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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.

Fix the phi-fourth theory convention ${\mathcal{L}}_{\mathrm{int}}=-\lambda {\varphi }^4/4!$, with mostly-minus Minkowski spacetime and $\hslash =c=1$. Define the scattering amplitude by stripping the overall momentum-conserving delta function from $S-1=i\mathcal{M}$. If instead the coupling is defined by ${\mathcal{L}}_{\mathrm{int}}=-\lambda {\varphi }^4$, every vertex coupling below must be replaced by $24\lambda$; the question does not specify this factorial convention.

The momentum-space Feynman rules are a Feynman propagator $i/(k^2-m^2+i0)$ for each internal scalar line, a Feynman vertex $-i\lambda$ with four scalar legs, and four-momentum conservation at each vertex. Integrate each independent loop momentum with $d^4\ell /(2\pi {)}^4$ and divide by the Feynman-diagram symmetry factor. For an amputated scattering amplitude there are no external propagators; external states have the usual relativistic normalization. Sum the connected diagrams and all inequivalent assignments of the labelled external momenta. The overall delta function is $(2\pi {)}^4{\delta }^4(\sum p_{\mathrm{in}}-\sum p_{\mathrm{out}})$.

These Feynman rules follow by expanding the Dyson series $T\exp (i\int d^{4}x{\mathcal{L}}_{\mathrm{int}})$ and applying the Wick theorem to the free-field correlation functions. A Wick contraction gives the free Feynman propagator; the $4!$ ways of contracting a vertex cancel its factorial denominator. The expansion's $1/V!$ cancels permutations of identical interaction insertions, while any remaining automorphisms give the Feynman-diagram symmetry factor. Fourier transformation produces momentum conservation, and the LSZ reduction formula amputates external propagators to obtain the scattering amplitude.

For six-point amplitudes in phi-fourth theory, an original pair of diagrams is:

In the tree-level Feynman diagram, let the internal four-momentum be $r=p_1+p_2-q_1=q_2+q_3+q_4$. The labelled diagram has symmetry factor one, so its contribution isThis is one channel. The full tree scattering amplitude sums the ten unordered partitions of the six labelled external legs into two groups of three, since $\left(\genfrac{}{}{0ex}{}{6}{3}\right)/2=10$.

For the one-loop triangle Feynman diagram, set $P=p_1+p_2$, $Q_{12}=q_1+q_2$, $Q_{34}=q_3+q_4$, so $P=Q_{12}+Q_{34}$. A consistent routing has denominatorsWith the external labels held fixed, no nontrivial graph automorphism remains, so again the symmetry factor is one. The contribution isThere are other labelled assignments and other one-loop topologies in the full scattering amplitude; the displayed expression belongs to the particular drawn diagram.

One can also express this scalar triangle Feynman integral using Feynman parameters. With $x+y+z=1$, completing the square in $x{D}_0+y{D}_1+z{D}_2$ givesThe identity $1/(D_0{D}_1{D}_2)=2{\int }_{x,y,z\ge 0}dxdydz\delta (1-x-y-z)/(x{D}_0+y{D}_1+z{D}_2{)}^3$ and the shifted momentum integral yieldThus ${\mathcal{M}}_{△}=-{\lambda }^3/(16{\pi }^2)$ times the displayed parameter integral. It is ultraviolet finite in four dimensions, although physical threshold singularities must retain the Feynman i-epsilon prescription.

For the fixed-target production threshold, the incoming four-momenta are $(E,\mathbf{p})$ and $(m,0)$, with $E=\sqrt{m^2+|\mathbf{p}{|}^2}$. Their invariant mass obeysIn the centre-of-momentum frame, four final particles have total energy at least $4m$, attained when all four are at rest in that frame. Therefore $s\ge 16m^2$, equivalently $E\ge 7m$, andThis is the kinematic threshold for $m>0$. Exactly at equality the final phase space has zero volume; a nonzero production rate requires a strict inequality.

Use the momentum-space Feynman rules with the scalar Feynman propagator $i/(p^2-m^2+i0)$. The four-leg vertex of a factorial-normalized scalar interaction has weight $i\lambda$ for the positive interaction sign printed here. There are $4!$ Wick contractions assigning four external legs to its four fields, cancelling the factorial in its coefficient. A negative interaction sign would give $-i\lambda$; its squared tree amplitude is the same.

At each vertex include $(2\pi {)}^4{\delta }^4(\sum p)$ with all incident momenta taken incoming. Assign an internal momentum to each line and integrate each independent loop with $\int d^4\ell /(2\pi {)}^4$. Divide a graph by its Feynman-diagram symmetry factor, sum the graphs at the chosen order, and omit disconnected vacuum graphs from normalized amplitudes. For an S-matrix element, amputate external propagators and put the external momenta on shell as in the LSZ reduction formula; the external one-particle residues are one at tree level.

