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.


2 Answers 2


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


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


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.

  • 2
    $\begingroup$ Could you spell out the manipulations to get the last line? $\endgroup$
    – lalala
    Mar 1, 2018 at 6:25
  • 1
    $\begingroup$ Explain last step a little bit more @Frederic Brunner $\endgroup$
    Apr 30, 2020 at 7:42
  • $\begingroup$ 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. $\endgroup$ Nov 22, 2021 at 5:48

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 $iW[J]$:

$$ \frac{Z[J]}{Z[0]} = 1 + iW[J] + \frac{1}{2!} {(iW[J])}^2 + \frac{1}{3!} {(iW[J])}^3 + ... = e^{iW[J]} $$


$$ iW[J] = ln \frac{Z[J]}{Z[0]} $$

This interpretation is just the same as Frederic's answer, but expressed in reverse order.


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