I am trying to understand a $1/2$ in the symmetry factor of the "cactus" diagram that appears in the bottom of page 92 In Peskin's book. This is the diagram in question (notice that we are in $\phi^4$ theory)

enter image description here

In the book it is claimed that the symmetry factor of the diagram is


where it says that the $3!$ comes from interchanging the vertices, the first $4\dot{}3$ from the placement of contractions in the $z$ vertex, the following $4\dot{}3\dot{}2$ from the placement of contractions in the $w$ vertex, the last $4\dot{}3$ from the placement of contractions in the $u$ vertex and the final $1/2$ from the interchange of $w-u$ contractions.

It is this last $1/2$ that I don't understand. Can you be more explicit on where this comes from?


3 Answers 3


We choose one of the $4$ z-fields to contract with the single x-field. We then choose one of the remaining $3$ z-fields to contract with one of the $4$ w-fields. The remaining two z-fields just contract with themselves. Now choose one of the remaining $3$ w-fields to contract with the single y-field.

(Here is where we have to be careful). There are $2$ choices for the w-field contraction with one of the $4$ u-fields, and then $3$ choices for the other w-u contraction. In computing this last combination we have over counted by a factor of $2$.

To see this more clearly, consider one of the contractions,

$\phi_a(w)\phi_b(w) \quad\phi_a(u)\phi_b(u)\phi(u)\phi(u)$

The subscript denote which fields are contracted with which other fields (I'm not sure how to express contractions in Latex).

There are two ways to get this particular contraction: we could either choose the first w-field to be contracted with the first u-field, and THEN choose the second w-field to be contracted with the second u-field; OR we could choose the second w-field to be contracted with the second u-field, and THEN choose the first w-field to be contracted with the first u-field.

Clearly both of these are equivalent. However, in the combinatorics we have counted both of them, and so we must divide by a factor of $2$. So the total number of different contractions giving the same expression as $(4.45)$ is

$3! \, \times \, 4 \cdot 3 \, \times \, 4 \cdot 3 \cdot 2 \, \times \, 4 \cdot 3 \, \times \, 1/2$

Where the $3!$ comes from the interchange of vertices.

EDIT: If that isn't clear, think about the following scenario. There are two boxes, in the first there are two objects, $A$, and $B$, and in the second there are two more, $C$, and $D$. How many different ways are there to pair off the objects so that each object in the first box has a partner in the second box? Clearly the answer is two: $A,C$ and $B,D$; and $A,D$ and $B,C$.

One might think the answer is $2\cdot 2$, but we can see that this produces duplicates

\begin{array}{|r|r|} \hline First Pair & Remaining Pair \\ \hline A,C & B,D\\ \hline A,D & B,C\\ \hline B,C & A,D\\ \hline B,D & A,C\\ \hline \end{array}

So we must multiply by a factor of $1/2$ to fix the overcounting.

Hope that helps.

  • 2
    $\begingroup$ For counting symmetry factors there's a neat phrase my QFT teacher used to say: When in doubt, back to Wick, i.e. write out the fields and figure out the combinatorics from there, which is often cleaner (if more cumbersome) than effective rules for Feynman diagrams. $\endgroup$
    – Neuneck
    May 7, 2015 at 9:59

I don't think the explanation in the book is clear, but you can just ignore it and get the correct factor $S=\frac{1}{8}$ as follows.

  1. Start by drawing 5 isolated vertices, two labelled ones with degree one and three unlabelled 4-valent vertices. Then $$ S=\frac{1}{3!}\left(\frac{1}{4!}\right)^3\times C $$ where $C$ is the number of contraction schemes that produce the given graph shape. The three vertices play different roles in the graph so one has $3!$ ways of choosing this role assignment (who is the nearest neighbor of $x$ etc.). Then a factor of $4$ for the $z$ leg that $x$ attaches to. Likewise, a factor of $3$ to choose which of the remaining $z$ legs moves on to $w$. Then at $w$ there is a factor of $4\times 3$ to choose the legs which receive the $z$ and $y$ edges. Then a factor of $6$ to pick which pair of legs at $u$ will form the tadpole. Finally there is a factor of $2$ to connect the remaining free legs at $u$ with the two remaining ones at $w$. Thus $C$ is number in your question.
  2. Alternatively $S=\frac{1}{|G|}$ where $G$ is the automorphism group of the graph. More precisely, let $E$ be your favorite set with $14$ elements (the half-edges in the picture). Equip $E$ with two set partitions $\mathcal{V}$ and $\mathcal{E}$. The latter is made of $7$ disjoint pairs corresponding to the edges. Whereas $\mathcal{V}$ is made of two singletons and three blocks of four elements. Here $G$ is the group of permutations of $E$ which preserve the set partitions $\mathcal{E}$ and $\mathcal{V}$ and also fix the external legs (which here is automatic because the picture is asymmetric in $x$ and $y$). It is generated by three commuting order $2$ elements. For each tadpole you can permute the corresponding half-legs. Finally there is the exchange of the two edges between $w$ and $u$.
  • $\begingroup$ Minor comment to the answer (v1): The symmetry factor is defined as the reciprocal in P&S p. 93-95, i.e. $S=|G|$. $\endgroup$
    – Qmechanic
    Jan 1, 2021 at 10:34
  • $\begingroup$ Well that's just a convention and I don't have the PS book in front of me. $\endgroup$ Jan 2, 2021 at 22:25

The 1/2 comes from the symmetry of the diagram. In the sense that if you look away and I switch the two propagators it is a "different" diagram, but you cannot tell. The numbers of ways to do this is 2.

If these were directed propagators it would not be the case.


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