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Consider the following system: the SM lagrangian somewhat below the EW transition, where we keep only bilinear terms, only the heaviest fermion -- $t$-quark, and plus the potential terms for a VEV $\phi$: $$ \mathcal{L} = \mathcal{L}_{\text{kinetic}}-V(\phi) - \frac{(m_{h}^{\phi})^{2}}{2}h^{2} + (m_{W}^{\phi})^{2}|W|^{2}+\frac{(m_{Z}^{\phi})^{2}}{2}Z^{2} - m_{t}^{\phi}\bar{t}t, $$ where $m_{X}^{\phi} \sim \phi$ is mass.

Having this Lagrangian, by integrating out fields $h,W,Z,t$ one may compute the thermal one-loop one-particle irreducible effective action $\Gamma[\phi] = \int d^{4}x V_{\text{eff}}(\phi)$. After tediuos dealing with thermal QFT, one arrives just to the conclusion that $$ \tag 1 V_{\text{eff}}(\phi) = V(\phi) - \sum_{i}P_{i}, $$ where $P_{i}$ denotes pressure of $i$-th species (free Fermi/Bose particles). My question is the following: is it possible to predict the result $(1)$ without calculating $\Gamma$ explicitly?

My attempt.

I expect that this follows from thermodynamic relations. E.g., $$ P =T\frac{\partial\log(Z)}{\partial V}, $$ where $Z$ is the partition function $Z = \text{Tr}[e^{-H - \mu N}]$ The only question is how to relate $Z$ to $\Gamma$. Let us introduce the generating functional $Z$: $$ \mathcal{Z}[j] = \text{Tr}[T_{c} e^{-H - \mu N}e^{i\int d^{4}x\phi j}] $$ We have $$ \mathcal{Z}[j] = \exp[-W[j]], \quad W[j] = \Gamma[\phi] - \int d^{4}x \frac{\delta W}{\delta j}j $$ It seems to be $$ Z = \left( \mathcal{Z}[j]\right)_{j = 0} = e^{-\Gamma[\phi]} $$ Is it correct?

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