quantum field theory - Proof of Connected Diagrams

If $Z[J]$ is the generating functional for the path-integral, could any prove (or more reasonably, refer me to a proof) that

$$W[J]\equiv\frac{\hbar}{i}\log\left(Z[J]\right)$$ "generates" only connected diagrams?

So far I've only seen theory-dependent "examples" (basically showing how in $\phi^4$ theory the two-point function from $W$ gives only connected contributions).

I'm looking for a generic systematic proof for a general field theory.

Answer

The logarithmic relationship is equivalent to

$$Z[J]=\exp[iW[J]]$$ where $W$ is the sum of connected diagrams. This formula is trivial to prove via Taylor expansion of the exponential $$\exp(X) = \sum_{n=0}^\infty \frac{X^n}{n!}$$ If we substitute $i$ times the sum of all connected diagrams $iW$ for $X$ in this formula, the term $X^n/n!$ will simply produce the products of $n$ components, i.e. all disconnected (for $n\gt 2$) diagrams with $n$ components.

The combinatorial factor will work, too. Recall that when we evaluate Feynman diagrams, we have to divide by the symmetry factor. The symmetry group of a disconnected, $n$-component diagram includes the permutation group of all the $n$ components if the components are the same, that's why there is $1/n!$ in front of a "fixed single 1-component diagram" to the $n$-th power.

The extra symmetry group from permuting the components is reduced to the product of $n_i!$ over all subgroups of the group of $n$ components that contain the same diagram. But

$$\prod_i \frac{1}{n_i!}$$ is exactly what we get if we calculate $1/n!$ times the coefficient from the expansion of the $n$th-power of the sum of the connected diagrams.