Chapter 9

Electronic Spectroscopy

9.1Introduction

Chapter 8 moved a molecule between rungs of one potential energy curve. This chapter moves it between curves: the electrons rearrange, and the nuclei find themselves on a surface with a different shape and a different minimum.

Three results come out, and all three are used in the final two chapters:

the Franck–Condon principle, which decides which vibrational levels of the upper state are populated and hence why a molecular absorption band is a band rather than a line;
photoelectron spectroscopy, the measurement performed by the research this course prepares for;
molecular Rydberg series, where the atomic physics of Chapters 4 and 5 returns with the molecule's several ionization thresholds each supporting a series of its own.

The last section shows that a series converging to a high threshold consists of states that are bound and simultaneously degenerate with a lower continuum. That is where the book has been going.

9.2What an Electronic Spectrum Measures

Electronic spectroscopy is read from three linked observables: transition energy, transition intensity, and excited-state decay.

The line position gives the energy gap,

\begin{equation} \label{eq:ELEC-gap} \Delta E=hc\tilde\nu=\frac{hc}{\lambda}. \end{equation}

In absorption experiments, intensity is reported through Beer–Lambert law,

\begin{equation} \label{eq:ELEC-beer-lambert} A(\tilde\nu)=\log_{10}\!\left(\frac{I_0}{I}\right)=\epsilon(\tilde\nu)cL, \end{equation}

where $\epsilon(\tilde\nu)$ is molar absorptivity, $$c$$ concentration, and $$L$$ path length.

The integrated intensity is often expressed by the oscillator strength $f_{ba}$ : strong allowed transitions have $f\sim1$ , while symmetry- or spin-forbidden transitions are much weaker.

9.3Factorising the Transition Dipole

Under the Born–Oppenheimer factorisation of Eq. (7.11), a molecular state is written as

\ket{i,\nu_i}\equiv\ket{\psi_i(\vec R)}\ket{\chi_{i,\nu_i}}, \qquad \Psi_{i,\nu_i}(\vec r,\vec R)=\psi_i(\vec r;\vec R)\chi_{i,\nu_i}(\vec R).

The transition dipole between two such states involves integration over both electronic and nuclear coordinates.

Theorem 9.3.1 - Franck–Condon factorisation

If the electronic transition dipole varies slowly with geometry over the range sampled by the nuclear wave functions, then

\begin{equation} \label{eq:ELEC-factorisation} \boxed{\; \bra{j,\nu_j}\hat{\vec d}\ket{i,\nu_i} \simeq \vec d_{ji}(R_e)\;\braket{\chi_{j,\nu_j}}{\chi_{i,\nu_i}}\;} \end{equation}

— an electronic factor times a nuclear overlap.

Write the total dipole operator as the sum of an electronic and a nuclear part, $\hat{\vec d}=\hat{\vec d}_e+\hat{\vec d}_N$ . For $j\neq i$ the nuclear part contributes nothing, because it does not act on the electronic coordinates and $\braket{\psi_j}{\psi_i}=0$ at each $\vec R$ . So

\bra{j,\nu_j}\hat{\vec d}\ket{i,\nu_i} =\int d^{3}R\;\chi^{*}_{j,\nu_j}(\vec R) \underbrace{\left[\int d^{3}r\;\psi^{*}_j\hat{\vec d}_e\psi_i\right]} _{\displaystyle \vec d_{ji}(\vec R)} \chi_{i,\nu_i}(\vec R).

The inner integral defines the electronic transition dipole as a function of geometry. Taylor-expanding it about the equilibrium geometry, $\vec d_{ji}(\vec R)\simeq\vec d_{ji}(R_e)+\dots$ , and keeping the leading term takes it outside the nuclear integral, leaving $\vec d_{ji}(R_e)\int\chi^{*}_{j,\nu_j}\chi_{i,\nu_i}\,d^{3}R$ .

The two factors are governed by quite different rules, and it is worth being explicit about which is which.

The electronic factor $\vec d_{ji}(R_e)$ obeys the symmetry rules of Chapter 7: Laporte's $g\leftrightarrow u$ in a centrosymmetric molecule, no change of spin ( $\Delta S=0$ , since the dipole operator does not act on spin), and $\Delta\Lambda=0,\pm1$ .

Spin-forbidden lines are not strictly impossible: spin–orbit coupling mixes states of different multiplicity and can make them weakly allowed. This is why phosphorescence exists at all.

