Chapter 6

Time-Dependent Potentials and Light–Matter Interaction

6.1Introduction

Up to now the Hamiltonian was time independent, so the main questions were energy levels and stationary states. Here the Hamiltonian depends on time because an external field drives the atom, and the central questions become: which transitions occur, how fast, and with what probability.

This chapter is the bridge from atomic structure (Chapters 4–5) to the time-domain spectroscopy model used in the final chapters. The logic is deliberately linear:

derive the exact coupled-amplitude equations from the TDSE;
insert the semiclassical dipole interaction and pulse model;
solve a near-resonant two-level system (Rabi oscillation);
derive first-order perturbation theory and Fermi's golden rule;
state clearly where each approximation fails.

If these five steps are clear, the light–matter formalism in the rest of the book is transparent rather than memorized.

6.2The Coupled Amplitude Equations

Let the Hamiltonian split into a solved part and a time-dependent part,

\begin{equation} \label{eq:TD-hamiltonian} \hat H(t)=\hat H_0+\hat V(t), \qquad \hat H_0\ket{n}=E_n\ket{n}, \qquad \braket{m}{n}=\delta_{mn} . \end{equation}

The eigenstates $\ket{n}$ form a complete basis, so any state expands in them. We pull out the free-evolution phase explicitly:

\begin{equation} \label{eq:TD-expansion} \ket{\Psi(t)}=\sum_n c_n(t)\,e^{-iE_nt/\hbar}\ket{n} . \end{equation}

Theorem 6.2.1 - Coupled amplitude equations

The coefficients in Eq. (6.2) obey, exactly,

\begin{equation} \label{eq:TD-coupled} \boxed{\; i\hbar\,\dot c_k(t) =\sum_n c_n(t)\,V_{kn}(t)\,e^{i\omega_{kn}t}\;} \end{equation}

where

\begin{equation} \label{eq:TD-matrix-elements} V_{kn}(t)=\bra{k}\hat V(t)\ket{n}, \qquad \omega_{kn}=\frac{E_k-E_n}{\hbar} . \end{equation}

Substitute Eq. (6.2) into $i\hbar\,\partial_t\ket{\Psi}=(\hat H_0+\hat V)\ket{\Psi}$ . The left-hand side, by the product rule, is

i\hbar\sum_n\left[\dot c_n e^{-iE_nt/\hbar} +c_n\left(-\frac{iE_n}{\hbar}\right)e^{-iE_nt/\hbar}\right]\ket{n} =\sum_n\left[i\hbar\dot c_n+E_nc_n\right]e^{-iE_nt/\hbar}\ket{n} .

The right-hand side is

\sum_nc_ne^{-iE_nt/\hbar}\left[E_n\ket{n}+\hat V\ket{n}\right].

The terms $$E_nc_n$$ appear on both sides and cancel — which is the whole point of extracting the phase in Eq. (6.2). What remains is

\sum_n i\hbar\dot c_ne^{-iE_nt/\hbar}\ket{n} =\sum_nc_ne^{-iE_nt/\hbar}\hat V\ket{n} .

Project onto $\bra{k}$ and use orthonormality on the left:

i\hbar\dot c_ke^{-iE_kt/\hbar} =\sum_nc_ne^{-iE_nt/\hbar}V_{kn} .

Multiplying by $e^{+iE_kt/\hbar}$ gives Eq. (6.3).

Three remarks on this result, since it is the skeleton of everything below.

Remark 6.2.1 - It is exact

No approximation has been made: one partial differential equation has been rewritten as an infinite set of coupled ordinary differential equations. Approximations enter only later, when we truncate the basis or simplify $\hat V(t)$ . That separation is useful: convergence can be tested by adding states and checking how much the observables change.

Remark 6.2.2 - The phase factor is where resonance lives

The factor $e^{i\omega_{kn}t}$ carries the energy difference between the two states. If $\hat V(t)$ oscillates at a frequency near $\omega_{kn}$ , the product $V_{kn}e^{i\omega_{kn}t}$ contains a slowly varying piece whose time integral accumulates; otherwise the product oscillates rapidly and averages to nearly nothing. That observation is resonance, and it is the mechanism behind every selection rule and every bandwidth argument in the rest of the book.

