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As the size of the lattice gets large, the superposition $$\vert \psi \rangle = \frac{1}{\sqrt{2}}\left(\vert \uparrow \uparrow \uparrow \cdots \rangle + \vert \downarrow \downarrow \downarrow \cdots \rangle\right)$$ in the Ising model becomes unstable to environmental perturbation, just like Schrodinger's cat. Like the superposition of a dead and live ...


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I thinking a lot of time, then I came in this assertion: The kinetic part of the Anderson's hamiltonian have a continuous spectrum. We want to work numerically with this part, so we need to find a some basis that diagonalizes approximately (the terms off-diagonal be small by controlling some discretization parameter) in such a way that now the "spectrum" is ...


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Your question is deep and to the best of my knowledge only partially resolved, see Sec. II of 2007 article in Review of Modern Phsyics. The trick is called "the Wilson chain" and is essential for NRG to work on strongly correlated models with Kondo-like behavior. The Wilson chain construction makes NRG a computationally efficient procedure. My ...


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Unfortunately, the premise is wrong. In general, lower dimension DOS can't converge to higher dimension DOS. That's one primary reason why 2D devices are interesting. For example, the van Hove singularities of the DOS can diverge in 2D, but only their derivatives can diverge in 3D.


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[As requested, I convert my comment into an answer, as it might also be useful for other people.] There is a very interesting series of works by Lieb and Yngvason on entropy and the second law of thermodynamics, based on the kind of axiomatic approach you seem to be interested in. You can start with this introductory paper, or this, this or this more ...


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The constant $C$ is not a part of the single-particle Hamiltonian. It is what is called the vacuum energy, and has no observable effect unless we look at gravitation. To be specific, adding this constant to the Hamiltonian makes $e^{-\beta H}\,\rightarrow e^{-\beta C}e^{-\beta H}$ and $Z=\rm{Tr}(e^{-\beta H})\, \rightarrow\, e^{-\beta C} Z$. That is, the ...


2

Ultimate physical motivation Strictly in the sense of physics, the entropy is less free than it might seem. It always has to provide a measure of energy released from a system not graspable by macroscopic parameters. I.e. it has to be subject to the relation $${\rm d}U = {\rm d}E_{macro} + T {\rm d} S$$ It has to carry all the forms of energy that cannot be ...


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Thermodynamically entropy is defined by \begin{equation} \mathrm{d}S = \frac{\mathrm{d}Q_{rev}}{T} \, ,\end{equation} where $\mathrm{d}Q_{rev}$ is the heat, transferred reversibly. As you point out it can be shown that this quantity is a function of state. This implies that the entropy of any thermodynamic system has, up to a constant, a well defined ...


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As you have in the commutation relations, $\sigma_i \sigma_j= \sigma_j\sigma_i$ e.g. spin operators on different sites commute, so there is no minus sign to pick up.


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The first thing to realise here, is that in your Hamiltonian formulation, the lattice constant is implicitly stated as being $b=1$ (I'm calling it b, not to confuse it with the lattice operators). Next you can identify, if you're working in two dimensions, $x=am$ and $y=bn$ yielding $$\hat{H}_0=-J(\sum_{(x/b),(y/b)} ...


1

Well, when did you want it to be written down? As it is, it was pretty much simultaneously and independently arrived at by Hubbard, Kanamori, and Gutzwiller an awful long time ago. Probably others too. The point is, it was written down when there were experimental phenomena that justified including interactions in the model. It wasn't some great conceptual ...


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min Zhang, The Hubbard model (offen reffered to as the U and J terms in ab-initio DFT or tight-binding models) is a little bit more complicated than it looked like. It is indeed an additional energy that you add locally to some states (d or f bands usually) to locallized them. Usually you want to do this because DFT tends to spouriously delocallize ...


1

Absorption isn't an instant event. At the level of simple quantum mechanics, this system can be described as follows. Evolution of electron in crystal is governed by Schrödinger's equation. External electromagnetic field, namely the light which we shine on the crystal, is a periodic addition to the Hamiltonian. When you start shining light at the crystal, ...


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I would say in the same way a thick beam in a structure "knows" that you are trying to tear it off: it is temporarily displaced somewhat due to your effort, but then returns to its stable state. So the electrons are not totally immune to the photon (for example, the crystal acquires some minuscule momentum due to light pressure, if the photon is reflected ...


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If you have the distance between atoms in a cubic lattice, then the (1,1,1) planes are the planes you get when you move one atom in X, one in Y, and one in Z. There is a nice presentation on this at https://nanohub.org/resources/5488/download/ch3_p3.pdf The following picture is taken from that presentation, and answers your question exactly:


0

Well I had an extensive discussion with experimentalist friend and he tells me that its the unfilled electrons i.e d-orbitals...... yea it can be more than spin half but we are discussing hopping in a lattice then it is a single electron with spin half and in condensed matter(since we cannot have more than one electron to hop from the same site) it is the ...


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A spin-half on a lattice site is a theoretical 'particle' with the property of having a spin of one half. It can be either 'up' or 'down' along the measuring-axis (in most textbooks the spin operator for assigning spin up-or down is the $\sigma_z$ operator, the third Pauli matrix. You could also use the $\sigma_x$ or $\sigma_y$. I believe the spin on a site ...


