Raman Scattering and Electronic Bands

Infrared absorption reaches only vibrations that change the electric dipole moment, so it is blind to the symmetric stretch of a homonuclear molecule such as $N_{2}$ or $O_{2}$ . Two other processes fill the gap. Raman scattering probes vibrations that change the polarizability, complementary to the infrared rule, and electronic excitation lifts the molecule to a new electronic state whose vibrational structure prints a band system on the spectrum. Both are governed by matrix elements between vibrational states, and both carry structural and thermal information that pure absorption cannot.

Rayleigh and Raman scattering

When light of frequency $ν_{0}$ falls on a molecule, its electric field $E = E_{0} cos (2 π ν_{0} t)$ induces a dipole moment $p = α E$ , where $α$ is the polarizability. The oscillating dipole reradiates at $ν_{0}$ ; this is Rayleigh scattering, elastic and much stronger than anything shifted. But $α$ is not constant: as the molecule vibrates with frequency $ν_{vib}$ along a normal coordinate $Q = Q_{0} cos (2 π ν_{vib} t)$ , the polarizability is modulated,

α (t) = α_{0} + (\frac{\partial α}{\partial Q})_{0} Q_{0} cos (2 π ν_{vib} t) .

Multiplying by the field and applying the product-to-sum identity gives three frequencies in the scattered light,

p = α_{0} E_{0} cos (2 π ν_{0} t) + \frac{1}{2} (\frac{\partial α}{\partial Q})_{0} Q_{0} E_{0} [cos (2 π (ν_{0} - ν_{vib}) t) + cos (2 π (ν_{0} + ν_{vib}) t)] .

The unshifted line is Rayleigh; the sideband at $ν_{0} - ν_{vib}$ is the Stokes line, where the molecule gains a vibrational quantum, and the sideband at $ν_{0} + ν_{vib}$ is the anti-Stokes line, where it gives one up. The shift $ν_{vib}$ is the vibrational frequency, read as a displacement from the exciting line regardless of what $ν_{0}$ is.

The classical picture gives the frequencies; the quantum picture gives the mechanism. The incident photon lifts the molecule to a virtual level — not an eigenstate, but a short-lived superposition — from which it drops back either to the original level (Rayleigh) or to a different vibrational level (Raman).

Rayleigh and Raman scattering through a virtual level: Rayleigh returns to the starting level, Stokes leaves the molecule one quantum higher, and anti-Stokes starts one quantum up and ends in the ground vibrational level.

Because anti-Stokes scattering starts from the thermally excited $v = 1$ level, its intensity is weaker than Stokes by the Boltzmann population ratio. Including the $ν^{4}$ scattering factor, the intensity ratio is

\frac{I _{anti-Stokes}}{I _{Stokes}} = (\frac{ν _{0} + ν _{vib}}{ν _{0} - ν _{vib}})^{4} e^{- h c ν_{vib} / k_{B} T} .

The ratio depends only on the vibrational shift and the temperature, so measuring the two sideband intensities returns the local temperature without any calibration — the basis of Raman thermometry in flames and plasmas.

The scattered spectrum: an intense Rayleigh line at the laser frequency flanked by weaker Raman sidebands, the Stokes line below and the fainter anti-Stokes line above, displaced by the vibrational frequency.

Complementary selection rules

Infrared and Raman activity ask different questions of a normal mode.

Infrared active if the mode changes the dipole moment: $(\partial μ / \partial Q)_{0} \neq = 0$ .
Raman active if the mode changes the polarizability: $(\partial α / \partial Q)_{0} \neq = 0$ .

For a molecule with a center of symmetry the two conditions are mutually exclusive: a mode symmetric under inversion changes $α$ but not $μ$ , and an antisymmetric mode changes $μ$ but not $α$ . This rule of mutual exclusion means no vibration of a centrosymmetric molecule appears in both spectra, so the two techniques together map every mode.

Mode	Dipole change	Polarizability change	Infrared	Raman
$CO_{2}$ symmetric stretch	none	yes	inactive	active
$CO_{2}$ antisymmetric stretch	yes	none	active	inactive
$CO_{2}$ bend	yes	none	active	inactive
$N_{2}$ stretch	none	yes	inactive	active

The homonuclear molecules $N_{2}$ and $O_{2}$ , invisible to the infrared, are the textbook Raman case: their single stretch modulates the polarizability and scatters, so Raman reads the vibrational frequency of a molecule with no dipole spectrum at all. Rotational Raman scattering follows $Δ J = 0, \pm 2$ , since the polarizability returns to itself twice per rotation, giving lines spaced by $4 B$ rather than the $2 B$ of dipole absorption.

