For <Phi-six theory>, let $D_0=D_F(0)$ and $D_{xy}=D_F(x-y)$. The <Dyson series> gives
$$
Z[0]=\langle0|T\{S\}|0\rangle
=1-\frac{i\lambda}{6!}\int d^4x\,\langle\phi(x)^6\rangle_0
+\frac{(-i\lambda)^2}{2(6!)^2}\int d^4x\,d^4y\,
\langle\phi(x)^6\phi(y)^6\rangle_0+O(\lambda^3).
$$
At one vertex, <Wick theorem> supplies $5!!=15$ pairings. At two vertices let $r$ be the number of propagators joining them. It must be $0,2,4,$ or $6$, and the number of contractions is
$$
N_r=\binom6r^2r!\bigl((5-r)!!\bigr)^2,
\qquad
(N_0,N_2,N_4,N_6)=(225,4050,5400,720).
$$
Therefore
$$
\boxed{
\begin{aligned}
Z[0]={}&1-\frac{i\lambda}{48}\int d^4x\,D_0^3\\
&+(-i\lambda)^2\int d^4x\,d^4y\left[
\frac{D_0^6}{4608}
+\frac{D_0^4D_{xy}^2}{256}
+\frac{D_0^2D_{xy}^4}{192}
+\frac{D_{xy}^6}{1440}
\right]+O(\lambda^3).
\end{aligned}}
$$
The four <Vacuum Feynman diagram> types are shown below. The $r=0$ term is two disconnected copies of the order-$\lambda$ three-tadpole graph; the other three are connected.

\Image[/media/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/iii/paper-301-vacuum-bubbles.png]
{title=Vacuum diagrams in phi-six theory through second order}
{description=At first order one six-valent vertex is paired into three tadpoles. At second order the two vertices can have two, four, or six connecting propagators, with the remaining legs paired into tadpoles.}
{height=650}

Define
$$
B_1=-\frac{i\lambda}{48}\int d^4x\,D_0^3
$$
and let $B_2$ be the sum of the $r=2,4,6$ terms in the second line. The disconnected $r=0$ contribution is exactly $B_1^2/2$. Hence
$$
\boxed{Z[0]=\exp\{B_1+B_2+O(\lambda^3)\},}
$$
which is the <linked-cluster theorem>: the logarithm of the vacuum amplitude is the sum of connected vacuum bubbles.

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 301 / 4 / iii / Solution

The perturbative expansion of $Z[J]$ contains arbitrary <Feynman diagrams>, including disconnected products. The <linked-cluster theorem> gives
$$
Z[J]=Z[0]\exp\left(\sum_{C\ \text{connected}}C[J]\right),
$$
because the factorials from repeated connected components reproduce the exponential series. Therefore $W[J]=-\hbar\log Z[J]$ is the sum of <connected Feynman diagrams>.

The Legendre transform removes diagrams that disconnect upon cutting one internal line. Equivalently, every connected diagram is a tree whose vertices are exact one-particle-irreducible vertices and whose edges are exact propagators. Thus
$$
\boxed{\Gamma[\Phi]=S[\Phi]+\text{the sum of loop-level one-particle-irreducible diagrams},}
$$
and its functional derivatives are the <one-particle-irreducible correlation functions>. Algebraically, differentiating the Legendre relations gives
$$
\int d^dz\,
\frac{\delta^2\Gamma}{\delta\Phi(x)\delta\Phi(z)}
\frac{\delta^2W}{\delta J(z)\delta J(y)}
=-\delta^{(d)}(x-y),
$$
so an exact <quantum field theory propagator> joining two proper vertices is precisely the inverse <Hessian matrix> needed to reconstruct connected diagrams.

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 304 / 1 / c / Solution

{wiki}

The linked-cluster theorem states that the logarithm of a vacuum generating functional is the sum of connected vacuum diagrams. Exponentiating that sum reconstructs arbitrary products of disconnected connected components with their correct factorials.


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 Incoming links: Linked-cluster theorem