Remark 6.2.3 - Probability conservation as a numerical test

$P_n(t)=|c_n(t)|^{2}$ is the probability of finding the system in $\ket{n}$ , and unitarity guarantees $\sum_n|c_n|^{2}=1$ for all time. When Eq. (6.3) is integrated numerically, monitoring that sum is the single most useful check available.

6.3How Light Couples to an Atom

We treat the field classically — a given $\vec E(t)$ , not a quantised photon field. This is the semiclassical approximation, excellent for lasers, where the photon number is enormous.

A charge distribution in a uniform field has interaction energy $-\hat{\vec d}\cdot\vec E$ , where for an electron at $\hat{\vec r}$ the dipole moment operator is $\hat{\vec d}=-e\hat{\vec r}$ . Hence

\begin{equation} \label{eq:TD-dipole-interaction} \boxed{\;\hat V(t)=-\hat{\vec d}\cdot\vec E(t) =e\,\hat{\vec r}\cdot\vec E(t)\;} \end{equation}

Definition 6.3.1 - Dipole approximation

The dipole approximation replaces the field at the electron's position by its value at the nucleus, $\vec E(\vec r,t)\to\vec E(t)$ . It is valid when the wavelength of the radiation greatly exceeds the size of the atom.

Example 6.3.1 - Checking the dipole approximation

Is Def. 6.3.1 safe for $$800$$ nm infrared light and for $$17.6$$ eV extreme ultraviolet light?

Solution.

The relevant atomic size is $\expval{r}=1.5\,a_0=0.079$ nm for a ground state (Chapter 4, Example 4.7.1).

For $$800$$ nm the ratio is $800/0.079\approx10^{4}$ : the field varies by one part in ten thousand across the atom. Entirely safe.

For $$17.6$$ eV, $\lambda=hc/E=1239.8/17.6=70.4$ nm, and the ratio is $70.4/0.079\approx900$ . Still safe, though by three orders of magnitude rather than four.

The approximation fails only for hard X-rays, where $\lambda$ approaches $$a_0$$ . Note however that a Rydberg state with $$n=20$$ has $\expval{r}\approx600\,a_0=32$ nm, comparable to the XUV wavelength — so “the atom is small compared with the wavelength” is a statement about the state, not just the atom.

Combining Eqs. (6.5) and (6.4), every transition is governed by the dipole matrix element

\begin{equation} \label{eq:TD-dipole-matrix} \vec d_{kn}=\bra{k}\hat{\vec d}\ket{n}=-e\bra{k}\hat{\vec r}\ket{n}, \end{equation}

and for fixed polarization this reduces to the scalar form $V_{kn}(t)=-d_{kn}E(t)$ .

This is exactly the object analysed in Chapter 3: $\hat{\vec r}$ is odd, so $\vec d_{kn}$ vanishes unless the states differ in parity (Thm. 3.6.1), giving $\Delta l=\pm1$ and $\Delta m=0,\pm1$ .

6.3.1What a Pulse Is

Definition 6.3.2 - Laser pulse

A pulse is a carrier wave inside an envelope,

\begin{equation} \label{eq:TD-pulse} \vec E(t)=\hat\epsilon\,E_0\,h(t)\cos(\omega t+\phi), \end{equation}

with $\hat\epsilon$ the polarization direction, $$E_0$$ the peak field amplitude, $$h(t)$$ a dimensionless envelope of peak value $$1$$ , $\omega$ the carrier frequency and $\phi$ the carrier–envelope phase. A common choice is the Gaussian

\begin{equation} \label{eq:TD-gaussian} h(t)=e^{-t^{2}/2\tau^{2}} . \end{equation}

Experiments report intensity, theory needs field; they are related by

\begin{equation} \label{eq:TD-intensity} I=\tfrac12\epsilon_0cE_0^{2}, \end{equation}

and one atomic unit of field ( $5.14\times10^{11}$ V/m) corresponds to $I=3.51\times10^{16}$ W/cm $^{2}$ .

Duration and bandwidth are conjugate. For Gaussian pulses the full-width-at-half-maximum values satisfy

\begin{equation} \label{eq:TD-bandwidth} \Delta E\,\Delta t\approx1.8~\mathrm{eV\,fs}, \end{equation}

a relation derived in Example 6.5.1 rather than assumed.

