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The dynamic electric polarizability relates the induced electric dipole moment at a driving frequency to the applied electric field at that frequency in linear response theory. A complex coefficient can encode both a dispersive phase and absorption. For a charged one-dimensional quantum harmonic oscillator with charge magnitude $e$, mass $m$, and natural frequency ${\omega }_0$, the undamped response isOne derivation uses the oscillator electric-dipole selection rule and first-order time-dependent perturbation theory. The same result follows from the classical Lorentz oscillator model: $m\ddot{z}+m{\omega }_0^{2}z=-eE$ and $d=-ez$ imply $\ddot{d}+{\omega }_0^{2}d=e^{2}E/m$. At exact resonance there is no bounded periodic response without damping; a finite switch-on produces secular growth rather than a finite steady-state value of the pole expression.

In the straight-trajectory dipole model, retain the Coulomb-coupling convention of $V(t)=Z{e}^2(bx+vty)/(b^2+v^2{t}^2{)}^{3/2}$. In SI units the coupling denoted $e^2$ here is $e^2/(4\pi {ϵ}_0)$. First-order time-dependent perturbation theory expresses a transition through the Fourier transform of this pulse at the transition angular frequency.

For positive $\omega$ put $\beta =\omega b/v$. The two Cartesian pulse transforms arewhere $K_{\nu }$ is a Modified Bessel function of the second kind. One may derive these from the Fourier transform of $(1+s^2{)}^{-1/2}$, differentiating in its scale and integrating its spatial derivative by parts. For $\beta \ll 1$, $\beta K_1(\beta )\to 1$ while $\beta K_0(\beta )\to 0$, so the transverse impulse gives the probability in the title. If the $x$ matrix element vanishes, the smaller longitudinal term can be the leading nonzero contribution.

The conditions include $b$ large compared with the atomic extent for the electric-dipole approximation, a small total transition probability for first-order theory, and a collision time $b/v$ short compared with $1/|\omega |$. Large $\beta$ instead gives exponentially suppressed excitation in this model through the large-argument behavior of $K_{\nu }$.

For $H(t)=\hslash \omega (a^{†}a+1/2)-Fx{e}^{t/\tau }$ on $t\le 0$, start in the unforced ground state at $t=-\infty$. Define $x_{01}=\sqrt{\hslash /(2m\omega )}$ and $\eta =F{x}_{01}/(\hslash \omega )$. Time-dependent perturbation theory givesThus the first excited-state probability in the original oscillator basis tends to zero for a rapid turn-on and to ${\eta }^2$ for slow turn-on, within the weak-force approximation $|\eta |\ll 1$.

This finite adiabatic limit is not a failure of the adiabatic theorem: the instantaneous ground state is displaced and has a nonzero projection onto the original excited states. In fact a linear force preserves a coherent state, whose exact original-basis displacement at $t=0$ is $\alpha =i\eta \omega \tau /(1+i\omega \tau )$. It follows either from $\dot{\alpha }=-i\omega \alpha +iF{x}_{01}{e}^{t/\tau }/\hslash$ or from the forced oscillator equations. Consequently the exact original-basis probability is $P_1^{(0)}=|\alpha {|}^2{e}^{-|\alpha {|}^2}$. The instantaneous oscillator basis has displacement $\eta$, so its mean excitation is instead $|\alpha -\eta {|}^2={\eta }^2/[1+(\omega \tau {)}^2]$, which vanishes adiabatically.

For a weak time-independent perturbation switched on at zero time, first-order time-dependent perturbation theory gives a transition probabilityThe frequency kernel has area $2\pi t$ and width of order $1/t$. Against a smoothly varying continuum of final states, it approaches $2\pi t\delta ({\omega }_f-{\omega }_i)$, giving the rate $2\pi |V_{fi}{|}^2\delta ({\omega }_f-{\omega }_i)/{\hslash }^2$. Using $E=\hslash \omega$ gives the energy form above. One integrates the Dirac delta function against the appropriate density of states; it is not an ordinary finite number for one isolated resonant level.

The time must be long enough to resolve the smooth continuum and short enough that the total transition probability remains small. For continuum variation scale $\Delta \omega$ and total rate $\Gamma$, the usual window is $(\Delta \omega {)}^{-1}\ll t\ll {\Gamma }^{-1}$. A discrete finite spectrum also limits this approximation through its recurrences.

The ladder identity $x=\sqrt{\hslash /(2m\omega )}(a+a^{†})$ proves the matrix element in the title. In time-dependent perturbation theory, a linear force excites only $n=1$ directly from $n=0$. Parity alone rules out even excited states; the ladder identity also rules out odd $n>1$. Higher-order evolution can reach other levels, so the driven oscillator is not an exact two-level system.

