Chapter 11

Time-Dependent Rydberg–Fano Dynamics in CO $$_2$$

11.1Introduction

This is where the book arrives.

Chapter 10 gave the Fano profile: the asymmetric line produced when a resonance and a direct route lead to the same continuum. That result is stationary. It describes the spectrum obtained after waiting long enough for everything to settle, and says nothing about how the profile came to be.

But the profile is not present at the instant of excitation. It builds up over the autoionization lifetime, while the Rydberg population that feeds it decays. With attosecond pulses that construction can be watched, and watching it requires abandoning the stationary picture and keeping the amplitudes and their phases — Chapter 6's coupled equations, with no perturbative expansion.

The chapter sets up the channel basis, derives the coupled equations by projection, states the approximations, and says what the calculation predicts. The equations of Sec. 11.4 are those of the research this course prepares for.(E. Saydanzad, V.-H. Hoang and U. Thumm, Time-resolved Rydberg dynamics in molecules, validated against the photoemission experiment of Biswas et al., Phys. Rev. A 110, 043106 (2024).) They look formidable; they are not. Every symbol has been defined in an earlier chapter, and the whole system is Theorem 6.2.1 with the terms written out.

We work in atomic units throughout, so $\hbar=1$ .

11.2The Physical Setup

An attosecond XUV pulse, centred near $$17.6$$ eV with bandwidth about $$0.4$$ eV, illuminates CO $$_2$$ in its ground state $X^{1}\Sigma_g^{+}$ . A near-infrared pulse follows after a controllable delay $$t_d$$ . Photoelectrons are collected and their energies recorded as a function of that delay.

Three things can happen, and the model is an accounting of them.

Pathway 1 — direct ionization. The XUV photon ionizes immediately, leaving CO $_2^{+}$ in $X^{2}\Pi_g$ (threshold $$13.78$$ eV, Example 7.5.2) plus a photoelectron. Since $$17.6>13.78$$ , this channel is always open.

Pathway 2 — resonant excitation and autoionization. The same photon instead excites a Rydberg state of the $(3\sigma_u)^{-1}nl\sigma_g$ series, converging to $B^{2}\Sigma_u^{+}$ at $$18.08$$ eV (Chapter 9). These states sit near $$17.5$$ eV, above the $$X$$ threshold, so they autoionize into the same continuum that pathway 1 reaches.

Pathways 1 and 2 end in the same final state. They interfere. That is Chapter 10's Fano problem, now happening in real time.

Pathway 3 — the probe. The delayed infrared pulse ionizes whatever Rydberg population survives, into the $B^{2}\Sigma_u^{+}$ continuum. No XUV route reaches this channel, so it is a clean measurement of the Rydberg population — and, because several Rydberg states were excited coherently by the broadband pump, of the phases between them.

That is why the experiment uses two colours. The coupled channel ( $\mathcal C$ , here $$X$$ ) shows interference between direct and resonant ionization; the uncoupled channel ( $\mathcal N$ , here $$B$$ ) shows quantum beats between coherently excited Rydberg states.

11.3The Channel Basis

Expand the molecular state in the available states, exactly as Theorem 10.2.1 prescribes — a sum over discrete states and an integral over each continuum — with the free-evolution phases extracted as in Chapter 6:

Following the notation of the manuscript Supplementary, $\mathcal C$ labels continua coupled to the Rydberg manifold by configuration interaction, and $\mathcal N$ labels continua that are not. For CO $$_2$$ in this chapter, $\mathcal C\equiv X^{2}\Pi_g$ and $\mathcal N\equiv B^{2}\Sigma_u^{+}$ .

\begin{equation} \label{eq:CO2-expansion} \boxed{ \begin{aligned} \ket{\Psi(t)} =\;& c_g(t)\,e^{-iE_gt}\ket{g} +\sum_n c_{\mathcal R_n}(t)\,e^{-iE_{\mathcal R_n}t}\ket{\mathcal R_n}\\ &+\int d\varepsilon_\alpha\, c_{\mathcal C}(t,\varepsilon_\alpha)\,e^{-i(E_{\mathcal C}+\varepsilon_\alpha)t} \ket{\mathcal C,\varepsilon_\alpha}\\ &+\int d\varepsilon_\beta\, c_{\mathcal N}(t,\varepsilon_\beta)\,e^{-i(E_{\mathcal N}+\varepsilon_\beta)t} \ket{\mathcal N,\varepsilon_\beta} \end{aligned}} \end{equation}

