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The partition function of QFT is defined as

$$Z=\int\mathcal{D}\varphi e^{iS[\varphi]}.$$

Now, it is a general fact that this formal path integral can be computed perturbatively as (sketchy)

$$Z=\sum_{\Gamma}\frac{\lambda^{V}}{\mathrm{sym}(\Gamma)}\mathcal{A}_{\Gamma}.$$

where the sum is over all Feynman amplitudes of closed (=vacuum) Feynman graphs and where $\lambda$ denotes the coupling. This equality is obtained by perturbing around the coupling $\lambda$ and computing Gaussian integrals, i.e.

$$Z=\int\mathcal{D}\varphi e^{iS[\varphi]}=\int\mathcal{D}\varphi e^{iS_{\mathrm{kin}}[\varphi]}\bigg (\sum_{k}\frac{1}{k!}\bigg[i\int\,\mathrm{d}^{4}x\mathcal{L}_{\mathrm{int}}\bigg ]^{k}\bigg )=\\\sum_{k}\int\,\mathrm{d}^{4}z_{1}\dots\mathrm{d}^{4}z_{k}\bigg [\text{apply propagator pairwise to fields }\mathcal{L}_{\mathrm{int}}(\varphi(z_{k}))\dots\mathcal{L}_{\mathrm{int}}(\varphi(z_{1}))\bigg ]=\sum_{\Gamma}\frac{\lambda^{V}}{\mathrm{sym}(\Gamma)}\mathcal{A}_{\Gamma}.$$

Now, in many text i have seen the claim that the last equality is only true for the free energy, i.e. the claim that

$$F=-i\cdot\mathrm{ln}(Z)=\sum_{\Gamma}\frac{\lambda^{V}}{\mathrm{sym}(\Gamma)}\mathcal{A}_{\Gamma}.$$

I am confused, which formula is correct? The formula for $Z$ or for $F$?

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    $\begingroup$ Could you provide a reference for this claim? My immediate guess is that there is an implicit difference: $Z$ is given by the sum over all diagrams while $F$ is given by the sum over all connected diagrams $\endgroup$
    – Níckolas Alves
    Commented Dec 10, 2021 at 15:15
  • $\begingroup$ I can provide references, but I do not know if this makes sense. The point is that I wrote my question in a more general fashion, such that it is understanable from a more general context. However, the question arises from some special quantum gravity models, like tensor models and GFTs. But if you want, the second claim of the free energy is for example contained in arxiv.org/abs/1109.4812 ... $\endgroup$
    – B.Hueber
    Commented Dec 10, 2021 at 15:19
  • $\begingroup$ But even if this is true that the second formula is only for connected diagrams, I can't see why this follows just by putting a logarithm in front... $\endgroup$
    – B.Hueber
    Commented Dec 10, 2021 at 15:20
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    $\begingroup$ Related: physics.stackexchange.com/q/107049/2451 , physics.stackexchange.com/q/324252/2451 and links therein. $\endgroup$
    – Qmechanic ♦
    Commented Dec 10, 2021 at 15:22
  • $\begingroup$ Thanks @Qmechanic! Thats indeed very helpful :-) $\endgroup$
    – B.Hueber
    Commented Dec 10, 2021 at 15:24

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