In this answer Proof of geometric series two-point function it is said:

  1. Now what about the coefficients in front of each Feynman diagram? Due to the combinatorics/factorization involved it becomes a geometric series $$G_c~=~G_0\sum_{n=0}^{\infty}(\Sigma G_0)^n\tag{A}$$

How can we prove this? My main concern is the combinatorics, for example in qed the symmetry factor for connected diagrams is 1.suppose that $\Sigma=A+B+...$ are irreducible diagrams of the photon propagator.

Since symmetry factor of $G_0AG_0AG_0=1$ we should have the symmetry factor of $A=1$.The same thing for $B$.

But we also have the factor $G_0AG_0BG_0+G_0BG_0AG_0$
The only way this to work is that $G_0AG_0BG_0=-G_0BG_0AG_0$

How can I prove this?

  • $\begingroup$ It's basically the argument that's used for perturbation theory in NRQM (using the operator resolvent, which makes the geometric series somewhat more transparent.) I think it should be possible to derive the general result without performing any Gaussian integrals or Wick contractions explicitly or referring to them, which is where the symmetry factors would come from. $\endgroup$
    – TLDR
    Commented May 23 at 4:36

1 Answer 1

  1. Recall

    • that the partition function $Z[J]$ is the generating functional of all$^1$ Feynman diagrams;

    • that $$W_c[J]~=~\frac{\hbar}{i}\ln Z[J]$$ is the generating functional of connected Feynman diagrams, cf. the linked cluster theorem;

    • and that the Feynman rules dictate that each such Feynman diagram should be divided by its symmetry factor.

  2. In contrast, for a connected $n$-point correlation function $$\langle\phi^{i_1}\ldots\phi^{i_n}\rangle^c_J~=~ \left(\frac{\hbar}{i}\right)^{n-1}\frac{\delta^n W_c[J]}{\delta J_{i_1} \ldots\delta J_{i_n}},$$ the $n$ external legs are considered distinguishable, and not symmetrized. We emphasize that it does not contain a $S_n$-symmetry factor of its $n$ external legs.

  3. In particular,

    • the full propagator/connected 2-pt function $$\langle\phi\phi\rangle^c_{J=0}~=~\frac{\hbar}{i}G_c$$ from eq. (A),

    • the bare/free propagator $\frac{\hbar}{i}G_0$, and

    • the self-energy $\frac{i}{\hbar}\Sigma$

    are not divided by the $\mathbb{Z}_2$-symmetry of their 2 external legs, independently of whether the propagator is directed or undirected/has or hasn't an arrow.

  4. Similarly, if the self-energy $$\frac{i}{\hbar}\Sigma~=~\sum_k\frac{F_k}{S_k}$$ is built from individual 1PI Feynman diagrams $F_k$ with 2 (amputated) external legs, we only divide with symmetry $S_k$ of $F_k$ as if the 2 external legs were distinguishable. Furthermore, the geometric series (A) therefore neatly generates connected Feynman diagrams weighted with appropriate symmetry factors.


  1. P. Etingof, Geometry & QFT, MIT 2002 online lecture notes; Chapter 3.


$^1$ For the perturbative expansion, see e.g. Ref. 1 or my Phys.SE answer here.

  • $\begingroup$ Where can i find the proof of these? $\endgroup$ Commented Apr 25, 2019 at 12:30

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