Why Does Raman Activity Require Anisotropic Polarizability?

[Dario56]

It's mentioned that the normal mode of molecule needs to involve the change in molecular polarizability to be Raman active.

Explanation is provided in Physical Chemistry textbook by Atkins on the example of the rotational Raman spectra. Only the frequency of the electric field $(f_i)$ occurs in the induced dipole formula if the polarizability is isotropic. If it's anisotropic, two additional frequenices occur $(f_i + 2f_R)$ and $(f_i - 2f_R)$ corresponding to Raman shift (Stokes and anti-Stokes lines), where $f_R$ is rotational frequency of the molecule. This explanation is clear, but it's quite math based without much intution.

Can you give more intuitive explanation of the relation between anisotropic polarizability and Raman activity?

 

[TeethWhitener]

I had typed out a long explanation but then realized it was far more math-based than your OP! The most intuitive explanation is probably that the Raman effect is a two-photon process. Since the EM field is represented by a vector, two-photon processes must be represented by a rank-2 tensor. We call that tensor the polarizability tensor. I don't know if that helps at all.