A teacher of mine told me once that there were no ninth gluon because such a one should be white and interact infinitely far, and no one has been observed. Is there also a theoretical reason?

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    $\begingroup$ A preprint with ninth gluon in the title: arxiv.org/abs/hep-ph/9606317 $\endgroup$ – Qmechanic Jan 25 '12 at 22:13
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    $\begingroup$ Interesting - it postulates that the color symmetry is broken $U(3)$ instead of $SU(3)$ and that the broken symmetry gives an extremely massive singlet gluon. $\endgroup$ – Mark Beadles Jan 25 '12 at 22:23
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    $\begingroup$ See also this question and its answer on "colorless" gluons. $\endgroup$ – Mark Beadles Jan 26 '12 at 2:17
  • $\begingroup$ @MarkBeadles Thanks for the flag, but they are not similar enough to be claimed duplicates. $\endgroup$ – user68 Jan 26 '12 at 15:05

There is one color singlet, which is colorless and, as your teacher said, does not correspond to any gluon interaction. If it did it would interact with other color singlet states.

$$(r\bar{r}+b\bar{b}+g\bar{g})/\sqrt{3} $$

And then the eight remaining independent color states corresponding to the eight colors of gluons. One way of representing these as linearly independent states is:

$$(r\bar{b}+b\bar{r})/\sqrt{2}$$ $$-i(r\bar{b}-b\bar{r})/\sqrt{2}$$

$$(r\bar{g}+g\bar{r})/\sqrt{2}$$ $$-i(r\bar{g}-g\bar{r})/\sqrt{2}$$

$$(b\bar{g}+g\bar{b})/\sqrt{2}$$ $$-i(b\bar{g}-g\bar{b})/\sqrt{2}$$

$$(r\bar{r}-b\bar{b})/\sqrt{2}$$ $$(r\bar{r}+b\bar{b}-2g\bar{g})/\sqrt{6} $$

The states are not, as might be assumed, just blue-antired, red-antiblue, etc.


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