= Solution
The first-order <Dyson series> contains $-ig\int d^4w\,\varphi^\dagger\varphi\phi$. For an amputated three-point <Feynman diagram>, each of the three distinct field types contracts in exactly one way. Writing each external plane wave in momentum space produces
$$
\int d^4w\,e^{-i(p_1+p_2+p_3)\cdot w}=(2\pi)^4\delta^4(p_1+p_2+p_3).
$$
Thus the momentum-space <Feynman rule>, with all <four-momenta> incoming, is
$$
\boxed{-ig\,(2\pi)^4\delta^4(p_1+p_2+p_3).}
$$
After stripping the conservation delta factor, it is simply $-ig$. There is no $3!$ combinatorial factor because the interaction has three distinct field slots. The charged lines carry opposite charge-flow orientations and the real heavy line is neutral.
Back to article page
