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The following diagram is given in Srednicki pg62 figure 9.7:

enter image description here

Srednicki gives this diagram a symmetry factor of $S=2^2=4$. But using a method which seems to work on every other diagram I get $S=2$ (the method is given below). Am I missing a subtly here or is the symmetry factor given in the book wrong?

My Method

We split each of the vertices into 3 and count the number of ways we can draw each line between two vertices (as shown below): enter image description here The number in front of the brackets gives the number of ways whilst the number in brackets gives the order which I chose them. On top of this we have a factor of $^4C_2 \times 2$ for swapping the $4$ vertices (taking into account that two vertices are identical). This thus gives us a symmetry factor of: $$S^{-1}=\frac{3\times 3\times 6 \times 6\times 2 \times 2 \times 1}{(3!)^4 \times 4!} \times{^4C_2}\times 2$$ $$=\frac{3\times 3\times 6 \times 6\times 2 \times 2 \times 1}{(3!)^4 \times 4!} \frac{4!}{2!2!}\times 2$$ $$=1/2$$ Thus $S=2$

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You are right. You are basically using method 1) from my answer to Problem understanding the symmetry factor in a feynman diagram and you handled it like a champ. If you use method 2) from that same answer you will also find there is only one nontrivial automorphism. Think of the graph as a "theta" taking a nap on a hammock attached to the axis given by the two external legs. This autmorphism is the rotation by 180 degrees around this axis.

PS: If you have seen two-stage experiments in elementary probability or more general counting methods based on decision trees, then this is basically what you are doing. I recommend to first choose the vertices: 4 choices for the neighbor of the left external leg times 3 choices for the right leg neighbor, and then looking at the internal line contraction count. Of course $4\times 3={}^{4}C_{2}\times 2$.


Edit as per AFT's comment: The above answer is based on the assumption that one is computing a two-point function rather than a vacuum diagram. Note that the issue of symmetry factors is a question that belongs to mathematics rather physics. The proper setting for handling these factors with rigor and accuracy is Joyal's theory of combinatorial species: https://en.wikipedia.org/wiki/Combinatorial_species . You can see how one can apply it to the specific context of Feynman diagrams in my article "Feynman Diagrams in Algebraic Combinatorics". The article focuses on a complex bosonic model with $\bar{\phi}\phi^n$ interaction but it is straightforward to transpose it to that of a real scalar $\phi^3$ model as in the OP's question.

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  • $\begingroup$ The symmetry factor in Srednicki is correct -- it is indeed $S=4$. This is easily confirmed by a straightforward computation in, say, Mathematica. Note that Srednicki takes external vertices to be unlabelled, so you must account for permutations thereof. Perhaps this factor of $2$ is what you (and OP) are missing. If you feel like it, expand the path integral explicitly, and check for yourself that the diagram carries a factor of $4$. If you expand the two-point function, it will carry a factor of $2$. This is because the diagrams in the latter have labelled leaves, unlike those in the former. $\endgroup$ – AccidentalFourierTransform Dec 26 '17 at 14:04
  • $\begingroup$ @AccidentalFourierTransform: I see. Srednicki does something a bit nonstandard which is: he computes vacuum diagrams of a theory with both $\phi^3$ and $\phi$ vertices. This indeed results in a symmetry factor $S=4$ because the two $1$-valent vertices on the left and right sides of the diagram are treated as internal. My answer was based on the assumption that one is trying to compute a two point function $\langle\phi(x_1)\phi(x_2)\rangle$ which gives $S=2$ because the counted automorphisms have to keep the external legs fixed. $\endgroup$ – Abdelmalek Abdesselam Dec 28 '17 at 18:05
  • $\begingroup$ +1 Now I agree with your answer. In the chapter OP is reading, Srednicki is explaining how to calculate the partition function $Z[j]$. At that point, leaves are indistinguishable and therefore you must take into account permutations thereof. When you start taking functional derivatives $\frac{\delta}{\delta j_1}\cdots \frac{\delta}{\delta j_n}$ to compute the $n$-point function, the leaves acquire a labelling (from $1$ to $n$), and therefore the automorphism group contains isomorphisms that fix the external labels only. $\endgroup$ – AccidentalFourierTransform Dec 28 '17 at 18:26

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