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

For

n = 2

, the coupling is a mass squared,

m^{2} = g

. After integration by parts, the quadratic action has kernel

K = - (□ + g - i 0)

. Completing the square in the Gaussian integral and choosing the source-independent normalization so that

Z_{2} [0] = 1

gives

Z_{2} [J] = exp [- \frac{i}{2} \int d^{d} x d^{d} y J (x) K^{- 1} (x - y) J (y)] .

Equivalently, in momentum space,

Z_{2} [J] = exp [- \frac{i}{2} \int \frac{d ^{d} p}{( 2 π ) ^{d}} \frac{J ( - p ) J ( p )}{p ^{2} - g + i 0}] .

The

i 0

prescription selects the Feynman propagator.

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