Amplitude	State	Introduced in
$$c_g$$	neutral ground state $X^{1}\Sigma_g^{+}$	Ch. 7
$c_{\mathcal R_n}$	Rydberg state $(3\sigma_u)^{-1}nl\sigma_g$	Ch. 9
$c_{\mathcal C}(t,\varepsilon_\alpha)$	coupled continuum (here $X^{2}\Pi_g$ ) + electron energy $\varepsilon_\alpha$	Ch. 9, 10
$c_{\mathcal N}(t,\varepsilon_\beta)$	uncoupled continuum (here $B^{2}\Sigma_u^{+}$ ) + electron energy $\varepsilon_\beta$	Ch. 9, 10

The two continua differ in one respect: $\mathcal C$ is coupled to the Rydberg series by configuration interaction — the states autoionize into it — and $\mathcal N$ is not. That is not an assumption but a consequence of energy: the Rydberg states lie below the $$B$$ threshold, so they cannot decay into it, only be driven into it by absorbing an infrared photon.

Because the expansion is over a complete orthonormal set, probability is conserved exactly:

\begin{equation} \label{eq:CO2-normalization} |c_g|^{2}+\sum_n|c_{\mathcal R_n}|^{2} +\int d\varepsilon_\alpha|c_{\mathcal C}|^{2} +\int d\varepsilon_\beta|c_{\mathcal N}|^{2}=1 . \end{equation}

This is the check to apply to any numerical solution (Remark 6.2.3).

11.3.1The Hamiltonian

\begin{equation} \label{eq:CO2-hamiltonian} \hat H(t)=\hat K_N+\hat V_{NN}+\hat H_e+\hat D(t), \end{equation}

with $\hat K_N$ and $\hat V_{NN}$ from Eq. (7.2). The essential step is to split the electronic Hamiltonian according to whether it acts within a channel or between channels. Introducing projectors $\hat P_a$ onto the individual channels,

\begin{equation} \label{eq:CO2-Hsplit} \hat H_e=\hat H_{\mathrm{ch}}+\hat V_{\mathrm{CI}}, \qquad \hat H_{\mathrm{ch}}=\sum_a\hat P_a\hat H_e\hat P_a, \qquad \hat V_{\mathrm{CI}}=\sum_{a\neq b}\hat P_a\hat H_e\hat P_b . \end{equation}

$\hat H_{\mathrm{ch}}$ gives each channel its energies. $\hat V_{\mathrm{CI}}$ — the configuration interaction — collects the off-diagonal blocks and couples the Rydberg configurations to the open $\mathcal C$ continuum. It is the autoionization coupling of Chapter 10, Eq. (10.4), under its research name.

The laser term is Chapter 6's dipole interaction with two pulses:

\begin{equation} \label{eq:CO2-dipole} \hat D(t)=\hat D_{\mathrm{XUV}}\,h_{\mathrm{XUV}}(t)\, e^{i(\omega_{\mathrm{XUV}}t+\phi_{\mathrm{XUV}})} +\hat D_{\mathrm{IR}}\,h_{\mathrm{IR}}(t-t_d)\, e^{i(\omega_{\mathrm{IR}}t+\phi_{\mathrm{IR}})} +\mathrm{h.c.}, \end{equation}

with envelopes $$h$$ , carrier frequencies $\omega$ , carrier phases $\phi$ , and the pump–probe delay $$t_d$$ appearing in exactly one place — the argument of the infrared envelope. That is the knob the experiment turns.

11.4Deriving the Coupled Equations

The derivation is Theorem 6.2.1 applied to Eq. (11.1), and it is worth following once because the result looks more complicated than the procedure.

Step 1: project. Substituting Eq. (11.1) into $i\partial_t\ket{\Psi}=\hat H\ket{\Psi}$ and projecting onto each basis state gives, exactly as in the proof of Theorem 6.2.1, one equation per amplitude. Each term carries three factors: the amplitude it comes from, the matrix element connecting the two states, and a phase $e^{i(E_{\mathrm{final}}-E_{\mathrm{initial}})t}$ .

