A phonon is a quantized unit of sound; they are encountered when quantizing lattice vibrations in solids. Now, even an ideal gas supports sound waves, but in this case, interactions between atoms are weak. That makes it hard to imagine what a quantized vibration would look like, since at small scales, the particles are free!

Is there a phonon picture for sound in an ideal gas? Is it ever useful?

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    $\begingroup$ @AnubhavGoel: Disordered solids are not the same as gases or liquids and the answers don't apply. $\endgroup$
    – CuriousOne
    May 5, 2016 at 9:08
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    $\begingroup$ I have not seen a quantum mechanical treatment of acoustic waves in thermodynamic gases, but that's probably not the end of the story. Did you look into excitations of Bose Einstein condensates? $\endgroup$
    – CuriousOne
    May 5, 2016 at 9:18
  • $\begingroup$ Ah! On googling I found highlights that someone picked this topic. But, unfortunately I did not see Keenan Pepper in his answer said he cannot answer that part. :p. $\endgroup$ May 5, 2016 at 9:22
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    $\begingroup$ I don't think a truly ideal gas supports sound waves. You need interactions between particles to support sound waves in a gas. In a quantum ideal gas I don't think it makes sense to talk about phonons either since the collective excitations are simply products of single particle excitations. I guess if you look in the literature you should be able to learn about phonons in imperfect gases. $\endgroup$
    – Ian
    May 8, 2016 at 3:40
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    $\begingroup$ I remember my undergraduate physics lab class when we excited a standing wave in a volume filled with water, and then shined a laser beam on this periodic structure. The laser beam interacted with it as with a diffraction grating; and scattering of photons could be described in terms of photon-phonon interaction, i.e., writing energy and momentum conservation for a two-particle collision gives the right answer for the laser beam scattering angle. Doesn't this say that the phonon paradigm is appropriate here? $\endgroup$ May 9, 2016 at 16:56

7 Answers 7


The only mention of this subject I can recall seeing is an aside in Xiao-Gang Wen's book, Quantum Field Theory of Many-Body Systems. Footnote on page 86:

A sound wave in air does not correspond to any discrete quasiparticle. This is because the sound wave is not a fluctuation of any quantum ground state. Thus, it does not correspond to any excitation above the ground state.

I'm not completely sure that I buy this, but it does certainly identify a crucial point. Plasmons or phonons in a condensed matter setting both have a restoring force, which lets one identify a minimum energy state to excite. In your typical view of an ideal gas, in which atoms mostly travel freely but occasionally collide with one another in some short-ranged way, this is not really true. You can make all sorts of density patterns in which the atoms are still not actually touching and thus the energy is not increased.

One might be tempted to get around this by taking a continuum limit somehow and considering a smooth quantum fluid, but then you are by definition trying to quantize a macroscopic field, which does not seem to make sense in even a formal way. In particular, since the field is a coarse-graining of the true system, one has necessarily thrown away some degrees of freedom, which means that the state of the field is never a pure quantum state and is more likely very close to a fully decohered statistical mixture.

In contrast, in a system with long-range interactions, and some boundary conditions, I would assume that phonon-like excitations are possible because the restoring force from mutual repulsion provides a well-defined ground state. This is a Coulomb crystal (1). But clearly this is very far from an ideal gas.

Edit: I should emphasize, as @Xcheckr has, that the above answer is interpreting the OP's question to refer to a Maxwell-Boltzmann ideal gas in a high-temperature state. There is of course no obstacle to defining the ground state of a BEC of an alkali gas, and such a ground state does indeed have phonon excitations (assuming a weak interaction). Similar remarks apply to a degenerate Fermi gas.


This is what I thought at first:

You need a restoring force in order to have a phonon: after all, phonons result from the quantization of the lattice energy written as a sum of harmonic oscillators. The hamiltonian of a gas cannot be written in this way: there are no restoring forces (you cannot "pull" a gas), so there can be no phonons.

As Rococo wrote, quoting Xiao-Gang Wen, a sound wave (SW for short) in a gas is different from a SW in a solid. If you think about it, a SW in a gas cannot be thought of as an "excitation" above some "fundamental vibrational state": it is just energy transiting through the medium. This is why you cannot quantize a sound wave in a gas, and hence you cannot introduce phonons in its description.

Then I talked with my professor of physics of liquids, and it seems like the subject is a bit more complicated than what I thought. He said that you can have collective excitations, hence phonons, in liquids, but there are some things to consider.

