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40

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 ...

24

Take for example an electron and a muon. The muon is unstable because it decays into an electron and two neutrinos in about 2$\mu$s. But a muon is not in some sense an excited electron. Both particles are excitations in a quantum field and they are both as fundamental as each other. The electron is stable only because there is no combination of lighter ...

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 ...

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 ...

12

Your question touches the question of ontology in particle physics. Historically we are used to be thinking of particles as tiny independent entities that behave according to some laws of motion. This stems from the atomistic theory of matter, which was developed some two thousand years ago from the starting point of what would happen if we could split ...

11

How can the unstable particles of the standard model be considered particles in their own right if they immediately decay into stable particles? Here I will only consider elementary, non composite particles. All the hadronic resonances are composite particles of quark antiquark combinations as well as the neutron . The standard model of particle ...

7

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, ...

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. ...

7

Ill answer the second part of your question about effective mass and quasiparticles, since I see that CuriousOne has answered the rest better than I could have. In a metal or semiconductor, the electron is not in the same free state it would be in a vacuum. It is bound to (although delocalised within) a lattice of positive ions. So its dispersion ...

7

The answer is Yes. See A physical understanding of fractionalization http://arxiv.org/abs/hep-th/0302201 Quantum order from string-net condensations and origin of light and massless fermions, Xiao-Gang Wen; Spin-1/2 and Fermi statistics from qubits http://arxiv.org/abs/hep-th/0507118 Quantum ether: photons and electrons from a rotor model, ...

6

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 ...

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

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

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 ...

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

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. ...

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

I think the most direct answer to this would be the fact that a heavier particle can decay into many different lighter particles for different reactions. The abundance of occurence of these relations are const. Again the same heavy particle can be created in multiple types of collisions of various different lighter particles. Thus we cannot say that the ...

4

Inmediately is not really true, there is some proportionality. Sorry I am not answering directly about "the standard model", this is quarks and leptons. But they will fit the general pattern, you will see. Let me first consider all the "particles" listed in the particle data group file. Most of the particles decaying via photons have a half-live about ...

3

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, ...

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

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 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

It depends if we look at particle as classical ones or quantum ones. In the first case, particles are usually following a Boltzmann statistics. However, things become more interesting when entering the quantum world. Here, the spin of the particles become crucial. We have that particles with integer spin follow a Bose-Einstein statistics. Whereas particles ...

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

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

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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