Solution (source code)

= Solution

Let $D_{xy}=\langle0|T\phi(x)\phi(y)|0\rangle_0$ be the free <Feynman propagator>, and abbreviate $v_3=-i\kappa$, $v_4=-i\lambda$. Expand the normalized <Dyson series>, so <vacuum> bubbles cancel between numerator and denominator, and retain the <connected correlation function>. <Wick's theorem> yields
$$
\begin{aligned}
G_c(x,y)={}&D_{xy}
+\frac{v_4}{2}\int d^dz\,D_{xz}D_{zz}D_{zy}\\
&+\frac{v_3^2}{2}\int d^dz\,d^dw\,D_{xz}D_{zw}^2D_{wy}\\
&+\frac{v_3^2}{2}\int d^dz\,d^dw\,D_{xz}D_{zy}D_{zw}D_{ww}
+O(\text{two loops}).
\end{aligned}
$$
These are respectively the free line, the quartic <tadpole diagram>, the two-cubic-vertex <bubble diagram>, and an <tadpole contribution to a cubic-theory two-point function>, attached by a bridge to the vertex carrying both external legs. The last graph is connected but is a <one-particle-reducible Feynman diagram>; removing its bridge disconnects it. It must be included when all connected graphs are requested. It disappears if an additional vanishing-one-point-function <renormalization condition> and its corresponding <tadpole subtraction> are imposed; that convention was not specified here.

The factors can be counted directly. The quartic term has $4\cdot3=12$ attachments for the two labelled external legs, divided by $4!=24$. At second order in the cubic interaction the expansion denominator is $2!(3!)^2=72$. There are 36 <Wick contractions> for the bubble, and 36 for the attached-tadpole graph, giving $1/2$ for each. Thus each loop topology shown has <symmetry factor> $S=2$. Disconnected products of one-point functions and <vacuum> bubbles are not part of $G_c$.<Quantum Field Theory by David Tong>[Sections 3.3–3.4, Wick contractions and diagrammatic perturbation theory]{ref}

The <one-loop cubic-quartic scalar vertex classification> follows without truncating arbitrarily in powers of the <couplings>. A connected graph with $E$ external legs, $V_3$ cubic vertices, $V_4$ quartic vertices and loop number $L$ satisfies
$$
V_3+2V_4=E+2L-2.
$$
For $E=2,L=1$, the only possibilities are $(V_3,V_4)=(2,0)$ or $(0,1)$, exactly those just listed.

\Image[/past-exam-of-the-quantum-field-theory-course-of-the-university-of-oxford/2017-real-scalar-one-loop-graphs.png]
{title=One-loop <Feynman diagrams> in the real cubic-plus-quartic <scalar field theory>, with <symmetry factors> and channel multiplicities}