---
title: Raman Scattering and Electronic Bands
module: Molecular Spectra
moduleNumber: 2
lessonNumber: 3
order: 203
summary: >
  Not every vibration absorbs in the infrared. Raman scattering reaches modes
  that modulate the polarizability, giving Stokes and anti-Stokes lines whose
  intensity ratio measures temperature, and the mutual-exclusion rule pairs it
  with infrared absorption. Electronic transitions add the vibronic structure of
  band spectra, governed by the Franck-Condon principle, and the radiative fates
  of an excited state are sorted by the Jablonski diagram into fluorescence and
  phosphorescence.
topics: [Molecular Spectra]
draft: false
sources:
  - book: Tipler & Llewellyn
    ref: "Ch. 9 — Molecular Structure and Spectra; §9-5 Scattering, Absorption, and Stimulated Emission"
  - book: Kittel
    ref: "Ch. 15 — Optical Processes and Excitons"
---

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
$\text{N}_2$ or $\text{O}_2$. 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 $\nu_0$ falls on a molecule, its electric field
$\vec E = \vec E_0\cos(2\pi\nu_0 t)$ induces a dipole moment $\vec p = \alpha
\vec E$, where $\alpha$ is the polarizability. The oscillating dipole reradiates
at $\nu_0$; this is **Rayleigh scattering**, elastic and much stronger than
anything shifted. But $\alpha$ is not constant: as the molecule vibrates with
frequency $\nu_{\text{vib}}$ along a normal coordinate $Q = Q_0\cos(2\pi\nu_{\text{vib}}t)$,
the polarizability is modulated,

$$
\alpha(t) = \alpha_0 + \left(\frac{\partial\alpha}{\partial Q}\right)_0 Q_0\cos(2\pi\nu_{\text{vib}} t).
$$

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

$$
\vec p = \alpha_0 \vec E_0\cos(2\pi\nu_0 t)
+ \tfrac12\left(\frac{\partial\alpha}{\partial Q}\right)_0 Q_0 \vec E_0
\Bigl[\cos\!\bigl(2\pi(\nu_0 - \nu_{\text{vib}})t\bigr) + \cos\!\bigl(2\pi(\nu_0 + \nu_{\text{vib}})t\bigr)\Bigr].
$$

The unshifted line is Rayleigh; the sideband at $\nu_0 - \nu_{\text{vib}}$ is the
**Stokes** line, where the molecule gains a vibrational quantum, and the sideband
at $\nu_0 + \nu_{\text{vib}}$ is the **anti-Stokes** line, where it gives one up.
The shift $\nu_{\text{vib}}$ is the vibrational frequency, read as a displacement
from the exciting line regardless of what $\nu_0$ 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).

$$
% caption: Rayleigh and Raman scattering through a virtual level: Rayleigh returns
% to the starting level, Stokes leaves the molecule one quantum higher, and
% anti-Stokes starts one quantum up and ends in the ground vibrational level.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  % vibrational levels v=0, v=1
  \draw[black, thick] (0,0) -- (9.6,0);
  \node[anchor=east, font=\scriptsize] at (-0.05,0) {$v=0$};
  \draw[black, thick] (0,0.9) -- (9.6,0.9);
  \node[anchor=east, font=\scriptsize] at (-0.05,0.9) {$v=1$};
  % virtual level (dashed)
  \draw[black, dashed] (0,3.6) -- (9.6,3.6);
  \node[anchor=east, font=\scriptsize, black] at (-0.05,3.6) {virtual};
  % Rayleigh
  \draw[acc, ->, thick] (1.0,0.02) -- (1.0,3.55);
  \draw[acc, ->, thick] (1.6,3.55) -- (1.6,0.02);
  \node[anchor=north, font=\scriptsize] at (1.3,-0.15) {Rayleigh};
  % Stokes: up from 0, down to 1
  \draw[acc, ->, thick] (4.0,0.02) -- (4.0,3.55);
  \draw[acc, ->, thick] (4.6,3.55) -- (4.6,0.92);
  \node[anchor=north, font=\scriptsize] at (4.3,-0.15) {Stokes};
  % anti-Stokes: up from 1, down to 0
  \draw[acc, ->, thick] (7.4,0.92) -- (7.4,3.55);
  \draw[acc, ->, thick] (8.0,3.55) -- (8.0,0.02);
  \node[anchor=north, font=\scriptsize] at (7.7,-0.15) {anti-Stokes};
\end{tikzpicture}
$$

