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A perturbative correlator factors into an exponential of connected Vacuum Feynman diagrams times a sum of Feynman diagrams attached to external insertions. Dividing its path integral by the vacuum generating functional at zero source cancels that exponential, including all disconnected products of vacuum components. The corresponding operator formalism uses the same denominator in the vacuum expectation of the Dyson series. This leaves externally disconnected correlators intact; the connected generating functional removes their products.

For a vacuum generating functional $Z[J]$, normalization gives $\hat{Z}[J]=Z[J]/Z[0]$ and $\hat{Z}[0]=1$. Functional derivatives of $\hat{Z}$ give normalized vacuum expectation values. The linked-cluster theorem shows that this ratio cancels Vacuum Feynman diagrams, while $-i\log \hat{Z}[J]$ gives connected correlators. Components attached to external insertions are not removed merely by the normalization.

Use ${\eta }_{\mu \nu }=\mathrm{diag}(1,-1,-1,-1)$ and $\hslash =c=1$. In the interaction picture, fields evolve with the free Hamiltonian, while interactions enter the state evolution through the Dyson series. Write the Klein-Gordon field as $\varphi ={\varphi }^{(+)}+{\varphi }^{(-)}$, with the positive-frequency part of a quantum field containing annihilation operators and ${\varphi }^{(-)}=({\varphi }^{(+)}{)}^{†}$ containing creation operators. The mode normalization in the PDF corresponds to $[a_{\mathbf{p}},a_{\mathbf{q}}^{†}]=(2\pi {)}^3{\delta }^3(\mathbf{p}-\mathbf{q})$ and $E_{\mathbf{p}}=\sqrt{|\mathbf{p}{|}^2+m^2}$.

The time-ordered product places fields with the latest time on the left. For two bosonic insertions,The normal-ordered product $:\varphi (x_1)\cdots \varphi (x_n):$ instead moves every creation operator to the left of every annihilation operator, without adding the commutators generated by that rearrangement. Its vacuum expectation value is zero whenever a nonempty set of free-field factors remains.

The mode commutator gives the positive-frequency Wightman functionDefine the Feynman propagator here to be the two-point function itself, rather than that function divided by $i$. ThenEquivalently,The positive-energy pole is below and the negative-energy pole above the real axis. Closing the energy contour integral below for positive time separation and above for negative time separation gives the two terms above. Thus this is the scalar Feynman propagator pole prescription, and $(\square +m^2){\Delta }_F=-i{\delta }^4$. In particular, $T\{\varphi (x)\varphi (y)\}=:\varphi (x)\varphi (y):+{\Delta }_F(x-y)$: the difference between time ordering and normal ordering is the Wick contraction.

For $n$ free bosonic fields, the Wick theorem stateswhere $P$ runs over all sets of disjoint pairs of indices, including the empty set, and $V(P)$ is the set of paired indices. The empty normal-ordered product equals one. This is an operator identity for free fields, understood with smearing or a regulator when products of operator-valued distributions require it.

Here is an inductive proof of the scalar Wick theorem. Sort the arguments chronologically so that $x_1^0$ is the latest. For any remaining set $I$, commuting the annihilation part of ${\varphi }_1$ through the creation parts in a normal-ordered product givesThe creation part of ${\varphi }_1$ is already in the correct position. Every commutator on the right is a scalar, so no further operator rearrangement is needed for it. Because $x_1$ is latest, that commutator is ${\Delta }_F(x_1-x_j)$. Assume the Wick theorem for the other $n-1$ fields and multiply its expansion by ${\varphi }_1$. The first term of the displayed identity gives every pairing in which index 1 is unpaired; the sum gives every pairing in which 1 is paired with exactly one $j$. Each final partial pairing has exactly one such construction. The one-field identity supplies the base case, proving the theorem by mathematical induction. Applying time ordering restores arbitrary argument order.

Taking the vacuum expectation value retains only complete pairings. Odd free correlators vanish; for example the free scalar four-point function is ${\Delta }_{12}{\Delta }_{34}+{\Delta }_{13}{\Delta }_{24}+{\Delta }_{14}{\Delta }_{23}$. These facts explain why propagators are the edges of Feynman diagrams.

For the interacting quartic scalar field theory, the interaction Hamiltonian density is $\lambda {\varphi }^4/4!$. With a single time ordering acting on external insertions and interaction vertices, the unnormalized numerator expands asApply the proved Wick theorem term by term. Each complete pairing connects external insertions and four-legged vertices by Feynman propagators. There are $4!$ assignments of four distinct incident legs at a vertex, canceling its factorial. Permuting identical integrated vertices cancels the corresponding part of $v!$. Permutations still leaving the labeled graph unchanged give its Feynman-diagram symmetry factor. For example,whereas the two-point tadpole diagram has coefficient $-i\lambda /2$, because choosing the two vertex legs contracted with the labeled external fields gives $4\cdot 3$ possibilities and the other two contract with each other.

Thus the position-space rules attach ${\Delta }_F$ to every edge, $-i\lambda \int d^{4}z$ to every four-legged interaction vertex, and divide by the graph's Feynman-diagram symmetry factor. Fourier transformation gives the momentum-space Feynman rules:Enforce momentum conservation at each vertex, retaining the overall four-momentum delta function. External lines of these correlation functions remain unamputated. The unnormalized expression includes Vacuum Feynman diagrams; dividing by $N_0=⟨0|S|0⟩$ cancels those disconnected vacuum factors. They should not be discarded without that normalization.

