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Phonons are a nice tool to simplify the quantum-mechanical description of lattice vibrations by identifying the ladder operator of normal modes as creation operators of a certain quasi-particle. In certain special cases, one could even formulate a certain displacement field $\phi_a(x)$, quantize it according to canonical field-quantization procedures and obtain the very same phonons as the "emergent" particles.

On the other hand, a Weinbergian picture of quantum field theory is that particles do not emerge from fields - on the contrary, fields are a necessary consequence of many-particle dynamics. One could then argue that the observation of single point-particles emerging from QFT-computed interactions is an experimental verification of a "Weinbergian" interpretation of modern particle physics. (Even though a lot of particles we observe are not fundamental, they are in the end composed of elementary particles and the fundamental-particle character extends to them.)

To phrase it differently: one could argue that there is no such physical entity as a fundamental quantum field, the so-called "causal fields" are merely useful operators and the only fundamental entity are many-particle states.

But if a clearly non-fundamental non-particle such as a phonon is experimentally indistinguishable from a "really real" particle, then the term "really real particle" has no physical meaning and there is no way to tell whether it is the field or the particle which is more fundamental. (And here by indistinguishable I mean indistinguishable within the cut-offs imposed by the discrete nature of the vibrating lattice.)


EDIT: To make clearer what I mean by "particle behaviour"; when an electron is flying out of a reaction or entering a double slit experiment we often describe it as a plane wave. Indeed, the electron then interacts with the double slit in a plane-wave-like manner and forms an interference pattern on the screen behind it. But in a single event, we see only a single dot on the screen. This is the usual "wave-function collapse" as described by non-relativistic quantum mechanics. But this very "wave-collapse" situation is implicitly assumed to be a part of QFT applied to particle physics. Hence, the plane waves flying off from a QFT reaction are in fact particle-waves which undergo the very same collapse into a single point upon measurement.

Now consider the phonons. As is, they seem to be true quantized waves, not particle-waves. A naive analysis would say that they do not undergo the "wave-function collapse" and never acquire a particle nature. That is, if we were to send them through a hypothetical double slit, they would form a true interference pattern, not one emerging from dots.

But what does it mean experimentally that they would form a true interference pattern? Is it maybe that any experimental method trying to measure a phonon will in principle not be able to assess the difference between a "particle-wave" and a "true quantum wave"? Or does the phonon plainly behave like a collapsing particle-wave upon measurements analogous to those of the electron?


So the question is:

Does a phonon behave like a point particle?

Is this experimentally verified and how?

Is this somehow self-evident from the lattice many-body Hamiltonian in a way I couldn't see?

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    $\begingroup$ You scatter off them, don't you? $\endgroup$ – dmckee Jun 13 '15 at 17:21
  • $\begingroup$ I'm not sure if Weinberg would agree with this caracterizations of particles as fundamental. I have never seen it in his QFT books, do you have a reference? In principle particles are representations of the Poincare group, but in a curved spacetime the Poincare group is not the real symmetry, so particles will behave differently in different spacetimes, and therefore should not be viewed as fundamental $\endgroup$ – cesaruliana Jun 13 '15 at 18:28
  • $\begingroup$ Weinberg is indeed neutral on the interpretation. I didn't want to go into nuances as the question is actually asking about something different but I am referring to e.g. page 2 of the 1995 ed. of his QFT1: "The reason that our field theories work so well is not that they are fundamental truths, but that any relativistic quantum theory will look like a field theory when applied to particles at sufficiently low energy. On this basis, if we want to know why quantum field theories are the way they are, we have to start with particles." $\endgroup$ – Void Jun 13 '15 at 18:40
  • $\begingroup$ @Void, I agree tha Weinberg's positions is somewhat neutral. I always though this statement to be a reflection of the effective field theory approach, that everything will look like standard QFT in the low energy limit, as long as you'd like to have particles in the infrared. As the phonon situation shows, the fields are the relevant DOFs, only if the QFT has a fixed point the particles will appear (whcih is usual for IR physics). But the question is relevant indeed $\endgroup$ – cesaruliana Jun 15 '15 at 1:30
  • $\begingroup$ @cesaruliana I am not arguing with you on a mathematical level, but my concern is more whether the Fock-like algebraic structure arising in QFT necessarily implies that there are corresponding particle-waves undergoing wave-function collapse and all (see edit). I.e., whether the label "quasi-particle" is given to something which is treated as a particle in every theoretical aspect but is in fact very different on the phenomenological side of things. $\endgroup$ – Void Jun 15 '15 at 9:41
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I don't know nearly enough QFT to address the background or implications of your question. However, I'd basically answer yes to your first two questions, but it depends a little on your definition. A single phonon mode is not localized in space. However a wave packet can in principle be built up of a small range of frequencies, giving a fairly well defined frequency and position. So I interpret your question to mean: have phonon wave packets with fairly well defined frequencies and positions been detected.

The answer is yes. They can be created and measured with superconducting tunnel junctions, which can produce nearly "monochromatic" phonons, each due to the decay of individual quasiparticles in the superconductor with well-defined energies.

I don't think there are detectors that can resolve individual phonons (the way there are single photon detectors), but the creation and detection of phonons with these superconducting tunnel junctions sure make it look like individual, particle-like phonons are being generated and absorbed even though the detectors currently aren't sensitive enough to directly see it happen.

The technique was originally developed in the '60s. See: http://journals.aps.org/prl/abstract/10.1103/PhysRevLett.18.125

Richard Robinson at Cornell is currently using the tecnique to study nanoscale heat transport: http://therobinsongroup.org/research/phonon-spectrometer-for-nanoscale-heat-transport/

As for the last question, I don't think so. Then again, I don't think that phonons are unique in this way. Pretty much everything in condensed matter physics is some type of quasiparticles (or collective excitations, if you want to call them that).

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In superfluid helium-4, the phonon excitation spectrum includes a mode which has the same energy and momentum as a neutron with a speed of about 440 m/s (wavelength $\lambda \approx 9\,Å$). You can create a neutron beam which contains only 9 Å neutrons by starting with cold neutrons and being clever with diffraction from crystals. If you send these neutrons into liquid helium, they essentially undergo "billiard ball" collisions with these magic resonant phonons and stop: the speed of the neutrons after the collision is typically about 5 m/s. The phonon energy corresponds to a temperature of about 11 K, much warmer than liquid helium, so the phonon runs off to your refrigerator and is removed from the liquid.

This process is known as "superthermal production of ultra-cold neutrons," and was used recently to fill a magnetic "bottle" with neutrons for a neutron lifetime measurement at NIST.

I don't know very much about this process from a field-theoretical perspective, but the phonon in this interaction is just as "particle-like" as the neutron is.

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