 Solution

The nonzero free scalar propagators join a field to its adjoint:Contractions $⟨T\varphi \varphi ⟩$, $⟨T\Phi \Phi ⟩$ and mixed-species contractions vanish. With all momenta incoming, the Feynman vertices areThe two identical light legs cancel the $1/2!$ in each cubic interaction vertex; there is no remaining factor $1/2$ in its Feynman rule. Multiply each Feynman vertex by $(2\pi {)}^4{\delta }^{(4)}(\sum p)$, and integrate an internal unconstrained four-momentum with $d^{4}p/(2\pi {)}^4$.

The arrows in the figure follow positive Noether charge: a light arrow carries one unit and a heavy arrow two. Consequently a heavy particle $\Phi$ travels opposite its charge arrow, whereas its antiparticle $\bar{\Phi }$ travels along it. At a ${\varphi }^2\Phi$ Feynman vertex, two unit-charge arrows enter and a two-unit arrow leaves; the conjugate Feynman vertex reverses every arrow. The quartic Feynman vertex has one incoming and one outgoing arrow of each weight. These rules prevent attaching a propagator between two unconjugated fields.