Chapter 8
Rotational and Vibrational Spectroscopy
8.1Introduction
Chapter 7 found the vibrational and rotational levels. A level scheme is not a spectrum: what a spectrometer records is a set of transitions, and which transitions appear is a separate question.
That question always has the same form. From Chapter 6, the rate of a one-photon transition is proportional to \(|\bra{f}\hat{\vec d}\ket{i}|^{2}\), so the whole of this chapter is one calculation repeated:
Nonzero means the line appears, with strength given by the square. Zero means it is invisible, however much light you shine. We evaluate the integrals rather than quoting the rules.
Two results are used later. A molecule can be entirely dark in one kind of spectroscopy and bright in another — symmetry decides, and CO\(_2\) is the standard example. And the pattern of a band gives the geometry, which is how a spectrum becomes structural information.
This chapter uses one workflow repeatedly:
identify initial and final states;
test if the transition moment is symmetry-allowed;
use level formulas to predict line positions;
use populations and matrix elements to interpret intensities.
8.2Radiative Processes and Einstein Coefficients
Before applying selection rules, it helps to separate the three elementary radiative processes between two levels \(1\) and \(2\):
where \(\rho(\nu)\) is the spectral energy density and \(N_i\) are level populations.
At thermal equilibrium, detailed balance gives
These relations are useful in practice: once an absorption strength is known, they connect it to spontaneous-emission lifetime \(\tau_{\mathrm{rad}}\sim1/A_{21}\) and therefore to fluorescence intensity.
8.3Rotational Spectroscopy
The rotational states of a diatomic are the spherical harmonics of Chapter 3, now describing the orientation of the molecular axis: \(\ket{J,M}\to Y_J^{M}(\theta,\varphi)\), with energies \(E_J=BJ(J+1)\).
The dipole moment of the molecule points along its axis. Writing its magnitude as \(d_0\) (a permanent, geometry-dependent constant) and its direction as the unit vector \(\hat n(\theta,\varphi)\),
The \(z\) component is \(d_0\cos\theta\), so the matrix element is
Two facts follow immediately, and both are physical rather than technical.
A molecule has a pure rotational spectrum only if \(d_0\neq0\), that is, only if it possesses a permanent electric dipole moment.
Equation (8.5) is proportional to \(d_0\). If the dipole moment vanishes, every rotational matrix element vanishes with it, whatever the angular integral does.
So HCl, CO and H\(_2\)O have microwave spectra. H\(_2\), N\(_2\), O\(_2\) and — importantly for this course — CO\(_2\) do not. A symmetric linear molecule O=C=O has its charge distribution balanced about the inversion centre, so \(d_0=0\) exactly by symmetry, and rotating it presents nothing for the field to grip.
The integral in Eq. (8.5) vanishes unless
Use \(\cos\theta=\sqrt{4\pi/3}\,Y_1^{0}\) from Eq. (4.31), so the integral becomes one over three spherical harmonics,
Two properties of that integral settle the matter. The \(\varphi\) dependence is \(e^{i(M-M'+0)\varphi}\), which integrates to zero unless \(M'=M\). And the product \(Y_1^{0}Y_J^{M}\) expands in spherical harmonics of rank \(|J-1|\) to \(J+1\) only, so orthogonality kills everything except \(J'=J\pm1\); the case \(J'=J\) is further excluded by parity, since \(Y_J^{M}\) and \(Y_1^{0}Y_J^{M}\) then have opposite parity \((-1)^{J}\) versus \((-1)^{J+1}\).
Combining Theorem 8.3.2 with Eq. (7.27), absorption from \(J\) to \(J+1\) costs
so the lines fall at \(2B, 4B, 6B,\dots\): a ladder of equally spaced lines separated by \(2B\). Measure the spacing, get \(B\), get \(I\), get the bond length — Example 7.7.1.
The lines are not equally strong. The population of level \(J\) at temperature \(T\) is proportional to the degeneracy times the Boltzmann factor,
which rises from the degeneracy and falls from the exponential, peaking at \(J_{\max}\simeq\sqrt{k_BT/2B}-\tfrac12\). A rotational band therefore has a characteristic intensity envelope, and its maximum is a thermometer.
8.4Vibrational Spectroscopy
Now the vibrational levels \(E_v=\hbar\omega_e(v+\tfrac12)\). The relevant dipole is again the molecule's own, but what matters is how it changes as the bond stretches. Expand about equilibrium in the displacement \(q=R-R_e\):
Within the harmonic approximation, the vibrational matrix element \(\bra{v'}d(R)\ket{v}\) vanishes unless
and even then only if \(\left.dd/dR\right|_{R_e}\neq0\): the dipole moment must change with the vibrational coordinate.
Insert Eq. (8.9) into \(\bra{v'}d(R)\ket{v}\) and take the terms in turn.
