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For an overall logarithmically divergent one-loop Feynman integral, its leading large-momentum term fixes its ultraviolet pole. Giving its denominators a positive auxiliary mass $M$ regulates the artificial small-momentum singularity produced by setting external scales to zero. The difference from the original leading expansion must be ultraviolet convergent. This justifies extracting its ultraviolet pole without mistaking a scaleless integral for an ultraviolet-finite one; multiloop subdivergences require additional treatment.

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.

The two-point function Feynman integral with two interaction vertices joined by five parallel scalar propagators has four independent loop momenta:This formula uses Euclidean momenta. In three dimensions the convolution theorem and $G(x)=e^{-m|x|}/(4\pi |x|)$ givewith cutoff regularization required at $\rho =0$. With a lower radial cutoff $a$, the coefficient of $p^2$ has ultraviolet divergence $-\log (1/(ma))/(1536{\pi }^4)$. It therefore generates a divergent wave-function renormalization in three-dimensional sextic scalar field theory. By contrast the three-line sunset diagram has only a momentum-independent logarithmic ultraviolet divergence in three dimensions. This illustrates why a superficial degree of divergence of two allows both a mass counterterm and a kinetic counterterm, while degree zero allows only the former in a two-point function.

The following scalar toy isolates one step in renormalization: removing a local divergence while retaining a finite dependence on a physical mass. For $M,\mu >0$ and a sharp cutoff regularization in Euclidean space with momentum bound $\Lambda$, define the dimensionless Feynman integralThis is a zero-external-momentum scalar bubble integral stripped of coupling and combinatorial factors. Using the three-sphere area $2{\pi }^2$ and $t=k_E^2$ reduces the integral toThe antiderivative follows by writing $t=(t+M^2)-M^2$. At large $\Lambda$ the divergent part is $\log {\Lambda }^2/(16{\pi }^2)$, with the logarithm's dimensionful argument understood only within the displayed ratios. It has no external-momentum dependence. In a four-scalar-field vertex this structure can therefore be removed by a local operator proportional to ${\varphi }^4$, rather than a momentum-dependent nonlocal interaction.

Choose the fixed reference mass $\mu$ and subtract $I_{\Lambda }(\mu )$. If the loop contribution to a chosen four-point vertex coefficient is $c{\lambda }^2{I}_{\Lambda }(M)$, choose the additive counterterm in that same coefficient as $\delta {\lambda }_{\Lambda }=-c{\lambda }^2{I}_{\Lambda }(\mu )$, where $c$ includes its stated diagram factors. This defines a finite subtraction prescription:The renormalization condition for this toy loop contribution is $I_R(\mu ;\mu )=0$. The reference mass stays fixed when $M$ is varied; finite dependence on $M$ survives. A different finite subtraction changes the renormalization scheme, and the defined renormalized coupling must change consistently.

For $M=2\mu$, the limiting result is $16{\pi }^2{I}_R=-\log 4\approx -1.38629436$. At $\Lambda =10\mu$, the exact regulated difference is about $-1.32846343$; at $\Lambda =100\mu$, it is about $-1.38569459$. Expanding the exact expression gives the leading residual $2(M^2-{\mu }^2)/{\Lambda }^2$ in $16{\pi }^2[I_{\Lambda }(M)-I_{\Lambda }(\mu )]$. This explicitly separates finite-cutoff error from the chosen finite renormalization condition.

A sharp cutoff regularization is convenient for this scalar demonstration, but shifts of a divergent loop momentum can change its integration domain. A gauge theory calculation cannot assume that it preserves the Ward identities. The toy contains no interacting gauge bosons, ghost loops or overlapping ultraviolet subdivergences, so its successful subtraction does not prove Non-Abelian gauge theory renormalizability.

An integral over momentum space. For example, state counting in $d$ spatial dimensions uses a measure proportional to $d^{d}p/(2\pi \hslash {)}^d$, whereas a relativistic Feynman integral commonly uses $d^{d}p/(2\pi {)}^d$ in natural units. The convention and any regulator must be specified. Independent loop momenta supply separate integration variables in a loop Feynman diagram; ultraviolet divergences and infrared divergences can require different regularizations.

