# Geometric series for two-point function

In deriving the expression for the exact propagator

$$G_c^{(2)}(x_1,x_2)=[p^2-m^2+\Pi(p)]^{-1}$$

for $\phi^4$ theory all books that i know use the following argument:

$$G_c^{(2)}(x_1,x_2)=G_0^{(2)}+G_0^{(2)}\Pi G_0^{(2)}+G_0^{(2)}\Pi G_0^{(2)}\Pi G_0^{(2)}+...$$

wich is a geometric series so the formula for the exact propagator.Here

$$G_0^{(2)}$$

is the free propagator and

$$\Pi=X+Y+Z+...+W$$

is the sum of all irreducible diagrams.Here the irreducible diagrams is represented by $X,Y,Z,...,W$.

Using the path integral i can see, that connected diagrams $D$ can be written in the form

$$D=G_0^{(2)}XG_0^{(2)}Z...G_0^{(2)}W$$

Question: But how to prove that there is no constant $C$ so that instead we have

$$D=G_0^{(2)}XG_0^{(2)}Z...G_0^{(2)}W$$

we would have

$$D=CG_0^{(2)}XG_0^{(2)}Z...G_0^{(2)}W~?$$

In the last case we would not have a geometrical serie. Can someone explain me i t please or give me another way to derive it.

• – Qmechanic Apr 25 at 12:47

It is probably not useful to think of $\Pi$ in terms of the separate diagrams. If you keep the $\Pi$ as it is you can see that the full propagator is like the summation of a geometric series, but where one treats $G_0$ and $\Pi$ like operators.

Take your second expression and multiply it by $G_0\Pi$. Then subtract this result from the second expression. The resulting equation reads

$G_c - G_0\Pi G_c = G_0 .$

Now one can operate on both sides with $(1-G_0\Pi)^{-1}$ to get

$G_c = (1-G_0\Pi)^{-1} G_0 = (G_0^{-1}-\Pi)^{-1} .$

Since $G_0^{-1}=p^2-m^2$, you get back the full propagator (accept for a minus sign in front of $\Pi$, which I think is a typo in your question).

• My main problem is how to problem is how to prove that reducible diagrams can be obtained from irreducible diagrams – Paul Dirac Aug 4 '16 at 5:28
• What i want is to prove that the second equation is correct – Paul Dirac Aug 4 '16 at 5:37