The constant \(d(R_e)\) gives \(d(R_e)\braket{v'}{v}=d(R_e)\delta_{v'v}\), which is zero for any transition: a constant dipole cannot drive a vibrational change.
The linear term gives \(\left.\frac{dd}{dR}\right|_{R_e}\bra{v'}\hat q\ket{v}\). From Chapter 2, Def. 2.7.1, the displacement operator is
so by Eq. (2.38)
which is nonzero only for \(v'=v\pm1\). That is Eq. (8.10), and it is proportional to the dipole gradient.
The rule is a property of the harmonic oscillator, not of molecules. Two things spoil it, both weakly. The quadratic term in Eq. (8.9) contributes \(\bra{v'}\hat q^{2}\ket{v}\), which connects \(v'=v\pm2\); and anharmonicity (Sec. 7.6.2) mixes the harmonic states. The resulting overtone bands are typically a hundred times weaker than the fundamental and are routinely observed.
Explain why the bulk of the atmosphere is transparent in the infrared.
Solution.
N\(_2\) and O\(_2\) are homonuclear. By symmetry their dipole moment is zero at every bond length — stretching a symmetric molecule keeps it symmetric — so \(d(R)\equiv0\) and hence \(\left.dd/dR\right|_{R_e}=0\). By Theorem 8.4.1 they have no infrared spectrum at all.
HCl is heteronuclear: the chlorine draws charge, so \(d(R)\neq0\), and stretching the bond changes the separation of the charges, so \(dd/dR\neq0\). It absorbs strongly.
Since N\(_2\) and O\(_2\) make up \(99\%\) of the atmosphere and neither absorbs infrared, the atmosphere's infrared behaviour is set entirely by trace species that do — H\(_2\)O, CO\(_2\), CH\(_4\). That is the physical basis of the greenhouse effect, and it is a selection rule.
8.4.1Vibration–Rotation Bands
A real infrared spectrum shows not one line per vibrational transition but a whole structure, because rotational spacings are a hundred times smaller (Sec. 7.10) and the molecule changes \(v\) and \(J\) together.
Combining Eqs. (8.6) and (8.10), and writing the band centre as \(\tilde\nu_0=\omega_e/2\pi c\), the transition wavenumbers for \(v\to v+1\) are
The result is a double comb of lines spaced \(2B\), with a gap of \(4B\) at the centre where the \(Q\) branch would be. For a diatomic in a \(\Sigma\) state the \(Q\) branch is forbidden — \(\Delta J=0\) requires the transition dipole to have a component along the molecular axis with \(\Delta\Lambda\neq0\) — so the gap is the signature of the band.
8.4.2Real Band Structure: \(B'\neq B''\)
Equation (8.11) assumes one rotational constant \(B\) for both vibrational states. Real bands use two constants: \(B''\) (lower) and \(B'\) (upper). Keeping only first-order vibration–rotation effects gives
If \(B'<B''\) (the usual case), lines in one branch crowd and can form a band head. The head condition is approximately
which is one of the quickest diagnostics for geometry change upon vibrational excitation.
8.5Normal Modes and CO\(_2\)
For a polyatomic molecule each normal mode of Sec. 7.8 is an independent oscillator in a collective coordinate \(Q_i\), and Theorem 8.4.1 applies to each separately: mode \(i\) is infrared active if and only if
CO\(_2\) has four modes (Sec. 7.8), and they do not all behave the same way.
Determine the infrared activity of the symmetric stretch, the antisymmetric stretch and the bend.
Solution.
Assign partial charges \(-q\), \(+2q\), \(-q\) to O, C, O, so that the molecule is neutral, and compute \(\vec d=\sum_iq_i\vec r_i\) during each motion.
Symmetric stretch \(\nu_1\). Both oxygens move outward together by \(+s\) and \(-s\) along the axis, carbon stationary. Then
identically, for every \(s\). The molecule stays symmetric about its centre throughout, so its dipole is zero at every instant, \(\partial d/\partial Q_1=0\), and the mode is infrared inactive.
Antisymmetric stretch \(\nu_3\). One bond lengthens by \(s\) while the other shortens by \(s\), so the carbon shifts relative to the centre of the oxygens. Now
an oscillating dipole along the axis. Active.
Bend \(\nu_2\). The carbon moves perpendicular to the axis while the oxygens move the other way, producing a dipole perpendicular to the axis. Active.
So CO\(_2\) absorbs at \(667\) and \(2349\) cm\(^{-1}\) and is transparent at \(1333\) cm\(^{-1}\); and by Theorem 8.3.1 it has no microwave spectrum at all. Its two active bands lie where the Earth radiates, which is what makes it a greenhouse gas — and its inactive mode is invisible to the same measurement.