Use Minkowski metric $g=\mathrm{diag}(1,-1,-1,-1)$ and the convention $S_{fi}={\delta }_{fi}+i(2\pi {)}^4{\delta }^{(4)}(p_f-p_i){\mathcal{M}}_{fi}$. Put $m=m_t$, $y={\lambda }_t/\sqrt{2}$, $Q=e_t=2/3$ andThe Feynman i-epsilon prescription is understood in subsequent denominators. Contracting the spinor indices around the fermion loop gives the traceHere $/p={\gamma }^{\rho }{p}_{\rho }$ is Feynman slash notation. The closed fermion loop sign is $-1$, the three interaction vertices contribute $(-iy)(-ieQ{)}^2$, and the three Dirac propagators contribute $i^3$. Their product, including the color charge trace, is $-N_{c}y{e}^2{Q}^2$. Thus one ordering hasMultiplying by the incoming polarization vectors and adding the exchanged Feynman diagram givesThe displayed four-dimensional Feynman integral is formal until regularized; subsequent manipulations use dimensional regularization, which permits the required loop momentum shifts.

Continue the loop momentum to $D=4-\epsilon$ dimensions, keeping $\mathrm{tr}1=4$ and the external physical polarization vectors four dimensional. Introduce Feynman parameters $x,y\ge 0$, $x+y\le 1$, denoting this integration triangle by $\mathcal{T}$. Completing the square givesOdd terms in $k$ integrate to zero. Lorentz invariance of dimensional regularization gives $k_{\mu }{k}_{\nu }\mapsto g_{\mu \nu }{k}^2/D$. The general projected numerator therefore becomes, under the Feynman integral,Only the $k^2$ term can give an ultraviolet divergence. Define, with the conventional scale factor implicit,The gamma function formula givesThus the pole of the numerator integral, including $2{\int }_{\mathcal{T}}dxdy=1$, isThe coefficients $(4,1,1)$ satisfy this condition. A nonzero local divergence would have to be removed by a counterterm. The Ward identity forbids a local $h{A}_{\mu }{A}^{\mu }$ term, and the first electromagnetic gauge-invariant coupling with one Higgs boson and two photons has the form $h{F}_{\mu \nu }{F}^{\mu \nu }$, of engineering dimension five. It is not an independent counterterm of the Standard Model, a renormalizable quantum field theory. The top quark contribution is separately electromagnetic gauge-invariant; no additional W boson loop Feynman diagram is needed to cancel this pole.

The finite remainder must still be retained. For $A=4C$,Putting $D=4$ before evaluating the divergent Feynman integral would incorrectly set this finite contribution to zero.

Use the Feynman parameters and $J_3$ defined in Solution. With $A=4C$, the finite term just found is $-C\Delta J_3$. Combining it with the remaining $g_{\mu \nu }$ coefficient givesThe other coefficient is $(B-4Cxy)J_3$. Thus, writing ${\mathcal{I}}_{\mu \nu }$ for the Feynman integral without the overall prefactor,For $T_{\mu \nu }=\frac{s}{2}{g}_{\mu \nu }-p_{2,\mu }{p}_{1,\nu }$, the two coefficients must have the corresponding ratio. The mismatch is simply $(B-C)J_3$, independent of $xy$. Therefore the required relation iswhich is satisfied. In the primitive normalization the result isand the full top quark scattering amplitude, in our stated $S$ convention, isIndeed $p_1^{\mu }{T}_{\mu \nu }=p_2^{\nu }{T}_{\mu \nu }=0$: replacing either polarization vector by its photon momentum makes the scattering amplitude vanish. This is the Ward identity, expressing electromagnetic gauge invariance. The result also is symmetric under exchange of the two photons. The finite term in Solution is essential for this Ward identity.

As an independent low-energy check, for $s/m^2\to 0$ the Feynman parameter integral isThe coefficient is consequently $-N_{c}y{e}^2{Q}^2/(6{\pi }^{2}m)$ at leading order. It has the inverse-mass dimension required for an effective $h{F}_{\mu \nu }{F}^{\mu \nu }$ coupling. If $y=m/v$, this coefficient stays finite in the large-$m$ limit at fixed $v$; that limit increases the Yukawa coupling, so it is not a fixed-coupling heavy-field decoupling limit.

