# Do I understand measurement of dispersion relation in a solid correctly?

I'm currently doing an introduction to solid state physics course and have a quick question about measurement of the dispersion relation of phonons in a solid:

The way I understood it is the following. One can look at the vibrations of the atoms in a solid in a quantum mechanical way and introduce phonons as a quasiparticle which is a boson. One can then derive the following laws of energy and momentum conservation for the interaction of photons with phonons:

$\hbar \omega(q) = \hbar \omega_0 \pm \hbar \omega_{Ph}$ (Energy conservation)

$\hbar q = \hbar q_0 \pm \hbar q_{Ph} + G$ (Momentum conservation)

Where the subscript ph denotes the frequency / q-value of a phonon, and the subscript 0 denotes the frequency/ q-value of the photon after the photon and phonon interacted with each other. G is a reciprocal lattice vector.

To sample the dispersion relation of the phonons in the solid one can now simply shoot photons at the solid and look at the out coming phonons.

Solving the second equation for $q_{Ph}$ and the first one for $\omega_{Ph}$ (assuming we can properly measure the frequency and impulses of the out coming and incoming photons) then give us a relation between the q-values of the phonons and the frequency which is exactly the dispersion relation.

Did I understand this correctly or are there any flaws in my reasoning?

Cheers!

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