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# Classical Theory of Raman Scattering

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The classical theory of Raman effect, also called the polarizability theory, was developed by G. Placzek in 1934. I shall discuss it briefly here. It is known from electrostatics that the electric field $E$ associated with the electromagnetic radiation induces a dipole moment $\mu$ in the molecule, given by
$\mu = \alpha E$ …….(1)
where $\alpha$ is the polarizability of the molecule. The electric field vector $E$ itself is given by
$E = E_0 \sin \omega t = E_0 \sin 2\pi \nu t$ ……(2)
where $E_0$ is the amplitude of the vibrating electric field vector and $\nu$ is the frequency of the incident light radiation.

Thus, from Eqs. (1) & (2),
$\mu= \alpha E_0 \sin 2\pi \nu t$ …..(3)
Such an oscillating dipole emits radiation of its own oscillation with a frequency $\nu$, giving the Rayleigh scattered beam. If, however, the polarizability varies slightly with molecular vibration, we can write
$\alpha =\alpha_0 + \frac {d\alpha} {dq} q$ …..(4)
where the coordinate q describes the molecular vibration. We can also write q as:
$q=q_0 \sin 2\pi \nu_m t$ …..(5)
Where $q_0$ is the amplitude of the molecular vibration and $\nu_m$ is its (molecular) frequency. From Eqs. 4 & 5, we have
$\alpha =\alpha_0 + \frac {d\alpha} {dq} q_0 \sin 2\pi \nu_m t$ …..(6)
Substituting for $\alpha$  in (3), we have
$\mu= \alpha_0 E_0 \sin 2\pi \nu t + \frac {d\alpha}{dq} q_0 E_0 \sin 2\pi \nu t \sin 2\pi \nu_m t$ …….(7)
Making use of the trigonometric relation $\sin x \sin y = \frac{1}{2} [\cos (x-y) -\cos (x+y) ]$ this equation reduces to:
$\mu= \alpha_0 E_0 \sin 2\pi \nu t + \frac {1}{2} \frac {d\alpha}{dq} q_0 E_0 [\cos 2\pi (\nu - \nu_m) t - \cos 2\pi (\nu+\nu_m) t]$ ……(8)
Thus, we find that the oscillating dipole has three distinct frequency components:

1• The exciting frequency $\nu$ with amplitude $\alpha_0 E_0$
2• $\nu - \nu_m$
3• $\nu + \nu_m$ (2 & 3 with very small amplitudes of $\frac {1}{2} \frac {d\alpha}{dq} q_0 E_0$. Hence, the Raman spectrum of a vibrating molecule consists of a relatively intense band at the incident frequency and two very weak bands at frequencies slightly above and below that of the intense band.

If, however, the molecular vibration does not change the polarizability of the molecule then $(d\alpha / dq )=0$ so that the dipole oscillates only at the frequency of the incident (exciting) radiation. The same is true for the molecular rotation. We conclude that for a molecular vibration or rotation to be active in the Raman Spectrum, it must cause a change in the molecular polarizability, i.e., $d\alpha/dq \ne 0$ …….(9)

Homonuclear diatomic molecules such as $\mathbf {H_2 \, N_2 \, O_2}$ which do not show IR Spectra since they don’t possess a permanent dipole moment, do show Raman spectra since their vibration is accompanied by a change in polarizability of the molecule. As a consequence of the change in polarizability, there occurs a change in the induced dipole moment at the vibrational frequency.