There are four topologies for the one-loop gluon self-energy, shown in the upper row of the figure:

The two internal-gluon graphs have symmetry factor $1/2$. The directed quark and ghost loops have symmetry factor one and a minus sign for a closed anticommuting loop. The four-gluon tadpole diagram must be included in the list even though its massless Feynman integral vanishes in dimensional regularization. Local counterterms are excluded here.

The Slavnov-Taylor identity implies transversality of the full perturbative gluon self-energy:The quark loop is separately transverse. This follows without evaluating its Feynman integral: contract its vector Feynman vertex with $k_{\mu }$ and use $/k=(/l+/k-m)-(/l-m)$ between its two Dirac propagators. One obtains a difference of two loop integrals that cancel after a shift of $l$ in dimensional regularization. The colour factor is proportional to $\mathrm{tr}(T^a{T}^b)$ and does not obstruct this argument.

The four-gluon tadpole diagram vanishes after removal of $m_g$. Thus the nontransverse parts of the two-three-gluon-vertex loop and the ghost loop must cancel. Contracting the preceding ghost-loop pole of the gluon self-energy givesFor QCD, set $C_A=3$. No evaluation of the remaining three-gluon Feynman diagram is required. Its uncontracted transverse part is not fixed by this identity.

Only the term proportional to $\alpha$ has an ultraviolet pole. Its $k^2$ cancels the massless denominator, leaving the standard Feynman integralIn the present dimensional regularization convention, $\Gamma (ϵ/2)=2/ϵ+O(1)$ and $\alpha =1+O(ϵ)$. ThereforeThe counterterm contribution is $-i{e}_D{\delta }_1{\gamma }^{\mu }$. Cancellation of the ultraviolet pole fixeswhere the displayed expression uses the minimal subtraction scheme; a different renormalization scheme may add finite $O(e^2)$ terms. In the frequently used convention $D=4-2\epsilon$, the same result is $Z_1=1-e^2/(16{\pi }^2\epsilon )+O(e^4)$. The sign agrees with $Z_1=Z_2$ from the Ward identity for QED renormalization constants in Feynman gauge.

Use the source's charge-conjugation matrix convention $C{\gamma }^{\mu }{C}^{-1}=-({\gamma }^{\mu }{)}^T$. It impliesA trace equals the trace of its matrix transpose, which reverses the order of all factors. Inserting the displayed identities givesThere is one minus sign from each of the three photon interaction vertices, and none from the transformed Dirac propagators. In the last line, change loop momentum to ${\ell }^{\prime }=-\ell -k_1$. Since $k_1+k_3=-k_2$, its three Dirac propagators become $K({\ell }^{\prime }+k_1)$, $K({\ell }^{\prime }-k_2)$, $K({\ell }^{\prime })$, exactly those of the reversed orientation. HenceThis is Furry's theorem. The regularization in quantum field theory matters: the individual Feynman integrals are superficially divergent, so an unregulated shift of a divergent integral is not a proof. A common dimensional regularization preserves charge conjugation and the vector Ward identity, making the change of variables legitimate. No chirality matrix occurs, so the axial triangle anomaly is not involved.

In the right panel of part a, $p$ enters the fermion line, $p^{\prime }=p+q$ leaves it, and a virtual photon of momentum $k$ bridges the external photon vertex. Write ${\Gamma }^{\mu }={\gamma }^{\mu }+{\Lambda }^{\mu }$ and $F_1=1+F_1^{(1)}+\cdots$. The one-loop amplitude isThis Feynman integral has a logarithmic superficial degree of divergence. Its ultraviolet pole is independent of the external momenta and $m$, because every such numerator insertion or denominator correction lowers the large-$k$ degree. Preserve an auxiliary positive mass $M$ while extracting the ultraviolet pole; setting everything to zero would create a scaleless integral and conceal the cancellation of ultraviolet and infrared divergences.

The leading numerator is ${\gamma }^{\alpha }/k{\gamma }^{\mu }/k{\gamma }_{\alpha }$. Symmetric integration replaces $k_{\rho }{k}_{\sigma }$ by $g_{\rho \sigma }{k}^2/d$. Two contractions of gamma matrices then giveFor the auxiliary-mass extraction of ultraviolet poles,The second term is ultraviolet convergent. ThusThe ultraviolet pole of the QED vertex is therefore ${\Lambda }^{\mu }{|}_{\mathrm{UV}}=e^2{\gamma }^{\mu }/(16{\pi }^2ϵ)$. The Pauli structure has zero trace against ${\gamma }_{\mu }$ and no ultraviolet pole at this order.