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
$\nu^4$ scattering factor, the intensity ratio is

$$
\frac{I_{\text{anti-Stokes}}}{I_{\text{Stokes}}} =
\left(\frac{\nu_0 + \nu_{\text{vib}}}{\nu_0 - \nu_{\text{vib}}}\right)^{\!4}
e^{-hc\,\nu_{\text{vib}}/k_B T}.
$$

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.

$$
% caption: The scattered spectrum: an intense Rayleigh line at the laser frequency
% flanked by weaker Raman sidebands, the Stokes line below and the fainter
% anti-Stokes line above, displaced by the vibrational frequency.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[->, black] (0,0) -- (10.4,0) node[below] {frequency};
  \draw[->, black] (0,0) -- (0,4.2) node[left] {intensity};
  % Rayleigh (tall) at center 5.2
  \draw[acc, very thick] (5.2,0) -- (5.2,3.9);
  \node[acc, anchor=south, font=\scriptsize] at (5.2,3.9) {Rayleigh};
  % Stokes (medium) at 3.0
  \draw[acc, very thick] (3.0,0) -- (3.0,1.7);
  \node[anchor=north, font=\scriptsize] at (3.0,-0.1) {Stokes};
  % anti-Stokes (short) at 7.4
  \draw[black, very thick] (7.4,0) -- (7.4,0.7);
  \node[anchor=north, font=\scriptsize] at (7.4,-0.1) {anti-Stokes};
  % shift markers
  \draw[black, <->] (3.0,2.1) -- (5.2,2.1) node[midway, above, font=\scriptsize] {shift};
  \draw[black, <->] (5.2,1.3) -- (7.4,1.3) node[midway, above, font=\scriptsize] {shift};
\end{tikzpicture}
$$

## Complementary selection rules

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

- **Infrared active** if the mode changes the dipole moment: $(\partial\mu/\partial Q)_0 \neq 0$.
- **Raman active** if the mode changes the polarizability: $(\partial\alpha/\partial Q)_0 \neq 0$.

For a molecule with a center of symmetry the two conditions are mutually
exclusive: a mode symmetric under inversion changes $\alpha$ but not $\mu$, and an
antisymmetric mode changes $\mu$ but not $\alpha$. 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 |
| --- | --- | --- | --- | --- |
| $\text{CO}_2$ symmetric stretch | none | yes | inactive | active |
| $\text{CO}_2$ antisymmetric stretch | yes | none | active | inactive |
| $\text{CO}_2$ bend | yes | none | active | inactive |
| $\text{N}_2$ stretch | none | yes | inactive | active |

The homonuclear molecules $\text{N}_2$ and $\text{O}_2$, 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
$\Delta J = 0, \pm 2$, since the polarizability returns to itself twice per
rotation, giving lines spaced by $4B$ rather than the $2B$ 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.

$$
% caption: The Franck-Condon principle: nuclei are fixed during the fast
% electronic jump, so the transition is vertical; with the upper curve displaced
% outward, the strongest line lands on the upper vibrational level whose turning
% point sits above the lower ground state.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[->, black] (0,-0.2) -- (0,5.4) node[left] {$U(r)$};
  \draw[->, black] (0,0) -- (7.2,0) node[below] {$r$};
  % lower electronic state well, min near x=2.0
  \draw[black, very thick, domain=1.1:4.3, samples=120, variable=\x]
    plot ({\x},{1.7*(\x-2.0)*(\x-2.0)});
  \node[black, anchor=east, font=\scriptsize] at (1.15,1.5) {ground};
  % ground v=0 level
  \draw[black] (1.55,0.35) -- (2.45,0.35);
  % upper electronic state well, displaced to x=3.2, raised
  \draw[acc, very thick, domain=2.2:5.6, samples=120, variable=\x]
    plot ({\x},{1.4*(\x-3.2)*(\x-3.2) + 2.9});
  \node[acc, anchor=west, font=\scriptsize] at (5.0,4.4) {excited};
  % upper vibrational levels
  \foreach \y in {3.25,3.75,4.25}
    \draw[acc] (2.55,\y) -- (3.85,\y);
  % vertical transition at r=2.0
  \draw[acc, ->, very thick] (2.0,0.4) -- (2.0,3.85);
  \draw[black, dashed] (2.0,0) -- (2.0,-0.2) node[below, font=\scriptsize] {frozen $r$};
\end{tikzpicture}
$$

