Chapter 10

Rydberg and Continuum States; Fano Resonances

10.1Introduction

Every chapter so far has been preparation for this one.

We have a molecule (Chapter 7) with several ionization thresholds (Chapter 9), supporting Rydberg series that converge to the higher ones and therefore lie above the lower ones. Such a state is bound to one ionic core and degenerate with the continuum of another, and the electron–electron repulsion connects them. The necessary tools are now in place: continuum normalization (Chapter 3), Fermi's golden rule (Chapter 6), and the $(n^{*})^{- 3}$ scaling of amplitude near the core (Chapter 4).

This chapter produces three things:

the width $Γ$ of an autoionizing state and hence its lifetime;
the Fano profile, the asymmetric line shape produced when a resonant and a direct route lead to the same final state, together with the asymmetry parameter $q$ ;
the scaling laws that parameterize a whole Rydberg series from measured level positions and a single anchor.

By the end of the chapter, numbers from a published research table are reproduced by hand. Chapter 11 then makes the whole thing time-dependent.

10.2Continuum Normalization

A bound state is normalizable, $⟨ ψ | ψ ⟩ = 1$ ; a free electron is not. Something must replace normalization, and the choice matters because factors of $2 π$ and densities of states hide inside it.

Definition 10.2.1Energy normalization

Continuum states labeled by energy are normalized to a Dirac delta,

\begin{matrix} ⟨ ε | ε^{'} ⟩ = δ (ε - ε^{'}) . \end{matrix}

Two consequences are used constantly.

Theorem 10.2.1Mixed completeness

For a system with both bound and free states,

\begin{matrix} | Ψ ⟩ = \sum_{n} c_{n} | n ⟩ + \int d ε c (ε) | ε ⟩, \sum_{n} | c_{n} |^{2} + \int d ε | c (ε) |^{2} = 1 . \end{matrix}

Equation (10.2) is the structure of the state vector in Chapter 11, and Def. 10.2.1 is what makes the two terms dimensionally compatible: $| c (ε) |^{2}$ is a probability density per unit energy, while $| c_{n} |^{2}$ is a probability.

The second consequence is Remark 6.5.1 of Chapter 6: the density of states is absorbed into the states themselves, so in atomic units Fermi's golden rule reduces to

\begin{matrix} Γ = 2 π | ⟨ ε | \hat{V} | i ⟩ |^{2} . \end{matrix}

This is the reason for the convention, since it removes one factor that would otherwise have to be tracked.

10.3A Discrete State Coupled to a Continuum

Here is the model, stripped to essentials. Two ingredients: a discrete state $| a ⟩$ at energy $E_{a}$ — an autoionizing Rydberg state — and a continuum $| ε ⟩$ spanning $E_{a}$ , normalized as in Def. 10.2.1. Neither is an eigenstate of the full Hamiltonian, because a coupling connects them:

\begin{matrix} V_{ε} = ⟨ ε | \hat{H} | a ⟩ . \end{matrix}

Physically $\hat{H}$ here is the electron–electron repulsion of Chapter 5, called configuration interaction in this context.

The model Hamiltonian is therefore

\begin{matrix} \hat{H} = E_{a} | a ⟩ ⟨ a | + \int d ε ε | ε ⟩ ⟨ ε | + \int d ε [V_{ε} | ε ⟩ ⟨ a | + V_{ε}^{*} | a ⟩ ⟨ ε |] . \end{matrix}

Theorem 10.3.1Autoionization width

Because the coupling is to a continuum rather than to a single state, $| a ⟩$ does not shift and split — it decays, with

\begin{matrix} Γ = 2 π | V_{ε} |^{2} |_{ε = E_{a}} \end{matrix}

and lifetime

\begin{matrix} τ = \frac{ℏ}{Γ}, i.e. Γ τ = ℏ . \end{matrix}

Apply Fermi's golden rule in the form Eq. (10.3), with $\hat{V}$ the coupling term of Eq. (10.5) and the final states the continuum at the same energy. Remark 6.5.2 of Chapter 6 established that the golden rule holds for any coupling, not only a radiative one.

In practical units, Eq. (10.7) reads

\begin{matrix} Γ [meV] \times τ [fs] = 658 \end{matrix}

so a state of width $37$ meV lives $18$ fs (helium's $2 s 2 p$ resonance from Chapter 5), and a state living a picosecond has a width under a microelectronvolt. Widths and lifetimes are the same information.

