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I'm studying QFT from Le Bellac's book, but I can't understand very well his proof for the generator of proper vertices. Can someone give a more readable and/or understandable proof?

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How does he show that that is the generator of disconnected diagrams? Ryder, for example, works out the 4point generator diagrammatically to show how the disconnected parts cancel. – innisfree May 12 '13 at 16:55

Very rougly, the logarithm has the function of transforming a product into a sum, therefore if the partition function is factorizable (i.e. disconnected), it will give a sum on connected parts.

For a slighly more rigorous discussion, consider the Hamiltonian of two non interacting systems, the partition function $z(j)=Z(j)/Z(0)$ (without vacuum fluctuactions) is obviously the product of the single partition functions.


Consider now the partition function generated by the connected diagrams $W(j)$, it is simply the sum of the the two connected partition functions, because if you want to create a connected diagram you can't select diagrams from both systems, since they are not interacting the diagram you created is disconnected.


Then, we have the fact that the partition function is function only of the connected diagrams, since any non-connected diagram can be formed by joining two connected diagrams, then $Z(j)$ is a function of $W(j)$, that we will note as $Z(W(j))$.

From the previous discussion we have:

$z(W(j)) = z(W_1(j)+W_2(j)) = z(W_1(j))z(W_2(j))$

the only function that satisfies this identity is the exponential, as you can easily check, so

$z(W(j))=exp(a W(j))$

expanding around $j=0$ gives $a=1$.

I couldn't find a proof simpler that the one given in Lebellac for the question you asked, you might try for example this Zinn-Justin

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This doesn't actually relate to the question. The OP is asking about the 1PI effective action. – Michael Brown May 12 '13 at 23:50
You are right, I'm going to provide the complete answer – Ikiperu May 13 '13 at 3:32

try the lecture notes by van Hees from Frankfurt. Check out p. 114

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Hi user24409, and welcome to Physics Stack Exchange! Although this may theoretically answer the question, it would be preferable to include the essential parts of the answer here, and provide the link only for further reading. – David Z May 12 '13 at 17:50

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