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Keep the mostly-plus Minkowski metric and the Fourier transform $e^{ip\cdot x}$. Write $/p={\gamma }^a{p}_a$, with $\{{\gamma }^a,{\gamma }^b\}=2{\eta }^{ab}$. To fix the otherwise unspecified phase of ${\gamma }^5$ and the Dirac adjoint, takewhere ${\Gamma }^a$ are ordinary mostly-minus gamma matrices. Then $({\gamma }^5{)}^2=-1$, $\{{\gamma }^5,{\gamma }^a\}=0$, and the fermionic time-derivative term is $i{\psi }^{†}{\partial }_t\psi$. In these conventions the interaction with real $g$ is Hermitian: in mostly-minus notation it is $ig{\bar{\psi }}_{\mathrm{std}}{\Gamma }^5\psi \varphi$. If one instead calls $i{\gamma }^5$ the square-one chirality matrix, its coefficient must be $-ig$ to represent the same interaction. These phase choices leave physical relativistic scattering cross-sections unchanged.

Expanding $e^{i{S}_{\mathrm{int}}}$ yields the following Feynman rules with relativistically normalized external states:

Strip the overall four-momentum conservation delta function when defining the scattering amplitude. There are no further bare interaction vertices, no gauge fixing and no Faddeev-Popov ghost fields in this theory. Renormalized higher-order calculations add the required counterterms; these are additional to the rules of the displayed classical Lagrangian density.

Use the momentum-space Feynman rules with the scalar Feynman propagator $i/(p^2-m^2+i0)$. The four-leg vertex of a factorial-normalized scalar interaction has weight $i\lambda$ for the positive interaction sign printed here. There are $4!$ Wick contractions assigning four external legs to its four fields, cancelling the factorial in its coefficient. A negative interaction sign would give $-i\lambda$; its squared tree amplitude is the same.

At each vertex include $(2\pi {)}^4{\delta }^4(\sum p)$ with all incident momenta taken incoming. Assign an internal momentum to each line and integrate each independent loop with $\int d^4\ell /(2\pi {)}^4$. Divide a graph by its Feynman-diagram symmetry factor, sum the graphs at the chosen order, and omit disconnected vacuum graphs from normalized amplitudes. For an S-matrix element, amputate external propagators and put the external momenta on shell as in the LSZ reduction formula; the external one-particle residues are one at tree level.

Define the invariant amplitude by the relativistically normalized matrix elementwith $⟨p|p^{\prime }⟩=2E_{\mathbf{p}}(2\pi {)}^3{\delta }^3(\mathbf{p}-{\mathbf{p}}^{\prime })$. The lowest-order connected four-point graph is one contact vertex:There is no exchange graph at this order because there is no three-field interaction.

For the elastic scattering from a quartic scalar contact interaction, write $\sqrt{s}$ for the total centre-of-mass energy and $E_p=\sqrt{s}/2$ for the energy of each incoming particle. The incoming and outgoing spatial momentum magnitudes both equal $k=\sqrt{E_p^2-m^2}$, with $E_p>m$. The invariant incident flux isThe Lorentz-invariant phase-space measure for two outgoing particles isIn the centre-of-mass frame, the spatial delta function sets ${\mathbf{p}}_4=-{\mathbf{p}}_3$, while the energy delta function has radial derivative $2k/E_p$. Therefore the relativistic two-body phase space satisfiesThe two outgoing real-scalar particles are identical. Integrating over the full solid angle counts each unordered pair twice, so include the identical final-state symmetry factor $1/2!$. This givesThis is isotropic. When $E$ denotes each particle's energy, $s=4E^2$ and the full-sphere event density is ${\lambda }^2/(512{\pi }^2{E}^2)$. If $E$ denotes the total energy of the pair, $s=E^2$ and it is ${\lambda }^2/(128{\pi }^2{E}^2)$. Stating the answer in $s$ removes that energy-label ambiguity.