6.3.2A Practical Workflow

For most light–matter calculations in this course, the steps are:

Choose the active basis states $\{\ket{n}\}$ and energies $$E_n$$ .
Compute the relevant dipole elements $d_{kn}$ and apply selection rules early to discard forbidden channels.
Specify the field $E(t)=E_0h(t)\cos(\omega t+\phi)$ and convert given intensity to $$E_0$$ with Eq. (6.9).
Build Eq. (6.3) and choose the model by regime: two-level near resonance $\to$ Rabi; weak excitation to many states $\to$ first-order perturbation; otherwise integrate the coupled equations numerically.
Extract populations $$P_n(t)=|c_n(t)|^2$$ and phases $\arg c_n(t)$ , then compare directly with measured line strengths, widths, and beats.

6.4Two Levels, Solved Exactly

Truncate Eq. (6.3) to two states, $\ket{1}$ and $\ket{2}$ , separated by $\hbar\omega_0=E_2-E_1$ . Since the dipole operator has no diagonal elements between parity eigenstates, $V_{11}=V_{22}=0$ , and writing $d\equiv d_{21}=d_{12}^{*}$ (real, by convention) with a constant envelope $$h=1$$ and $\phi=0$ , Eq. (6.3) reads

\begin{equation} \label{eq:TD-two-level-raw} \begin{aligned} i\hbar\dot c_1&=d\,E_0\cos(\omega t)\,e^{-i\omega_0t}\,c_2 ,\\ i\hbar\dot c_2&=d\,E_0\cos(\omega t)\,e^{+i\omega_0t}\,c_1 . \end{aligned} \end{equation}

6.4.1The Rotating-Wave Approximation

Expand the cosine as $\cos\omega t=\tfrac12\bigl(e^{i\omega t}+e^{-i\omega t}\bigr)$ and look at the second equation:

\begin{equation} \label{eq:TD-RWA-expansion} i\hbar\dot c_2 =\frac{dE_0}{2} \Bigl[ \underbrace{e^{i(\omega_0+\omega)t}}_{\text{fast}} +\underbrace{e^{i(\omega_0-\omega)t}}_{\text{slow near resonance}} \Bigr]c_1 . \end{equation}

Define the detuning

\begin{equation} \label{eq:TD-detuning} \Delta=\omega_0-\omega . \end{equation}

Near resonance $|\Delta|\ll\omega_0$ , so the two exponents are wildly different: the first oscillates at about $2\omega_0$ and the second at the small frequency $\Delta$ .

Definition 6.4.1 - Rotating-wave approximation

The rotating-wave approximation (RWA) discards the counter-rotating term $e^{i(\omega_0+\omega)t}$ , retaining only the term oscillating at the detuning.

The justification is that integrating $e^{2i\omega_0t}$ over any interval long compared with an optical cycle gives a contribution smaller by $\sim\Delta/\omega_0$ than the retained term. It fails when the field is strong enough that $\Omega$ (below) approaches $\omega_0$ , or when the detuning is comparable to the carrier frequency.

Definition 6.4.2 - Rabi frequency

\begin{equation} \label{eq:TD-rabi-def} \Omega=\frac{dE_0}{\hbar} . \end{equation}

Within the RWA, Eq. (6.11) becomes

\begin{equation} \label{eq:TD-RWA-equations} i\dot c_1=\frac{\Omega}{2}e^{-i\Delta t}c_2 , \qquad i\dot c_2=\frac{\Omega}{2}e^{+i\Delta t}c_1 . \end{equation}

6.4.2Solving the Two-Level Problem

Theorem 6.4.1 - Rabi oscillation

With $$c_1(0)=1$$ and $$c_2(0)=0$$ , the solution of Eq. (6.15) gives

\begin{equation} \label{eq:TD-rabi-solution} \boxed{\; P_2(t)=|c_2(t)|^{2} =\frac{\Omega^{2}}{\Omega^{2}+\Delta^{2}} \,\sin^{2}\!\left(\frac{\Omega_R t}{2}\right), \qquad \Omega_R=\sqrt{\Omega^{2}+\Delta^{2}} \;} \end{equation}

$\Omega_R$ is the generalized Rabi frequency.

Differentiate the second equation of Eq. (6.15):

i\ddot c_2=\frac{\Omega}{2}\left[i\Delta e^{i\Delta t}c_1 +e^{i\Delta t}\dot c_1\right].

