Past exam of the quantum field theory course of the University of Oxford / 2021 / 1 / e / Solution

Put $Q=M-3m$. For $Q<0$ the decay width is zero by conservation of energy. Just above the three-particle decay threshold, all three daughters are nonrelativistic. The decay amplitude of iv remains finite: the internal scalar propagator denominator approaches $(2m{)}^2-m^2=3m^2$, and the Dirac scalar bilinear does not vanish at rest. Thus, for nonzero fixed $g$ and $\lambda$, the leading suppression comes entirely from three-body threshold phase space.

Use the spatial Dirac delta function to eliminate one daughter momentum. The remaining integral has six independent momentum coordinates and one conservation of energy constraint,Rescaling both momenta by $\sqrt{mQ}$ gives $Q^3$ from the measure and $Q^{-1}$ from the Dirac delta function. The smooth factors $1/(2E)$ do not change the power. Consequentlywith a finite positive leading coefficient when $g\lambda \ne 0$; $\Gamma =0$ for $m\ge M/3$. The $1/2!$ for the identical bosons changes only this coefficient. Equivalently, at fixed $M$ the opening is quadratic in $M/3-m$. If $g\lambda =0$, this leading tree-level Feynman diagram vanishes, so the stated leading coefficient is zero.