The nuclear factor $\braket{\chi_{j,\nu_j}}{\chi_{i,\nu_i}}$ is not governed by $\Delta v=\pm1$ . That rule came from $\bra{v'}\hat q\ket{v}$ with both states on the same curve (Thm. 8.4.1). Here the two vibrational wave functions belong to different potentials, so they are not members of one orthonormal set and no orthogonality forces the overlap to vanish. Any pair $(\nu_i,\nu_j)$ can contribute.

Definition 9.3.1 - Franck–Condon factor

\begin{equation} \label{eq:ELEC-FC-factor} \mathrm{FC}_{\nu_j\nu_i}=\bigl|\braket{\chi_{j,\nu_j}}{\chi_{i,\nu_i}}\bigr|^{2}, \end{equation}

the squared overlap of the two vibrational wave functions. The intensity of the vibrational line is proportional to it.

Remark 9.3.1 - Sum rule

Because $\{\chi_{j,\nu_j}\}$ is a complete set on the upper curve,

\begin{equation} \label{eq:ELEC-sum-rule} \sum_{\nu_j}\mathrm{FC}_{\nu_j\nu_i} =\sum_{\nu_j}\braket{\chi_{i,\nu_i}}{\chi_{j,\nu_j}}\braket{\chi_{j,\nu_j}}{\chi_{i,\nu_i}} =\braket{\chi_{i,\nu_i}}{\chi_{i,\nu_i}}=1 . \end{equation}

The total intensity of the band is fixed by the electronic factor alone; the Franck–Condon factors only decide how it is distributed among the vibrational lines. This is the check to apply to any computation of them.

9.4The Franck–Condon Principle

The physical content of Theorem 9.3.1 is the timescale hierarchy of Sec. 7.10. An electronic transition is driven in attoseconds to femtoseconds; nuclear motion takes tens of femtoseconds. The nuclei therefore do not move during the transition, and on a diagram of potential curves the transition is a vertical arrow: the molecule arrives on the upper curve at the bond length it had on the lower one.

Mathematically that is exactly what Eq. (9.4) says. The overlap $\int\chi^{*}_{j,\nu_j}\chi_{i,\nu_i}\,dR$ is largest when the two functions have amplitude at the same $$R$$ .

What follows depends on the geometry of the two curves.

Same minimum. If the upper curve has the same $$R_e$$ , then $\chi_{j,0}$ and $\chi_{i,0}$ are nearly the same function, so $\braket{\chi_{j,0}}{\chi_{i,0}}\simeq1$ and, by Eq. (9.5), everything else is nearly zero. The band is a single strong line.

Displaced minimum. The usual case: promoting or removing a bonding electron weakens the bond, so the upper curve sits at larger $$R$$ . The vertical arrow from $$R_e$$ of the lower state now lands on the wall of the upper well, well above its minimum. The upper vibrational functions with large $$v'$$ have their greatest amplitude near their classical turning points, which is precisely where the arrow lands — so those are the ones with good overlap. The band shows a long progression peaking at some $$v'>0$$ .

Displaced beyond dissociation. If the arrow lands above the upper curve's dissociation limit there are no discrete levels to reach, and the absorption is continuous rather than a set of lines. The molecule falls apart.

Example 9.4.1 - Reading a band shape

An electronic absorption band shows a long vibrational progression peaking at $$v'=4$$ . A second band, in the same molecule, shows a single strong line at $$v'=0$$ . What do they say?

Solution.

The first: a peak away from $$v'=0$$ means the vertical transition lands high on the upper well, so the two curves have appreciably different equilibrium bond lengths. The longer the progression and the higher its peak, the larger the displacement. If the upper state was reached by removing an electron, that electron came from a bonding orbital — weaken a bond and it lengthens.

The second: a single line at $$v'=0$$ means the geometry barely changed, so the electron came from a nonbonding orbital — a lone pair, say, that was not holding the molecule together.

A band shape is therefore a measurement of how bonding an orbital was. This is the standard way photoelectron spectra are assigned, and it is worth knowing before looking at one.

9.5Excited-State Relaxation Pathways

After absorption, population does not stay where it was created. It relaxes through competing channels:

vibrational relaxation within one electronic state (typically fs–ps);
internal conversion ( $S_n\to S_m$ , same spin);
fluorescence ( $S_1\to S_0$ , radiative, usually ns);
intersystem crossing ( $S\to T$ , nonradiative spin change);
phosphorescence ( $T_1\to S_0$ , radiative, often $\mu$ s–s).

