This is closely related to my previous post Double line notation - gluon propagator

I'm trying to understand the double-line vertices for the gluon in the case of a $U(N)$ gauge group. Normally, the vertex factors are given by: $$ \mathrm{Three} \ \to \ g f^{a}_{\ bc} \left[ h^{\mu \nu} ( k - p )^{\rho} + h^{\nu \rho} ( p - q )^{\nu} + h^{\rho \mu} ( q - k )^{\nu} \right] \\ \mathrm{Four}\ \ \to \ - i g^{2} \left[ f^{a}_{\ be}\ f^{c}_{\ de} \left( h^{\mu \rho} h^{\nu \sigma} - h^{\mu \sigma} h^{\nu \rho} \right) + f^{a}_{\ ce}\ f^{b}_{\ de} \left( h^{\mu \nu} h^{\rho \sigma} - h^{\mu \sigma} h^{\nu \rho} \right) + f^{a}_{\ de}\ f^{b}_{\ ce} \left( h^{\mu \nu} h^{\rho \sigma} - h^{\mu \rho} h^{\nu \sigma} \right) \right] $$

Since the gauge group is $U(N)$, if you write the group indices in the form $a = (\bar{j}, k)$, then the structure constants are going to be given by $$ f^{( \bar{m}, n )}_{ ( \bar{j},k )(\bar{p},q) } = i \big( \delta_{jq} \delta_{kn} \delta_{pm} - \delta_{jm} \delta_{kp} \delta_{qn} \big) $$ This inspires the use of the double line notation.

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The way I understand it though, the three vertex factor now has TWO pieces with $\delta\delta\delta$'s - which means that I need to consider two double line diagrams to represent just one ordinary three gluon feynman diagram?

And futhermore the factors $ff$ yield six combinations of $\delta \delta \delta \delta$'s - which would mean I need 6 double line diagrams to represent just one ordinary four gluon feynman diagram?

I believe my understanding of this is correct (please call me out on this if its not).

My question is; how is this a simplification at all? Now instead of my original two diagrams, I've got 8 that I'm looking at? It seems to me that the double line notation is making the problem more convoluted rather than helping to simplify it? What is the point of this?

  • 1
    $\begingroup$ Probably some people do use this for actual calculations, but personally, the only thing I really use the double line notation for is power counting in N. Loops give traces over color indices, so it's simpler to do the power counting if every loop counts the same rather than having to work with the gluon's index structure. In particular you can conclude that many diagrams (in particular nonplanar ones) can be ignored at large N. $\endgroup$
    – Logan M
    Dec 16, 2016 at 23:30

2 Answers 2


First, it is probably useful to point out that the double line notation does not only affect the propagators, but also the vertices. In the conventional Feynman rules, these vertices that represent the coupling of the gauge field with the fermion fields, contain the generators of the gauge group. What the double line notation effectively does, is to remove these generators from the vertices and attach them to the ends of the propagators. The resulting vertex rule is now much simpler. It then also allows one to use the properties of the $f_{abc}$'s and the generators to simplify the expression of the propagator, by replacing the contraction of the $f_{abc}$'s and the generators by a combination of the $\delta$'s. Although the propagator now contains a factor that consists of a sum of different combinations of the $\delta$'s, it does not mean that you need a different diagram. It would all be part of the calculation for one diagram. Moreover, it is much simpler to work with the $\delta$'s than to work with the $f_{abc}$'s and the generators. Effectively, the group factors that one would have calculated in evaluating the diagram has now already been done and it is just a matter of applying the contractions via the $\delta$'s.


You are correct: there are now two and six different diagrams for the three and four-point function, where before there was only one for each instance. This is because now non-cyclic permutations of the vertices yield different diagrams.

However, this is not really a complication. Let's limit ourselves to talking about tree-level n-gluon scattering, and to drawing planar diagrams, that is, diagrams for which no lines cross. Each external line contains a factor of $T^{a_i}$, where $T^a$ are generators of the Lie algebra of $U(n)$. Since each line encodes a Kronecker delta contraction, as we go around the diagram we see it is proportional to $Tr(T^{a_1}\dots T^{a_n})$. For example:four-gluon interaction

Thus we can write the four-gluon tree-level scattering amplitude as

$$A_4=A[1234] Tr(T^{a_1}T^{a_2}T^{a_3}T^{a_4})+\text{non-cyclic permutations of {1,2,3,4}},$$

where $A[1234]$ is the amplitude from Feynman diagrams with this ordering of external lines, after having stripped it from all the $U(n)$ algebra. These are often called color-ordered amplitudes. Note the four-gluon vertex needed to calculate color-ordered amplitudes is simply $V^{\mu_1 \mu_2\mu_3\mu_4}=\eta^{\mu_1\mu_3}\eta^{\mu_2\mu_4}$, and you only need to calculate one of these $A[ijkl]$ to obtain all others from permutations.

More complicated amplitudes will have a similar structure, although you might need more traces depending on how general diagrams you want to consider.

In conclusion, although you are correct that there are more diagrams, you usually only need one of them to arrive at the full answer, and the group algebra gets taken care of.



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