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In this answer Proof of geometric series for connected 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?

Qmechanic
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amilton moreira
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1 Answers1

1

  1. Firstly, let us 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, in the geometric series mentioned by OP $$ G_c~=~G_0\sum_{n=0}^{\infty}(\Sigma G_0)^n.\tag{A}$$

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

    • 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.

References:

  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.

Qmechanic
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