For fluorescence from $$S_1$$ , a simple kinetic model gives

\begin{equation} \label{eq:ELEC-fluorescence-yield} \Phi_F=\frac{k_F}{k_F+k_{\mathrm{IC}}+k_{\mathrm{ISC}}+k_q}, \qquad \tau_F=\frac{1}{k_F+k_{\mathrm{IC}}+k_{\mathrm{ISC}}+k_q}, \end{equation}

with $$k_q$$ collecting external quenching processes.

Remark 9.5.1 - Kasha's rule (working form)

For most molecules in condensed phase, fluorescence is emitted mainly from the lowest excited singlet state $$S_1$$ , even if absorption initially populates higher singlets. Fast internal conversion funnels population down before emission occurs.

9.6Photoelectron Spectroscopy

Raise the photon energy above an ionization threshold and the electron does not move to another bound orbital — it leaves. What remains is a molecular ion in one of the states catalogued in Example 7.5.2, plus a free electron.

Energy conservation is the whole of the technique:

\begin{equation} \label{eq:ELEC-PES} \boxed{\;\hbar\omega=E_{\mathrm{th}}+\varepsilon\;} \end{equation}

with $\hbar\omega$ the known photon energy, $E_{\mathrm{th}}$ the threshold for producing that particular ionic state, and $\varepsilon$ the kinetic energy of the departing electron. Measure $\varepsilon$ and you have measured $E_{\mathrm{th}}$ .

Because a molecule has one threshold per orbital, a spectrum taken at a single photon energy shows one band per accessible orbital, each at its own electron energy and each carrying the Franck–Condon vibrational structure of the corresponding ionic state. One measurement gives the orbital energies and, from the band shapes, how bonding each orbital was.

Example 9.6.1 - Reading a photoelectron spectrum

A molecule is ionized with $$21.2$$ eV photons. Bands appear at electron kinetic energies $$7.4$$ , $$3.9$$ and $$3.1$$ eV; the first is a single sharp line and the second a long progression. Interpret.

Solution.

From Eq. (9.7), $E_{\mathrm{th}}=\hbar\omega-\varepsilon$ :

21.2-7.4=13.8~\mathrm{eV}, \qquad 21.2-3.9=17.3~\mathrm{eV}, \qquad 21.2-3.1=18.1~\mathrm{eV}.

Three orbitals, at $$13.8$$ , $$17.3$$ and $$18.1$$ eV. Note that the fastest electron corresponds to the lowest threshold: the spectrum is energy-reversed relative to the orbital diagram, which is a standard source of confusion.

The band shapes assign the orbitals. The sharp first band means removing that electron barely changed the geometry — a nonbonding orbital. The long progression in the second means a substantial geometry change — a bonding orbital.

These three numbers are, in fact, CO $$_2$$ : $$13.78$$ eV for $X^{2}\Pi_g$ (from the nonbonding $1\pi_g$ lone-pair orbital, hence sharp), $\approx17.3$ eV for $A^{2}\Pi_u$ (from the bonding $1\pi_u$ , hence a progression) and $$18.08$$ eV for $B^{2}\Sigma_u^{+}$ .

Remark 9.6.1 - The continuum and its normalization

The final state here is not bound: the electron may leave with any $\varepsilon>0$ , so the final states form a continuum and must be normalized as flagged at the end of Chapter 3, $\braket{\varepsilon}{\varepsilon'}=\delta(\varepsilon-\varepsilon')$ . The transition rate into them is Fermi's golden rule (Thm. 6.5.1), which with this normalization reduces to $\Gamma=2\pi|\bra{\varepsilon}\hat V\ket{i}|^{2}$ (Remark 6.5.1). Chapter 10 makes this quantitative.

Two further pieces of vocabulary, because the literature uses them without explanation. The departing electron is described by partial waves — the angular momentum components $\varepsilon s$ , $\varepsilon p$ , $\varepsilon d$ of its wave function — and in a linear molecule these carry an axial projection too, so a channel is written $\varepsilon p\pi_u$ or $\varepsilon s\sigma_g$ : a free electron of energy $\varepsilon$ , in a $$p$$ wave, with $\Lambda=1$ , odd under inversion. These are Chapter 3's atomic labels plus Chapter 7's molecular ones, and the selection rules apply to them unchanged.

