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If a time-dependent perturbation remains diagonal in a fixed eigenbasis, Hamiltonians at different times commute. Each component therefore acquires the exact phase $\exp [-i\int E_{\pm }(t)dt/\hslash ]$, with no time ordering required.

For $H(t)={\Omega }_1(t)A+{\Omega }_2(t)B$ with fixed noncommuting Hermitian operators, $[H(t),H(s)]=[{\Omega }_1(t){\Omega }_2(s)-{\Omega }_2(t){\Omega }_1(s)][A,B]$. A constant ratio of the envelopes is sufficient to remove time ordering. Independently changing that ratio generally produces commutator corrections in the Magnus expansion, and a product of successive pulses need not equal the exponential of their summed areas.

The Dyson series solves interaction picture evolution by keeping noncommuting Hamiltonian operators in chronological order. With $\hslash =1$, integrating $i{\partial }_t{U}_I=H_I(t)U_I$ and iterating gives

Time ordering puts later times on the left. Integrating over all time labels instead of the ordered region introduces $1/n!$. The formal scattering operator limit isAn ordinary matrix exponential suffices when all the $H_I(t)$ commute. For quantum fields, the displayed expansion requires a prescription for singular products and need not converge. ^[ref]

For two odd fermionic operators, time ordering contributes a minus sign whenever the chronological rearrangement exchanges them. For several fields the sign is the sign of a permutation restricted to the odd fields. This convention makes the free Dirac propagator a Green function of the Dirac equation with its contact term fixed by the canonical anticommutation relations.

The Feynman propagator is a vacuum time-ordered two-point function, with the conventional overall factors stated. For a free real scalar field, $⟨0|T\varphi (x)\varphi (y)|0⟩$ has momentum-space form $i/(p^2-m^2+i0)$ in the usual Minkowski convention.

Time ordering places the later field to the left, and the pole prescription selects the vacuum boundary condition. The propagator is the Green function used for internal lines in perturbative amplitudes; it is not a directly measured probability for a particle to follow a particular path. It can be nonzero at spacelike separation without enabling superluminal signalling. Causal response is instead organized by the retarded Green function and local commutators, so these different two-point functions must not be interchanged.

An impulsive quantum kick is the limit of a pulse with finite integrated Hamiltonian operator, rather than a bounded finite step. For a bounded self-adjoint operator $K$ with units of action, a rectangular pulse $H_{\tau }=K/\tau$ lasting $\tau$ has exact unitary time evolution $e^{-iK/\hslash }$ for every $\tau$. In the limiting notation $H(t)=K\delta (t)$, $\delta$ is the Dirac delta function andThe integral estimate for quantum state continuity at a finite Hamiltonian step no longer tends to zero: the pulse height diverges. A bounded background Hamiltonian operator contributes vanishing evolution during this limit. A more general pulse with noncommuting generators can require time ordering; integrated area alone then need not specify the result.

For example, $K=\hslash \varphi {\sigma }_y/2$, with a $y$ Pauli matrix, sends a qubit in $|0⟩$ to $\cos (\varphi /2)|0⟩+\sin (\varphi /2)|1⟩$. At $\varphi =\pi$, the Born rule predicts $|1⟩$ with probability one, however short the regularized pulse is. Its finite rotation is precisely what a bounded finite step cannot accomplish at the switching instant.

For a Hermitian scalar field in a positive-norm Hilbert space with a Lorentz-invariant vacuum, the spectrum condition and completeness give the connected vacuum two-point function as a superposition of free mass-shell functions. Define the Källén–Lehmann spectral density by $\tilde{W}(p)=2\pi \theta (p^0)\rho (p^2)$, where $W(x)=⟨0|\varphi (x)\varphi (0)|0⟩$ and its vacuum expectation value has been subtracted if nonzero. Squared intermediate-state matrix elements make $\rho$ a nonnegative measure supported on nonnegative mass squared. Applying time ordering to each free mass shell gives the displayed Feynman propagator representation, using $i\Delta =⟨T\varphi \varphi ⟩$ and signature $(+---)$. Its positivity does not automatically extend to gauge-variant fields in an indefinite auxiliary space.

An operator equation in a path integral $\mathcal{O}(x)=0$ asserts that insertions of $\mathcal{O}(x)$ vanish in every time-ordered correlation function with the other insertion points separated from $x$. Derivatives of time ordering can produce Ward identity contact terms at coincidence. For example, conservation of a Dirac current gives a vanishing separated-point divergence, while a charged insertion contributes a delta-function contact term fixed by its transformation. A vanishing one-point expectation value alone is insufficient.

