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Sketched proof:

1. In general we know that a connected diagram is a tree of bare propagators $G_0$ and (amputated) 1PI vertices, cf. Lemma 3.11 in Ref. 1.

2. In particular, the full propagator/connected 2-pt function $G_c$ must be strings of bare propagators $G_0$ and (amputated) 2-pt vertex $\Sigma\equiv \Pi$, which we call self-energy.

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

4. We can isolate the (amputated) 2-pt vertices in eq. (A) $$\Sigma~=~G_0^{-1}-G_c^{-1}.\tag{B}$$

5. In general the self-energy $\Sigma$ consists of connected diagrams with 2 amputated legs such that the 2 legs cannot be disconnected by cutting a single internal line.

6. If there are no tadpoles, the self-energy $\Sigma$ is 1PI, cf. my Phys.SE answer here.

References:

1. P. Etingof, Geometry & QFT, MIT 2002 online lecture notes; Sections 3.11 & 3.12.

Sketched proof:

1. In general we know that a connected diagram is a tree of bare/free propagators $G_0$ and (amputated) 1PI vertices, cf. Lemma 3.11 in Ref. 1.

2. In particular, the full propagator/connected 2-pt function $G_c$ must be strings of bare/free propagators $G_0$ and (amputated) 2-pt vertex $\Sigma\equiv \Pi$, which we call self-energy.

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

4. We can isolate the (amputated) 2-pt vertices in eq. (A) $$\Sigma~=~G_0^{-1}-G_c^{-1}.\tag{B}$$

5. In general the self-energy $\Sigma$ consists of connected diagrams with 2 amputated legs such that the 2 legs cannot be disconnected by cutting a single internal line.

6. If there are no tadpoles, the self-energy $\Sigma$ is 1PI, cf. my Phys.SE answer here.

References:

1. P. Etingof, Geometry & QFT, MIT 2002 online lecture notes; Sections 3.11 & 3.12.

Sketched proof:

1. In general we know that a connected diagram is a tree of bare propagators $G_0$ and (amputated) 1PI vertices, cf. Lemma 3.11 in Ref. 1.

2. In particular, the full propagator/connected 2-pt function $G_c$ must be strings of bare propagators $G_0$ and (amputated) 2-pt vertex $\Sigma\equiv \Pi$, which we call self-energy.

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

4. We can isolate the (amputated) 2-pt 1PI vertices in eq. (A) $$\Sigma~=~G_0^{-1}-G_c^{-1}.\tag{B}$$

5. In general the self-energy $\Sigma$ consists of connected diagrams with 2 amputated legs such that the 2 legs cannot be disconnected by cutting a single internal line.

6. If there are no tadpoles, the self-energy $\Sigma$ is 1PI, cf. my Phys.SE answer here.

References:

1. P. Etingof, Geometry & QFT, MIT 2002 online lecture notes; Sections 3.11 & 3.12.

Sketched proof:

1. In general we know that a connected diagram is a tree of bare propagators $G_0$ and (amputated) 1PI vertices, cf. Lemma 3.11 in Ref. 1.

2. In particular, the full propagator/connected 2-pt function $G_c$ must be strings of bare propagators $G_0$ and (amputated) 2-pt vertex $\Sigma\equiv \Pi$, which we call self-energy.

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

4. We can isolate the (amputated) 2-pt vertices in eq. (A) $$\Sigma~=~G_0^{-1}-G_c^{-1}.\tag{B}$$

5. In general the self-energy $\Sigma$ consists of connected diagrams with 2 amputated legs such that the 2 legs cannot be disconnected by cutting a single internal line.

6. If there are no tadpoles, the self-energy $\Sigma$ is 1PI, cf. my Phys.SE answer here.

References:

1. P. Etingof, Geometry & QFT, MIT 2002 online lecture notes; Sections 3.11 & 3.12.

Sketched proof:

1. In general we know that a connected diagram is a tree of bare propagators $G_0$ and (amputated) 1PI vertices, cf. Lemma 3.11 in Ref. 1.

2. In particular, the full propagator/connected 2-pt function $G_c$ must be strings of bare propagators $G_0$ and (amputated) 2-pt 1PI vertices $\Sigma\equiv \Pi$.

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

4. We can isolate the (amputated) 2-pt 1PI vertices in eq. (A) $$\Sigma~=~G_0^{-1}-G_c^{-1},\tag{B}$$ which shows that $\Sigma$ is the self-energy. $\Box$

References:

1. P. Etingof, Geometry & QFT, MIT 2002 online lecture notes; Sections 3.11 & 3.12.

Sketched proof:

1. In general we know that a connected diagram is a tree of bare propagators $G_0$ and (amputated) 1PI vertices, cf. Lemma 3.11 in Ref. 1.

2. In particular, the full propagator/connected 2-pt function $G_c$ must be strings of bare propagators $G_0$ and (amputated) 2-pt vertex $\Sigma\equiv \Pi$, which we call self-energy.

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

4. We can isolate the (amputated) 2-pt 1PI vertices in eq. (A) $$\Sigma~=~G_0^{-1}-G_c^{-1}.\tag{B}$$

5. In general the self-energy $\Sigma$ consists of connected diagrams with 2 amputated legs such that the 2 legs cannot be disconnected by cutting a single internal line.

6. If there are no tadpoles, the self-energy $\Sigma$ is 1PI, cf. my Phys.SE answer here.

References:

1. P. Etingof, Geometry & QFT, MIT 2002 online lecture notes; Sections 3.11 & 3.12.