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## Hot answers tagged phonons

11

I read that phonons are (the quantum mechanical analog of) normal modes of vibration in a crystalline system of atoms or molecules, so I guess a superposition, i.e. a general vibration should also be a phonon. Is that so? Why would they then be described as normal modes? The reason that phonons are described in terms of normal modes is because the ...

8

Because a phonon is a quantum of "sound" and "sound" is a longitudinal wave while a photon is a quantum of "light" and "light" is a transverse wave (an electromagnetic wave). For example, if two waves are moving in the $z$ direction, the sound wave moves the molecules of the medium in the $z$ direction as well, up and down, one possible direction. ...

6

As I understand this occurs when a cystalline solid has more than one atom in the primitive cell of the lattice. For simplicity I'll consider the case of two types of atoms, distinguished by their position in the primitive cell (but which may or may not be otherwise identical). This gives rise to multiple modes of vibration. The acoustical phonon ...

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Yes. The Seebeck effect, for example, is the direct conversion of thermal gradient to electric voltage. It is used in the small scale in thermocouples to measure temperatures electronically, or in the larger scales in thermoelectric generators for power generation. In fact, many of the long-range space probes launched by NASA get power this way. ...

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I'm not sure if a Condensed Matter book is going to give you what you want: as pointed out by commenters, you cannot derive a Lagrangian, you can only justify it because it represents the correct physics. But here is a simple interpretation of the 3rd order term. For small deformations, Hooke's law holds and the restoring force $F_{a}=−k_{ab}q_b$. (For ...

3

My opinion: Phonons are collective excitations of the crystal lattice vibration. They are massless Goldstone bosons resulting from the violation of the continuous translational symmetry of free space, by the crystal lattice. Phonons must carry momentum because they interact with electrons and change the momentum of the latter. An example is the electronic ...

3

Phonon dispersions are generally indicated by a linear spectrum, so the second derivative is 0. Thus phonons are effectively massless, and have a velocity given by the first derivative.

3

Answer to title question because the spatial volume is infinite--- the p values in a finite volume are discrete, but as you take the infinite volume limit, you get a continuum of p's. The p's are Fourier variables, they are inside-out with respect to the space variable. To see this consider a periodic function in one dimension with period L: $$... 3 You made it a challenge by asking so many different questions, but since no one else has attempted an answer, I'll do my best. :) I read that phonons are (the quantum mechanical analog of) normal modes of vibration in a crystalline system of atoms or molecules, so I guess a superposition, i.e. a general vibration should also be a phonon. Is that so? Why ... 3 In Raman it should say "optical phonon" instead of vibrational oscillation. I'm not sure if you can call the vibrations in a molecule phonons, since phonons are the vibrations of the lattice in a solid (even when the mathematical description, in first approximation are the same). 3 Actually the framework where one can describe electromagnetic fields is a classical framework. When one is talking of photons phonons etc one is in the quantum mechanical regime where the concept field, is different. A classical field in physics: A field can be classified as a scalar field, a vector field, a spinor field or a tensor field according ... 2 A photon is an excitation of a quantum field, which is classically A_{\mu} governed by the Lagrangian,$$\mathcal{L}=-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}$$which we may quantize canonically by the usual procedure, i.e. expanding the field as a plane wave, promoting the Fourier coefficients to operators, etc. A virtual photon does not necessarily satisfy, ... 2 The distinction isn't actually as sharp as it may seem. If you get really pedantic about it, you can create borderline cases that blur the line as much as you like. The first answer, that it's just the difference between acoustic phonons and optical phonons, is correct. But then that just leads to the next question: What's the difference between those? In a ... 2 Brillouin scattering is caused by an interaction between light and lattice phonon modes. Raman scattering is caused by an interaction between light and molecular vibrations. The key difference is that phonon modes are a collective, long-range phenomenon involving billions or more atoms, whereas molecular vibrations are localized vibrations of a single ... 2 The Debye temperatures of some steel alloys can be found in the thesis of Rajevac, Vedran: "Lattice dynamics in Hydrogenated Austenitic Stainless Steels and in the Superionic Conductor Cu 2-δ Se" on p. 43. In general searching for material specific data I recommend starting with a search in Google, Google Scholar and Landolt-Boernstein. If neither turn ... 2 When you solve for such Hamiltonian for the harmonic oscillator you get a set of eigenstates which by definition are orthogonal and thus you have phonons that don't interact. When you include a non harmonic term to the Hamiltonian and treating it as a perturbation you get new eigenstates that are a mixture of those the simple harmonic oscillator. This ... 2 In classical mechanics, you can describe a crystal (in some approximation) by a Hamiltonian that is a quadratic form in coordinates and momenta of atoms. After you diagonalize this quadratic form, you obtain a Hamiltonian of a set of independent, rather than coupled, oscillators (modes). Then you can quantize this system and you do get zero-point energy for ... 2 This is a great question. The point is we've made a huge structural change in going from Lehmann representation to the, um, other one. In Lehmann representation, we've chosen to write G as a sum over an infinite number of real poles, which is fine of course, but when you start adding infinite numbers of delta functions together, things can get tricky. ... 2 1) The Pauli exclusion principle says that either 0 or 1 identical fermion, but not more, may occupy the same quantum state. It's been originally induced from the properties of atoms (periodic table). Today, we may derive it from the first principle out of the antisymmetric wave functions for many fermions, those obeying$$\psi(\vec x_1,\vec x_2) = ...

