Past exam of the theoretical physics major option course of the University of Oxford / 2015 / 6 / d / Solution

The leading decay is into one complex particle and its antiparticle, which are distinguishable. Using $|\mathcal{A}{|}^2=g^2$ in a givesIn four space-time dimensions a canonically normalized scalar has mass dimension one, so $g$ has mass dimension one. The amplitude starts at order $g$, a rate is quadratic in that amplitude, and the only mass scale in the massless-daughter decay is $m$. Those facts already require $\Gamma \propto g^2/m$, leaving only a dimensionless numerical factor to calculate.

For low-energy charged-scalar scattering through a neutral scalar, label the incoming particle and antiparticle as $1,2$ and their outgoing counterparts as $3,4$. Charge flow allows $s$-channel annihilation and $t=(p_1-p_3{)}^2$ exchange of the real massive scalar. It forbids a separate $u$-channel graph for this fixed assignment of distinguishable final charges. The tree amplitude isAt low energy it is approximately angle independent. The massless two-body scattering formula then givesIn particular the requested scale estimate is $\sigma \sim g^4/(m^{4}s)$, not $g^4/m^6$ with the incoming-energy dependence omitted. Integrating out $\varphi$ gives ${\mathcal{L}}_{\mathrm{eff}}=g^2({\phi }^{†}\phi {)}^2/(2m^2)$ at this order, whose complex-scalar four-leg combinatorics reproduce the same $2g^2/m^2$ amplitude. These are weak-coupling perturbative results about the specified vacuum; possible ultraviolet stabilizing interactions are not determined by the cubic term alone.