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The disjoint Fourier supports make the quadratic cross term $\int \partial \varphi \cdot \partial \chi$ vanish. ThereforewhereDefine the high-mode free generating functional with source conventionThen inserting $\chi =-\delta /\delta J$ reproduces every high-field factor, andExpanding the interaction exponential and applying Wick theorem evaluates the Wilsonian effective action as a sum of connected diagrams whose internal lines are restricted to the high-momentum shell.

An interaction depending on the field can be represented by substituting $(1/i)\delta /\delta J$ into its action and acting on the free generating functional. Expanding that exponential produces Feynman diagrams, contractions and vertex weights. Normalize by the resulting value at $J=0$ to cancel vacuum diagrams.

In a finite-dimensional integral, $u_i{e}^{b\cdot u}={\partial }_{b_i}{e}^{b\cdot u}$. Expand $e^{V(u)}$ as a formal perturbation series to obtain $\int du{e}^{-u^{T}Au/2+b\cdot u+V(u)}=e^{V({\partial }_b)}{I}_0(b)$. With a finite-mode regulator, a Minkowski source insertion satisfies $\varphi (x)e^{i\int J\varphi }=(1/i)\delta e^{i\int J\varphi }/\delta J(x)$; this gives the displayed interacting generating functional. The substitution is a perturbative differential-operator identity, not an assertion that the interaction exponential defines a convergent series. Divide the resulting functional by $Z[0]$ for normalized correlation functions.

An external source is a prescribed field coupled to a dynamical field or operator in the action, for example $S_J=S_0+\int J_i(x){\varphi }_i(x)d^{d}x$. Functional derivatives of the generating functional with respect to $J$ insert the corresponding fields into correlation functions. The source is varied to generate these insertions, rather than integrated over as a dynamical degree of freedom. Transforming it in the appropriate representation maintains an invariant source coupling when deriving a Ward identity.

For $d=4-ϵ$, write the bare field as ${\varphi }_0=Z^{1/2}\varphi$ and the renormalized source coupling as ${\mu }^{-ϵ/2}h\varphi$. In bare variables this is $h_0{\varphi }_0$, with the displayed source. At fixed bare parameters, the source running is $\mu d\log h/d\mu =ϵ/2+\frac{1}{2}\hat{\beta }{\partial }_{\lambda }\log Z$. Normalizing the generating functional by its zero-source value cancels source-independent vacuum diagrams and the constant measure factor from field rescaling. It does not remove source-dependent ultraviolet divergences or infrared divergences; the former require the ordinary field, mass and interaction counterterms.

Use the Minkowski generating functional $Z=e^{iW}$, source $+J\varphi$, mean field $\phi =\delta W/\delta J$, and quantum effective action $\Gamma =W-\int J\phi$. Then $W^{(2)}=i{G}_2^c$, $W^{(3)}=-G_3^c$, and ${\Gamma }^{(2)}=-(W^{(2)}{)}^{-1}$. Differentiating this inverse gives ${\Gamma }^{(3)}=(W^{(2)}{)}^{-1\otimes 3}{W}^{(3)}=-i(G_2^c{)}^{-1\otimes 3}{G}_3^c$, with integration over the repeated spacetime arguments. Thus amputation of external propagators gives $i{\Gamma }^{(3)}$ in this convention. The full connected two-point functions, rather than bare propagators, perform the exact amputation.

A translation-invariant connected generating functional with a constant external source is extensive in spacetime volume ${\mathcal{V}}_d$. Its total logarithm is dimensionless, while its density has mass dimension $d$. Consequently its dimensionless representation at fixed physical volume must retain ${\mu }^d{\mathcal{V}}_d$. For a free massive scalar with source coupling ${\mu }^{-ϵ/2}h\varphi$, completing the square gives $F={\mathcal{V}}_d{\mu }^{-ϵ}{h}^2/(2m^2)$. This already disproves a literal rewriting of the full uniform-source logarithm using only $\lambda ,m^2/{\mu }^2,h/{\mu }^3$ at fixed volume. In infinite volume the total functional is generally infinite; use the density, a source of compact support, or an explicit volume regulator. After writing the density as ${\mu }^{d}f$, its renormalization group equation has the engineering term $+df$. Ultraviolet subtraction alone does not establish infrared finiteness for a massless constant source.