 Solution

For an allowed decay channel in iii, $|\mathcal{M}{|}^2=h^2$. Integrating the spatial Dirac delta distribution in the printed Lorentz-invariant phase-space measure sets ${\mathbf{p}}_2=-{\mathbf{p}}_1$. With $p=|{\mathbf{p}}_1|$ and $E_p=\sqrt{p^2+m^2}$, its remaining value isThe positive root is $p_{\ast }=(M/2)\beta$, where $\beta =\sqrt{1-4m^2/M^2}$, and the Dirac delta distribution contributes the Jacobian determinant $E_{\ast }/(2p_{\ast })$. HenceIt follows that the literal printed integral evaluates toThe printed integration ranges count each final configuration twice, once as $({\mathbf{p}}_1,{\mathbf{p}}_2)$ and once in the opposite order. For two identical final bosons, the physical decay width therefore requires an additional identical final-state symmetry factor $1/2!$:There is one allowed decay channel per heavy particle, not an extra factor of two from summing the two charge-conjugate heavy species. Both expressions vanish below the two-body decay threshold $M=2m$, and the limiting value at threshold is zero. For the literally forbidden labels $\Phi \to \varphi \varphi$, the decay width is instead zero because its decay amplitude vanishes, as explained in iii.