= Solution
For a single species both panels of the figure in <2/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 <2/c>. The exchange of identical external <fermionic operators> gives a relative minus sign:
$$
\boxed{\mathcal M=-g^2\left(\frac{(\bar v_2u_1)(\bar u_3v_4)}{s-M^2+i0}-\frac{(\bar u_3u_1)(\bar v_2v_4)}{t-M^2+i0}\right).}
$$
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:
$$
\boxed{\mathcal M\longmapsto-\mathcal M,\qquad \overline{|\mathcal M|^2}(s,t)=\overline{|\mathcal M|^2}(t,s).}
$$
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, <2/c> fixes the square of the first diagram. <Crossing symmetry> fixes the second by $s\leftrightarrow t$. Consequently the strongest immediate inference is
$$
\boxed{\overline{|\mathcal M|^2}=g^4\left[\frac{(s-4m^2)^2}{(s-M^2)^2}+\frac{(t-4m^2)^2}{(t-M^2)^2}+\frac{C(s,t)}{(s-M^2)(t-M^2)}\right],\quad C(s,t)=C(t,s).}
$$
The <interference> function is not determined by the separate diagram squares. In particular \b[adding the two squares and dropping <interference> is unjustified]. The <Cauchy-Schwarz inequality> applied to the arrays of spin amplitudes also gives $|C(s,t)|\leq2|(s-4m^2)(t-4m^2)|$ on the physical scattering region.
For completeness, its value can be checked with one contracted <trace>, without another diagram calculation. The cross product is
$$
K=\frac14\operatorname{tr}[(\not p_2-m)(\not p_1+m)(\not p_3+m)(\not p_4-m)]
=-\frac12st-2m^2(s+t)+8m^4.
$$
The relative minus in the <invariant scattering amplitude> gives $C=-2K$. Thus the fully evaluated result, beyond the symmetry inference alone, is
$$
\boxed{\overline{|\mathcal M|^2}=g^4\left[\frac{(s-4m^2)^2}{(s-M^2)^2}+\frac{(t-4m^2)^2}{(t-M^2)^2}+\frac{st+4m^2(s+t)-16m^4}{(s-M^2)(t-M^2)}\right].}
$$
Every 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.
Back to article page
