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Split the free interaction picture field as $\varphi ={\varphi }^{(+)}+{\varphi }^{(-)}$, where the first part contains annihilation operators and the second creation operators. Time ordering places the operator at the later time on the left, while normal ordering places all creation operators on the left. For $x^0>y^0$, commuting the annihilation part of the first field through the creation part of the second givesThe commutator is the c-numberFor $y^0>x^0$, the same calculation interchanges $x,y$. Combining the two time orders proves the two-field Wick contraction identitywhereThe equality with the four-dimensional integral follows by closing the energy contour at the positive-energy pole for $x^0>y^0$ and the negative-energy pole for the reverse order. This fixes the scalar Feynman propagator pole prescription; in this convention $D_F$ itself includes the factor $i$.

Because these interactions have no derivatives, $H_{\mathrm{int}}=-\int {\mathcal{L}}_{\mathrm{int}}{d}^{3}x$. Expand the Dyson seriesApply the Wick theorem to each term together with the external insertions. Every paired field contributes a propagator, and the remaining normal-ordered vacuum expectation vanishes unless no fields remain. Representing the propagators by Fourier integrals and integrating each vertex position gives the four-momentum conservation delta function. The $3!$ and $4!$ contractions into labelled vertex slots cancel the factorials in the interaction density. The resulting scalar-field cubic and quartic Feynman vertices areEach vertex also has $(2\pi {)}^4{\delta }^{(4)}(\sum p)$, each independent internal momentum is integrated with $d^{4}p/(2\pi {)}^4$, and repeated equivalent contractions give the usual Feynman-diagram symmetry factor. For the unamputated correlation functions, retain the propagators joining external insertions to vertices. The displayed unnormalized numerator includes disconnected vacuum bubbles; dividing by $⟨0|S|0⟩$ cancels them. Amputated scattering amplitudes additionally use the LSZ reduction formula.

For stability, examine the classical scalar potentialThe tree-level boundedness criterion is $\lambda >0$, with arbitrary real cubic coupling. If $\lambda =0$, boundedness requires $\mu =0$ and $m^2\ge 0$; if $\lambda <0$, or $\lambda =0$ with $\mu \ne 0$, the potential is unbounded below. Thus a positive quartic coupling permits a stable ground-state vacuum, although it need not be the perturbative vacuum at $\varphi =0$.

For $\lambda >0$ and $m^2\ge 0$, the stability of a cubic-quartic scalar potential determines when zero is a global minimum:The square bracket is nonnegative for every field value exactly whenFor a nontrivial equality case there is a second degenerate minimum at $\varphi =2\mu /\lambda$. If $m^2>0$ and ${\mu }^2>3\lambda m^2$, zero remains a local minimum but is a false vacuum, with a lower minimum elsewhere. If $m^2=\mu =0$ and $\lambda >0$, the origin is the quartic minimum despite its vanishing quadratic curvature. These are classical or tree-level stability statements; renormalized couplings and the quantum effective potential refine the quantum analysis.

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.