First, while the structure of a solid remains unchanged when a SW passes through it, this is not true for a liquid: the liquid can "restructure" itself locally with a certain speed. So, if the perturbation's frequency is greater than the inverse of the time it takes for the system to restructure itself, we will have a coherent wave. But if the frequency is smaller, the liquid will restructure faster than the propagation of the wave and the wave will undergo decoherence. The time it takes for the liquid to restructure can be estimated from the time it takes for the dynamic structure factor $S(\vec k, t)$ to go to $0$ (see figure below)

enter image description here

Another thing is that the wave vector $\vec k$ could be a bad quantum number to describe this kind of collective motion, since the liquid has no periodic long-range order and it is difficult to define for example a Brillouin zone. We can talk of normal modes of vibration, but we cannot apply the same formalism we use on crystals. This is also true for glasses and other amorphous solids.

I think many of these insights can be applied to gases too. The situation there is of course even more different from a crystal because short-range order (which is present in liquids) is lacking, too. So we can have collective motion only at very low $\vec k$, i.e. on large scales where the medium can be seen as a continuum.

  • $\begingroup$ Given the update, shouldn't you remove the initial sentence ''No, they haven't''? $\endgroup$ May 11, 2016 at 12:57
  • $\begingroup$ Yes, you're right. Even if I'm not really sure how much you can apply what I said about liquids to gases... $\endgroup$
    – valerio
    May 11, 2016 at 17:07
  • $\begingroup$ Where does your figure come from? $\endgroup$
    – Rococo
    May 11, 2016 at 17:16
  • $\begingroup$ pubs.rsc.org/en/content/articlelanding/2008/sm/… it was just to show that $S$ goes to $0$ as a function of time in a non-crystalline material. Any other example would be ok, but maybe I should have specified where it came from. $\endgroup$
    – valerio
    May 11, 2016 at 17:30
  • $\begingroup$ Yes, it was fine just as a general illustration of your point. But I was curious whether it was referring to a study where quantum physics was applied or (as it appears) a purely classical study. $\endgroup$
    – Rococo
    May 12, 2016 at 5:01

As you mentioned, a phonon is a quantized sound wave. Let us ask then what is the difference between a quantum and classical sound wave in a crystal? At room temperature, for most solids, the specific heat is pretty close to $3Nk$. This is the classical Dulong-Petit answer.

However, at low temperature, the specific heat deviates away from the Dulong-Petit answer and exhibits a $T^3$ relation. By this measure, one could then say that phonons are really observable at low temperature. The problem then is that most gases at room temperature become solids at these low temperatures. However, there are important exceptions. Helium remains a liquid at low temperatures and indeed quantized sound waves are observed in this case. See for instance:


Probably the most well-known kind of low temperature gas is the Bose-Einstein condensate, and here phonons have been observed as well:


I borrowed a lot of this response from the link below where the answer is discussed more thoroughly:



I'm tempted to answer in the affirmative, depending upon BOUNDARY CONDITIONS. Without getting into it too far

Suppose the ideal gas is contained within (for simplicity) a cubic box of length L. One has that the wavenumbers $$k_{n}=\frac{n\pi}{2L}$$ are then quantized, such that there are phonon-like modes of propagation.

One will then obtain something like:

$$k_{nml}=\sqrt{k_{n}^{2}+k_{m}^{2}+k_{l}^{2}}=\frac{\pi}{2L}\sqrt{n^{2}+m^{2}+l^{2}} $$

with momenta:

$$p_{nml}=\hslash k_{nml}$$

and other various properties depending upon your dispersion relation. Actually without boundary conditions, I would say no phonon modes. which is very interesting when you think about it. In any practical application there are always boundary conditions, especially with sound.

In a sense then, normal modes of sound are phonons.

EDIT: a quick search of this yields the same thing: https://en.m.wikipedia.org/wiki/Gas_in_a_box It is known as the Thomas-Fermi approximation and is used for massive or massless non or weakly interacting particles

It's also worth noting that, for phonons in a crystal, it is in actuality, the boundary of the whole crystal that cause quantization of phonon wavevectors http://users.physik.fu-berlin.de/~pascual/Vorlesung/SS09/slides/EPIV-09SS-SolSt_K3-Lattice%20vibrations.pdf (page 2) For example, in an infinite crystal, the allowed wavevectors would be continuous, and momentum wouldn't be quantized at all! Any good solid state text explains this, I like Kittel's intro to solid state physics


It doesn't make sense to talk about phonons for sound propagating through an ideal gas.

You might think that the normal modes should have their energy quantised in the usual way. After all, they exhibit sinusoidal oscillations, so we might expect that there is some way of parameterizing the degrees of freedom in the gas that reveal it to contain a set of harmonic oscillators (alongside more complex microscopic dynamics). These harmonic oscillators would have discrete energy levels in a quantum description, which we could describe using the language of phonons.

This argument seems convincing, but in fact, the normal mode oscillations for sound in an ideal gas are not perfectly sinusoidal. There will also be thermal fluctuations on top. And these thermal fluctuations make it impossible to specify the energy of a sound wave precisely enough for the usual quantisation procedure to be relevant.

