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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.

There is an interpretation issue in this clause. A proper quartic vertex, defined as the fourth derivative of the quantum effective action, has no additional tree or reducible contribution in the bare cubic theory: the three boxes exhaust its one-loop diagrams. The strict local zero-momentum vertex of a Wilsonian effective action also has no high-mode exchange bridge, since the momentum carried by that bridge would be zero, outside the shell. Thus $\boxed{\text{for a local proper or strict Wilsonian quartic vertex, there are no additional diagrams}.}$

If “other diagrams” means the amputated connected correlation function, there are additional one-particle-reducible Feynman diagrams. Specify full external-propagator amputation and a vacuum with tadpole subtraction. The three tree exchange channels each contribute $g^2/m^2$ at zero momentum. At one loop, a triangle Feynman diagram can replace either cubic end of any exchange, giving six diagrams; a bubble diagram can be inserted in the internal exchange line, giving three diagrams with symmetry factor $1/2$. The proper lower-point vertices at zero momentum, from the same trace-log expansion, areTherefore the reducible part of the connected four-point amplitude isAdding the three boxes gives ${\mathcal{A}}_4=3g^2/m^2+3g^4{I}_4+6g^4{I}_3/m^2+3g^4{I}_2/(2m^4)$ in this convention. This is the cubic-scalar exchange corrections at zero momentum result, obtained equivalently from ${\mathcal{A}}_4=-{\Gamma }^{(4)}+3({\Gamma }^{(3)}{)}^2/{\Gamma }^{(2)}$.

If the tadpole has not been subtracted, it is an additional one-loop subgraph that must not be forgotten. The proper one-point function is $t=g{I}_1/2$. Its attachment to the internal exchange line gives another three reducible diagrams, contributing $3g^4{I}_1/(2m^6)$; equivalently the stationary background shifts by $v=-g{I}_1/(2m^2)$ and changes the exchanged mass by $gv$. A linear counterterm enforcing $v=0$ cancels this contribution. External self-energy or tadpole diagram attachments are removed by the stated full-propagator amputation; using bare external amputation would retain them instead. In that convention the four external-line decorations add $6g^4{I}_2/m^4+6g^4{I}_1/m^6$ to the displayed full-amputated result, with the $I_1$ part absent after tadpole subtraction. If the expansion is expressed in renormalized parameters, also include the corresponding counterterm diagrams: replacing either cubic exchange vertex gives $6g\delta g/m^2$, and a mass insertion on the bridge gives $-3g^2\delta m^2/m^4$. A linear counterterm adds $3g^3\delta h/m^6$ through the same tadpole attachment and cancels it when $\delta h=-g{I}_1/2$. The kinetic counterterm has zero insertion at zero exchanged momentum. At one loop these exhaust the possibilities: the single cycle has length four, three, two or one, corresponding respectively to box, triangle, bubble or tadpole, with tree branches attached.

The mass dimensions are $[g]=0$, $[I_r]=6-2r$, $[m]=1$. Accordingly $g^2/m^2$, $g^4{I}_4$, $g^4{I}_3/m^2$, $g^4{I}_2/m^4$ and $g^4{I}_1/m^6$ all have dimension $-2$. These connected exchanges are generally nonlocal functions of external momentum before taking the zero-momentum value; they should not be silently reclassified as an independent local quartic coupling.

Continue with the cubic scalar field theory of c, and use precisely its counterterm signs. Take $m^2>0$ to separate ultraviolet divergences from infrared divergences. Make a Wick rotation and impose the rotationally invariant cutoff regularization $|q_E|<\Lambda$. PutFor example, spherical coordinates give $I_1=C{\int }_0^{{\Lambda }^2}{t}^2/(t+m^2)dt$. Polynomial division and then differentiation with respect to $m^2$ give the divergent partsThe derivative relations are $I_2=-{\partial }_{m^2}{I}_1$ and $I_3=\frac{1}{2}{\partial }_{m^2}^2{I}_1$; differentiation also generates finite terms, which are irrelevant here. The first supplied integral identity is consistent with these divergent parts after Wick rotation.

The second supplied expansion is incorrect. Write $A=p^2-m^2$. Direct multiplication of the two denominators gives $(A+k^2{)}^2-4(p\cdot k{)}^2$, so the correct Taylor expansion isA concrete counterexample is $p=0$, $m=1$, $k=(ϵ,0)$: the exact expression is $(1-{ϵ}^2{)}^{-2}=1+2{ϵ}^2+O({ϵ}^4)$, whereas the printed expansion gives $1+{ϵ}^2/2+O({ϵ}^4)$. We therefore compute the self-energy without that identity.

For Euclidean external momentum $P$, define the scalar bubble integralA Feynman parameter $x$ combines its denominators into one squared denominator with $K=m^2+x(1-x)P^2$. Applying the radial result above and integrating ${\int }_0^{1}x(1-x)dx=1/6$ yieldsOne can instead expand with the ultraviolet cutoff fixed on the unshifted loop momentum: the $P^2$ coefficient of the integrand is $-(q_E^2+m^2{)}^{-3}+\frac{2}{3}{q}_E^2(q_E^2+m^2{)}^{-4}$, using rotational averaging in six dimensions. Its leading radial term is $-1/(3q_E^6)$, giving the same logarithm. Shifting a sharp integration boundary changes finite terms in this scalar bubble integral, and does not change the displayed divergent coefficients.

