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Split the free interaction picture field as $\varphi ={\varphi }^{(+)}+{\varphi }^{(-)}$, where the first part contains annihilation operators and the second creation operators. Time ordering places the operator at the later time on the left, while normal ordering places all creation operators on the left. For $x^0>y^0$, commuting the annihilation part of the first field through the creation part of the second givesThe commutator is the c-numberFor $y^0>x^0$, the same calculation interchanges $x,y$. Combining the two time orders proves the two-field Wick contraction identitywhereThe equality with the four-dimensional integral follows by closing the energy contour at the positive-energy pole for $x^0>y^0$ and the negative-energy pole for the reverse order. This fixes the scalar Feynman propagator pole prescription; in this convention $D_F$ itself includes the factor $i$.

Because these interactions have no derivatives, $H_{\mathrm{int}}=-\int {\mathcal{L}}_{\mathrm{int}}{d}^{3}x$. Expand the Dyson seriesApply the Wick theorem to each term together with the external insertions. Every paired field contributes a propagator, and the remaining normal-ordered vacuum expectation vanishes unless no fields remain. Representing the propagators by Fourier integrals and integrating each vertex position gives the four-momentum conservation delta function. The $3!$ and $4!$ contractions into labelled vertex slots cancel the factorials in the interaction density. The resulting scalar-field cubic and quartic Feynman vertices areEach vertex also has $(2\pi {)}^4{\delta }^{(4)}(\sum p)$, each independent internal momentum is integrated with $d^{4}p/(2\pi {)}^4$, and repeated equivalent contractions give the usual Feynman-diagram symmetry factor. For the unamputated correlation functions, retain the propagators joining external insertions to vertices. The displayed unnormalized numerator includes disconnected vacuum bubbles; dividing by $⟨0|S|0⟩$ cancels them. Amputated scattering amplitudes additionally use the LSZ reduction formula.

For stability, examine the classical scalar potentialThe tree-level boundedness criterion is $\lambda >0$, with arbitrary real cubic coupling. If $\lambda =0$, boundedness requires $\mu =0$ and $m^2\ge 0$; if $\lambda <0$, or $\lambda =0$ with $\mu \ne 0$, the potential is unbounded below. Thus a positive quartic coupling permits a stable ground-state vacuum, although it need not be the perturbative vacuum at $\varphi =0$.

For $\lambda >0$ and $m^2\ge 0$, the stability of a cubic-quartic scalar potential determines when zero is a global minimum:The square bracket is nonnegative for every field value exactly whenFor a nontrivial equality case there is a second degenerate minimum at $\varphi =2\mu /\lambda$. If $m^2>0$ and ${\mu }^2>3\lambda m^2$, zero remains a local minimum but is a false vacuum, with a lower minimum elsewhere. If $m^2=\mu =0$ and $\lambda >0$, the origin is the quartic minimum despite its vanishing quadratic curvature. These are classical or tree-level stability statements; renormalized couplings and the quantum effective potential refine the quantum analysis.

Split the free scalar field into its annihilation part ${\varphi }^{(+)}$ and creation part ${\varphi }^{(-)}$. Normal ordering moves every creation operator to the left of every annihilation operator, without adding the commutators produced by that rearrangement. For two fields,Its vacuum expectation is zero. Time ordering instead places the field at the later time on the left:The equal-time convention is immaterial away from coincident singularities; these expressions are understood as operator-valued distributions.

The only nonzero vacuum contraction uses $a_{\mathbf{p}}{a}_{\mathbf{q}}^{†}$. The oscillator commutator therefore gives the Wightman functionPut $\tau =x^0-y^0$ and $\mathbf{r}=\mathbf{x}-\mathbf{y}$. Combining the two time orders, and reversing $\mathbf{p}$ in the second when necessary, givesTo recover the four-dimensional Fourier transform, use the Feynman i-epsilon prescription in the energy variable:The positive-energy pole lies just below the real axis, and the negative-energy pole just above it. For $\tau >0$, close below clockwise: the residue at $E_{\mathbf{p}}-i0$ is $i{e}^{-i{E}_{\mathbf{p}}\tau }/(2E_{\mathbf{p}})$, and the clockwise factor $-i$ after dividing by $2\pi$ gives the required positive coefficient. For $\tau <0$, close above counterclockwise: the residue at $-E_{\mathbf{p}}+i0$ is $-i{e}^{i{E}_{\mathbf{p}}\tau }/(2E_{\mathbf{p}})$, and the factor $+i$ gives the other time order. Equivalently the energy contour passes above the positive-energy pole and below the negative-energy pole. HenceThe limit is distributional. The scalar Feynman propagator pole prescription is essential; omitting it from the denominator must be accompanied by the specified contour.

