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Source PDF: mmathphys.physics.ox.ac.uk/sites/g/files/vzsiac2991/files/mmathphys/documents/media/a15088h1.pdf

The solutions use natural units $\hslash =c=1$ and the Minkowski metric $\eta =\mathrm{diag}(1,-1,-1,-1)$. In question 1, ${\gamma }^5={\gamma }^0{\gamma }^1{\gamma }^2{\gamma }^3$ is exactly the paper's convention: it contains no factor $i$. The qualifications in iii, d and d distinguish the printed expressions from the results of charge conservation, identical final-state symmetry factors and fermionic time ordering.

For the complex scalar fields, use the conventional particle names: $\varphi$ annihilates a light $\varphi$ particle and creates a light antiparticle $\bar{\varphi }$; $\Phi$ annihilates a heavy $\Phi$ particle and creates $\bar{\Phi }$. These choices make the distinction between particle labels and field symbols explicit throughout ii and iii. The couplings $g,h$ are real as required by the printed Hermitian Lagrangian density.

Keep the particle labels specified under 2. There are two connected tree-level Feynman diagrams: the quartic contact interaction vertex and an $s$-channel light antiparticle exchange. The initial Noether charge is $1-2=-1$, exactly the charge of $\bar{\varphi }$. At the first cubic Feynman vertex, the incoming $\varphi$ and $\Phi$ leave the second $\varphi$ leg internal; the conjugate Feynman vertex supplies both outgoing particles.

ThusHere $s=(p_1+p_2{)}^2$ for incoming $\varphi (p_1),\Phi (p_2)$. Both terms have the same sign in the chosen S-matrix convention.

The distinction from antiparticles matters. If the paper's heavy “$\Phi$ particle” is instead intended to mean the charge-$+2$ state $\bar{\Phi }$, so that its printed decay into $\varphi \varphi$ is allowed, the exchange diagram is crossed. For that alternative physical process,This is a consistent alternative naming convention only if it is also applied to the decay in iii; one cannot use the same heavy particle label for both charge assignments.

For an allowed decay channel in iii, $|\mathcal{M}{|}^2=h^2$. Integrating the spatial Dirac delta distribution in the printed Lorentz-invariant phase-space measure sets ${\mathbf{p}}_2=-{\mathbf{p}}_1$. With $p=|{\mathbf{p}}_1|$ and $E_p=\sqrt{p^2+m^2}$, its remaining value isThe positive root is $p_{\ast }=(M/2)\beta$, where $\beta =\sqrt{1-4m^2/M^2}$, and the Dirac delta distribution contributes the Jacobian determinant $E_{\ast }/(2p_{\ast })$. HenceIt follows that the literal printed integral evaluates toThe printed integration ranges count each final configuration twice, once as $({\mathbf{p}}_1,{\mathbf{p}}_2)$ and once in the opposite order. For two identical final bosons, the physical decay width therefore requires an additional identical final-state symmetry factor $1/2!$:There is one allowed decay channel per heavy particle, not an extra factor of two from summing the two charge-conjugate heavy species. Both expressions vanish below the two-body decay threshold $M=2m$, and the limiting value at threshold is zero. For the literally forbidden labels $\Phi \to \varphi \varphi$, the decay width is instead zero because its decay amplitude vanishes, as explained in iii.