The time-domain <Feynman rules> are:

* each internal line joining times $t,t'$ contributes $D(t-t')$;
* each cubic vertex contributes $-i\lambda$ and an integration $\int dt$;
* divide by the graph's <Feynman-diagram symmetry factor>;
* attach the external times to the corresponding lines and retain connected graphs.

There is no order-$\lambda$ connected two-point correction. Through order $\lambda^2$, the two connected topologies are the two-vertex fish graph and the one-particle-reducible tadpole graph. With the displayed vertex convention,
$$
\begin{aligned}
\langle T x(t_1)x(t_2)\rangle_{\mathrm{conn}}
={}&D(t_1-t_2)\\
&+\frac{(-i\lambda)^2}{2}\int dt\,dt'\,
D(t_1-t)D(t-t')^2D(t'-t_2)\\
&+\frac{(-i\lambda)^2}{2}\int dt\,dt'\,
D(t_1-t)D(t_2-t)D(t-t')D(0)
+O(\lambda^4),
\end{aligned}
$$
up to the common factors of $i$ associated with the propagator convention. Vacuum normalization removes disconnected vacuum bubbles.

These integrals are ultraviolet finite in one time dimension: a harmonic-oscillator propagator behaves as $E^{-2}$ at large frequency, and the loop-frequency integrals have negative superficial degree of divergence. They are also infrared finite because $\omega>0$ supplies a gap. The cubic instability affects nonperturbative convergence but does not create a divergence in these fixed-order integrals.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 304 / 1 / c / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-304/1/c/solution