= Solution
The interaction-picture Hamiltonian density is $\mathcal H_I=\lambda\phi\Phi^\dagger\Phi$. The first-order <Dyson series> term cannot connect the four external particles, so the leading contribution is
$$
S^{(2)}
=\frac{(-i\lambda)^2}{2}
\int d^4x\,d^4y\,
\mathcal T\{\phi\Phi^\dagger\Phi(x)\,
\phi\Phi^\dagger\Phi(y)\}.
$$
By the <Wick theorem>, the nonzero terms annihilate the incoming complex particle using the annihilation part of $\Phi$, create the outgoing complex particle using the creation part of $\Phi^\dagger$, annihilate the incoming real particle using the annihilation part of $\phi$, and create the outgoing real particle using its creation part. The remaining $\Phi$ and $\Phi^\dagger$ form an internal <Feynman propagator>. Interchanging which vertex absorbs the incoming real scalar gives the two contractions; this factor of two cancels the Dyson factor $1/2$.
There are therefore an $s$-channel internal momentum $p+k$ and a crossed channel internal momentum $p-k'$. With covariantly normalized external states,
$$
\langle p',k'|S|p,k\rangle_{\rm conn}
=i(2\pi)^4\delta^{(4)}(p+k-p'-k')\mathcal M,
$$
where
$$
\boxed{
i\mathcal M=(-i\lambda)^2i
\left[
\frac1{(p+k)^2-M^2+i\epsilon}
+\frac1{(p-k')^2-M^2+i\epsilon}
\right]}.
$$
Equivalently,
$$
\mathcal M=-\lambda^2
\left[
\frac1{s-M^2+i\epsilon}
+\frac1{u-M^2+i\epsilon}
\right],
$$
with $s=(p+k)^2$ and $u=(p-k')^2$.
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