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If you go through the definitions for "phonon" (or for an explanation of phonon), most of the text or articles make the analogy to photons:

A photon is a discrete quantum of light.

A phonon is a discrete quantum of vibration.

Further, in this text, the author says: Phonon are particles (just like photons), then we see that we can put many phonons in this same state. Like photons, phonons are bosons.

So this text seems to put the two particles on equal footing in some way.

Now if so, can I catch a phonon just as a photon using some device like in photons case its photomultiplier? I found photons to be real particles (not an analogy) while phonons to be some kind of pseudo particle, an analogy that makes things simpler to think and doing calculations. After all, you can find photons in the standard model, but not phonons.

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    $\begingroup$ Does this answer your question? Is differentiating particle and quasiparticle meaningless? $\endgroup$ – Vadim Apr 12 at 12:41
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    $\begingroup$ @Vadim although that question is related, it doesn't ask whether single phonons have been measured, while this one does. $\endgroup$ – Ruslan Apr 12 at 13:03
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    $\begingroup$ @Ruslan This is routinely done in nanomechanics, although this is probably not what the OP means by phonons. Observing phonon sidebands in all kind of spectra (optical or tunneling) is also not uncommon. Then there is also Mandelstam-Brillouin scattering. $\endgroup$ – Vadim Apr 12 at 13:08
  • $\begingroup$ @Vadim these observations are still far from one-by-one detection like PMTs do for photons. $\endgroup$ – Ruslan Apr 12 at 13:41
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One has to keep in mind when studying the use of quantum field theory in modeling physics observations that quantum field theory is a mathematical tool that can be used where quantum phenomena appear, not only for fundamental particle physics , as in the standard model that fits the observations of elementary particle interactions.

In the particular acoustic model, they have successfully used a similar to Quantum Electrodynamics (QED) field theory, that is why the gauge exchanged "particle" is called a phonon.

The "equal footing" is in an equal mathematical representation. Photons are part of the SU(3)xSU(2)xU(1) model of particle physics , whereas phonons only model vibrations.

It seems that individual quanta of vibrations have been measured

Now, the same group of researchers has improved their previous design enough to detect individual phonons.

You ask in the comments:

Are phonons' particles similar to other particles we know like photons or electrons?

No, they are particular quantized states of vibrations in solids.

If not, then they are just mathematical entities. Aren't they?

No they are physically observable discrete excitations, as seen in the link above, which are modeled with a mathematical QFT similar to the QFT that models photons so well.

If they are, Do they have all properties that particles have like mass (if massless then they must move with a speed of light which I don't think they do), spin, etc. Why they aren't in the standard model?

Yes, they will have a four vector defined mass, the addition of all the four vectors that make up the excitation. They must have spin 1 to obey a similar QFT to the photon one.

Even the protons and neutrons are not in the particle table of the standard model, which has elementary point particles at its base. Phonons, in an even higher level of compositeness, (they are composite excitations on composite molecules which are composite of protons neutrons and electrons) also are not elementary.

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    $\begingroup$ Thanks for your answer but you see, I'm currently in my bachelor's and not much familiar with QFT or QED. But If I ask in yes and no, and if possible you can give a little less rigorous explanation. Are phonons' particles similar to other particles we know like photons or electrons? If not, then they are just mathematical entities. Aren't they? If they are, Do they have all properties that particles have like mass (if massless then they must move with a speed of light which I don't think they do), spin, etc. Why they aren't in the standard model? $\endgroup$ – Young Kindaichi Apr 12 at 18:22
  • $\begingroup$ see my edit to the answer $\endgroup$ – anna v Apr 12 at 18:42
  • $\begingroup$ No fundamental particle is safe from becoming either a composite particle or a quasiparticle as the science progresses on. $\endgroup$ – fraxinus Apr 13 at 7:36
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    $\begingroup$ @fraxinus the answer is within the mainstream standard model of particle physics at the moment, as we are supposed to do on this site. $\endgroup$ – anna v Apr 13 at 9:48
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A part of the question is the distinction between a quasiparticle, such as phonon and a fundamental particle. In this respect it is necessary to note that phonons are not the only quasiparticles in the solid state - there are quite a few of them, starting with the "free" electrons and holes. Quantum field theory provides a systematic way of defining quasiparticles and studying their properties. A more detailed discussion is given in my answer to a this question.

The other part of the question is a suggestion that phonons are just a useful computational tool, but not really existing entities, i.e., one can detect a photon, but not a phonon. In this connection let me point out the following:

  1. Quantized mechanical oscillations are detectable:
    • Within the context of nanomechanics, where macroscopic mechanical modes can be populated and detected a quantum by quantum, just like photons.
    • The bulk mechanical oscillations, which are more conventionally meant by phonons, are also detectable - they appear as distinct peaks in all kinds of spectra - whether optical or conductance.
    • There are also phenomena such as Mandelstam-Brillouin scattering - which is the equivalent of Raman scattering, but with one of the photons replaced by a phonon
  1. Other quasipartcles Once we extend the definition of quasiparticles beyond phonons, to all of the excitations in the solid state, many of them are detectable. As one remarkable example one can cite the Bose-Eisntein condensation of excitons, which are complex many-body excitations.
  2. Standard model The fact that phonons do not obey standard model is also not surprizing - standard model defines particles as irreducible representations of certain group. Phonon are excitations of a crystal lattice with different symmetry - their properties are derivable from the crystal symmetry in the same way as the properties of fundamental particles.

