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?
Answer
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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