21
$\begingroup$

Do sound waves in a gas consist of phonons?

What about a glass? Or other non-crystalline materials such as quasicrystals?

How does the lack of translational symmetry affect the quantization of the displacement field?

All the answers so far have treated this question at a much more elementary level than I was expecting. I am already quite familiar with the properties of phonons in crystals. Therefore, do not explain the well-known derivations of the dispersion relation and second quantization of phonons in crystal lattices in your answer (and especially don't get them wrong!).

Improve this question
$\endgroup$
14
  • $\begingroup$ Yes, yes, yes, and interesting question... $\endgroup$
    – Carl Brannen
    Commented Mar 18, 2011 at 4:48
  • $\begingroup$ @Carl Marek seems to disagree $\endgroup$
    – Tobias Kienzler
    Commented Mar 18, 2011 at 10:51
  • 1
    $\begingroup$ The answers so far seem to completely ignore the fact that people study phonons in disordered systems! I even know some of them! Of course one needs to reinterpret what is a phonon, but clearly the correspondence needs to be elastic waves in the continuum approximation, which has very little to do with symmetry breaking or quantum mechanics. I invite those who think otherwise to simply do a quick google search, which will reveal both research papers and whole books on the subject. $\endgroup$
    – genneth
    Commented Mar 18, 2011 at 14:20
  • $\begingroup$ @genneth: I don't think there is any argument about the fact that unordered systems do carry sound and that this should be quantized. Rather, it's about whether the object you obtain is the same thing you get in lattice. And as you say yourself, you need to modify (or reinterpret) the definition of phonon. So the answer to this question obviously depends on what does one consider a phonon. In any case, I deleted my answer as you were right that it was a little too simplistic (perhaps even wrong). $\endgroup$
    – Marek
    Commented Mar 18, 2011 at 14:33
  • 2
    $\begingroup$ @anna v: for disordered materials, one talks of different realisations of the disorder, and average over them. For each realisation, it should be clear that one gets coupled oscillators and the vibration modes of these are quantised. The question is then what survives the averaging over disorder. The answer, unfortunately, is not universal, and does depend on the precise geometry and degree/type of disorder. Transport properties such as thermal conduction nevertheless can usually be described adequately with field theory or replica theory. E.g.: prb.aps.org/abstract/PRB/v48/i17/p12581_1 $\endgroup$
    – genneth
    Commented Mar 18, 2011 at 18:02

3 Answers 3

Reset to default
5
$\begingroup$

Consider a solid rod made of a glassy substance, and model it as a set of atoms in random locations, held together by randomly oriented harmonic springs (enough so that the graph is rigid). The Hamiltonian of this system is in principle diagonalizable, and since all the springs are harmonic, the potential energy is quadratic in all the atomic coordinates, so the whole system is equivalent to a set of uncoupled harmonic oscillators, which are of course the normal modes. The lowest frequency mode will be rod flexing back and forth with two nodes, the second will be the second overtone with three nodes, and so on, but importantly this set of normal modes goes all the way up to modes with a number of nodes on the order of the number of atoms in the rod. None of these normal modes has a precisely defined wavevector, because of the lack of periodicity in the glass, but they do have approximate wavevectors, and of course they have precise frequencies corresponding to the eigenvalues of the Hamiltonian.

One weird thing about this that seems completely different from the usual treatment of phonons in crystals is that all these modes are standing waves - they don't propagate in a particular direction, and their approximate wavevectors are only defined up to sign. This is actually the case for a finite-size crystal as well - no propagating waves can actually be eigenstates, and instead the boundary conditions at the ends of the crystal cause them to mix into standing waves that are the exact physical eigenstates. The only reason we introduce periodic boundary conditions and talk about the propagating waves in crystals is that it's so much more convenient. Of course, if you create a wave packet at one end of the material, either crystal or glass, you can always express that in whatever basis of eigenstates you want, and as they evolve the packet will end up moving through the crystal and spreading out according to some dispersion relation.

I don't know how you would actually calculate that dispersion relation for a glass (other than brute-force computation), but it's possible in principle.

The same considerations also apply to quasicrystals, but with the interesting addition that there are now diffusively propagating modes called phasons with long relaxation times. 2

Phonons in a gas is a really weird thing to think about because in an ideal gas, the particles are assumed to be non-interacting, and they have to be in a thermal distribution of single-particle quantum states for it to be a gas (rather than a Bose-Einstein condensate or something). If the gas particles are delocalized, and don't interact with each other, then what the heck is a phonon? Yet sound waves in a gas obviously exist, so the question remains whether they're quantized or not. I can't answer this part of the question.

Improve this answer
$\endgroup$
2
  • 1
    $\begingroup$ The lack of translational symmetry does not affect the quantization. The normal modes can be quantized into localized phonons in the same manner as the crystal case. $\endgroup$
    – Everett You
    Commented Feb 19, 2013 at 12:11
  • $\begingroup$ About phonons in gases: physics.stackexchange.com/questions/254088/… $\endgroup$
    – valerio
    Commented Feb 5, 2018 at 12:04
2
$\begingroup$

Phonons were named after photons and have the same functionality quantum mechanically. They are both waves and particles. Nobody disputes the wave nature of sound in general. In non ordered materials there is no way that a sound wave will give all its energy to an atom, or a cluster of atoms, as an example. The reason is because all these atoms are in an incoherent state and no pure quantum mechanical state function can be defined. An atom can have a pure state function but the wavelengths of sound are way larger than the wavelengths that an atom can absorb.

