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With a single quartic interaction vertex, the two unused line ends close into one tadpole diagram. With a single sextic interaction vertex, the four unused ends pair into two tadpole diagrams. These are the only one-vertex one-particle irreducible Feynman diagrams for the two-point function.

Define the Minkowski Feynman integralsThe amputated two-point function contributions arewhere $C_4,C_6$ are the reciprocals of the corresponding Feynman-diagram symmetry factors. Both are independent of external momentum, and hence affect the mass counterterm rather than wave-function renormalization. No numerical Feynman-diagram symmetry factors are needed here.

There are three independent nonfactorized momentum integrals, namely a tadpole integral, the three-line sunset diagram integral, and the five-line banana integral. The six displayed Feynman diagrams involve six complete integral expressions:The distinction between independent building blocks and their products removes an ambiguity in counting “different momentum integrals”. For the strict divergent-wave-function renormalization interpretation of ii, the graphs in c reduce to $T,T^2,B_5$, comprising three full integral expressions and two independent nonfactorized building blocks.

Make a Wick rotation and use $m>0$ to separate ultraviolet divergences from infrared divergences. Write $D_E(k)=1/(k^2+m^2)$, and let $T,B_r(P)$ be the Euclidean versions of the integrals above, with their harmless overall factors of $i$ removed. A spherical ultraviolet cutoff givesThus the one-vertex quartic graph is linearly divergent, and the one-vertex sextic graph is quadratically divergent as $T_{\Lambda }^2$.

For the other two building blocks, the spatial Fourier transform is especially transparent. The three-dimensional Euclidean scalar propagator isThe convolution theorem converts the $r$ parallel lines into $G(x{)}^r$. Angular integration gives, with a short-distance cutoff $a={\Lambda }^{-1}$,This is a different convenient cutoff regularization for the connected block; power-divergent coefficients depend on the cutoff convention, whereas the conclusions about powers and the presence of a kinetic logarithm do not. The sinc function is understood to be $1$ at $P=0$.

For $r=3$, the integrand starts as ${\rho }^{-1}+O(1)$, soIts divergent part is independent of external momentum. In particular,The finite difference and its finite derivative show why the pure three-line sunset diagram does not need a divergent wave-function renormalization in three dimensions.

For $r=5$, the small-distance Taylor expansion isConsequentlyIn particular, the nonzero $P^2\log \Lambda$ term requires a genuine divergent wave-function renormalization, in addition to the mass counterterm. The coefficients here refer to the displayed position-space cutoff and have not included Feynman-diagram symmetry factors or bare couplings.

The comparison with the superficial degree of divergence isThus the actual leading power agrees with $\omega$ in every case. The additional logarithms in products come from ultraviolet subdivergences, which uniform power counting alone does not resolve.

For $T{B}_3$ and $T^2{B}_3$, differentiation with respect to $P^2$ removes the logarithmic divergence of $B_3$, but leaves divergent factors $T$ or $T^2$ multiplying a finite, nonpolynomial function of $P^2$. These are subdivergent contributions, not new nonlocal counterterms: subtracting each tadpole diagram subdivergence and the momentum-independent $B_3$ subdivergence removes them. For example, in a factorized subtraction scheme $T{B}_3\mapsto (T-T_{\mathrm{div}})(B_3-B_{3,\mathrm{div}})$, and likewise for $T^2{B}_3$. The remaining $B_5$ kinetic logarithm explains the strict interpretation in ii. A dimensional regularization scheme can suppress power divergences; it does not invalidate this cutoff-based comparison with $\omega$.

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.

The one-loop one-particle irreducible correction is a tadpole diagram: two external lines meet one quartic interaction and the other two legs are joined. Interchanging the two ends of its loop gives the symmetry factor $2$.

Define the self-energy by an insertion $-i\Sigma (p^2)$ and a full scalar propagator $i/[p^2-m^2-\Sigma (p^2)+iϵ]$ before counterterms. The supplied cutoff regularization integral givesThusIt has no external momentum dependence. The correction to the unamputated two-point correlation function in momentum space isThe positive sign of ${\Sigma }_1$ raises the unrenormalized squared pole mass for $g_4>0$.