Step 2: identify the matrix elements. Only three kinds are nonzero, and each was established earlier:

$\bra{g}\hat D_{\mathrm{XUV}}\ket{\mathcal R_n}$ and $\bra{g}\hat D_{\mathrm{XUV}}\ket{\mathcal C,\varepsilon_\alpha}$ — XUV excitation and direct ionization, both allowed by Example 9.7.1;
$\bra{\mathcal R_n}\hat D_{\mathrm{IR}}\ket{\mathcal N,\varepsilon_\beta}$ — infrared probe ionization;
$\bra{\mathcal C,\varepsilon_\alpha}\hat V_{\mathrm{CI}}\ket{\mathcal R_n} =\sqrt{\Gamma_{nn}/2\pi}$ — autoionization, from Eq. (10.9).

Step 3: apply the rotating-wave approximation to the XUV pulse (Def. 6.4.1), discarding the counter-rotating terms.

Step 4: eliminate the coupled continuum. The $\mathcal C$ continuum appears in the Rydberg equation through the autoionization coupling. Solving the $c_{\mathcal C}$ equation formally and substituting back — a Laplace transform, with the Markov approximation that the coupling is flat across the narrow resonance — converts that coupling into two terms: a decay $-i\Gamma_{nn}c_{\mathcal R_n}/2$ and a principal-value shift, which is absorbed into the measured resonance energy exactly as $$F(E)$$ was in Sec. 10.4. This is the Markov–Fano reduction, and it is where $\Gamma$ enters the time domain.

The result is four equations.

The ground state loses population to the Rydberg states and to the $\mathcal C$ continuum:

\begin{equation} \label{eq:CO2-cg} \begin{aligned} i\dot c_g =\;&\sum_n c_{\mathcal R_n}\,h_{\mathrm{XUV}}(t)\, e^{i\phi_{\mathrm{XUV}}}\, e^{-i(E_{\mathcal R_n}-E_g-\omega_{\mathrm{XUV}})t}\, \bra{g}\hat D_{\mathrm{XUV}}\ket{\mathcal R_n}\\ &+\int d\varepsilon_\alpha\, c_{\mathcal C}(t,\varepsilon_\alpha)\,h_{\mathrm{XUV}}(t)\, e^{i\phi_{\mathrm{XUV}}}\, e^{-i(E_{\mathcal C}+\varepsilon_\alpha-E_g-\omega_{\mathrm{XUV}})t}\, \bra{g}\hat D_{\mathrm{XUV}}\ket{\mathcal C,\varepsilon_\alpha} . \end{aligned} \end{equation}

Each Rydberg amplitude is pumped from the ground state, decays by autoionization, and is depleted by the probe:

\begin{equation} \label{eq:CO2-cR} \begin{aligned} i\dot c_{\mathcal R_n} +i\frac{\Gamma_{nn}}{2}c_{\mathcal R_n} =\;&\rho_n\,c_g\,h_{\mathrm{XUV}}(t)\, e^{-i\phi_{\mathrm{XUV}}}\, e^{i(E_{\mathcal R_n}-E_g-\omega_{\mathrm{XUV}})t}\\ &+\int d\varepsilon_\beta\, c_{\mathcal N}(t,\varepsilon_\beta)\,h_{\mathrm{IR}}(t-t_d)\, e^{i\phi_{\mathrm{IR}}}\, e^{-i(E_{\mathcal N}+\varepsilon_\beta-E_{\mathcal R_n}-\omega_{\mathrm{IR}})t}\, \bra{\mathcal R_n}\hat D_{\mathrm{IR}}\ket{\mathcal N,\varepsilon_\beta} . \end{aligned} \end{equation}

The coupled continuum receives amplitude from both routes:

\begin{equation} \label{eq:CO2-cX} \begin{aligned} i\dot c_{\mathcal C}(t,\varepsilon_\alpha) =\;& c_g\,h_{\mathrm{XUV}}(t)\,e^{-i\phi_{\mathrm{XUV}}}\, e^{i(E_{\mathcal C}+\varepsilon_\alpha-E_g-\omega_{\mathrm{XUV}})t}\, \bra{g}\hat D_{\mathrm{XUV}}\ket{\mathcal C,\varepsilon_\alpha}\\ &+\sum_n c_{\mathcal R_n}\, e^{i(E_{\mathcal C}+\varepsilon_\alpha-E_{\mathcal R_n})t}\, \sqrt{\frac{\Gamma_{nn}}{2\pi}} . \end{aligned} \end{equation}