An equally valid angular convention selects one outgoing particle in a hemisphere, so each event is represented once. In that convention omit $1/2!$ and use $d\sigma /d\Omega ={\lambda }^2/(64{\pi }^{2}s)$ on the hemisphere, or ${\lambda }^2/(256{\pi }^2{E}_p^2)$. Both conventions give ${\sigma }_{\mathrm{event}}={\lambda }^2/(32\pi s)$ at this order. The identical-state factor concerns counting final states and is separate from the vertex factorial.

The cubic scalar field theory has a one-particle-irreducible Feynman diagram with two cubic vertices, one external leg at each vertex, and two internal lines joining them:

The Euclidean Feynman rule for each cubic interaction vertex is $-g{\mu }^{(6-d)/2}$: the minus sign comes from expanding $e^{-S_{\mathrm{int}}}$, and $1/3!$ in the action cancels the permutations of its three fields. The mass dimension of the cubic coupling is $(6-d)/2$, so ${\mu }^{(6-d)/2}$ permits $g$ to remain dimensionless. Two vertices supply $(-g{)}^2{\mu }^{6-d}$.

Each internal scalar propagator contributes $1/(p^2+m^2)$ with its own momentum. Conservation leaves one independent loop momentum; choose the two propagator momenta to be $p$ and $p+k$. The remaining Fourier integration measure is $d^{d}p/(2\pi {)}^d$. Finally, the factor $1/2$ is the Feynman-diagram symmetry factor for exchanging the two identical internal lines. Direct Wick contraction counting gives the same factor: two choices for which vertex receives a labeled external leg, $3^2$ choices of the incident fields and $2!$ pairings of the remaining fields, divided by $2!(3!{)}^2$, yield $1/2$.

The displayed loop integral is the amputated two-point insertion. For the correction to the full propagator, multiply it by the external scalar propagators, giving $D_0(k{)}^{2}I(k)$ with $D_0(k)=1/(k^2+m^2)$.

Use units $\hslash =c=1$, the Minkowski metric ${\eta }_{ab}=\mathrm{diag}(1,-1,-1,-1)$, and the Fourier transform convention $f(x)=\int d^{4}p{e}^{-ip\cdot x}\tilde{f}(p)/(2\pi {)}^4$. Here gamma matrices satisfy $\{{\gamma }^a,{\gamma }^b\}=2{\eta }^{ab}{I}_4$, and $\bar{\psi }={\psi }^{†}{\gamma }^0$ is the Dirac adjoint. A single Dirac field describes both Electrons and Positrons; these are the particle and antiparticle sectors of that field.

The free Maxwell Lagrangian and Dirac action, with a linear covariant gauge condition for the photon, giveBefore gauge fixing, the Maxwell Lagrangian has zero modes in field theory along $A_a\mapsto A_a+{\partial }_a\lambda$, so its quadratic operator cannot be inverted on all potentials. In the linear Lorenz gauge, the Faddeev-Popov determinant is $\det \square$. It is independent of $A$ and may be absorbed into the normalization; the corresponding Faddeev-Popov ghost fields have no interacting vertices in this Abelian linear gauge. A residual gauge symmetry is removed by the specified boundary conditions.

The free generating functional isThe Dirac field variables and their sources are independent Grassmann fields in this path integral. They anticommute; replacing them by ordinary commuting fields would give the wrong statistics and the wrong functional determinant. The bosonic Gaussian functional integral contributes an inverse square root of a determinant, and the Grassmann Gaussian integral contributes a determinant. If $K_A$ is the photon quadratic operator and $K_D=i{\gamma }^a{\partial }_a-m$, completing the square givesIn the last expression $D_F=i{K}_A^{-1}$ and $S_F=i{K}_D^{-1}$ are the photon propagator and Dirac propagator, with Feynman i-epsilon prescription. The products include the appropriate spacetime integrals and index contractions. Functional derivatives with respect to $J$, and consistently ordered left or right Grassmann derivatives with respect to the fermionic sources, generate the time ordering of the corresponding fields. In Feynman gauge, $\alpha =1$, the momentum-space two-point functions areThe Feynman i-epsilon prescription specifies vacuum boundary conditions rather than an arbitrary inverse of the differential operator.

