# Tag Info

0

It depends on what group you consider the starting point, and that depends on context. One context is the mathematical one (forget all you know about spins etc) where we start with the vector space $\mathbb C^2$ and the natural action of $SU(2)$ on it. If we then look at the projective vector space $\mathbb C^2/\mathbb C^* = \mathbb CP^1$, then the action ...

1

1) Are we able to perform perturbative analysis and use diagrammatic expansion, Green function etc. – all these field-theoretical stuff [for bosons]? In general, the field-theoretic methods (at finite or zero temperature) can be applied to both bosons and fermions with slight differences which originate from the Fermi-Dirac and Bose-Einstein ...

1

When are spin exchanges said to be isotropic or anisotropic? I agree with the other answer, that the precise meaning depends on the context. My answer concentrates on the specific context of ab initio calculations in condensed matter physics/chemistry. Calculations are often performed using density functional theory (DFT) with on-site Coulomb corrections, ...

4

It's true that "the perturbative series is valid only when the perturbed state is qualitatively similar to the unperturbed state". Generally perturbation theory is acceptable when the coupling is weak, in which case the coupling can be treated as a small perturbation of the free field theory at all energies (for example Yukawa theory and $\phi^{4}$ theory. ...

0

This is a really good question. I have also thought about this quite a bit. At present, it does not seem like it is possible with a standard experimental probe. Inelastic x-ray scattering and electron energy loss spectroscopy only measure the longitudinal response function at finite frequency and momentum. As you said though, if you are only seeking small ...

3

1) I should note that most perturbative expansions that are of interest in physics are not formally convergent (and more often than not, not Borel-resummable either). 2) There are many examples of useful perturbative calculations for bosons. The oldest example (probably) in Many-Body physics is the calculation of the energy per particle of the weakly ...

-2

I am looking for the name of a theory positing that "empty" space itself is not empty Rather surprisingly, it's called General Relativity. See the Einstein digital papers where in his 1920 Leyden Address Einstein said this: "This space-time variability of the reciprocal relations of the standards of space and time, or, perhaps, the recognition of the ...

1

You take the problem upside-down. Once you got a Hamiltonian, if it has a particle-hole symmetry $P$ which by definition verifies $\left\{ H,P\right\} =0$ with an anti-unitary operation $P$, you can construct it explicitly, and then you know how it applies on the operator basis and so on. For instance, a so-called s-wave superconductor can be described by ...

3

Expanding on what Couchyam said: To some extent, this is actually the definition of what it is to be a "particle." Intuitively, a particle should have some particular definite energy, and be stable enough that it can exist in its own right. These two requirements are linked by energy-time uncertainty. The natural "clock" to compare against when asking if ...

1

No. The elasticity of the solids is a liquid kind of character. The non newtonian fluids provide a solid kind of character. As everything is just electromagnetism, you can split your thoughts down the an single atom level, and play it with magnets on real scale. You can push your hands between magnets, and you can make them flow. In the wikipedia ...

0

In STM, you tunnel from states of your tip into states of the sample. Electrons can tunnel into states within the whole bulk and the matter beneath. However, their spectral weight decays rather quickly. As a rule of thumb in STM, your current increases by one order of magnitude per Angstroem that you reduced the tip-sample distance. So, the DOS of matter 1 ...

5

Metals are good conductors of electricity because the outer (valence) electrons of the metal atoms are only loosely bound to the nucleus and form molecular orbitals known as the conduction band. Electrons can move more or less freely through the conduction band and so metals conduct electricity generally well. When a metal is chemically oxidised its outer ...

0

It's because valence electrons are bounded. For example, consider Si and SiO2. While Si is semiconductor, SiO2 is insulator because it has no free valence electrons. BTW, many metal oxides ARE NOT in fact insulators - for example ZnO, Fe3O4 are all conductors. But it's true that oxides of metals have lower conductivity than pure metals.

2

After x-rays hit a substance they will be scattered in all directions; if the material is a crystal then you will obtain a diffraction pattern where each point is created by the constructive interference of the scattered rays. The connection between the diffraction pattern and the reciprocal space is readily found: take a crystal and consider an atom ...

0

I think the notion of ‘band structure’ is deeply related to a “quasi-particle view” of an interacting system – even, an strongly interacting one. This means that although the original elementary excitations of the system (e.g., single electrons in a metal) do not provide a good and efficient description of the states and energies of the interacting system ...

1

I don't believe that the thermal conductivity of most metals is very sensitive to magnetic fields. Yes, there will be some field-induced band shifting in the case of an itinerant ferromagnet which, in principle, leads to a change in the density of states at the Fermi level, but that will typically be a very small effect. If the magnetic field induced ...

2

Magnetic fields certainly can influence thermal conductivity. This shows up, not surprisingly, when there is a strong influence of the magnetic field on other properties, particularly electronic ones. One (non-metal) example is 'Thermal conductivity tensor in YBa$_{2}$Cu$_{3}$O$_{7-x}$: Effects of a planar magnetic field' by R. Ocana and P. Esquinazi, Phys ...

2

Since the question is rather vague, I will just give you some key points: Debye's model treats oscillation modes of a solid as sound waves (phonons) with frequency $\omega(\mathbf{k})=v|\mathbf{k}|$ ($v$ the sound velocity). As a result, with this model, Debye shows how the heat capacity is directly related to the rate of change of the energy expectation ...