Form-factor qualification. The two electron electromagnetic vertex form factors in the source describe matrix elements between external on shell Dirac spinors. A general off shell vertex has more structures, so the unprojected trace is not a projector for the complete finite on-shell $F_1$. It does correctly isolate the divergent local ${\gamma }^{\mu }$ coefficient, which is all that is required here.

Adopt the phase convention $CP|K^0⟩=-|{\bar{K}}^0⟩$ and $CP|{\bar{K}}^0⟩=-|K^0⟩$. Changing both phases changes the convention for the mixing parameter, not physical predictions. The flavor contents are $K^0=d\bar{s}$ and ${\bar{K}}^0=s\bar{d}$. A pair of charged weak currents changes strangeness by two units through the following box contractions, with $u_i,u_j=u,c,t$ summed over.

The drawing uses unitary gauge: the W boson gauge-boson propagator contains its longitudinal part. In a gauge of a renormalizable quantum field theory, charged Goldstone boson boxes must also be included to obtain the same gauge-independent result. The external lines represent the valence quarks; the meson matrix element also contains strong interaction binding effects.

Put ${\lambda }_i=V_{is}^{\ast }{V}_{id}$, $x_i=m_i^2/m_W^2$ and $P_L=(1-{\gamma }^5)/2$. The charged weak box contribution to kaon mixing has the effective four-fermion interaction structureHere $G_F$ is the Fermi constant, $V$ is the CKM matrix, and $S$ is a dimensionless loop function from a Feynman integral; the estimate suppresses convention-dependent numerical factors. Four weak interaction Feynman vertices and a one-loop Feynman integral give the coupling and loop order suppression. Unitarity of the CKM matrix gives ${\sum }_i{\lambda }_i=0$, so the GIM mechanism cancels the mass-independent part. With a negligible up quark mass, the leading charm quark term has $S(x_c,x_c)\sim x_c$. Define $f_K$ by $⟨0|\bar{s}{\gamma }_{\mu }{\gamma }^{5}d|K^0(p)⟩=i{f}_K{p}_{\mu }$. Estimating the hadronic matrix element by $f_K^2{m}_K{B}_K$, with the dimensionless kaon bag parameter $B_K$ of order unity, givesThe hadronic normalization includes division by $2m_K$ when relativistically normalized states are converted to the effective Hamiltonian for neutral-meson mixing. Since $|V_{cs}^{\ast }{V}_{cd}|\simeq \sin {\theta }_C\cos {\theta }_C$, the Cabibbo suppression is explicit. Illustrative scales $G_F\sim 10^{-5}{\mathrm{GeV}}^{-2}$, $m_c\sim 1\mathrm{GeV}$, $f_K\sim 0.1\mathrm{GeV}$ and $m_K\sim 0.5\mathrm{GeV}$ give an order $10^{-16}$ to $10^{-15}\mathrm{GeV}$ estimate. The full sum includes charm quark-top quark and top quark-top quark terms, important for CP violation; long-distance strong interaction contributions prevent this dimensional estimate from being a precision prediction. A meson matrix element necessarily needs hadronic input in addition to electroweak parameters.

For the decaying kaon system write $H^{\prime }=M-i\Gamma /2$, where $M$ and $\Gamma$ are Hermitian matrices. The entries denoted $M_{ij}$ in the question are entries of this effective Hamiltonian for neutral-meson mixing; they need not themselves form a Hermitian matrix. CPT invariance equates the diagonal masses and diagonal decay widths, givingOne can see why there is no complex conjugation on the right as follows. In the flavor subspace let $J=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$ and represent the antiunitary CPT operation by $J$ followed by complex conjugation. CPT changes a projected forward propagator into its adjoint. Indeed, if $P$ projects onto the flavor subspace and $H_{\mathrm{full}}$ is the Hermitian microscopic generator, its invariance and the antiunitary reversal of $i$ give $\Theta P{e}^{-i{H}_{\mathrm{full}}t}P{\Theta }^{-1}=P{e}^{+i{H}_{\mathrm{full}}t}P=(P{e}^{-i{H}_{\mathrm{full}}t}P{)}^{†}$. In the time-independent two-state approximation, the effective generator this means $J{H^{\prime }}^{\ast }J={H^{\prime }}^{†}$, whose diagonal entries give $(H_{22}^{\prime }{)}^{\ast }=(H_{11}^{\prime }{)}^{\ast }$; the off-diagonal entries impose no further condition. Equivalently, apply CPT separately to the dispersive and absorptive Hermitian matrices. Treating the decaying effective Hamiltonian for neutral-meson mixing as a Hermitian matrix would incorrectly discard the absorptive part.

