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I know that in the $\phi^4$ theory the tree-level diagram for two in-going and two out-going particles is simply a 'cross' where all the external legs meet at one point.

I'm now interested in a slightly more complicated case where I have 4 rather than 2 outgoing particles and I want not only the tree level diagram but also the one loop diagram. I think I know what the tree-level diagram looks like, but I have some trouble drawing the one loop one. I know that each vertex needs to have 4 lines connected to it. I managed to come up with something but it was just trial and error and I'd like someone to verify whether what I've done is right.

My attempt is:

enter image description here

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Yes, that is the correct one loop topology that appears assuming no snail and/or one particle reducible contributions (inclusion of these gives you a plethora of other diagrams, such as ones where you decorate the tree level contribution with snails etc). With a labelling of the external momenta in place, you can show by simple combinatorics the number of inequivalent permutations of the external legs you have. The contributing diagrams are essentially generated by attaching two legs to each vertex of a triangle.

Naively there are $6!$ permutations of the external legs but to avoid overcounting due to equivalent diagrams related by vertex relabelling we have to thereby divide out by the cardinality of the symmetry group of the triangle which is $|S_3| = 3!$ Now, we also need to divide out by the permutation of two legs at each of the three vertices. So the number of contributing diagrams is $6!/(3! \cdot (2!)^3) = 15$.

The same argument can be applied to e.g the more familiar one loop contribution to $2 \rightarrow 2$ scattering within $\phi^4$ (the so-called dinosaur diagram discussed in many QFT books in the pursuit of renormalisation of the theory at one loop). By exactly the same argument the number of contributing diagrams is $4!/(2! \cdot (2!)^2) = 3$, the $s, t$ and $u$ like channel contributions.

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  • $\begingroup$ On the combinatorics: what counts as an equivalent diagram? Because I can, for example, interchange two particles on outgoing legs; would that be a separate contribution I need to include? I think yes. But then what if by interchanging external legs I arrive on a diagram that is the same as the original one up to, say, a reflection (by which I mean: is the diagram with the outgoing legs, top to bottom, corresponding to particles 1, 2, 3 and 4 the same or distinct from a one that has them as 4, 3, 2 and 1 and a similar flip for in-going legs? Because that's the same diagram only up-side-down)? $\endgroup$ – Piotr May 12 '17 at 15:52
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    $\begingroup$ Hi @Piotr In general, it's a nasty business trying to write down all inequivalent diagrams without overcounting at any point. Label the six external momenta from $1,\dots, 6$ and in each diagram have them originate from the same spacetime point (just means to say we fix the external momenta in place). Now, label the vertices $1,2,3$. Any diagram related by a permutation of the vertices labels are equivalent :) $\endgroup$ – CAF May 13 '17 at 10:02
  • $\begingroup$ Try this out on some simple examples first, maybe the one loop contribution to $2 \rightarrow 2 $ scattering (the so called dinosaur diagram) or even the familiar $s,t,u$ tree level $\phi^3$ interactions. $\endgroup$ – CAF May 13 '17 at 10:03

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