Electronic transitions and the Franck-Condon principle

An electronic transition moves the molecule to a different electronic state, one with its own potential curve, equilibrium separation, and vibrational ladder. The photon energies are of order electron-volts, so these bands lie in the visible and ultraviolet. Because each electronic state carries a vibrational structure, a single electronic transition splits into many closely spaced vibronic lines, turning a would-be line into a band.

Which vibronic lines are strong is fixed by the Franck-Condon principle: the electrons rearrange so much faster than the nuclei that the internuclear distance is frozen during the transition. On a potential-energy diagram the transition is therefore vertical, a straight up-arrow at fixed $r$ . If the upper electronic state has a larger equilibrium separation than the lower — as it usually does, since promoting a bonding electron weakens the bond — the vertical line from the bottom of the lower well reaches the upper curve high on its inner wall, near a classical turning point of an excited vibrational level.

The Franck-Condon principle: nuclei are fixed during the fast electronic jump, so the transition is vertical; with the upper curve displaced outward, the strongest line lands on the upper vibrational level whose turning point sits above the lower ground state.

Quantitatively, the intensity of a vibronic line from lower level $v^{''}$ to upper level $v^{'}$ is proportional to the square of the vibrational overlap, the Franck-Condon factor,

I_{v^{'} \leftarrow v^{''}} \propto ⟨ ψ_{v^{'}} ∣ ψ_{v^{''}} ⟩^{2},

the electronic transition moment being nearly constant across the band. From the ground level $v^{''} = 0$ , whose wavefunction peaks at the well center, the overlap is largest with the upper level whose wavefunction also has large amplitude there — which for a displaced upper curve is a level of nonzero $v^{'}$ . The band intensity therefore rises through the first few members of the progression, peaks, and falls, tracing a Franck-Condon envelope whose maximum shifts to higher $v^{'}$ the more the upper curve is displaced.

A vibronic absorption progression: lines at the upper vibrational levels v' from the ground level, their intensities following the Franck-Condon factors so the band peaks at an intermediate v' set by the curve displacement.

The mirror image appears in emission. A molecule excited to the upper electronic state relaxes to its lowest vibrational level and then emits downward to the spread of lower-state vibrational levels, so the emission progression is the reflection of the absorption progression about the shared $0 \to 0$ line, and the spacing of the emission lines gives the ground-state vibrational frequency while the absorption spacing gives the excited-state one.

Fluorescence, phosphorescence, and the Jablonski diagram

After absorption an excited molecule has several routes back down, radiative and nonradiative, organized by the Jablonski diagram. The relevant distinction is electron spin: the ground state is a spin singlet $S_{0}$ , and the accessible excited states are singlets $S_{1}$ (spin allowed) or triplets $T_{1}$ (spin forbidden).

Jablonski diagram: absorption to a singlet, fast nonradiative relaxation to the lowest excited singlet, then either spin-allowed fluorescence back to the ground state or intersystem crossing to a triplet followed by slow phosphorescence.

The two emissions differ in rate and delay:

Fluorescence is the spin-allowed $S_{1} \to S_{0}$ emission. It is fast, with a lifetime of order $1 0^{- 9}$ - $1 0^{- 7} s$ , and stops promptly when the excitation is removed. Because the molecule relaxes to the bottom of $S_{1}$ before emitting (Kasha's rule), the fluorescence is red-shifted from the absorption by the vibrational relaxation energy, the Stokes shift.
Phosphorescence follows an intersystem crossing from $S_{1}$ to the lower triplet $T_{1}$ , a spin flip made possible by spin-orbit coupling. The $T_{1} \to S_{0}$ emission is spin-forbidden and therefore slow, with lifetimes from milliseconds to seconds, so a phosphor glows after the light is switched off. The triplet lies below $S_{1}$ , so phosphorescence is red-shifted still further.

The same triplet bottleneck that makes phosphorescence slow makes it useful: because molecules dwell in $T_{1}$ , that state is where photochemistry and, in $O_{2}$ , the reactive singlet-oxygen chemistry of photosensitizers begin.

Electronic band systems close the survey of molecular spectroscopy. A band spectrum is a progression of vibronic bands, each itself a rotational envelope, so a single electronic transition resolves under high dispersion into thousands of rotational lines — the fingerprint by which a diatomic molecule is identified in a stellar atmosphere or a discharge. The coherent counterpart of stimulated emission between such levels builds the laser, where a population inversion between molecular or atomic states turns spontaneous emission into an amplified beam.