From the second equation, $e^{i\Delta t}c_1=2i\dot c_2/\Omega$ ; from the first, $\dot c_1=-i\tfrac{\Omega}{2}e^{-i\Delta t}c_2$ . Substituting both,

i\ddot c_2 =\frac{\Omega}{2}\left[i\Delta\cdot\frac{2i\dot c_2}{\Omega} +e^{i\Delta t}\left(-\frac{i\Omega}{2}e^{-i\Delta t}c_2\right)\right] =-i\Delta\dot c_2-\frac{i\Omega^{2}}{4}c_2 .

Dividing by $$i$$ gives a linear equation with constant coefficients,

\ddot c_2+i\Delta\dot c_2+\frac{\Omega^{2}}{4}c_2=0 .

Trying $c_2\propto e^{i\lambda t}$ gives $-\lambda^{2}-\Delta\lambda+\Omega^{2}/4=0$ , hence

\lambda=\frac{-\Delta\pm\sqrt{\Delta^{2}+\Omega^{2}}}{2} =\frac{-\Delta\pm\Omega_R}{2} .

The general solution is therefore $c_2=e^{-i\Delta t/2}\bigl[A\sin(\Omega_Rt/2)+B\cos(\Omega_Rt/2)\bigr]$ . The initial condition $$c_2(0)=0$$ forces $$B=0$$ , and $\dot c_2(0)=-i\Omega/2$ (from the second equation with $$c_1(0)=1$$ ) fixes $A=-i\Omega/\Omega_R$ . Hence

c_2(t)=-\frac{i\Omega}{\Omega_R}e^{-i\Delta t/2} \sin\!\left(\frac{\Omega_Rt}{2}\right),

whose squared modulus is Eq. (6.16).

Three consequences follow, each worth stating separately.

On resonance the transfer is complete. With $\Delta=0$ , $\Omega_R=\Omega$ and $P_2=\sin^{2}(\Omega t/2)$ , which reaches $$1$$ at $t=\pi/\Omega$ . The atom is then certainly excited. Continue driving and it returns. This is coherent, reversible oscillation, not decay: a pulse delivering $\Omega t=\pi$ — a “ $\pi$ pulse” — inverts the population exactly.

Off resonance the transfer is incomplete. The prefactor $\Omega^{2}/(\Omega^{2}+\Delta^{2})$ is a Lorentzian in the detuning of half-width $\Omega$ . For $|\Delta|\gg\Omega$ the maximum transfer is only $\Omega^{2}/\Delta^{2}$ , however long one waits.

Strong driving broadens the resonance. The width of that Lorentzian is $\Omega\propto E_0$ : a strong field drives a transition it is badly detuned from. “Resonant” is a statement about field strength as well as frequency.

Example 6.4.1 - Field required for a

\pi

pulse

A transition has dipole matrix element $$d=1$$ a.u. What peak field and intensity invert the population in $$10$$ fs?

Solution.

A $\pi$ pulse requires $\Omega t=\pi$ , so with $$t=10$$ fs $$=10/0.02419=413.4$$ a.u.,

\Omega=\frac{\pi}{413.4}=7.60\times10^{-3}~\text{a.u.}

With $$d=1$$ , Eq. (6.14) in atomic units ( $\hbar=1$ ) gives $E_0=\Omega=7.60\times10^{-3}$ a.u., that is

E_0=7.60\times10^{-3}\times5.14\times10^{11}=3.91\times10^{9}~\mathrm{V/m}.

The intensity follows from Eq. (6.9):

I=\tfrac12(8.854\times10^{-12})(3\times10^{8})(3.91\times10^{9})^{2} =2.0\times10^{16}~\mathrm{W/m^{2}}=2.0\times10^{12}~\mathrm{W/cm^{2}}.

This is a readily available laboratory intensity, and it is well below the $3.5\times10^{16}$ W/cm $^{2}$ at which the field reaches one atomic unit — so the RWA and the two-level truncation are both defensible here.

6.5Perturbation Theory and the Golden Rule

When many states participate, the two-level truncation fails. The alternative is to assume the transfer out of the initial state is small and iterate Eq. (6.3).

Set $c_1\approx1$ and all other $c_n\approx0$ on the right-hand side. Then for any $k\neq1$ ,

\begin{equation} \label{eq:TD-first-order-int} i\hbar\dot c_k\simeq V_{k1}(t)e^{i\omega_{k1}t} \end{equation}

which integrates immediately:

\begin{equation} \label{eq:TD-first-order} \boxed{\; c_k(t)\simeq\frac{1}{i\hbar}\int_{-\infty}^{t} V_{k1}(t')\,e^{i\omega_{k1}t'}\,dt' \;} \end{equation}

The formula is physically direct: the amplitude to reach state $$k$$ is the Fourier component of the perturbation at the transition frequency. This is the basic selection mechanism of ultrafast spectroscopy.