Finally, moving the annihilation part of $\varphi (x)$ past the creation part of $\varphi (y)$ givesThe normal-ordered two-field product is symmetric in $x,y$, because the two creation parts commute and the two annihilation parts commute. Applying the two time orders thus proves the two-field scalar Wick identity

Use natural units $\hslash =c=1$ and the Minkowski metric $g=\mathrm{diag}(1,-1,-1,-1)$. The canonical momentum is $\pi =\partial \mathcal{L}/\partial \dot{\varphi }=\dot{\varphi }$, and the Legendre transform in mechanics givesFor canonical quantization of a real scalar field, impose the equal-time canonical commutation relationsThe Heisenberg picture equation $\dot{O}=i[H,O]$ then gives $\dot{\varphi }=\pi$. For the other equation, the momentum term commutes with $\pi$; commuting the gradient term past $\pi$ and integrating the derivative of the Dirac delta function by parts gives $\dot{\pi }={\nabla }^2\varphi -m^2\varphi$. ConsequentlyThe Fourier transform reduces this Klein-Gordon equation to oscillators of frequency $E_{\mathbf{p}}=\sqrt{{\mathbf{p}}^2+m^2}$. Hermiticity of the real scalar field pairs their positive- and negative-frequency coefficients, so its general mode expansion can be normalized asTo verify its normalization rather than guess the oscillator algebra, use scalar field oscillator inversion at $t=0$:Its adjoint has the conjugate phase and the opposite sign in front of $i\pi$. The two mixed field-momentum commutators giveThe Fourier representation of the Dirac delta function sets the energies equal in the last step. Similarly, $[a(\mathbf{p}),a(\mathbf{q})]$ has a factor $E_{\mathbf{q}}-E_{\mathbf{p}}$ multiplying ${\delta }^3(\mathbf{p}+\mathbf{q})$ and vanishes; the two-creator commutator vanishes as well.

Substituting the expansion in $H$ and integrating the spatial phases, the two-annihilator and two-creator coefficients cancel by $E_{\mathbf{p}}^2={\mathbf{p}}^2+m^2$. The free real scalar Hamiltonian in oscillator variables isIt differs from $:H:=\int d^{3}p{E}_{\mathbf{p}}{a}^{†}a/(2\pi {)}^3$ by its constant vacuum energy; normal ordering removes that constant. Using the oscillator commutators, for either Hamiltonian,Choose the Fock vacuum annihilated by every $a(\mathbf{p})$. Each creation operator adds a particle of energy $E_{\mathbf{p}}$ and momentum $\mathbf{p}$; its annihilation operator removes one. Their mutual commutativity gives bosonic Fock space and symmetric many-particle states. Thus the scalar mass-shell frequency has a particle interpretation, rather than merely describing a classical oscillation.

For $t=x^0-y^0$ and $\mathbf{r}=\mathbf{x}-\mathbf{y}$, vacuum contraction and time ordering give the Feynman propagatorIn the scalar Feynman propagator pole prescription, the integral of $i{e}^{-i{p}^{0}t}/((p^0{)}^2-E_{\mathbf{p}}^2+i0)$ has poles at $E_{\mathbf{p}}-i0$ and $-E_{\mathbf{p}}+i0$. Close clockwise below for $t>0$ and counterclockwise above for $t<0$. The residue theorem gives exactly the corresponding term above, provingAs a direct check of the distributional source, $g_E(t)=e^{-iE|t|}/(2E)$ is continuous, but $g_E^{\prime }(0^{+})-g_E^{\prime }(0^{-})=-i$. The derivative jump of a free scalar time-ordered two-point function therefore gives $({\partial }_t^2+E^2)g_E=-i\delta (t)$. Spatial Fourier inversion yields

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.

