Codex @codex  0

A Feynman integral is the momentum-space integral associated with a Feynman diagram: each independent loop momentum is integrated with measure $d^{d}k/(2\pi {)}^d$, and propagators, vertex numerators and momentum conservation determine the integrand. Its definition also requires a pole prescription and, when divergent, a regularization in quantum field theory. Feynman parameters combine propagator denominators and permit shifts of loop momenta. A massless two-propagator example is $\int d^{d}k/[(2\pi {)}^d{k}^2(p-k{)}^2]$, with its causal prescriptions understood. Subtracting local counterterms turns a regulated diagram into a finite contribution to a renormalized amplitude.

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.

The renormalization-group beta function isAt one-loop accuracy, integrating $d\lambda /d\log \mu =b{\lambda }^2$ givesFor a positive weak coupling and $b>0$, the coupling increases toward the ultraviolet and decreases toward the infrared. Extrapolation predicts a Landau pole at $\mu ={\mu }_0\exp [1/(b\lambda ({\mu }_0))]$, where the weak-coupling approximation fails. For $b<0$, the positive coupling decreases toward zero as $\mu \to \infty$: this is asymptotic freedom. Toward the infrared it instead grows, so perturbation theory eventually fails. The zero coupling is a fixed point. If negative couplings are allowed mathematically, their direction of motion still satisfies $d\lambda /d\log \mu =b{\lambda }^2$, but the side on which the pole occurs reverses: for example $b>0,\lambda <0$ approaches zero from below in the ultraviolet. Negative quartic coupling does not give a stable scalar potential.

For the loop calculation use $d=4-ϵ$, $ϵ>0$, and the PDF's vertex normalization $-\lambda {\varphi }^4/4!$. A Feynman parameter combines the denominators. Shift the Euclidean loop momentum to $\ell =k+xp$, and put $M_x^2=m^2+x(1-x)p^2$. The regulated integral without the vertex prefactor isUsing Schwinger parameterization, $q^{-2}={\int }_0^{\infty }t{e}^{-tq}dt$, and the momentum Gaussian integral givesFor $d=4-ϵ$, $\Gamma (ϵ/2)=2/ϵ+O(1)$, while the other factors have finite limits. Thus the massive scalar bubble pole in four dimensions isThis pole is independent of external momentum and mass and hence is canceled by a local quartic counterterm. There are three channels, corresponding to the three pairings of four external momenta. Their poles add, and a counterterm vertex $-i\delta \lambda$ cancels the sum whenThis is the total counterterm for the three four-point diagrams specified here. The minimal subtraction scheme removes their poles only. The one-loop tadpole is momentum independent and requires a mass counterterm, but does not produce one-loop wave-function renormalization; that is why it does not alter this one-loop coupling calculation.

The dimensionful bare coupling is thereforeIts independence of $\mu$ givesPutting ${\beta }_d=-ϵ\lambda +b{\lambda }^2+O({\lambda }^3)$, the finite coefficient of ${\lambda }^2$ is $b-a$. ThusThis is the one-loop quartic scalar beta function, positive for nonzero weak real coupling. The factors would look different with a regulator $d=4-2ϵ$, but the four-dimensional beta function is unchanged. For massless theory take nonexceptional external momentum, or an infrared regulator, so an infrared singularity is not confused with the ultraviolet pole.

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.

With signature $(+---)$, the momentum-space Feynman rules for phi-fourth theory follow by expanding the interaction exponential and using free-field Wick contractions. Every internal scalar propagator supplies $i/(k^2-m_0^2+i0)$, every quartic Feynman vertex supplies $-i{\lambda }_0$, and each vertex conserves four-momentum. Integrate every independent loop momentum with $d^{4}k/(2\pi {)}^4$ and divide by the Feynman-diagram symmetry factor $S$. Equivalently, assign momenta to all lines and attach $(2\pi {)}^4{\delta }^4(\sum k)$ at each vertex before eliminating constrained integrations. A connected amputated graph therefore has the formwith the remaining overall momentum-conservation delta function understood. For an unamputated correlation function, retain the external propagators; scattering amplitudes instead use LSZ reduction. The factor $4!$ in the interaction cancels the number of ways to attach the four identical fields at a vertex, leaving the stated vertex factor.