It is often convenient to read Eq. (10.6) backwards: the width is what experiments measure, the coupling is what a calculation needs. Taking the coupling real and constant across the narrow resonance,

\begin{matrix} ⟨ ε | \hat{H} | a ⟩ = \sqrt{\frac{Γ}{2 π}} \end{matrix}

This expression appears verbatim in the research equations of Chapter 11.

10.4Fano's Diagonalization

Now solve Eq. (10.5) exactly. The eigenstates of the full Hamiltonian are not $| a ⟩$ and $| ε ⟩$ but mixtures of them. Write the eigenstate at energy $E$ as

\begin{matrix} | Ψ_{E} ⟩ = α (E) | a ⟩ + \int d ε β_{E} (ε) | ε ⟩, \end{matrix}

and impose $\hat{H} | Ψ_{E} ⟩ = E | Ψ_{E} ⟩$ .

Projecting onto $⟨ a |$ and onto $⟨ ε |$ gives two equations:

\begin{aligned} E_{a} α + \int d ε V_{ε}^{*} β_{E} (ε) & = E α, \\ ε β_{E} (ε) + V_{ε} α & = E β_{E} (ε) . \end{aligned}

Equation (10.12) rearranges to $(E - ε) β_{E} (ε) = V_{ε} α$ , which cannot simply be divided through because $E - ε$ vanishes somewhere in the integration range. The correct general solution of such an equation adds a homogeneous piece proportional to $δ (E - ε)$ and interprets the quotient as a Cauchy principal value:

\begin{matrix} β_{E} (ε) = [𝒫 \frac{1}{E - ε} + z (E) δ (E - ε)] V_{ε} α, \end{matrix}

with $z (E)$ fixed by substituting back into Eq. (10.11):

\begin{matrix} z (E) = \frac{E - E_{a} - F (E)}{| V_{E} |^{2}}, F (E) = 𝒫 \int d ε \frac{| V_{ε} |^{2}}{E - ε} . \end{matrix}

The function $F (E)$ is a level shift: it displaces the resonance from $E_{a}$ to $E_{a} + F$ , and in practice it is absorbed into the measured resonance position. What remains is the interesting part.

Theorem 10.4.1The modified discrete state

The admixture of the discrete state in the exact eigenstate at energy $E$ is

\begin{matrix} | α (E) |^{2} = \frac{| V_{E} |^{2}}{[E - E_{a} - F (E)]^{2} + π^{2} | V_{E} |^{4}} = \frac{Γ / 2 π}{(E - E_{r})^{2} + Γ^{2} / 4}, \end{matrix}

a Lorentzian of full width $Γ = 2 π | V_{E} |^{2}$ centered on the shifted position $E_{r} = E_{a} + F$ .

The discrete state is therefore not destroyed — it is spread over a range of exact eigenstates of width $Γ$ . That is the spectral statement of a finite lifetime, and Eq. (10.15) is where Theorem 10.3.1 comes from in an exact treatment rather than a perturbative one.

10.5The Fano Profile

Now add the laser. Light can reach the continuum from the ground state $| g ⟩$ in two ways:

directly, $| g ⟩ \to | ε ⟩$ , with amplitude $⟨ ε | \hat{D} | g ⟩$ ;
resonantly, $| g ⟩ \to | a ⟩ \to | ε ⟩$ , via the discrete state, which then autoionizes.

Both end in the same final state at the same energy, so they are indistinguishable and their amplitudes add before squaring. The cross term is the entire Fano effect.

The transition rate is $| ⟨ Ψ_{E} | \hat{D} | g ⟩ |^{2}$ , and from Eq. (10.10) the matrix element has one piece from each route. Carrying the algebra through with Eqs. (10.13) and (10.14) gives Fano's 1961 result.

Theorem 10.5.1Fano profile

\begin{matrix} σ (E) = σ_{b} \frac{(q + ϵ)^{2}}{1 + ϵ^{2}} \end{matrix}

where $σ_{b}$ is the background cross section from the direct route alone,

\begin{matrix} ϵ = \frac{E - E_{r}}{Γ / 2} = \frac{2 (E - E_{r})}{Γ} \end{matrix}

is the detuning in units of the half-width, and

\begin{matrix} q = \frac{⟨ \tilde{a} | \hat{D} | g ⟩}{π V_{E}^{*} ⟨ ε | \hat{D} | g ⟩} \end{matrix}

is the asymmetry parameter: the ratio of the amplitude for exciting the (shifted, “dressed”) discrete state $| \tilde{a} ⟩$ to that for direct ionization.

