When it comes to measuring the phonon dispersion at the edge of a Brillouin zone, why are neutrons chosen and not photons?

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    $\begingroup$ Photon dispersion or phonon dispersion? $\endgroup$ – probably_someone Jul 7 at 18:10
  • $\begingroup$ Thanks, I corrected it. $\endgroup$ – MathIsFun Jul 7 at 18:18
  • $\begingroup$ Getting the required $k$ out of a photon is much harder. $\endgroup$ – Jon Custer Jul 7 at 18:24

Phonons have energies in the range 10-100 meV. The wave vector of photons with energies close to the energies of phonons is very small compared to the typical size of a Brillouin zone. Therefore, photons can mostly probe zone-center phonons (e.g. Raman or IR spectroscopies). To probe zone-boundary phonons you need a large wave vector with an energy in the 10-100 meV scale, and this is what neutrons provide.

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  • $\begingroup$ Hi @ProfM. If we take X-ray photons (which have a range of their wave length from $10^{-11}$ to $10^{-8}$), then the wave vector isn't much smaller than the typical size of a BZ, is it? Also: Why do zone-center phonons have a smaller energy than zone-boundary phonons? $\endgroup$ – MathIsFun Jul 7 at 20:18
  • $\begingroup$ Phonons have typical energies in the scale 10-100 meV, so even though X-rays may have a momentum that would allow them to explore further afield in the BZ, their energies (of the order of 100 eV) are way off, so are not useful to probe phonons. $\endgroup$ – ProfM Jul 7 at 20:26
  • $\begingroup$ As to your question of why zone-center phonons have a smaller energy than zone-boundary phonons, I don't think this is generally true. Many systems have zone-center (optical) phonons that are higher in energy than the corresponding zone-boundary phonons. $\endgroup$ – ProfM Jul 7 at 20:27

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