Define the invariant amplitude by the relativistically normalized matrix elementwith $⟨p|p^{\prime }⟩=2E_{\mathbf{p}}(2\pi {)}^3{\delta }^3(\mathbf{p}-{\mathbf{p}}^{\prime })$. The lowest-order connected four-point graph is one contact vertex:There is no exchange graph at this order because there is no three-field interaction.

For the elastic scattering from a quartic scalar contact interaction, write $\sqrt{s}$ for the total centre-of-mass energy and $E_p=\sqrt{s}/2$ for the energy of each incoming particle. The incoming and outgoing spatial momentum magnitudes both equal $k=\sqrt{E_p^2-m^2}$, with $E_p>m$. The invariant incident flux isThe Lorentz-invariant phase-space measure for two outgoing particles isIn the centre-of-mass frame, the spatial delta function sets ${\mathbf{p}}_4=-{\mathbf{p}}_3$, while the energy delta function has radial derivative $2k/E_p$. Therefore the relativistic two-body phase space satisfiesThe two outgoing real-scalar particles are identical. Integrating over the full solid angle counts each unordered pair twice, so include the identical final-state symmetry factor $1/2!$. This givesThis is isotropic. When $E$ denotes each particle's energy, $s=4E^2$ and the full-sphere event density is ${\lambda }^2/(512{\pi }^2{E}^2)$. If $E$ denotes the total energy of the pair, $s=E^2$ and it is ${\lambda }^2/(128{\pi }^2{E}^2)$. Stating the answer in $s$ removes that energy-label ambiguity.

An equally valid angular convention selects one outgoing particle in a hemisphere, so each event is represented once. In that convention omit $1/2!$ and use $d\sigma /d\Omega ={\lambda }^2/(64{\pi }^{2}s)$ on the hemisphere, or ${\lambda }^2/(256{\pi }^2{E}_p^2)$. Both conventions give ${\sigma }_{\mathrm{event}}={\lambda }^2/(32\pi s)$ at this order. The identical-state factor concerns counting final states and is separate from the vertex factorial.

Use units $\hslash =c=1$, the Minkowski metric ${\eta }_{ab}=\mathrm{diag}(1,-1,-1,-1)$, and the Fourier transform convention $f(x)=\int d^{4}p{e}^{-ip\cdot x}\tilde{f}(p)/(2\pi {)}^4$. Here gamma matrices satisfy $\{{\gamma }^a,{\gamma }^b\}=2{\eta }^{ab}{I}_4$, and $\bar{\psi }={\psi }^{†}{\gamma }^0$ is the Dirac adjoint. A single Dirac field describes both Electrons and Positrons; these are the particle and antiparticle sectors of that field.

The free Maxwell Lagrangian and Dirac action, with a linear covariant gauge condition for the photon, giveBefore gauge fixing, the Maxwell Lagrangian has zero modes in field theory along $A_a\mapsto A_a+{\partial }_a\lambda$, so its quadratic operator cannot be inverted on all potentials. In the linear Lorenz gauge, the Faddeev-Popov determinant is $\det \square$. It is independent of $A$ and may be absorbed into the normalization; the corresponding Faddeev-Popov ghost fields have no interacting vertices in this Abelian linear gauge. A residual gauge symmetry is removed by the specified boundary conditions.

The free generating functional isThe Dirac field variables and their sources are independent Grassmann fields in this path integral. They anticommute; replacing them by ordinary commuting fields would give the wrong statistics and the wrong functional determinant. The bosonic Gaussian functional integral contributes an inverse square root of a determinant, and the Grassmann Gaussian integral contributes a determinant. If $K_A$ is the photon quadratic operator and $K_D=i{\gamma }^a{\partial }_a-m$, completing the square givesIn the last expression $D_F=i{K}_A^{-1}$ and $S_F=i{K}_D^{-1}$ are the photon propagator and Dirac propagator, with Feynman i-epsilon prescription. The products include the appropriate spacetime integrals and index contractions. Functional derivatives with respect to $J$, and consistently ordered left or right Grassmann derivatives with respect to the fermionic sources, generate the time ordering of the corresponding fields. In Feynman gauge, $\alpha =1$, the momentum-space two-point functions areThe Feynman i-epsilon prescription specifies vacuum boundary conditions rather than an arbitrary inverse of the differential operator.

The covariant photon propagator uses four potential components. Its operator formalism counterpart is Gupta-Bleuler quantization: impose $({\partial }_a{A}^a{)}^{(+)}|\mathrm{phys}⟩=0$ and take the Gupta-Bleuler null-state quotient. This leaves a positive physical state space with two transverse photon polarization vectors. The temporal and longitudinal oscillator components occur in intermediate covariant expressions; the Ward identity removes their dependence from physical amplitudes. The free Dirac field uses the canonical anticommutation relations, producing the same fermionic signs as its Grassmann fields in the path integral.

