# Why does Raman Activity Require Anisotropic Polarizability?

Normal mode of a molecule needs to involve the change in molecular polarizability (anisotropic 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, $$f_i$$ is occurs in the induced dipole formula if the polarizability is isotropic. If it's anisotropic, two additional frequencies occur, $$f_i - 2f_R$$ and $$f_i + 2f_R$$ corresponding to Raman shift (Stokes and anti-Stokes lines) where $$f_R$$ is the rotational frequency of the molecule. In another words, induced dipole of the molecule has two additional frequencies of oscillation. This explanation is clear, but it's quite math based without much intuition.

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

The Raman effect involves scattering of a photon with a change in energy due to vibrations or rotations of the molecule. It is changes in the polarisability ellipsoid (either size, shape or orientation) that causes the effect. You can think of polarisability as the shape of the electron distribution in a molecule, we usually imagine this as an ellipsoid. In the case of a linear molecule such as $$H_2$$ the ellipsoid is prolate (cigar-shaped, rather than plate-shaped which is oblate) and so when it rotates end to end the polarisability changes in orientation, however, the same shape is presented to the electric field of the radiation twice in one rotation, this means that the $$J$$ quantum number changes by 2 in a rotational transition. In a classical model, the electric field of the radiation and the rotating polarisability can interact and then energy can be taken from the radiation or given to it in the Raman transition. Rotation about the bond axis shows no change in the shape of the polarisability and so there is Rayleigh scattering only and at the same frequency as the radiation. If the molecule is 'cold' and $$J=0$$ is the only level populated only one rotational frequency can be measured ($$J=0\to 2$$) even though two lines appear in the spectrum, Stokes and anti-Stokes. (In a real spectrum many rotational levels are populated as rotational quanta are usually small wrt thermal energy so many rotational levels may be populated and the spectrum has many lines.)

Suppose that the radiation moves along x direction with polarisation in the z direction and polarises the molecule forming an ellipsoid pointing in some direction $$x',y',z'$$ and the ellipsoid has axis $$\alpha_s$$ and $$\alpha_L$$ for short and long axes. The long axis makes the angle $$\theta$$ to the radiation polarisation or $$z$$ axis. This angle changes as the molecule rotates. If the components of the polarisation are worked out on the z and x axis they are
$$M_z=(\alpha_s+\alpha_L)E_z/2+(\alpha_s-\alpha_L)\cos( 2\pi 2v_{rot}tE_z )/2$$
$$M_x=(\alpha_s-\alpha_L)\sin( 2\pi 2v_{rot}tE_z )/2$$
what matters here is the term $$(\alpha_s-\alpha_L)$$ which is zero if the ellipsoid is spherical, i.e. the polarisation is not zero if the ellipsoid is prolate or oblate. The term $$(\alpha_s+\alpha_L)$$ causes Rayleigh scattering due to rotational motion.
(In passing vibrational Raman spectra can also be observed. In this case it is $$d\alpha/dr$$ that matters so a change in size of the polarisability is all that is needed and thus all diatomic molecules have a vibrational Raman spectrum.)