# Tag Info

19

Phonons are indeed massless, as you can see from their dispersion relation or from the fact that they are Goldstone bosons. The phonon dispersion relation that you wrote down tells us that we can excite a phonon mode, with some finite momentum, using an arbitrarily small amount of energy, hence they have no rest mass (in condensed matter language, they are ...

15

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 ...

9

The only mention of this subject I can recall seeing is an aside in Xiao-Gang Wen's book, Quantum Field Theory of Many-Body Systems. Footnote on page 86: A sound wave in air does not correspond to any discrete quasiparticle. This is because the sound wave is not a fluctuation of any quantum ground state. Thus, it does not correspond to any excitation ...

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 ...

6

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. ...

5

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. ...

5

Phonons follow a wave equation, which is at least in first approximation simply a standard wave equation, the only difference to relativistic particles is that the speed of the waves is not c but the speed of sound $c_s$. But this does not change the mathematics of the equation, so that in general there may be phonons which follow a massless wave equations ...

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 ...

4

This is what I thought at first: You need a restoring force in order to have a phonon: after all, phonons result from the quantization of the lattice energy written as a sum of harmonic oscillators. The hamiltonian of a gas cannot be written in this way: there are no restoring forces (you cannot "pull" a gas), so there can be no phonons. As Rococo wrote, ...

4

Lets start with the bare ions. You set up a linear lattice of protons where the boundaries are held in place by some means. The protons want to get as far apart from each other as possible. So the minimum energy state has the protons spaced on the line with equal distances between them. So yes, the ions form a regular lattice without the electrons. Now ...

4

@GerryHarp's answer contains the gist of the main idea, but there is a point that doesn't quite make sense: There is no such thing as bare ions crystal in solid state physics. To answer your last question: The "bare phonon frequencies" definitely do not refer to phonon frequencies in the absence of electrons. In fact, an ion crystal without electrons is ...

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 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

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 ...

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

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

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

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

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

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

The term vibron isn't a standard accepted word in solid state physics. It's simply a synonym of a phonon, which was probably coined before phonon. 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

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 ...

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) = ...

Only top voted, non community-wiki answers of a minimum length are eligible