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In a six-dimensional cubic scalar field theory, three labelled box Feynman diagrams give the one-loop local quartic vertex at zero external momentum. With an effective-action term $g_4{\varphi }^4/4!$, their contribution is $\delta g_4=-3g^4{I}_4$. To check its sign and multiplicity, expand the one-loop scalar effective action as $\frac{1}{2}\mathrm{Tr}\log (D+g\varphi )$, where $D=-{\partial }^2+m^2$. Its fourth-order term is $-g^4\mathrm{Tr}(D^{-1}\varphi {)}^4/8$, giving the stated coefficient. The associated amputated connected diagram insertion has the opposite sign.

With $Z[J]=e^{iW[J]}$, the logarithm $W$ is the connected generating functional. Its derivative $\phi =\delta W/\delta J$ is the mean field. The specified Legendre transform satisfies $\delta \Gamma /\delta \phi =J$, and the quantum effective action generates amputated one-particle-irreducible vertices by further functional derivatives. In this question $\Gamma =-W+\int J\phi$ is the negative of the commonly used Minkowski effective-action convention; consequently its tree term is $-S[\phi ]$. Its diagrammatic content is unchanged if the convention is used consistently.

Expand the action around a stationary background in the presence of the source. For $\varphi =\phi +f$, integration by parts givesThe linear term vanishes at the tree-level saddle, $\delta S/\delta \phi +J=0$. Keeping the Gaussian fluctuation approximation, and fixing the field-independent normalization, givesto one-loop order. The shift from the tree saddle to the exact mean field cancels from the tree Legendre transform at this order. Hence the one-loop scalar effective action in the question's convention isThe functional determinant has a Feynman i-epsilon prescription before Wick rotation. Higher loops come from the omitted cubic and higher fluctuation vertices.

To see the one-loop one-particle-irreducible Feynman diagrams, write $\Delta ={\Delta }_0+U(\phi )$. Its determinant expansion isEach trace closes a single scalar loop, with propagators ${\Delta }_0^{-1}$ and background insertions $U$. No internal edge is a bridge, so these graphs are one-particle irreducible. Expanding the background factors produces their external legs.

For a constant $\phi$, set $A=V^{\prime \prime }(\phi )>0$. The quantum effective potential is defined here by $\Gamma [\phi ]=\int d^{d}x{V}_{\mathrm{eff}}(\phi )$. A plane-wave basis diagonalizes $\Delta$, and the supplied Wick rotation converts its correction per unit volume toHere the power of the renormalization scale keeps the continued integral in four-dimensional units. The Gamma function recurrence and expansion giveConsequentlyThe divergence is local, so subtract it with a potential counterterm. In the modified minimal subtraction scheme, remove its associated $\gamma -\log 4\pi$ constants as well. The one-loop scalar effective potential becomesup to an arbitrary field-independent vacuum-energy constant. Another allowed finite subtraction replaces the nonlogarithmic constant. Restoring a loop-counting parameter would multiply the displayed correction by $\hslash$. The logarithm assumes positive curvature $A$; a background with negative curvature requires an analytic continuation and can have an imaginary part signalling instability.

The quartic case is special because of quartic closure of scalar effective-potential counterterms. If $V$ has degree at most four, then $V^{\prime \prime }$ has degree at most two and $(V^{\prime \prime }{)}^2$ has degree at most four: its pole is removable by the original polynomial couplings and a constant. For example, $V=m^2{\phi }^2/2+\lambda {\phi }^4/4!$ givesThe last two terms reproduce the one-loop massive phi-fourth counterterms, and the first is a vacuum-energy subtraction. A degree-$r$ potential with $r>4$ instead generates degree $2r-4>r$, forcing new operators. This is the potential-level manifestation of power counting in quantum field theory and four-dimensional renormalizability, even though the finite effective potential itself need not remain polynomial.

Put $I_r={\int }_{\Lambda <|q|<{\Lambda }_0}{d}^{6}q(2\pi {)}^{-6}(q^2+m^2{)}^{-r}$, the scalar shell integral in six dimensions. Each of the three boxes is $g^4{I}_4$ as a connected insertion. To fix the effective-action sign and avoid a factorial ambiguity, use the one-loop scalar effective action:For a constant background the fourth-order density is $-g^4{I}_4{\varphi }^4/8$. Thus the cubic scalar box contribution to a quartic coupling is $\delta g_4=-3g^4{I}_4$, whose vertex insertion is $-\delta g_4=+3g^4{I}_4$.

Using the supplied area of the unit five-sphere,where $G(u)=u^{-1}-m^2{u}^{-2}+m^4/(3u^3)$. This follows by substituting $u=q^2+m^2$ and integrating $(u-m^2{)}^2/(2u^4)$. HenceAs a check, the massless limit at positive $\Lambda$ is $-3g^4({\Lambda }^{-2}-{\Lambda }_0^{-2})/(128{\pi }^3)$. Both versions have mass dimension minus two, as a six-dimensional local quartic coupling must.