Example 6.5.1 - The pulse spectrum decides what is excited

Evaluate Eq. (6.18) for the Gaussian pulse of Def. 6.3.2, and deduce Eq. (6.10).

Solution.

With $V_{k1}=d_{k1}E_0h(t)\cos\omega t$ and the RWA retaining only the $e^{-i\omega t}$ half of the cosine,

c_k(\infty)\simeq\frac{d_{k1}E_0}{2i\hbar} \int_{-\infty}^{\infty}e^{-t^{2}/2\tau^{2}} e^{i(\omega_{k1}-\omega)t}\,dt .

The integral is the Fourier transform of a Gaussian, a standard result:

\int_{-\infty}^{\infty}e^{-t^{2}/2\tau^{2}}e^{i\Delta_kt}\,dt =\tau\sqrt{2\pi}\,e^{-\Delta_k^{2}\tau^{2}/2}, \qquad \Delta_k=\omega_{k1}-\omega .

Hence

P_k=|c_k(\infty)|^{2} =\frac{|d_{k1}|^{2}E_0^{2}\tau^{2}\pi}{2\hbar^{2}} \,e^{-\Delta_k^{2}\tau^{2}} .

The probability falls as a Gaussian in the detuning, with $$1/e$$ half-width $\Delta_k=1/\tau$ in angular frequency. Converting to an energy FWHM and pairing it with the intensity FWHM of the pulse (itself $2\tau\sqrt{\ln2}$ ) gives the product quoted in Eq. (6.10), $\Delta E\,\Delta t\approx1.8$ eV fs.

Two lessons. A short pulse populates a band of states, each weighted by its own dipole matrix element times the pulse spectrum at its own frequency — so two states with equal dipoles but different detunings receive different amplitudes. And this is why a research calculation must use the measured pump bandwidth: it weights every state that gets excited.

Example 6.5.2 - Reading a pulse specification

An attosecond XUV pulse has bandwidth $$0.4$$ eV centred at $$17.6$$ eV. How long is it, and how many states can it excite at once if the levels are spaced by $$0.1$$ eV?

Solution.

From Eq. (6.10),

\Delta t\approx\frac{1.8~\mathrm{eV\,fs}}{0.4~\mathrm{eV}}=4.5~\mathrm{fs}.

A bandwidth of $$0.4$$ eV covers states within roughly $\pm0.2$ eV of the carrier, so with $$0.1$$ eV spacing it reaches four or five levels simultaneously and coherently.

This inverts the usual spectroscopic picture. A narrowband laser selects one level and measures it; a broadband pulse creates a superposition of several, whose relative phases then advance at different rates so that the system beats. Chapter 11 is about reading those beats.

6.5.1Transitions into a Continuum

Now let the final states form a continuum — the situation whenever the atom ionizes, since the freed electron may carry away any energy. Chapter 3 flagged the normalization this requires, $\braket{\varepsilon}{\varepsilon'}=\delta(\varepsilon-\varepsilon')$ .

Theorem 6.5.1 - Fermi's golden rule

For a perturbation of constant strength $\hat V$ switched on at $$t=0$$ , the transition rate from a discrete state $\ket{i}$ into a continuum of final states is

\begin{equation} \label{eq:TD-golden-rule} \Gamma=\frac{2\pi}{\hbar}\bigl|\bra{f}\hat V\ket{i}\bigr|^{2}\rho(E_f), \end{equation}

evaluated at the energy-conserving final energy, with $\rho$ the density of final states per unit energy.

From Eq. (6.18) with $V_{fi}$ constant for $$0<t'<t$$ ,

c_f(t)=\frac{V_{fi}}{i\hbar}\int_0^{t}e^{i\omega_{fi}t'}dt' =\frac{V_{fi}}{i\hbar}\, \frac{e^{i\omega_{fi}t}-1}{i\omega_{fi}} ,

P_f(t)=|c_f|^{2} =\frac{|V_{fi}|^{2}}{\hbar^{2}} \cdot\frac{4\sin^{2}(\omega_{fi}t/2)}{\omega_{fi}^{2}} =\frac{|V_{fi}|^{2}t^{2}}{\hbar^{2}} \,\mathrm{sinc}^{2}\!\left(\frac{\omega_{fi}t}{2}\right).