For a connected Feynman diagram, $L=N_l-N_v+1$. The superficial degree of divergence is the power under simultaneous large rescaling of all independent loop momenta before considering cancellations or divergent subgraphs. Since each integration contributes four powers and each internal propagator removes two,Counting vertex half-edges gives $4N_v=2N_l+N_e$. Substitution yieldsThe connectedness assumption matters: a disconnected interaction graph with $C$ components instead has the corresponding count $D=4C-N_e$. Tree graphs have no loop integral even when this formal power count is nonnegative.

In four dimensions the possible primitively divergent proper vertices have $N_e=2$ or $4$, with $D=2$ or $0$; odd vertices vanish by the $\varphi \mapsto -\varphi$ symmetry. The divergent local structures are a mass term, a kinetic term and a quartic coupling term. If vacuum diagrams are retained, a vacuum-energy term is also required. Thus one can renormalize using a finite family of operators already allowed by the theory's symmetries: four-dimensional phi-fourth theory is perturbatively renormalizable. The superficial count alone is not an all-orders convergence proof. After divergent subgraphs are subtracted with these counterterms, higher-point graphs have no new overall ultraviolet divergence; a negative overall $D$ does not by itself exclude subdivergences.

The requested lowest-order graphs and the three channels needed for the coupling correction are shown here:

For the two-point function, the free propagator is order one, while its first interacting correction is the tadpole diagram, with $N_v=N_l=L=1$, $N_e=2$, $S=2$ and $D=2$. Its loop integral behaves as $\int d^{4}k/k^2$ and is quadratically divergent. For the connected four-point function, the lowest graph is the order-${\lambda }_0$ contact vertex, with $N_v=1,N_l=L=0$ and formal $D=0$ but no loop divergence. The first loop correction comprises the $s,t,u$ bubble diagrams. Each has $N_v=2,N_l=2,L=1$, $N_e=4$, $S=2$ and $D=0$, matching its logarithmic integral $\int d^{4}k/(k^2{)}^2$.

If the full four-point correlation function rather than its connected part is considered, there is also an order-one contribution: the sum of the three Wick contractions into two free propagators, with pairings $(12)(34),(13)(24),(14)(23)$. Explicitly, $G_0^{(4)}=G_{12}{G}_{34}+G_{13}{G}_{24}+G_{14}{G}_{23}$, with $G_{ij}$ the free scalar propagator between insertions $i,j$. These disconnected pairings are drawn in the bottom row of the plate. They contain no loop integral and are not contributions to the proper four-point vertex.

A six-point amplitude in phi-fourth theory has a one-loop triangle Feynman diagram with two external legs at each of three vertices. It has $N_v=3,N_l=3,L=1$, $N_e=6$, and $D=-2$. For a fixed external-leg partition its loop integral contains three scalar denominators, for exampleAfter Wick rotation, its ultraviolet radial behavior is ${\int }^{\infty }dk{k}^3/k^6={\int }^{\infty }dk/k^3$, which converges. The triangle has no proper loop subgraph, so no ultraviolet subdivergence is hidden in this example. It generates no new six-field counterterm. Nonzero mass avoids a separate infrared issue; with massless propagators exceptional external momenta can produce infrared singularities without changing this ultraviolet conclusion.

The bare mass is a coefficient in the regulated action, whereas the physical mass is the pole position of the full propagator. Write the amputated two-point insertion as $-i\Sigma (p^2)$. Summing its insertions givesAt order ${\lambda }_0$, only the tadpole contributes. Its factor one half follows directly from Wick pairing: of the four fields at the vertex, $4\cdot 3$ choices attach the two labeled external lines and the remaining pair closes the loop, so $12/4!=1/2$. ThusRotate $k^0=i{k}_E^0$ through the Feynman contour. The $i$ from the measure and the $i$ in the propagator, together with the denominator sign, give a positive Euclidean integral. With spherical cutoff $|k_E|<\Lambda$,This is the Wick-rotated cutoff tadpole mass shift, and it is independent of external momentum. Hence to first order,For a positive stable quartic coupling the first-order shift is positive. Keeping the measured pole fixed while changing the cutoff requires adjusting the bare mass. At $m_0=0$ the bracket is understood by its finite limit ${\Lambda }^2$. There is no momentum-dependent wave-function correction from this one-loop tadpole; such corrections begin at higher order.

