Past exam of the quantum field theory course of the University of Oxford / 2025 / 2 / c

Let $p_1,p_2$ be the incoming and $p_3,p_4$ the outgoing four-momenta, and set $s=(p_1+p_2{)}^2$, $t=(p_1-p_3{)}^2$, $u=(p_1-p_4{)}^2$, the Mandelstam variables. The only connected tree-level Feynman diagram is the $s$-channel exchange of $\bar{\Phi }$, with Noether charge $+2$. The two incoming light particles attach to ${\varphi }^2\Phi$, and the two outgoing light particles to its conjugate. There is no four-light contact Feynman vertex, nor a charge-allowed $t$ or $u$ exchange with these external labels.

Define the invariant scattering amplitude by $⟨f|S-1|i⟩=i(2\pi {)}^4{\delta }^{(4)}(P_f-P_i)\mathcal{M}$. ThenThe Feynman vertices already contain the identical-leg combinatorics. No extra factor of two multiplies this invariant scattering amplitude; the distinct identical final-state symmetry factor enters a cross section integral.

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.

With the conventional particle labels, the allowed cubic decay channels areEach 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.

Taken literally with these labels, the printed $\Phi \to \varphi \varphi$ 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 $\bar{\Phi }\to \varphi \varphi$. The following decay width calculation concerns either allowed channel, whose widths are equal by charge conjugation.