Write ${\varphi }_B=Z^{1/2}\varphi$, $Z=1+{\delta }_Z$, and define the mass counterterm ${\delta }_m$ as a squared-mass coefficient:With the self-energy convention in b, the inverse scalar propagator is $p^2-m^2+{\delta }_Z{p}^2-{\delta }_m-\Sigma (p^2)$. Requiring the pole mass to equal $m$ therefore gives the on-shell renormalization scheme conditionAt one loop order, ${\Sigma }_1^{\prime }=0$; hence a unit pole residue also gives ${\delta }_Z^{(1)}=0$, andThe mass counterterm Feynman rule is $-i{\delta }_m$, cancelling the tadpole diagram. A convention defining a positive mass subtraction by $m_B^2=m^2-{\delta }_m^{\mathrm{sub}}$ instead has ${\delta }_m^{\mathrm{sub}}=-{\delta }_m$; specifying this convention avoids a spurious sign disagreement.

There is no one-loop wave-function renormalization because the tadpole diagram is independent of momentum. Each one-loop four-point bubble diagram has two scalar propagators, so its large Euclidean loop momentum integral behaves as $\int d^2{k}_E/k_E^4$ and is ultraviolet finite. Thus no divergent quartic interaction counterterm is required. Finite coupling redefinitions remain possible. At higher loop orders, finite wave-function renormalization may be needed if unit pole residue is imposed; power counting only rules out its ultraviolet divergence.

Let $I,L,V$ count internal lines, independent loops, and interaction vertices of a connected Feynman diagram. The graph identity $L=I-V+1$ and the two inverse powers from each scalar propagator give its superficial degree of divergence:The same formula applies to every connected subgraph. Graphs with $V\ge 2$ have negative superficial degree of divergence, but this alone does not remove a possible ultraviolet subdivergence. A one-vertex subgraph with external legs in the quartic scalar field theory can diverge only by a tadpole diagram, whose subtraction was fixed in c. Subtract every such tadpole diagram; every remaining loop subgraph then has $V\ge 2$ and negative superficial degree of divergence. Its ultraviolet divergence is absent. Equivalently, normal ordering the quartic interaction removes all self-contractions at a single interaction vertex. Thus no new divergent field-dependent counterterm structure is needed, and in a normal-ordering subtraction scheme in two dimensions the one-loop mass subtraction exhausts these divergent coefficients.

The literal assertion that there are no further counterterm contributions in the on-shell renormalization scheme is stronger and generally false. A concrete counterexample is the two-loop sunset diagram. After analytic continuation below its three-particle threshold, the magnitude of its self-energy at $p^2=m^2$ contains the strictly positive finite integralThe integrable corners do not spoil convergence. Its nonzero value shifts the pole mass, requiring an additional finite mass counterterm if $m$ is to remain the exact pole mass; its derivative can likewise change the pole residue. These are compatible with the power counting result. If vacuum diagrams are retained rather than divided out in normalized correlation functions, a separate constant vacuum energy counterterm is also required: the one-vertex double-loop vacuum diagram is already divergent. The claim about the mass subtraction concerns field-dependent counterterms and normalized correlation functions.

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.

Define the free Feynman propagators byThe interaction preserves the global phase symmetry $\phi \mapsto e^{i\alpha }\phi$, with the real $\varphi$ neutral. Therefore the charge-two correlator vanishes in a symmetry-invariant perturbative vacuum:In Wick contractions, a complex field contracts only with its conjugate; one vertex has one of each and cannot absorb both external $\phi$ fields. The symmetry shows this vanishing holds to all perturbative orders, not only first order.

The neutral connected correlator has one allowed contraction to each external point at order $g$:The propagators are even functions of their separation; their orientations encode the conjugate-field contractions. Equivalently, with every external momentum Fourier transformed as incoming,No factor of two accompanies the vertex: $\phi$ and ${\phi }^{†}$ are distinct fields.