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.

Use units $\hslash =c=1$ and the Minkowski metric $g_{\mu \nu }=\mathrm{diag}(1,-1,-1,-1)$, so $p\cdot x=E_{p}t-\mathbf{p}\cdot \mathbf{x}$. The canonical momentum of the real scalar field is $\pi =\dot{\varphi }$, and the Hamiltonian operator isCanonical quantization of a real scalar field promotes $\varphi$ and $\pi$ to operator-valued distributions satisfying the equal-time canonical commutation relationProducts and operator commutators can first be interpreted after spatial smearing. An additive vacuum constant in $H$ does not affect the following equations. In the Heisenberg picture, $\dot{O}=i[H,O]$. Applying the canonical commutation relation to the quadratic Hamiltonian operator, and integrating the gradient term by parts, givesThus the Klein-Gordon equation is an operator equation. The spatial Fourier transform reduces it to ${\ddot{\varphi }}_{\mathbf{p}}+E_p^2{\varphi }_{\mathbf{p}}=0$, with $E_p=\sqrt{{\mathbf{p}}^2+m^2}$. Its two frequency branches, together with the Hermitian adjoint condition ${\varphi }^{†}=\varphi$, give the mode expansion of a free fieldThe measure $d\mu (p)$ is Lorentz invariant; its normalization fixes that of the creation and annihilation operators rather than imposing any extra physical assumption.

For invariant-normalized scalar mode extraction, take the Klein-Gordon inner product of $f_p$ with the field. The positive-frequency contribution has coefficient $E_p+E_q$ and spatial integral $(2\pi {)}^3{\delta }^3(\mathbf{p}-\mathbf{q})$. The negative-frequency contribution has coefficient $E_p-E_q$ and spatial integral $(2\pi {)}^3{\delta }^3(\mathbf{p}+\mathbf{q})$, so it vanishes because $E_{-p}=E_p$. ConsequentlyTaking the Hermitian adjoint, with the order of the scalar factors immaterial, yieldsEquivalently $a(\mathbf{p})=\int d^{3}x{f}_p^{\ast }(E_p\varphi +i\pi )$ and $a^{†}(\mathbf{q})=\int d^{3}y{f}_q(E_q\varphi -i\pi )$. At a common time their operator commutator therefore isThe same calculation gives $[a(\mathbf{p}),a(\mathbf{q})]=[a^{†}(\mathbf{p}),a^{†}(\mathbf{q})]=0$: the surviving factor is an energy difference supported at opposite spatial momenta. These are the bosonic canonical commutation relations in relativistic normalization of a one-particle state.

Choose the Fock vacuum by $a(\mathbf{p})|0⟩=0$ and $⟨0|0⟩=1$. Only the annihilator-creator product contributes to the Wightman function, givingUsing time ordering and the reverse product for $y^0>x^0$, the Feynman propagator, with the convention $D_F=i{\Delta }_F$ in this calculation, isIn particular the denominator in the spatial measure is $2E_p$ in both terms.

To verify the four-dimensional Fourier transform representation, let $\tau =x^0-y^0$ and integrate the energy variable first:The Feynman i-epsilon prescription puts the positive-energy pole at $E_p-i0$ and the negative-energy pole at $-E_p+i0$. For $\tau >0$, the contour integral closes in the lower half-plane with clockwise orientation. The residue theorem multiplies the residue $i{e}^{-i{E}_p\tau }/(2E_p)$ by $-i$, giving $e^{-i{E}_p\tau }/(2E_p)$. For $\tau <0$, it closes counterclockwise above: multiplying the residue $-i{e}^{i{E}_p\tau }/(2E_p)$ by $i$ gives $e^{i{E}_p\tau }/(2E_p)$. Thus $I(\tau )=e^{-i{E}_p|\tau |}/(2E_p)$. After inserting the spatial exponential, reversing $\mathbf{p}$ in the negative-time term gives exactly the time ordering expression above. HenceThis equality is understood as a distribution limit; the same prescription gives $(\square +m^2)D_F(x-y)=-i{\delta }^4(x-y)$.

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.

Use a real scalar field and Minkowski action $S[\varphi ]$, with the vacuum selected by a Feynman propagator pole prescription or adiabatic vacuum boundary conditions. The normalized correlation functions areHere $|\Omega ⟩$ is the interacting vacuum and $\mathcal{T}$ denotes time ordering. These are the canonical vacuum expectation values, rather than unordered products of noncommuting fields. The functional integral formula assumes a regulated measure and the same vacuum prescription on both sides.