9.7Molecular Rydberg Series

Now the pieces come together.

Excite one electron to an orbital of high principal quantum number, short of ionizing it. That electron is far outside everything else, and what it sees — by Eq. (5.15) of Chapter 5, which knew nothing about molecules — is a net charge of $$+1$$ . It is a Rydberg electron, and the core it orbits is a molecular ion.

Everything from Chapters 4 and 5 therefore carries over unchanged:

\begin{equation} \label{eq:ELEC-rydberg-formula} E_n=E_{\mathrm{th}} -\frac{\mathcal{E}_{\mathrm{Ryd}}}{(n-\delta_l)^{2}}, \qquad n^{*}=n-\delta_l , \end{equation}

with the same quantum defect, the same penetration argument, and the same ordering $\delta_s>\delta_p>\delta_d$ .

One thing is genuinely new.

Theorem 9.7.1 - One series per threshold

A molecule with several ionization thresholds supports a separate Rydberg series converging to each. The $E_{\mathrm{th}}$ appearing in Eq. (9.8) is the energy of the specific ionic state the series converges to.

The reason is immediate from the construction: the Rydberg electron is bound to a particular ionic core, and removing it entirely leaves that core in that state. Which core is specified by the notation:

Definition 9.7.1 - Molecular Rydberg label

\begin{equation} \label{eq:ELEC-rydberg-label} (\text{core orbital})^{-1}\,nl\lambda \end{equation}

names both the hole left in the core and the orbital the excited electron occupies. Thus $(3\sigma_u)^{-1}nd\sigma_g$ means: one electron removed from $3\sigma_u$ , one placed in a Rydberg orbital of principal number $$n$$ , $$d$$ character, axial projection zero, even parity.

That series converges to the ionic state produced by emptying $3\sigma_u$ — for CO $$_2$$ , $B^{2}\Sigma_u^{+}$ at $$18.08$$ eV, not the lowest threshold at $$13.78$$ eV. Using the wrong threshold is the most common error in this arithmetic, and it produces an $n^{*}$ that is nonsense.

9.7.1Bound States Above a Threshold

Here is the point the last two chapters turn on.

A series converging to a high threshold has members lying just below that threshold — and therefore well above the lower ones. For CO $$_2$$ concretely, the $(3\sigma_u)^{-1}nl\sigma_g$ series converges to $$18.08$$ eV, so its members sit between about $$17.0$$ and $$18.0$$ eV. The lowest ionization threshold is $$13.78$$ eV. Every one of those Rydberg states is more than three electron volts above the energy at which CO $$_2$$ can already ionize.

Each is therefore simultaneously:

bound, with respect to the ionic core it belongs to, $B^{2}\Sigma_u^{+}$ ; and
degenerate with a continuum, namely $X^{2}\Pi_g+\varepsilon$ , belonging to a different ionic core.

An autoionizing molecular state is bound to one ionic core and embedded in the continuum of another.

This is the situation Chapter 5 met in helium's doubly excited states. There it required exciting two electrons at once and was something of a curiosity; in a molecule, with several thresholds available, it is the normal state of affairs.

The coupling connecting the two is the electron–electron repulsion $\hat V_{ee}$ — the same term that made helium unsolvable in Chapter 5. In orbital language: the Rydberg electron drops into the $3\sigma_u$ hole while a $1\pi_g$ electron takes up the released energy and departs. No photon is involved, so no dipole selection rule applies; what constrains it is conservation of energy, parity and angular momentum.

Example 9.7.1 - Parity of a molecular Rydberg state

Show that the $(3\sigma_u)^{-1}nl\sigma_g$ states have total parity $$u$$ , and deduce two consequences.

Solution.

Use the product rule of Remark 7.5.1. Start from the ground configuration Eq. (7.18), whose total parity is $$g$$ (Example 7.5.1).

Removing one electron from $3\sigma_u$ removes one factor of $$u$$ from the product, so the remaining core has parity $g\times u^{-1}=u$ (parities multiply as $\pm1$ , so dividing and multiplying are the same). Adding a Rydberg electron in a $$g$$ orbital multiplies by $$g=+1$$ . The total is therefore $$u$$ .

First consequence. The ground state is $$g$$ and these states are $$u$$ , so by Laporte's rule a one-photon XUV transition into them is allowed. This is how they are populated.

