= Solution
Use the <Fourier transform> convention $\phi(x)=\int d^4p/(2\pi)^4\,e^{-ip\cdot x}\widetilde\phi(p)$, and similarly for $\Phi$. All <four-momenta> at an <interaction vertex> are incoming. The quadratic terms are diagonal, so the <scalar propagators> are
$$
\boxed{D_\phi(p)=\frac{i}{p^2-m^2+i0},\qquad
D_\Phi(p)=\frac{i}{p^2-M^2+i0},\qquad D_{\phi\Phi}=0.}
$$
The <field interaction terms> are $\lambda\phi\Phi^2/2$ and $g\Phi\partial_\mu\phi\partial^\mu\phi/2$. The two identical field slots cancel the factor $1/2$ at either cubic <interaction vertex>. Expanding $\exp(iS_{\rm int})$ therefore gives
$$
\boxed{\begin{aligned}
\phi(p)\Phi(k)\Phi(\ell):&\quad i\lambda,\qquad p+k+\ell=0,\\
\Phi(k)\phi(p)\phi(q):&\quad -ig\,p\cdot q,\qquad k+p+q=0.
\end{aligned}}
$$
In the second rule $p,q$ are specifically the two $\phi$ <four-momenta>; there is no derivative acting on the $\Phi$ leg. Indeed the two derivatives give $(-ip_\mu)(-iq^\mu)=-p\cdot q$, fixing the minus sign. This is a <mixed scalar derivative cubic vertex>. For arbitrary off-shell legs one may equivalently write $-ig(k^2-p^2-q^2)/2$, but may not replace the squared <four-momenta> by squared <masses> unless those legs are on their <mass shells>.
Each <interaction vertex> also carries $(2\pi)^4\delta^{(4)}(\sum p)$ for <momentum conservation>. Integrate independent <loop momenta> with $d^4p/(2\pi)^4$ and divide by the <Feynman-diagram symmetry factor>. An amputated external scalar leg has factor 1. The <Feynman i-epsilon prescription> fixes the propagator poles. As a dimensional check in four spacetime dimensions, $[\phi]=[\Phi]=1$, $[\lambda]=1$ and $[g]=-1$ in <mass dimensions>; the two powers of <momentum> in the derivative rule restore <mass dimension> one for its cubic <interaction vertex>.
Back to article page
