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Influence of Magnetic Domain Walls and Magnetic Field on the Thermal Conductivity of Magnetic Nanowires http://pubs.acs.org/doi/abs/10.1021/nl502577y H.T. Huang et al., “Influence of magnetic domain walls and magnetic field on the thermal conductivity of magnetic nanowires,” Nano Letters, vol. 15, pp. 2773-2779, 2015.

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They are referring to the terrace-ledge-kink (TLK) model of crystal growth (Google images may help you). The terraces are the flat surfaces of the crystal. When a new layer of atoms is partially formed, there is a step (usually, but not always, a single atom in height) that separates the existing terrace from the new terrace that's forming on top, and you ...

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The expected effects at elementary level would be that the positive gate potential would attract the electrons into the well, and also that it would counteract the built -in field orthogonal to the surface. Orthogonal field reduces mobility, so reducing it increases mobility. This elementary view may not, of course, be correct.

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$\hat{X} = i\partial_{k_x}$ in momentum space is wrong. The left hand side is an operator, the right hand side is the representation of an operator when acting on functions, taken with respect to the scalar product with the basis $|\ p\rangle$. In particular we have $$\langle x\,|\,\hat{x}\,|\,\phi\rangle = x\phi(x)$$ and $$\langle p\,|\,\hat{x}\,|\,\... 1 Charge density waves and Wigner crystals are two ways of understanding the same broken symmetry. In the Wigner crystal picture we imagine the electrons sitting on lattice sites; the electronic charge density thus has broken translation and rotation symmetry. In the CDW picture we imagine an essentially uniform distribution of electrons that develops a ... 1 I could only find this poor-quality picture. It should give you an idea, anyway. For example, consider the first picture in the first row: (l,m,n)=(1,0,0) in that case, and it is easy to verify that the distance between the grey planes is$$d=a$$In the second case, (l,m,n)=(1,1,0), and you can see that$$d=\frac a {\sqrt 2}$$etc. 0 I think that finally understood why you are confused. The density of states which your question gave is for a 2D electron gas, and not a 3D gas. Therefore, your first answer of E_F/2 is correct! You should get the same result even if you integrate with respect to E rather than k. The probably got a different answer because of an algebraic mistake. ... 0 The density of states for a 3D gas is \underbrace{2}_{\text{spin degeneracy}} \times \underbrace{4\pi k^2}_{\text{surface of sphere in momentum space}} \underbrace{\frac{V}{\pi^3}}_{\text{volume of single k-state}}. 2 The expression that you wrote for the density of states is for a free electron gas in 2D. Your answer is right, if you have a 2D free electron gas. However, for a 3D gas, the density of states per volume is given by$$D(k)=\frac{k^2 }{ \pi^2}$$You get the above expression as follows:$$D(k)\ dk\times Volume= \underbrace{2}_{\mbox{spin degeneracy}}\times ...

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Fermi level is an energy in which the electron distribution probability is 1/2. Due to Pauli Exclusion, the electrons will pile up so the Fermi level for electrons will move up. Chemical potential is the same with Fermi level. To calculate the value, you have to integral the Fermi-Dirac distribution function times the density of states. For a rough ...

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The standard approach is that there is bosonization. Suppose that we have the fields $\psi_1(x)$ and $\psi_2(x')$ for two fermions, or usually electrons. Let us consider the product of states $\bar\psi_1\psi_2$ and $\bar\psi_2\psi_1$ $=~(\bar\psi_1\psi_2)^\dagger$ with $\psi_i~=~\psi_i^\dagger\gamma^0$, and the positions implicit. The anticommutator of ...

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In superconducting materials two electrons are bound together such that they have up and down spin and make usually a spin 0 (sometimes spin 1 as in Helium-3 super fluid) particle (a boson), this bonded pair of electrons is known as cooper pair. This pairing is due to electron-phonon interaction in which at low temperature the positive lattice of ions is ...

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First of all, you can't expect to recover general classical mechanics by simply making averages in quantum mechanics. Apart from very special cases, you can recover it only in the limit $\hslash\to 0$. In such limit, something similar of what you expect can be proved. In particular, it holds when considering (squeezed) coherent states $C_{\hslash}(q,\xi)$ ...

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The question is always what is a 'decent' level... There are many books that appear to deal with semiconductor device physics, but generally deal more with semiconductor technology. I have a number of those gathering dust on my shelf. The one that has stood the test of time since my graduate device physics course is S.M. Sze's Physics of Semiconductor ...

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Technically speaking all solids have crystalline structure. Anything that is truly amorphous is known as supercooled liquid (such as glass). Depending on how well crystals are oriented they can be divided into two categories, as pointed out by @lemon, single crystals and polycrystals. Single crystals have nearly perfect orientation of sizes few mm long. ...

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Solids are rigid and resist a change in volume. There are three general types of solid: As you can see from the picture, a crystal is a solid for which the atoms/molecules are highly ordered to form a periodic lattice. Examples of (poly)crystalline solids include diamond and table salt. Examples of non-crystalline solids include glass and amorphous ...

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Conclusion: The particles are, in-fact, magnetic This is doubtful. There are two possibilities here: If the objects used to grind (let's call them grinders) the dust grains were made of a ferromagnetic material like iron/steel and/or nickel, then it is possible that the grinders deposited some material onto the dust. There is no reason to think that ...

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No way, amount of dark matter is estimated to be 4 times the amount of normal matter. Even if we ignore all other arguments, it still does not restore the symmetry. Dark matter (if there is such a thing), is actually transparent matter (we can not see it) and cold matter (it does not absorb/emit any heat/radiation). That also means that you can not measure ...

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No, we know enough of the "bulk properties" of antimatter to rule this out. Antimatter interacts with the electromagnetic field in exactly the same way as regular matter, just with the opposite charge. Therefore, antimatter should be detectable using most of the techniques we use to detect regular matter in astronomy. This works even if the antimatter is a ...

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I think we can pretty safely say this is not the case. The main reason is that we have a pretty good idea of where dark matter is--to some degree, it can be reconstructed from the gravitational influence it has on surrounding matter. The dark matter appears to be distributed evenly throughout the galaxy. Thing is, a galaxy is pretty "dirty," as far as space ...

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One reason is that the volume of your third BZ is twice the volume of your first BZ. To see it another way, notice that the black dot on the far right (the middle of the three vertical dots) is a BZ center, a $\Gamma$ point. The black dot directly above that one (far upper right corner) is also a $\Gamma$ point, but it must be in the next BZ. Half-way ...

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