= 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 <1/b/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,
$$
\int d^3p\,d^3q\,
\delta\!\left(Q-\frac{p^2+q^2+(\mathbf p+\mathbf q)^2}{2m}\right).
$$
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. Consequently
$$
\boxed{\Gamma=C(M,g,\lambda)\,(M-3m)^2[1+O((M-3m)/M)]\quad\text{as }m\uparrow M/3,}
$$
with a finite positive leading coefficient when $g\lambda\ne0$; $\Gamma=0$ for $m\geq M/3$. 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.
Back to article page