Insert the expansion in the time-dependent Schrödinger equation, cancel the unperturbed evolution, and project onto $⟨{\psi }_n|$. With $V_{nm}(t)=⟨{\psi }_n|V(t)|{\psi }_m⟩$,In first-order time-dependent perturbation theory, put $c_m(t)={\delta }_{m0}$ on the right and use $c_n(-\infty )=0$ for $n\ne 0$. Integration givesThe initial state must be the specified unperturbed eigenstate, the integral must exist under the switching prescription, and total transition probability must stay small. The diagonal term produces a first-order phase; setting $c_0=1$ in the off-diagonal integral changes those amplitudes only at second order. A weak peak perturbation alone is insufficient if its accumulated action invalidates this expansion.

For the infinite square well, retain the source's zero-based labels:An electron in a field along $+x$ has perturbation $V(t)=+e{E}_{0}x{e}^{-t/\tau }$ for $t\ge 0$. About the well centre, $x-w/2$ is odd, whereas ${\psi }_0$ and ${\psi }_2$ are both even. The constant $w/2$ has zero off-diagonal matrix element, so the first-order parity selection in an infinite well gives $⟨{\psi }_2|x|{\psi }_0⟩=0$. Excitation to that state is forbidden at first order, not necessarily at higher orders.

For the first excited state, direct integration or the supplied integral givesThe Fourier amplitude of the exponentially decaying electric-field pulse is ${\int }_0^{\infty }{e}^{-t/\tau +i\omega t}dt=\tau /(1-i\omega \tau )$. ThusThis is the leading, quadratic-in-field transition probability; the total probability of all excitations must remain small.

Starting instead in ${\psi }_1$, which is odd about the centre, the pulse can transfer amplitude to the ground state and to even-labelled excited states ${\psi }_2,{\psi }_4,\dots$. Other odd-labelled states have zero first-order dipole amplitude. Both downward and upward transitions are possible; their relative probabilities depend on dipole matrix elements and the pulse's spectral weight at the corresponding energy gaps. Most population remains in ${\psi }_1$, with a small depletion equal to the summed leading transition probabilities. The common diagonal dipole phase in an infinite well is $-e{E}_{0}w\tau /(2\hslash )$ and has no population effect. After the pulse, the remaining coherent superposition evolves under the unperturbed Hamiltonian; no spontaneous relaxation is implied by this closed-system model.

With $y=x-x_0$ and oscillator length $a_0=\sqrt{\hslash /(m\omega )}$, the first two harmonic oscillator eigenfunctions and energies areThe ground state is even and nodeless; the first excited state is odd with one node of a wavefunction. Their rms widths are $a_0/\sqrt{2}$ and $\sqrt{3/2}{a}_0$. Also, $m{\omega }^2{a}_0^2\sim \hslash \omega$ gives the characteristic extent directly.

For the shaken harmonic trap, expanding about the unshifted Hamiltonian givesThe scalar term gives only a common phase and no transitions. In time-dependent perturbation theory, initially $c_0=1$, $c_{n\ne 0}=0$ and the oscillatory factor for $0\to 1$ is $e^{i(E_1-E_0)t/\hslash }=e^{i\omega t}$. ThusThe odd operator $x$ makes transitions to even $n>0$ vanish by parity. In fact the first-order selection rule for a linearly driven oscillator is stronger: since $x=a_0(a+a^{†})/\sqrt{2}$,All excited $n\ne 1$ amplitudes vanish at first order, including odd $n>1$. This assertion does not apply to the surviving $c_0$ or to higher perturbative orders.

Put $\delta =\omega -\Omega$, $\Sigma =\omega +\Omega$. The exact first-order time integral isNear resonance, the second term oscillates rapidly at approximately $2\omega$ and is bounded by $1/\Sigma$, whereas the first grows as $T/2$ on resonance. Thus $\omega T\gg 1$ justifies the rotating-wave approximation, $\cos \Omega t\mapsto e^{-i\Omega t}/2$. At zeros of the resonant term, the omitted small contribution can dominate the residual; the approximation has no uniform relative-error bound there.

