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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]

On a connected transition chain, independently selected resonant pulses realize embedded two-level rotations with tunable angle and phase. Eliminate target amplitudes from one end of the chain using complex Givens rotations, and invert those rotations to prepare the target from a basis state. The pulse angles come from amplitude ratios and the phases from relative complex arguments. Spectator components are unchanged by each embedded rotation. Work in the interaction picture and compensate the known drift phases at the chosen final time when the target is specified in the laboratory frame.

For a localized quantum harmonic oscillator along $z$, let the transverse magnetic vector potential be $A_z(t)=2A_0\cos (\theta -\omega t)$ for $t>0$, with zero potential before the switch. The Coulomb gauge form of minimal coupling gives $H_1=e{A}_z{p}_z/m$. The creation operator matrix element $⟨1|p_z|0⟩=i\sqrt{m\hslash {\omega }_0/2}$ means that only $|1⟩$ is reached at first order from the ground state. Its interaction picture amplitude isIn particular $F_0=t$, so resonant finite-time expressions are nonsingular. The induced electric dipole moment is $-2e\sqrt{\hslash /(2m{\omega }_0)}\mathrm{Re}(e^{-i{\omega }_{0}t}{b}_1)$. It contains a driven component with the dynamic electric polarizability and free-frequency transients required by the initial conditions. The discontinuity in $A_z$ gives an impulsive electric field and an initial mechanical velocity $e{A}_z(0^{+})/m$ even though the initial canonical momentum has zero mean.

The interaction picture divides quantum dynamics between an exactly handled reference Hamiltonian and an interaction. Write $H(t)=H_0+V(t)$ with time-independent $H_0$, and definewhere $\hslash$ is the reduced Planck constant and the subscript $S$ denotes the Schrödinger picture. A linear operator transforms as $A_I(t)=U_0^{†}{A}_S(t)U_0$. Differentiating the transformed quantum state and using the Schrödinger equation cancels the $H_0$ terms and givesThe resulting unitary time evolution isThe Dyson series expands this time-ordered expression in powers of the interaction. ^[ref]

This is useful when $H_0$ has a known eigenbasis of energy eigenstates and the interaction is weak enough for perturbation theory. The transformation itself is exact and preserves expectation values: $⟨{\psi }_S|A_S|{\psi }_S⟩=⟨{\psi }_I|A_I|{\psi }_I⟩$. For time-dependent $H_0(t)$, use its exact unitary time evolution in place of the simple matrix exponential.

For $H=-\gamma B_z{S}_z-\gamma B_x{S}_x\cos (\omega t)$, choose ${\omega }_z=\gamma B_z>0$ and nominal resonance $\omega ={\omega }_z$. Starting from up spin, the rotating-wave approximation gives the normalized Bloch vector in the title, with signed $\Omega =\gamma B_x/2$. It follows by transforming the spinor $(\cos (\Omega t/2),i\sin (\Omega t/2){)}^T$ from the interaction picture back to the Schrödinger picture. Thus weak transverse driving combines slow meridional Rabi oscillations with fast Larmor precession.

This result requires $|B_x|\ll |B_z|$ and cannot be extrapolated as a controlled strong-drive formula. The exact equation is $\dot{\mathbf{s}}=\gamma \mathbf{s}\times (B_x\cos \omega t,0,B_z)$. For $|\gamma B_z|t\ll 1$, neglecting the bias gives the controlled leading solutionThe fixed-axis driving generators commute at all times, making this solution exact when the bias is zero at any prescribed drive frequency. At nominal resonance in a strong-drive experiment, the initial rotation scale is $\Theta \simeq \gamma B_{x}t$, twice the incorrectly extrapolated rotating-wave rate. Bias effects must be retained on longer times.

Expand the time-dependent quantum state in unperturbed energy eigenstates, factoring out free dynamical phases:The Schrödinger equation gives the exact interaction picture coefficient equationsInitially $c_n(-\infty )={\delta }_{ni}$. To first order in time-dependent perturbation theory, replace the coefficients on the right by these zeroth-order values and integrate. ThusFor a transition $j\ne i$, this is the desired leading probability amplitude, and $P_{ji}=|{\gamma }_{ji}^{(1)}{|}^2$ to leading order. The formula is understood with free asymptotic phases removed. For $j=i$, the full survival amplitude also contains the zeroth-order $1$; the displayed integral alone is its first-order correction, not the whole amplitude. This specifies the usual scattering operator meaning of the paper's notation at infinite times.