For every interacting $n\ge 3$, the one-loop scalar effective action already contains counterexamples. On a constant background, with a positive auxiliary mass $\mu$ to control infrared divergences, its field-dependent part isThe term of order ${\lambda }^V$ has $E=(n-2)V$ external scalar fields and is a one-loop polygon one-particle-irreducible Feynman diagram. Choose $V>n/(n-2)$ and $2V>d$. Then $E>n$, its loop momentum integral is ultraviolet convergent, and its coefficient is nonzero. Removing the ultraviolet cutoff does not remove this finite higher-point scalar vertex. For $d>2$, generic nonexceptional external momenta give the same conclusion without retaining the auxiliary mass. In $d\le 2$, the massless theory still needs an infrared prescription; removing the ultraviolet cutoff does not remove that need or the induced vertices. Hence there are no interacting pairs $(n,d)$ under the literal wording. For $n=2$ the functional determinant is field independent; $n=1$ merely shifts a Gaussian integral when an infrared prescription exists.

There is a different conventional interpretation: absence of new independent divergent counterterms. The superficial degree of divergence of a connected diagram isThe standard perturbative counterterm closure criterion is $[\lambda ]\ge 0$, with the same pairs as in part ii, provided one includes all symmetry-allowed lower-degree potential terms, the kinetic term, and the vacuum energy. A pure monomial family need not itself have counterterm closure: a sextic interaction in three dimensions generates a quartic counterterm, and a quartic interaction generates a mass counterterm. This interpretation concerns the local divergent part or the continuum defining action, not the complete quantum effective action with its finite interaction vertices and derivative expansion.

For the massive four-dimensional quartic scalar field theory, write ${\lambda }_B$ for the bare quartic coupling, ${\lambda }_R$ for its renormalized value, and $m>0$ for the mass held fixed by a renormalization condition. Define the scalar bubble integralThe quartic one-particle-irreducible correlation function, defined as a derivative of the Euclidean quantum effective action, isThere are three bubble diagrams, one for each pairing of external momenta, and each has Feynman-diagram symmetry factor $2$. The minus sign follows equivalently from the quadratic term in the functional determinant $\frac{1}{2}\mathrm{Tr}\log [1+({\lambda }_B/2)(-{\partial }^2+m^2{)}^{-1}{\varphi }^2]$.

Radial integration at zero external momentum givesThe bare quartic vertex is not finite at fixed bare coupling. Its logarithmic ultraviolet divergence must be subtracted. Impose the momentum-subtraction scheme condition ${\Gamma }_R^{(4)}(0,0,0,0)={\lambda }_R$. To one loop this requiresFor the finite difference, a Feynman parameter and a shift of loop momentum giveThe boundary error due to shifting a sharp cutoff vanishes in this limit. Substitution yields the finite renormalized quartic scalar vertexThus finiteness requires holding the renormalized coupling fixed and allowing the bare coupling to depend on ${\Lambda }_0$. The unsubtracted assertion would be false.

The same one-loop scalar effective action explains the other effective interactions. For a constant background its expansion isThe $r=1$ tadpole diagram gives a mass correction $\delta m^2={\lambda }_R{I}_1/2$, withIt requires a mass counterterm. The field-independent vacuum energy also requires subtraction. The $r=2$ term is the quartic logarithm already treated. There is no one-loop wave-function renormalization from the quartic tadpole diagram.

For $r\ge 3$ the polygon diagrams generate finite higher-point scalar vertices, withFor example, the induced sextic term in the effective potential is $+{\lambda }_R^3{\varphi }^6/(768{\pi }^2{m}^2)$, or a six-point interaction vertex $15{\lambda }_R^3/(16{\pi }^2{m}^2)$ when normalized by $6!$. These finite terms survive the removal of the ultraviolet cutoff at fixed $m,{\lambda }_R$.

The external-momentum dependence of the bubble diagrams also generates a derivative expansion of quartic interactions. At small $P$, the subtracted scalar bubble integral is $-P^2/(96{\pi }^2{m}^2)+O(P^4/m^4)$, giving finite derivative couplings. At order ${\lambda }_R^2$ these are the new field-dependent interactions beyond the mass and quartic terms; the sextic and higher interaction vertices require higher powers of ${\lambda }_R$, though they still occur at one loop. Such coefficients are suppressed by powers of the physical mass or external momentum, not by powers of ${\Lambda }_0$. The calculation establishes an order-by-order perturbative limit; it does not establish a nonperturbative interacting continuum limit of a quantum field theory in four dimensions.