Wave-particle duality of phonons?

[Demystifier]

As every condensed-matter physicist knows, a phonon is a quantum of sound (or more precisely, of oscillations in a condensed-matter material). Clearly, sound is a wave. But quantum theory teaches us that, in a certain sense, sound is also a particle. Here I want to understand how far the view of a phonon as a particle is justified. To clarify what I mean, here are some more specific questions.

- It is often said that phonon is not a particle but a pseudoparticle. What exactly does it mean and how an experiment can determine whether something is a particle or a pseudoparticle?

- Are there experiments showing that phonons are objects localized in space?
(Like for photons and electrons.)

- Are there double-slit experiments with SINGLE phonons, showing interference of a phonon with itself on the ensemble level, but pointlike-particle nature of a phonon on the individual level?
(Like for photons and electrons.)

- Does a classical pointlike phonon particle make any physical sense?
(Like for electrons and gamma-rays.)

- Does first quantization of the classical pointlike phonon particle make sense?
(Like for electrons and to certain extent to photons as well.)

P.S. Please don't move this thread to the "condensed matter" forum! This topic is about foundations of quantum mechanics and condensed matter physicists usually don't know much about it.

 

[ZapperZ]

Certain parts of your questions do not quite make sense because a "phonon" is a collective excitation, and in many cases, it is simply a quantum field representation of an interaction with that collective excitation.

If we assume that QFT is correct and can represent ALL types of interaction (and not just EM, strong, and weak), then one can represent any and all interactions via such exchange of a gauge boson. So when a particle interacts with the lattice vibrations, for example, one represents that lattice vibrations via a quantized field. That quantized field is the phonon.

Note that this isn't that unique of a representation. Other "quantized field" of interaction in condensed matter includes magnons, spinons, polarons, etc... etc. In fact, one can even argue that the charge on a particle (or in this cases, the Landau quasiparticle) is also an "excitation", since a "chargon" or "holon" can in fact have a different dispersion than say the spin, as exhibited by the spin-charge separation in a Luttinger Liquid.

BTW, note that one could argue that the "simplest" phonon quantization is the quantization one obtain in the quantum SHO, which is relevant for a single particle in a SHO potential.

I'm not sure to what level of "simplicity" you want this. At one of the simplest level, I would recommend reading this paper:

S. Johnson and T. Gutierrez "Visualizing the phonon wave function" AJP v.70, p.227 (2002).

Edit: BTW, there are double-slit like interference observation of phonons. See, for example, T. Brandes and B. Kramer PRL v.83, p.3021 (1999).

Zz.

 

[Demystifier]

Certain parts of your questions do not quite make sense because a "phonon" is a collective excitation, and in many cases, it is simply a quantum field representation of an interaction with that collective excitation.

Does it mean that the answer to all my yes/no questions is - no?

 

[Dmitry67]

Demistifier, what is your hidden agenda? You definitely know the answers :)

 

[Demystifier]

I'm not sure to what level of "simplicity" you want this. At one of the simplest level, I would recommend reading this paper:

S. Johnson and T. Gutierrez "Visualizing the phonon wave function" AJP v.70, p.227 (2002).

Thanks, it has been illuminating. Now I think I can answer some of my questions by myself.

As an example, consider a system of 2 particles interacting with a harmonic force. The positions of these 2 particles are x1, x2. Now let us quantize their DIFFERENCE
q=x1-x2
This is a prototype of a phonon. The phonon can appear as a "particle" with a definite position in the sense that q may have a definite value in the experiment. However, a definite value of q is actually a definite value of TWO positions x1 and x2. Nevertheless, it is conceivable that a measuring apparatus reveals the value of q without revealing the separate values of x1 and x2. So even though we actually have 2 particles, the response of the measuring apparatus looks like a response of a single-particle detector.
Does it make sense to you?

 

[Demystifier]

Demistifier, what is your hidden agenda? You definitely know the answers :)

I have no hidden agenda. Nevertheless, I do have some hidden questions that I didn't want to ask explicitly. (Like, does a Bohmian interpretation of phonons makes sense? Or does Unruh effect of phonons makes sense?)
By the way, if you are convinced that I know the answers, does it mean that you know them too?

 

[Dmitry67]

I have no hidden agenda. Nevertheless, I do have some hidden questions that I didn't want to ask explicitly. (Like, does a Bohmian interpretation of phonons makes sense? Or does Unruh effect of phonons makes sense?)
By the way, if you are convinced that I know the answers, does it mean that you know them too?

You definitely know that individual photons still show the same interference pattern.
But I would definitely like to know how BM explains the Unruh effect.

 

[Dmitry67]

ooops, phonons, not photons... sorry :) as an excuse, I browsed it from XPERIA X1

 

[Demystifier]

You definitely know that individual photons still show the same interference pattern.

Sure I do, but I asked about phonons, not photons.

But I would definitely like to know how BM explains the Unruh effect.

That's not the topic here, and I've already answered to you several times. You obviously didn't catch my answer, so let me try again, in a simpler way.
As I have shown in
http://xxx.lanl.gov/abs/0904.2287 [to appear in Int. J. Mod. Phys. A]
the vacuum actually contains an infinite number of particles, but they are all dead. A dead particle is a particle the 4-velocity of which vanishes, so such a particle does not have a trajectory. Instead, it lives for an infinitesimally short time. However, if we have an accelerated particle detector, then its interaction with the vacuum makes some dead particles alive by changing their 4-velocity. In other words, the detector creates particles.
The point is that the number of created particles does NOT depend on the observer. Instead, it depends on whether a particle detector exists and on how does it move. In the case of accelerated particle detector, the created particles exist for any observer, not only for the observer comoving with the detector.
Does it make sense to you? Or will you ask me the same question again?

 

[Dmitry67]

But is the number of Bohmian particles objective or not?
I know that in BM there are 'empty waves'. But these point like beasts - is the number of them invariant for different observers?

 

[Demystifier]

But is the number of Bohmian particles objective or not?
I know that in BM there are 'empty waves'. But these point like beasts - is the number of them invariant for different observers?

Yes, their number is invariant for different observers. Yet, their number depends on acceleration of the measuring apparatus.

 

[Dmitry67]

The point is that the number of created particles does NOT depend on the observer. Instead, it depends on whether a particle detector exists and on how does it move. In the case of accelerated particle detector, the created particles exist for any observer, not only for the observer comoving with the detector.
Does it make sense to you? Or will you ask me the same question again?

Wait, I don't understand.

In Unruh effect particles comes not from the detector, but from apparent cosmological horizons.

http://en.wikipedia.org/wiki/Unruh_effect

The Unruh effect could only be seen when the Rindler horizon is visible. If a refrigerated accelerating wall is placed between the particle and the horizon, at fixed Rindler coordinate ρ0, the thermal boundary condition for the field theory at ρ0 is the temperature of the wall. By making the positive ρ side of the wall colder, the extension of the wall's state to ρ > ρ0 is also cold. In particular, there is no thermal radiation from the acceleration of the surface of the Earth, nor for a detector accelerating in a circle[citation needed], because under these circumstances there is no Rindler horizon in the field of view.

How your theory can explain that?

 

[Demystifier]

Wait, I don't understand.

In Unruh effect particles comes not from the detector, but from apparent cosmological horizons.

http://en.wikipedia.org/wiki/Unruh_effect



How your theory can explain that?

Well, there are actually two inequivalent versions of the Unruh effect. The one you mention is not measurable. But that's really not the topic of this thread, can you ask further questions here?:
https://www.physicsforums.com/showthread.php?t=306788&page=5