= Solution
Use <Minkowski spacetime> with signature $(+,-,\ldots,-)$ and $\hbar=1$, with a vacuum time-ordering prescription. At a regulator, define the normalized <generating functional>
$$
Z[J]=\frac{\int\mathcal D\phi\;e^{i(S[\phi]+\int J\phi)}}{\int\mathcal D\phi\;e^{iS[\phi]}},\qquad Z[0]=1.
$$
The normalized vacuum <correlation functions> of a <time-ordered product> are
$$
G_n(x_1,\ldots,x_n)=\left.\frac1{i^n}\frac{\delta^n Z}{\delta J(x_1)\cdots\delta J(x_n)}\right|_{J=0}.
$$
The <connected generating functional> is
$$
\boxed{W[J]=-i\log Z[J],\qquad
C_n=\left.i^{1-n}\frac{\delta^n W}{\delta J(x_1)\cdots\delta J(x_n)}\right|_{J=0}.}
$$
The 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>:
$$
\boxed{G_{1234}=C_{1234}+C_{12}C_{34}+C_{13}C_{24}+C_{14}C_{23}.}
$$
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 $G_{1234}=iW_{1234}-W_{12}W_{34}-W_{13}W_{24}-W_{14}W_{23}$, since $C_2=-iW_2$ and $C_4=iW_4$.
Let $\varphi(x)=\delta W/\delta J(x)$ be the source-dependent mean field. On a locally invertible source-to-field branch, define the <quantum effective action> by the Minkowski <Legendre transform>
$$
\boxed{\Gamma[\varphi]=W[J]-\int d^dx\,J(x)\varphi(x).}
$$
Its variation is $\delta\Gamma=-\int J\delta\varphi$, so $\delta\Gamma/\delta\varphi=-J$. Differentiating the two inverse source-field maps yields the <Minkowski inverse-Hessian relation for an effective action>:
$$
\boxed{\int d^dz\,\frac{\delta^2\Gamma}{\delta\varphi(x)\delta\varphi(z)}
\frac{\delta^2W}{\delta J(z)\delta J(y)}=-\delta^{(d)}(x-y).}
$$
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 write
$$
S[\phi]=\frac12\phi A\phi,\qquad A=-\Box-m^2,
\qquad A_\epsilon=A+i\epsilon,
$$
where repeated spacetime variables are integrated. Specify the <propagator> convention explicitly:
$$
\Delta_F(x-y)=\int\frac{d^dp}{(2\pi)^d}\frac{i\,e^{-ip(x-y)}}{p^2-m^2+i0},\qquad \Delta_F=iA_\epsilon^{-1}.
$$
Completing the <Gaussian path integral> around $\phi=-A_\epsilon^{-1}J$ cancels the source-independent <determinant> between numerator and denominator. Therefore
$$
Z[J]=\exp\left[-\frac i2JA_\epsilon^{-1}J\right],\qquad
\boxed{W[J]=-\frac12JA_\epsilon^{-1}J=\frac i2J\Delta_FJ.}
$$
The positive imaginary term in $A_\epsilon$ 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 $\varphi=-A_\epsilon^{-1}J$, hence $J=-A_\epsilon\varphi$. Substitution into the <Legendre transform> gives the <vacuum-normalized effective action of a free scalar field>:
$$
\boxed{\Gamma[\varphi]=\frac12\varphi A_\epsilon\varphi\longrightarrow S[\varphi]\quad(\epsilon\downarrow0).}
$$
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.
Back to article page
