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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 the connected tree scattering amplitude of one particle and one antiparticle of a complex scalar field, the Dyson series inserts $-i\lambda {\varphi }^{†2}{\varphi }^2/4$. There are $2!$ assignments of the particle and antiparticle external attachments to each pair of identical field factors, giving $2!2!=4$ and hence the displayed Feynman vertex. With relativistic normalization of a one-particle state, integrating the vertex position gives $i\mathcal{T}(2\pi {)}^4{\delta }^4(p_1+q_1-p_2-q_2)$ with $\mathcal{T}=-\lambda$. Vacuum diagrams and one-particle self-energy insertions are separate disconnected contributions; normal ordering eliminates the single-vertex vacuum and tadpole contractions.

For bounded norm-continuous Hamiltonian operators on $[t_0,t]$, the Dyson series has an $n$-th term bounded by $M^n|t-t_0{|}^n/n!$, so it converges in operator norm. In its nested-integral form, differentiating the outer endpoint puts $H_I(t)$ on the left of the $(n-1)$-fold term. Summing yields ${\partial }_{t}U=-i{H}_I(t)U$ and $U(t_0,t_0)=I$. If $H_I$ is Hermitian, differentiating $U^{†}U$ proves unitarity. For unbounded field operators the same algebra is a formal perturbative identity unless domain and convergence assumptions or a regulator justify the steps. Evolution backward uses inverse, anti-time-ordered evolution.

For a real scalar field with interaction $-\lambda {\varphi }^4/4!$, the first Dyson series term is $-i\lambda \int {\varphi }^4{d}^{4}x/4!$. Four labeled external legs can attach to the four fields in $4!$ ways, or in $\left(\genfrac{}{}{0ex}{}{4}{2}\right)2!2!=24$ ways after separating incoming and outgoing legs. The factorial cancels and spacetime integration supplies the four-momentum conservation delta function. This gives the connected tree scattering amplitude $-\lambda$. The separate identical final-state symmetry factor belongs in the event phase-space integral and is not another vertex factor.

For $H=H_0+H_{\mathrm{int}}$, the interaction picture moves the free evolution into operators and leaves states to evolve with $H_I(t)=e^{i{H}_{0}t}{H}_{\mathrm{int}}{e}^{-i{H}_{0}t}$. Its evolution operator is the time-ordered Dyson series.

The real scalar field has canonical momentum $\pi =\dot{\varphi }$. Its Legendre transform givesThe first three terms constitute $H_0$, the free massive Hamiltonian operator. The remaining term is $H_I=\lambda \int {\varphi }^4{d}^{3}x/4!=-\int {\mathcal{L}}_I{d}^{3}x$. This sign is a consequence of having no time derivatives in the interaction: the canonical momentum is unchanged, while the interaction enters the Hamiltonian density with the opposite sign to the Lagrangian density.

In the interaction picture, operators evolve with $H_0$ and states evolve with $H_I(t)$ constructed from the free interaction-picture field. Their evolution operator satisfiesIntegrating once gives $U_I=I-i{\int }_{t_0}^t{H}_I(t_1)U_I(t_1,t_0)d{t}_1$. Repeated substitution orders the integrations as $t\ge t_1\ge \cdots \ge t_n\ge t_0$. Replacing the ordered simplex by the full integration cube introduces $1/n!$ and a time-ordered product, which proves the Dyson seriesTaking the asymptotic limits with the usual adiabatic or regulated scattering prescription yieldsThe field in this expression is the free field evolving in the interaction picture, not an independently prescribed classical field. For quantum fields the expansion is a regulated perturbation series; the time-ordering symbol cannot be dropped because the operators at different times need not commute.

