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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.

Take $\varphi (p_1)+\bar{\varphi }(p_2)\to \varphi (p_3)+\bar{\varphi }(p_4)$ with the Mandelstam variables of i. The allowed tree-level Feynman diagram is the crossed $u$-channel $\Phi$ exchange in panel (c) of b's figure. At one end the incoming particle and outgoing antiparticle transfer charge $+2$; the other end absorbs that charge. The zero-charge $s$- and $t$-channel assignments are forbidden by the interaction vertices. ThusIn the center-of-mass frame with $|\mathbf{p}|\ll m$, $s=4m^2+O(|\mathbf{p}{|}^2)$ and $u=O(|\mathbf{p}{|}^2)$. For $M\gg m$, both scattering amplitudes approach $g^2/M^2$. The elastic two-body Lorentz-invariant phase space and incident flux agree for the two experiments. Usingfor distinguishable final particles, and an extra factor $1/2!$ when integrating the identical-particle final state over the whole sphere, givesThere is no additional division for identical incoming particles in the definition of a beam scattering cross-section. Before taking $M/m\to \infty$, the zero-momentum ratio is $\frac{1}{2}[M^2/(M^2-4m^2){]}^2$, provided the threshold is away from the heavy pole.

Near $s=M^2$, the bare $s$-channel scalar propagator makes fixed-order perturbation theory fail: repeated self-energy insertions contain powers of ${\Sigma }_{\Phi }(s)/(s-M^2)$. The $\Phi \to \varphi \varphi$ channel is open for $M>2m$, giving the leading decay widthResumming the self-energy gives a finite Breit-Wigner resonance cross section, with denominator $s-M_{\mathrm{pole}}^2+i{M}_{\mathrm{pole}}{\Gamma }_{\Phi }$ near the pole, rather than an infinite scattering cross-section. Also, $s\simeq M^2$ with $M\gg m$ lies outside the assumed nonrelativistic incoming kinematics; the ratio $1/2$ is not a resonance prediction.

Label the incoming four-momenta $p_1,p_2$ and the outgoing ones $p_3,p_4$. Define the Mandelstam variables $s=(p_1+p_2{)}^2$, $t=(p_1-p_3{)}^2$, $u=(p_1-p_4{)}^2$. Two incoming $\varphi$ particles have conserved charge $+2$, so the only tree-level Feynman diagram is an $s$-channel $\Phi$ exchange, shown in panel (b) of b's figure. Each end converts two charge-one lines into one charge-two line. There are no $t$- or $u$-channel exchanges: an incoming and an outgoing particle at such an end would transfer zero charge, for which this theory has no cubic interaction vertex.The single Feynman diagram is already symmetric under exchanging the two outgoing particles. This does not add another copy of the scattering amplitude; the identical final-state symmetry factor enters the integrated scattering cross-section instead.

Cross the second incoming heavy particle and one outgoing heavy particle in i to obtain $\Phi \bar{\Phi }\to \Phi \bar{\Phi }$. Panel (e) of b's figure shows the resulting box diagram. Its four light-charge arrows again alternate source and sink corners; it is the only connected order-$g^4$ topology. There is no tree exchange because every cubic interaction vertex has only one heavy leg. With $|\mathbf{p}|\ll M\ll m$, crossing does not affect the zero-external-momentum loop integral, soAn independent check is to integrate out the complex scalar field $\varphi$ at constant $\Phi$. Its two real-component squared masses are $m^2\pm g|\Phi |$. The one-loop quantum effective potential is thereforeExpanding the logarithms to fourth order in $\Phi$ givesThus the effective Lagrangian has the positive local coefficient $g^4/(384{\pi }^2{m}^4)$. Its $2!2!$ external attachments give the common scattering amplitude $g^4/(96{\pi }^2{m}^4)$ in both channels. This also checks the box multiplicity and sign.

Four external $\Phi$ lines require four cubic interaction vertices, since each such interaction vertex contains only one heavy field. Their other eight endpoints must form four light internal lines. A connected Feynman diagram therefore has $V=I=4$ and $L=1$: the leading contribution is a one-loop box diagram, of order $g^4$, with four $\varphi$ scalar propagators. Charge-two incoming and outgoing heavy lines attach alternately around the box, as shown in panel (d) of b's figure. Each incoming heavy line supplies the two outgoing light-charge arrows at its corner, and each outgoing heavy line absorbs them. With external labels fixed there is one such box and its symmetry factor is one.

For $|\mathbf{p}|\ll M\ll m$, the external energies are also small compared with $m$, so expand the loop integrand in all external four-momenta. Its leading term isA Wick rotation and the Euclidean massive loop integral giveHence, more precisely than just its requested scaling,The mass dimension $[g]=1$ makes $g^4/m^4$ dimensionless, as a four-dimensional scattering amplitude should be. The leading amplitude has no extra identical-particle multiplicity.