This function of $\omega_{fi}$ has height $\propto t^{2}$ and width $\propto1/t$ , so its integral grows as $$t$$ . Summing over the continuum, $\sum_f\to\int\rho(E_f)dE_f=\hbar\int\rho\,d\omega_{fi}$ , and using

\int_{-\infty}^{\infty}\frac{4\sin^{2}(\omega t/2)}{\omega^{2}}d\omega =2\pi t ,

the total probability is

P(t)=\frac{|V_{fi}|^{2}}{\hbar^{2}}\,\hbar\,\rho(E_f)\,2\pi t =\frac{2\pi}{\hbar}|V_{fi}|^{2}\rho(E_f)\,t ,

assuming $\rho$ and $V_{fi}$ vary slowly over the narrow width of the sinc. Since $$P$$ grows linearly, the rate $\Gamma=dP/dt$ is constant and equal to Eq. (6.19).

Two features of this result are used directly at the end of the book.

Remark 6.5.1 - Energy normalization absorbs

\rho

If the continuum states are normalized as $\braket{\varepsilon}{\varepsilon'}=\delta(\varepsilon-\varepsilon')$ then the density of states is already carried by the states themselves, $\rho=1$ , and in atomic units Eq. (6.19) reduces to

\begin{equation} \label{eq:TD-golden-normalized} \Gamma=2\pi\bigl|\bra{\varepsilon}\hat V\ket{i}\bigr|^{2} . \end{equation}

This is why the research literature uses that convention: it removes a bookkeeping factor that is otherwise easy to get wrong.

Remark 6.5.2 - The perturbation need not be light

Nothing in the proof assumed $\hat V$ was a laser field. Any coupling between a discrete state and a degenerate continuum gives a rate of this form — including the electron–electron repulsion that autoionized helium's $$2s2p$$ state in Chapter 5. Reading Eq. (6.20) backwards, $|\bra{\varepsilon}\hat V\ket{i}|=\sqrt{\Gamma/2\pi}$ , converts a measured width into a coupling strength. That is exactly how autoionization enters the model in Chapter 10.

6.5.2Where These Methods Fail

Perturbation theory assumed $c_1\approx1$ ; the golden rule assumed further that the initial state persists while the rate acts. Both fail, and the two failures define the research frontier this course points at.

Strong fields. When $$E_0$$ approaches one atomic unit ( $5.14\times10^{11}$ V/m, $I\sim10^{16}$ W/cm $^{2}$ ) the field rivals the atom's own binding and no expansion in it converges. The atom no longer absorbs one photon at a time; the barrier confining the electron is bent down until the electron tunnels through. Describing that requires the Keldysh parameter, tunnelling rates, and classical propagation of the freed electron — a subject of its own, not treated in these notes.

Decaying states. The golden rule gives a constant rate, hence exponential decay and a Lorentzian line. But if the state being driven is itself decaying into the same continuum the laser ionizes into, the two routes to that continuum interfere. The line is then not Lorentzian and no rate equation describes it: the amplitudes must be kept and their phases tracked. That is Eq. (6.3) with no perturbative expansion at all, and it is the subject of Chapters 10 and 11.

6.6Summary

The chapter reduces light–matter interaction to one working pipeline:

Start from Eq. (6.3): it is an exact rewrite of the TDSE in the stationary basis of $\hat H_0$ .
Insert the dipole interaction, Eq. (6.5), and evaluate dipole matrix elements to determine allowed channels.
Encode the driving pulse with amplitude, envelope, carrier frequency, and phase; use Eq. (6.9) to connect theory and experiment.
Use a two-level truncation near resonance to obtain Rabi oscillation and $\pi$ -pulse control (Thm. 6.4.1).
Use first-order perturbation theory for weak excitation of many states; for continua this gives Fermi's golden rule (Thm. 6.5.1).
Check validity every time: perturbation theory fails at large population transfer, and rate equations fail when coherent interference between decay paths is important.

