Why are two gluons needed for Feynman diagram?

Question

Why do we need two gluons for the decay $$\pi^-+ p\rightarrow\pi^-+n+\pi^+\:\:?$$ If we have always the gluon $$\frac{1}{\sqrt{2}}(r\bar{r}-g\bar{g})$$ it should be possible with only one gluon

instead of 2 gluons

Answer 1

The pion and proton are colourless, so if only one gluon was exchanged, that gluon would also have to be colourless, but there is no colourless gluon so at least two coloured gluons are required for a net colour-neutral gluon exchange. (More formally, one should replace "colourless" with "colour singlet" and "coloured" with "colour octet".)

There is no unique way to write down the colour-octet gluons, but any allowable $r\bar{r}$, $g\bar{g}$, $b\bar{b}$ combination must be a linear combination of the commonly chosen basis states: $$\frac{r\bar{r}-g\bar{g}}{\sqrt{2}}$$ $$\frac{r\bar{r}+g\bar{g}-2b\bar{b}}{\sqrt{6}}$$ There is no way to combine these two combinations to give $r\bar{r}$. The only way to get $r\bar{r}$ is to include some of the colour singlet $(r\bar{r}+g\bar{g}+b\bar{b})/\sqrt{3}$, i.e. $$ r\bar{r} = \frac{1}{\sqrt{2}}\frac{r\bar{r}-g\bar{g}}{\sqrt{2}} +\frac{1}{\sqrt{6}}\frac{r\bar{r}+g\bar{g}-2b\bar{b}}{\sqrt{6}} +\frac{1}{\sqrt{3}}\frac{r\bar{r}+g\bar{g}+b\bar{b}}{\sqrt{3}} $$ But the colour-singlet gluon does not exist, so the $r\bar{r}$ state is also not allowed.

The answers to "Is there such a thing as red - anti red gluon? …" and "Do color-neutral gluons exist?" may also be helpful.

Update: Gluon "Colours" are not Visible Colours (in response to a comment)

One must be careful not to stretch the analogy between visible colours and QCD "colour" too far.
There is obviously no simple way to use visible colours to draw the QCD "colour" flow for the $\lambda_3$ and $\lambda_8$ gluons, but just because one can simply draw the other 6 gluons as colour-anticolour combinations does not mean that is only unique way to do so. In fact, none of the gluons in what I believe is the most common gluon Gell-Mann matrix representation can be represented simply by visible colours:

\begin{array}{lll} \lambda_1 = \frac{g\bar{r}+r\bar{g}}{\sqrt{2}}=\begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} & \lambda_2 = \frac{g\bar{r}-r\bar{g}}{\sqrt{2}}=\begin{pmatrix} 0 & -i & 0 \\ i & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}\\ \lambda_3 = \frac{r\bar{r}-g\bar{g}}{\sqrt{2}}=\begin{pmatrix} 0 & -i & 0 \\ i & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}\\ \lambda_4 = \frac{r\bar{b}+b\bar{r}}{\sqrt{2}}=\begin{pmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 1 & 0 & 0 \end{pmatrix} & \lambda_5 = \frac{r\bar{b}-b\bar{r}}{\sqrt{2}}=\begin{pmatrix} 0 & 0 & -i \\ 0 & 0 & 0 \\ i & 0 & 0 \end{pmatrix}\\ \lambda_6 = \frac{g\bar{b}+b\bar{g}}{\sqrt{2}}=\begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix} & \lambda_7 = \frac{g\bar{b}-b\bar{g}}{\sqrt{2}}=\begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & i \\ 0 & -i & 0 \end{pmatrix}\\ \lambda_8 = \frac{r\bar{r}+g\bar{g}-2b\bar{b}}{\sqrt{6}}=\frac{1}{\sqrt{3}}\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -2 \end{pmatrix} \end{array}

If you naively interpreted $r$, $g$, and $b$ as visible colours in the above representation, all the gluons would be invisible in your diagram. For interaction diagrams, drawing $r\bar{g}$, $g\bar{r}$, $r\bar{b}$, $b\bar{r}$, $b\bar{g}$, and $g\bar{b}$ with their visible colours may be convenient, but when doing actual calculations, the above more symmetric basis is simpler.

Answer 2

You are indulging in impressionistic color tracing by isolated examples, so, then, skipping diagrams. In your unrepresentative model, you might further add the missing diagram where the colors of the $\pi^-$ quark are $g\bar g$ instead of $r\bar r$, (the color singlet wavefunction of it is $\propto r\bar r + g\bar g + b\bar b$) and then your one gluon amplitude would enter with a minus sign exactly cancelling the $r\bar r$ piece one you wrote down.

In practice, an indefinite number of gluons is exchanged to ensure conservation of color. To count gluons, where it matters, you might go back to the simpler decay of the J/ψ to hadrons, where you may explicitly see your argument failing the way you framed it. Recall the trace of each of all 8 color generators individually vanishes.