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For a vertex $2i{g}^2{\eta }_{\mu \nu }$ and two identical outgoing real scalar particles, the longitudinal-pair amplitude is $g^2(s-2m_A^2)/m_A^2$. The high-energy cross section is the displayed expression, including the final-state factor $1/2!$. An unpolarized incoming pair averages over nine spin states and has leading cross section $g^{4}s/(288\pi m_A^4)$. A written interaction $g^2{A}_{\mu }{A}^{\mu }{h}^2$ instead gives vertex $4i{g}^2{\eta }_{\mu \nu }$ and four times these cross sections. The distinction separates a vertex normalization from polarization averaging and identical-particle phase-space counting.

For longitudinal-vector annihilation through a scalar contact interaction, first use the supplied vertex $2i{g}^2{\eta }_{\mu \nu }$. Let $s=(p_1+p_2{)}^2$ and work in the center-of-mass frame. The amplitude is $\mathcal{M}=2g^2{ϵ}_1\cdot {ϵ}_2$, up to its overall phase. For two longitudinal incoming vectors,This grows as $s^2/m_A^4$, whereas transverse amplitudes stay of order $g^2$ and mixed longitudinal-transverse amplitudes vanish in this frame. Consequently the leading high-energy contribution is longitudinal.

For an unpolarized total cross section, average over the $3\times 3$ incoming polarizations. The massive-vector polarization sum givesThe identical final scalars require a factor $1/2!$ in the two-body phase space. With ${\beta }_A=\sqrt{1-4m_A^2/s}$ and ${\beta }_h=\sqrt{1-4m_h^2/s}$, the angle-independent result isFor a prepared longitudinal pair there is no incoming polarization average, giving insteadThere is no extra initial-state factor $1/2$: a cross section uses the incoming beam flux, while the final-state factor removes double counting of indistinguishable outgoing states.

Define the energy in the stated bound to be the total center-of-mass energy, $E=\sqrt{s}$. Equating the leading expressions to $4\pi /E^2$ givesfor $m_A=100\mathrm{GeV}$ and $g=0.4$. The second value shows that a longitudinal channel reaches the supplied perturbative bound earlier. If the source's unspecified $E$ denotes the energy of one incoming particle and the bound is literally $4\pi /E^2$, substitute $s=4E^2$ instead: the corresponding values are $1.83\mathrm{TeV}$ unpolarized and $1.05\mathrm{TeV}$ longitudinal. Stating the energy and polarization conventions is necessary for a numerical answer.

Interaction-normalization qualification: the written ${\mathcal{L}}_{\mathrm{int}}=g^2{A}_{\mu }{A}^{\mu }{h}^2$ differentiates to $4i{g}^2{\eta }_{\mu \nu }$, because there are $2!$ vector assignments and $2!$ scalar assignments. For example, differentiating $g^2(A^0{)}^2{h}^2$ twice in each variable gives $4g^2$. The supplied $2i{g}^2{\eta }_{\mu \nu }$ would instead follow from ${\mathcal{L}}_{\mathrm{int}}=g^2{A}_{\mu }{A}^{\mu }{h}^2/2$. Thus all cross sections above follow the explicitly requested rule. Using the literal written interaction multiplies them by four and divides every quoted threshold energy by $\sqrt{2}$. The energy growth itself is the physical issue: a contact interaction of massive longitudinal vectors needs additional dynamics before its perturbative high-energy extrapolation becomes reliable.