If we take a region within an ideal gas that contains N particles, on average, then the root mean square deviation in particle number is $\sqrt{N}$. This means if we calculate the density as a function of position by averaging over small volumes of order $V_d$, the thermal fluctuations in the density, $\rho$, will be of order

$\sqrt{m \rho / V_d}$

where $m$ is the mass of a single gas particle.

To describe a sound wave of wavelength $\lambda$, we need $V_d << \lambda^3$. So the thermal fluctuations in the amplitude of a sound wave of this wavelength must be much larger than

$\sqrt{m \rho / \lambda^3}$.

On the other hand, consider an approximately planar sound wave in a large box of volume $V$. It's energy is

$\frac{1}{2} V \rho^{-1} c_{s}^{2} a^2$

where $c_s$ is the speed of sound, and $a$ is the amplitude of the sound wave (in terms of density).

For the energy to be quantised in jumps of $\hbar \omega$, a single phonon would correspond to density fluctuations of order

$\sqrt{\hbar \omega \rho / V c_{s}^2}$

To describe an approximately planar sound wave of wavelength $\lambda$, we need a box much bigger than $\lambda^3$. So the amplitude fluctuations in a single phonon must be much less than

$\sqrt{\hbar \rho / \lambda^4 c_s}$

For phonons to make sense, we need the density fluctuations from one phonon to be much larger than the thermal fluctuations. So, using the above results, we would need:

$\sqrt{m \rho / \lambda^3} << \sqrt{\hbar \rho / \lambda^4 c_s}$

where the speed of sound in a monatomic ideal gas is $\frac{5}{3} \frac{k T}{m}$.

This simplifies to

$\lambda << \frac{\hbar}{\sqrt{k T m}}$

In physical terms: to get phonons, we need to consider sound waves with a wavelength much less than the wavelength of a single gas particle.

In air at room temperature and pressure, this would be much less than the typical inter atomic distance, so it's impossible to get sound waves with wavelengths this short. This means that there are no phonons for sound waves travelling through air.

What if we try to bring the gas particles closer together, or reduce the temperature, so that their wave functions overlap? We can do this, but then we will get a phase transition, and we won't have an ideal gas any more. In a realistic gas, it would condense to a liquid or solid. But even in an idealized gas with no inter-atomic interactions outside of collisions, we still get a phase transition once the wave functions of the gas particles overlap: Bose Einstein condensation.

The upshot is: there is no way to observe phonons in an ideal gas. Sound in an ideal gas is a purely classical phenomenon.

  • $\begingroup$ Fantastic answer, and welcome to Phys.SE! $\endgroup$
    – knzhou
    Nov 15 at 0:45

Light, passing through a transparent medium, is scattered when it interacts with that medium's spatial & temporal variations producing the medium's refractive index. This is called Brillouin scattering and can be described as interaction of a photon with a phonon representing the medium's compressional deformation. Although mostly studied in the context of crystalline lattice waves, Brillouin scattering has been observed and analyzed in liquids, e.g., "Brillouin scattering in liquids at 4880 ${\buildrel _{\circ} \over {\mathrm{A}}}$" by Shapiro et al, IEEE Transaction on Quantum Mechanics, v. QE02, No. 5 (1966). As Brillouin scattering in liquids is interpreted in terms of photon-phonon interaction, similar to what is done in solids, clearly the concept of phonons in liquids proves to be useful, and the same should in principle apply to gases.


My intuition is that ideal gasses must have some form of quantized quasiparticle/collective excitation (I'll just call it a quasiparticle), but it usually isn't (or never is) a useful concept. (Just like a basketball at room temperature has quantized energy levels, but it's not useful to think of them.) Whether you call that quasiparticle a "phonon" is a matter of semantics. Quantized sound waves in liquids are called phonons, so I guess quantized sound waves in gasses could be called phonons too.

So here are some random thoughts:

What we think of as a conduction electron in a solid is really a quasiparticle involving the other electrons and ions in the system; a conduction electron is not simply one free electron.

Likewise, I guess that at some level an atom in a gas is really a quasiparticle shaped by the other atoms in the gas. The difference between conduction electrons in solids and atoms in gases is that the latter interact much more weakly, so an atom quasiparticle (at room temperature) will look almost exactly like a lone atom. That means the quasiparitcle picture isn't useful, but it isn't wrong either (like the quantized basketball).

While, at room temperature, that atom quasiparticle will look nothing like a phonon, my guess is that if the thermal energy of the atoms in the gas is similar to the "binding" energy between the atoms in the gas, then the atom quasiparticle could start to look like a phonon as we normally think of it (or at least look like a phonon in a liquid). That said, if the thermal energy is comparable to the "binding" energy, then you probably won't have a gas any more; it'll probably be in another phase (solid, liquid, Bose-Einstein condensate, etc.). Maybe that means that the concept of a gas phonon isn't wrong, but it is never useful either.


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