Using the requested order-one approximation $⟨1|x|0⟩\simeq a_0$ and $m{\omega }^2{a}_0/\hslash =\omega /a_0$ gives the square-pulse resonance profileThe exact harmonic oscillator matrix element instead gives half this probability. The finite central maximum is at $\Omega =\omega$; first zeros are at $|\Omega -\omega |=2\pi /T$, with smaller side lobes and a $1/{\delta }^2$ envelope. Increasing $T$ narrows the peak as $1/T$ and increases its height as $T^2$, as long as weak resonant driving keeps $p_1\ll 1$. Small displacement alone does not ensure this for arbitrarily long $T$; near resonance one also needs $\epsilon \omega T/a_0\ll 1$.

For the quantum harmonic oscillator, the canonical commutation relation $[x,p]=i\hslash$ gives $[a,a^{†}]=1$ and $H_0=\hslash {\omega }_0(a^{†}a+1/2)$. Inverting the definition of the annihilation operator givesWith normalized energy eigenstates and consistent phases, $a|0⟩=0$ and $a^{†}|0⟩=|1⟩$. Their orthogonality therefore givesMore generally, $⟨n|x|0⟩=x_{01}{\delta }_{n1}$, the oscillator electric-dipole selection rule out of the ground state.

To derive the transition amplitude, insert the expansion in stationary energy eigenstates into the Schrödinger equation with $H=H_0+V_1(t)$. The derivatives of the explicit phases cancel the $H_0$ terms. Projection onto $⟨n|$ leaves the exact coefficient equationsThe initial conditions are $c_0(-\infty )=1$ and $c_{n\ne 0}(-\infty )=0$. To first order in the perturbation, replace the amplitudes on the right by these initial values. Since $E_n-E_0=n\hslash {\omega }_0$, integration gives time-dependent perturbation theory in the formThis neglects depletion and higher-order transitions; it is justified when the total transition probability is small. Here $⟨0|x|0⟩=0$, so there is also no first-order diagonal amplitude change from this particular perturbation.

For the exponential electric field, the electric-dipole interaction is $V_1=-qEx{e}^{t/\tau }$, with $\tau >0$. Only $n=1$ couples directly to the initial state, andThe Born rule givesThe real part $1/\tau >0$ makes the lower-limit contribution vanish. ThusSet $u={\omega }_0\tau$ and $\eta =qE{x}_{01}/(\hslash {\omega }_0)$. The dimensionless form is $p_1/{\eta }^2=u^2/(1+u^2)$, increasing monotonically from zero to one, with half the asymptotic value at $u=1$. A sufficient weak-field condition for the entire curve is $|\eta |\ll 1$.

For ${\omega }_0\tau \ll 1$, the force has acted for too little time to appreciably change the state at the instant of measurement. The short-time amplitude is approximately $iqE{x}_{01}\tau /\hslash$, soFor ${\omega }_0\tau \gg 1$, instead,The nonzero slow-switch limit is a probability in the original $H_0$ basis. It does not contradict the adiabatic theorem. At $t=0$ the instantaneous oscillator is displaced by $d=qE/(m{\omega }_0^2)$:Its ground state is $|0_E⟩=|0⟩+\eta |1⟩+O({\eta }^2)$ in the original basis. Adiabatic following therefore predicts exactly the finite original-basis limit found above.

The exponential switch-on of a driven quantum oscillator also permits an exact check. A linear force preserves a coherent state, with original-basis displacement $\alpha (0)=i\eta u/(1+iu)$. Thus the exact original-basis probability is $|\alpha {|}^2{e}^{-|\alpha {|}^2}$, whose weak-field limit is the expression above. In the instantaneous displaced basis the mean excitation is instead $|\alpha -\eta {|}^2={\eta }^2/(1+u^2)$ and does tend to zero for slow switching. The two bases measure different notions of excitation.

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.

In first-order time-dependent perturbation theory, the resonant time factor before taking the Fermi's golden rule limit isIts central frequency width is of order $1/t$, and its total area is $2\pi t$. To replace it by $2\pi t\delta (\nu )$, the oscillator density of states and relevant occupations must be approximately constant over that width. Thus $t\Delta \omega \gg 1$. At the same time, the perturbative flip probability must remain much smaller than one.

Define ${\Gamma }_0=2\pi {\lambda }^{2}g(\Omega )/{\hslash }^2$, so the two flip rates are ${\Gamma }_0(n_{\Omega }+1)$ and ${\Gamma }_0{n}_{\Omega }$. A sufficient common time window for both initial spins isFor a specified initial spin, the upper condition is simply its own flip rate times $t\ll 1$. Such a window requires weak enough coupling. The continuum approximation also presumes times shorter than recurrences from a resolved finite oscillator spectrum, and long enough to resolve the positive-frequency resonance from its nonresonant counterpart.