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**,

$$
I_{v' \leftarrow v''} \propto \bigl|\langle \psi_{v'} \,|\, \psi_{v''}\rangle\bigr|^2,
$$

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.

$$
% caption: A vibronic absorption progression: lines at the upper vibrational
% levels v' from the ground level, their intensities following the Franck-Condon
% factors so the band peaks at an intermediate v' set by the curve displacement.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[->, black] (0,0) -- (9.6,0) node[below] {photon energy};
  \draw[->, black] (0,0) -- (0,3.6) node[left] {absorption};
  % progression lines with FC-envelope heights
  \foreach \x/\h/\lab in {1.2/0.6/0,2.4/1.6/1,3.6/2.8/2,4.8/3.1/3,6.0/2.4/4,7.2/1.4/5,8.4/0.6/6}{
    \draw[acc, very thick] (\x,0) -- (\x,\h);
    \node[anchor=north, font=\scriptsize] at (\x,0) {\lab};
  }
  \node[anchor=north, font=\scriptsize] at (4.8,-0.5) {upper vibrational level};
  % envelope
  \draw[black, densely dotted] (1.2,0.6) .. controls (3.0,3.4) and (5.6,3.4) .. (8.4,0.6);
\end{tikzpicture}
$$

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).

$$
% caption: Jablonski diagram: absorption to a singlet, fast nonradiative relaxation
% to the lowest excited singlet, then either spin-allowed fluorescence back to the
% ground state or intersystem crossing to a triplet followed by slow
% phosphorescence.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  % S0 ground
  \draw[black, thick] (0.5,0) -- (3.0,0);
  \node[anchor=east, font=\scriptsize] at (0.45,0) {$S_0$};
  % S1
  \draw[acc, thick] (0.5,3.2) -- (3.0,3.2);
  \node[anchor=east, font=\scriptsize, acc] at (0.45,3.2) {$S_1$};
  % higher singlet S2
  \draw[black, thick] (0.5,4.2) -- (3.0,4.2);
  \node[anchor=east, font=\scriptsize, black] at (0.45,4.2) {$S_2$};
  % triplet T1 (to the right, lower than S1)
  \draw[black, thick] (5.0,2.4) -- (7.2,2.4);
  \node[anchor=west, font=\scriptsize, black] at (7.25,2.4) {$T_1$ triplet};
  % absorption
  \draw[acc, ->, very thick] (1.1,0.05) -- (1.1,4.15);
  \node[anchor=east, font=\scriptsize] at (1.05,2.0) {absorb};
  % internal conversion S2->S1
  \draw[black, ->, thick] (1.6,4.15) -- (1.6,3.25);
  \node[anchor=west, font=\scriptsize, black] at (1.62,3.7) {relax};
  % fluorescence S1->S0
  \draw[acc, ->, very thick] (2.4,3.15) -- (2.4,0.05);
  \node[anchor=west, font=\scriptsize, acc] at (2.42,1.6) {fast emit};
  % intersystem crossing S1->T1
  \draw[black, ->, thick] (3.05,3.1) -- (4.95,2.5);
  \node[anchor=south, font=\scriptsize, black] at (4.0,2.85) {crossing};
  % phosphorescence T1->S0
  \draw[black, ->, very thick] (6.0,2.35) -- (6.0,0.05);
  \node[anchor=west, font=\scriptsize, black] at (6.05,1.2) {slow emit};
  \draw[black, thick] (5.0,0) -- (7.2,0);
\end{tikzpicture}
$$

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 $10^{-9}$-$10^{-7}\ \text{s}$, 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
$\text{O}_2$, 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](/condensed-matter/molecular-spectra/lasers-and-masers),
where a population inversion between molecular or atomic states turns spontaneous
emission into an amplified beam.