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25

Let me give a second, more technical answer. Observable particles. In QFT, observable (hence real) particles of mass $m$ are conventionally defined as being associated with poles of the S-matrix at energy $E=mc^2$ in the rest frame of the system (Peskin/Schroeder, An introduction to QFT, p.236). If the pole is at a real energy, the mass is real and the ...


12

Elementary particles, like photons and electrons, are not more elementary in the sense that there are underlying theories, such as quantum spin model on lattice, from which they can be derived as an effective approximation (see for example arXiv:hep-th/0302201). In particular, the string-net condensation provides a unified origin for gauge interactions and ...


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

All observed particles are real particles in the sense that, unlike virtual particles, their properties are verifiable by experiment. In particular, W and Z bosons are real but unstable particles at energies above the energy equivalent of their rest mass. They also arise as unobservable virtual particles in scattering processing exchanging a W or Z boson, ...


6

Photons are force carrying bosons and come in both virtual and real varieties. There is nothing wrong with that. Virtual means off-shell, and real means on-shell. Even on-shell weak bosons decay very quickly, however, because there are plenty of modes with the right quantum numbers and much lower total mass (and thus lots of phase space). I want to ...


6

They are more elementary in the sense that there is no accepted underlying theory from which they can be derived as an effective approximation. On the other hand, what is elementaty changes with time. At some time, protons and neutrons were considered to be elementary particles, whily they are now considered to be composed of quarks. There are various ...


5

Not all phonons are Nambu-Goldstone bosons and not all Nambu-Goldstone bosons are phonons. Nambu-Goldstone bosons are (usually) gapless excitations that arise from spontaneous symmetry breaking. For instance, in a spinless Bose-Einstein condensate, the NG boson is indeed a phonon, with a linear dispersion at low energy. However, in a ferromagnet the NG boson ...


5

The spin of a quasiparticle can be determined from a number of ways: If the quasi-particle is a "compound" object, you just add the individual spins according to the appropriate rules for adding angular momenta. An example would be the polaron, which is an electron dressed with a bunch of phonons. The electron has spin $1/2$, the phonons have spin $0$, so ...


4

If you integrate out the fermions in the quantum Hall system (or the Chern insulator), you will end up with an effective $U(1)$ gauge theory, with a Chern-Simons term. The Chern-Simons term is originated from the non-zero Chern number of the occupied fermion bands, and reflects the Hall response of the system. This Chern-Simons term makes a great ...


4

As you mention fractional quantum Hall effect, let me consider a system of $N$ electrons typical of a condensed matter system. Now think of your Hamiltonian as having two parts $\hat{H} =\hat{H}_0+\hat{H}_{int}e^{-\zeta t}$ with $t >0$ so that you gradually switch-off the interacting part so that at large times you can map your complete Hamiltonian ...


4

Superfluid Helium-4 has a very well studied excitation structure -- at very low momenta, there is a low energy excitation, the phonon, that corresponds to a periodic density fluctuation in the superfluid with well defined wave-number and an energy $E = c \hbar k$ (c being the speed of sound in the superfluid). Though others might quibble with me over ...


3

You may think this way: take a perfect infinite crystal where Bloch theorem perfectly work and add potential which makes real crystal finite. Next question you may ask how this potential is "seen" by quasiparticles which have been obtained from infinite crystal consideration. This procedure is perfectly self-consistent and is applicable in all cases. Also, ...


3

Dear Robert, the answer to your question is trivial and your statement holds pretty much by definition. You know, the Green's functions contain terms such as $$G(\omega) = \frac{K}{\omega-\omega_0+i\epsilon}$$ where $\epsilon$ is an infinitesimal real positive number. The imaginary part of it is $$-2\Im(G) = 2\pi \delta(\omega-\omega_0)$$ So it's the Dirac ...


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 ...


2

I was wondering something similar few month ago. Then I concluded that most of the topological staffs appear at the boundary between two different topological sector. A sector being characterised by a Chern number, or if you prefer a topological charge, one needs a boundary / an interface between two systems characterised by different topological charge. ...


2

Hole as a particle First, hole can really be treated as a particle. For electrons, there are Pauli exclusion principle, so there are only one electron per state(state can be described by momentum $\vec p$, band index and spin). In semiconductors, there are valence band and conduction band. In ground state, valence band is completely occupied by electrons, ...


2

In the context of condensed matter physics: To make it short, and with the caveat that it is not a universally accepted definition, an elementary excitation may be called a quasiparticle if it is fermionic (e.g. dressed electron), and collective excitation if it is bosonic in nature (e.g. phonon, magnon). But there is not clear cut and absolute divide ...


2

Dictionary for this answer: Excitation = particle; collective excitation = quasi-particle. Short answer: Elementary particles are never quasi-particles, by definition of elementary. This does not mean that what now it is thought as an elementary particle could be a quasi-particle of entities to be discovered. Mathematical answer: Elementary particles ...


2

A quasiparticle is an elementary collective nonlocal excitation of a quantum medium (such as a crystal); examples are phonons (describing sound waves) or Cooper pairs (describing superconductivity). In contrast, a particle is (in the context of high energy physics) the elementary excitation of a local quantum field; examples are the electron or the photon. ...


2

Bloch's theorem generalizes nicely to a finite size crystal if we take periodic boundary conditions (pbc). If we have pbc than the a translation by one unit cell is still a symmetry of the system and so Bloch's theorem will apply. The only difference will be that the quasimomentum $q$ will only be allowed to take certain discrete values since the ...


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

They can be real, no problem with that. However, all Z and W observed are virtual. And yes, they are off-shell, whatever that means to you. We actually measure their width, for instance, http://arxiv.org/pdf/0909.4814 (it's more or less 2 GeV around a mass of more or less 80-90 GeV). What is observed is a pole in the S matrix for some final states in ...


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

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

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

The answer to the apparent contradiction of these two transformations (the excitations seem to be either bosonic or fermionic) comes the fact that the spins are not equivalent to the fermions, because they have string attached to them, to respect the commutative nature of spins on different sites, see JW transformation on wiki. Therefore, even though the ...


1

I should say that you have 3 related questions, namely 1) To what extent can we trust the approximations based on HP and Jw transformations, 2) The nature of the low excitation spectrum and 3) The relation with Goldstone modes. We shall look first at the Holstein-Primakoff method. The spin ladder operators for at a site $j$ are given by $S^-_j = ...



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