In the chosen flavor phase convention CP is represented by $-J$. Its invariance requires $J{H}^{\prime }J=H^{\prime }$. Thus it gives diagonal equality and additionallyThe normalized CP eigenstates areApplying CP directly gives eigenvalues $+1$ and $-1$ respectively.

Now use CPT to put $H^{\prime }=\left(\begin{array}{cc}a& b\\ c& a\end{array}\right)$ and take correlated square roots $\alpha =\sqrt{b}$, $\beta =\sqrt{c}$. Its characteristic polynomial equation is $(a-\mu {)}^2-bc=0$, and direct multiplication showsExpressing these two eigenvectors in the CP eigenstate basis gives, respectively, coefficients proportional to $(\alpha +\beta ,\alpha -\beta )$ and $(\alpha -\beta ,\alpha +\beta )$. Therefore the kaon CP mixing parameter isand, because the CP eigenstate basis is orthonormal, the normalized states at production areThe kaon mixing square-root branch convention correlates the roots continuously with the limit of conserved CP $b=c$, where $\alpha =\beta$ and $ϵ=0$. Reversing one root interchanges the two eigenstates. The labels short and long are fixed by their decay widths: their complex eigenvalues are ${\mu }_{S,L}=m_{S,L}-i{\Gamma }_{S,L}/2$, and subsequent evolution supplies $e^{-i{\mu }_{S,L}t}$. The displayed formula assumes nonzero mixing and $\alpha +\beta \ne 0$, as in the physical kaon system; degenerate or defective matrices require a separate limiting treatment.

Without gauge fixing, gauge-equivalent configurations are integrated repeatedly, and the quadratic gauge-field operator has zero modes along infinitesimal gauge transformations. It therefore has no inverse with which to define perturbative propagators. A gauge condition removes these directions locally in the perturbative expansion. The associated Faddeev-Popov determinant can be represented by anticommuting Faddeev-Popov ghost fields and Faddeev-Popov antighost fields.

First make the normalization explicit. For the fields literally occurring in the action with overall $1/g^2$, putThe resulting kinetic terms are canonical, and the interaction terms carry powers of $g$. We use signature $(-,+,+,+)$ and incoming Fourier modes $e^{ipx}$, consistent with the displayed $p^2-i0$ denominators. The quadratic action in the canonical fields isColor indices are diagonal. Introduce the longitudinal and transverse projectors $P_{L,\mu \nu }=p_{\mu }{p}_{\nu }/p^2$ and $P_T=\eta -P_L$. The gauge kinetic matrix is $-p^2(P_T+{\xi }^{-1}{P}_L)$, with inverse $-(P_T+\xi P_L)/p^2$. The ghost kinetic operator is $-p^2$. The Gaussian functional integral and Grassmann integral therefore giveThe longitudinal formula is understood as the Feynman-prescribed inverse of the kinetic operator; away from its poles the projector inversion is ordinary algebra. The propagators for the original unrescaled fields are $g^2$ times these expressions. Using $P_{T}p=0$ and $P_{L}p=p$ giveswith either normalization. This is the longitudinal gauge propagator contraction.

The canonical ghost interaction comes directly from the covariant derivative:For incoming antighost momentum $p$, the derivative supplies $i{p}^{\mu }$; the expansion of $e^{iS}$ supplies another $i$. Thus the ghost-gluon vertex is $g{p}^{\mu }{f}_{abc}$, as requested. For unrescaled fields its coefficient is instead $p^{\mu }{f}_{abc}/g^2$. The printed order-$g$ rule presupposes canonical normalization; keeping the overall $1/g^2$ and an order-$g$ vertex for the same unrescaled variables would mix conventions. Choosing Fourier modes $e^{-ipx}$ instead reverses the vertex sign and must be done consistently for all momentum assignments.