For the ion pulse, write $x_{ji}=⟨j|x|i⟩$ and $y_{ji}=⟨j|y|i⟩$. ThenConsider an excitation with ${\omega }_{ji}>0$ and put $\beta ={\omega }_{ji}b/v$, $u={\omega }_{ji}t$. The first real pulse is even in time and the second is odd. Their respective surviving Fourier components are therefore cosine and imaginary sine:Since $\beta \log (1/\beta )\to 0$, the longitudinal term is smaller for comparable position matrix elements. Hence the electric-dipole excitation by a passing ion hasThis leading limit extends to either sign of ${\omega }_{ji}$ using $|{\omega }_{ji}|b/v\ll 1$. If $x_{ji}=0$ by a selection rule, the retained leading term vanishes and the smaller longitudinal term may give the first nonzero probability. The conditions also include a small total transition probability, approximately straight constant-velocity motion, and $b$ large compared with atomic size for the electric-dipole approximation. The paper uses a Coulomb-coupling convention in which the factor is written $Z{e}^2$; in SI units replace $e^2$ throughout by $e^2/(4\pi {ϵ}_0)$.

The collision duration is of order $b/v$. Excitations with angular frequency ${\omega }_{ji}$ receive an unsuppressed impulse when this is short relative to $1/{\omega }_{ji}$. In contrast, for ${\omega }_{ji}b/v\gg 1$, cancellations in the Fourier transform suppress the transition: a slowly varying distant field approximately preserves the atomic energy eigenstate. The exact dipole-pulse transforms make this explicit:where $K_{\nu }$ is a Modified Bessel function of the second kind. Their large-argument exponential decrease explains the slow-passage cutoff. These transforms are optional refinements of the small-$\beta$ derivation above, not a separate assumption about the target.

To connect the microscopic transition probability to energy loss, suppose the atom begins in a low-energy state. Exciting bound electronic states and causing ionization transfer positive energies $E_j-E_i$ from the ion to the material. For target atomic number density $n_{\mathrm{at}}$ and path length $d$, a schematic independent-encounter sum iswith a continuum integral as well as a sum for ionized final states. The short-collision formula gives $P_{ji}\propto Z^2/(b^2{v}^2)$, so the impact-parameter integral has logarithmic dependence on its cutoffs. Its upper effective cutoff is of order $v/{\omega }_{ji}$, where the transition ceases to be sudden; the lower cutoff must exclude regions where the dipole or weak-collision approximation fails. Thus the leading high-speed charge and isotope scaling of electronic stopping isAlthough a faster ion can access more energetic transitions and a wider range of distant encounters, its shorter interaction time reduces each ordinary excitation probability. In the fast-ion regime shown, the $1/v^2$ decrease dominates the slower variation of the excitation factor. The energy loss consequently falls with increasing incident kinetic energy. The original figure has $\delta E$ increasing downward, so the curves' rise toward the top of the page means decreasing loss, not increasing loss. A low-speed extrapolation of the inverse-square law is invalid because excitation thresholds, slow-passage suppression, changing ion charge and nuclear recoil can then matter.

For nonrelativistic ion mass $M$, $E=M{v}^2/2$. At fixed incident energy the leading prediction becomesThe greater atomic number of nitrogen than carbon than boron therefore gives larger loss; at the same velocity the leading species factor is $Z^2$. Within an isotopic family, the heavier ion moves more slowly at the same $E$ and loses more energy. Equivalently, a heavier isotope needs proportionally greater kinetic energy to achieve the same velocity and approximately the same electronic loss. Comparing at fixed energy per nucleon, rather than fixed total $E$, approximately removes this leading isotope mass dependence. This accounts for the separate isotope bands in the original graph without assigning different electronic excitation spectra to isotopes of the same projectile element.