Use first-order time-dependent perturbation theory, setting $c_1=1$ in the equation for $c_2$ and retaining the generally complex matrix element $d=⟨{\psi }_2|{\hat{D}}_0|{\psi }_1⟩$. With $\delta ={\omega }_0-\omega$ and $\Sigma ={\omega }_0+\omega$, the probability amplitude isFor $t>0$, split the integral at the pulse maximum. For either $x=\delta$ or $x=\Sigma$,Thus the full first-order answer, without discarding the counter-rotating contribution, isAt long times it givesThis full expression is not exactly one squared Lorentzian line shape. To obtain the intended near-resonance spectral line profile, additionally assume $b,|\delta |\ll {\omega }_0$, so the pulse has a narrow positive-frequency lobe and $F(\Sigma ,t)$ is negligible. ThenThe weak-excitation condition near resonance is $|d|/(\hslash b)\ll 1$. The squared Lorentzian line shape occurs because the symmetric exponential pulse has a Lorentzian line shape amplitude Fourier transform, which is then squared to obtain probability.

At half the maximum, $[1+(\delta /b{)}^2{]}^2=2$, soThe full width at half maximum refers to varying the drive frequency through the positive-frequency resonance. For a concrete failure of the unqualified squared-Lorentzian line shape claim, set $b=\omega ={\omega }_0$ while keeping $|d|/(\hslash b)\ll 1$. The full first-order resonant probability is $36/25$ times the one-lobe prediction. Weak radiation alone therefore does not justify dropping the second Fourier transform lobe.

Expand the state in the energy eigenstates of $H_0$ asSubstitution into the Time-dependent Schrödinger equation and projection onto $⟨k|$ gives the exact interaction picture equationsIn first-order time-dependent perturbation theory, put $c_j(t)={\delta }_{j0}$ on the right-hand side and integrate from $0$ to $t$:

For a uniform electric field, the electric-dipole interaction is $\Delta (t)=e{\mathcal{E}}_{0}z{e}^{-t/\tau }$ up to an irrelevant overall sign. The supplied hydrogen atom wavefunctions giveFor $t\gg \tau$, the time integral isThe long-time transition fraction is consequentlyBoth $|100⟩$ and $|200⟩$ have even parity, whereas $z$ is odd, so the electric-dipole selection rule gives $⟨200|z|100⟩=0$. Thus the requested fraction in $|200⟩$ is zero.

In the interaction picture, first-order time-dependent perturbation theory gives the transition amplitudeTaking its squared modulus and using $|e^{i\theta }-1{|}^2=4{\sin }^2(\theta /2)$ gives

Let $H_0|r⟩=E_r|r⟩$. In the interaction picture,and the state obeysThe first term of the Dyson series, or equivalently first-order time-dependent perturbation theory, givesThus, if $|\psi (0)⟩={\sum }_r{a}_r|r⟩$,

First-order time-dependent perturbation theory givesThe lowest nontrivial transition probability is $|c_1(t){|}^2$, with errors of higher order in $V$.

To first order in $F$, time-dependent perturbation theory replaces the state on the right-hand side by its unperturbed value $e^{-i{E}_{0}t/\hslash }|0⟩$. The bound state is even, while the position operator $x$ is odd. The parity selection rule therefore givesso ${\dot{b}}_k=0$ and, with no initial continuum component,

Projection onto the odd state givesWriting ${\Omega }_k=(E_k-E_0-\hslash \omega )/\hslash$ and integrating from zero to $t$ yields

At first order in time-dependent perturbation theory set $c_1\simeq 1$ and use the parity result $V_{22}=0$. Decomposing the cosine into its two exponentials givesBoth exponentials contribute to the exact first-order result:Each apparent zero denominator has a continuous limit. In particular the near-resonant fraction tends to $it$ as $\omega \to {\omega }_0$, while the counter-rotating fraction remains bounded on the optical timescale. One may equivalently write the expression asThis form makes $c_2(0)=0$ explicit and avoids cancellation at resonance. At extremely short time both sinc functions tend to one, so $c_2\simeq -i\Omega t$, agreeing with the initial derivative of the original cosine-driven equation. No rotating-wave approximation has yet been made. The separate assumption of weak excitation remains necessary: the first-order expression must not be extended until it predicts probabilities comparable to or larger than one.

Perturbation theory expands a quantum problem around a solvable Hamiltonian. Time-independent perturbation theory expands stationary energies and eigenstates; time-dependent perturbation theory expands dynamical amplitudes and transitions.

An additional time-dependent operator in a quantum Hamiltonian. It may couple stationary states or shift their phases. It is the physical operator being treated, whereas time-dependent perturbation theory is a method for approximating the response.