To summarize: there is a case to make that phonons are less real/fundamental than photons, however the distinction is rather subtle.

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  • $\begingroup$ Thanks for your answer; I really don't understand much of the rigorous detail of your answer due to a lack of knowledge of QFT. But I suppose when you get into QFT where particles themselves in a crazy way, I just lose sight of particles. Now still have a few questions, Can phonon exist in free space like other particles? Do they have other particle properties like mass, spin (they are bosons so it should be an integer)? Can I condensate (BEC) them? $\endgroup$ – Young Kindaichi Apr 12 at 18:33
  • $\begingroup$ Phonons can exist only on the lattice. The point I sm making is that it is rather difficult to meaningfully distinguish them ftom the real particles, which transform into each other, consist of quarks, etc. Bose-einstein condensation of phonons is a subject under investigation - there are ideas of "phonon maser" and some experiments... the difficulty is creating a cavity. $\endgroup$ – Vadim Apr 12 at 19:00
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The analogy holds because phonons are collective excitation of lattice vibrations, and photons quantum excitation of electromagnetic field; both described in terms of harmonic oscillators. But phonons are quasi particles, not elementary particles and require a medium for exist, they can't exist in vacuum. Allthought one can think it's only a mathematical description of collective behavior, it's possible to manipulate and detect it. There are some papers talking about single phonon. I am not an expert, but it seems to involve interaction of lattice vibration with fast pulse laser.

Phonons are bosons like photons. In thermodynamic equilibrium, in a cavity, there is thermal radiation, where photons are created and absorbed by the cavity. The distribution of the mean number is the Bose Einstein distribution. Phonons are described by the same distribution. They are both gas of bosons.

Phonons have polarization, transverse and longitudinal while for photons is only transverse. They can have a mass, depending on the system. It's also possible to define a pseudo momentum.

An interesting thing, in my opinion, is phononic crystal. This is an analogy from photonic crystal. Phononic crystals are metamaterials which can be used to manipulate phonons, like trapping or changing the band gap for phonons in a way to amplifying or suppressing sound waves below or above certain cutoff frequencies. Changing the properties of the material like bulk modulus, density or others, it's possible to make the acoustic refractive index periodic, like in photonic crystal where the refractive index is periodic. This implies a modification of the speed of the waves in the two cases.

Finally, I want to express a thought. Even if phonons don't belong to the Standard Model, this doesn't imply that are not real. Even the definition of photon sometimes can be misleading, see other questions here, or in the literature. After all, these are descriptions of what we observe, and if we can measure something, then it's real.

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Phonons are quantum states of the harmonic oscillator formed by a material lattice, for example a salt crystal. Photons are quantum states of postulated harmonic oscillators that as far as we know have no material origin. If they had we might call that the aether. So the have a lot in common but not everything.

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I would like to add a few interesting facts that I believe are not mentioned in the other answers and are important.

  1. On the one hand, photons do not interact with each other (except for extreme high energy levels), while on the other hand, phonons do interact with each other.

https://news.mit.edu/2010/explained-phonons-0706#

  1. Photons and phonons both can be transformed through Parametric Down Conversion.

As well as photons, phonons can interact via parametric down conversion[17] and form squeezed coherent states.[18]

https://en.wikipedia.org/wiki/Phonon

  1. while we have lasers and single photon emitters, we do have now SASERs, that is Sound Amplification by Stimulated Emission of Radiation.

In a SASER device, a source (e.g., an electric field as a pump) produces sound waves (lattice vibrations, phonons) that travel through an active medium. In this active medium, a stimulated emission of phonons leads to amplification of the sound waves, resulting in a sound beam coming out of the device. The sound wave beams emitted from such devices are highly coherent.

https://en.wikipedia.org/wiki/Sound_amplification_by_stimulated_emission_of_radiation

  1. Phonons (and rotons) both have gravitational effects, just like photons (which do bend spacetime, since they have stress-energy).

Recent research has shown that phonons and rotons may have a non-negligible mass and be affected by gravity just as standard particles are.[19] In particular, phonons are predicted to have a kind of negative mass and negative gravity.[20] This can be explained by how phonons are known to travel faster in denser materials. Because the part of a material pointing towards a gravitational source is closer to the object, it becomes denser on that end. From this, it is predicted that phonons would deflect away as it detects the difference in densities, exhibiting the qualities of a negative gravitational field.[21] Although the effect would be too small to measure, it is possible that future equipment could lead to successful results.

This is a very interesting point, because based on this even quasiparticles can bend spacetime, since they do possess stress-energy.

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    $\begingroup$ Thank you for adding these points, I get more interested in phonons than before. $\endgroup$ – Young Kindaichi Apr 14 at 17:17
  • $\begingroup$ On your first point: while the statement "photons do not interact with each other" might be patently obvious to many, it may not be so to less-advanced students of physics like myself. I found this post and its answers very illuminating - do you think it's worth adding as a link to your post? PS +1 for the useful additional information :-) $\endgroup$ – Chappo Hasn't Forgotten Monica Apr 16 at 7:40
  • $\begingroup$ @ChappoHasn'tForgottenMonica thank you so much! $\endgroup$ – Árpád Szendrei Apr 16 at 15:40

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