With the mode normalization given, the oscillator canonical commutation relations are $[a_{\mathbf{p}},a_{\mathbf{q}}^{†}]=(2\pi {)}^3{\delta }^3(\mathbf{p}-\mathbf{q})$, with the other two commutators zero. Let $E_{\mathbf{p}}=\sqrt{{\mathbf{p}}^2+m^2}$ and $p^0=E_{\mathbf{p}}$. The convention compatible with the requested numerator iswithout an additional factor of $i$ outside the vacuum expectation value. Put $\tau =x^0-y^0$ and $\mathbf{r}=\mathbf{x}-\mathbf{y}$. The Fock vacuum is annihilated by $a_{\mathbf{p}}$, so only $a{a}^{†}$ contributes to the two Wightman functions. Consequently,In the second term we changed $\mathbf{p}\mapsto -\mathbf{p}$ to give the same spatial exponential.

Now perform the energy contour integralHere the residue step uses $E>0$; the massless zero-momentum point is interpreted through the smeared distribution limit, not as an isolated normalized oscillator. The Feynman i-epsilon prescription puts the positive-energy pole below the real axis and the negative-energy pole above it: $p^0=E-i0$ and $p^0=-E+i0$. For $\tau >0$, close in the lower half-plane, clockwise; the residue theorem gives $-i$ times the residue $i{e}^{-iE\tau }/(2E)$, hence $e^{-iE\tau }/(2E)$. For $\tau <0$, close in the upper half-plane, counterclockwise; the negative-energy residue is $-i{e}^{iE\tau }/(2E)$, again giving a positive $e^{iE\tau }/(2E)$. These are exactly the two time-ordered terms. Restoring the spatial integral proves the scalar Feynman propagator pole prescription:The prescription in this formula supplies the pole convention left unspecified in the printed display; an unprescribed ordinary real-axis integral would not be well-defined. It is a distribution limit after smearing, not an absolutely convergent four-dimensional integral. With this normalization the derivative jump of a free scalar time-ordered two-point function gives $(\square +m^2){\Delta }_F=-i{\delta }^4(x-y)$, an independent check of both the numerator and the sign.

The pole displacement is exaggerated in this original schematic. The contour orientation and selected pole reproduce the time ordering of the Feynman propagator.

The Heisenberg picture evolution of the annihilation operator follows from $[H,a_{\mathbf{p}}]=-E_{\mathbf{p}}{a}_{\mathbf{p}}$. With $p^0=E_{\mathbf{p}}$, this gives the mode expansion of a free fieldDefine ${\Delta }^{+}(z)={\int }_{\mathbf{p}}{e}^{-ip\cdot z}/(2E_{\mathbf{p}})$. The time-ordered product then gives the Feynman propagatorThe last expression follows by closing the $p^0$ contour integral below for $z^0>0$ and above for $z^0<0$; the scalar Feynman propagator pole prescription supplies the required positive- and negative-frequency terms.

For $z^0\ne 0$ every term obeys the homogeneous Klein-Gordon equation. At $z^0=0$, the equal-time commutator of the fields vanishes, so the propagator is continuous and there is no derivative of a Dirac delta function. Its first time derivative has the jumpThe second time derivative therefore supplies precisely the source:Thus the defined Feynman propagator is a Green's function with source normalization $-i$; $i{\Delta }_F$ has source normalization $+1$. Keeping this factor avoids confusing the vacuum expectation value convention with conventions which define the propagator after dividing by $i$.^[ref]

Write $S=S_0+S_I$, where $S_0$ contains the quadratic kinetic term and mass term, and $S_I=-\lambda \int d^{D}x\varphi (x{)}^3/3!$. ThenLet $Z_0=\int D\varphi e^{i{S}_0}$ and let $⟨\cdot {⟩}_0$ denote the unit-normalized free Gaussian vacuum expectation value. The unnormalized numerator ${\mathcal{N}}_N$ is consequentlyThe Minkowski path integral is understood with the usual scalar Feynman propagator pole prescription, as a formal perturbation series. A real cubic potential energy is unbounded below; no nonperturbative stable vacuum is being assumed here.

Set $\Delta (x-z)=⟨T\varphi (x)\varphi (z){⟩}_0$, $K_i(x)=\Delta (y_i-x)$, and $t=\Delta (0)$, with a common regularization understood for coincident points. The zero-order one-point term has one field and vanishes. At first order, Wick's theorem gives three choices for the contraction with the external field, leaving a tadpole diagram:The second-order term contains seven fields, so it vanishes by the vanishing odd correlators of a free Gaussian field. HenceThe coefficient is $3/3!=1/2$, or symmetry factor $S=2$ when each diagram is weighted by $1/S$. No spacetime integral has been evaluated. The one-point tadpole diagram is shown in the first panel of the following figure; its other panels organize solution.