6.7Exercises

The coupled equations. Reproduce the proof of Theorem 6.2.1, showing explicitly where the $$E_nc_n$$ terms cancel. Then explain in one sentence what would change if the phase $e^{-iE_nt/\hbar}$ had not been extracted in Eq. (6.2).
The RWA, term by term. Starting from Eq. (6.11), expand both cosines and write out all four resulting terms with their exponents. Identify which two are discarded by Def. 6.4.1 and estimate the size of the error for $\omega_0=0.4$ a.u. and $\Omega=0.01$ a.u.
Rabi, checked.
1. Verify that Eq. (6.16) satisfies $$P_2(0)=0$$ , reaches $$1$$ on resonance, and has maximum $\Omega^{2}/\Delta^{2}$ for $\Delta\gg\Omega$ .
2. Show that $$P_1+P_2=1$$ at all times, using the explicit $$c_2(t)$$ from the proof and the first equation of Eq. (6.15).
3. Repeat Example 6.4.1 for a $$100$$ fs pulse. By what factor does the required intensity change, and why?
The golden rule. Fill in the two steps of Theorem 6.5.1's proof that are quoted: the elementary integral giving the sinc form, and $\int 4\sin^{2}(\omega t/2)/\omega^{2}\,d\omega=2\pi t$ . Then explain why the rate is constant even though $$P_f(t)$$ for a single final state oscillates.
Bandwidth in practice. An XUV pulse has a Gaussian envelope with $\tau=2$ fs.
1. Using Example 6.5.1, find the spectral width in eV over which it can excite states.
2. Two states lie $$0.15$$ eV apart, both within the bandwidth and with equal dipole matrix elements. Compute the ratio of their final amplitudes if the carrier is tuned exactly to the lower one.
3. How long after the pulse do the two amplitudes return to their initial relative phase?

6.8Project: Driving a Two-Level Atom

The problem. Integrate the coupled amplitude equations numerically for a two-level atom driven by a Gaussian pulse, and map out where perturbation theory works. This project builds the tool the last chapters need: the research code is this same integration with more amplitudes.

Work in atomic units. The exact equations, with the envelope restored, are

i\dot c_1=d\,E_0\,h(t)\cos(\omega t)\,e^{-i\omega_0t}\,c_2 , \qquad i\dot c_2=d\,E_0\,h(t)\cos(\omega t)\,e^{+i\omega_0t}\,c_1 ,

with $h(t)=e^{-(t-t_0)^{2}/2\tau^{2}}$ . Take $\omega_0=0.4$ , $$d=1$$ , $\tau=200$ , and $$t_0$$ large enough that the pulse begins from nothing.

On paper. Apply the RWA by hand as in Sec. 6.4.1 and write down the resulting pair of equations.
On paper. Predict the peak field for a $\pi$ pulse from $\int\Omega(t)\,dt=\pi$ with $\Omega(t)=dE_0h(t)$ and $\int h\,dt=\tau\sqrt{2\pi}$ . This is the number the simulation must confirm.
On the computer. Integrate the exact equations from $$c_1=1$$ , $$c_2=0$$ . Plot $$P_1(t)$$ and $$P_2(t)$$ through the pulse and confirm $$P_1+P_2=1$$ ; report the actual deviation, since it is your error bar for everything else.
On the computer. Scan $$E_0$$ from far below your $\pi$ -pulse value to several times it, and plot $P_2(\infty)$ against $$E_0$$ . Overlay the first-order prediction from Example 6.5.1. Identify the field at which the two part company by more than ten per cent.
On the computer. Fix $$E_0$$ weak and scan the detuning. Plot $P_2(\infty)$ against $\Delta$ and compare the width with both $1/\tau$ and $\Omega$ . Which sets it at weak field, and which takes over as $$E_0$$ grows?
On the computer. Integrate the RWA equations alongside the exact ones and plot the difference. Where does the RWA visibly fail — in detuning, in field strength, or in the fast wiggles?
What has to move. One animation: the two populations against time with the pulse envelope drawn behind, played for increasing $$E_0$$ so the flopping speeds up. Caption it with what to watch.

The check. Part (c)'s norm conservation is the test that matters. Drift in $$P_1+P_2$$ almost always means the time grid does not resolve the optical period $2\pi/\omega$ — you must resolve the carrier, not merely the envelope.

Be ready to answer. Perturbation theory predicts $$P_2$$ growing as $E_0^{2}$ without limit, so it must eventually exceed $$1$$ . Point at the specific step in the derivation of Eq. (6.18) that allows a probability greater than one, and say what the exact calculation does that the perturbative one cannot.