The formula is most easily understood by taking limits.

Far from resonance ( $ϵ \to \pm \infty$ ), $σ \to σ_{b}$ : the undisturbed background.
There is an exact zero at $ϵ = - q$ . The two amplitudes cancel completely at one energy. Absorption does not merely dip — it vanishes. A Lorentzian has no such feature.
At line center ( $ϵ = 0$ ) the ratio to background is $q^{2}$ . This is the practical handle: a measured peak-to-background ratio at the resonance position gives $q^{2}$ directly.
The true maximum sits slightly off center. Differentiating Eq. (10.16),
$\frac{d}{d ϵ} \frac{(q + ϵ)^{2}}{1 + ϵ^{2}} = \frac{2 (q + ϵ) (1 - q ϵ)}{(1 + ϵ^{2})^{2}},$
which vanishes at $ϵ = 1 / q$ , where the ratio is $1 + q^{2}$ .

Two limits of $q$ deserve names:

$| q | \to \infty$ : the resonant route dominates, the direct one is negligible, and Eq. (10.16) reduces to a Lorentzian peak. No interference, no asymmetry.
$q = 0$ : the discrete state is not excited at all, yet it still autoionizes and so still interferes. The profile becomes $ϵ^{2} / (1 + ϵ^{2})$ — a dip to zero with no peak, a “window resonance”.

So $q$ interpolates between a peak and a hole. Nothing about the state changes; only the balance of the two routes to it.

Remark 10.5.1Asymmetry means interference

A symmetric line — Lorentzian or Gaussian — can always be read as a rate: something decays, and the spectrum is the Fourier transform of an exponential. An asymmetric line cannot. Asymmetry requires an amplitude whose phase varies across the resonance, interfering with something else. An asymmetric profile in a spectrum is therefore a signature of interference, and the asymmetry measures a relative phase.

Example 10.5.1Extracting

Γ

and

q

from a measured line

A photoabsorption line shows a minimum at $59.9$ eV and a maximum at $60.2$ eV, with the peak four times the far-off background. Find $q$ , $E_{r}$ and $Γ$ .

Solution.

The peak-to-background ratio is $1 + q^{2} = 4$ , so $q^{2} = 3$ and $q = \pm \sqrt{3} = \pm 1.73$ . Take $q > 0$ for now.

The zero is at $ϵ = - q$ and the maximum at $ϵ = 1 / q$ , so in energy

E_{min} = E_{r} - \frac{q Γ}{2}, E_{max} = E_{r} + \frac{Γ}{2 q} .

Subtracting,

E_{max} - E_{min} = \frac{Γ}{2} (q + \frac{1}{q}) = \frac{Γ}{2} \cdot \frac{q^{2} + 1}{q} = \frac{Γ}{2} \cdot \frac{4}{1.73} = 1.156 Γ .

With $E_{max} - E_{min} = 0.3$ eV,

Γ = \frac{0.3}{1.156} = 0.26 eV,

and by Eq. (10.8) the lifetime is $τ = 658 / 260 = 2.5$ fs. Then

E_{r} = E_{min} + \frac{q Γ}{2} = 59.9 + \frac{1.73 \times 0.26}{2} = 60.1 eV .

Note that the resonance position is at neither the peak nor the dip: it lies between them. Quoting the peak as the resonance energy is a common and avoidable error, and for small $q$ it can be badly wrong.

10.6A Rydberg Series of Autoionizing States

Put the resonance into a Rydberg series and everything becomes predictable, because Chapter 4 already established how Rydberg quantities scale.

