It is known that the number of loops in $\lambda\phi^4$ theory is given by the formula


where $L$ is the number of loops, $I$ the number of internal lines and $V$ the number of vertices. I would like to know the proof of this statement.

  • 1
    $\begingroup$ It's combinatorics, and has really nothing to do with physics. It's Euler's formula for planar graphs (note that "loop" = "face"). $\endgroup$
    – ACuriousMind
    Apr 16 '15 at 10:57
  • 1
    $\begingroup$ you can prove the result by induction. @ACuriousMind isn't "loop" = "face" - 1? because the space outside a graph is a face? sorry if that's pedantic, you might be just making a point about vocabulary $\endgroup$
    – innisfree
    Apr 16 '15 at 11:37
  • $\begingroup$ @innisfree: Indeed, #loops = #faces - 1. That's pedantic, but correct :) $\endgroup$
    – ACuriousMind
    Apr 16 '15 at 11:42
  • $\begingroup$ @silvrfück, it's (surprising) straight-forward, if I haven't made a mistake with what I've scribbled down. i) Prove it for $I=0$. ii) Assuming it's true for $I=N$, prove it's true for $I=N+1$, by thinking about what happens when you break an internal line into two external lines (you remove a loop and an internal line). I encourage you to answer your own question $\endgroup$
    – innisfree
    Apr 16 '15 at 11:47
  • $\begingroup$ As @tparker points out in his/her answer, this result is easy to see considering momentum conservation at each of the vertices. See these notes for a detailed explanation for the $\phi^4$ case. Also, related: https://math.stackexchange.com/q/2368908/404320. $\endgroup$ Jul 23 '17 at 15:38

This formula is actually Euler's formula for planar graphs, and holds for all Feynman diagrams regardless of what theory we are in.

The proof proceeds by induction and is easy if we first disregard the case of crossing lines:

  1. Observe that a one-loop graph has two vertices, one loop, and two internal lines, so the formula holds.

  2. Observe that a $(n+1)$-loop graph is produced from a $n$-loop graph by either drawing one additional line between two already existing vertices, which doesn't change $L-I$, or by adding a new vertex and connecting it to two other vertices, which doesn't change $L-I+V$.

  3. By induction, the formula holds for all graphs with finitely many loops.

More formally, we can say that

A Feynman diagram is called planar if the adjoint graph obtained by connecting all external lines to a single vertex is planar.

and then we have proven up to now that the formula holds for all planar Feynman graphs. Interestingly, not even all $\phi^4$ graphs are planar. Consider $2\to 2$ (or $1\to 3$)-scattering with a box diagram, where each external line is connected to its own vertex, and each vertex is connected with each other vertex. The adjoint graph is the complete graph on five vertices, which is known to be not planar.

Nevertheless, the "Feynman-Euler formula" $$ L-I+V = 1$$ still holds because of the way loops are formally counted. By the general Euler formula, $$ \#\{\mathrm{vertices}\} - \#\{\mathrm{edges}\} + \#\{\mathrm{faces}\} = 2 - 2g$$ where $g$ is the genus of the surface on which the graph can be drawn without intersections, and "faces" are all regions bounded by edges. A "face" does not have to have a vertex at every corner, so when you get two crossing lines in a Feynman graph, you get two additional faces that you do not count as loops - the above boxy $\phi^4 $ diagram has four faces inside the box, but only two loops.

Since every crossing of lines that cannot be eliminated by deforming the graph (and is hence a "true crossing" and not just us being too dumb to draw the graph properly) increases the genus on which you could draw the graph without crossings by $1$, the "Feynman-Euler formula" for all graphs follows from the general Euler formula.

  • $\begingroup$ do you know any other ways to prove this result? $\endgroup$
    – innisfree
    Apr 16 '15 at 11:57
  • $\begingroup$ @innisfree: Cracking the nut with the sledgehammer: Planar finite graphs are the CW-complexes of the polyhedra whose Schlegel diagram they are. Since polyhedra are homotopically equivalent to the sphere, their Euler characteristic is 2, and since the cells of the CW complexes are precisely the vertices, edges and faces (without the "outer" face), the alternating sum "#vertices - #edges + #faces" must be equal to two. $\endgroup$
    – ACuriousMind
    Apr 16 '15 at 12:05
  • $\begingroup$ @ACuriousMind I have two questions. Why do you say that the box diagram has only 2 loops, what is your criteria to define the number of loops in a diagram? and second could you give a way to understand that every crossing requires another genus knowing that I know no topology? $\endgroup$
    – Yossarian
    Apr 28 '15 at 19:15
  • $\begingroup$ I feel that there is a step missing as it is not entirely clear to me how the Euler formula gives the correct equation. In a simple situation (ignoring lines having to cross) we should use g=0 since we can put the diagram on a sphere. Then we need to get +1 on the RHS somewhere. Does this come from the identification F=L+1, due to the outer face of the diagram that we miss by just counting the number of loops? $\endgroup$
    – Kvothe
    Nov 19 '17 at 23:31
  • $\begingroup$ @Kvothe I'm not sure what you mean by "outer face", but the "+1" comes from me applying Euler's formula to the adjoint graph, not the Feynman graph itself. $\endgroup$
    – ACuriousMind
    Nov 20 '17 at 12:45

Page 140 of Srednicki's QFT textbook provides a much simpler proof:

This can be seen by counting the number of internal momenta and the constraints among them. Specifically, assign an unfixed momentum to each internal line; there are [$I$] of these momenta. Then the $V$ vertices provide $V$ constraints. One linear combination of these constraints gives overall momentum conservation, and so does not constrain the internal momenta. Therefore, the number of internal momenta left unfixed by the vertex constraints is $[I] − (V−1)$, and the number of unfixed momenta is the same as the number of loops L.


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