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Interpret the repeated species label in the Yukawa interaction as summed over $i=1,2$: each Feynman vertex is diagonal in species and equals $-ig$. Define the invariant scattering amplitude by $S-1=i(2\pi {)}^4{\delta }^{(4)}(P_f-P_i)\mathcal{M}$. There is one tree-level Feynman diagram, an $s$-channel scalar exchange. In the left panel, $j=2$ for this part; the right panel only occurs in d. Arrows show fermion-number flow, so an antifermion arrow opposes its physical momentum.

Writing $k=p_1+p_2=p_3+p_4$ and $s=k^2$, the momentum-space Feynman rules giveThe scalar carries no spin label. Average over the $2\times 2$ incoming spin states and sum the four outgoing combinations. For the real scalar Yukawa interaction, the conjugate of ${\bar{v}}_2{u}_1$ is ${\bar{u}}_1{v}_2$. The fermion spin sums then yieldThe bar denotes the initial spin average, not a Dirac adjoint. The gamma matrix trace identities giveand the second trace has the same value because $p_3+p_4=p_1+p_2$ and all external masses are $m$. ThereforeThis is the fixed-order expression away from $s=M^2$. If that pole is accessible, a physical scattering resonance requires a scalar self-energy and decay width; the tree formula alone is not finite there.

For a single species both panels of the figure in c contribute, now with $j=1$. They are the $s$-channel annihilation and $t$-channel scalar exchange, where the Mandelstam variables are $s=(p_1+p_2{)}^2$, $t=(p_1-p_3{)}^2$, $u=(p_1-p_4{)}^2$ and $s+t+u=4m^2$. Fix the external-state order consistently with c. The exchange of identical external fermionic operators gives a relative minus sign:Changing the convention for ordering all external states can change the overall sign, but not this relative fermionic sign.

The exchange $p_1\leftrightarrow -p_4$ must include crossing symmetry the associated external spinors, as well as continuing the four-momenta. It interchanges $s$ and $t$, leaves $u$ fixed, and interchanges the two spinor pairings. Hence the analytically continued invariant scattering amplitude is antisymmetric:The first relation is understood with consistently chosen spinor crossing phases. It is not a substitution of negative-energy arguments into the original positive-energy $u$ spinors without changing external states.

Without repeating the spin sum calculation, c fixes the square of the first diagram. Crossing symmetry fixes the second by $s\leftrightarrow t$. Consequently the strongest immediate inference isThe interference function is not determined by the separate diagram squares. In particular adding the two squares and dropping interference is unjustified. The Cauchy-Schwarz inequality applied to the arrays of spin amplitudes also gives $|C(s,t)|\le 2|(s-4m^2)(t-4m^2)|$ on the physical scattering region.

For completeness, its value can be checked with one contracted trace, without another diagram calculation. The cross product isThe relative minus in the invariant scattering amplitude gives $C=-2K$. Thus the fully evaluated result, beyond the symmetry inference alone, isEvery term respects the inferred $s,t$ symmetry. In the massless limit the interference numerator reduces to $st$, which is nonzero at a generic scattering angle.

Let $p_1,p_2$ be the incoming and $p_3,p_4$ the outgoing four-momenta, and set $s=(p_1+p_2{)}^2$, $t=(p_1-p_3{)}^2$, $u=(p_1-p_4{)}^2$, the Mandelstam variables. The only connected tree-level Feynman diagram is the $s$-channel exchange of $\bar{\Phi }$, with Noether charge $+2$. The two incoming light particles attach to ${\varphi }^2\Phi$, and the two outgoing light particles to its conjugate. There is no four-light contact Feynman vertex, nor a charge-allowed $t$ or $u$ exchange with these external labels.

Define the invariant scattering amplitude by $⟨f|S-1|i⟩=i(2\pi {)}^4{\delta }^{(4)}(P_f-P_i)\mathcal{M}$. ThenThe Feynman vertices already contain the identical-leg combinatorics. No extra factor of two multiplies this invariant scattering amplitude; the distinct identical final-state symmetry factor enters a cross section integral.

For $g=0$, the invariant scattering amplitude from the tree-level Feynman diagram in i has a pole at $s=M^2$. The intermediate heavy antiparticle can go on shell and live for a time of order its mean lifetime. Near this scattering resonance, inserting a single self-energy is no longer a uniformly small correction. Dyson resummation gives, to leading accuracy in a narrow decay width,Here $M$ is understood as the renormalized resonance mass; the real self-energy shifts the original pole position. The imaginary term removes the unphysical divergence of the fixed-order invariant scattering amplitude.

For equal incoming and outgoing light masses, the two-body momentum ratio cancels in the cross section. The angle-independent invariant scattering amplitude, integrated over the full solid angle with the identical final-state symmetry factor $1/2!$, gives $\sigma =|\mathcal{M}{|}^2/(32\pi s)$. Thus the Breit-Wigner resonance cross section isAt $s\gg M^2$ it reproduces $h^4/(32\pi s^3)$. At the resonance peak,This is also $8\pi /k_{\ast }^2$ for $k_{\ast }=M\beta /2$, the elastic identical-boson S wave partial-wave unitarity limit when this is the only open decay channel. The approximation assumes ${\Gamma }_{\mathrm{phys}}\ll M$ and a resonance sufficiently far from threshold for the width to vary slowly across it.

If one mechanically uses the unsymmetrized ${\Gamma }_{\mathrm{printed}}=2{\Gamma }_{\mathrm{phys}}$ from d, the same displayed Breit-Wigner resonance cross section would have that doubled width and predict a peak $8\pi /(M^2{\beta }^2)$. This is the literal-width substitution, rather than the consistent physical result: the given high-energy cross section already includes the identical-final-particle factor that the printed decay width integral omits.