In deriving the expression for the exact propagator


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


is the free propagator and


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


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


we would have


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.


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).

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

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