Past exam of the quantum field theory course of the University of Oxford / 2019 / 2

Use the Fourier transform convention $\varphi (x)=\int d^{4}p/(2\pi {)}^4{e}^{-ip\cdot x}\tilde{\varphi }(p)$, and similarly for $\Phi$. All four-momenta at an interaction vertex are incoming. The quadratic terms are diagonal, so the scalar propagators areThe field interaction terms are $\lambda \varphi {\Phi }^2/2$ and $g\Phi {\partial }_{\mu }\varphi {\partial }^{\mu }\varphi /2$. The two identical field slots cancel the factor $1/2$ at either cubic interaction vertex. Expanding $\exp (i{S}_{\mathrm{int}})$ therefore givesIn the second rule $p,q$ are specifically the two $\varphi$ four-momenta; there is no derivative acting on the $\Phi$ leg. Indeed the two derivatives give $(-i{p}_{\mu })(-i{q}^{\mu })=-p\cdot q$, fixing the minus sign. This is a mixed scalar derivative cubic vertex. For arbitrary off-shell legs one may equivalently write $-ig(k^2-p^2-q^2)/2$, but may not replace the squared four-momenta by squared masses unless those legs are on their mass shells.

Each interaction vertex also carries $(2\pi {)}^4{\delta }^{(4)}(\sum p)$ for momentum conservation. Integrate independent loop momenta with $d^{4}p/(2\pi {)}^4$ and divide by the Feynman-diagram symmetry factor. An amputated external scalar leg has factor 1. The Feynman i-epsilon prescription fixes the propagator poles. As a dimensional check in four spacetime dimensions, $[\varphi ]=[\Phi ]=1$, $[\lambda ]=1$ and $[g]=-1$ in mass dimensions; the two powers of momentum in the derivative rule restore mass dimension one for its cubic interaction vertex.

Let $p_1,p_2$ denote the incoming $\varphi ,\Phi$ four-momenta and $p_3,p_4$ the outgoing $\varphi ,\Phi$ four-momenta. Thus $p_1+p_2=p_3+p_4$, $p_1^2=p_3^2=m^2$ and $p_2^2=p_4^2=M^2$. Define the Mandelstam variables and exchanged four-momenta byThe leading connected four-point scattering amplitude uses two cubic interaction vertices. There are four tree-level Feynman diagrams: $s$ and $u$ exchange of $\Phi$ with two $\lambda$ vertices, and $s$ and $u$ exchange of $\varphi$ with two $g$ vertices.

There is no term proportional to $\lambda g$: one vertex of each kind has three field slots of each species altogether, whereas the external state uses two of each. This leaves one $\varphi$ and one $\Phi$ to contract internally, and their mixed scalar propagator is zero. There is also no $t$ channel: grouping both external $\varphi$ legs at one vertex would require a $g$ vertex with an internal $\Phi$, while grouping both external $\Phi$ legs at the other would require a $\lambda$ vertex with an internal $\varphi$.

Define $i\mathcal{M}$ after removing the overall momentum conservation delta function. The nonderivative exchange givesFor the derivative $s$ channel, the incoming $\varphi$ four-momenta at the first vertex are $p_1,-q_s$, and at the second they are $q_s,-p_3$. The two interaction vertices are therefore $ig{p}_1\cdot q_s$ and $ig{p}_3\cdot q_s$. The same reasoning with $q_u$ applies to the $u$ channel, givingThe mass shell conditions and momentum conservation implyCombining the four Feynman diagrams therefore yieldsEach displayed tree-level Feynman diagram has Feynman-diagram symmetry factor 1 for fixed external species and labels. The result is of order ${\lambda }^2,g^2$; if both couplings are scaled by a common parameter $\epsilon$, the next connected corrections begin at order ${\epsilon }^4$. A four-point cubic Feynman diagram has $3V=2I+4$, so its vertex count $V$ is even. The internal lines remain off shell throughout: imposing an on-shell relation on an exchanged four-momentum would incorrectly change the derivative vertices.