Autoionization happens at the core: the Rydberg electron must be close in for ${\hat{V}}_{e e}$ to transfer energy. Chapter 4's central result, Eq. (4.62), was that the probability of being near the core falls as $(n^{*})^{- 3}$ . Combining with Theorem 10.3.1:

Theorem 10.6.1Rydberg scaling of widths and dipoles

\begin{matrix} Γ_{n} \propto | V_{ε} |^{2} \propto (n^{*})^{- 3}, τ_{n} \propto (n^{*})^{3}, \end{matrix}

and, since a dipole matrix element is linear in the wave function where the width is quadratic,

\begin{matrix} ⟨ ℛ_{n} | \hat{D} | ε ⟩ \propto (n^{*})^{- 3 / 2} . \end{matrix}

But $n^{*}$ is not what an experiment measures — level positions are. Eliminate it using $n^{*} = [ℰ_{Ryd} / (E_{th} - E_{n})]^{1 / 2}$ from Chapter 5, Eq. (5.18). Since $(n^{*})^{3} \propto (E_{th} - E_{n})^{- 3 / 2}$ ,

\begin{matrix} τ_{n} = τ_{ref} {(\frac{E_{th} - E_{ref}}{E_{th} - E_{n}})}^{3 / 2} \end{matrix}

and since $(n^{*})^{- 3 / 2} \propto (E_{th} - E_{n})^{3 / 4}$ ,

\begin{matrix} ⟨ ℛ_{n} | \hat{D} | ε ⟩ = ⟨ ℛ_{ref} | \hat{D} | ε ⟩ {(\frac{E_{th} - E_{n}}{E_{th} - E_{ref}})}^{3 / 4} \end{matrix}

Notice what has dropped out: the quantum defect. It cancels between the two applications of the $n^{*}$ formula. Measured level positions plus one anchor give every other member's lifetime and coupling, with no structure calculation at all.

Example 10.6.1Reproducing a published parameter table

The CO $_{2}$ Henning sharp series, $(3 σ_{u})^{- 1} n d σ_{g}$ , converges to $E_{B} = 18.08$ eV. The $5 d$ member lies at $17.5395$ eV with a measured lifetime $372.6$ fs. Predict the lifetime and relative infrared dipole of the $4 d$ member at $17.2831$ eV.

Solution.

Binding energies below threshold:

E_{B} - E_{5 d} = 18.08 - 17.5395 = 0.5405 eV, E_{B} - E_{4 d} = 18.08 - 17.2831 = 0.7969 eV .

Before computing, predict the direction. The $4 d$ state is more deeply bound, hence more compact, hence closer to the core — so it should autoionize faster and couple to the continuum more strongly.

Lifetime, from Eq. (10.21):

τ_{4 d} = 372.6 {(\frac{0.5405}{0.7969})}^{3 / 2} = 372.6 \times (0.6782)^{3 / 2} = 372.6 \times 0.5585 = 208.1 fs .

Dipole, from Eq. (10.22), taking the $5 d$ value as $1$ :

⟨ ℛ_{4 d} | {\hat{D}}_{IR} | ε ⟩ = {(\frac{0.7969}{0.5405})}^{3 / 4} = (1.4744)^{3 / 4} = 1.338 .

Both agree with the published values to every digit quoted. The lifetime fell and the dipole rose, as predicted.

Two arithmetic warnings. The exponents $\frac{3}{2}$ and $\frac{3}{4}$ come with the ratio the other way up — invert one and you get a plausible but wrong answer, the commonest error here. And as a sanity check, neighboring members near $n^{*} \approx 5$ should differ by factors of order $(5 / 4)^{3} \approx 2$ , which $0.56$ and $1.34$ bracket.

10.6.1How $q$ Varies Along a Series

From Eq. (10.18), $q$ is the ratio of resonant to direct excitation amplitude. The denominator involves the continuum coupling, fixed by the width through Eq. (10.9), so

\begin{matrix} q_{n} ≃ \frac{⟨ g | {\hat{D}}_{XUV} | ℛ_{n} ⟩}{\sqrt{π Γ_{n} / 2} ⟨ g | {\hat{D}}_{XUV} | ε ⟩} \propto ⟨ g | {\hat{D}}_{XUV} | ℛ_{n} ⟩ \sqrt{τ_{n}} . \end{matrix}

A longer-lived state — a narrower resonance — therefore has a larger $q$ , hence a more peak-like, less asymmetric line. That is intuitive in hindsight: a narrow resonance concentrates the resonant amplitude into a small energy range where it dominates the smooth background.

Notice which quantity obeys no scaling law: the bound–bound dipole $⟨ g | {\hat{D}}_{XUV} | ℛ_{n} ⟩$ in the numerator. It depends on overlap with the ground state and on core details that no $n^{*}$ power captures, and in practice it is read off measured photoabsorption peak heights state by state. Knowing which quantities scale and which must be measured is most of the skill in using a model like this.

