I am reading this paper by Dixon, where it is mentioned that one can diagrammatically compute the colour factor of certain Feynman graphs, such as shown in figure 1, where they expand a three gluon vertex as a difference of two diagrams with fermion loops.

enter image description here

Now, my question is, is this elimination of three gluon vertex a "trick" to just calculate the colour amplitude, or is it a valid factorization of the complete diagram? More precisely, can I write down:

$\displaystyle f^{mnp}A^m(x)A^n(x)A^p(x)=\left(f^{abm}f^{bcn}f^{cap}\partial CA^m(x)\partial CA^n(y)\partial CA^p(z)-f^{abm}f^{bcn}f^{cap}\partial CA^m(x)\partial CA^p(z)\partial CA^n(y) \right)$

where, the three colour trace is defined as

$Tr[\partial CA(x)\partial CA(y)\partial CA(z)] \equiv f^{abm}f^{bcn}f^{cap}\partial CA^m(x)\partial CA^n(y)\partial CA^p(z)$


No, of course you can't do this. For starters, you can't just conjure up these two new position variables. Also, the loop diagrams would involve integration with respect to the loop momentum, leading you to a completely different class of functions. Finally, if you include the coupling constant, you realise that these loop diagrams are higher order corrections in the perturbative series.

In other words: this is merely a rewriting of the colour structure of the diagram which does not extend to the kinematic dependence.

  • $\begingroup$ Why different order? The three gluon vertex should carry three factors of the coupling from $g^2A^2 g\partial A$ $\endgroup$ – lux Dec 21 '19 at 16:06
  • $\begingroup$ @lux fermion loop diagrams are proportional to $g^3$, and the 3-gluon vertex is proportional to $g$. $\endgroup$ – Prof. Legolasov Dec 21 '19 at 19:11
  • $\begingroup$ Yes, you're right - I miscounted the three gluon vertex. $\endgroup$ – lux Dec 21 '19 at 21:38

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