The uncoupled continuum is fed only by the probe:

\begin{equation} \label{eq:CO2-cB} i\dot c_{\mathcal N}(t,\varepsilon_\beta) =\sum_n c_{\mathcal R_n}\,h_{\mathrm{IR}}(t-t_d)\, e^{-i\phi_{\mathrm{IR}}}\, e^{-i(E_{\mathcal R_n}-E_{\mathcal N}-\varepsilon_\beta+\omega_{\mathrm{IR}})t}\, \bra{\mathcal R_n}\hat D_{\mathrm{IR}}\ket{\mathcal N,\varepsilon_\beta} . \end{equation}

11.4.1Reading the Terms

Four observations account for everything above.

Remark 11.4.1 - Every coupling is envelope

\times

phase

\times

dipole

That is the structure of Eq. (6.3): $$h(t)$$ says when the pulse is present, the exponential carries the energy mismatch, and the matrix element says how strongly the states are connected. By Remark 6.2.2 a term matters only when its exponent is small — when the photon energy nearly matches the transition.

Remark 11.4.2 - The decay term is not from the laser

The term $i\Gamma_{nn}c_{\mathcal R_n}/2$ on the left of Eq. (11.7) carries no envelope and no dipole. It is autoionization, present whether or not any light is on, and it makes the effective Hamiltonian non-Hermitian in the Rydberg block. The factor $\tfrac12$ is because amplitudes decay at half the rate of populations: $|c|^{2}\sim e^{-\Gamma t}$ requires $c\sim e^{-\Gamma t/2}$ .

Remark 11.4.3 - The two ends of the same process

The second term of Eq. (11.8) is $\sqrt{\Gamma_{nn}/2\pi}$ — Eq. (10.9) exactly. It and the decay term of Remark 11.4.2 are the same physics viewed from opposite ends: population leaving the Rydberg state arrives in the continuum. Together they conserve Eq. (11.2).

Remark 11.4.4 - Interference lives in Eq. (11.8)

Its two terms add amplitudes into the same final state $\ket{\mathcal C,\varepsilon_\alpha}$ . Squaring the sum produces a cross term, and that cross term is the Fano asymmetry — now time-dependent, because the two contributions arrive on different schedules. The direct term follows the pump envelope and stops when the pulse does; the resonant term keeps arriving for as long as the Rydberg states take to decay.

11.4.2The Effective Pump Amplitude

One symbol in Eq. (11.7) remains: $\rho_n$ , the effective strength with which the XUV pumps the Rydberg state. It is not simply the bound dipole, because the Markov elimination of the continuum folds the direct route into the resonant one. The result is

\begin{equation} \label{eq:CO2-rho} \boxed{\; \rho_n=(q_n-i)\sqrt{\frac{\pi\Gamma_{nn}}{2}}\; \bra{g}\hat D_{\mathrm{XUV}}\ket{\mathcal C,\varepsilon_\alpha} \Big|_{\varepsilon_\alpha=E_{\mathcal R_n}-E_{\mathcal C}}\;} \end{equation}

evaluated at the on-shell photoelectron energy — the energy the electron must carry for energy to be conserved when the resonance decays.

Chapter 10's asymmetry parameter appears here in the combination $$(q_n-i)$$ , and that complex factor is the whole time-dependent Fano effect in one bracket. The real part is resonant excitation; the imaginary part is interference with the direct route; and $$q_n$$ measures their ratio, exactly as in Eq. (10.18). Setting $q_n\to\infty$ leaves the resonance alone; setting $$q_n=0$$ leaves only the interference term, which is the window resonance of Chapter 10.

11.5The Approximations

A model is only as good as its stated assumptions. Six are made, and you should now be able to judge every one.

