CAF123 said:
'Show that 2 mesons in the colour singlet state ##\frac{1}{\sqrt{3}} (r \bar r + g \bar g + b \bar b)## experience a potential V=-4/3 1/r'
Not between two mesons. That is the
short range potential between
quark and
anti-quark in a (colour singlet)
meson, i.e., the QCD (colour) analogue of the attractive QED (electrostatic) potential. It is this potential that makes mesons the mesons we
observe.
At short distances QCD is
asymptotically free. This means that the q-q interaction gets
weaker at short inter-quark distances, and at r \sim 0.1 \mbox{fm} the lowest order (
one-gluon exchange) diagrams
dominate. So, it is not unreasonable to expect a Coulomb-like potential analogous to that arising from
one-photon exchange in QED. Indeed, we can show that \langle q\bar{q}|V(r) |q\bar{q}\rangle_{\mbox{singlet}} = -\frac{4}{3} \frac{\alpha_{s}}{r} , \ \ \ r \sim 0.1 \mbox{fm} . The factor 4/3 arises from summing the colour factors of all possible
lowest-order q_{i}\bar{q}_{k} \to q_{j}\bar{q}_{l}
partonic processes in the
colour singlet meson. The colour factor for such partoinc processes arises when we calculate the Feynman amplitude \mathcal{M}\left( q_{i}\bar{q}_{k} \to q_{j}\bar{q}_{l} \right). It is defined by C \left( q_{i}\bar{q}_{k} \to q_{j}\bar{q}_{l} \right) = \left( \frac{\lambda^{a}}{2}\right)_{ji} \left( \frac{\lambda^{a}}{2}\right)_{kl} . Now, group theory comes to the rescue because of the following identity \left( \frac{\lambda^{a}}{2}\right)_{ji} \left( \frac{\lambda^{a}}{2}\right)_{kl} = \frac{1}{2} \left( \delta_{jl} \delta_{ik} - \frac{1}{3}\delta_{ji}\delta_{kl} \right) . So, you find the following colour factors C(x\bar{x} \to x\bar{x}) = \frac{1}{3}, \ \ \ x = r, g, b , C(x\bar{x} \to y\bar{y}) = \frac{1}{2} , \ \ \ x \neq y , C(x\bar{y} \to x\bar{y}) = - \frac{1}{6} \ \ \ x \neq y .
So, “inside” the colour singlet meson, we have
3 partonic processes (diagrams) of the form x\bar{x} \to x\bar{x} and
6 diagrams of the form x\bar{x} \to y\bar{y}. Thus C(q\bar{q} \to q\bar{q})_{\mbox{singlet}} = (\frac{1}{\sqrt{3}})^{2} \left( 3 \times (1/3) + 6 \times (1/2) \right) = \frac{4}{3} .
In similar but more complicated way, you can calculate the colour factor for two quarks exchanging a gluon within a colour-singlet Baryon qqq. However, one can deduce the value C(qqq)_{[1]} = 2/3 by simple group theory arguments.