In ordered materials like crystals this can happen because phases are defined so that there can be large dimensional coherent scattering: the order allows a quantum mechanical state function for the crystal to be defined, which can interact with a sound wave so that the whole energy of that sound is absorbed by the crystal, thus a phonon.

Edit: In the comments to the question it becomes clear that there is a confusion on the use of the term "phonon". I am using the definition in wikipedia.

Edit2: Copying from Carl's comment in the question, I would add that the quantization is the familiar E=hν=ℏω, and that this applies to fluids as well. But without a translational symmetry, this cannot display quantization but instead can take any value of ω.

If we expand the definition of a phonon to a continuous spectrum, it seems that the answer to the question above finally is yes, yes, yes. Though I guess that in disordered media the particle nature is not manifest. Actually this is also true about photons in ambient light, as an example. Maybe somebody should expand the Wikipedia article.

Improve this answer
$\endgroup$
9
  • 1
    $\begingroup$ as a comment, photon comes from the greek root phos=light, hence phosphor . Phonon comes from the greek root phone=sound, hence all the derivatives like telephone, phonetic etc. $\endgroup$
    – anna v
    Commented Mar 18, 2011 at 12:00
  • $\begingroup$ Nice answer, tackling the problem from the perspective of scale and coherence. +1 $\endgroup$
    – Marek
    Commented Mar 18, 2011 at 12:08
  • $\begingroup$ When you say "in disordered media, $\omega$ is continuous", that doesn't make sense because possible frequencies of a phonon in a crystal are also continuous. There is no requirement that the phonon wavelength be commensurate with the lattice constant. The difference is that in a crystal I know how to calculate a (again, already continuous even for crystals) phonon dispersion relation. How do I get the phonon dispersion relation in a glass? $\endgroup$
    – Keenan Pepper
    Commented Mar 20, 2011 at 0:57
  • $\begingroup$ @Keenan Pepper I will change the statement, since I just quoted Carl. Googling "dispersion relations in disordered media" one gets a list of papers, so it has been done. If one is serious in studying the subject the papers exist. $\endgroup$
    – anna v
    Commented Mar 20, 2011 at 4:51
  • $\begingroup$ @Keenan Pepper even wikipedia has dispersion relations for amorphous media hidden under the "density of states" article . en.wikipedia.org/wiki/Density_of_states $\endgroup$
    – anna v
    Commented Mar 20, 2011 at 5:09
1
$\begingroup$

It has to be remembered there are two types of phonons. Phonons with transverse modes are optical phonons. These possess oscillatory motion of charged ions in a lattice, where that motion is perpendicular to the wave motion direction along the lattice. These are phonons responsible for Raman scattering and related physics. The other type of phonon has longitudinal modes which are parallel to the direction of motion. These are acoustical phonons, which for large wavelength are sound waves.

These two types of phonons are distinct in a number of ways. Optical phonons occur with a lattice structure. This tends to be because amorphous solids do not often have a conduction band of electrons in a medium of ions. Ions in some array are necessary for the electric field oscillation in an optical phonon. Disordered solids are most often not of this nature. In fact it is one reason light can pass through them, such as glass, for a photon does not couple to a lattice of ions. Acoustical phonons can exist in disordered solids, and these are responsible for sound waves. Acoustical phonons of a very short wave length are attenuated in disordered solids. Compare trying to drive a car down New York streets or any city laid out on a grid when compared to driving in Boston. However, long wavelength acoustical phonons that are 10 or more times the average atom spacing can pass through a disordered solid.

Improve this answer
$\endgroup$
6
  • 1
    $\begingroup$ In solids phonons can be transversal, longitudional and "mixed", because crystals are often very anisotropic. "Optical" is reserved for near IR up to UV, The phonons You speak of are IR transitions in crystals. There is no difference beteween such elastic waves but wavelenght! BTW Raman is a nonlinear effect of all IR transitions (molecular AND crystal vibrations!) $\endgroup$
    – Georg
    Commented Apr 19, 2011 at 15:36
  • $\begingroup$ You are right and the two modes can be exchanged into each other. An optical phonon can evolve into an acoustical one. Indeed I am talking about long wavelength or IR phonons. My details may be basic, but I am drawing on an Ashcroft & Mermen course taken quite some time ago. $\endgroup$
    – Lawrence B. Crowell
    Commented Apr 20, 2011 at 0:57
  • $\begingroup$ This answer has parts that are correct, parts that I understand, and parts that are new to me, but none of those properties in combination! $\endgroup$
    – Keenan Pepper
    Commented Apr 21, 2011 at 15:10
  • $\begingroup$ "Phonons with transverse modes are optical phonons." - This is definitely incorrect because there are transverse acoustic phonons as well. The four different kinds of phonon branches - TA, LA, TO, and LO - all exist. $\endgroup$
    – Keenan Pepper
    Commented Apr 21, 2011 at 15:11
  • $\begingroup$ "In fact it is one reason light can pass through them, such as glass, for a photon does not couple to a lattice of ions." This sentence makes no sense to me at all. A glass doesn't have a lattice, and photons do couple to lattices of ions. I have studied the coupling of photons to lattices of ions in my solid state physics course. Basically this answer is just terrible. $\endgroup$
    – Keenan Pepper
    Commented Apr 21, 2011 at 15:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.