Two channels. One coupled continuum ( $\mathcal C$ , here $X^{2}\Pi_g$ ) and one uncoupled ( $\mathcal N$ , here $B^{2}\Sigma_u^{+}$ ); the $$A$$ and $$C$$ ionic states of Example 7.5.2 are dropped. Justified because the experiment resolves photoelectron energy and can separate the channels.
One vibrational level and one partial wave per channel. Nuclear motion is frozen — the timescale argument of Sec. 7.10: the pump is a few femtoseconds, nuclear motion takes tens. This is the weakest assumption for the longest-lived resonances, which survive for picoseconds.
Configuration interaction restricted to the resonant coupling. Only $\bra{\mathcal R_n}\hat V_{\mathrm{CI}}\ket{\mathcal C,\varepsilon_\alpha}$ is kept, set to $\sqrt{\Gamma_{nn}/2\pi}$ ; continuum–continuum interactions are dropped.
Independent resonances. Cross-widths $\Gamma_{nm}=0$ for $m\neq n$ , so each state decays at its own rate without coherent mixing through the shared continuum. Exercise E3 tests this.
No infrared coupling within the Rydberg manifold. $\bra{\mathcal R_m}\hat D_{\mathrm{IR}}\ket{\mathcal R_n}=0$ , the parity argument of Example 9.7.1: both series are built on the same $3\sigma_u$ hole with $$g$$ Rydberg orbitals, so both have total parity $$u$$ , and one photon cannot connect two states of equal parity. Not an approximation so much as a symmetry.
Predissociation neglected. The sharp states can also fall apart into neutral fragments, competing with autoionization. Omitting it makes the calculated autoionization yields upper bounds.

Assumption (v) illustrates what this course was for. It removes an entire block of the Hamiltonian and costs nothing — no fitting, no parameter, no error — because it follows from the inversion symmetry of a linear centrosymmetric molecule. A symmetry argument proved in Chapter 3, applied to labels earned in Chapter 7, simplifies a research calculation in Chapter 11.

11.6What the Model Predicts

Solving Eqs. (11.6)–(11.9) for each delay $$t_d$$ gives $|c_{\mathcal C}(\infty,\varepsilon_\alpha)|^{2}$ and $|c_{\mathcal N}(\infty,\varepsilon_\beta)|^{2}$ — the photoelectron spectrum in each channel as a function of delay. The two observables answer different questions.

The coupled channel (C, here X): a Fano profile assembling itself

Both pathways reach this continuum, so its spectrum carries the interference. Early in the pulse only the direct route has contributed and the spectrum is smooth. As the Rydberg states decay — over hundreds of femtoseconds for the sharp series, tens for the diffuse — the resonant amplitude accumulates and the asymmetric Fano profile of Chapter 10 grows in where a featureless background used to be.

The Fano line shape is not a property a state has; it is something that happens, on the timescale $\tau=\hbar/\Gamma$ .

Chapter 10's stationary formula, Eq. (10.16), is the $t\to\infty$ limit of this.

The uncoupled channel (N, here B): quantum beats

No XUV route reaches this continuum, so its population is proportional to the Rydberg population surviving when the probe arrives. The broadband pump excited several Rydberg states coherently (Example 6.5.2), and their amplitudes carry phases advancing at different rates, $e^{-iE_{\mathcal R_n}t}$ . Any pair therefore beats at

\begin{equation} \label{eq:CO2-beat} \omega_{nm}=E_{\mathcal R_n}-E_{\mathcal R_m}, \end{equation}

and the $\mathcal N$ -channel yield (here the $$B$$ channel) oscillates with $$t_d$$ at that frequency, superposed on the overall exponential decay.

Fourier transforming the delay dependence therefore recovers the level spacings directly. Because the photoelectron energy is resolved as well, pairs whose beat frequencies nearly coincide but whose electron energies differ can be told apart — which is what makes the technique more than a lifetime measurement.

Example 11.6.1 - Which beats are observable

Two members of the sharp series lie at $$17.5395$$ eV ( $$5d$$ , $\tau=372.6$ fs) and $$17.7018$$ eV ( $$6d$$ , $\tau=1457.5$ fs). Find the beat period and decide whether it can be seen.

Solution.

The spacing is

\Delta E=17.7018-17.5395=0.1623~\mathrm{eV},

and with $2\pi\hbar=4.136$ eV fs the beat period is

T=\frac{2\pi\hbar}{\Delta E}=\frac{4.136}{0.1623}=25.5~\mathrm{fs}.