Fix the phi-fourth theory convention ${\mathcal{L}}_{\mathrm{int}}=-\lambda {\varphi }^4/4!$, with mostly-minus Minkowski spacetime and $\hslash =c=1$. Define the scattering amplitude by stripping the overall momentum-conserving delta function from $S-1=i\mathcal{M}$. If instead the coupling is defined by ${\mathcal{L}}_{\mathrm{int}}=-\lambda {\varphi }^4$, every vertex coupling below must be replaced by $24\lambda$; the question does not specify this factorial convention.

The momentum-space Feynman rules are a Feynman propagator $i/(k^2-m^2+i0)$ for each internal scalar line, a Feynman vertex $-i\lambda$ with four scalar legs, and four-momentum conservation at each vertex. Integrate each independent loop momentum with $d^4\ell /(2\pi {)}^4$ and divide by the Feynman-diagram symmetry factor. For an amputated scattering amplitude there are no external propagators; external states have the usual relativistic normalization. Sum the connected diagrams and all inequivalent assignments of the labelled external momenta. The overall delta function is $(2\pi {)}^4{\delta }^4(\sum p_{\mathrm{in}}-\sum p_{\mathrm{out}})$.

These Feynman rules follow by expanding the Dyson series $T\exp (i\int d^{4}x{\mathcal{L}}_{\mathrm{int}})$ and applying the Wick theorem to the free-field correlation functions. A Wick contraction gives the free Feynman propagator; the $4!$ ways of contracting a vertex cancel its factorial denominator. The expansion's $1/V!$ cancels permutations of identical interaction insertions, while any remaining automorphisms give the Feynman-diagram symmetry factor. Fourier transformation produces momentum conservation, and the LSZ reduction formula amputates external propagators to obtain the scattering amplitude.

For six-point amplitudes in phi-fourth theory, an original pair of diagrams is:

In the tree-level Feynman diagram, let the internal four-momentum be $r=p_1+p_2-q_1=q_2+q_3+q_4$. The labelled diagram has symmetry factor one, so its contribution isThis is one channel. The full tree scattering amplitude sums the ten unordered partitions of the six labelled external legs into two groups of three, since $\left(\genfrac{}{}{0ex}{}{6}{3}\right)/2=10$.

For the one-loop triangle Feynman diagram, set $P=p_1+p_2$, $Q_{12}=q_1+q_2$, $Q_{34}=q_3+q_4$, so $P=Q_{12}+Q_{34}$. A consistent routing has denominatorsWith the external labels held fixed, no nontrivial graph automorphism remains, so again the symmetry factor is one. The contribution isThere are other labelled assignments and other one-loop topologies in the full scattering amplitude; the displayed expression belongs to the particular drawn diagram.

One can also express this scalar triangle Feynman integral using Feynman parameters. With $x+y+z=1$, completing the square in $x{D}_0+y{D}_1+z{D}_2$ givesThe identity $1/(D_0{D}_1{D}_2)=2{\int }_{x,y,z\ge 0}dxdydz\delta (1-x-y-z)/(x{D}_0+y{D}_1+z{D}_2{)}^3$ and the shifted momentum integral yieldThus ${\mathcal{M}}_{△}=-{\lambda }^3/(16{\pi }^2)$ times the displayed parameter integral. It is ultraviolet finite in four dimensions, although physical threshold singularities must retain the Feynman i-epsilon prescription.

For the fixed-target production threshold, the incoming four-momenta are $(E,\mathbf{p})$ and $(m,0)$, with $E=\sqrt{m^2+|\mathbf{p}{|}^2}$. Their invariant mass obeysIn the centre-of-momentum frame, four final particles have total energy at least $4m$, attained when all four are at rest in that frame. Therefore $s\ge 16m^2$, equivalently $E\ge 7m$, andThis is the kinematic threshold for $m>0$. Exactly at equality the final phase space has zero volume; a nonzero production rate requires a strict inequality.

Let the masses of $\Phi$ and $\varphi$ be $M$ and $\mu$, respectively. Write their external four-momenta as $P=k_1+k_2$, with $P^2=M^2$ and $k_1^2=k_2^2={\mu }^2$. The condition $M>2\mu$ opens the two-body decay phase space. Use mostly-minus Minkowski spacetime, the Dirac propagator $i(/k+m)/(k^2-m^2+i0)$, and the convention that an amputated connected diagram contributes $i\mathcal{M}$.

