# I Combinatorics of a phi4 interaction

1. May 12, 2017

### CAF123

Consider the one loop correction to a 2 -> 4 scattering process in phi4 theory.The only IPI/non snail contributions is that shown in the attachment. I have an automated package that will do all the feynman diagram generation for me and for this process it returns 15 diagrams, which means to say there are 15 inequivalent permutations of the external momenta. My question is basically how to derive this number?

If we draw the diagrams with 2 always incoming and 4 outgoing, then there are just 2! permutation of the incoming and 4! permutation of the outgoing. There are 6! permutation of all external lines naively and it seems that 6!/(2! x 4!) = 15, but I don't understand how this makes sense. Also, 5!/((2!)^3 = 15 which I can sort of make an argument for but I don't think it's a solid argument.

So,
a) what counts as an inequivalent diagram?
b) what is the correct combinatoric argument that gives rise to 15?

#### Attached Files:

• ###### phi4_2_6.png
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2. May 13, 2017

### vanhees71

The different diagrams of this topology are characterized by how the 6 external points are paired at each of the three vertices. For the pair at vertex 1 you have $\binom{6}{2}=6!/(4! 2!)=15$ possibilities. For the pair at vertex 2 then there remain $\binom{4}{2}=4!/(2! 2!)=3$ possibilities and then the other pair is fixed, but now we've over counted the possibilities since all diagrams which only differ by a cyclic permutation of the vertices are in fact the same diagram, i.e., we have to devide by the number of cyclic permutations of the three vertices, i.e., by $3!/2=3$. So you get 15 different topologies and thus 15 diagrams.

3. May 13, 2017

### CAF123

Thanks! Yes, I just realised we can allow for permutation of the vertices, so that the number of diagrams is 6!/(3! x (2!)^3) = 15 after dividing throughout by the size of the symmetry group of triangle and the permutations of each of the two legs at the three vertices. I suppose this 'cyclic permutation of vertices' argument doesn't work beyond scalar theories because then we can allow for our propagators to carry arrows so the number of diagrams increases? e.g the 2-> 4 scattering for two incoming fermions (psi/psibar) and four outgoing fermions(psi,psi,psibar,psibar) has 48 contributions with gluon propagators.

4. May 13, 2017

### vanhees71

Then it gets indeed more complicated, exactly for the reason you mention!