# Combinatorics geometric series two-point function

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?

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 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.

References:

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

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

• Where can i find the proof of these? Apr 25, 2019 at 12:30