Setup

The refractive index of most optical materials is monotonically decreasing, sideways sigmoidal in shape, with its highest value (for the purposes of this discussion) at the short wavelength (UV/Vis) side and its lowest value at the long wavelength (IR) side. An example of Silica is shown in the attached figure as a reference.

We typically refer to the short wavelength cutoff as the electron-transition edge and the long wavelength as the multi-phonon edge. At each of these edges, the photon begins to interact with the lattice of the optical material: on the short (high energy) side, by exciting an electronic transition in one of the constituent atoms, and at the long (low energy) side by exciting a multi-atom vibration called a phonon. In each of these cases the energy of the incident photon is transferred to the lattice and the photon is absorbed.

Question

Since the Refractive Index is nothing but the speed of light in the optical material (n=c/v , where c is the speed in vacuum and v is the phase velocity in the material), why does the Refractive Index decrease (implying an increased phase velocity) at the multi-phonon edge?

Shouldn't the fact that the photon is interacting with the lattice through the excitation of phonons cause it to have a decreased phase velocity and thus a higher refractive index?

What am I missing?

Thanks in advance for your thoughtful responses.

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