Use relativistically normalized two-particle states $|p_1,p_2⟩=a^{†}(p_1)a^{†}(p_2)|0⟩$, with $⟨p|q⟩=(2\pi {)}^{3}2E_p{\delta }^3(\mathbf{p}-\mathbf{q})$. The connected first-order scattering contribution isHere the connected S-matrix uses a normalized vacuum and physical external one-particle states. At this order, normal ordering the vertex separates the four-external-leg contribution from vacuum and tadpole pieces; those disconnected or external-mass contributions are not the two-to-two scattering amplitude.

Two fields annihilate the incoming particles and two create the outgoing ones. There are $\left(\genfrac{}{}{0ex}{}{4}{2}\right)$ choices of the annihilating fields, two assignments of the incoming momenta and two assignments of the outgoing momenta. Hence the total factor is $\left(\genfrac{}{}{0ex}{}{4}{2}\right)2!2!=4!$, the external-leg factorial cancellation at a quartic scalar vertex. Each attachment contributes its plane-wave factor, soThe spacetime integral is the four-momentum conservation delta function, givingIn particular, the tree scattering amplitude is independent of angle.

For completeness, derive the relativistic two-body phase space in the centre-of-mass frame. Write $\sqrt{s}=2E_{\ast }$, $p_{\ast }^2=E_{\ast }^2-m^2$. The spatial delta function sets the two final momenta opposite, and the energy delta function is $\delta (\sqrt{s}-2\sqrt{p^2+m^2})$. Its radial Jacobian is $E_{\ast }/(2p_{\ast })$. ThusThe invariant flux factor is $\mathcal{F}=4p_{\ast }\sqrt{s}$ for equal initial masses. Using $d\sigma =|T{|}^{2}d{\Phi }_2/\mathcal{F}$, the labeled angular density is ${\lambda }^2/(64{\pi }^{2}s)$. The outgoing quanta belong to one real scalar field, so they are identical. A full-sphere integration counts each event twice and requires the identical final-state symmetry factor $1/2!$. ConsequentlyEquivalently one may integrate the labeled density over a hemisphere representing each event once. The final-state factor is separate from the vertex factorial: applying it at the vertex instead would give the wrong elastic scattering from a quartic scalar contact interaction cross-section.

For the complex scalar field, the canonical momenta are $\pi ={\dot{\varphi }}^{†}$ and ${\pi }^{†}=\dot{\varphi }$. Split $H=H_0+H_I$. The interaction picture is obtained from the Schrödinger picture by $|{\Psi }_I(t)⟩=e^{i{H}_{0}t}|{\Psi }_S(t)⟩$ and $O_I(t)=e^{i{H}_{0}t}{O}_S{e}^{-i{H}_{0}t}$, choosing a common time origin. The fields therefore obey the free Klein-Gordon equation, while the states evolve through $H_I(t)$. The canonical quantization of a complex scalar field givesThe adjoint expansion contains $a^{†}$ and $b$. Particle and antiparticle modes are independent; their nonzero commutators areAll cross-species commutators, and all pairs of creation operators or of annihilation operators, vanish. These formulas reproduce $[\varphi (\mathbf{x}),\pi (\mathbf{y})]=[{\varphi }^{†}(\mathbf{x}),{\pi }^{†}(\mathbf{y})]=i{\delta }^{(3)}(\mathbf{x}-\mathbf{y})$ and the remaining vanishing equal-time commutators.

There are no derivative interactions, so the interaction Hamiltonian is $H_I(t)=-\int d^{3}x{\mathcal{L}}_I(x)$. Differentiating the interaction picture state gives $i{\partial }_t|{\Psi }_I⟩=H_I(t)|{\Psi }_I⟩$. Its time-evolution operator satisfiesIteration gives ordered integration regions $t_1>\cdots >t_n$. Each is $1/n!$ of the full region with a time-ordered product, yielding the Dyson seriesWith the usual asymptotic scattering prescription, taking $t_0\to -\infty$ and $t\to +\infty$ provesTime ordering is essential because different interaction Hamiltonians need not commute.

