Replace each occurrence of $x(t)$ in the interaction by the <functional derivative> that inserts it. In standard Minkowski source conventions,
$$
Z[J]
=\exp\left[
-\frac{i\lambda}{3!}\int dt
\left(\frac1i\frac{\delta}{\delta J(t)}\right)^3
\right]Z_0[J]
$$
and hence
$$
Z[J]
=\sum_{n=0}^\infty\frac1{n!}
\left[-\frac{i\lambda}{3!}\int dt
\left(\frac1i\frac{\delta}{\delta J(t)}\right)^3\right]^n
Z_0[J].
$$
The factors of $i$ are adjusted together if one uses the source convention of part a directly.

This is a <perturbation series>, generally an asymptotic rather than convergent series because the number of <Wick theorem>[Wick contractions] grows factorially. For real cubic coupling the potential is also unbounded on one side, so the real-axis theory does not possess a stable nonperturbative ground state without a contour prescription or further stabilizing interactions.

Solution

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

A perturbation series expands a correlation function or scattering amplitude in powers of interaction couplings. In quantum field theory it is commonly asymptotic because diagram counts grow factorially and nonperturbative effects are invisible to every finite order.


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