To fix all signs and symmetry factors together, expand the one-loop Euclidean effective action around a background scalar field:These terms are the tadpole diagram, the two-point bubble diagram, and the triangle Feynman diagram. The tadpole diagram has factor $1/2$; the two-point bubble diagram also has factor $1/2$ after two differentiations with respect to the background scalar field. The local three-point interaction vertex is $g^3{I}_3$. Thus their divergent contributions to the one-, two-, and three-point Euclidean one-particle irreducible vertices are respectivelyThe Euclidean counterterm Lagrangian is ${\delta }_Z({\partial }_E\varphi {)}^2/2+{\delta }_m{\varphi }^2/2+{\delta }_g{\varphi }^3/6+{\delta }_h\varphi +{\delta }_{\mathcal{E}}$. Cancelling each divergent one-particle irreducible vertex therefore givesFor example, the two-point sum is ${\delta }_Z{P}^2+{\delta }_m-g^{2}B(P)/2$ and its divergent coefficients cancel separately. Finite counterterms depend on the chosen renormalization conditions. The zero-mass limit can use a positive subtraction scale in the logarithms after infrared regularization.

If unnormalized vacuum functionals are retained, specify the determinant measure byExpanding its radial integrand at large $q_E$ fixes the additional vacuum energy density counterterm:The purely mass-independent terms depend on the determinant normalization. Normalized Green functions eliminate this entire constant.

Set the renormalized $g_3$ to zero. For a six-scalar vertex, b gives $\omega =-n_2$, so a new overall logarithmic divergence can only occur in graphs with $n_2=0$. With $b=6,f=0$, the line counts become $I_f=n_1$ and $I_b=n_1-3$, hence $L=n_1-2$.

At one loop this forces $n_1=3$ and $I_b=0$: the only candidate is a closed triangle Feynman diagram with two external scalar legs at each vertex. Its numerator at large loop momentum is proportional toThis vanishes for the specified two-component gamma matrices, each of which is traceless. Replacing one leading $/k$ by the fermion mass or an external momentum lowers the superficial degree of divergence from zero to at most $-1$. Thus the apparent logarithm cancels. Graphs with a $g_2$ vertex already have negative overall degree; the pure-scalar one-loop triangle, for instance, has $\omega =-3$. There is no one-loop ultraviolet subdivergence inside these proper one-loop vertices. ThereforeThe six-point correlation function need not vanish: finite terms are allowed. Nor is it correct to say that all odd gamma-matrix traces vanish in $2+1$ dimensions. In this representation,The cancellation above follows from repeating the same leading loop momentum, not from a false four-dimensional trace rule.

At two loops, $n_1=4$, $I_f=4$ and $I_b=1$ are possible. A closed fermion box with a scalar line joining two of its vertices leaves six external scalar legs. It has $L=2$, $\omega =0$, and four leading gamma matrices; the one-loop odd-numerator argument no longer applies.

A background-field check demonstrates that its logarithm is actually nonzero. Put $\varphi =v+\eta$ with constant $v$. Define the background fermion mass $\mu =m+g_1{v}^2/2$, the background scalar squared mass ${\mathcal{B}}^2=M^2+g_2{v}^2/2$, and the induced three-leg Yukawa coupling $y=g_{1}v$. The two-loop quantum effective potential contains the sunset diagram with two fermion lines and one scalar line. After a Wick rotation, its relevant integral, apart from a nonzero overall combinatorial factor, isHere the dots and gamma-matrix traces are Euclidean. With $A=k^2+{\mu }^2$, $B=l^2+{\mu }^2$, $C=(k-l{)}^2+{\mathcal{B}}^2$, its numerator is exactlyThe first three terms reduce to products of one-loop tadpole integrals, which have no logarithmic pole near three dimensions in dimensional regularization. The remaining term isFor positive regulator masses, the large-momentum region of $H$ has a positive, nonzero logarithmic divergence. One can see this directly by restricting the relative angles and the ratio $|k|/|l|$ to compact intervals away from zero: the remaining common radial integral is proportional to ${\int }^{\Lambda }dr/r$. Its proper one-loop subgraphs are ultraviolet convergent.

Finally, the coefficient multiplying $H$ containsThe two-loop sextic counterterm in quartic Yukawa theory follows because its $v^6$ coefficient is $g_1^4\ne 0$ for a nonzero coupling. Other two-loop vacuum topologies reduce to products of finite-in-dimensional-regularization tadpoles and do not cancel this logarithmic pole. ThusA sextic counterterm is required even if the renormalized sextic coupling is set to zero at a chosen renormalization scale. A theory constrained to $g_3=0$ at every scale lacks counterterm closure when $g_1\ne 0$. The decoupled exception $g_1=0$ is the three-dimensional scalar quartic interaction, which needs no sextic counterterm by power counting.

Expanding $e^{i{S}_{\mathrm{int}}}$ gives the Feynman vertexThere is one scalar leg and a continuous oriented fermion line; no extra factor $i$ is absorbed into ${\gamma }^5$. The Dirac propagator is $S_F(k)=i(/k+M)/(k^2-M^2+i0)$, where $/k={\gamma }^{\mu }{k}_{\mu }$. Each closed fermion loop supplies a minus sign.

The tadpole diagram is proportional tobecause both numerator traces vanish by a. This is zero before the loop momentum integration.

For either orientation of the triangle Feynman diagram, let $a,b,c$ be its three internal four-momenta. Its numerator trace isAfter using cyclic property of the trace, each expanded term is a trace of ${\gamma }^5$ times at most three ordinary gamma matrices. Such traces vanish in four dimensions. For an odd number of ordinary gamma matrices, conjugation by ${\gamma }^5$ changes the sign. For two ordinary gamma matrices, the contracted Clifford algebra identity ${\gamma }_{\rho }{\gamma }^{\mu }{\gamma }^{\nu }{\gamma }^{\rho }=4{\eta }^{\mu \nu }I$ givesThe zero-matrix case is a. Hence both the one-point and three-point one-loop bosonic vertices vanish, independently of their external momenta. No cancellation between different loop orientations or delicate shift of a divergent integral is needed.