The covariant photon propagator uses four potential components. Its operator formalism counterpart is Gupta-Bleuler quantization: impose $({\partial }_a{A}^a{)}^{(+)}|\mathrm{phys}⟩=0$ and take the Gupta-Bleuler null-state quotient. This leaves a positive physical state space with two transverse photon polarization vectors. The temporal and longitudinal oscillator components occur in intermediate covariant expressions; the Ward identity removes their dependence from physical amplitudes. The free Dirac field uses the canonical anticommutation relations, producing the same fermionic signs as its Grassmann fields in the path integral.

The operator-path-integral equivalence can be seen directly with a regulator. Divide time into small intervals and insert complete sets of field-coordinate states for bosons, and resolutions in fermionic coherent states for fermions. The bosonic matrix elements produce the phase-space factor $\exp i\sum [\pi \Delta \varphi -H\Delta t]$; integrating out the quadratic canonical momentum produces the bosonic action. The fermionic coherent state overlaps produce the first-order term $i{\psi }^{†}\dot{\psi }$ and the Berezin integral measure. Multiplying the short-time kernels recovers the path integral. Projecting the remote endpoints onto the Fock vacuum with an infinitesimal damping selects the same Feynman propagators as the operator formalism. Field insertions become time-ordered products under this construction. Conversely, their quadratic generating functional obeys the canonical free-field equations and has precisely the oscillator two-point functions, so its higher free correlators agree by the Wick theorem. This establishes the equivalence for the regulated free theory and order by order in the perturbation series.

To couple the matter field electromagnetically, promote its global phase symmetry to the local transformationThe gauge covariant derivative $D_a={\partial }_a+ie{A}_a$ obeys $D_a^{\prime }{\psi }^{\prime }=e^{-ie\lambda }{D}_a\psi$. Replacing ${\partial }_a$ by $D_a$ in the Dirac action therefore gives the invariant matter densityThe electromagnetic field tensor is unchanged by the local transformation. Thus the unfixed quantum electrodynamics action is gauge-invariant. The Dirac current $j^a=\bar{\psi }{\gamma }^a\psi$ is conserved by the matter equations, and the Electron and Positron excitations carry opposite charges. The added gauge fixing density selects a representative and is not itself invariant under arbitrary local transformations; it does not change gauge-invariant observables. At a free fermion vertex,For external on-shell Dirac spinors the corresponding current contraction vanishes. The quantum extension is the Ward identity, which makes physical amplitudes insensitive to adding a multiple of the photon momentum to its polarization vector. A gauge-compatible regularization preserves this vector-current identity.

With ${\mathcal{L}}_{\mathrm{int}}=-e\bar{\psi }{\gamma }^a\psi A_a$, expand $\exp (i\int {\mathcal{L}}_{\mathrm{int}})$ in powers of $e$. A term of order $n$ contains $n$ spacetime integrations and $1/n!$. Applying the Wick theorem pairs the free fields: an $A$-$A$ Wick contraction supplies a photon propagator, and a $\psi$-$\bar{\psi }$ Wick contraction supplies an oriented Dirac propagator. Each insertion supplies an interaction vertex. The permutations of contractions cancel the expansion factorials except for the Feynman-diagram symmetry factor. Interchanging Grassmann fields produces the fermionic sign, including a minus sign for every closed fermion loop. This is how Feynman diagrams arise from expectation values, rather than an extra dynamical assumption.

The normalized vacuum generating functional removes components with no external insertions. In the operator formalism, for an interacting-vacuum expectation value of an inserted product $\mathcal{O}$, the same cancellation appears asVacuum projection and the Feynman i-epsilon prescription are implicit. The linked-cluster theorem exponentiates all connected vacuum bubbles into the same factor in numerator and denominator, so it cancels. Normalize by $Z[0]$ to remove every vacuum component. This cancellation of vacuum bubbles still leaves products of disconnected diagrams that each contain external insertions. If only connected correlators are wanted, differentiate the connected generating functional $W=-i\log [Z[J,\bar{\eta },\eta ]/Z[0]]$.

