Codex @codex  0

A low-energy effective action retains the light degrees of freedom and describes the effects of heavier ones through local interactions and a derivative expansion. It can be obtained by matching an effective field theory or by a low-energy compactification. A Wilsonian effective action implements integration of high-energy modes with a scale cutoff. The quantum effective action that generates one-particle-irreducible vertices is a different construction; the two need not coincide.

Use metric signature $(+,-,\dots ,-)$ and initially set $\hslash =1$. For an internal scalar line with momentum $p$, the Feynman propagator is $i/(p^2-m^2+i0)$. An $r$-leg Feynman vertex, for $r=3,4,6$, contributes $-i{\lambda }_r$, with all incident momenta incoming. Its $1/r!$ in the Lagrangian density is canceled by the permutations of its identical fields. Each vertex conserves momentum and supplies $(2\pi {)}^d{\delta }^{(d)}(\sum p)$. Integrate each independent loop momentum with $d^{d}k/(2\pi {)}^d$, multiply the line and vertex factors, and divide by the Feynman-diagram symmetry factor for labeled external legs. Sum the distinct graphs and external-momentum assignments without counting the same assignment twice.

An unamputated momentum-space correlation function also contains the external propagators. Dividing the generating functional by its zero-source value removes disconnected vacuum bubbles; taking its logarithm selects connected correlation functions. Amputation of external propagators removes the external line factors when computing vertex amplitudes. Translational invariance leaves one overall momentum-conservation delta function for each connected graph.

The loop order is the number of independent graph cycles, equivalently the number of independent momentum integrations left after vertex conservation. A spanning tree through $V$ interaction vertices contains $V-1$ internal edges; hence, with $I$ internal lines,Restoring $\hslash$ in $\exp (iS/\hslash )$ gives $\hslash$ for each internal propagator and ${\hslash }^{-1}$ for each interaction vertex. Thus an amputated connected amplitude hasThe same counting holds for a diagram in $\log Z[J]$ when the external source is coupled as $J\varphi /\hslash$: every external propagator brings $\hslash$, canceled by the source insertion's ${\hslash }^{-1}$. This is Planck-constant loop counting with source normalization. An ordinary connected expectation of $n$ fields, with no such source factors, instead scales as ${\hslash }^{L+n-1}$; multiplying $\log Z$ by $-i\hslash$ to form $W$ shifts its source-kernel counting to ${\hslash }^L$. The normalization must be specified when applying the printed loop rule.

Let $V_r$ count $r$-valent interaction vertices. Counting their incident half-edges gives $2I+n=3V_3+4V_4+6V_6$. Substituting the loop identity yieldsEach coefficient on the right lies between $1/2$ and $2$, proving the vertex-count bounds for bounded scalar valenceThe free two-point term, which has no interaction vertex, separately satisfies the bounds with $V=L=0$ and $n=2$.

A one-particle-irreducible Feynman diagram is connected and remains connected after any single internal line is cut. The left graph below contains two quartic vertices and two parallel internal lines: cutting either line leaves the vertices connected by the other. The right graph contains two cubic vertices joined by one internal line, whose removal disconnects it, so it is a one-particle-reducible Feynman diagram. Both have four external legs.

The quantum effective action generates the one-particle-irreducible vertices. Their ultraviolet divergences and those of their proper divergent subgraphs are removed by local counterterms. Reducible connected graphs are reconstructed by joining these vertices with full propagators; they do not introduce independent local interaction counterterms merely because they contain bridges. This makes the one-particle-irreducible Feynman diagram decomposition useful for organizing renormalization. Overall power counting still has to be supplemented by subtraction of divergent subgraphs.

For a scalar with a two-derivative kinetic term, mass dimensions obeyThe general power-counting criterion used here is that interactions of nonnegative coupling dimension can be included in a perturbatively renormalizable quantum field theory, with all symmetry-allowed required local counterterms included; negative-dimensional couplings generally require an unbounded tower of higher-dimensional counterterms. Indeed, the scalar graph power-counting identity givesfor the superficial degree of divergence. Nonnegative coupling dimensions prevent this degree from increasing indefinitely with the vertex count.

The coupling dimensions areThus all three are power-counting renormalizable in three dimensions; cubic and quartic in four; cubic only in six. Positive-dimensional interactions are super-renormalizable, while the zero-dimensional ones are marginal by classical power counting. Usual kinetic, mass, vacuum and, when no symmetry forbids it, linear counterterms must also be admitted. This conclusion concerns perturbative ultraviolet renormalizability; it does not assert global stability of a cubic scalar potential.

