How to prove that $\ln(Z(J))$ generates only connected Feynman diagrams? I can't find the proof of this statement, and have only met its demonstrations for case of 2- and 4-point.
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$\begingroup$ math.upenn.edu/~wilf/DownldGF.html $\endgroup$– Ryan ThorngrenCommented Mar 1, 2018 at 5:46
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2 Answers
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Assume that the generating functional is given by a sum of all possible diagrams, i.e.
$$Z(J)=\sum_{n_i} D_{n_i}.$$
Furthermore, assume that each diagram D is given by a product of connected diagrams $C_i$, i.e. a diagram D can be disconnected. We will write this as
$$D_{n_i}=\prod_i\frac{1}{n_i!}C_i^{n_i},$$
where dividing by $n_i!$ amounts for a symmetry factor coming from exchanges of propagators and vertices between different diagrams. Combining this with our first expression, we get
$$Z(J)=\sum_{n_i}\prod_i\frac{1}{n_i!}C_i^{n_i}.$$
With some manipulation, this can be shown to be equivalent to
$$Z(J)=\exp\left(\sum_i C_i\right).$$
Taking the logarithm on both sides gives you the desired expression.
AccidentalFourierTransform
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answered Apr 5, 2014 at 23:19
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2$\begingroup$ Could you spell out the manipulations to get the last line? $\endgroup$– lalalaCommented Mar 1, 2018 at 6:25
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1$\begingroup$ Explain last step a little bit more @Frederic Brunner $\endgroup$ Commented Apr 30, 2020 at 7:42
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$\begingroup$ It looks like you have made a mistake in your second formula. $i$ is defined in the sum on the RHS, so it shouldn't exist on the LHS as well. Unless these are two different $i$s, but in that case that's very confusing. $\endgroup$ Commented Nov 22, 2021 at 5:48
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An intuitive interpretation from Timo Weigand's lecture notes:
Suppose $iW[J]$ contains all connected diagrams, then all possible connected and disconnected diagrams can be showed as products of $iW[J]$:
$$ \frac{Z[J]}{Z[0]} = 1 + iW[J] + \frac{1}{2!} {(iW[J])}^2 + \frac{1}{3!} {(iW[J])}^3 + ... = e^{iW[J]} $$
So
$$ iW[J] = ln \frac{Z[J]}{Z[0]} $$
This interpretation is just the same as Frederic's answer, but expressed in reverse order.
