Why do we have two different types of phonons, optical and acoustical? Why are they called optical phonons?


As I understand this occurs when a cystalline solid has more than one atom in the primitive cell of the lattice. For simplicity I'll consider the case of two types of atoms, distinguished by their position in the primitive cell (but which may or may not be otherwise identical). This gives rise to multiple modes of vibration. The acoustical phonon corresponds to both types of atoms moving in phase in the long wavelength limit (that is, near the center of the 1st Brillouin zone), whereas the optical phonon corresponds to the two types moving out of phase, such that adjacent atoms move in opposite directions in the long-wavelength limit.

(For the case of only one atom in the primitive cell, you can also get adjacent atoms moving in opposite directions, but this happens at the edge of the 1st Brillouin zone, in the limit where the wave becomes a standing wave.)

They are called "optical" because in ionic crystals they can be excited by EM radiation, with the positive ions moving one way while their negatively charged neighbors move the other way.

We can see the two modes from the dispersion relation, which for two atoms in the primitive cell (in the 1-dimensional case) is given by:

$$\omega^2 = \beta \left( \frac{1}{m} + \frac{1}{M} \right) \pm \sqrt{\left( \frac{1}{m} + \frac{1}{M} \right)^2 - \frac{4 \sin^2 ka}{Mm}}$$

Here m and M are the masses of the two types of atoms. There are two branches, the lower of which is the acoustical branch and the higher of which is the optical branch, with a gap between them. (Note that the two branches still exist for equal masses, but the gap closes.)

Equation copied from this site, which also has this nice animation of the two modes.

  • $\begingroup$ To clarify, I believe they'd be 180 degrees out of phase if the wavelength of the radiation is large compared to the lattice spacing. $\endgroup$ Sep 23 '13 at 17:42
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    $\begingroup$ why couldn't the atoms in a one-type-of-atom solid move out of phase also? Wouldn't those be optical phonons also? $\endgroup$
    – BeauGeste
    Sep 24 '13 at 1:54
  • $\begingroup$ @BeauGeste I had seen some references to "two types of atoms", but on further review I think this really means "two atoms in a unit cell of the lattice". I.e., the two atoms are distinguished only by the geometry of the lattice. For instance, this page chembio.uoguelph.ca/educmat/chm729/phonons/optical.htm shows how the dispersion relations give rise to acoustical and optical branches -- the calculations are for different masses but you can see that the branches still exist when the masses are the same, just without the gap between them. $\endgroup$ Sep 24 '13 at 14:34
  • $\begingroup$ That should say "I.e. the two atoms need only be distinguished by the geometry of the lattice, they can of course also have different charge, mass, etc. With one atom per unit cell you can also get neighboring atoms moving in opposite directions, but I think they only go 180 degrees out of phase in the limit that the wave becomes a standing wave. This is different than the optical mode, which is more like two propagating waves (one for each "type" of atom) that are out of phase with each other. $\endgroup$ Sep 24 '13 at 14:48
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    $\begingroup$ @Seanny123 Yes, it's a very important fundamental concept in solid state physics. I'm not sure if I can do justice to it in a comment, perhaps you should post a separate question on that? (Or have a look at an introductory text on solid state.) The short version is that it's a cell of the reciprocal lattice (the lattice of k-vectors that arises from the Fourier transforms of the states in your original lattice). It is relevant both to the case where the primitive cell contains one atom or multiple atoms. $\endgroup$ Apr 25 '14 at 15:06

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