The operator-path-integral equivalence can be seen directly with a regulator. Divide time into small intervals and insert complete sets of field-coordinate states for bosons, and resolutions in fermionic coherent states for fermions. The bosonic matrix elements produce the phase-space factor $\exp i\sum [\pi \Delta \varphi -H\Delta t]$; integrating out the quadratic canonical momentum produces the bosonic action. The fermionic coherent state overlaps produce the first-order term $i{\psi }^{†}\dot{\psi }$ and the Berezin integral measure. Multiplying the short-time kernels recovers the path integral. Projecting the remote endpoints onto the Fock vacuum with an infinitesimal damping selects the same Feynman propagators as the operator formalism. Field insertions become time-ordered products under this construction. Conversely, their quadratic generating functional obeys the canonical free-field equations and has precisely the oscillator two-point functions, so its higher free correlators agree by the Wick theorem. This establishes the equivalence for the regulated free theory and order by order in the perturbation series.

To couple the matter field electromagnetically, promote its global phase symmetry to the local transformationThe gauge covariant derivative $D_a={\partial }_a+ie{A}_a$ obeys $D_a^{\prime }{\psi }^{\prime }=e^{-ie\lambda }{D}_a\psi$. Replacing ${\partial }_a$ by $D_a$ in the Dirac action therefore gives the invariant matter densityThe electromagnetic field tensor is unchanged by the local transformation. Thus the unfixed quantum electrodynamics action is gauge-invariant. The Dirac current $j^a=\bar{\psi }{\gamma }^a\psi$ is conserved by the matter equations, and the Electron and Positron excitations carry opposite charges. The added gauge fixing density selects a representative and is not itself invariant under arbitrary local transformations; it does not change gauge-invariant observables. At a free fermion vertex,For external on-shell Dirac spinors the corresponding current contraction vanishes. The quantum extension is the Ward identity, which makes physical amplitudes insensitive to adding a multiple of the photon momentum to its polarization vector. A gauge-compatible regularization preserves this vector-current identity.

With ${\mathcal{L}}_{\mathrm{int}}=-e\bar{\psi }{\gamma }^a\psi A_a$, expand $\exp (i\int {\mathcal{L}}_{\mathrm{int}})$ in powers of $e$. A term of order $n$ contains $n$ spacetime integrations and $1/n!$. Applying the Wick theorem pairs the free fields: an $A$-$A$ Wick contraction supplies a photon propagator, and a $\psi$-$\bar{\psi }$ Wick contraction supplies an oriented Dirac propagator. Each insertion supplies an interaction vertex. The permutations of contractions cancel the expansion factorials except for the Feynman-diagram symmetry factor. Interchanging Grassmann fields produces the fermionic sign, including a minus sign for every closed fermion loop. This is how Feynman diagrams arise from expectation values, rather than an extra dynamical assumption.

The normalized vacuum generating functional removes components with no external insertions. In the operator formalism, for an interacting-vacuum expectation value of an inserted product $\mathcal{O}$, the same cancellation appears asVacuum projection and the Feynman i-epsilon prescription are implicit. The linked-cluster theorem exponentiates all connected vacuum bubbles into the same factor in numerator and denominator, so it cancels. Normalize by $Z[0]$ to remove every vacuum component. This cancellation of vacuum bubbles still leaves products of disconnected diagrams that each contain external insertions. If only connected correlators are wanted, differentiate the connected generating functional $W=-i\log [Z[J,\bar{\eta },\eta ]/Z[0]]$.

The resulting momentum-space QED Feynman rules for the bare theory can be stated in Feynman gauge as follows:

The illustrated interaction vertex has incoming fermion momentum $p$, incoming photon momentum $k$, and outgoing fermion momentum $p^{\prime }=p+k$. Its solid-line arrows indicate fermion flow. The photon wavy line has no fermion-flow arrow. The propagators, the vertex $-ie{\gamma }^a$, momentum conservation, and the fermionic signs determine the perturbative amplitudes.

Let $G_5$ be the connected time-ordered five-point function. The LSZ reduction formula gives, up to one factor $Z^{-1/2}$ per external leg,Translation invariance factors this as

Define asymptotic states $|\varphi (k)⟩=a_k^{†}|0⟩$ and $|\psi (p,s)\bar{\psi }(q,s^{\prime })⟩=b_{p,s}^{†}{c}_{q,s^{\prime }}^{†}|0⟩$. The LSZ reduction formula reduces the decay matrix element from the connected three-point functionIn momentum space, amputate the scalar leg with $i(k^2-M^2)$ and the fermion legs with the inverse Dirac propagators, contract them with ${\bar{u}}^s(p)$ and $v^{s^{\prime }}(q)$, and take $k^2=M^2$, $p^2=q^2=m^2$. The spacetime integrations produce $(2\pi {)}^4{\delta }^{(4)}(k-p-q)$.

The S-matrix maps incoming asymptotic states to outgoing asymptotic states. Its connected transition matrix elements contain an overall energy-momentum Dirac delta function and an invariant scattering amplitude. In quantum field theory, the LSZ reduction formula obtains these elements from amputated field correlators.