8.5.1Raman Scattering
A mode invisible in absorption need not be invisible altogether. In Raman scattering the molecule is not asked to absorb resonantly: a photon is scattered, and in the process leaves behind or removes one vibrational quantum, so the scattered light returns shifted by the vibrational frequency.
The relevant quantity is not the permanent dipole but the polarizability \(\alpha\) — how easily the electron cloud distorts in a field. The field induces a dipole \(\vec d_{\mathrm{ind}}=\alpha\vec E\), and expanding \(\alpha\) in the normal coordinate as in Eq. (8.9) gives
A vibrational mode is Raman active if and only if
For the symmetric stretch of CO\(_2\) the polarizability certainly changes — the molecule grows and a larger electron cloud is easier to polarise — even though the dipole does not. The mode that infrared spectroscopy cannot see, Raman scattering sees clearly.
In a molecule with a centre of inversion, no vibrational mode is both infrared and Raman active.
Under inversion the dipole moment is odd (\(\hat\Pi\vec d\hat\Pi^{\dagger}=-\vec d\)) while the polarizability, a second-rank tensor built from two factors of position, is even. A normal coordinate \(Q_i\) is itself either even (\(g\)) or odd (\(u\)) under inversion. For \(\partial\vec d/\partial Q_i\) to be nonzero, \(Q_i\) must be odd; for \(\partial\alpha/\partial Q_i\) to be nonzero, \(Q_i\) must be even. No mode can be both.
The two techniques are therefore complementary by symmetry rather than by accident, and for a centrosymmetric molecule both are needed to see all the modes.
8.5.2Stokes and Anti-Stokes Lines
Raman spectra contain two shifted sets of lines around the laser frequency: Stokes (red-shifted) and anti-Stokes (blue-shifted). Their intensity ratio is
The Boltzmann factor dominates: anti-Stokes lines are weaker because they start from thermally excited vibrational states. This ratio is therefore a direct temperature probe.
8.6Summary
Whether a line appears is decided by one question: is \(\bra{f}\hat{\vec d}\ket{i}\) zero? Everything here is that question asked of different operators.
Absorption, stimulated emission, and spontaneous emission are connected by Einstein coefficients, Eqs. (8.2) and (8.3); they link line strengths and radiative lifetimes.
Rotation: \(\Delta J=\pm1\) (Thm. 8.3.2), lines equally spaced by \(2B\), and only if the molecule has a permanent dipole (Thm. 8.3.1). CO\(_2\) has none. Intensities follow the Boltzmann envelope, peaking near \(J_{\max}\simeq\sqrt{k_BT/2B}\).
Vibration: \(\Delta v=\pm1\) from \(\bra{v'}\hat q\ket{v}\propto\sqrt{v}\delta_{v',v-1}+\sqrt{v+1}\delta_{v',v+1}\), and only if \(dd/dR\neq0\) (Thm. 8.4.1). Homonuclear diatomics are infrared-dark, which is why the atmosphere is transparent except for its trace gases.
A vibrational band carries rotational structure: \(P\), \(Q\) and \(R\) branches, Eq. (8.11), with lines spaced \(2B\) and a central gap of \(4B\) in the simplest model. In real spectra, \(B'\neq B''\) causes branch asymmetry and can create a band head, Eq. (8.13).
Polyatomics: mode \(i\) is infrared active if \(\partial\vec d/\partial Q_i\neq0\). CO\(_2\)'s symmetric stretch is inactive because the molecule stays symmetric throughout the motion; the bend and antisymmetric stretch are active.
Raman scattering is governed by \(\partial\alpha/\partial Q_i\) instead, and in a centrosymmetric molecule the two techniques are mutually exclusive by parity (Thm. 8.5.2). The Stokes/anti-Stokes ratio, Eq. (8.16), gives temperature.
8.7Exercises
The rotational integral. Evaluate \(\int Y_1^{0*}\cos\theta\,Y_0^{0}\,d\Omega\) explicitly using Eq. (4.31), and confirm it is nonzero. Then evaluate \(\int Y_0^{0*}\cos\theta\,Y_0^{0}\,d\Omega\) and confirm it vanishes. Which part of Theorem 8.3.2 does each illustrate?
Which molecules are active? For H\(_2\), HCl, CO\(_2\), H\(_2\)O and N\(_2\)O (linear but N–N–O, hence not symmetric), state whether each has (a) a pure rotational spectrum and (b) at least one infrared-active vibration. Give the one-line symmetry reason in each case.
The vibrational matrix element. Reproduce the proof of Theorem 8.4.1. Then compute \(\bra{2}\hat q^{2}\ket{0}\) using \(\hat q\propto\hat a+\hat a^{\dagger}\) and Eq. (2.38), and confirm it is nonzero — this is the first overtone of Remark 8.4.1.
