= Solution
Assume $m^2>0$ and take $\lambda\to0^+$. The free <Gaussian integral> gives $Z_0=\sqrt{2\pi}/|m|$, and the normalized free <Gaussian measure> has <covariance> $C=m^{-2}$. Expanding only the interaction exponential gives the <asymptotic expansion>
$$
\frac{Z}{Z_0}=\sum_{V\geq0}\frac{(-\lambda)^V}{V!(6!)^V}\langle\phi^{6V}\rangle_0.
$$
In this <zero-dimensional scalar field theory>, the <path integral> has become an ordinary <integral>, but its combinatorics are precisely those of <sextic scalar field theory>.
The free <generating functional> is $\langle e^{J\phi}\rangle_0=e^{CJ^2/2}$. Differentiating it proves <Wick theorem>: odd <moments> vanish, while each even <moment> is the sum over pairings of its factors. Each <Wick contraction> contributes $C$, so
$$
\langle\phi^{2r}\rangle_0=(2r-1)!!\,C^r.
$$
Thus the <Feynman rules> are a <quantum field theory propagator> $C$ for each internal line and a factor $-\lambda$ for each six-valent <interaction vertex>. The <factorials> $V!$ and $(6!)^V$ remove vertex and half-edge labels; the remaining weight is $(-\lambda)^V C^I/S$, where $S$ is the <Feynman-diagram symmetry factor>, the order of the <automorphism group> of the resulting diagram, including its half-edge symmetries. A <tadpole diagram> has an additional interchange symmetry of the two ends of its loop.
There is one empty <Vacuum Feynman diagram>, contributing $1$. At one <interaction vertex>, the six half-edges form three loops. Its <Feynman-diagram symmetry factor> is $2^3 3!=48$, so its contribution is $-\lambda C^3/48$.
At two <interaction vertices> let $r$ count the lines joining them. Each <interaction vertex> has $(6-r)/2$ self-loops, so $r$ can only be $0,2,4,6$. These four possibilities exhaust all <Vacuum Feynman diagrams> at this order. Their <Feynman-diagram symmetry factors> are
$$
S_r=2\,r!\left[2^{(6-r)/2}\left(\frac{6-r}{2}\right)!\right]^2,
\qquad
\begin{array}{c|rrrr}
r&0&2&4&6\\\hline
S_r&4608&256&192&1440
\end{array}.
$$
Here the leading $2$ exchanges the two <interaction vertices>, $r!$ permutes the joining lines, and the remaining factors permute and reverse the self-loops. The $r=0$ graph is disconnected and must be included in $Z/Z_0$; only a <connected generating functional> would discard it.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2018/iii/paper-304-phi6-vacuum-diagrams.png]
{title=All <Vacuum Feynman diagrams> with at most two <interaction vertices>}
{description=The empty diagram, the one-<interaction vertex> diagram, and the four two-<interaction vertex> diagrams, with their <Feynman-diagram symmetry factors>.}
Consequently,
$$
\boxed{\frac{Z(m,\lambda)}{Z(m,0)}=1-\frac{\lambda}{48m^6}+\frac{77\lambda^2}{7680m^{12}}+O(\lambda^3).}
$$
Indeed $\sum_r S_r^{-1}=77/7680$, independently agreeing with $11!!/[2(6!)^2]$ from <Wick theorem>.
This is an <asymptotic expansion>, rather than a convergent <perturbation series>: the coefficient of $\lambda^V$ contains $(6V-1)!!/[V!(6!)^V]$ and grows too rapidly for a nonzero <radius of convergence>. For any fixed truncation order, the <Taylor expansion> remainder bound for $e^{-x}$ on $x\geq0$, followed by the finite Gaussian <moment> <integral>, proves the stated remainder estimate. For negative real $\lambda$ the original integral diverges.
Back to article page