For the full correlator as literally written, a disconnected scalar three-point tadpole is also possible. Contract the two external complex fields with each other and contract the complex pair at the vertex into a loop. Unless normal ordering is imposed on the interaction or a tadpole counterterm fixes $⟨\varphi ⟩=0$, addThe coincident propagator is regulator dependent; in dimensional regularization the massless scaleless tadpole diagram vanishes. More generally this is $⟨T{\phi }^{†}(x)\phi (y){⟩}_0⟨\varphi (z)⟩$ to this order and is removed by taking the connected correlator or imposing the stated vacuum-renormalization condition. Normalizing by vacuum bubbles alone does not remove an external-leg disconnected tadpole diagram. The charged correlator remains zero in every one of these conventions.^[ref]

Let ${\Delta }_M$ be the free real-scalar Feynman propagator and $D_m(x-y)=⟨0|T\phi (x){\phi }^{†}(y)|0{⟩}_0$ the free complex-scalar one. Expanding the interaction exponential once and applying Wick's theorem gives the connected decay Green's functionThere is one allowed contraction of each external field with the appropriate vertex field, so no factorial remains. Fourier transformation gives the momentum-conserving delta factor times $(-ig){\tilde{\Delta }}_M(p){\tilde{D}}_m(q_1){\tilde{D}}_m(q_2)$, and amputation of external propagators gives the decay vertex $-ig$. An on-shell physical decay additionally requires $M>2m$.

If one literally asks for the full three-point correlation function without a connected restriction or a tadpole condition, it also containsthe product of the induced heavy-field one-point function and the free light-field two-point function. This disconnected tadpole diagram does not describe a three-particle decay and drops out of its connected amputated amplitude.

For two equally charged daughters the corresponding Green's function vanishes to every perturbative order:Every interaction contains equal numbers of $\phi$ and ${\phi }^{†}$, so the two excess conjugate fields can never be paired. Equivalently the exact global U(1) symmetry $\phi \mapsto e^{i\alpha }\phi$, ${\phi }^{†}\mapsto e^{-i\alpha }{\phi }^{†}$ leaves the vacuum and action invariant, whereas this insertion has charge $-2$. Charge conservation forbids the decay of a neutral particle into two particles of the same nonzero charge.

Every interaction contains an even number of $b$ fields, so the exact discrete symmetry $b\mapsto -b$ forbids $\Phi \to bbb$ to every perturbative order. There is no such prohibition for $\Phi \to bb$.

The latter decay channel is absent at tree level in the displayed interactions, but contracting two $b$ legs at a $\Phi b^2(\partial b{)}^2$ vertex produces a tadpole diagram with two external $b$ legs. For example, contracting the undifferentiated pair leaves the local operator $\Phi (\partial b{)}^2$ multiplied by an integral proportional to $\int d^4\ell /({\ell }^2+i0)$. Contracting the differentiated pair instead gives a local $\Phi b^2$ contribution. Both are allowed effective field theory operators. With a ultraviolet cutoff, these local terms generically carry quadratic or quartic cutoff dependence and require matching coefficients or counterterms.

There is a relevant qualification to a claim that the one-loop decay width must be nonzero. Since $b$ is exactly massless here, the one-vertex tadpole integrals have no external scale and are scaleless integrals. They vanish in dimensional regularization. Thus the correct conclusion is allowed, and generically generated by quantum corrections or matching, not a regulator-independent nonzero one-loop prediction from ${\lambda }_b$ alone. No exact symmetry protects the two-$b$ channel; its numerical rate requires the renormalized coefficients of the allowed operators.

For $\mathcal{L}=|\partial \varphi {|}^2-m^2|\varphi {|}^2-{\mu }^{ϵ}g|\varphi {|}^4/4$ in $d=4-ϵ$, the directed tadpole diagram has symmetry factor one. Its divergent Feynman contribution to $i{\Gamma }^{(2)}$ is $ig{m}^2/(8{\pi }^2ϵ)$. A mass insertion $-i{\delta }_{m^2}$ therefore gives the displayed minimal subtraction scheme counterterm. Differentiating at fixed bare mass and coupling gives the displayed ${\gamma }_m=\frac{1}{2}\mu d\log m^2/d\mu$. The equivalent two-real-field formulation has vertex coefficients $3g/2$ for a pure species and $g/2$ for two of each, giving total tadpole weight $(3g/2+g/2)/2=g$.

