= Solution
With the conventional <particle> labels, the allowed cubic <decay channels> are
$$
\boxed{\Phi\to\bar\phi\bar\phi,\qquad\bar\Phi\to\phi\phi,\qquad\mathcal M=-h\quad\hbox{for either channel}.}
$$
Each has one <tree-level Feynman diagram>, consisting of one cubic <Feynman vertex>. The final identical-leg contractions supply $2!$ and cancel the $1/2!$ in the <Lagrangian density>.
\Image[/past-exam-of-the-quantum-field-theory-course-of-the-university-of-oxford/2025-heavy-scalar-decay.png]
{title=The two charge-conjugate heavy-scalar decays, with explicit antiparticle labels and conserved charge arrows}
Taken literally with these labels, \b[the printed $\Phi\to\phi\phi$ channel is forbidden]: its initial <Noether charge> is $-2$ and its final <Noether charge> is $+2$. Its <decay amplitude> is zero, at every order preserving the <global phase symmetry>. If the source uses “$\Phi$” to name the heavy <antiparticle> instead, its intended answer is the nonzero <decay amplitude> $-h$ for $\bar\Phi\to\phi\phi$. The following <decay width> calculation concerns either allowed channel, whose widths are equal by <charge conjugation>.
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