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Brillouin Zones, Phonons, Energy Propagation through a Crystal

  1. Oct 8, 2013 #1
    Hello, I am having quite a bit of trouble really grasping Brillouin Zones and their relation to phonons, energy propagation, etc. I've got a few questions, and there will probably be a number of misconceptions, but I figure they'll clarify what I exactly don't understand. I think a lot of the formalism is lost on me, and while I can with it to get the answers to problems, conceptually I don't get it.

    1) So the first Brillouin Zone (BZ) has a range of k's from -pi/a to pi/a. Within this range, w (frequency) varies, and in the diagram in my book for a 1-D chain, it is maximal at the edges of the BZ. Now when we talk about w's and k's, are we basically just saying that certain wavelengths of light (with a given k), will cause phonon propagation with a given w through the sample? Or is w referring to something else, like lattice vibrations?

    2) When we say "the edge of a BZ, what does that mean exactly? Just the range of allowed k's that will propagate through the lattice? Does it have any relation at all to physical space? I ask because my text shows an image of atoms (connected via springs in a harmonic oscillator-like fashion) immediately above the k vs w graph of the 1st BZ. These atoms obviously get displaced as energy propagates through them, and it is tempting for me to associate something about a BZ with this displacement, but I can't make sense of it.

    3) dw/dk=0 at the edges of the BZ (that is where w peaks). Apparently, that means that phase velocity = 0, and therefore energy is not propagating. But atoms are still moving as a result of the light passing through (when k=+/-pi/a)-- is that right? Also, if there are standing waves on the edge of a BZ, is there any other consequence I should be aware of?

    4) My book also says that the shortest possible wavelength must be 2a. This is the wavelength of incoming light? Or is it of the lattice vibrations? I ask because my professor also mentioned that as k approaches zero, the wavelength goes to infinity (and I'm not sure what is going to infinity here, but he was talking about acoustic and optical phonons).

    5) Acoustic & Optical Phonons. From what I can gather, acoustic phonons are the movement of atoms fo a lattice out of their equilibrium position, while optical phonons are movements within a lattice (when there are 2+ atoms per basis). In the case of NaCl, wouldn't Na & Cl individually generate acoustic phonons? And if so, wouldn't the "intra-lattice description" (optical phonons) be redundant?

    6) Finally, Diamond is made up of just carbon, so the masses and charges are equal for all atoms within the basis. So why would there be any optical phonons (wouldn't a k-vector have the same effect on all atoms within a basis?)?

    Any insight on any portion of this would be greatly appreciated, as you can see I have a lot of confusion. Thank you.
  2. jcsd
  3. Oct 8, 2013 #2


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    1) the wave vectors are for the vibrational states of the crystal
    2) the Brillouin Zone (BZ) is located in reciprocal space - a Fourier transform of the crystal's configuration space
    3) acoustic phonons are the lowest vibrational modes; optical phonons are all others. There are a number of differences, but acoustic phonons show dispersion like sound waves - for more description and images see:
    http://www.ece.rochester.edu/course...SC426 Workshop06/term papers 06/Mathew_06.pdf

    4) 2a is the shortest wavelength for a lattice vibration
    5) The degrees of freedom to describe a vibrational mode come from a single unit cell ...
    6) Diamond has optical phonons because the unit cell has more than one atom; the masses do not enter the degrees of freedom

    This set of graduate notes is a good summary for many of your questions:
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