In Feynman gauge, $\xi =1$, the one-loop ghost two-point graph has one internal ghost line and one internal gauge line joined by two such vertices. The external ghost line is open, so it does not carry a closed ghost-loop minus sign. With an orthonormal adjoint basis, the structure constants are totally antisymmetric, and the color product along this line is $f_{adc}{f}_{cdb}=-C{\delta }_{ab}$. The two internal propagators supply $(-i{)}^2=-1$, leaving a positive color-and-vertex factor for the amputated insertion. DefineThe amputated insertion is $i{\delta }_{ab}{g}^2{I}_d(p)C$. Restoring the two external canonical ghost propagators gives the one-loop contributionThe literal unrescaled correlator in the question is $g^2{G}^{(1)}$, hence of order $g^4$, whereas its tree term is $-i{\delta }_{ab}{g}^2/(p^2-i0)$. In $d\ne 4$ the dimensionless renormalized coupling convention adds ${\mu }^{4-d}$ multiplying $g^2$ in the canonical loop term.

To evaluate the Feynman integral, combine the denominators with a Feynman parameter:Shift $\ell =k-(1-\alpha )p$. The denominator becomes $[{\ell }^2+\alpha (1-\alpha )p^2-i0{]}^2$, and the numerator is $p\cdot \ell +(1-\alpha )p^2$. The odd term vanishes in translation-invariant dimensional regularization. For spacelike $p$, a Wick rotation converts the remaining integral into a Euclidean one and cancels the prefactor $1/i$. The needed radial integral isIndeed, spherical coordinates and $u={\ell }_E^2/M^2$ leave a beta integral ${\int }_0^{\infty }{u}^{d/2-1}(1+u{)}^{-2}du=\Gamma (d/2)\Gamma (2-d/2)$; the angular factor cancels $\Gamma (d/2)$. Establish this first in a convergent range and then analytically continue in $d$. Substituting $M^2=\alpha (1-\alpha )p^2$ yieldsThis is the printed result, including the factor $1/i$ in its Minkowski-space definition. Equivalently the parameter integral is $B(d/2-1,d/2)$. Symmetry under $k\leftrightarrow p-k$ also gives $I_d(p)$ as $p^2/2$ times the massless scalar bubble, independently checking the numerator factor.

Write $ϵ=4-d$. Since $\Gamma (ϵ/2)=2/ϵ+O(1)$ and the parameter integral tends to ${\int }_0^1(1-\alpha )d\alpha =1/2$, the pole isHere $p$ is off shell and nonzero, separating the ultraviolet pole from massless infrared issues. The unrescaled amplitude has an extra factor $g^2$. The divergence is proportional to the existing ghost kinetic operator, so the one-loop ghost kinetic counterterm in Feynman gauge is local:It inserts $-i\delta Z_c{p}^2$ into the inverse propagator, cancelling the divergent insertion $+i{g}^{2}C{I}_d(p)$. With the two external propagators attached, its contribution is $+i{\delta }_{ab}\delta Z_c/p^2$, cancelling the displayed connected amplitude pole. In the original variables the same action term is $-\delta Z_c\partial \bar{c}\cdot \partial c/g^2$. If instead one defines dimensional regularization by $d=4-2ϵ$, the identical counterterm is written $\delta Z_c=g^{2}C/(32{\pi }^2ϵ)$; the factor of two is purely the regulator convention.

For the amputated one-particle-irreducible vertex, every internal bridge is forbidden. Define the Feynman integralsFor $E=3,L=1$, the counting identity in part b gives $V_3+2V_4=3$. The only one-particle-irreducible Feynman diagrams are a triangle of three cubic vertices and a two-propagator bubble joining one cubic and one quartic vertex. Write ${\mathcal{V}}_3$ for the amputated Feynman vertex, including its overall factor of $i$. With all external momenta incoming,The triangle has $S=1$ for labelled external legs. The mixed bubble has $S=2$ because its two internal lines may be interchanged. There are three distinct terms, according to which external leg is attached to the cubic vertex; the other two attach to the quartic vertex. No tadpole attached by a bridge is one-particle irreducible.