There is a time-ordering convention in the printed correlation-function notation worth making explicit. The usual interacting correlator uses the single global time ordering in $N_n$ above. If the displayed $T\{{\varphi }_1\cdots {\varphi }_n\}S$ is read literally with $S$ outside that ordering, the external block and the vertex block are separately time ordered: their cross-contractions are $D^{+}(x_i-z)$ because each external field stands to the left of every vertex field. The same pairing construction then uses ${\Delta }_F$ for lines within either block and $D^{+}$ for lines between blocks; in momentum space the latter is $2\pi \theta (p^0)\delta (p^2-m^2)$. This literal product is different from the conventional globally time-ordered interacting correlator, and does not have the ordinary external Feynman propagators derived above.

Use the normalized Gaussian integral with action $\mathrm{tr}(M^2)/2$. Its Wick contraction isThere is no factor $1/N$ here: this part uses $V$, whereas the following part uses $NV$. Expanding the normalized Hermitian matrix model integral to first order givesThe subtracted term removes Vacuum Feynman diagrams disconnected from the external pair. Since the one-point function vanishes by $M\mapsto -M$, the resulting two-point function is a connected correlation function.

Write the vertex as $M^a{}_b{M}^b{}_c{M}^c{}_d{M}^d{}_a$. Each external field must contract with a different vertex field, leaving the other two to form a tadpole diagram. Eight of the twelve connected pairings attach the external fields at adjacent cyclic positions. The remaining index loop gives $N{\delta }^i{}_l{\delta }^k{}_j$ in each case. The other four attach them at opposite positions and give ${\delta }^i{}_j{\delta }^k{}_l$, with no free index loop. ThereforeBoth index structures are required at finite $N$. For $N=1$ the answer is $1-3g+O(g^2)$, agreeing with the ordinary zero-dimensional quartic integral. This is a formal perturbative quantum field theory expansion; the real integral is convergent for $g\ge 0$.

Let the shell-restricted scalar propagator beWrite $\Delta S=S_0[\hat{\varphi }]+V[\varphi ,\hat{\varphi }]$ and average with the normalized Gaussian shell measure. The cumulant expansion givesThe subtraction removes disconnected Feynman diagrams; the logarithm retains connected Feynman diagrams made from shell contractions. This is the connected shell-contraction expansion. An external line denotes the background low field, not an additional low-momentum propagator.

At order $g$, panel A is the existing four-field interaction vertex. The new shell contractions are B, a tadpole diagram contributing a two-field vertex, and C, a Vacuum Feynman diagram contributing a constant. The Gaussian functional determinant in panel L is another field-independent term, of order $g^0$; the original quadratic kinetic term and mass term are retained as well.

For completeness, all two-vertex topologies from Wick contractions at order $g^2$ are shown. Let $r$ be the number of shell lines joining the vertices and $t_1,t_2$ their numbers of self-contractions. The external-field counts areInterchanging the two vertices identifies the same topology. Enumerating these conditions gives exactly the following eight possibilities:

The support qualification is important. In E or F, a vertex with one external low field and one bridge also has a tadpole whose two momenta cancel. Conservation forces the bridge momentum to equal that single low momentum, outside the shell. Equivalently, convolution by $C_{>}$ annihilates $\varphi$. This is momentum-support exclusion for Wilsonian bridge diagrams. D is different: its bridge carries the sum of three low momenta, which can lie in the shell. Its kernel is proportional to ${\varphi }^3(x)C_{>}(x-y){\varphi }^3(y)$ and need not vanish. Thus one-particle-reducible Feynman diagrams must not be excluded merely because the object being computed is an effective action.

These kernels need not be local before a derivative expansion. Their momentum dependence generates derivative interactions where such an expansion is valid. If all external momenta are sufficiently far below ${\Lambda }^{\prime }$, D vanishes too; in particular it does not produce a zero-momentum local ${\varphi }^6$ coupling at order $g^2$. A local six-field term is allowed and is generated at higher orders, for example by a three-vertex shell triangle.

To all orders, expect every scalar interaction allowed by the original symmetries: arbitrary even powers of the field and their allowed derivative couplings, together with vacuum terms. The Z2 symmetry $\varphi \mapsto -\varphi$ excludes odd-field vertices. The original quartic form is therefore not closed under exact Wilsonian integration.

Assume $m^2>0$ and take $\lambda \to 0^{+}$. 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 expansionIn 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 $⟨e^{J\varphi }{⟩}_0=e^{C{J}^2/2}$. Differentiating it proves Wick theorem: odd moments vanish, while each even moment is the sum over pairings of its factors. Each Wick contraction contributes $C$, soThus 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 areHere 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.

Consequently,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 $x\ge 0$, followed by the finite Gaussian moment integral, proves the stated remainder estimate. For negative real $\lambda$ the original integral diverges.

For Phi-six theory, let $D_0=D_F(0)$ and $D_{xy}=D_F(x-y)$. The Dyson series givesAt 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 isThereforeThe 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.

Defineand let $B_2$ be the sum of the $r=2,4,6$ terms in the second line. The disconnected $r=0$ contribution is exactly $B_1^2/2$. Hencewhich is the linked-cluster theorem: the logarithm of the vacuum amplitude is the sum of connected vacuum bubbles.

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:

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

Inside the generating functional, multiplication by a field can be replaced by a functional derivative of the source factor:Expanding the interaction exponential, making this replacement in every term, and resumming givesThis 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.