Keep $Z=e^{iW}$, define the mean scalar field $\phi (x)=\delta W/\delta J(x)$, and locally invert the source-field relation. The quantum effective action is the Legendre transformIts value at zero source is expanded around ${\phi }_0=⟨\varphi ⟩$, which need not vanish. The kernels ${\Gamma }^{(n)}={\delta }^n\Gamma /\delta {\phi }^n{|}_{{\phi }_0}$ are the one-particle-irreducible vertices, with the Minkowski diagram vertex factor $i{\Gamma }^{(n)}$. A one-particle-irreducible Feynman diagram remains connected after any one internal line is cut. Amputation of external propagators removes the full two-point functions on its external legs. For higher-point connected graphs amputation alone need not remove internal one-particle-reducible Feynman diagrams; the derivatives of $\Gamma$ select precisely the irreducible kernels.

To derive the three-point relation, use condensed indices that include spacetime integration. Put $D_{ij}=W_{,ij}=i{G}_{2,ij}^c$ and $K={\Gamma }^{(2)}$. Differentiating ${\Gamma }_{,i}=-J_i$ gives $K_{ia}{D}_{aj}=-{\delta }_{ij}$, so $K=-D^{-1}$. A further derivative, using $\delta D_{ab}/\delta {\phi }_k=W_{,ab\ell }(D^{-1}{)}_{\ell k}$ and the derivative of an inverse, givesSince $W^{(3)}=-G_3^c$ and $D^{-1}=-i(G_2^c{)}^{-1}$, the three-point amputation in the connected-action convention isEquivalently, attach a full two-point correlation function to each leg of $i{\Gamma }^{(3)}$ to reconstruct $G_3^c$. The connected functions are evaluated at the same zero-source background. The displayed factors of $i$ follow from the declared Minkowski convention; Euclidean generating functionals redistribute these factors.

In phi cubed theory, a two-point bubble diagram has two cubic vertices, two internal scalar propagators and symmetry factor $1/2$. Its connected amputated insertion is $+{\mu }^{ϵ}{g}^2{B}_d(p)/2$. The inverse scalar propagator, or quadratic one-particle-irreducible vertex, therefore receives the negative insertion:Equivalently, the connected two-point correction is $G_0(p{)}^2{\mu }^{ϵ}{g}^2{B}_d(p)/2$, and expansion of its inverse changes the sign. The same sign follows from the second term of $\frac{1}{2}\mathrm{Tr}\log (K+{\mu }^{ϵ/2}g\varphi )$.

Using the six-dimensional residue and ${\int }_0^{1}x(1-x)dx=1/6$, the divergent inverse-propagator correction isIts momentum term requires wave-function renormalization, unlike the massive quartic tadpole. The minimal-subtraction two-point counterterms in cubic scalar theory areThese are additive coefficients in a Lagrangian expressed in terms of the renormalized field. If ${\varphi }_0=Z^{1/2}\varphi$, $Z=1+\delta Z$, and the bare mass is separately written $m_0^2=m^2+\Delta m^2$, then $\delta m^2=m^2\delta Z+\Delta m^2$ at this order, so $\Delta m^2=-5g^2{m}^2/[6(4\pi {)}^3ϵ]$. Distinguishing these conventions prevents a spurious disagreement between mass counterterms. The local Minkowski counterterms are $\delta Z(\partial \varphi {)}^2/2-\delta m^2{\varphi }^2/2$. A complete cubic theory also has one-point and three-point subtractions; they are not the specified two-point bubble poles.

Let $p$ be the total external Euclidean momentum crossing one four-point channel. The two internal scalar propagators produce the scalar bubble integralUsing the Feynman parameter identity $1/(ab)={\int }_0^{1}dx[xa+(1-x)b{]}^{-2}$ and shifting the loop momentum yieldsIn phi-fourth theory the bubble diagram has two quartic vertices and a symmetry factor $1/2$ from exchanging its two identical internal lines. There are three partitions of four labelled external legs into two pairs, the $s,t,u$ channels. The connected amputated loop diagram has positive coefficient ${\lambda }^{2}B/2$, whereas the Euclidean one-particle-irreducible vertex in the effective action has the opposite sign. ThusOne can check this sign without external-line conventions by expanding the one-loop quantum effective action $\frac{1}{2}\mathrm{Tr}\log (K+{\mu }^{ϵ}\lambda {\varphi }^2/2)$. Its second logarithmic term is $-{\mu }^{2ϵ}{\lambda }^2\mathrm{Tr}(K^{-1}{\varphi }^2{K}^{-1}{\varphi }^2)/16$; four field derivatives give precisely the three negative bubble terms above.