The resulting momentum-space QED Feynman rules for the bare theory can be stated in Feynman gauge as follows:

The illustrated interaction vertex has incoming fermion momentum $p$, incoming photon momentum $k$, and outgoing fermion momentum $p^{\prime }=p+k$. Its solid-line arrows indicate fermion flow. The photon wavy line has no fermion-flow arrow. The propagators, the vertex $-ie{\gamma }^a$, momentum conservation, and the fermionic signs determine the perturbative amplitudes.

Use the Euclidean path integral with weight $e^{-S}$ and Fourier transform convention $\varphi (x)=\int d^{d}p(2\pi {)}^{-d}{e}^{ip\cdot x}\varphi (p)$. The momentum-space Feynman rules are: an internal scalar propagator $1/(p^2+m^2)$; a quartic vertex $-\lambda$ together with $(2\pi {)}^d{\delta }^{(d)}(\sum p_i)$ for incoming momenta; and an integral $\int d^{d}k/(2\pi {)}^d$ over each independent loop momentum. Multiply by the Feynman-diagram symmetry factor; external propagators are retained for a full correlation function and removed for an amputated connected correlation function.

To enumerate the requested one-particle-irreducible Feynman diagrams, let $V$ be the number of quartic vertices and $I$ the number of internal edges. Four external legs imply $4V=2I+4$, while one loop order implies $I-V+1=1$. Thus $V=I=2$. Both internal edges must connect the two vertices: an alternative with a tadpole and one connecting edge would disconnect on cutting that edge. The only graphs are therefore the three pairings of labelled external legs, each with Feynman-diagram symmetry factor $1/2$.

The corresponding channel momenta are $q_s=p_1+p_2$, $q_t=p_1+p_3$, and $q_u=p_1+p_4$, with all external momenta incoming. This exhausts the connected one-loop four-point one-particle-irreducible Feynman diagrams.

For the Euclidean massive loop integral, first take $m^2>0$ and an integer $n>d/2$, where the integral converges. Schwinger parameterization givesThe supplied Gaussian integral then yieldsIntegrating over $s$, and using the Gamma function recurrence to obtain $\Gamma (n)=(n-1)!$, provesOutside its initial convergence range, the right-hand side defines the meromorphic continuation used in dimensional regularization; the divergent ordinary integral is not being assigned a convergent value.

For the scalar bubble integral, the Feynman parameter identity $1/(AB)={\int }_0^{1}dx[xA+(1-x)B{]}^{-2}$ followed by a translation of loop momentum givesWith $d=4-ϵ$, the Gamma function satisfies $\Gamma (ϵ/2)=2/ϵ+O(1)$, and the parameter integral tends to one. Hence the ultraviolet pole is independent of external momentum:One can also see why the same pole occurs without introducing a Feynman parameter: at large $k$, the difference between this integrand and $(k^2+m^2{)}^{-2}$ is ultraviolet integrable near four dimensions, so both have the same dimensional regularization pole. The positive mass avoids an infrared ambiguity in this argument.

In the quantum effective action convention, the tree-level four-point vertex is $+\lambda$, the negative of the amputated connected correlation function tree vertex. The three bubble corrections giveThus the pole counterterm is $\delta \lambda =3{\lambda }^2/(16{\pi }^2ϵ)$. The modified minimal subtraction scheme also subtracts the conventional finite $\log 4\pi -{\gamma }_E$ combination, or equivalently absorbs it into the subtraction-scale convention; that does not change the one-loop renormalization-group beta function. Using that scale convention, writeHere $\lambda$ is dimensionless, $[{\lambda }_B]=ϵ$, and the conventional constant scale factor is implicit. The one-loop tadpole is momentum independent, so wave-function renormalization does not contribute at this order. Differentiating at fixed ${\lambda }_B$, with $a=3/(16{\pi }^2)$, gives $0=ϵ(\lambda +a{\lambda }^2/ϵ)+\beta (1+2a\lambda /ϵ)+O({\lambda }^3)$. ThereforeFor the stable quartic scalar field theory, with $\lambda >0$, the one-loop quartic scalar beta function is positive: the running coupling increases toward the ultraviolet and decreases toward the infrared. The theory is not asymptotically free; extrapolating the one-loop flow gives a Landau pole.