Now the question that decides whether the experiment works. The beating survives only as long as both states do, so the shorter lifetime governs:

\frac{372.6}{25.5}\approx15~\text{beat periods}.

Fifteen oscillations before the envelope decays is ample — the Fourier transform will show a clean, narrow peak.

Contrast the diffuse series, where the $$5s$$ state lives $$15.9$$ fs. Its beats against a neighbour would be damped within a single period, so no oscillation is observable at all: that channel reports a decay and nothing more.

The same experiment therefore measures a frequency for one series and only a lifetime for the other, and the difference traces back through $\Gamma\tau=\hbar$ to states differing only in $$l$$ — that is, to the centrifugal barrier of Chapter 3.

11.7Summary

An XUV pump reaches the coupled continuum ( $\mathcal C\equiv X^{2}\Pi_g$ ) both directly and through the $(3\sigma_u)^{-1}nl\sigma_g$ Rydberg states, which autoionize into it; those two routes interfere. A delayed infrared probe ionizes the surviving Rydberg population into the uncoupled continuum ( $\mathcal N\equiv B^{2}\Sigma_u^{+}$ ), which no XUV route reaches.
The state is expanded as Eq. (11.1): two discrete kinds of amplitude and two continuum integrals, with probability conserved by Eq. (11.2).
$\hat H_e$ splits into channel-diagonal blocks and the configuration interaction $\hat V_{\mathrm{CI}}$ , which is the autoionization coupling of Chapter 10.
Projecting, applying the RWA and eliminating the coupled continuum in the Markov–Fano reduction gives Eqs. (11.6)–(11.9): Chapter 6's $i\dot c_k=\sum_nc_nV_{kn}e^{i\omega_{kn}t}$ written out. Every coupling is envelope $\times$ phase $\times$ dipole; the decay term $i\Gamma_{nn}/2$ carries neither and is autoionization.
Interference sits in Eq. (11.8), whose two terms add amplitudes into one final state. The complex factor $$(q_n-i)$$ in $\rho_n$ carries the balance of the two routes.
Of the six approximations, the one removing most work — no infrared coupling within the Rydberg manifold — is a parity argument, not a fit.
The $\mathcal C$ channel (here $$X$$ ) shows the Fano profile building up over $\tau=\hbar/\Gamma$ : the line shape is an event, not a property. The $\mathcal N$ channel (here $$B$$ ) shows quantum beats at the Rydberg level spacings, Eq. (11.11), so a Fourier transform of the delay dependence returns the spectrum of the wave packet.

11.8Exercises

Account for every term. Take Eq. (11.7). For each of its three terms state (a) which physical process it represents, (b) which pulse if any drives it, and (c) which earlier chapter defined the quantity in it. Then explain why the second term contains $c_{\mathcal N}$ rather than $c_{\mathcal C}$ .
Beats and lifetimes. Repeat Example 11.6.1 for $$7d$$ ( $$17.8000$$ eV, $\tau=3378.5$ fs) and $$8d$$ ( $$17.8650$$ eV, $\tau=1485.2$ fs). Give the beat period and the number of observable periods. Then find the pair from Chapter 10's table whose beat would be hardest to observe, and say why.
Testing assumption (iv). Independent resonances requires that neighbouring resonances not overlap: widths small compared with spacing.
1. Using $\Gamma_n\propto(n^{*})^{-3}$ (Thm. 10.6.1) and the Rydberg level spacing $\Delta E\propto(n^{*})^{-3}$ (Eq. (4.61)), show that the ratio $\Gamma/\Delta E$ is independent of $n^{*}$ .
2. Evaluate that ratio for the $$5d$$ state using its width and its spacing to $$6d$$ .
3. Comment: does the assumption get better or worse as $$n$$ increases, and why is that answer surprising?
The pump amplitude. From Eq. (11.10), compute $|\rho_n|$ in terms of $$q_n$$ and $\Gamma_{nn}$ , and find the phase of $\rho_n$ . What happens to both as $q_n\to\infty$ , and what physical situation does that limit describe?
Norm conservation. Show analytically that with the probe off and the pump over,
$\frac{d}{dt}\left[\sum_n|c_{\mathcal R_n}|^{2} +\int d\varepsilon_\alpha|c_{\mathcal C}|^{2}\right]=0 .$
using Eqs. (11.7) and (11.8). Identify which two terms cancel, and relate the result to Remark 11.4.3.