The four-dimensional pole of $I_d$ is independent of its mass argument, so each channel contributes $2/[(4\pi {)}^2ϵ]$. The divergence of the proper four-point vertex is $-3{\lambda }^2/[(4\pi {)}^2ϵ]$. The minimal subtraction scheme therefore requiresFor a Minkowski interaction $-{\mu }^{ϵ}\lambda {\varphi }^4/4!$, the corresponding local interaction counterterm is $-{\mu }^{ϵ}\delta \lambda {\varphi }^4/4!$. No momentum-dependent four-point counterterm is needed for these logarithmic poles. This calculation concerns the requested four-point bubble contributions; a massive two-point tadpole and vacuum diagrams require their own mass and vacuum-energy subtractions if those functions are also renormalized.

Use the mostly-plus Minkowski metric, so $p^2=-p_0^2+{\mathbf{p}}^2$, and the source term $+\int J\varphi$. Define the vacuum-normalized generating functional, connected generating functional, and quantum effective action byThe Legendre transform is taken after expressing $J$ in terms of $\phi$, locally where this relation is invertible. Varying it gives $\delta \Gamma /\delta \phi =J$ and ${\Gamma }^{(2)}=(W^{(2)}{)}^{-1}$. The logarithm generates connected graphs; the Legendre transform removes graphs joined by a single propagator, leaving one-particle-irreducible vertices. This sign convention is important: its classical limit is $\Gamma =-S$, consistent with the positive vertices in this question.

For the free real scalar field, integration by parts puts the action into the form $S=-\frac{1}{2}\varphi A\varphi$, where $A=-\square +m^2$ has Fourier symbol $p^2+m^2$. Include the Feynman i-epsilon prescription $A\mapsto A-i0$. Completing the square yieldsand consequentlyThe physical scalar propagator is $-i{A}^{-1}$, whereas the inverse quadratic kernel appearing in $W_0$ is $A^{-1}$. Differentiating this quadratic functional and Fourier transforming givesThe infinitesimal prescription specifies propagators and is dropped when quoting the local vertex kernel.

At the order represented by tree-level Feynman diagrams, stationary phase gives $W_{mtree}=S[\phi ]+\int J\phi$, with the classical field equation $\delta S/\delta \phi +J=0$. The Legendre transform cancels the source term, soThusOther connected tree graphs with several quartic vertices are one-particle reducible and belong to $W$, not to the proper vertices of $\Gamma$. This explains why no extra tree correction is hidden in these formulas.

The one-loop proper two-point graph is the tadpole diagram. The one-loop proper four-point graphs are the three bubble diagrams, with pairings $(12|34)$, $(13|24)$ and $(14|23)$ of the external legs. Each has Feynman-diagram symmetry factor $1/2$. The figure shows the topologies and every four-point channel; external momenta are incoming.

The tadpole factor can also be counted directly: there are $4\cdot 3=12$ assignments of two labeled external legs to one quartic vertex, while its $4!$ normalization leaves $12/24=1/2$. For a fixed four-point partition, the interchange of the two internal lines gives the same factor $1/2$; there are three distinct partitions. Graphs with an external self-energy insertion are one-particle reducible and are not additional contributions to ${\tau }_4$.

For the two-point divergence, perform a Wick rotation. The Euclidean tadpole contribution to the proper inverse kernel isAssume $m^2>0$ to separate the ultraviolet calculation from infrared issues. A Schwinger parameterization gives $1/(k_E^2+m^2)={\int }_0^{\infty }ds{e}^{-s(k_E^2+m^2)}$. Using the supplied Gaussian integral, initially in the convergence region, givesThe last expression provides the meromorphic dimensional regularization continuation. Set $d=4-ϵ$. Since $\Gamma (-1+ϵ/2)=-2/ϵ+O(1)$,If the coupling is written as ${\mu }^{ϵ}\lambda$, that factor multiplies the intermediate integral but does not change the displayed pole residue. The negative pole is a property of analytic continuation; it must not be inferred from the positive cutoff integrand without carrying out the continuation.

