= Solution
The nonzero free <scalar propagators> join a field to its adjoint:
$$
\boxed{\langle T\phi\phi^\dagger\rangle(p)=\frac{i}{p^2-m^2+i0},\qquad \langle T\Phi\Phi^\dagger\rangle(p)=\frac{i}{p^2-M^2+i0}.}
$$
Contractions $\langle T\phi\phi\rangle$, $\langle T\Phi\Phi\rangle$ and mixed-species contractions vanish. With all <momenta> incoming, the <Feynman vertices> are
$$
\boxed{V_{\phi\phi^\dagger\Phi\Phi^\dagger}=-ig,\qquad V_{\phi\phi\Phi}=-ih,\qquad V_{\phi^\dagger\phi^\dagger\Phi^\dagger}=-ih.}
$$
The 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 $d^4p/(2\pi)^4$.
\Image[/past-exam-of-the-quantum-field-theory-course-of-the-university-of-oxford/2025-charged-scalar-rules.png]
{title=Charged scalar propagators and vertices, with arrows following positive conserved charge and heavy arrows carrying two units}
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 $\phi^2\Phi$ <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.
Back to article page