The charge in this scaling is the effective ion charge if screening by bound projectile electrons is important. The interpretation assumes the same traversed material and thickness and uses the paper's small-fractional-loss regime, $\delta E/E\ll 1$; points with loss comparable to incident energy cannot be described quantitatively by treating $v$ as constant. The plotted curves support the qualitative ordering and high-energy decrease, rather than determining a universal normalization from the supplied single-atom model.

Start with the Schrödinger equation $i\hslash {\partial }_t|\psi ⟩=(H_0+V)|\psi ⟩$ and differentiate the definition of the interaction picture state:The free Hamiltonian operator terms cancel because $H_0$ commutes with its own matrix exponential. ThereforeThis interaction picture transformation is exact.

Write ${\omega }_z=\gamma B_z$ and use the ordered spin basis $(|\uparrow ⟩,|\downarrow ⟩)$. The free Hamiltonian operator is $H_0=-\hslash {\omega }_z{\sigma }_z/2$, soResolving the cosine into its two complex exponentials gives the exact upper off-diagonal factorwith the complex conjugate in the lower entry. Choose the field orientation with ${\omega }_z>0$, as implicit in the printed near-resonance condition, and put $\delta =\omega -{\omega }_z$. The rotating-wave approximation discards the terms at $\omega +{\omega }_z$, retaining the slow terms at $\delta$. The resulting interaction picture equations areThe factor $1/4$ contains $1/2$ from $S_x=\hslash {\sigma }_x/2$ and $1/2$ from the cosine decomposition. This is a Driven two-level Rabi model, with signed on-resonance coupling $\Omega =\gamma B_x/2$ and generalized Rabi angular frequency ${\omega }_R=\sqrt{{\delta }^2+{\Omega }^2}$. Here the Rabi detuning uses the paper's convention, drive minus free frequency.

Small detuning alone is insufficient to derive the printed equations. One also requires $|\gamma B_x|,|\delta |\ll \omega +{\omega }_z$, and an evolution time over which neglected terms stay small. At resonance this is the weak transverse-field regime $|B_x|\ll |B_z|$. The supplied solution is thus an approximate solution for the linearly oscillating field, not an exact one at arbitrary drive strength. If $\gamma B_z<0$, the other exponential is resonant and the physical positive resonance frequency is $|\gamma B_z|$.

At resonance, retain the signed coupling $\Omega =\gamma B_x/2$. In the interaction picture, the given solution reduces to $c_1=\cos (\Omega t/2)$ and $c_2=i\sin (\Omega t/2)$. Returning to the Schrödinger picture givesFor a normalized spinor $(a,b{)}^T$, the Pauli matrices give $⟨\mathbf{\sigma }⟩=(2\mathrm{Re}{a}^{\ast }b,2\mathrm{Im}{a}^{\ast }b,|a{|}^2-|b{|}^2)$. Therefore the spin expectation value within the rotating-wave approximation isIts magnitude is $\hslash /2$: the Bloch vector stays on the unit sphere. In a frame rotating about $z$ at angular velocity $-{\omega }_z$, it makes a uniform rotation in the $yz$ plane. For $|B_x|\ll |B_z|$, this gives slow Rabi oscillations between the poles, accompanied by rapid laboratory Larmor precession about $z$.

If the same formula is formally used for $|B_z|\ll |B_x|$, it predicts fast meridional rotations whose plane turns slowly about $z$. However, the original PDF really does ask for this opposite limit, and it lies outside the assumption used in part (b). That formal description is not a controlled prediction for the stated linearly driven Hamiltonian operator.

A valid strong-drive description starts from the exact equation for the Bloch vector $\mathbf{s}=2⟨\mathbf{S}⟩/\hslash$:For times $|\gamma B_z|t\ll 1$, neglecting the weak bias gives a controlled leading approximation, since the time integral of the omitted Hamiltonian operator has small norm. The remaining field has a fixed $x$ axis, so its generators commute at different times andAt the nominal resonance $\omega =\gamma B_z$ and on the early spin-rotation scale $t\sim 1/|\gamma B_x|$, this becomes $\Theta \simeq \gamma B_{x}t$. The initial rotation rate is consequently $\gamma B_x$, twice the rate predicted by extrapolating the rotating-wave approximation. For longer times the exact differential equation must be retained. The following original calculation illustrates both regimes rather than asserting an exact strong-drive Rabi formula.