10.7Two Series, and Why They Differ

CO $_{2}$ has two Rydberg series converging to $B^{2} Σ_{u}^{+}$ : the Henning sharp series $(3 σ_{u})^{- 1} n d σ_{g}$ and the Henning diffuse series $(3 σ_{u})^{- 1} n s σ_{g}$ . They interleave in the same energy range, autoionize into the same coupled continuum ( $𝒞$ , here $X^{2} Π_{g}$ ), and have lifetimes differing by an order of magnitude — hundreds to thousands of femtoseconds for the sharp states, tens for the diffuse.

Everything needed is already in hand. From Chapter 5, the quantum defects: $δ_{d} \approx 0$ and $δ_{s} \approx 1.3$ , because the centrifugal barrier of Chapter 3 keeps a $d$ electron out of the core while an $s$ electron dives through it. From this chapter, the width is proportional to the probability of being at the core. An $s$ Rydberg electron is there far more often, so it autoionizes far faster.

One mechanism — penetration — fixes the quantum defects, the autoionization widths and the asymmetry parameters of both series.

The names are historical and inverted: the sharp series is the $d$ series, because a long-lived state gives a narrow line by $Γ τ = ℏ$ . Sharp means slow.

10.8Summary

Continuum states are energy-normalized (Def. 10.2.1), which lets a sum over bound states and an integral over the continuum appear in one expansion (Thm. 10.2.1) and absorbs the density of states so that $Γ = 2 π | ⟨ ε | \hat{V} | a ⟩ |^{2}$ .
A discrete state degenerate with a continuum and coupled to it acquires a width $Γ = 2 π | V_{ε} |^{2}$ and decays with $Γ τ = ℏ$ ; in practice $Γ [meV] \times τ [fs] = 658$ . Conversely, a measured width gives the coupling $\sqrt{Γ / 2 π}$ .
Diagonalizing the coupled system exactly (Sec. 10.4) shows the discrete state is not destroyed but spread over the exact eigenstates in a Lorentzian of width $Γ$ , shifted by the principal-value integral $F (E)$ .
When light reaches the same continuum both directly and resonantly, the amplitudes interfere and give the Fano profile $σ = σ_{b} (q + ϵ)^{2} / (1 + ϵ^{2})$ with $ϵ = 2 (E - E_{r}) / Γ$ . There is an exact zero at $ϵ = - q$ ; the ratio to background at line center is $q^{2}$ ; the maximum is $1 + q^{2}$ at $ϵ = 1 / q$ . $| q | \to \infty$ gives a Lorentzian, $q = 0$ a window resonance.
An asymmetric line is evidence of interference; a symmetric one can be explained by a rate.
Along a Rydberg series $Γ_{n} \propto (n^{*})^{- 3}$ and bound–continuum dipoles $\propto (n^{*})^{- 3 / 2}$ (Thm. 10.6.1); in terms of measured positions these become Eqs. (10.21) and (10.22), with the quantum defect cancelling. The asymmetry parameter follows, $q_{n} \propto ⟨ g | \hat{D} | ℛ_{n} ⟩ \sqrt{τ_{n}}$ .
CO $_{2}$ 's sharp ( $d$ ) and diffuse ( $s$ ) series differ by an order of magnitude in lifetime for one reason: penetration of the core.