1

The spin referred in condensed matter is the spin of the electrons least bound to the atoms (usually valance electrons). The atoms reside on the lattice sites. A spin half problem means the atoms have only one valance electron. But there are other possibilities like spin 1, 3/2 and all. As qeb has already mentioned it can also be used for nuclear spins also. ...


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Basically, one needs to have a non-zero polarization when computing the Berry phase in traversing the Brillouin zone. It is difficult to say immediately if inversion-symmetry breaking is the only way to get a non-zero polarization, but at the present time this is the only way that we can conceive. That is not to say that in the future someone won't come up ...


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There may be a few options for describing the crystalline order. Personally, I love the one used by Anderson in his book 'Basic Notions of Condensed Matter Physics'. Following him, one writes the atomic density as $\rho(\vec{r}) = \sum_{\vec{G}}\rho_{\vec{G}}e^{i\vec{G}\vec{r}}$. The appearance of Crystalline order is signified by a set of finite ...


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You need to define an appropriate "order parameter" for your system, one that takes into account the symmetries in the configuration as well (rotational, transnational, etc). There are many ways you could define such "correlator" as you call it, it depends on the system. For example the nematic order parameter in liquid crystals is taken with respect to a ...


1

If time-reversal symmetry is broken at the surface of a topological insulator, a gap could open at the Dirac point of the topological surface state. The Dirac point, where forward- and backward-moving electrons have the same energy, is located at a time-reversal invariant momentum point (also called a Kramer's point) in the reciprocal space (the crystal ...


1

In the image above, you can see a series of Bragg planes drawn in the crystal. This is called one "set of planes". Another "set of planes" would be if one would just draw a series of horizontal lines through the atoms. (Of course by lines I mean planes, but they are projected here onto a 2D image). The planes are those formed by the atoms, so in that ...


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and 2. Crystal planes contain sets of atoms, which occupy identical positions in the primitive cell. So to say, the planes are made out of atoms (and nothing in between). Or if you want a more quantum mechanical picture, you would have electron orbitals in between, which are responsible for the chemical bonding. X-rays get scattered by individual atoms. If ...


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I have found that the review by Nayak et al. on topological quantum computation (http://arxiv.org/abs/0707.1889) to be helpful in getting acquainted with topological phases, Chern-Simons theory and many related issues. It is really a wonderful review and definitely should be on your reading list.


2

Tungsten has been known to bait gold bars (historically). There are a few methods we use to determine if something in front of use is gold or if it is alloyed, or if its plate, fill or scrap. You can cut the bar in half...You will then know immediately of you got bunk gold. You can do a specific gravity check of your gold. There are scales designed for ...


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There is an ambiguity. Although I did not understand your analysis of the problem completely, charge carriers certainly can run against the (averaged) electric force due to difference in available bands and other particle statistics effects. The gauge freedom is irrelevant. There are two cases for the “ubiquity”. First, these non-Maxwellian deviations ...


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Sorry to disappoint you, but the second "article" you are citing is the usual pseudo-science nonsense published by the crank crowd. I would suggest you pick up half a dozen good textbooks on relativity and study the introductory chapters until you understand the experimental difference between relativity and ether models. It's the only way to get past the ...


2

No, “this conclusion” is based on the topological properties of rotation groups. Namely, for any $n > 2$ $\mathrm{Spin}(n)$ is the universal cover of $\mathrm{SO}(n)$, whereas for $n = 2$ it is not. That’s why in $n > 2$ any thing has to be controlled by a representation of the Spin group, whereas in $n = 2$ it has not.


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The answer is 'no' in condensed matter also. As anyons are neither bosons nor fermions, they can follow some statistics other than BE or FD, but it has nothing to do with fractional charge. just-learning has already given you a perfect example.


5

Not sure about the condensed matter context, but in general the answer is NO. For instance, quarks have fractional charge but are regular fermions.


2

At present, there is a belief (though obviously not verifiable) by solid-state physicists that a metal cannot exist at absolute zero. The Fermi surface of the metal will be unstable to order of some sort such as superconductivity, charge density waves, magnetic ordering, etc. With that said, let us concentrate on your scenario though. If there are no ...


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In a metal the Fermi energy is somewhere in an unfilled band. At any temperature above absolute zero (which you can never reach) there are states available for electrons to get to and result in conduction at the Fermi surface. This will occur in any metal. Superconductivity is a separate phenomena that I won't touch on here.


0

After going through number of articles for temperature reaching absolute zero, I find that it is difficult to attain absolute zero, which may mean that it is very difficult to stop interatomic movements or energy exchanges, and thus absolute zero is near theroretical. As far as superconductivity is concerned it must be a critical point, minimum energy ...


1

Have you heard of superconductivity? This is a phenomenon where a material exhibits zero resistivity near absolute zero: it clearly contradicts your assertion that thermal excitation is needed for conductivity near absolute zero. For a semiconductor, it is true that electrons need to be kicked into the conduction band by thermal fluctuations - but for a ...


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The experiment on kxy has not been done yet, to my knowledge, but there are several (recent) experiments which are closely related and probe the heat flow through the edge channels of a Hall bar in the fractional regime: http://www.nature.com/nature/journal/v466/n7306/abs/nature09277.html or ...



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