1

When a liquid or solid evaporates, it turns into a gas. In a closed container, pressure builds as gas accumulates. There are two competing processes. In the solid or liquid, the higher energy atoms at the surface fly off. In the gas, the slower atoms stick to the surface and condense. The number of atoms available to condense is proportional to the gas ...

1

First of all, the SSH model does not have particle-hole symmetry. Particle-hole symmetry is an exact symmetry (at mean field level) reserved for superconductors and is an anti-unitary symmetry. A symmetric spectrum does not mean particle-hole symmetry. SSH model and end states Let me first explain a simple way to understand the topological end states. The ...

2

Are fermions non-local objects, in a sense in which gauge bosons are not? As far as I understand, the answer is definitely NO. Fermionic particles are local objects as bosonic ones. Based merely on the non-local form of the bosonized Jordan-Wigner fermions, one cannot conclude that fermions are non-local. Jordan-Wigner transformation, like any ...

-1

Okay, I think I have a semi-convincing picture of this in my head. Both of the other answers contain at least part of the story I wanted; I will put the whole thing here in hopes of feedback and that it is useful to someone else. As SM Kravec points out, fermionic parity is a non-local symmetry of a fermionic system. This suggests, as various people have ...

0

This is a very interesting point and it is somehow related to universality classes. A typical examples are the universality classes defined by time reversal symmetry. This is at the basis of Random Matrix Theory and has application for the analysis of the atomic emission spectra and in mesoscopic physics as well. There is a direct connection between the ...

0

Turns out you just use Gauss' law for an infinite sheet!

0

Not an expert in this but I can try to answer some parts of your question. As you pointed out, one can find the optical conductivity to measure the current-current correlation. For that if you use linear response then you can conclude the following (for an isotropic material): $$\sigma^{ij} = \sigma_L \frac{q^i q^j}{q^2} + \sigma_T \left( \delta^{ij} - ... 0 Are fermions intrinsically non-local? Yes, definitely. It's quantum field theory, not quantum point-particle theory. An electron's field is what it is. And that field doesn't have a surface. From a great distance it will be swamped and undetectable, but there is no defineable place where that field stops. When one studies quantum mechanics of more ... 1 TL;DR In general, no. A longer but possibly irrelevant discussion follows. Consulting the classic review RevModPhys.58.323 by Rammer and Smith, the quantities you are considering are defined as (Eq. 2.5):$$G^{<}(\boldsymbol x_1,t_1,\boldsymbol x_{1'},t_{1'})=\mp i\langle \psi^\dagger_{\mathcal H}(\boldsymbol x_1,t_1) \psi_{\mathcal H}(\boldsymbol ...

1

These lectures on the QHE and FQHE given at the Les Houches summer school are a great start imho.

4

Friction causes the chalk to stay on the chalkboard. While the chalkboard appears smooth, under a microscope its surface is rough. Chalk is a much weaker material than the chalk board. When it is forced across the chalk board, small parts of chalk ('dust') are broken and remain trapped by friction in the surface asperities of the chalk board. The rougher ...

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So what people mean by 'non-local' varies from context to context and person to person. Wen has a very particular meaning to this. 1) In fermionization in $D=1+1$ the Jordan-Wigner fermions are, in the bosonic language, operators supported over many sites. The emergent (mutual)-fermions in the toric code are also supported at the ends of strings. 2) ...

0

This can be derived using diagrammagic technique (see eg Altland and Simons' book). The more fundamental reason is that the diffuson is a Goldstone mode, which has to be massless (also see Altland and Simons). But to my knowledge, the direct answer to your question is "no" -- there is no "simple" way because the math underlying the above is quite involved. ...

1

(1) For anyons to be created locally in a physical model they must be created in groups such that the local excitation is a boson or a fermion. However, the local excitation can fractionalize into anyonic parts which can propagate independently. In terms of second quantized operators the expectation is that the the local fermionic/bosonic degree of freedom ...

2

The surface plasmons are bosons. The bosonic nature of photons is preserved. Plasmons are hybridizations of photons and excitons. Although electrons are fermions, their particle-hole excitations (excitons) are bosonic. Because to create a exciton, one needs to move an electron from one state to another, which is implemented by a fermion bilinear operator ...

1

First you need to bring it into the following form: $H=\Psi^\dagger h \Psi$ Here $\Psi$ is a big column vector: $\Psi=(\dots, c_{m,n}, \dots, c_{m,n}^\dagger, \dots)^T$ Basically, the first half of $\Psi$ are all annihilation operators, and the second half are all creation ones. If the number of sites is $N$, the size of $\Psi$ is $2N$. So $h$ is a ...

2

So: I assume you want to diagonalize this problem by rewriting the Hamiltonian as $H=\sum E_id_i^\dagger d_i$, where $d_i$ are quasiparticle operators which obey the Fermionic commutation relations. If we only had $c^\dagger c$ terms, we would be able to write H as $$H=H_{ij}c_i^\dagger c_j$$ We could then prove that if $\{c_i\}$ obey the Fermion ...

-1

From an asymmetrical peak one can determine both a- and c- lattice constant. \begin{align}a=\frac{\lambda\cdot h}{(q_\text{par} \cdot \sqrt3)} \\c=\frac{\lambda\cdot l}{(q_\text{ort} \cdot 2)}\end{align} Where $\lambda$ is the wave length of your X-rays, $q_\text{par}$ (q_parallel) is the peak position alongside the in-plane direction ($q_x$ or $q_y$) ...

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