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.

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