For a connected Feynman diagram with cubic vertices with four external legs and one loop, $3V=2I+4$ and $1=I-V+1$ give $V=I=4$. In a one-particle-irreducible Feynman diagram, every internal edge lies on the single cycle. Each of the four cubic vertices therefore has two internal and one external leg, giving a box Feynman diagram. With labelled external momenta there are three inequivalent cyclic orderings, modulo rotation and reversal, represented by $1234$, $1243$ and $1324$:

Take all momenta incoming with ${\sum }_i{p}_i=0$. The Feynman rules for the Euclidean action give a propagator $(q^2+m^2{)}^{-1}$ and a cubic insertion $-g$. For order $1234$, the amputated connected insertion iswhere $\chi (q)=1_{\{\Lambda <|q|<{\Lambda }_0\}}$ is the high-mode projector for a strict Wilsonian effective action. Each labelled box has Feynman-diagram symmetry factor one. A common fixed-loop-cutoff convention instead restricts only the chosen $q$ integration to the shell; both prescriptions give the same zero-external-momentum value. The effective-action vertex has the opposite sign to this connected insertion. Thus $\boxed{\text{three labelled boxes, with one cyclic topology}.}$

Assume $m^2>0$ and take $\lambda \to 0^{+}$. The free Gaussian integral gives $Z_0=\sqrt{2\pi }/|m|$, and the normalized free Gaussian measure has covariance $C=m^{-2}$. Expanding only the interaction exponential gives the asymptotic expansionIn this zero-dimensional scalar field theory, the path integral has become an ordinary integral, but its combinatorics are precisely those of sextic scalar field theory.

The free generating functional is $⟨e^{J\varphi }{⟩}_0=e^{C{J}^2/2}$. Differentiating it proves Wick theorem: odd moments vanish, while each even moment is the sum over pairings of its factors. Each Wick contraction contributes $C$, soThus the Feynman rules are a quantum field theory propagator $C$ for each internal line and a factor $-\lambda$ for each six-valent interaction vertex. The factorials $V!$ and $(6!{)}^V$ remove vertex and half-edge labels; the remaining weight is $(-\lambda {)}^V{C}^I/S$, where $S$ is the Feynman-diagram symmetry factor, the order of the automorphism group of the resulting diagram, including its half-edge symmetries. A tadpole diagram has an additional interchange symmetry of the two ends of its loop.

There is one empty Vacuum Feynman diagram, contributing $1$. At one interaction vertex, the six half-edges form three loops. Its Feynman-diagram symmetry factor is $2^{3}3!=48$, so its contribution is $-\lambda C^3/48$.

At two interaction vertices let $r$ count the lines joining them. Each interaction vertex has $(6-r)/2$ self-loops, so $r$ can only be $0,2,4,6$. These four possibilities exhaust all Vacuum Feynman diagrams at this order. Their Feynman-diagram symmetry factors areHere the leading $2$ exchanges the two interaction vertices, $r!$ permutes the joining lines, and the remaining factors permute and reverse the self-loops. The $r=0$ graph is disconnected and must be included in $Z/Z_0$; only a connected generating functional would discard it.

Consequently,Indeed ${\sum }_r{S}_r^{-1}=77/7680$, independently agreeing with $11!!/[2(6!{)}^2]$ from Wick theorem.

This is an asymptotic expansion, rather than a convergent perturbation series: the coefficient of ${\lambda }^V$ contains $(6V-1)!!/[V!(6!{)}^V]$ and grows too rapidly for a nonzero radius of convergence. For any fixed truncation order, the Taylor expansion remainder bound for $e^{-x}$ on $x\ge 0$, followed by the finite Gaussian moment integral, proves the stated remainder estimate. For negative real $\lambda$ the original integral diverges.

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.

