# Question on irreducible versus reducible feynman graphs

 P: 969 Consider the functional: $$(1) \mbox{ }e^{iW[J]} = \int d \hat{\phi} \mbox{ }e^{i\int d^4x \mbox{ } \mathcal L(\hat{\phi})+J\hat{\phi}}$$ Define a Legendre transformation to get a functional in $$\phi(x)$$ instead of $$J(x)$$: $$(2) \mbox{ }\Gamma[\phi]=W[J(\phi)]- \int d^4x \mbox{ } J(\phi) \phi$$ where $$J(\phi)$$ is found by solving $$\frac{\partial W[J]}{\partial J}=\phi$$ for J in terms of $$\phi$$ and substituting this expression in for the value $$J(\phi)$$. Also, by differentiating eqn (2) with respect to $$\phi$$, one can show: $$\frac{\partial \Gamma[\phi]}{\partial \phi}+J(\phi)=0$$ To calculate $$\Gamma[\phi]$$ by diagrammatic methods instead, exponentiate it and substitute the earlier result for $$e^{iW[J]}$$: $$(3) \mbox{ } e^{i\Gamma[\phi]}= e^{i(W[J(\phi)]- \int d^4x \mbox{ } J(\phi) \phi )} =\int d \hat{\phi} \mbox{ }e^{i\int d^4x \mbox{ } \mathcal L(\hat{\phi})+J(\phi)(\hat{\phi}-\phi)}$$ Now here is what I don't understand. The author of the paper now says: "A saddle-point evaluation of eqn. (1) gives W[J] as the sum of all connected graphs that are constructed using vertices and propagators built from the classical lagrangian, L, and having the currents, J, as external lines. But $$\Gamma[\phi]$$ just differs from W[J] by subtracting $$\int d^4x \mbox{ } J\phi$$, and evaluating the result at the specific configuration $$J(\phi) = -\frac{\partial \Gamma}{\partial \phi}$$. This merely lops off all of the 1-particle reducible graphs, ensuring that $$\Gamma[\phi]$$ is given by summing 1-particle irreducible graphs." How does one see that adding all irreducible graphs is equivalent to evaluating eqn. (3)? In other words, how does doing all that "merely lops off all the 1-particle reducible graphs"?