Reading a band. An infrared band shows a comb of lines spaced \(0.72\) cm\(^{-1}\) on either side of a gap centred at \(2143\) cm\(^{-1}\).
Identify the branches and the band centre.
Find \(B\), and hence the bond length, taking \(\mu=1.139\times10^{-26}\) kg.
The central gap is \(4B\) wide rather than \(2B\). Explain why, using Eq. (8.11).
Boltzmann envelope. Using Eq. (8.8) with \(B=1.931\) cm\(^{-1}\) for CO at \(300\) K, find \(J_{\max}\). (Note \(k_BT=208\) cm\(^{-1}\) at \(300\) K.) Sketch the expected intensity pattern of the \(P\) and \(R\) branches.
Mutual exclusion. Reproduce the proof of Theorem 8.5.2. Then determine, for each of CO\(_2\)'s four modes, whether the normal coordinate is \(g\) or \(u\) under inversion, and hence confirm the activity assignments of Example 8.5.1.
Einstein coefficients and lifetime. A transition has frequency \(\nu=6.0\times10^{13}\) Hz and \(B_{21}=2.5\times10^{20}\) in SI units.
Use Eq. (8.3) to find \(A_{21}\).
Estimate the radiative lifetime \(\tau_{\mathrm{rad}}\).
Explain qualitatively how the lifetime changes for a higher-frequency transition with similar dipole strength.
Real branch formulas. For a band with \(\tilde\nu_0=2143\) cm\(^{-1}\), \(B''=1.93\) cm\(^{-1}\), and \(B'=1.88\) cm\(^{-1}\):
Raman thermometry. A Raman spectrum collected with \(\tilde\nu_0=18797\) cm\(^{-1}\) (532 nm laser) shows a mode at \(\tilde\nu_v=1000\) cm\(^{-1}\) with \(I_{\mathrm{AS}}/I_{\mathrm{S}}=0.11\). Estimate the sample temperature using Eq. (8.16).
8.8Project: The Normal Modes of CO\(_2\)
The problem. Compute the normal modes of CO\(_2\) from a ball-and-spring model and predict which are infrared active. This is the classical normal-mode analysis that underlies the quantum result, and it is an eigenvalue problem.
Model the molecule as three masses on a line — \(m_\mathrm{O}\), \(m_\mathrm{C}\), \(m_\mathrm{O}\) — joined by two identical springs of constant \(k\). Restrict attention to motion along the axis, which captures the two stretches; the bend needs the perpendicular coordinates.
On paper. Write Newton's equations for the three axial displacements \(x_1,x_2,x_3\). Put them in the form \(\ddot{\vec x}=-\mathbf{M}^{-1}\mathbf{K}\vec x\) and write both \(3\times3\) matrices out explicitly.
On paper. Find the three eigenvalues by hand. One is zero: identify which motion it is and explain why it must be there. Show that the other two are
\[ \omega_1^{2}=\frac{k}{m_\mathrm{O}}, \qquad \omega_3^{2}=k\left(\frac{1}{m_\mathrm{O}}+\frac{2}{m_\mathrm{C}}\right), \]and identify which is the symmetric and which the antisymmetric stretch.
On paper. Show from these that \(\omega_3>\omega_1\) always, and compute the predicted ratio \(\omega_3/\omega_1\) using \(m_\mathrm{O}=16\) and \(m_\mathrm{C}=12\) u. Compare with the measured \(2349/1333=1.76\).
On the computer. Diagonalise the matrix numerically and confirm your eigenvalues and eigenvectors. Choose \(k\) so the antisymmetric mode comes out at \(2349\) cm\(^{-1}\), then report what the model predicts for the symmetric stretch. Comment on the discrepancy with (c): what has a two-spring model left out?
On the computer. For each mode compute the dipole moment as a function of time, assigning charges \(-q,+2q,-q\) and summing \(\sum_iq_ix_i(t)\). Plot both. One should oscillate and the other be identically zero, reproducing Example 8.5.1 numerically rather than by symmetry argument.
What has to move. One animation per stretching mode: the three atoms oscillating with the relative amplitudes and phases from your eigenvectors, with the instantaneous dipole drawn as an arrow beneath. The arrow should stay dead at zero for the symmetric mode and swing for the antisymmetric one. That contrast is the whole of infrared selection rules in one figure.
The check. The zero eigenvalue in (b) is the test that your matrices are right: rigid translation of the whole molecule costs no energy, so it must appear as a zero mode. If it does not come out exactly zero numerically, your force-constant matrix has an error — each of its rows must sum to zero.
Be ready to answer. Your part (e) shows the symmetric stretch has no oscillating dipole and so cannot absorb infrared light. Yet the mode exists and carries energy. How would you detect it experimentally, and what property of the molecule does that technique couple to instead?