Second consequence. Both the $ns\sigma_g$ and the $nd\sigma_g$ series are built on the same $3\sigma_u$ hole with a $$g$$ Rydberg orbital, so both have total parity $$u$$ . A one-photon transition between two states of the same parity vanishes, so an infrared photon cannot move population from one Rydberg state to another, whatever $$n$$ and whatever $$l$$ .

That second result removes an entire block of the Hamiltonian in Chapter 11, and it costs nothing — no fitting, no parameter, no error. It is a symmetry argument, made in Chapter 3, applied to labels earned in Chapter 7.

9.8Summary

Electronic line positions follow $\Delta E=hc\tilde\nu$ and intensities in absorption follow Beer–Lambert law, Eq. (9.2). Strong transitions have large oscillator strength; forbidden ones are weak.
The transition dipole factorises (Thm. 9.3.1) into an electronic part, governed by the symmetry rules of Chapter 7, and a nuclear overlap — the Franck–Condon factor, Def. 9.3.1. The factors sum to one (Rem. 9.3.1), so they redistribute intensity rather than create it.
Electronic transitions are vertical: the nuclei cannot move during them. The band shape therefore reports the change in equilibrium geometry, and hence how bonding the orbital was. Same minimum gives one line; displaced gives a progression; displaced past dissociation gives a continuum.
Photoelectron spectroscopy uses $\hbar\omega=E_{\mathrm{th}}+\varepsilon$ : one band per orbital, each with its own vibrational structure, and the fastest electron corresponds to the lowest threshold.
Excited-state pathways compete: fluorescence, internal conversion, intersystem crossing, and phosphorescence. Fluorescence yield and lifetime are set by rate competition, Eq. (9.6).
The free electron is described by partial waves labelled $\varepsilon s\sigma_g$ , $\varepsilon p\pi_u$ and so on.
Molecular Rydberg states obey the atomic formula Eq. (9.8), but there is one series per ionization threshold (Thm. 9.7.1) and $E_{\mathrm{th}}$ must be the right one. They are labelled $(\text{core})^{-1}nl\lambda$ .
A series converging to a high threshold consists of states bound to that core and degenerate with the continuum of a lower one. They autoionize through $\hat V_{ee}$ . For CO $$_2$$ the $(3\sigma_u)^{-1}nl\sigma_g$ series near $$17.5$$ eV sits above the $X^{2}\Pi_g$ threshold at $$13.78$$ eV, and all its members have total parity $$u$$ .

