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The cumulant expansion of the shell integral keeps precisely connected Feynman diagrams built from shell-restricted scalar propagators. The external low fields are backgrounds, not propagators. Connected graphs with bridges can contribute, so the expansion is not limited to one-particle-irreducible Feynman diagrams.

The quantum effective action is the Legendre transform of the generating functional of connected correlation functions. Its functional derivatives are the one-particle-irreducible vertex functions.

The self-energy bubble diagram at loop order one in six-dimensional cubic scalar field theory has Feynman-diagram symmetry factor two. With positive Euclidean scalar propagators, its amputated Green's function is the displayed integral. Its superficial degree of divergence is two. Expanding at large internal momentum and taking the angular average shows that its divergent part is a constant plus a multiple of $p^2$, precisely the mass counterterm and wave-function renormalization structures.

For a spherical ultraviolet cutoff $|k|\le \Lambda$ and $m^2>0$, its value at $p=0$ isThe divergent $p^2$ coefficient is $-g^2\log ({\Lambda }^2/m^2)/(768{\pi }^3)$. Finite terms depend on the regularization convention. Tadpole diagram insertions are excluded from the one-particle-irreducible Feynman diagram, and are removed from the connected two-point function by imposing a vanishing one-point function.

For a connected real-scalar Feynman diagram, degree counting and $L=I-V+1$ give $V_3+2V_4=E+2L-2$. A one-loop one-particle-irreducible Feynman diagram is a cycle: every vertex has two internal half-edges. For three labelled external legs, the possibilities are one cubic triangle with symmetry factor one and three mixed cubic-quartic bubbles with symmetry factor two. For four labelled external legs, there are three quartic bubbles with symmetry factor two, six mixed triangles with symmetry factor one, and three cubic boxes with symmetry factor one. Cyclic ordering modulo reversal explains the three boxes.

In Euclidean signature, for a canonically normalized real scalar field with potential $V$, the one-loop contribution to the quantum effective action is ${\Gamma }_1[\varphi ]=\frac{1}{2}\mathrm{Tr}\log [-{\partial }^2+V^{\prime \prime }(\varphi )]$, with a suitable regularization in quantum field theory. Expanding this functional determinant generates one-particle-irreducible Feynman diagrams with arbitrarily many external fields.

A one-particle-irreducible Feynman diagram is a connected Feynman diagram that remains connected after any one internal line is cut. The vertices of the quantum effective action generate precisely these diagrams.

These Feynman diagrams generate the vertices of the quantum effective action.

The one-particle-irreducible scalar two-point function has three one-loop topologies: a scalar-photon exchange loop built from two cubic vertices; a photon tadpole at one seagull vertex; and a scalar tadpole at one quartic vertex. The last is absent when $\lambda =0$. Reducible chains of self-energy insertions belong to the Dyson expansion and are not additional one-loop self-energy graphs. The scalar-QED photon-tadpole symmetry factor is $1/2$ because its two identical photon legs are joined; the oriented complex-scalar tadpole has no such identical-real-line factor.

For the proper four-point interaction vertex, retain one-particle-irreducible Feynman diagrams with two labelled external photons and the two ends of a directed scalar line. Put $C$ for a cubic scalar–photon vertex and $S$ for a seagull vertex. The one-loop counting identities require $V_C+2V_S=4$. There are three possibilities, giving the nine proper scalar-QED seagull-vertex diagrams:

Each graph has Feynman-diagram symmetry factor one with these external labels: the internal photon and scalar are different species, and the charged scalar endpoints cannot be exchanged. The list is complete because the internal photon must join the ends of the open scalar chain for the graph to remain connected after any one internal line is cut. If it joined two interior points, an outside scalar segment would be a bridge and the graph would be one-particle reducible.

External-leg self-energy insertions contribute to an unamputated four-point correlation function, not this proper vertex. The local four-point counterterm from i is included separately when constructing the renormalized answer; it is not another one-loop topology. Abelian Faddeev-Popov ghosts do not interact in this linear gauge and add no diagram.

Source paper: www.whatdotheyknow.com/request/oxford_physics_past_papers_2/response/2340696/attach/8/Physics%20Papers%202019.pdf?cookie_passthrough=1

Official standalone paper: mmathphys.physics.ox.ac.uk/sites/default/files/mmathphys/documents/media/aqft_2020.pdf

The annual bundle's filename says 2019, but this paper's cover says Trinity Term 2020 and identifies A15272W1. These are solutions to that 2020 paper. Use natural units, the Minkowski metric $g=\mathrm{diag}(1,-1,-1,-1)$ and the Feynman i-epsilon prescription. Write $/p={\gamma }^{\mu }{p}_{\mu }$. In question 1, $D=4-ϵ$, $e_D={\mu }^{ϵ/2}e$, and $\mu$ is the renormalization scale; keeping the dimensionful coupling $e_D$ explicit prevents a factor-of-two ambiguity in the ultraviolet pole. In question 2(a), the factors of $i$ in the stated generating functional fix the normalization of its kernels, as explained in the solution.