For the requested oscillator response, treat $y$ as the fixed transverse position of a one-dimensional quantum harmonic oscillator along $z$, and put $\theta =ky$. The supplied radiation is transverse and already satisfies the Coulomb gauge. Its perturbation isThe annihilation operator and creation operator giveThus the only first-order transition out of the ground state is to $|1⟩$. More generally, the oscillator electric-dipole selection rule is $\Delta n=\pm 1$. This is a rule for matrix elements; exact energy conservation is replaced by a finite-time resonance profile during a switched finite-duration interaction.

To display both the resonant and nonresonant terms without spurious singularities, defineThe interaction picture coefficient in the given expansion is thenThe full Schrödinger picture coefficient of $|1⟩$ is $e^{-3i{\omega }_{0}t/2}{b}_1(t)$, while that of $|0⟩$ is $e^{-i{\omega }_{0}t/2}$ at this order. The normalization changes only at order $A_0^2$. In particular, at $\omega ={\omega }_0$ the first term is $C{e}^{i\theta }t$, finite for every finite time, and the perturbative result requires $|b_1|\ll 1$.

The fixed-$y$ interpretation is needed for the displayed target formulae. If $p^2$ were instead a three-dimensional kinetic operator with no transverse confinement, that model would have no normalizable transverse ground state, and $e^{\pm iky}$ would transfer transverse momentum $\pm \hslash k$. A treatment of that different model needs specified transverse states and recoil energies; they are absent from the question. The following parts use the intended localized-oscillator response.

For this spin-boson model, take $H_0=H_1+H_2$. Since $[H_0,{\sigma }_z]=0$, the interaction picture observable isThe normalized vectors $|{\varphi }_{\uparrow }⟩$ and $|{\varphi }_{\downarrow }⟩$ belong to different eigenspaces of the Hermitian operator ${\sigma }_z$, and are therefore orthogonal. This remains true even though each vector includes the oscillator degrees of freedom. Hence all cross terms in the expectation value vanish, while the diagonal terms have eigenvalues $+1$ and $-1$:Normalization also gives $|c_{\uparrow \sigma }{|}^2+|c_{\downarrow \sigma }{|}^2=1$, so these are the probabilities of the two spin outcomes.

The orthogonal projection onto the full spin-$\eta$ subspace isHere the identity acts on all oscillator states. The finite-time scattering operator $S(t)$ in the question is the interaction picture time-evolution operator, so $|{\psi }_{\sigma }^I(t)⟩=S(t)|\sigma ,\{n\}⟩$. Projecting its two orthogonal spin components yieldsExpand the right-hand side in the orthonormal basis $|\eta ,\{n^{\prime }\}⟩$. Taking its squared norm, and using the normalization of $|{\varphi }_{\eta }⟩$, givesThus one sums probabilities over unobserved final oscillator states, not their probability amplitudes. Unitarity of $S(t)$ makes the sum over both $\eta$ equal to one.

Expand the wavefunction in unperturbed energy eigenstates,The $c_j$ are interaction picture probability amplitudes, so $|c_j{|}^2$ are level populations and $|c_1{|}^2+|c_2{|}^2=1$ in this closed two-level system. A classical field ${\mathbf{E}}_0\cos \omega t$ supplies the electric-dipole interaction $-\mathbf{d}\cdot {\mathbf{E}}_0\cos \omega t$. Choosing the relative phase of the energy eigenstates makes $\Omega =-⟨1|\mathbf{d}\cdot {\mathbf{E}}_0|2⟩/\hslash$ real. This is the resonant Rabi angular frequency.

Combining the field's two exponentials with the unperturbed quantum phases produces terms oscillating at $\omega -{\omega }_0=\delta$ and at $\omega +{\omega }_0$. The rotating-wave approximation drops the latter, leaving exactly the supplied equations with coupling $\Omega /2$. It requires $|\Omega |,|\delta |\ll {\omega }_0$ and a field envelope slow enough on the carrier timescale. Spontaneous emission and dephasing must be negligible during the sequence. For a ground-state hyperfine level transition, the analogous direct drive is a magnetic dipole transition; the same equations hold with the magnetic matrix element replacing the electric one.