Couple an external source $J$ linearly to the scalar field and choose the normalized vacuum generating functionalThen functional derivatives insert the fields:In the convention used below, the connected generating functional isThe logarithm removes products of disconnected components. For example the connected correlation function $G_2^c=G_2-G_1{G}_1$; at three points one subtracts all three $G_2{G}_1$ products and adds $2G_1{G}_1{G}_1$. Normalization has already removed vacuum bubbles.

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.

Use signature $(+---)$ and covariant one-particle state normalization $⟨p^{\prime }|p⟩=(2\pi {)}^{3}2E_p{\delta }^3(\mathbf{p}-{\mathbf{p}}^{\prime })$. The future scattering condition uses the outgoing asymptotic scalar field:As with the incoming condition, this is understood in suitably smeared scattering matrix elements, not as unrestricted pointwise strong convergence of interacting and free local operators.

The free incoming field has expansionThe creation operator defines $|p⟩=a_{\mathrm{in}}^{†}(p)|0⟩$, with $a_{\mathrm{in}}|0⟩=0$, so $⟨p|{\varphi }_{\mathrm{in}}(x)|0⟩=e^{ipx}$. On the other hand, translation invariance givesMatching the stable one-particle state component in the asymptotic condition fixes the constant overlap. Choose its state phase so that it is real and nonnegative. ThenThe same stable particle appears in the outgoing description, with compatible one-particle phases. Thus $Z$ is the squared overlap measured by wave-function renormalization.

Insert a complete set of four-momentum eigenstates between the fields. For $x^0>0$, translation invariance and Hermiticity giveFor $x^0<0$, reversing the time-ordered product gives the same sum with $e^{i{p}_{\alpha }x}$. Therefore, with continuous-state integrals implicit in the completeness sum,Here $\theta$ is the Heaviside step function.

For the usual spectral representation, take a Lorentz-invariant vacuum, a positive-norm Hilbert space and the future-cone spectrum condition. Also take $⟨0|\varphi |0⟩=0$, as for the unbroken real scalar field, or subtract this constant and use the connected correlation function. Otherwise the vacuum intermediate state adds a disconnected constant not represented by the ordinary mass-shell measure below.

Let $W(x)=⟨0|\varphi (x)\varphi (0)|0⟩$ for that centered field, and use the Fourier transform convention $W(x)=\int d^{4}p(2\pi {)}^{-4}{e}^{-ipx}\tilde{W}(p)$. Define the Källén–Lehmann spectral density by the distributional equalityLorentz invariance makes the weight on each future mass shell a function or measure of $p^2$; the spectrum condition gives $p^2\ge 0$. Its nonnegativity follows from the displayed squared matrix elements. ThusApplying time ordering to this expression mass shell by mass shell proves the Källén–Lehmann spectral representation:The factor $i$ is outside ${\Delta }_F$ under the convention in the question.

The stable one-particle state part of completeness is $\int d^3\mathbf{p}[(2\pi {)}^{3}2E_p{]}^{-1}|p⟩⟨p|$. Its overlap is $\sqrt{Z}$, so its contribution to $W$ is $Z$ times the free mass-$m$ two-point function. HenceOther isolated particles as well as multiparticle states can belong to ${\rho }_{\mathrm{rest}}$. The full scalar propagator accordingly has isolated pole $Z/(p^2-m^2+i0)$, making $Z$ its pole residue.

Finally use canonical field normalization: the canonical momentum is $\dot{\varphi }$ and the equal-time canonical commutation relation is $[\dot{\varphi }(t,\mathbf{x}),\varphi (t,\mathbf{y})]=-i{\delta }^3(\mathbf{x}-\mathbf{y})$. The vacuum commutator $C(x)=W(x)-W(-x)$ has the same spectral decomposition. For each free mass shell,independently of $\sigma$. Comparing with the interacting equal-time commutator gives the canonical scalar spectral sum rule:For the particle genuinely interpolated by the stated asymptotic field, $Z>0$; the general nonnegative bound allows a field with zero overlap. The total weight and the upper bound require this canonical normalization and a well-defined distributional equal-time limit. They do not follow from the asymptotic condition alone: rescaling $\varphi$ rescales both $Z$ and the entire spectral measure.

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 $[V(t),V(s)]=0$.