The added mass counterterm contributes $+B$ to ${\tau }_2$, because ${\Gamma }_{mtree}=-S$ and its Lagrangian term is $-B{\varphi }^2/2$. Hence cancellation requiresup to an arbitrary finite subtraction specifying the renormalization scheme. No external momentum occurs in the tadpole, so no one-loop wave-function renormalization is needed. At $m=0$ the unregulated tadpole is scaleless and vanishes in dimensional regularization; the massive calculation identifies the mass-dependent ultraviolet pole without conflating it with a massless infrared limit.

For completeness, the signs of the four-point graphs in this effective-action convention can be read from the Euclidean one-loop functional $\frac{1}{2}\mathrm{Tr}\log (A_E+\lambda {\phi }^2/2)$. Its expansion has terms $\lambda \mathrm{Tr}(A_E^{-1}{\phi }^2)/4$ and $-{\lambda }^2\mathrm{Tr}(A_E^{-1}{\phi }^2{A}_E^{-1}{\phi }^2)/16$. Differentiating four times givesThis is the Euclidean form, analytically continued to the physical external momenta. It confirms the three channel diagrams and their symmetry factors; evaluating its divergence is not needed for the requested mass counterterm.

For $m^2>0$ and initially $\Re n>d/2$, Schwinger parameterization and the supplied Gaussian integral giveSubstitute $u=\alpha m^2$ to obtain the Euclidean massive loop integralFor a positive integer $n$, $\Gamma (n)=(n-1)!$. Outside the convergence region this expression defines its meromorphic continuation, used in dimensional regularization.

Combine the two denominators with a Feynman parameter and shift $k$ by $(1-x)p$. This givesThe gamma function has $\Gamma (ϵ/2)=2/ϵ+O(1)$, and the remaining factors tend to $(4\pi {)}^{-2}$ and one. Hence the massive scalar bubble pole in four dimensions isThe pole is independent of the external momentum, as required for a local quartic counterterm.

Use the mostly-plus Minkowski metric and expand $e^{iS}$. The momentum-space Feynman rules are: an internal scalar propagator $i/(-p^2-m^2+i0)$; a four-scalar interaction vertex $-i{\mu }^{ϵ}\lambda$; a momentum conservation delta function at each vertex; and $\int d^{d}k/(2\pi {)}^d$ per independent loop momentum. Divide by the Feynman-diagram symmetry factor. All momenta can be taken incoming. External propagators are omitted when computing an amputated one-particle-irreducible vertex.

The connected one-loop one-particle-irreducible Feynman diagrams are the two-point tadpole diagram and the three four-point bubble diagrams, in the $s,t,u$ channels. Each has symmetry factor $1/2$. A tadpole attached to an external leg through a single bridge would be reducible, so is excluded here.

The Euclidean massive loop integral at $n=1$ has $\Gamma (-1+ϵ/2)=-2/ϵ+O(1)$, so $I_{d,1}=-m^2/(8{\pi }^2ϵ)+O(1)$. Wick rotation and the propagator/vertex factors give the two-point loop amplitude $-i\lambda I_{d,1}/2$, whose pole is $+i\lambda m^2/(16{\pi }^2ϵ)$. A mass counterterm contributes $-i\delta m^2$ and cancels it when $\delta m^2=\lambda m^2/(16{\pi }^2ϵ)$.

Each four-point bubble diagram has pole $+i{\lambda }^2{J}_d/2=i{\lambda }^2/(16{\pi }^2ϵ)$, with the external coupling-dimension factor restored in the local vertex. Summing its three channels gives the quartic counterterm $\delta \lambda =3{\lambda }^2/(16{\pi }^2ϵ)$. Therefore the one-loop massive phi-fourth counterterms are exactlyThere is no one-loop wave-function renormalization, since the tadpole is independent of external momentum. Vacuum graphs would additionally require a vacuum-energy subtraction.

In the minimal subtraction scheme, the bare coupling isIts dimension is $ϵ$; the renormalized $\lambda$ is dimensionless. Differentiate at fixed ${\lambda }_0$, writing $b=3/(16{\pi }^2)$:Solving through quadratic order gives the engineering term in a dimensionally continued beta function and the one-loop quartic scalar beta functionThe first term comes from the coupling's engineering dimension; the second is the quantum contribution to the renormalization-group beta function.

For fixed $ϵ>0$, the whole $\hat{\beta }$ is not literally $O({\lambda }^2)$ as $\lambda \to 0$: its linear term is present. Interpret the final instruction as keeping terms through quadratic order. Set ${\lambda }_{\ast }=ϵ/b=16{\pi }^2ϵ/3$. Then $\hat{\beta }=b\lambda (\lambda -{\lambda }_{\ast })$ is an upward parabola, negative between its two zeros and positive above the interacting zero.