There is one one-particle-irreducible Feynman diagram: an internal photon joins two points of the open fermion line, with the external photon attached to that line between them. It is the upper-left panel of the figure. Route internal photon four-momentum $k$ from the incoming side to the outgoing side; the adjacent internal fermion four-momenta are $p-k$ and $p^{\prime }-k$. The symmetry factor is one.

The vertex renormalization counterterm is an additional local contribution when forming the renormalized Feynman vertex. External-leg self-energy insertions belong to the complete three-point correlation function, rather than this amputated one-particle-irreducible Feynman diagram.

The proper quark-gluon vertex colour factors come from two one-particle-irreducible Feynman diagrams, shown in the upper-middle and upper-right panels. One has a virtual gluon joining the open quark line on opposite sides of the external gluon insertion. The other has two virtual gluons meeting the external gluon at a three-gluon Feynman vertex, with their other ends joined by an internal quark. A Faddeev-Popov ghost field has no direct quark coupling, so it supplies no additional one-loop Feynman vertex graph here.

Let $T^a$ be the Hermitian generators of the fundamental representation of $SU(N)$, with $\mathrm{tr}(T^a{T}^b)={\delta }^{ab}/2$, $[T^a,T^b]=i{f}^{abc}{T}^c$, ${\sum }_b{T}^b{T}^b=C_{F}1$, and $f^{abc}{f}^{dbc}=C_A{\delta }^{ad}$. Here $C_F=(N^2-1)/(2N)$ and $C_A=N$ are quadratic Casimir operators evaluated in the fundamental representation and the Adjoint representation.

For the first graph, the ordered product is ${\sum }_b{T}^b{T}^a{T}^b$. Expand the double commutator:Thus its colour factor isFor the second graph choose the orientation of the three-gluon Feynman vertex so that its group tensor is $f^{abc}$ and the ordered product along the quark line is $T^b{T}^c$. Antisymmetry removes the anticommutator part:The corresponding real group coefficient is $C_A/2=3/2$ after the conventional factor of $i$ is incorporated into the remaining Feynman rules. If the vertex is written with $i{f}^{abc}$, its raw matrix product is instead $-C_A{T}^a/2$. Reversing the ordered product reverses its sign together with the orientation-dependent three-gluon Feynman rule; these conventions describe the same scattering amplitude. External-leg corrections and local counterterms are separate from these two proper one-loop graphs.

The first panel is the one-particle-irreducible Feynman diagram for photon vacuum polarization. The two internal Dirac propagators carry $k$ and $k+q$, and the external photons carry $q$. The other panels fix the routings used in parts c and d.

The closed fermion loop sign contributes a minus sign; there is no extra symmetry factor. Applying the given Feynman rules, with the external propagators amputated,Equivalently, the required amplitude isThe Feynman slash notation is $/k={\gamma }^{\rho }{k}_{\rho }$ and $\mathrm{tr}1=4$ in the chosen dimensional regularization convention.

The middle panel in part a has an internal photon of momentum $k$ joining two points on a fermion line; the internal fermion has momentum $p-k$. It is a one-particle-irreducible Feynman diagram with no closed fermion loop sign. In Feynman gauge,orThe decomposition in the source gives $\mathrm{tr}(/p\Sigma )=4p^{2}S(p^2)$. In particular,For a direct ultraviolet divergence extraction, use a Feynman parameter $x$ multiplying the massive denominator, shift $r=k-xp$, and define $\Delta =x{m}^2-x(1-x)p^2$. Then $/p-/k=(1-x)/p-/r$ andOdd $r$ terms integrate to zero. Taking the trace explicitly supplies the weight $(2-d)(1-x)$, whose integral tends to $-1$. HenceAs a check on the same fermion self-energy calculation,This is the ultraviolet pole of QED fermion self-energy. It is safe to extract it at generic external momentum; an on shell or massless limit must distinguish any infrared divergence from this ultraviolet divergence.

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.

Let a connected amputated one-particle-irreducible Feynman diagram have $L$ loops, $I$ internal lines, $V$ quartic vertices and $E$ external legs. Each loop integration contributes four momentum powers, each scalar propagator removes two, and the interaction adds none. Thus its superficial degree of divergence is $D=4L-2I$. Counting half-edges and independent loops givesThis is simultaneous large-momentum scaling of the whole Feynman diagram. It need not describe every subregion of loop-momentum space.

For a concrete counterexample, start with a six-point triangle of three quartic vertices, with two external legs at each vertex. Insert a one-loop tadpole diagram self-energy in one internal line. The resulting Feynman diagram has $V=4$, $I=5$, $L=2$, $E=6$, and $D=-2$. It is still one-particle irreducible: each edge around the triangle has an alternative route, and cutting the tadpole loop does not disconnect the Feynman diagram. Nevertheless the tadpole subgraph diverges when its loop momentum tends to infinity while the triangle momentum is held fixed. In massive theory the divergent mass insertion is nonzero. This demonstrates that a superficially convergent graph can have a divergent subgraph.