2

Lattice vibration excitations, i.e. phonons, are typically at energies of $\sim 25 meV$ which corresponds to room temperature. Therefore, neutron scattering with thermal neutrons is done. Thermal neutrons are neutrons from a nuclear power plant which are moderated to room temperature by sending them through a big tank of water, so they also have energies on ...

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It means that they have the same energy, that basically they are indistinguishable from each other. This stems from the symmetry of the crystal used for the example.

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Your answer is indeed what's going on in the lecture, but it doesn't explain what was wrong with your initial argument: you'd expect a model with $l_0>0$ to be a closer representation of reality than one with $l_0=0$, wouldn't you? Actually, your initial reasoning was correct: transverse displacements of springs under zero tension do indeed result in ...

1

I haven't studied polarons in detail so can't comment on this, but they very probably relate to electron-phonon interactions. I have somewhat more to say about 2.: Phonons are indeed bosons, and Goldstone bosons to boot, resulting from the spontaneous breaking of Galilean, translational, and rotational symmetries in the phase transition producing the solid. ...

1

Question: Are you completely sure that the phonon wavevector spectrum is continuous? For any finite body, phonon modes are (nominally) quantized in much the same fashion as the fermion seas of electrons in metals. It's quasi-continuous, sure, but a true continuum spectrum for phonons is necessarily a bit of an abstraction.

1

Suppose we take a classical approach and model the crystal lattice as a series of coupled simple harmonic oscillators. if the number of individual lattice sites $N$ is large (but not infinite), we may assume periodic boundary conditions, which result in the quantization of allowed wavevectors. Using Ashcroft and Mermin pg 430 as a reference, the above ...

1

How about the Piezo-electric effect? if I'm not mistaken, pressure on the crystal is in essence equal to long-wavelength phonons. If that's not what you're after, perhaps read this, or else this paper (both the result of a 5 minute Google scholar session). They're don't seem to contain exactly what you're after (haven't read them thoroughly though), ...

1

Unstable particles are concepts of effective field theories (or few-particle systems) in reduced descriptions where the decay products are ignored. In these reduced descriptions, they appear as particles with complex masses, and their Green's functions have complex poles. In an unreduced description, unstable particles appear as poles of the analytically ...

1

Different types of phonons can not be considered as separate systems. They are oscillations of the same crystal and interact with each-other. For example, LO phonon lifetime is about $10^{-12}$ - $10^{-11}$ seconds while the period of the oscillations is about $10^{-13}$ seconds (GaAs). At the end it turns into two LA phonons that run opposite directions. ...

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This figure shows the phonon dispersions of ZnO. It is clear that while some phonons have very roughly linear dispersions many do not (especially close to zone center). The second derivative would be non zero in these regions. I hope this has added to the conversation.

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Found something today. "Actually the concept of polaritons has been described in words as early as 1946 [F. Bloch, Phys. Rev., 70, 460 (1946)] in nuclear paramagnetic resonance, however without using the then still unkown term 'polariton'." Klingshirn, Claus F., Jul o6 2012, Semiconductor Optics Springer Berlin Heidelberg, Dordrecht, ISBN: 9783642283628

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