For positive initial coupling ${\lambda }_r=\lambda ({\mu }_r)$, the quartic running coupling below four dimensions follows by solving the linear equation for $1/\lambda$:If $0<{\lambda }_r<{\lambda }_{\ast }$, then $\lambda (\mu )\to {\lambda }_{\ast }$ as $\mu \to 0$ and $\lambda (\mu )\to 0$ as $\mu \to \infty$. If ${\lambda }_r={\lambda }_{\ast }$ it stays fixed; the zero solution also stays fixed. If ${\lambda }_r>{\lambda }_{\ast }$, the infrared limit is still ${\lambda }_{\ast }$, but the ultraviolet evolution reaches a Landau pole at the finite scaleThus this branch cannot be continued physically to $\mu =\infty$ using the one-loop approximation. These are statements about the quartic coupling direction. The Wilson-Fisher fixed point of the full scalar theory also has a thermal relevant direction at the Wilson-Fisher fixed point; approaching the critical theory requires mass tuning. For negative initial coupling, the quartic potential is unstable: the same formula tends to zero from below in the ultraviolet and reaches a pole at finite scale toward the infrared.

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.

Write $\epsilon ={\epsilon }_a{t}_a$ and $A_{\mu }=A_{\mu a}{t}_a$. The proposed gauge-field transformation law becomes $\delta A_{\mu }=-{\partial }_{\mu }\epsilon +[\epsilon ,A_{\mu }]$. ConsequentlyBoth $\varphi$ and its gauge covariant derivative therefore transform as the same multiplet, with no derivative of the local parameter left over. The original algebraic invariance of the Lagrangian density now applies pointwise to $\mathcal{L}(\varphi ,D\varphi )$, so this constructs the local symmetry.

Also transform $J$ by $\delta J=\epsilon J$. Antisymmetry gives $(t_{a}J)\cdot \varphi +J\cdot (t_a\varphi )=0$, so the source coupling is invariant. Let $S_{A_{\mu a}}$ denote the functional derivative at a point. Vary the action functional and integrate the derivative of ${\epsilon }_a$ by parts, taking the local parameter to have compact support:Since every ${\epsilon }_a(x)$ is arbitrary, its coefficient vanishes. This proves the requested local identity, including the derivative sign and the order of the structure-constant indices.

For the generating functional, the integral of the first term is zero as a change-of-variable identity:The divergence of this vector field in field space is proportional to $\mathrm{tr}{t}_a=0$. Thus an invariant regulated functional measure, with no gauge anomaly, leaves no Jacobian term. Inserting the classical identity into the integral givesThe same first-order differential operator annihilates the connected generating functional $W$, because $Z=e^{iW}$. Its source term is $(t_{a}J)\cdot \phi$.

For the specified Legendre transform, vary $\Gamma =-W+\int J\phi$. Since $W_J=\phi$, the terms involving $\delta J$ cancel, leavingAntisymmetry also gives $(t_{a}J)\cdot \phi =-J\cdot (t_a\phi )$. Substitution in the identity for $W$, followed by multiplication by $-1$, proves the background gauge invariance of a scalar effective action:

The derivatives of the quantum effective action are proper, amputated one-particle-irreducible vertices. At each loop order they receive contributions from one-particle-irreducible Feynman diagrams with the indicated external scalar legs, including tree vertices and the required counterterms. Diagrams which disconnect after cutting one internal line are not additional proper-vertex contributions. Translation invariance supplies the overall momentum-conserving delta function factored out in the definition. The signs of these vertices must follow the chosen overall convention for $\Gamma$.

Differentiate the effective-action identity twice with respect to ${\phi }_i(y)$ and ${\phi }_j(z)$ and set $\phi =A=0$. In an invariant zero-field vacuum this gives the Ward identity contact termsThe term proportional to $A$ vanishes. With the printed Fourier factors $e^{i{p}_r{x}_r}$, integration by parts turns ${\partial }_{\mu }$ into $-i{p}_{1\mu }$. The first delta function combines $p_1$ with $p_2$, and the second combines $p_1$ with $p_3$. Using $(t_a{)}_{ki}=-(t_a{)}_{ik}$ and symmetry of the scalar two-point kernel then yieldsThis derivation fixes the relative signs independently of any vertex rule.