Choose the relative phases of the two energy eigenstates so that $d=⟨{\psi }_2|{\hat{D}}_0|{\psi }_1⟩$ is real and nonnegative. Define the resonant Rabi angular frequency $\Omega =d/\hslash$, the Rabi detuning $\delta ={\omega }_0-\omega$, and $\Sigma ={\omega }_0+\omega$. Substitution into the Schrödinger equation, followed by projection onto the two eigenstates, gives the exact interaction picture equationsKeep both probability amplitudes dynamical: replacing $c_1$ by one would incorrectly impose weak excitation and miss complete population transfer.

For $|\delta |,\Omega \ll {\omega }_0$, apply the rotating-wave approximation to discard the rapidly oscillating terms at $\Sigma$. Differentiating the second remaining equation and using the first givesThe initial conditions are $c_2(0)=0$ and ${\dot{c}}_2(0)=-i\Omega /2$. With ${\Omega }_R=\sqrt{{\Omega }^2+{\delta }^2}$, their solution isConsequently,This result permits order-one excitation, including full population transfer on resonance, even though the drive is small compared with the optical angular frequency.

For an average from switch-on to time $\tau$, integrate the squared sine:On resonance this limiting average is $1/2$, although the instantaneous probability repeatedly reaches one. The zero-coupling, zero-Rabi detuning case is understood separately as $P_2=0$.

Near resonance alone does not make this an exact result for arbitrarily strong driving. The stated sinusoidal Hamiltonian also needs the rotating-wave approximation. A concrete counterexample is $\omega ={\omega }_0$ and $\Omega ={\omega }_0$: the exact equations give $P_2(t)={\Omega }^2{t}^2+O(t^4)$ at switch-on, while the claimed rotating-wave sinusoid gives ${\Omega }^2{t}^2/4+O(t^4)$. Thus “without assuming weak radiation” can mean without assuming small excitation, but cannot remove the drive-to-carrier scale separation. The infinite-time average above belongs to the rotating-wave model; omitted corrections can accumulate over sufficiently long times.

Use the ordered computational basis $(|00⟩,|01⟩,|10⟩,|11⟩)$, with $Z|0⟩=|0⟩$ and $Z|1⟩=-|1⟩$. Write $X,Y,Z$ for the Pauli matrices, $I$ for the identity operator, $J={\omega }_{12}$, $\Sigma ={\omega }_1+{\omega }_2$, and $\Delta ={\omega }_1-{\omega }_2$. As in the question, the Hamiltonian operators have angular frequency units, so the time-evolution operator is $e^{-iHt}$. The Heisenberg coupling matrix for two spins isConsequently the full Heisenberg model Hamiltonian operator and its proposed Ising coupling approximation are

For nondegenerate perturbation theory, choose $H_{\mathrm{I}}$ as the unperturbed Hamiltonian operator andas the perturbation. Only $|01⟩$ and $|10⟩$ are coupled, and their unperturbed energy levels differ by $\Delta$. All first-order energy corrections vanish. The first-order eigenvectors and second-order energy corrections areThe states $|00⟩,|11⟩$ are exact eigenvectors throughout. Hence the weak coupling condition for two NMR spins isA large common Zeeman energy alone does not ensure this condition. Crossings involving the uncoupled parallel-spin states do not invalidate the argument: the relevant denominator is the separation of the two states connected by $V$.

As an exact check, the middle $2\times 2$ matrix has eigenvaluesFor $\Delta \ne 0$, labeling by the limiting computational basis states reproduces the shifts above. The quantum transition probability from $|01⟩$ to $|10⟩$ isThus unwanted exchange has a uniformly small amplitude when $|J/\Delta |\ll 1$. If accumulated phases must also be approximated over a duration $t$, neglecting the second-order shifts additionally requires $J^{2}t/|\Delta |\ll 1$, or those shifts must be retained in an improved effective Hamiltonian operator.

