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To obtain the energy gap in the thermodynamic limit, one should take $N\rightarrow\infty, V\rightarrow\infty$ where $N$ is the number of atoms and $V$ the system size, but hold $N/V$ (i.e. density) fixed. In your case, it means simply taking $N$ to $\infty$ is enough and keep all $t, u, \alpha$ fixed for now. This tight-binding model can be solved exactly. ...

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A zero-field-cooled/field-cooled split in the magnetic susceptibility vs. temperature doesn't have to be superparamagnetism. In the case of superconductors, if we apply a field to the material and cool past T$_c$, some flux can be trapped inside, but if we cool first and then apply field, that flux will be shielded away, resulting in greater diamagnetism.

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As I expected, this simple question calls for no more than a little sleight of hand, as pointed out by the sole comment above. It looks as if @Meng Cheng won't make it an answer. Thanks to him. And here I confirm that it works well.

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The poster above is incorrect: FCC metals do not have a ductile to brittle transition temperature and instead remain ductile at low temperatures. This is because the stress required to move dislocations is not strongly temperature-dependent in FCC metals, and thus failure occurs by plastic flow instead of crack propagation. In BCC metals, the stress ...

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There is a very nice demonstration that you can treat holes as positively charged carriers in the Hall effect. As you may know, to observe the Hall effect we place a semiconductor in a magnetic field and pass a current through it. We observe a polarization (voltage) at right angles to both the current and the magnetic field as a consequence of the Lorentz ...

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Holes are technically just vacancies, but when an electron jumps into this vacancy, one more hole is created, but this constant filling up of a hole and creation of a hole appears as if the hole is moving, that is it appears as if there are these +ve carriers which are moving in the direction of the current (opposite to that of the direction of electron) ...

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Because $$n_\downarrow n_\downarrow = n_\downarrow$$ and similarly for $n_\uparrow$. Why? Because $n_\downarrow$ can only take on the values 0 or 1 and $$0^2=0$$ and $$1^2=1$$

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Here's a way to think of it: In electric equilibrium, the doping atoms, such as boron or arsenic, have an extra electron or hole, but it's not doing anything. When a photon comes in, it brings up the quantum energy level of the extra charge carrier, and the electron or hole moves to the conduction zone. Because it is in the presence of an electromagnetic ...

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The issue here is how much the refractive index $n$ tells you about dissipation. As you rightly said, the imaginary part of $n$, which depends on both real and imaginary parts of $\epsilon$, leads to an imaginary part in k which describes an exponentially decaying electric field. However, this doesn't necessarily correspond to dissipation (i.e. a drop in ...