At first order, use the connected tree scattering amplitude, equivalently the normal-ordered quartic vertex with the vacuum and one-particle mass contributions removed. This qualification matters if the local product is interpreted literally without normal ordering: single-vertex tadpole diagrams and vacuum diagrams are not the requested connected two-particle amplitude. The complex scalar quartic contact vertex is obtained fromIn a fully connected matrix element, an incoming particle and outgoing antiparticle attach to the two $\varphi$ factors; an outgoing particle and incoming antiparticle attach to the two ${\varphi }^{†}$ factors. Each pair has two assignments, giving $2!2!=4$. The external relativistic normalization of a one-particle state cancels each mode measure: for example $[\varphi (x),a^{†}(p_1)]=e^{-i{p}_{1}x}$ and $[b(q_2),\varphi (x)]=e^{i{q}_{2}x}$. Consequently the connected integrand is $4e^{-i(p_1+q_1-p_2-q_2)\cdot x}$, andThus the invariant connected amplitude isThe Dirac delta function expresses four-momentum conservation.

For the identity contribution, commute $a(p_2)$ through $a^{†}(p_1)$ and independently $b(q_2)$ through $b^{†}(q_1)$; all terms leaving a creation operator against the left Fock vacuum or an annihilation operator against the right Fock vacuum vanish. HenceThere is no exchange term between the two distinct particle and antiparticle species.

All three fields have mass $m$. In the interaction picture the quantum fields evolve with the free Hamiltonian operator, so their equations areThe interactions evolve the states instead. For $\chi \in \{\varphi ,\psi ,\Phi \}$ the appropriate mode expansion of a free field isIndependent species have commuting creation and annihilation operators. The full set of canonical commutation relations isThese follow from the separate equal-time canonical commutation relations for the three real scalar fields. For comparison, in the Heisenberg picture the full Euler-Lagrange field equations contain interactions: $(\square +m^2){\varphi }_H=-g{\varphi }_H{\Phi }_H$, $(\square +m^2){\psi }_H=-g{\psi }_H{\Phi }_H$, and $(\square +m^2){\Phi }_H=-\frac{g}{2}({\varphi }_H^2+{\psi }_H^2)$. These are distinct from the free equations used in the interaction picture.

There are no derivative couplings, so the interaction Hamiltonian is $H_I(t)=-\int d^{3}x{\mathcal{L}}_I(x)$. The interaction picture evolution operator obeysIterating its integral equation gives the Dyson seriesThe ordered integration region is equivalent to the unrestricted region with time ordering and the factor $1/n!$. Taking the scattering limit $t_0\to -\infty$, $t\to \infty$, with the usual adiabatic prescription for perturbation theory, therefore yields the S-matrixAll fields inside this expression are interaction picture fields.