For incoming momenta, the Euclidean momentum-space Feynman rules are

The one-loop one-particle-irreducible diagrams with at most three external legs can be classified by their external species. There is a one-point $\chi$ tadpole diagram made from a $\varphi$ loop. There are two two-point bubble diagrams: the $\varphi$ self-energy has one internal $\varphi$ and one internal $\chi$, while the $\chi$ self-energy has two internal $\varphi$ lines and symmetry factor $1/2$. For three external legs, three vertices make a triangle: one triangle corrects the $\varphi \varphi \chi$ vertex and contains two internal $\varphi$ lines and one internal $\chi$ line; another has three external $\chi$ legs and a $\varphi$ loop, generating a ${\chi }^3$ interaction. The symmetry $\varphi \mapsto -\varphi$ forbids amplitudes with an odd number of external $\varphi$ legs. The classical $\varphi \varphi \chi$ vertex is the corresponding tree-level three-point diagram.

Adopt the self-energy conventionThis follows by summing the geometric series of exact propagators separated by amputated one-particle-irreducible two-point insertions. At one loop, after writing ${\lambda }_0={\mu }^{ϵ/2}\lambda +O({\lambda }^3)$,Introduce a Feynman parameter and shift the loop momentum. Withdimensional regularization givesUsing $\Gamma (-1+ϵ/2)=-2/ϵ+\gamma -1+O(ϵ)$ produceswhereand one convenient integral form of the finite part isChanging the definition of the dimensional-regularization scale only moves a finite constant between $C$ and the counterterm.

To make the two-point function finite, write ${\varphi }_0=Z_{\varphi }^{1/2}\varphi$, express $m_0^2$ in terms of a renormalized mass and a mass counterterm, and choose the pole parts of $\delta Z_{\varphi }$ and $\delta m^2$ to cancel $B{p}^2/ϵ$ and $A/ϵ$. In the minimal subtraction scheme no additional finite pieces are removed. The physical mass is the pole mass, so with the self-energy convention above it obeysThe explicit $\mu$ dependence of the finite self-energy cancels the running of $m(\mu )$, leaving $m_{\mathrm{phys}}$ independent of the renormalization scale.

The superficial degree of divergence counts the ultraviolet power before subdivergences and symmetry cancellations are considered. At $d=6$, a connected graph made from cubic vertices haswhere $E$ is its number of external legs. Hence one-, two-, and three-point functions can have quartic, quadratic, and logarithmic superficial divergences, whereas graphs with more external legs are superficially convergent.

Full renormalization also requires the $\chi$ mass and wave-function counterterms from its two-point function, a linear $\chi$ counterterm cancelling the tadpole, a coupling counterterm from the divergent $\varphi \varphi \chi$ triangle, and a ${\chi }^3$ counterterm from the three-$\chi$ triangle. A vacuum-energy counterterm removes divergent vacuum diagrams. These are precisely the local operators allowed by power counting in quantum field theory and the exact $\varphi \mapsto -\varphi$ symmetry.

For the Euclidean quartic scalar field theory, expansion of $e^{-S}$ gives the momentum-space rules

At one loop, the four-point one-particle-irreducible correlation function receives the three bubble diagrams in the $s$, $t$, and $u$ channels. If $P$ is the momentum through one channel, its contribution isUsing a Feynman parameter, shifting the loop momentum, and writing ${\Delta }_x=m^2+x(1-x)P^2$ gives, up to terms caused by shifting the boundary of a hard cutoff,Thus every channel has the logarithmic ultraviolet divergenceThe complete one-loop vertex isLet $I_{m,\mathrm{os}}$ denote the bracket evaluated at the chosen on-shell kinematic point. The on-shell renormalization scheme requires $V_a^{(4)}{|}_{\mathrm{os}}=-{\lambda }_{\mathrm{phys}}$, hence the perturbative solution isThe cutoff dependence of ${\lambda }_a$ is the coupling counterterm needed to hold the measured coupling fixed.

The time-domain Feynman rules are:

There is no order-$\lambda$ connected two-point correction. Through order ${\lambda }^2$, the two connected topologies are the two-vertex fish graph and the one-particle-reducible tadpole graph. With the displayed vertex convention,up to the common factors of $i$ associated with the propagator convention. Vacuum normalization removes disconnected vacuum bubbles.