# Understanding the intermediate field method for the $\phi^4$ interaction

In Rivasseau's and Wang's How to Resum Feynman Graphs, on page 11 they illustrate the intermediate field method for the $$\phi^4$$ interaction and represent Feynman graphs as ribbon graphs. I had to read up about ribbon graphs as I've never heard of them before and still don't fully understand their use for the intermediate field method. For the $$\phi^4$$ interaction, we have:

In that case each vertex has exactly four half-lines. There are exactly three ways to pair these half-lines into two pairs. Hence each fully labeled (vacuum) graph of order $$n$$ (with labels on vertices and half-lines), which has $$2n$$ lines can be decomposed exactly into $$3^n$$ labeled graphs $$G'$$ with degree $$3$$ and two different types of lines

• the $$2n$$ old ordinary lines
• $$n$$ new dotted lines which indicate the pairing chosen at each vertex (see Figure 5).

The extension illustrated in Figure 5 looks as follows: I'm not sure I correctly understand the way to create such "3-body extensions". I understand it in that way that we "split up" the given vertex into two vertices and add an edge between them. Then, each vertex has two additional half-edges and there are three ways to pair them together. Joining the upper half-edges together, we get the second term on the RHS and joining the two half-edges on one side together, we will get the first term on the RHS (the dashed line would be the new edge between the two vertices). The only case left is the one of joining the upper left and the lower right (and the upper right and lower left) half-edge together. However, when drawing that specific case, I don't see any way to stretch or rotate it such that the third term on the RHS results.

Am I misunderstanding the way 3-body extensions are intended to get or am I just missing a way to stretch the edges to get the third term on the RHS? Also, where exactly do we need the cyclic ordering characterising ribbon graphs? And why are there two dashed lines in the third term on the RHS?

• I am not sure but maybe this question would recieve better answers on the site : physics.stackexchange.com Nov 21, 2013 at 20:31
• @DanielSoltész I already tried my luck there to no avail (so far).
– Huy
Nov 22, 2013 at 8:37
• @DanielSoltész: Nope, high-level questions generally get largely ignored there these days. Nov 26, 2013 at 14:40

The half-edges incident to a 4-valent vertex can be labeled 1,2,3,4. There are three ways to split them into pairs: 12-34, 13-24, and 14-23. Another way to think of the three pairings is by considering the 6 edges of a tetrahedron (with vertices labeled 1,2,3,4), and matching them into three skew pairs of edges.

The twisty shape in the picture indicates that you should join the pairs of half-edges that are diagonal to each other. I think the illustrator chose to move the dotted line to the side of the crossing to improve visibility.

• But how come the shape on the left of the crossing is not twisted either? After all, we'd join two pairs of half-edges diagonally, so it should be symmetric, shouldn't it?
– Huy
Nov 23, 2013 at 10:05
• @Huy Instead of explaining again, I suggest you try drawing this by yourself. Put the ends of the half-edges at the corners of a square, and calculate the three ways of joining them using two trivalent vertices. You will find that it is difficult to draw the diagonal crossing in a symmetric way. Nov 23, 2013 at 11:29
• If I draw the three ways to pair the four half-edges, I get those three graphs: i.imgur.com/EpZQyRL.jpg The third is the "problematic" one. Is the red one the correct one? I can see that we could stretch half-edges 1/3 such that both are "below" the newly added horizontal edge. But why does the other edge (2/4) twist, when we put it on the top side? Is that just by convention?
– Huy
Nov 23, 2013 at 13:06
• Your drawings are slightly off, because you didn't preserve the connections 1-2, 3-4 outside of the 4-valent vertex. Nov 24, 2013 at 3:20
• I'm sorry, but I just don't get what you mean. How are my connections 1-2, 3-4 not outside of the 4-valent vertex? And isn't the 4-valent vertex supposed to be replaced by two 3-valent vertices and an edge between them (hence "3-body extension")?
– Huy
Nov 24, 2013 at 10:03