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If you notice, the target equation has $r_0$ in it. So in equation 8, instead of solving for $r_0$, solve for C. Taking the first derivative of u and assuming that it goes to zero at $r_0$ gives $$\frac{\alpha e^2r_0^{m-1}}{m} = C .$$ Next, take the second derivative of u with respect to r to get $$\frac{d^2u}{dr^2}= \frac{-2\alpha e^2}{r^3} + ... 4 The mean free path can be meaningful quantity in quantum mechanics, although usually only in a semi-classical regime. It is particularly useful in the kinetic theory of quantum liquids at low temperature, where the excitations of the system can be described as quasiparticles that propagate approximately ballistically and interact only rarely. You can define ... 5 First, for clarification: As Jon Custer already mentioned, THz radiation is not difficult to generate. It's simply part of the black body radiation. What is in fact difficult is to generate coherent or at least narrow-band THz radiation. Regarding emission from semiconductor material: There is not so many materials, which would offer a bandgap in the THz ... 0 I will expand on my comment to more fully answer the question. A reactor is a fine source of neutrons, but they come out at a spread of energies particularly since 'moderating' within the reactor, and 'thermalizing' outside of the reactor both involve multiple scattering events which, even if perfectly elastic, result in a distribution of neutron energies. ... 3 You are in luck. Alexander Gavriliuk, Ivan Trojan, and Viktor Struzhkin observed the anticipated transition in 2012. Their paper is in Physical Review Letters 109 086402 (2012) (link to the paper). They mention Mott's prediction in the abstract, and discuss the various prediction of where it would occur in comparison to what they observed. Now, they had ... 0 Just in case somebody else is asking himself the same question: The sum has to performed over all 8 bands, so including the spin. And for the second question, yes, the coefficients is what you get as the eigenvector. 1 Measuring an off-axis peak simply means that you're looking for crystal planes, which are not parallel to the sample surface. Therefore, they have an in-plane component (in addition to the perpendicular lattice plane spacing, which is commonly measured), which is the part, you're looking for. 1 In some sense, you have answered your own question. All this means is that you cannot determine a from a series of reflections along the same axis (e.g. (001), (002), (003), etc.). These are all along that same direction in reciprocal space, so therefore they are all on the same axis. If you measure an off-axis reflection, you will able to calculate ... 2 From this step,$$U_G=\frac1a\int_{-a/2}^{a/2}\mathrm{d}x U(x)e^{-iGx}=\frac1a\int_{-a/2}^{a/2}\mathrm{d}x \sum^\infty_{n=-\infty}A\cdot \delta(x-na)e^{-iGx}$$note that the summation is an impulse train with spacing a. Since the integral is from -\frac{a}{2} to \frac{a}{2}, just the impulse at x = 0 is integrated over so only the n=0 term ... 0 I would start with the derivation in the following way: Your Hamiltonian has additionally to its kinetic part an extension by a Zeeman term:$$ H_Z = -\frac{g \mu _B }{\hbar} \sum_{i=1}^N s_i^z H$$with g=2 the gyromagnetic ratio and \mu _B the Bohr magneton. One can absorb the energy difference of the parallel and antiparallel spins into a ... 0 The claim here is actually false. If one looks at tetragonal crystals from the copper oxide superconducting class, they can have differences in in-plane conductivity and out of plane conductivity which can be as high as four orders of magnitude. These are an extreme example as they are the most anisotropic crystals known, but they illustrate the point well. ... 1 This is a very good question. It turns out that the phase transition occurs precisely when the chemical potential becomes equal to zero (assuming that the ground state energy is at zero). The order parameter in the BEC is the "macroscopic wave function" or rather the square root of the single-particle reduced density matrix. The broken symmetry is usually ... 1 How atomic orbitals merge into crystal band structure is, well, complicated to capture in simple models. As seen in, say, Ashcroft and Mermin (chapter 28), the energy surfaces for Si have symmetry along the <100> directions. In contrast, the surface for Ge have symmetry along the <111> directions, with the band minimum at the zone edge. A comparison ... 5 Can this frozen form freeze further? Or can it become more solid? (for example, by exposing to colder temperatures and/or a higher pressure). Can ice freeze further by transforming into a different crystalline form? The ice will remain solid while lowering temperature or pressure but might change in state, or phase, as you mention. But you should ... 1 Those other (lower) bands will be filled. The most interesting bands are the ones that are only just barely full, or only partially filled, as well as the empty bands just above the filled ones. Usually when people speak of the band gap, they mean between the highest filled band and the lowest empty band. 2 The density of states \rho(E) \propto E^{1/2} is valid only for free electrons. Electrons in a solid are certainly not free, and the density of states is complicated. Certainly: the density of states is zero inside the band gap. 1 I think that you make a mistake. The operator (i)  \ \hat U = −\hat {\vec {\mu}} \cdot \vec B = −g\ \mu _B \ \hat {\vec J} \cdot \vec B doesn't give energy levels, but an additional energy term in the Hamiltonian. Let's take for simplicity the direction z of our system of axes in the direction of \vec B. Then my formula (i) becomes (ii)  \ \hat U = ... 0 In a semiconductor, current is produced in two different ways. There are the electron current and the hole current. The electron current is produced when electrons are pushed from the negative terminal into the semiconductor. Holes are positions in the semiconductor atoms that can be but are not occupied by electrons. An atom with a hole can "rob" the ... 1 Recombination A useful way to think about recombination rate in semiconductors follows from the equation,$$ R = An + Bnp + Cpn^2 - G + I/q $$where n and p are the electron and hole density. The first term deals with the non-radiative recombination. In your question you mentioned Shockley-Reed-Hall (SRH), think of this term as a very simple ... 0 The Hubbard Model is a parametric model - that is U and t are adjustable parameters which can be tuned with experimental data. However, there are so many uses for this, that the effective value of these parameters depend not only on the material, but also the type of information that you want to get out of the model. So, from doping to long range ... 4 Imagine the single energy-bands as a set of "places" (\rightarrow states) which can be ingested by an electron for example. Between these sets of places there are gaps (at least for non-free electrongas models). So if the band is not filled up, electrons can "swap seats" and conductivity is given. If your band is full you need a quite "huge" amount of ... 0 Suppose the rod expands less than the cylinder, and suppose we unstick the end of the rod. Then after we've heated the system it will look like this: Because the rod expanded less than the cylinder there will be a gap x between the end of the rod and the plate. So if you want to re-attach the rod to the plate you have stretch the rod by a distance x_r ... 1 I think the relationships result from an interplay between the dependence of energy on quantum numbers and the volume element in the space of quantum numbers. For an m dimensional particle in a box,we can write an eigenvalue for the energy as$$ E_{n} = k \sum^{i=m}_{i=1}{{n_i}^2 = E n^2}$$where k is some physical constant and and n_i are the quantum ... 0 Some people like using angular wavevectors, while others like using ordinary wavevectors. For example, angular wavevectors are more common in most area of physics, except for crystallography ... ordinary wavevectors are common in signal processing, etc. There should be a personality test... If you like writing \cos(\mathbf{k}\cdot\mathbf{r}) or ... 0 You understand this in one dimension and then three dimensions is obvious. Consider a function f(x). If we define the Fourier transform of this function by$$ \tilde{f}(k) = \int \, dx \, f(x) e^{- i k x}$$then the original function f(x) is can be written as$$ f(x) = \int \frac{dk}{2\pi} \tilde{f}(k) e^{i k x} \, . The $2 \pi$ is there because the ...

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They are not necessary, but with this choice you can interpret them as a basis for the wave numbers $\mathbf k$ (as in $e^{i\mathbf k\cdot\mathbf r}$).

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When an electron is ejected from a metal surface it experiences an attraction due to the image force. However the attraction falls off rapidly with distance. A back of the envelope calculation tells me that (relative to infinity) the potential energy due to the image force is only around -0.001eV at a distance of a micron. My understanding is that the image ...

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