The two identical photon fields at a scalar electrodynamics seagull give a vertex multiplicity two. Joining them into one closed photon line gives Feynman-diagram symmetry factor $1/2$. Directly expanding $-e^2{A}^2{\varphi }^{†}\varphi$ gives only one photon Wick contraction, independently checking the compensation. In a transverse gauge the remaining tensor trace is $d-1$; a massless photon tadpole is a scaleless integral in dimensional regularization.

With canonical field $\phi =\varphi /\beta$, the quartic interaction has coupling $-m^2{\beta }^2$. Cancelling the vacuum tadpole diagram requires the divergent mass counterterm $\delta m^2=(m^2{\beta }^2/4){\int }_{-\Lambda }^{\Lambda }dk/[2\pi \sqrt{k^2+1}]$. Its kink energy is $\Delta M_{\mathrm{ct}}=4\delta m^2/(m{\beta }^2)$ because $\int (1-\cos {\varphi }_K)dx=4$. This cancels the logarithmic ultraviolet divergence in the one-loop soliton mass correction. Finite parts depend on the renormalization condition.

For interaction ${\lambda }_{2k}{\varphi }^{2k}/(2k)!$, one vertex with two labelled external legs has $k-1$ independently paired tadpole diagrams. Counting contractions gives symmetry factor $S=2^{k-1}(k-1)!$. In a normalized Euclidean path integral, its amputated contribution is $-{\lambda }_{2k}{T}_D(m{)}^{k-1}/S$, where $T_D$ is the tadpole integral. It is linear in the coupling but has $k-1$ loops and a highest four-dimensional ultraviolet pole of order $k-1$ for nonzero mass.

A linear scalar counterterm fixes the one-point function or the expansion point of the scalar potential. An unbroken parity symmetry in quantum field theory forbids this term when the scalar is a pseudoscalar. A vanishing one-loop tadpole diagram alone would not prove the all-orders restriction without the symmetry argument.

A tadpole diagram contains an internal line that begins and ends at the same interaction vertex. In a scalar theory it commonly gives a momentum-independent correction to a mass or lower-order coupling.

The one-propagator Euclidean loop momentum integral associated with a tadpole diagram. Dimensional regularization gives $T_d(m)=\Gamma (1-d/2)(m^2{)}^{d/2-1}/(4\pi {)}^{d/2}$. Near $d=3$ this has no pole and $T_3(m)=-m/(4\pi )$ for $m>0$. A momentum-cutoff definition instead contains a linear ultraviolet divergence; the absence of a dimensional-regularization pole here concerns logarithms, not the positivity or convergence of the unregularized integral.

A linear counterterm can be chosen so that the renormalized one-point function, or equivalently the chosen vacuum expectation value, is zero. This cancels attached tadpole diagrams in the perturbative expansion about that vacuum. In a massive six-dimensional cubic scalar field theory, the one-loop proper tadpole is $t=g{I}_1/2$; without subtraction the stationary background shifts by $v=-t/m^2+O(g^3)$ at one-loop order. One must specify this convention when comparing reducible exchange diagrams or mass corrections.

An ultraviolet subdivergence arises when loop momenta internal to a proper subgraph tend to infinity while the momenta entering that subgraph are held fixed. It can occur even if simultaneous scaling of all momenta of the full Feynman diagram is convergent. Recursive local counterterms subtract these subgraph divergences before the remaining overall divergence is analyzed. A tadpole diagram inserted into an otherwise convergent six-point phi-fourth theory Feynman diagram is a simple massive example.

With interaction $-{\lambda }_0{\varphi }^4/4!$, the tadpole diagram has symmetry factor one half. After Wick rotation and a rotationally invariant Euclidean cutoff, its self-energy is $({\lambda }_0/2){\int }_{|k|<\Lambda }{d}^{4}k(2\pi {)}^{-4}(k^2+m_0^2{)}^{-1}$. The sphere area $2{\pi }^2$ and radial integral ${\int }_0^{\Lambda }{k}^{3}dk/(k^2+m_0^2)=\frac{1}{2}[{\Lambda }^2-m_0^2\log (1+{\Lambda }^2/m_0^2)]$ give the displayed shift. It is independent of external momentum, so no one-loop momentum-dependent wave-function renormalization arises from this graph. Different regulator schemes can treat power divergences differently while defining the same chosen pole mass.