= Solution
The interaction $-g\phi\Psi_i^*\Psi_i$ gives one cubic vertex $-ig$ joining a real-scalar line to a particle-antiparticle pair of either complex scalar species. The second-order term of the <Dyson series> is
$$
\frac{(-i)^2}{2!}\int d^4x\,d^4y\,
T\{\mathcal H_{\mathrm{int}}(x)\mathcal H_{\mathrm{int}}(y)\}.
$$
<Wick theorem> contracts the incoming $\Psi_1\overline\Psi_1$ pair at one vertex, the outgoing $\Psi_2\overline\Psi_2$ pair at the other, and the two $\phi$ fields with each other. The two assignments of $x,y$ cancel the factor $2!$. Thus there is one connected <tree-level Feynman diagram>, the $s$-channel exchange
$$
\Psi_1(p)+\overline\Psi_1(q)
\longrightarrow\phi^*(p+q)
\longrightarrow\Psi_2(p')+\overline\Psi_2(q').
$$
With standard relativistic external-state normalization,
$$
i\mathcal M=(-ig)^2\frac{i}{(p+q)^2-\mu^2+i\epsilon},
\qquad
\mathcal M=-\frac{g^2}{s-\mu^2+i\epsilon},
$$
up to the physically irrelevant common sign convention for $\mathcal M$.
In the <center-of-momentum frame>, $q=(E_1,-\mathbf p)$ and $p=(E_1,\mathbf p)$, so
$$
s=(p+q)^2=4E_1^2=4(M_1^2+|\mathbf p|^2).
$$
Pair creation is kinematically possible exactly when $s\geq4M_2^2$. The threshold incident momentum is therefore
$$
|\mathbf p|_{\min}
=\sqrt{\max\{0,M_2^2-M_1^2\}}.
$$
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