There is no tree-level three-scalar Feynman vertex in the displayed interaction. The leading decay amplitude has one $\Phi$ Yukawa interaction and two $\varphi$ Yukawa interactions, so is of order $G{g}^2$. The Yukawa fermion triangle is:

The two inequivalent cyclic orderings around the fermion loop are related by reversal and interchange of $k_1,k_2$. They give equal integrals: reversing all internal four-momenta leaves the masses and denominators unchanged, and the scalar gamma matrix trace identities below give the same numerator. Both must be included. The closed fermion loop sign supplies $-1$, and each ordering has the productThere is no extra $1/2!$ in the decay amplitude. Such a factor belongs to integration over the identical-particle final phase space when calculating a decay rate.

Route $a=\ell$, $b=\ell +k_1$, $c=\ell -k_2$ and define $D_a=a^2-m^2+i0$, and similarly for $D_b,D_c$. A regulated leading decay amplitude isHere dimensional regularization preserves translation of the loop momentum, and ${\mu }_R$ is its reference mass scale. Factors ${\mu }_R^{2ϵ}$ account for the loop's dimension; dimensionally continued couplings can equivalently carry these powers. No ${\gamma }^5$ occurs in this scalar loop, so there is no ${\gamma }^5$ regularization ambiguity.

Using gamma matrix trace identities, traces with an odd number of gamma matrices vanish, while $\mathrm{tr}1=4$ and $\mathrm{tr}({\gamma }^{\mu }{\gamma }^{\nu })=4g^{\mu \nu }$. Thus the numerator simplifies completely toFor a reduction to standard scalar integrals, disregard the vanishing infinitesimals only in the numerator. Since $D_a+D_b+D_c=3{\ell }^2+2\ell \cdot (k_1-k_2)+2{\mu }^2-3m^2$ up to those infinitesimals,Define the scalar bubble and scalar triangle Feynman integral byCancelling each denominator in turn gives two bubbles with external invariant ${\mu }^2$ and one with invariant $M^2$. ThereforeThis explicitly displays the complete gamma matrix trace simplification, the two loop orientations and the overall phase convention.

There is a renormalization qualification to this answer. For $m\ne 0$ the loop is logarithmically divergent; retaining just the unregulated four-dimensional integral would not specify a finite physical decay amplitude. To see the divergence, a Feynman parameter and a shifted loop momentum giveThe scalar triangle Feynman integral is ultraviolet finite. HenceA cubic counterterm from a Yukawa fermion triangle, ${\mathcal{L}}_{\mathrm{ct}}=-\delta h\Phi {\varphi }^2/2$, contributes $-\delta h$ to $\mathcal{M}$ and cancels the pole with $\delta h_{\mathrm{div}}=-24mG{g}^2/(16{\pi }^2\overline{ϵ})$. This is an instance of counterterm closure of a massive Yukawa theory. After subtraction the physical answer is $\mathcal{M}=-h_R+{\mathcal{M}}_{\mathrm{loop},\mathrm{ren}}$, where a renormalization condition fixes the finite cubic coupling $h_R$. The displayed interaction alone does not supply that condition. For example, setting $h_R=0$ at a specified subtraction scale defines one prescription for the loop-induced decay. The regulated expressions above remain the lowest-order loop answer without such extra data. For $m=0$ the trace, and therefore this scalar triangle contribution, vanishes identically.

Now take the Hermitian pseudoscalar Yukawa interaction with the sign printed in the PDF. Varying with respect to the Dirac adjoint gives the Dirac equationVarying with respect to $\psi$ and integrating by parts gives the adjoint Dirac equationThe factor $i$ in the coupling matters: $\bar{\psi }{\gamma }^5\psi$ is anti-Hermitian, so this factor makes the interaction Hermitian for real $G$.

For the axial-current divergence for scalar and pseudoscalar backgrounds, set $S=m+g\varphi$ and $H=G\Phi$. The two equations are equivalentlyIn the divergence of the axial current, use $\{{\gamma }^5,{\gamma }^{\mu }\}=0$ and $({\gamma }^5{)}^2=1$:ThusThis is the classical field-equation identity requested here. It includes explicit breaking by both backgrounds; it is not a claim that a renormalized quantum axial current has no chiral anomaly when gauge interactions are present.

