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For a single quark, the colour vectors are given by $$r=\begin{pmatrix} 1 \\ 0 \\0 \end{pmatrix} \qquad g=\begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} \qquad b=\begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}$$

and the $F_3$ and $F_8$ colour charge operators are $$F_3 = \frac{1}{2} \begin{pmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 0 \end{pmatrix} \\ F_8= \frac{1}{2\sqrt{3}} \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -2 \end{pmatrix}.$$

The color charges $F_3$ and $F_8$ for the color vectors are given by this table:

enter image description here

For example, acting $F_3$ on vectors $r, g ,b$ give

$$F_3r=\frac{1}{2} r, \quad F_3 g = -\frac{1}{2} g, \quad F_3 b = 0$$

My question is what would be the colour vectors $\bar{r} , \bar{g}, \bar{b}$?

From the above table, I can see that we need to have $$F_3 \bar{r} = - \frac{1}{2} \bar{r}, \ \ F_3 \bar{g} = \frac{1}{2} \bar{g} \ \ F_3 \bar{b} = 0.$$

This gives me $$\bar{r} = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} \qquad \bar{g} = \begin{pmatrix} 1 \\ 0 \\0\end{pmatrix} \qquad \bar{b} = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}$$

But these are wrong since the $F_8$ charges are not satisfied: $$F_8 \bar{r} = \frac{1}{2\sqrt{3}} \bar{r} \neq -\frac{1}{2\sqrt{3}} \bar{r}$$

$$F_8 \bar{g} = \frac{1}{2\sqrt{3}} \bar{g} \neq -\frac{1}{2\sqrt{3}} \bar{g}$$

$$F_8 \bar{b} = -\frac{1}{\sqrt{3}} \bar{b} \neq \frac{1}{\sqrt{3}} \bar{b}$$

What am I doing wrong?

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    $\begingroup$ SU(3) has nothing to do with spin. Elements of representation spaces of SU(3) are not called spinors. $\endgroup$
    – Ghoster
    Commented Jul 2, 2023 at 19:08
  • $\begingroup$ See Dual Representation. The operators are different on different representations, so the dual space operators are given by the negative transpose of the defining representations operators. This is what cosmos answer is doing, just in the standard way where we treat dual vectors as row vectors. $\endgroup$
    – Craig
    Commented Jul 2, 2023 at 21:35

1 Answer 1

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From the above table, I can see that we need to have $F_3 \bar{r} = - \frac{1}{2} \bar{r}, \ \ F_3 \bar{g} = \frac{1}{2} \bar{g} \ \ F_3 \bar{b} = 0.$ ... What am I doing wrong?

No. That's what you are doing wrong. Color transformations act on the transposed antiquarks from the right, and complex conjugated; since these Hermitian color generators are real, $$ q\to e^{i(\theta_3 F_3+\phi_8 F_8)}q,\\ \bar{q}\to \bar{q} e^{-i(\theta_3 F_3+\phi_8 F_8)} .$$ Plugging in this "minus transpose" feature, you obtain the table values.

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