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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

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

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 ...


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 ...


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 ...


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

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 ...


1

The vertical axis is energy and the horizontal axis is k-vector (crystal momentum). $E_L$ is the energy of the incoming photon, not the phonon. The vertical arrow on the left describes the process where light excites an electron. Photons have almost no momentum (on the scale of this plot), so the arrow is essentially vertical. The downward-tilting right ...


1

Here are some wordy, math-free answers. a phonon is the minimal amount of energy which can be stored in an lattice vibration in a given mode Sounds good. I.e. that when a crystal vibration interacts with matter it does so by the creation/destruction of whole phonons at a time, which may also get absorbed at more or less precise locations, e.g. the ...


1

J Phys Condens Matter. 2010 Feb 10;22(5):055402 : "The Debye temperature, Θ(D), of Fe(100-x)Cr(x) disordered alloys with 0 ≤ x ≤ 99.9 was determined from the temperature dependence of the centre shift of (57)Fe Mössbauer spectra recorded in the temperature range of 60-300 K. Its compositional dependence shows an interesting non-monotonous behaviour. For 0 ...


1

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 ...


1

I am sure there is a deeper explanation, but here is my take at a heuristic one: Spin is intrinsic angular momentum, and phonons do not have that. The spin of the photon is “fundamental” in the sense that you have to postulate it. But the phonon is a collective excitation, so any spin you would assign to it would have to come from its constituents, and the ...


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

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

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

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

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

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


1

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.


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), ...



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