10.9Exercises

Widths and lifetimes. Verify Eq. (10.8) from $Γ τ = ℏ$ with $ℏ = 0.658$ eV fs. Then: a state of width $1.8$ meV lives how long? A state living $372.6$ fs has what width, in meV and $μ$ eV?
Anatomy of the profile. For Eq. (10.16):
1. Confirm the zero at $ϵ = - q$ and the value $q^{2}$ at $ϵ = 0$ .
2. Carry out the differentiation in Sec. 10.5 and confirm the maximum is at $ϵ = 1 / q$ with value $1 + q^{2}$ .
3. Sketch the profile for $q = 0$ , $1$ , $3$ and $q \to \infty$ on one set of axes.
Extracting parameters. Repeat Example 10.5.1 for a line whose minimum is at $60.05$ eV, maximum at $60.20$ eV, with peak-to-background $8$ . Find $q$ , $Γ$ , $τ$ and $E_{r}$ .
Building a parameter table. Using $E_{B} = 18.08$ eV, the $5 d$ anchor ( $E = 17.5395$ eV, $τ = 372.6$ fs, dipole $\equiv 1.000$ ) and Eqs. (10.21)–(10.22), compute $n^{*}$ , $τ_{n}$ and the relative dipole for the sharp-series members at $17.8650$ eV ( $8 d$ ) and $17.9407$ eV ( $10 d$ ). Then, with $q_{5 d} = 2.00$ and XUV dipoles $2.0196 \times 10^{- 2}$ ( $5 d$ ) and $1.3687 \times 10^{- 2}$ ( $8 d$ ), use Eq. (10.23) to predict $q_{8 d}$ . The published value is $2.71$ ; if you disagree, find the arithmetic error rather than adjusting anything.
The diffuse series. Repeat for the diffuse series anchored at $6 s$ ( $E = 17.4579$ eV, $τ = 33.0$ fs) for the $9 s$ state at $17.8457$ eV. Then answer: $9 s$ and $9 d$ lie only $0.06$ eV apart yet their lifetimes differ by more than tenfold. Which quantity in Eq. (10.21) carries the difference, and which does not? Explain the physical origin in two sentences.

10.10Project: Fano Profiles from a Measured Series

The problem. Build the absorption spectrum of a whole Rydberg series of autoionizing states from measured level positions and one anchor, and watch the interference structure emerge.

The data are the CO $_{2}$ Henning sharp series converging to $E_{B} = 18.08$ eV:

State	$4 d$	$5 d$	$6 d$	$7 d$	$8 d$	$10 d$
$E_{n}$ (eV)	17.2831	17.5395	17.7018	17.8000	17.8650	17.9407

with $τ_{5 d} = 372.6$ fs measured and $q_{5 d} = 2.00$ as anchor.

On paper. Compute $n^{*}$ for each state from Eq. (5.18), hence the quantum defect. Confirm $δ_{d}$ is small and roughly constant — that constancy is Chapter 5's claim, and this is the test of it.
On paper. Using Eq. (10.21), compute $τ_{n}$ for all six, then convert to widths $Γ_{n}$ in meV using Eq. (10.8).
On the computer. Plot $Γ_{n}$ against $n^{*}$ on log–log axes and fit a power law. Report the fitted exponent; it should be $- 3$ .
On the computer. Take $q_{n}$ from Eq. (10.23), assuming for simplicity a common XUV bound dipole — an approximation the real calculation avoids, and one you should comment on. Plot the total spectrum
$σ (E) = σ_{b} \sum_{n} \frac{(q_{n} + ϵ_{n})^{2}}{1 + ϵ_{n}^{2}}, ϵ_{n} = \frac{2 (E - E_{n})}{Γ_{n}},$
over $17.2$ to $18.0$ eV. Work out how many energy points are needed to resolve the narrowest resonance before plotting.
On the computer. Zoom in on one resonance and mark the zero at $ϵ_{n} = - q_{n}$ , the peak at $ϵ_{n} = 1 / q_{n}$ and the center $E_{n}$ . Then vary $q$ for that resonance from $0$ to $10$ and show the shape morphing from a window to a peak.
What has to move. One animation: sweep $q$ upward from $0$ for the whole series at once, so every line turns from a hole into a peak together. Caption it with what to watch — the zero crossing sliding through the resonance as $q$ grows.

The check. Part (c) is the real test, and it is a consistency check rather than a measurement: since (b) was derived from a $(n^{*})^{- 3}$ law, recovering $- 3$ confirms the arithmetic, not the physics. Say so plainly in your write-up.

Be ready to answer. Your spectrum shows the higher members crowding together as they approach $18.08$ eV while their widths shrink. Both trends are powers of $n^{*}$ . Which power governs each — and what therefore happens to the ratio of width to spacing as $n$ grows? What does that tell you about whether neighboring resonances can be treated as independent, an assumption the research model makes explicitly?

10.1Introduction

10.2Continuum Normalization

10.3A Discrete State Coupled to a Continuum

10.4Fano's Diagonalization

10.5The Fano Profile

10.6A Rydberg Series of Autoionizing States

10.6.1How q Varies Along a Series

10.7Two Series, and Why They Differ

10.8Summary

10.9Exercises

10.10Project: Fano Profiles from a Measured Series

10.6.1How $q$ Varies Along a Series