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If I'm correct a quark can change color by emitting a gluon. For example a blue up quark $u_b$ can change into a red up quark by emitting a gluon: $$u_b \longrightarrow u_r + g_{b\overline{r}}$$ (Here, the subscript indicates color and $\overline{r}$ means anti-red). This is needed to keep the color balance (left hand side: $b$, right hand side $r+b+\overline{r}=b$).

My question is then, do color-neutral gluons exist? Eg a gluon that is blue-anti-blue? If it would, it could be created by any quark then: $$u_b \longrightarrow u_b + g_{b\overline{b}}$$

I'm learning about the Standard Model in school, but the text isn't always that clear.

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up vote 6 down vote accepted

Color-neutral gluons that have the component blue-antiblue do exist, much like red-antired and green-antigreen. However, the sum of these three possible kinds of gluons is unphysical, so there are only two "diagonal" types of gluons. None of these two types of gluons are "genuinely color-blind" or "completely color-neutral".

This is more manifest if you realize that the color dependence of the gluon field may be written as a traceless $3\times 3$ Hermitian matrix. It is traceless because the gauge group is $SU(3)$ rather than $U(3)$ whose dimension is 8 rather than 9. (There are 3 complex entries strictly above the diagonal, which are copied in the complex conjugate way beneath the diagonal, plus 2 or 3 real entries on the diagonal, depending on whether we require the trace to vanish.)

Completely color-neutral gluons, if they were added, would be proportional to the identity matrix and they would couple to all three colors of the quarks equally. In other words, the interactions mediated by such gluons would only depend on the baryon number of the quarks. Experimentally, this interaction doesn't exist. In beyond-the-Standard-Model physics, one may try to extend $SU(3)$ to $U(3)$ in this way (this is very common in braneworld models) but because no new baryon-charge long-distance interaction is seen, the $U(1)$ in the $U(3)$ has to be spontaneously broken at a pretty high energy scale.

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@Lubosh: Your statement "This is more manifest if..." reminds me a little of when physics text books often say things like 'this easy proof is left to the reader' when in fact the proof would only be easy if the reader happened to be a professional in that particular subject. I've been meaning to ask about the physical significance of diagonal and off-diagonal matrix entries for some time. My reason for not doing so is that I feel it would be a bit too mathsy for physics.SE and visa versa. I would therefore be hugely greatful if you could elaborate on your second paragraph a little. –  qftme Jun 13 '11 at 17:12
    
ok, so a $u_b$ can turn into a $u_{\overline{b}}$ by emitting a $g_{b\overline{b}}$, with no problem at all? –  romeovs Jun 13 '11 at 21:51
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@romeovs Nope, a $u_b$ stays a $u_b$ by radiating a $g_{b\bar{b}}$, just like how an electron stays an electron by radiating a photon. –  QuantumDot Aug 1 '12 at 4:34
    
Dear @qftme, the middle paragraph just says that there's no "color neutral" generator in $SU(3)$, this could mean "a generator in the center" i.e. one that commutes with all generators of $SU(3)$. The group is simple, we say. The other statements about the dimension 8 or 9 are even simpler. The paragraph is supposed to be the full-fledged proof. Nothing is missing there! To understand it, one needs to know some linear algebra. But if you don't know e.g. what SU(3) and U(3) is, you can't understand any proper answer about "color non-blindness" of QCD. –  Luboš Motl Feb 27 '13 at 6:41
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