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Use Minkowski spacetime with signature $(+,-,\dots ,-)$ and $\hslash =1$, with a vacuum time-ordering prescription. At a regulator, define the normalized generating functionalThe normalized vacuum correlation functions of a time-ordered product areThe connected generating functional isThe logarithm selects connected source diagrams, or equivalently cumulants. With the assumed vanishing one-point function, derivatives of $Z=e^{iW}$ give the centered four-point cumulant decomposition:In detail, four differentiations either act on one connected block or form one of three pair partitions. Contributions with blocks of size one vanish. If $W_{ij}$ and $W_{1234}$ denote source derivatives at zero, the same relation is $G_{1234}=i{W}_{1234}-W_{12}{W}_{34}-W_{13}{W}_{24}-W_{14}{W}_{23}$, since $C_2=-i{W}_2$ and $C_4=i{W}_4$.

Let $\phi (x)=\delta W/\delta J(x)$ be the source-dependent mean field. On a locally invertible source-to-field branch, define the quantum effective action by the Minkowski Legendre transformIts variation is $\delta \Gamma =-\int J\delta \phi$, so $\delta \Gamma /\delta \phi =-J$. Differentiating the two inverse source-field maps yields the Minkowski inverse-Hessian relation for an effective action:Thus ${\Gamma }^{(2)}=-(W^{(2)}{)}^{-1}$. The inverse is understood at the regulator and on a nonsingular fluctuation sector; it is not an assertion that every source-field map is globally invertible.

For a free real scalar field, integrate by parts to writewhere repeated spacetime variables are integrated. Specify the propagator convention explicitly:Completing the Gaussian path integral around $\varphi =-A_{ϵ}^{-1}J$ cancels the source-independent determinant between numerator and denominator. ThereforeThe positive imaginary term in $A_{ϵ}$ damps the oscillatory Gaussian. Together with vacuum projection at the time boundaries, it selects the Feynman poles and the time-ordered, in-out correlation function rather than a retarded inverse. It can also be obtained by continuation from the Euclidean vacuum integral. A denominator convention without the numerator $i$ redistributes the displayed factors of $i$.

Finally $\phi =-A_{ϵ}^{-1}J$, hence $J=-A_{ϵ}\phi$. Substitution into the Legendre transform gives the vacuum-normalized effective action of a free scalar field:All connected functions beyond order two vanish and all proper vertices beyond the classical quadratic kernel vanish. Vacuum normalization fixes the additive constant; without it, the quantum effective action equals the classical action up to a field-independent determinant contribution.