Use relativistic normalization of a one-particle state and take $|i⟩=a_{\varphi }^{†}(\mathbf{p})a_{\psi }^{†}(\mathbf{q})|0⟩$, $|f⟩=a_{\varphi }^{†}({\mathbf{p}}^{\prime })a_{\psi }^{†}({\mathbf{q}}^{\prime })|0⟩$. Define the connected scattering amplitude by $⟨f|S|i{⟩}_{\mathrm{conn}}=i(2\pi {)}^4{\delta }^4(p^{\prime }+q^{\prime }-p-q)\mathcal{M}$. The identity contribution describes no scattering; the term linear in $g$ cannot connect these four external particles. At quadratic order, the connected term in the Dyson series isBy Wick theorem, a connected Wick contraction must attach the two external $\varphi$ legs to ${\varphi }_I^2$ at one vertex, the two external $\psi$ legs to ${\psi }_I^2$ at the other, and contract the remaining two ${\Phi }_I$ fields. The two assignments of species to $x,y$ cancel the Dyson factor $1/2!$. At either vertex, the $2!$ assignments of its identical fields to the incoming and outgoing particle cancel the interaction factor $1/2$. Equivalently, the coefficient $-g^2/8$ is multiplied by $2\times 2\times 2$, giving $-g^2$. Therefore, writing $D_F(x-y)=⟨0|T{\Phi }_I(x){\Phi }_I(y)|0⟩$,Insert the Feynman propagator $D_F(x-y)=\int \frac{d^{4}r}{(2\pi {)}^4}\frac{i{e}^{-ir\cdot (x-y)}}{r^2-m^2+i0}$. The two vertex integrals impose $r=p^{\prime }-p$ and $p^{\prime }+q^{\prime }=p+q$. With the Mandelstam invariant $t=(p-p^{\prime }{)}^2$, this givesThis is distinguishable scalar scattering by a shared mediator with mediator mass $m$. In physical elastic kinematics $t\le 0$, so the denominator has no pole for $m>0$. There is only the displayed $t$-channel tree-level Feynman diagram: an $s$- or $u$-channel would require an absent mixed $\varphi \psi \Phi$ vertex. Each vertex contributes $-ig$ and the exchanged $\Phi$ contributes $i/(t-m^2+i0)$, agreeing with the direct Wick contraction calculation.

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.

An antilinear involution need not preserve norms, so the required physical assumption is that momentum reversal maps a complete normalized basis to a normalized basis. For a spinless sector take $\hat{T}|\mathbf{p}⟩=\eta (\mathbf{p})|-\mathbf{p}⟩$ with $|\eta |=1$ and relativistic normalization $⟨\mathbf{p}|\mathbf{q}⟩=(2\pi {)}^{3}2E_p{\delta }^3(\mathbf{p}-\mathbf{q})$. Any extra degeneracy labels must likewise be mapped by a norm-preserving permutation or unitary matrix. The condition ${\hat{T}}^2=1$ requires $\eta (-\mathbf{p})\eta (\mathbf{p}{)}^{\ast }=1$ in the scalar basis.

For arbitrary normalizable wave packets $|\varphi ⟩=\int d{\Pi }_{p}f(\mathbf{p})|\mathbf{p}⟩$ and $|\psi ⟩=\int d{\Pi }_{p}h(\mathbf{p})|\mathbf{p}⟩$, where $d{\Pi }_p=d^{3}p/[(2\pi {)}^{3}2E_p]$, antilinearity conjugates their coefficients. Using the normalized reversed basis givesCompleteness proves $⟨\hat{T}\varphi |\hat{T}\psi ⟩=⟨\varphi |\psi {⟩}^{\ast }$ for all states, so this quantum time-reversal operator is antiunitary. The same proof with sums over extra labels uses the unitarity of their reversal matrix.

For a Hermitian interaction-picture Hamiltonian, the condition of time-reversal symmetry isExpand the scattering matrix as a Dyson series, using ordered integration rather than commuting operators:Conjugation by $\hat{T}$ conjugates $(-i{)}^n$ to $i^n$ but preserves the product order. Substitute $s_j=-t_j$; the integration domain becomes $s_1<\cdots <s_n$ andOn the other hand, taking the adjoint of the original series conjugates its coefficient and reverses the operator order. Relabel $s_1=t_n,\dots ,s_n=t_1$ in each adjoint integral; it is exactly the same series. ThusEquivalently, time ordering turns into anti-time ordering after reversing the time arguments. One can first use a finite symmetric time interval and the same time-symmetric switching prescription before taking the scattering limit; the argument never assumes $[V(t),V(s)]=0$.

Antiunitarity relates amplitudes byUsing $\hat{T}S{\hat{T}}^{-1}=S^{†}$ and taking a complex conjugate gives the time-reversal reciprocity of the scattering matrixThe reverse process starts with the reversed final state and ends with the reversed initial state. Intrinsic state phases can appear when these reversed states are rewritten in a separately chosen basis, but the corresponding squared amplitudes agree. Equal amplitudes do not mean identical numerical cross sections when the reversed process has different flux or phase-space factors.

