Colour decomposition of $n-$gluon tree amplitude

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I have here a $SU(N_c)$ Yang-Mill's theory and let the index $i$, label the $n$-gluons, and $\{k_i, \lambda_i, a_i\}$ be its momenta, helicity and colour index and $\cal{A}_n^{tree/1-loop}(\{k_i, \lambda_i, a_i\})$ be the tree/1-loop-level amplitude for their scattering. Then apparently the following two equations hold,

• ${\cal A}_n^{tree}(\{k_i, \lambda_i, a_i\}) = g^{n-2}\sum_{\sigma \in S_n/\mathbb{Z}_n} Tr[T^{a_\sigma (1)}\ldots T^{a_\sigma (n)}] A_{n}^{tree}(\sigma(1^{\lambda_1})\ldots\sigma(n^{\lambda_n}))$

• ${\cal A}_n^{1-loop}(\{k_i, \lambda_i, a_i\}) = g^n [ \sum_{\sigma \in S_n/\mathbb{Z}_n} N_c Tr[T^{a_\sigma(1)}\ldots T^{a_\sigma(n)}] A_{n;1}^{tree}(\sigma(1^{\lambda_1})\ldots\sigma(n^{\lambda_n})) +$ $\sum _ {c=2} ^{[\frac{n}{2}] +1} Tr[T^{a_\sigma (1)}\ldots T^{a_\sigma(c-1)}] Tr[T^{a_\sigma (c)}\ldots T^{a_\sigma(n)}] A_{n;c}^{tree}(\sigma(1^{\lambda_1})\ldots\sigma(n^{\lambda_n}))]$

I want to know the proof for the above two equations.

It seems that this lecture note tries to sketch some argument for the first of the above two expressions but then its not very clear.

• Thought I haven't seen this clearly written anywhere but I guess that the factors of $A_n^{tree}$ and $A_{n;1}$ and $A_{n;c}$ that occur on the RHS of the above two equations are what are called "colour ordered amplitudes". It would be great if someone can say something about this idea too. (..i do plan to put up another separate question later focusing on that aspect..)

{..my LaTeX seems all garbled! It would be great if someone can edit that and put in a line as to what has gone wrong..}

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retagged Apr 19, 2014

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You can show that the amplitude has that form by thinking about the Feynman rules. I'll only discuss particles in the adjoint representation (this is suitable for supersymmetric theories), but this can be done more generally. Note that if you have particles transforming in the fundamental representation of $SU(N)$ then the amplitude does not have the form in your question.

Think about the three-gluon vertex. It contains a $f^{a b c}$ factor which, up to a constant, can be written as $\operatorname{tr}(T^a [T^b, T^c]) = \operatorname{tr}(T^a T^b T^c) - \operatorname{tr}(T^a T^c T^b)$. So this can be written as a combination of traces of products of Lie algebra generators. Now, think about joining together two such triple vertices. We have to compute quantities like $\operatorname{tr}(T^a T^b T^c) \operatorname{tr}(T^c T^d T^e)$, where the index $c$ is summed over.

Now, for $SU(N)$ we have that $$(T^a)_i^j (T^a)_k^l = \delta^j_k \delta^l_i - \frac 1 N \delta^i_j \delta^k_l,$$ where we sum over the index $a$. For $U(N)$ the last term in the right hand side is absent. Using this, you'll see that the color factor for joining two triple vertices can also be written as a single trace. The same can done for the quartic vertex.

For a tree level scattering, your can do this recursively. You start with some triple vertex and you keep joining other vertices and by using the two identities above you can always rewrite the answer as a single trace.

At loop level, this doesn't work because you can get things like $\operatorname{tr}(T^a \cdots T^b T^c) \operatorname{tr}(T^c T^d \cdots T^b)$, with sum over $b$ and $c$. Using the $SU(N)$ identity you get a contribution $\operatorname{tr}(T^a \cdots T^b T^d \cdots T^b)$. Now using the identity again you get $\operatorname{tr}(T^a \cdots) \operatorname{tr}(T^d \cdots)$. In general, at $\ell$ loops you can have $\ell+1$ traces if you have a large enough number of particles.

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answered Feb 12, 2012 by (160 points)
If i understand you right then you are saying that the third equation on the second page of my linked lecture notes reduces to the usual Feynman rules when written for the adjoint representation? Can you make it a little more explicit as to what you have in mind as the "adjoint" and the "fundamental" representation? Its getting confusing. The usual normalization stated on the top of the first page of my linked notes when used on the trace factors of the third equation of the second page of my linked notes then it seems to produce an extraneous factor of "-2".

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For the purely gluonic vertices (the 3-point and the 4-point ones) in Yang-Mill's theory does one have any choice about the matrices $T^a$? I thought they are bound to be in the adjoint i.e $bc$ element of the matrix $T^a$ is forced to be the structure constant $f^{abc}$. I have seen that review by Dixon but that only states the two factorization theorems that I quoted. I haven't seen the Mangano-Park review - its quite a large one - can you point out as to where in it does it prove these factorizations?

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@Anirbit, the fermions can be in an arbitrary representation. In my answer I considered fermions in the adjoint for simplicity. If they are in the fundamental of $SU(N)$, then you get a color factor of $(T^a)_i^j$ for an interaction with gauge fields, while you had an $f^{abc}$ in the case of the adjoint representation. As a consequence, you can now also get strings of gauge algebra generators, $(T^a T^b \cdots)_i^j$. For references you can try this [review article](http://arxiv.org/abs/hep-th/0509223) or these [lecture notes](http://arxiv.org/abs/hep-ph/9601359).

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[Ignore the garbled type above!] Can you kindly expand a bit more on how the constraint on the representation of $SU(N)$ comes in? Like if you could write some explicit equations for what you call the fundamental representation - in standard Yang-Mill's theory is there a constraint on which representation should the fermions be in? (..like the gauge fields are constrained to be in the adjoint representation..) The last equation on the first page of my linked lecture note already gives a plausibility argument for this decomposition.

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Can you kindly explain as to how does the third equation on the second page of my linked notes match against the usual Feynman rules? I did not get the part of your argument at the 1-loop level. Can you kindly give a reference for what you are saying?

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