The same result is the secular approximation for scalar-coupled spins. In the interaction picture of the Zeeman energy terms,These terms average rapidly on the coupling timescale $1/|J|$ when $|\Delta |\gg |J|$, while $JZ\otimes Z/2$ survives.

For fixed nuclei and electronic environments, the individual NMR angular frequencies scale with the static magnetic field $B_0$: schematically ${\omega }_i={\kappa }_i{B}_0$, where ${\kappa }_i$ includes the gyromagnetic ratio and magnetic shielding. The NMR scalar coupling $J$ in angular frequency units is approximately independent of $B_0$. Therefore a nonzero separation $|\Delta |=|{\kappa }_1-{\kappa }_2|B_0$ increases with magnetic field, improving the approximation. In particular, a fixed difference of chemical shifts gives a larger separation in hertz at higher magnetic field. If the two resonances remain exactly equal, increasing their common frequency does not suppress their exchange.

Conservation of $\mathcal{N}$ confines evolution from $|e,0⟩$ to $|e,0⟩,|g,1⟩$. On this subspace,The interaction matrix squares to the identity, so its exponential separates into cosine and sine terms. The state isThe overall phase can be discarded or removed by the interaction picture. The relative phase $-i$ follows from Schrödinger evolution with the positive coupling in a. The atom and field are generally entangled; complete excitation transfer occurs at $gt=\pi /2$.

Quantize a single electromagnetic normal mode of angular frequency ${\omega }_c$ and represent one isolated atomic transition as a two-level system with raising operator ${\sigma }_{+}=|e⟩⟨g|$. The bare Hamiltonian operator isThe first main approximation is the electric-dipole approximation: the field varies negligibly across an atom, $k{a}_{\mathrm{atom}}\ll 1$, and the leading coupling is $H_d=-\mathbf{d}\cdot \mathbf{E}({\mathbf{r}}_0)$. For states of definite opposite parity, the relevant dipole part is off-diagonal. After choosing operator phases, this gives $H_d=\hslash g(a+a^{†})({\sigma }_{+}+{\sigma }_{-})$, with $g$ fixed by the transition dipole and the one-photon field wave amplitude.

In the interaction picture, $a{\sigma }_{+}$ and $a^{†}{\sigma }_{-}$ oscillate at the difference ${\omega }_a-{\omega }_c$, whereas $a{\sigma }_{-}$ and $a^{†}{\sigma }_{+}$ oscillate at the sum ${\omega }_a+{\omega }_c$. The second main approximation is the rotating-wave approximation: near resonance and for $g\sqrt{n+1}\ll {\omega }_a+{\omega }_c$, the latter terms average away on the interaction time scale. The resulting Jaynes-Cummings model hasThe two-level system and single-normal mode truncations are additional assumptions: other atomic transitions and optical normal modes must be sufficiently far off resonance or weakly coupled. Dissipation is neglected over the interaction time. The retained terms exchange one atomic excitation for one photon, conserving $a^{†}a+{\sigma }_{+}{\sigma }_{-}$.

Use the resonant interaction picture and define $g>0$ by the one-atom Jaynes-Cummings model in part (a). Equal coupling to both atoms gives the Tavis-Cummings modelThe initial state has two excitations and is symmetric under atom exchange. Excitation conservation and this symmetry confine the evolution toThe antisymmetric atomic spin singlet state is uncoupled and unpopulated. Applying the annihilation operator and creation operator givesThe first off-diagonal entry combines two indistinguishable atomic emission paths; the second also includes the $\sqrt{2}$ enhancement for creating the second photon. Its eigenvalues are $0,\pm \sqrt{6}g$. Put $\Omega =\sqrt{6}g$ and $K=H_I/\hslash$, so $K^3={\Omega }^{2}K$. Its exponential isApplying this to $|A⟩$ yieldsAll three states have the same bare energy on resonance, so restoring the free Schrödinger-picture evolution adds only a common global phase. The expression starts at $|ee;0⟩$, has initial derivative $-i\sqrt{2}g|B⟩$, and its three squared probability amplitudes sum to one.

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.