9.9Exercises

The factorisation. Reproduce the proof of Theorem 9.3.1, stating clearly why the nuclear part of the dipole operator drops out for $j\neq i$ . Then derive the sum rule Eq. (9.5).
Why not $\Delta v=\pm1$ ? Explain in three or four sentences why $\braket{\chi_{j,\nu_j}}{\chi_{i,\nu_i}}$ can be nonzero for $\nu_j\neq\nu_i$ , when vibrational states of the same curve are orthogonal. What property of the two sets is different?
Reading a spectrum. Repeat Example 9.6.1 for a measurement made with $$40.8$$ eV photons on the same molecule.
1. What kinetic energies would the same three bands appear at?
2. A fourth band now appears at $$21.4$$ eV kinetic energy. What threshold does it correspond to, and why was it invisible at $$21.2$$ eV?
Which threshold? The CO $$_2$$ state $7d\sigma_g$ of the $(3\sigma_u)^{-1}$ series lies at $$17.80$$ eV.
1. Compute $n^{*}$ with the correct threshold $$E_B=18.08$$ eV, and hence $\delta_d$ .
2. Compute it wrongly with $$13.78$$ eV and report the result.
3. Explain in two sentences why the lowest threshold is wrong here, in terms of which ion the Rydberg electron is bound to.
Parity, end to end. Reproduce Example 9.7.1, then determine whether each of the following one-photon processes is allowed, with a one-line reason:
1. $X^{1}\Sigma_g^{+}\to(3\sigma_u)^{-1}ns\sigma_g$ ;
2. $(3\sigma_u)^{-1}ns\sigma_g\to(3\sigma_u)^{-1}nd\sigma_g$ ;
3. $(3\sigma_u)^{-1}nd\sigma_g\to B^{2}\Sigma_u^{+}+\varepsilon p\pi_u$ ;
4. $X^{1}\Sigma_g^{+}\to X^{2}\Pi_g+\varepsilon s\sigma_g$ .
Autoionization is not a photon process.
1. Which term in the electronic Hamiltonian Eq. (7.2) drives autoionization?
2. Describe in terms of orbitals which electron goes where when $(3\sigma_u)^{-1}nd\sigma_g$ autoionizes into $X^{2}\Pi_g+\varepsilon$ .
3. Explain why $\Delta l=\pm1$ does not restrict this process, and state what does.
Beer–Lambert and oscillator strength. A chromophore has $\epsilon_{\max}=1.8\times10^{4}$ L mol $^{-1}$ cm $^{-1}$ at $\tilde\nu=2.00\times10^{4}$ cm $^{-1}$ .
1. For $c=2.0\times10^{-5}$ mol L $^{-1}$ and $$L=1.00$$ cm, compute $$A$$ and transmitted fraction $$I/I_0$$ .
2. Explain why doubling concentration and halving path length leaves $$A$$ unchanged.
3. A second transition at similar energy has $\epsilon_{\max}=120$ L mol $^{-1}$ cm $^{-1}$ . Which is more strongly allowed, and what symmetry reason is likely?
Fluorescence yield and lifetime. For a molecule, $k_F=2.5\times10^{8}$ s $^{-1}$ , $k_{\mathrm{IC}}=1.0\times10^{8}$ s $^{-1}$ , $k_{\mathrm{ISC}}=3.0\times10^{7}$ s $^{-1}$ , $k_q=2.0\times10^{7}$ s $^{-1}$ .
1. Compute $\Phi_F$ and $\tau_F$ from Eq. (9.6).
2. If oxygen quenching doubles $$k_q$$ , what happens to $\Phi_F$ and $\tau_F$ ?
3. Which single rate constant would you try to reduce first to brighten fluorescence, and why?

9.10Project: Franck–Condon Factors

The problem. Compute Franck–Condon factors between two displaced potential curves and reproduce the band shapes of Example 9.4.1 from scratch. This is the calculation that turns a pair of potential curves into a predicted spectrum.

Use two Morse curves as in Chapter 7: a lower state with minimum at $$R_e$$ , and an upper state with minimum at $R_e+\Delta R$ , with $\Delta R$ under your control.

On paper. Explain why $\braket{\chi_{j,\nu_j}}{\chi_{i,\nu_i}}$ is not zero for $\nu_j\neq\nu_i$ when the two curves differ. Then derive the sum rule Eq. (9.5) and say what it will let you check.
On paper. Predict, without computing, which $$v'$$ carries the strongest line when $\Delta R$ is large, using the classical turning-point argument of Sec. 9.4.
On the computer. Using the Chapter 7 solver, obtain $\chi_{i,0}$ on the lower curve and $\chi_{j,\nu_j}$ for $\nu_j=0\dots15$ on the upper one, on a common $$R$$ grid. Compute the overlaps and hence $\mathrm{FC}_{\nu_j,0}$ .
On the computer. Verify $\sum_{\nu_j}\mathrm{FC}_{\nu_j,0}\approx1$ and report how close you get. Falling short means the upper-state basis is truncated or the grid is too small — diagnose which, and fix it.
On the computer. Plot the simulated band — a stick spectrum of $\mathrm{FC}_{\nu_j,0}$ against transition energy — for $\Delta R=0$ , $$0.1$$ , $$0.2$$ and $$0.4$$ Å. Confirm the progression lengthens and its peak moves to higher $$v'$$ , and find the $\Delta R$ that peaks at $$v'=4$$ . Compare with your prediction in (b).
What has to move. One animation: increase $\Delta R$ continuously, showing side by side the two potential curves with a vertical arrow between them, and the resulting stick spectrum. The arrow climbs the wall of the upper well while the progression stretches out.

The check. At $\Delta R=0$ the two curves are identical, so the vibrational states are the same orthonormal set and $\mathrm{FC}_{v'0}$ must be $$1$$ for $$v'=0$$ and zero otherwise. If it is not, the grids are misaligned or the wave functions are not normalized. Fix that before trusting any displaced case.

Be ready to answer. Your simulated band gives the geometry change of the excited state. Suppose an experiment hands you a measured band and asks for $\Delta R$ . Describe how you would extract it, say what else about the upper curve you would need to know, and identify what could go wrong if you simply assumed it.