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I will refomulate my question(Geometric series for two-point function) because it seems that i did not make it clear. In order to have

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

the reducible diagram must be of some form. for example the 2 loop in $\phi$ theory is of the form

png

and the 3 loop is

png

how can we prove that there is not multiplicative constant say for example a 7 times the 2 loop and a 8 times the 3 loop that would invalidate the geometrical series?

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Paul Dirac
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  • Where exactly do you think a factor of 7 or 8 could come from? Feynman diagrams don't get extra factors for no reason. – knzhou Aug 04 '16 at 06:32
  • for example there are a symmetry factor in a 1 loop diagram. how can we prove that the symmetry factor of a two loop is a multiplication of two symmetry factor of one loop – Paul Dirac Aug 04 '16 at 06:37
  • If the one-loop diagram has symmetry factor $N$, each half of the two-loop diagram has $N$ equivalent contractions, so a symmetry factor of $N^2$. You don't get any 'extra' equivalent contractions, because the two halves are connected by a single propagator, which has symmetry factor $1$. – knzhou Aug 04 '16 at 06:39
  • where i can get the prove of this – Paul Dirac Aug 04 '16 at 06:41
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    This question is essentially a duplicate of http://physics.stackexchange.com/q/272505/2451 – Qmechanic Aug 04 '16 at 07:47
  • This is still the same question, isn't it? If you are unhappy with the answer given, update that question - please don't post revised versions of the same question. – sammy gerbil Aug 04 '16 at 18:43

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One way to find out what the symmetry factors (or whatever other factors) are with which different diagrams should be multiplied is to expand the Feynman path integral. One would add source terms for all the fields in the Lagrangian. Then one can pull all the interaction terms out of the path integral by replacing the fields in these terms with functional derivatives with respect to the sources. What remains inside the integrals are only the kinetic terms. These can be integrated out by completing the squares, so that only the propagators, connected with sources, remain. The latter now serve as generating functions on which the exponential function containing the interaction terms act. One can expand these exponential functions and then let the individual terms act on the generating function containing the propagators. In this way one would obtain all the possible Feynman diagrams together with the correct numerical (symmetry) factors.

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