Raman Scattering: Molecular Polarizability

Introduction

When a laser beam passes through a transparent liquid, almost all the light goes straight through. A tiny fraction is scattered sideways, and most of that has exactly the incident colour. Yet about one scattered photon in ten million returns with a slightly different energy, carrying a precise record of the molecule's vibrations. This Raman effect , discovered by C. V. Raman and K. S. Krishnan in 1928, rests not on a dipole moment but on how easily an electron cloud can be distorted: its polarisability .

Core explanation

Induced dipoles. An electric field E pushes a molecule's electrons one way and its nuclei the other, creating an induced dipole

μ ind = αE

where α is the polarisability. Large, diffuse electron clouds (iodine, sulfur, extended π systems) have large α; compact clouds (fluorine) have small α.

Oscillating field, oscillating dipole. Light of frequency ν₀ has a field E = E₀ cos(2πν₀t). The induced dipole oscillates at ν₀ and re-radiates light at the same frequency. This elastic process is Rayleigh scattering .

Adding vibration. A vibrating molecule changes shape at its vibrational frequency ν vib, and its polarisability changes with it:

α = α₀ + (∂α/∂Q)₀ Q₀ cos(2πν vib t)

Multiplying α by the oscillating field and using a product-to-sum identity gives three terms: one at ν₀ (Rayleigh), one at ν₀ − ν vib ( Stokes Raman ) and one at ν₀ + ν vib ( anti-Stokes Raman ). The Raman lines exist only if the second term is non-zero.

Selection rule. A vibration is Raman active if the polarisability changes during the vibration: (∂α/∂Q)₀ ≠ 0. This is quite different from infrared, where the dipole moment must change. Homonuclear diatomics such as N₂ (2331 cm⁻¹), O₂ (1556 cm⁻¹) and H₂ (4161 cm⁻¹) have no dipole moment at all and are IR-inactive, but stretching changes the size of their electron clouds, so they are Raman active.

Quantum view. A photon raises the molecule to a short-lived virtual state , not a real electronic level; the molecule immediately re-emits. If it returns to a higher vibrational level, the photon has lost energy (Stokes); if it started in an excited level and returns lower, the photon has gained energy (anti-Stokes).

Raman shift. Spectra are plotted against Raman shift Δν̃ = ν̃₀ − ν̃ scattered in cm⁻¹, so the band positions are independent of the laser colour. Scattered intensity scales as ν⁴, so blue lasers give stronger signals but often more fluorescence.

Polarisation. With a polarised laser, totally symmetric vibrations give strongly polarised bands (depolarisation ratio ρ < 3/4), while all other modes give ρ = 3/4. This is a direct experimental test of vibrational symmetry.

Formulae

μ ind = αE; Raman active if (∂α/∂Q)₀ ≠ 0; Raman shift Δν̃ = ν̃₀ − ν̃ s (cm⁻¹); ν̃ (cm⁻¹) = 10⁷ / λ (nm); scattered intensity ∝ ν⁴.

Step-by-step reasoning

To decide whether a vibration is Raman active:

1. Picture the molecule at the two extremes of the vibration. 2. Ask whether the electron cloud is larger, smaller or differently shaped at the two extremes. 3. If the polarisability differs between the extremes, the mode is Raman active. 4. If the two extremes are mirror images with identical polarisability, the first-order change is zero and the mode is inactive.

Visual explanation

Draw the polarisability of a molecule as an ellipsoid surrounding it. For the symmetric stretch of CO₂, the ellipsoid swells and shrinks as the bonds lengthen and shorten: Raman active. For the antisymmetric stretch, the ellipsoid at one extreme is identical to that at the other, so the mode is Raman inactive to first order.

Real-world analogy

Think of a trampoline struck by rain. A rigid, unchanging surface bounces each drop back with the same speed. If the trampoline is itself flexing up and down, drops leave slightly faster or slower depending on its motion. The oscillating polarisability is the moving surface that alters the energy of scattered light.

Real-world example

Portable Raman analysers are used at borders and by pharmaceutical quality control to identify substances through sealed glass or plastic packaging. Glass and water scatter weakly, so a laser can probe the contents without opening the container, and the fingerprint is matched against a spectral library in seconds.

Why?

Why is Raman scattering so weak? Only molecules whose polarisability is modulated by vibration contribute, and the modulation is a small fraction of the total polarisability. The Raman term is therefore orders of magnitude smaller than the Rayleigh term, which itself is only a small fraction of the transmitted light.

Common misconception

"Raman lines are fluorescence." Fluorescence involves absorption into a real excited electronic state followed by emission after nanoseconds; Raman scattering is essentially instantaneous via a virtual state, and its shifts move exactly with the laser frequency.

Worked example

Question: A 532 nm laser excites a band with a Raman shift of 1000 cm⁻¹. At what wavelength does the Stokes line appear?

Reasoning: ν̃₀ = 10⁷/532 = 18 797 cm⁻¹. Stokes line: 18 797 − 1000 = 17 797 cm⁻¹. λ = 10⁷/17 797 = 561.9 nm.

Answer: About 562 nm.

Quick check

1. Is the stretching vibration of N₂ infrared active, Raman active, or both? Explain briefly. Answer: Raman active only: the stretch changes the polarisability, but N₂ has no dipole moment to change.

Exam focus

Examiners reward a clear derivation of the three frequencies from μ = αE, the selection rule (∂α/∂Q)₀ ≠ 0, conversion of Raman shift to wavelength, and the contrast with infrared selection rules. Remember that Raman shifts are independent of the excitation wavelength.

Advanced insight

Polarisability is really a second-rank tensor, relating the three components of the field to the three components of the induced dipole. In group theory a mode is Raman active if it transforms like a quadratic function such as x², xy or x² − y² in the character table. Totally symmetric modes change the trace of the tensor, which is why they alone can give depolarisation ratios below 3/4.

Summary

Light induces a dipole μ = αE. Scattering at the incident frequency is Rayleigh; modulation of α by a vibration adds weak Stokes (ν₀ − ν vib) and anti-Stokes (ν₀ + ν vib) lines. A mode is Raman active if its polarisability changes, (∂α/∂Q)₀ ≠ 0, so homonuclear diatomics are Raman active though IR-inactive. Spectra are plotted as Raman shifts in cm⁻¹, and polarisation measurements identify totally symmetric modes.

Practice questions

1. State the selection rule for a vibration to be Raman active. Answer: The polarisability must change during the vibration, (∂α/∂Q)₀ ≠ 0. 2. Explain why Rayleigh scattering is much stronger than Raman scattering. Answer: Rayleigh scattering depends on the whole equilibrium polarisability α₀, whereas Raman scattering depends only on the small change in α caused by the vibration. 3. Using 785 nm excitation, find the wavelength of the Stokes band for a 1600 cm⁻¹ vibration. Answer: ν̃₀ = 10⁷/785 = 12 739 cm⁻¹; Stokes at 12 739 − 1600 = 11 139 cm⁻¹, so λ = 10⁷/11 139 ≈ 897.7 nm. 4. A Raman band has depolarisation ratio 0.05. What does this reveal about the vibration? Answer: The band is strongly polarised (ρ < 3/4), so the vibration is totally symmetric.