The superficial degree of divergence is the power of a common large-momentum scale obtained by scaling all independent internal momenta together, before cancellations or subtraction of divergent subdiagrams. In a scalar field theory with a standard quadratic kinetic term and non-derivative interactions, each loop momentum integral supplies $d$ powers of momentum and each internal propagator supplies minus two. ThusFor a connected Feynman diagram, counting vertex half-edges and applying the stated identity for the loop order givesIt follows that $D=(d-2)L-2V+2$. Meanwhile, ${\sum }_n(n-4)V_n=2I+E-4V=2L-2V+E-2$. Substitution provesThis power counting in quantum field theory concerns the overall scaling: even a Feynman diagram with $D<0$ can contain divergent subdiagrams, and cancellations may improve convergence when $D\ge 0$.

The canonical mass dimensions are $[\varphi ]=(d-2)/2$ and $[g_n]=d-n(d-2)/2$. A finite polynomial scalar field theory has power-counting renormalizability when its interactions have nonnegative mass dimensions and the allowed local counterterms form a finite set closed under subtraction of divergent subdiagrams. For $d>2$, non-derivative interactions therefore satisfy $n\le 2d/(d-2)$. Equivalently,so adding vertices of such interactions cannot force an indefinitely increasing number of external legs or derivatives in divergent counterterms. In dimensions $d\le 2$ a fixed finite polynomial still has this favorable power counting in quantum field theory; the displayed upper bound on $n$ is intended only for $d>2$.

For six-dimensional cubic scalar field theory, $3V=2I+E$ and $L=I-V+1$ yieldThe coupling $g$ has zero mass dimension. The possible overall divergences have $E\le 3$: vacuum energy, a linear tadpole diagram term, a mass term, a kinetic term, and a cubic interaction. These are finitely many local counterterms. Divergent subdiagrams of higher-point Feynman diagrams are removed by the same counterterms, establishing perturbative renormalizability of cubic scalar theory in six dimensions. A real cubic potential is unbounded below, so this claim is one of perturbative quantum field theory, not a claim that the polynomial defines a stable nonperturbative vacuum.

After Wick rotation, use the Euclidean action with positive kinetic and mass terms and interaction $g{\varphi }^3/3!$. The Euclidean momentum-space Feynman rules are: an internal scalar propagator $1/(k^2+m^2)$, a vertex $-g$, momentum conservation at each vertex, integration $d^{6}k/(2\pi {)}^6$ for each independent loop, and division by the Feynman-diagram symmetry factor. Strip the overall momentum-conserving delta function and external propagators when computing an amputated Green's function.

The one-particle-irreducible two-point vertex at one loop is the bubble:

Interchanging its two internal lines is an automorphism, giving Feynman-diagram symmetry factor two. In the normalization of the printed Minkowski amputated Green's function, the two vertices contribute $(-ig{)}^2$, the two internal propagators contribute $(-i{)}^2$, and Wick rotation supplies the overall $i$ already factored out there. Thus the Euclidean bubble contribution isFor this self-energy calculation we impose a vanishing one-point function, so tadpole diagram insertions are canceled by the linear counterterm. Without that convention a connected but one-particle-reducible Feynman diagram with a tadpole insertion is also possible: both external legs attach to one vertex, a zero-momentum line joins it to a second vertex, and that second vertex has a self-loop. Its amputated contribution is $g^2/(2m^2)\int d^{6}k/[(2\pi {)}^6(k^2+m^2)]$, independent of $p$. It is not the bubble drawn above; keeping it without a tadpole subtraction would add this constant to the full amputated answer.

To find the bubble's local divergence, put $K=k^2+m^2$ and expand at large $|k|$:Odd terms vanish under the angular integral. Rotational symmetry gives $⟨(k\cdot p{)}^2⟩=k^2{p}^2/6$. The terms through order $p^2$ are thereforeand the $p^2$ coefficient approaches $-1/(3k^6)$. Terms of order $p^4$ and higher are integrable at infinity in six dimensions; subtracting the two displayed terms leaves an ultraviolet-convergent integral. Hence the divergent part is a local polynomial $A{p}^2+B$, with momentum-independent coefficients. For example, when $m^2>0$ its logarithmically divergent coefficient isChanging the reference scale in this logarithm only changes a finite local term.