Gluons are always colored. A hadron is color neutral because it can be in a state where all three colors are present. Think of the Hydrogen atom. It is neutral because the charge - neutralizes the charge + of the proton. With three charges neutrality can come only with all three color combinations with their SU(3) group structure to ensure that this can only happen on more than one particle combination. The one particle, gluon, cannot be neutral (because of what we observe in nature of course) because of the SU(3) algebra $\endgroup$
    – anna v
    Jun 5 '14 at 4:00
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    $\begingroup$ @annav Mesons may be color-neutral while their quark content contains one color and one anticolor. $\endgroup$
    – rob
    Jun 5 '14 at 6:02
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There is no fundamental difference between the gluons $(r\bar{r}-b\bar{b})/\sqrt{2}$ and $(r\bar{b} + b\bar{r})/\sqrt{2}$. The first one is represented by the matrix $$ Z = \frac{1}{\sqrt{2}}\left(\begin{array}{rrr} 1&0&0 \\ 0& -1 & 0 \\ 0 & 0 & 0\end{array}\right)$$ and the second by the matrix $$ X = \frac{1}{\sqrt{2}}\left(\begin{array}{rrr} 0&1&0 \\ 1&0&0 \\ 0 & 0 & 0\end{array}\right).$$ However, these two matrices are related by the change of basis $$ H = \left(\begin{array}{rrr}1/\sqrt{2}&1/\sqrt{2}&0 \\ 1/\sqrt{2}& -1/\sqrt{2} & 0 \\ 0 & 0 & 1\end{array}\right). $$ It is easy to check this by multiplying matrices and seeing that $HZH^\dagger = X$.

Thus, if you call $(r\bar{r}-b\bar{b})/\sqrt{2}$ "color-neutral" and $(r\bar{b} + b\bar{r})/\sqrt{2}$ "non-color-neutral", it is clear that "color-neutral" is not a property that is invariant under change of basis, and thus is not a meaningful property in quantum chromodynamics.

Actually, neither of these gluons is color-neutral.

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  • $\begingroup$ Peter Shor: "neither of these gluons is color-neutral." -- Could you please write down an explicit "charge operator", e.g. some particular matrix for which the already given gluon matrices $Z$ and $X$ are both eigenstates, and each with some particular non-zero eigenvalue representing its "color-charge"? (I'm not really sure if that's easily done, or easily found in some suitable literature, or whether my request is even sensible. But at least I know enough matrix multiplication to check the operator matrix myself if you'd provide it; and to me that would make your answer especially useful). $\endgroup$
    – user12262
    Jun 5 '14 at 21:58
  • $\begingroup$ Peter, you are right. 'Color-neural' does not exist. Just as there is no 'spin-neutral' state for spin 1/2 particles. $\endgroup$
    – NoEscape
    Jun 23 '14 at 9:39
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One could state your problem as " why are there only eight gluons accepted as physically existing from the SU3 symmetry . This link gives the answer simply, and it says that if the ninth all color neutral gluon existed, then as you guess, it would act similarly to the photon with interactions between hadrons that have not been observed.

So the answer is: so that the SU(3) color model should fit the data , i.e. strong interactions between hadrons are confined because we have not observed colorless gluons in the interactions between hadrons.

In the matrix analysis the link goes on to show this :

But, no matter how we take complex linear combinations of trace-zero hermitian matrices, we cannot get

1 0 0

0 0 0

0 0 0

since this has trace 1. (I.e., the sum of the diagonal entries is 1: that's what the trace means, the sum of the diagonal entries.) So we cannot really get red anti-red. The closest we can get are things like

1 0 0

0 -1 0

0 0 0

or

1 0 0

0 0 0

0 0 -1

or

0 0 0

0 1 0

0 0 -1

But note, these three are not linearly independent: any one of them is a linear combination of the other two. So we can get stuff like

(red anti-red) − (blue anti-blue)

and so on, but not 3 linearly independent things of this sort, only 2: one less than you might expect.

It is thus fortunate that SU(3) color can be used to described what has been observed with hadrons, i.e. no neutral photon like gluons

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  • $\begingroup$ No, sorry, you misunderstood the question. I was asking about the two remaining 'neutral' gluons... $\endgroup$
    – NoEscape
    Jun 4 '14 at 18:30
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    $\begingroup$ for each gluon there are red anti red, green anti green, blue antiblue possible color combinations in SU(3) there do not exist neutral individual gluons as far as I know. All three color combinations carried by one gluon would be neutral. When only two are tagged on a gluon bb-rr-, then the green antigreen is missing. When you have two charges + and - they cancel.When you have three all three should be there to cancel $\endgroup$
    – anna v
    Jun 4 '14 at 19:24
  • $\begingroup$ The question seems to be how do we know there are eight (instead of six), not how do we know there are eight (instead of nine). $\endgroup$ Jun 5 '14 at 21:53

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