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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.

Introduce a renormalization scale $\mu$ and a quartic counterterm. With $\lambda$ denoting a dimensionless renormalized coupling constant,After factoring out ${\mu }^{\epsilon }$ from the vertex, each loop contains ${\mu }^{\epsilon }{f}_d$. The counterterm contributes $-\delta \lambda$ to ${\hat{\tau }}_4$ and cancels the pole from all three channels. More explicitly,In modified minimal subtraction, replace the pole-only counterterm bywhich subtracts the first three terms inside the brackets. The finite renormalized quartic scalar vertex is thenAnother renormalization condition can change the finite constant, but not the pole cancellation or the momentum-dependent logarithms. The corresponding timelike expressions follow by the same Feynman continuation.

Normalize the vacuum path integral by $Z[0]$. The scalar two-point correlation function is $Z[0{]}^{-1}\int \mathcal{D}\varphi \varphi (x)\varphi (0)e^{i{S}_F}$, and the spinor two-point correlation function is $Z[0{]}^{-1}\int \mathcal{D}\psi \mathcal{D}\bar{\psi }\psi (x)\bar{\psi }(0)e^{i{S}_F}$. Vacuum boundary conditions make these time-ordered Feynman propagators. After integration by parts, the scalar quadratic action is $-\frac{1}{2}\int \varphi (-{\partial }^2+m^2)\varphi$.

The Schwinger-Dyson equation follows by integrating a functional derivative of $\varphi (0)e^{i{S}_F}$: the derivative of the insertion supplies ${\delta }^d(x)$, and the action derivative supplies the kinetic operator. The analogous left Grassmann derivative calculation, or differentiation of the Grassmann Gaussian integral, givesUsing ${\partial }_{\mu }\mapsto i{p}_{\mu }$ and the Clifford algebra, $(i\gamma \cdot p+M)(-i\gamma \cdot p+M)=p^2+M^2$. ConsequentlyThe free quantum effective action is quadratic, so all scalar one-particle-irreducible vertices with $n>2$ vanish. Its two-point vertex is the stated inverse kinetic form $-p^2-m^2$.

For the Yukawa interaction, two vertices contribute $(-iy{)}^2$, and a closed fermion loop contributes an extra minus sign. Tracing the two spinor numerators gives $4[M^2-k\cdot (k-p)]$, since the one-gamma traces vanish. Removing the overall $i$ from the amplitude gives the displayed loop integral. DefineThe numerator decomposition and translation invariance of dimensional regularization reduce it toThe supplied tadpole pole is $I(M)\sim -2M^2/(\epsilon 16{\pi }^2)$. A Feynman parameter combines the two bubble denominators. Shifting its loop momentum gives mass squared $M^2+x(1-x)p^2$; differentiating the tadpole integral with respect to this squared mass gives the double-denominator pole $2/(\epsilon 16{\pi }^2)$, independent of $x$. Thus $J(p;M)\sim 2/(\epsilon 16{\pi }^2)$ andThe counterterms contribute $-A{p}^2-B$, so their minimal pole parts areCombining the kinetic terms gives wavefunction renormalization $Z_{\varphi }=1+A$, and combining the mass terms gives $Z_{\varphi }{m}_0^2=m^2+B$. ThereforeThese are the Yukawa scalar self-energy pole coefficients; finite parts depend on the chosen renormalization condition.

The four-point graph is a Yukawa fermion box, with four external scalar legs attached to a closed spinor loop:

Each high-momentum dirac propagator is $O(k^{-1})$. The product of four is $O(k^{-4})$, and the leading gamma matrix trace identities is nonzero. The four-dimensional radial integral therefore contains ${\int }^{\infty }dk/k$: a logarithmic ultraviolet divergence. This local four-scalar divergence cannot be absorbed by scalar mass or field normalization. Add $-\lambda {\varphi }^4/4!$, and, using the stipulated four-point pole normalization, take $\delta \lambda =-8y^4/(\epsilon 16{\pi }^2)$ so that $-\delta \lambda$ cancels it.

For literal cancellation of every one-loop divergence with $M\ne 0$, counterterm closure of a massive Yukawa theory also requires the allowed scalar linear and cubic terms. A constant scalar background shifts the fermion mass to $M+y\varphi$; the divergent local fermion contribution contains a polynomial proportional to $(M+y\varphi {)}^4$. Its linear and cubic terms are not forbidden by a symmetry when the fermion mass is nonzero. A closed renormalizable family therefore haswith field, mass and coupling redefinitions for both scalar and spinor fields. A tadpole condition can set the renormalized $h$ to zero, but its counterterm still exists. The vacuum constant is needed if vacuum energy is retained. If an exact discrete chiral symmetry is imposed with $M=0$, the scalar potential can be even and the odd terms are forbidden; the essential new interaction is then the quartic one.

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.

The kinetic terms determine the engineering dimensions in $d$ dimensions:Requiring $e\bar{\psi }{\gamma }^{\mu }{A}_{\mu }\psi$ to have dimension $d$ givesThus the electric charge is dimensionless only in four dimensions. In dimensional regularization one writes the bare interaction with $e_0={\mu }^{ϵ/2}{Z}_{e}e$, where $e$ is dimensionless and the renormalization scale $\mu$ supplies the missing dimension.

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 $L[A^a]={\partial }^{\mu }{A}_{\mu }^a$, eliminating the Nakanishi-Lautrup field gives the covariant gauge LagrangianThe first term supplies the gluon kinetic term, three-gluon vertex, and four-gluon vertex. The second makes the quadratic gauge-field operator invertible. The last supplies the ghost propagator and ghost-antighost-gluon vertex.

The nonzero one-loop contributions to the gluon propagator are a gluon bubble with two three-gluon vertices and a closed ghost bubble with two ghost-antighost-gluon vertices. A four-gluon tadpole is also present with a cutoff regulator; for massless fields it is a scaleless integral and vanishes in dimensional regularization.

The amputated connected four-point function through one loop contains the tree quartic vertex, the quartic-counterterm vertex, and three bubble diagrams in the $s$, $t$, and $u$ channels. The unamputated connected function also has one-loop self-energy and two-point-counterterm insertions on each external propagator, which do not determine $\delta g$.

For channel momentum $P$, the bubble has symmetry factor $1/2$ and the Feynman parameter identity gives an integral proportional toIn dimensional regularization, $\Gamma (ϵ/2)=2/ϵ+O(1)$, so the pole from one channel is $g^2/(16{\pi }^2ϵ)$. Summing the three crossing channels gives $3g^2/(16{\pi }^2ϵ)$. The minimal subtraction scheme cancels only this pole, hence

Use dimensional regularization with $d=4-ϵ$ and choose Feynman gauge. Combining the electron and photon denominators with a Feynman parameter, shifting the loop momentum, and discarding the odd term leaves the numeratorThe pole of the rotationally symmetric integral consequently givesWith the inverse-propagator conventionthe minimal subtraction scheme chooses the counterterm pole to equal ${\Sigma }_{\mathrm{pole}}$. Hence

For one four-component Dirac field with scalar interaction $-y\bar{\psi }\psi \varphi$ and the mostly-plus propagator convention, the fermion loop numerator is $4[M^2-k\cdot (k-p)]$. Reduce it to a bubble and tadpole: $-4y^2[(2M^2+p^2/2)J-I]$. Dimensional regularization gives pole parts $J=2/(\epsilon 16{\pi }^2)$ and $I=-2M^2/(\epsilon 16{\pi }^2)$. The local pole then requires both wavefunction renormalization and a scalar mass counterterm.