11.9Project: Time-Resolved Fano Buildup

The problem. Solve a reduced version of Eqs. (11.6)–(11.9) numerically and watch both signatures appear: the Fano profile assembling itself in the coupled channel ( $\mathcal C$ , here $$X$$ ), and quantum beats in the uncoupled channel ( $\mathcal N$ , here $$B$$ ). This is the capstone of the course, and a small version of the research code.

Reduce as follows: keep the ground state, two Rydberg states (so there is something to beat), and both continua on energy grids. Take the pump weak enough that $c_g\approx1$ throughout, which removes Eq. (11.6).

Use real parameters: $E_{\mathcal N}=E_B=18.08$ eV, and for the two Rydberg states the $$5d$$ and $$6d$$ values — energies $$17.5395$$ and $$17.7018$$ eV, lifetimes $$372.6$$ and $$1457.5$$ fs, infrared dipoles $$1.000$$ and $$0.765$$ , $q_{5d}=2.00$ and $q_{6d}=3.69$ . Take a Gaussian XUV pulse of $$0.4$$ eV bandwidth centred at $$17.6$$ eV and a longer infrared pulse. Work in atomic units.

On paper. Write the reduced equations explicitly for two Rydberg states with $$c_g=1$$ . Confirm by inspection that with the probe off each $c_{\mathcal R_n}$ decays as $e^{-\Gamma_{nn}t/2}$ once the pump has passed, so $|c_{\mathcal R_n}|^{2}$ decays at $\Gamma_{nn}$ .
On paper. Convert every parameter to atomic units and state the time step you will need. It must resolve the fastest phase in the problem — identify which exponent in Eqs. (11.7)–(11.9) that is.
On the computer. With the probe off, integrate the $c_{\mathcal R_n}$ and $c_{\mathcal C}$ equations and plot the $\mathcal C$ -channel spectrum $|c_{\mathcal C}(t,\varepsilon_\alpha)|^{2}$ at several times: during the pulse, at $\tau/4$ , $\tau$ and $4\tau$ of the shorter-lived state. The asymmetric profile should grow out of a smooth background. Compare the final spectrum with Eq. (10.16) using the same $$q$$ and $\Gamma$ .
On the computer. Verify that $\sum_n|c_{\mathcal R_n}|^{2}+\int d\varepsilon_\alpha|c_{\mathcal C}|^{2}$ is conserved once the pump is over — the analytic result of Exercise E5. Report the drift; this tests whether your continuum grid is dense and wide enough, and you should state what you needed.
On the computer. Switch the probe on and scan $$t_d$$ over several hundred femtoseconds. Plot the total $\mathcal N$ -channel yield $\int d\varepsilon_\beta|c_{\mathcal N}(\infty,\varepsilon_\beta)|^{2}$ against $$t_d$$ : exponential decay with oscillations superposed.
On the computer. Fourier transform that delay trace and identify the beat frequency. Compare with $E_{6d}-E_{5d}$ from Eq. (11.11) and with the $$25.5$$ fs period of Example 11.6.1.
What has to move. Two animations. First, the $\mathcal C$ -channel spectrum (here $$X$$ ) against time with the stationary Fano profile drawn behind as a dashed target — the moving curve should converge onto it. Second, the two Rydberg populations and the $\mathcal N$ -channel yield (here $$B$$ ) as the delay is scanned, so the beating is visible as it is measured.

The check. Part (c) is the one to get right and part (d) is how you know you have. If the built-up profile does not converge to Eq. (10.16), the usual causes are a continuum grid too narrow to hold the resonance wings, or a grid spacing coarser than $\Gamma$ . Both show up first as drift in (d).

Be ready to answer. Your part (c) spectra converge to the stationary Fano profile, which Chapter 10 derived without any reference to time. Explain what the time-dependent calculation tells you that the stationary one cannot — and identify one experimental observable that would be identical in both descriptions, and one that would not.