# Tag Info

## Hot answers tagged phonons

14

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

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 ...

7

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

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 ...

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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. ...

4

The zero phonon line is the wavelength at which an excitation/relaxation is not phonon assisted.

4

What are phonons? Phonons aren't particles like electrons or protons are, phonons are quasi particles, these type of particles are just used to describe excitations of a field: in phonons case, phonons are used to describe elementary lattice vibrations which have certain frequency. Electron-Phonon Interaction: Basically Cooper pairs are just pairs of ...

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 ...

3

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, ...

3

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 ...

3

Careful! The essence of your question is a good thought but I think you are having several misconceptions. First of all, it doesn't make sense to talk about a phonon being directed towards a single atom. Phonons are delocalized. Secondly, the "input of temperature required to eject an electron" is a dangerous idea. You need to input energy to eject an ...

3

While a cavity can have modes of arbitrarily high frequency, a phonon can't have a wavelength smaller than the separation between atoms in the solid (well, it could, but it'd be degenerate with phonons of smaller frequency because the discrete Fourier transform is periodic). Hence, there is a maximum possible wavenumber. A physical (or sometimes ...

3

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 ...

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

Phonon scattering goes up a lot as temperature increases -- faster than electron numbers increase in the conduction band. Keep in mind that phonons obey the Bose-Einstein distribution, so their numbers scale like $$N_{BE}=\frac{1}{e^{\frac{\hbar\omega}{k_b T}}-1}$$ In the large $T$ limit, this becomes $$\frac{k_b T}{\hbar\omega}$$ So their numbers ...

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

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

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. ...

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

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 ...

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

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 ...

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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

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

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

The term vibron isn't a standard accepted word in solid state physics. It's simply a synonym of a phonon, which was probably coned before phohon. As far as localization is concerned, phonons can be localized too. For example, in a crystal lattice, the vibrations can propagate throughout the crystal, or there can be vibrations in only a small localized part ...

2

Yes, and no. Since the group of rotations is not a continuous group in real crystals, it is not possible to define spin in a meaningful way. It is only in an isotropic ideal medium that is possible to define spin for a phonon (quantized accoustic wave). Equivalently it is only possible to define a spin if the the wavelength of the phonon is long or if one ...

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

If the potentials are all quadratic, then the force varies linearly with displacement. That means the superposition principle holds. (Don't confuse that with quantum superposition!) In other words the lattice waves then behave like electromagnetic waves: you put two of them together and they simply add together but don't otherwise affect one another; they ...

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