There is an overall-sign inconsistency in the final printed example. At tree level, stationarity of the source integral gives $W=S_0[\phi ,A]+\int J\phi$, hence the specified transform gives ${\Gamma }_{\mathrm{tree}}=-S_0$. For the free scalar multiplet this implies, after factoring out the delta function,The mixed result follows by expanding $-S_0$ to first order in $A$: its contribution is $\int A_{\mu a}({\partial }^{\mu }{\phi }_i)(t_a{)}_{ij}{\phi }_j$. The two scalar derivatives yield $i{p}_2^{\mu }(t_a{)}_{ij}+i{p}_3^{\mu }(t_a{)}_{ji}$. Momentum conservation givesso both sides of the Ward identity equal $i(p_3^2-p_2^2)(t_a{)}_{ij}$. Because the scalar theory is Gaussian for prescribed $A$, its fluctuation determinant depends on $A$ but not on $\phi$; it cannot add further two-scalar vertices. Thus this quadratic verification is sufficient, not merely the first term of an omitted scalar-loop correction.

The printed mixed vertex instead has a minus sign. It satisfies the same Ward identity if one uses the opposite effective-action convention ${\Gamma }_{\mathrm{alt}}=W-\int J\phi =S_0$ at tree level, because its two-point vertex is then $-(p^2+m^2){\delta }_{ij}$ as well. Either overall convention works, but the printed Legendre transform and printed mixed vertex cannot both be used unchanged. For example, with $p_2^2\ne p_3^2$ and a nonzero generator, they give opposite sides of the claimed identity.

Use the metric signature $(-,+,\dots ,+)$, so $p^2=-p_0^2+{\mathbf{p}}^2$, and the weight $e^{iS}$. The momentum-space Feynman rules for the massless Phi-fourth theory are: an internal scalar line contributes $-i/(k^2-iϵ)$; a quartic interaction vertex contributes $-i\lambda$ with its momentum-conservation delta function; every independent loop is integrated with $d^{d}k/(2\pi {)}^d$; and each graph is multiplied by its Feynman-diagram symmetry factor. All external momenta can be taken incoming. These rules follow by expanding the interaction exponential and contracting free fields with the Wick theorem.

For a connected graph with $I$ internal lines, $V$ interaction vertices and $L$ loops,Here $D$ is the superficial degree of divergence. In four dimensions,Odd $n$ vanish by Z2 symmetry. Thus only the nonvacuum two-point and four-point one-particle-irreducible Feynman diagrams can have overall ultraviolet divergences: degree two for $n=2$, degree zero for $n=4$. Vacuum graphs with $n=0$ may also diverge but cancel from normalized correlators. This is an overall power-counting statement; subgraphs must be subtracted before using it to conclude finiteness for $n\ge 6$.

For the full one-particle-irreducible vertices, the tree-level quadratic kernel and quartic interaction giveIf ${\hat{\tau }}_2$ is instead reserved for interaction-generated self-energy insertions, its tree value is zero and $-p^2$ is displayed separately as the free inverse kernel. The distinction is only that convention; the loop insertions below are unchanged.

The requested one-loop diagrams are the two-point tadpole diagram and the three four-point bubble diagrams. Each has symmetry factor $1/2$. The three bubble channels correspond to $P_s=p_1+p_2$, $P_t=p_1+p_3$, $P_u=p_1+p_4$.

The tadpole insertion obeysIn massless dimensional regularization this scaleless integral is zero. A mass or cutoff regulator exposes the ultraviolet divergence allowed by power counting; its vanishing in the scaleless convention does not remove the need to include this diagram.

For a bubble, the vertex and quantum field theory propagator factors give $(-i\lambda {)}^2(-i{)}^2/2={\lambda }^2/2$. The PDF defines $f$ with the factor $1/i$ outside the Minkowski integral, so the integral itself is $if(P^2)$. ThereforeThe converted TeX misplaces this $i$ as though it were part of the exponent of $(2\pi )$; keeping the printed $1/i$ is essential to this real pole coefficient.

For nonexceptional Euclidean external momentum, Wick rotation cancels that factor $1/i$. Use a Feynman parameter and translate the integration variable:For example, represent the inverse square as ${\int }_0^{\infty }d\alpha \alpha e^{-\alpha ({\ell }^2+\Delta )}$, do the Gaussian momentum integral, and then use the Gamma function. This also derives the formula rather than assuming a loop-integration table. At $d=4-\epsilon$, $\Gamma (\epsilon /2)=2/\epsilon +O(1)$ and the parameter integral tends to one, givingTake $P^2>0$ before continuation back to Minkowski momenta. At $P=0$, a massless scaleless integral mixes ultraviolet and infrared issues, so it cannot be used to read off this pole.