Write renormalized perturbation theory withThe last term is a field-independent vacuum counterterm, needed if vacuum energies are retained. A normalized correlation functional removes that sector. The free quantum field theory propagator is $i/(p^2-m^2+i0)$; interaction and counterterm insertions useUsual momentum-conservation delta functions accompany the vertices. The coefficients depend on the regulator and renormalization conditions and are expanded order by order in the renormalized coupling.

The reason this list closes is locality of ultraviolet subtraction. After divergent proper subgraphs have been subtracted recursively, the remaining overall divergent part is a polynomial in external momenta of degree at most $D$. For $E=2$, $D=2$, Lorentz symmetry allows only a constant and a $p^2$ term, supplied by the mass and wavefunction counterterms. For $E=4$, $D=0$, only the constant quartic counterterm is needed. Odd-point terms vanish by the field's $\varphi \mapsto -\varphi$ symmetry. For $E>4$, the overall degree is negative and there is no new overall counterterm after subdivergences have been removed. Vacuum graphs are handled separately.

Counterterm insertions subtract each subgraph's local divergence, including overlapping subgraphs, and then the overall divergence. At each perturbative order the regulated sum has a finite ultraviolet limit for renormalized correlation functions at suitable nonexceptional momenta. This is why superficial power counting of every relevant subgraph determines the allowed counterterms, even though it does not alone establish that a bare Feynman diagram diverges or converges. Possible infrared or on-shell singularities are a separate issue; ultraviolet renormalization is not a promise of finiteness in every kinematic limit.

Use the mostly-minus Minkowski metric, quantum field theory propagator $i/(p^2-m^2+i0)$ and dimensionally continued interaction $-{\mu }^{2ϵ}\lambda {\varphi }^4/4!$, with $d=4-2ϵ$. Take all four external momenta incoming, so ${\sum }_j{p}_j=0$, and set $s=(p_1+p_2{)}^2$, $t=(p_1+p_3{)}^2$, $u=(p_1+p_4{)}^2$.

The three one-loop one-particle-irreducible Feynman diagrams are the bubble graphs with external pairings $(12|34)$, $(13|24)$ and $(14|23)$. Their external quantum field theory propagators are omitted by amputation. Each Feynman diagram has symmetry factor $1/2$ from exchanging the two internal lines; the fixed external labels are already accounted for by the three distinct channels.

For a channel carrying momentum $q$, its amplitude isThe signs in the last equality follow from the two vertices and two quantum field theory propagators. Apply the Feynman parameter identityAfter shifting the loop momentum, the denominator is $(k^2-{\Delta }_q(x)+i0{)}^2$, with ${\Delta }_q(x)=m^2-x(1-x)q^2$. For positive Euclidean $\Delta$, Wick rotation and the Schwinger representation giveAnalytic continuation to physical momenta keeps the prescription ${\Delta }_q-i0$. Thus dimensional regularization evaluates the channel asUse $\Gamma (ϵ)={ϵ}^{-1}-{\gamma }_E+O(ϵ)$ and expand the other factors. The minimal-subtraction scalar four-point bubble channels result through the finite part isThe parameter integral is left unevaluated, as requested. The three channel contributions are $B(s)+B(t)+B(u)$. Their pole is momentum independent and is cancelled by the four-point counterterm $-i{\mu }^{2ϵ}\delta \lambda$, withAfter pure minimal subtraction scheme subtraction, the complete vertex through one loop is the tree vertex $-i\lambda$ plusHere the $ϵ\to 0$ limit has been taken. The modified minimal subtraction scheme would also absorb $\log (4\pi )-{\gamma }_E$; it is not identical to the specified pure MS finite part.

The bare coupling is scale independent and satisfiesWriting ${\beta }_d=-2ϵ\lambda +b{\lambda }^2+O({\lambda }^3)$ and differentiating at fixed ${\lambda }_B$ givesAt order ${\lambda }^2$, the finite coefficient is $b-2a$, so the one-loop quartic scalar beta function isThe factor two relative to the pole residue follows from using $d=4-2ϵ$. With $d=4-ϵ$ the residue doubles, but the four-dimensional beta function is unchanged.

For pure nonabelian gauge theory, the corresponding pure Yang-Mills one-loop beta function has the opposite sign:Gauge-boson self-interactions and the gauge-consistency ghost contributions give antiscreening. For a compact nonabelian simple factor $C_A>0$, the coupling decreases at high energy: asymptotic freedom. In contrast, positive scalar $\lambda$ increases toward the ultraviolet. Its one-loop extrapolation has a Landau pole, indicating breakdown of weak-coupling perturbation theory, not a proof by itself of nonperturbative triviality. Ultraviolet poles should be extracted at infrared-safe mass or nonexceptional kinematics; a scaleless massless zero-momentum integral cannot be used to read off this ultraviolet counterterm without separating its infrared behavior.

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: