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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.

Qmechanic
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Andrew McAddams
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2 Answers2

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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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Frederic Brünner
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    Could you spell out the manipulations to get the last line? – lalala Mar 01 '18 at 06:25
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    Explain last step a little bit more @Frederic Brunner – ROBIN RAJ Apr 30 '20 at 07:42
  • 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. – HelloGoodbye Nov 22 '21 at 05:48
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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.

Sean
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