 Solution

For the interaction sign in this paper, the cubic Feynman rule is $+i\lambda$, while the scalar Feynman propagator is $i/(k^2-m^2+iϵ)$. Each exchange graph therefore contributes $-i{\lambda }^2/(k^2-m^2+iϵ)$ to $i\mathcal{M}$. The three graphs giveThe equal-mass Mandelstam variables satisfy $s+t+u=4m^2$, so $u-m^2=3m^2-s-t$. In the physical elastic region with $m>0$, none of these denominators vanishes, and the tree amplitude is real up to its irrelevant overall convention. ThusThe square includes interference between all three channels; summing their individual squared contributions would omit physical cross terms. If a denominator approaches a pole, retain the Feynman i-epsilon prescription in the modulus before taking a limit.