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Experimental dispersion relation of phonons

  1. Jul 7, 2009 #1
    Hi guys,

    I don't understand how one would exactly determine a dispersion relation
    of phonons experimentally.

    There are two equations, one for momentum and one for energy conservation:

    [tex]\vec{k} - \vec{k^{'}} = \vec{G} + \vec{K} [/tex]

    [tex]\omega - \omega ^{'} = \omega(K) [/tex]

    where [tex]\omega(K)[/tex] is the energy difference of the scattered neutrons,
    [tex]\vec{k} [/tex], [tex] \vec{k^{'}} [/tex] are the wave vectors of the neutrons before and
    after scattering, [tex]\vec{K} [/tex] is the created phonon and [tex]\vec{G} [/tex] a
    reciprocal lattice vector.

    The question is, how is the difference [tex]\vec{k} - \vec{k^{'}} [/tex] respectively [tex]\vec{K} [/tex] determined experimentally?

    Of course I also have to know [tex]\vec{G} [/tex].

  2. jcsd
  3. Jul 7, 2009 #2
    You can't. Phonons simply do not have momentum, but *lattice*-momentum, which is lives on a torus (or some other compact manifold) rather than a plane. This is because the underlying crystal violates translational symmetry, reducing it to a discrete one. Similarly, electrons in crystals only have lattice momentum.

    But, you will say, what about when I shoot something into it? Outside, it's got definite momentum, but inside it doesn't?

    Answer: Yep. It's momentum will get broken into two components, one that lies in the Brillouin zone, and a component that is a multiple of the reciprocal lattice vector. The latter is transferred to the entire crystal as a whole. Upon emission, the opposite happens, one component from within the Brillouin zone is added to a crystal momentum to give the exiting particle its true momentum.

    So when determining the dispersion of phonons, one would shoot in some neutrons, and observe the change of (true) momentum. This curve should repeat after translation by one reciprocal lattice vector, from which the zone can be determined. Then folding the curve back on itself gives the desired relationship.
  4. Jul 8, 2009 #3
    Hi genneth,

    thanks for your (very good) explanation. It helps a lot.
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