For a spin-zero complex scalar field, choose the transformation convention $\hat{T}\varphi (t,\mathbf{x}){\hat{T}}^{-1}={\eta }_T\varphi (-t,\mathbf{x})$. Its mode expansion of a free field isApply $\hat{T}$ to the expansion. Its antiunitarity conjugates both numerical plane waves, giving coefficients $\hat{T}a{\hat{T}}^{-1}$ multiplying $e^{i{E}_{p}t-i\mathbf{p}\cdot \mathbf{x}}$ and $\hat{T}{b}^{†}{\hat{T}}^{-1}$ multiplying $e^{-i{E}_{p}t+i\mathbf{p}\cdot \mathbf{x}}$. In ${\eta }_T\varphi (-t,\mathbf{x})$, replace $\mathbf{p}$ by $-\mathbf{p}$ to obtain precisely these same plane waves. Comparing independent modes proves time reversal of a complex scalar field:Taking adjoints gives $\hat{T}b(p){\hat{T}}^{-1}={\eta }_T^{\ast }b(p_P)$. Positive energy is preserved, and an annihilation operator remains an annihilation operator. Time reversal does not exchange particles with antiparticles.

The coefficient ${\eta }_T$ is an intrinsic time-reversal phase. Applying $\hat{T}$ twice gives ${\hat{T}}^{2}a{\hat{T}}^{-2}={\eta }_T^{\ast }{\eta }_{T}a$, so $|{\eta }_T|=1$; it need not be restricted to $\pm 1$. Write ${\eta }_T=e^{i\vartheta }$ and redefine $a^{\prime }=e^{i\vartheta /2}a$. Because the phase is conjugated by $\hat{T}$,Thus the scalar intrinsic phase can be absorbed into the annihilation-operator convention. Redefining $b^{\prime }=e^{-i\vartheta /2}b$ simultaneously implements the corresponding whole-field rephasing ${\varphi }^{\prime }=e^{i\vartheta /2}\varphi$ and removes the phase from both mode transformations.

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.

The interaction contains no time derivatives, so $\pi =\dot{\varphi }$. The Legendre transform in mechanics givesThusHere $H_0=\int {\mathcal{H}}_0$ is the free Hamiltonian operator of mass $m$, and $H_I=\int {\mathcal{H}}_I$ is the interaction Hamiltonian.

In the interaction picture, operators evolve with $H_0$ and the states with $H_I(t)=e^{i{H}_{0}t}{H}_I{e}^{-i{H}_{0}t}$. With $U_I(t_0,t_0)=I$, the state evolution satisfies $i{\partial }_t{U}_I(t,t_0)=H_I(t)U_I(t,t_0)$. Integrating once and iterating gives the Dyson seriesEquivalently it is the time ordering of the exponential. Taking the infinite-time scattering limit, with adiabatic switching understood, gives the scattering matrixThe field in the interaction is the free interaction-picture field, which is essential for applying the free-field contractions.

Use relativistic normalization of a one-particle state, $|\mathbf{p}⟩=\sqrt{2E_{\mathbf{p}}}{a}^{†}(\mathbf{p})|0⟩$, so each external scalar contraction contributes its plane-wave phase without an extra $1/\sqrt{2E}$ factor. For the connected on-shell two-to-two scattering amplitude, the first Dyson series term is $-i\lambda \int d^{4}x{\varphi }_I(x{)}^4/4!$. The Wick theorem gives $4!$ ways to attach the two incoming and two outgoing external particles to the four fields, canceling the denominator. ConsequentlywhereThe delta function expresses four-momentum conservation. The connected scattering convention removes spectator, vacuum and external self-energy contributions; equivalently at this order one can normal-order the interaction and keep the physical external mass fixed. These contributions should not be mistaken for additional connected four-point diagrams.