A single quintic vertex has five half-edges, so <Wick theorem> cannot contract all of them in pairs. The leading connected <Vacuum Feynman diagrams> therefore have two quintic vertices and are of order $g^2$. If $c$ propagators join the two vertices, each vertex has $5-c$ half-edges left for tadpoles, so $c$ must be odd. This gives exactly three topologies:

* $c=1$: one line joins the vertices and each vertex carries two tadpole loops.
* $c=3$: three lines join the vertices and each vertex carries one tadpole loop.
* $c=5$: all five lines join the vertices.

The three diagrams are connected; every other pairing is either isomorphic to one of them or disconnected.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 304 / 1 / a / Solution

Inside the <generating functional>, multiplication by a field can be replaced by a <functional derivative> of the source factor:
$$
\phi(x)e^{i\int J\phi}=\frac1i\frac{\delta}{\delta J(x)}e^{i\int J\phi}
=-i\frac{\delta}{\delta J(x)}e^{i\int J\phi}.
$$
Expanding the interaction exponential, making this replacement in every term, and resumming gives
$$
\boxed{
Z_n[J]=\exp\!\left[-ig\frac{(-i)^n}{n!}
\int d^dx\,\frac{\delta^n}{\delta J(x)^n}\right]Z_0[J]}.
$$
This formal identity assumes a common regulator, a source-independent normalization, and permission to interchange the <path integral>, power series, and functional derivatives. A normalized functional with $Z_n[0]=1$ requires division by the same expression evaluated at $J=0$, which removes connected <Vacuum Feynman diagrams>.

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 304 / 1 / c / Solution

{c}

A vacuum Feynman diagram has no external lines and contributes a source-independent factor to the partition function. Taking the logarithm retains only connected vacuum diagrams, while normalizing $Z[0]=1$ removes all of them.


Codex @codex  0

 Incoming links: Vacuum Feynman diagram