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Electronic Structure: Basic Theory and Practical Methods by Richard M. Martin is a good book. But, it takes time to read completely through it. In this primer, the first chapter gives a good introduction to DFT. You can probably start with this.

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The same way any substance can have transnational energy. It arises because the energy of motion is proportional to the square of the velocity so even if the average is zero, the average of the square is not. If the mean velocity of the particles is $\bar v$ then the instantaneous transnational energy would be: $$\frac12 m\sum_i (v_i-\bar v)^2$$ ...

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Note that the authors define $\vec{b}_k=[b_{B,k},b_{A,k}]^T$ and write the Hamiltonian $$H_\text{kin}=\sum_k \vec{b}_k ^\dagger\left[\begin{array}{cc} 2t\cos(ka) & t'(1+e^{ika})\\ t'(1+e^{-ika}) & 0 \end{array}\right]\vec{b}_k.$$ The $1,1$ element therefore corresponds to hopping between $B$ sites, which, by Fig. 1(d), is just the tight-binding ...

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The BCS wave function is a superposition of components with definite particle number $N=0,2,4,\ldots$. In the infinite volume limit, the particle number is sharply peaked around $N=nV$, where $V$ is the volume and $n$ is the partcle density. It is intuitively clear that projecting the BCS state on a definite particle number should not make a difference. This ...

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$u_k(r)$ has the periodicity of the crystal. Therefore, it's Fourier expansion only includes reciprocal lattice vectors. $u_k(r)= \sum_GC_{k-G}e^{iG.r}$ Therefore, $u_{k′}(r)exp(iGr)=u_k(r)$

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Referring to your figure: Each corner atom contribute, 1/18. Top, bottom, left and right atoms on the faces each contribute, 1/9. The closest and furthest atoms on the faces each contribute, 2/9. To calculate these numbers one needs to find angles which are nothing but 60 or 120 degrees. Here is the method explicitly:

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I'm guessing that the question is asking how you work out how many lattice points are in the cell. If so the standard procedure is to displace the cell a small distance along each of the lattice vectors than count the number of points the cell contains. I'll illustrate this in 2D since my abilities to draw convincing 3D diagrams are limited. Consider this ...

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The lattices differ from each other in the amount of symmetry they have. You are correct that lattice $\mathbf{3}$ is basically the same as $\mathbf{1}$, but then all the lattices are basically the same as lattice $\mathbf{1}$ just with extra symmetry. For example the square lattice is derived from $\mathbf{1}$ by requiring that $a_1 = a_2$ and $\phi = ... 1 Well, The FCC's (along with BCC's) are conventional unit cells, not primitive unit cells. As for the contribution of points, it is different for the corners and face centers. Each corner of a unit cell in a lattice is joined to 7 other unit cells. so the corner point is shared equally between 8 unit cells. Hence the corner contributes only$\frac{1}{8}$th of ... 1 This is because we want to avoid extensive mathematical calculations of pairwise interactions with all particle inside or outside the box. MIC is a way of providing a cutoff distance over which we are not calculating pairwise potential. The cutoff is usually half the box length. This means if distance between particle i and particle j is more than L/2, you ... 1 I can't speak specifically for organic polymers, but I will try my best for polymers in general. Every bulk polymer is made of thousands polymer chains, which is made of many "mers" (Greek for unit). Consequently we have the name polymer . For many polymers at room temperature these chains are able to rotate, and because the bonds are not 180 degrees apart ... 2 The Young's modulus of steel doesn't change significantly between say 10ºC and 20ºC (I'm guessing this is roughly the range of temperature between morning and midday). So the stiffness of the steel won't be changing. However I would guess that the steel wire has a polymer binding it together, and possibly a polymer coating on the outside of the wire as ... 1 From the diagrams on the webpages you linked, it appears that other ice phases begin to form at around 200 MPa of pressure, and about$-20\text{C}^{\circ}$. Keep in mind normal water freezes at$0\text{C}^{\circ}$, and air pressure at sea level is around 0.101MPa. That means an ice cube made of one of these phases would sublimate or explode very quickly. If ... 1 As you can see from the phase diagram plot in the first link you provided, the only other ice phase which is stable at atmospheric pressure is ice XI, and its density is about the same as that of the most familiar ice phase (ice Ih). The other denser ice phases that you see on the phase diagram are only stable at pressures significantly above 1 atmosphere. ... 0 Roughly speaking, the conductivity is inversely proportional to the scattering rate$1/\tau$(in the Drude model, the longitudinal conductivity is$\sigma_0=\frac{n e^2 \tau}{m}$). From Fermi's golden rule,$1/\tau$is proportional to the number of density of states a quasiparticle can scatter into, which is proportional to the density of states near the ... 1 The Fermi velocity is related to the Fermi energy$\epsilon_F=\frac12m_{\text e}v_F^2$. Actually, the Fermi energy$\epsilon_F$is the chemical potential of the electron gas (that is the minimum energy required to add an extra electron to the gas). The energies of the electrons are distributed, according to the Fermi-Dirac distribution, as$\$ ...

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You may find the following paper useful: A Symbolic Solution of the Hubbard Model for Small Clusters, by J. Yepez. You may also want to review group theory for condensed matter physics, because your questions essentially span the basics of group and representation theory. Many texts give good overviews of the fundamentals of group theory as applied to ...

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A generic procedure to obtain the particle-hole conjugate of a quantum Hall wavefunction is described in http://journals.aps.org/prb/abstract/10.1103/PhysRevB.29.6012. It is not clear whether this is the most "general" form. But even for the Pfaffian state, we do not know the most general form either.

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Perhaps the answers and comments to this related question, Is crystal momentum really momentum?, can help with the conceptual picture. The crystal momentum arises from a discrete translational symmetry of the total Hamiltonian and in that sense it is a good quantum number. It is in principle the eigenvalue of the generator of the discrete symmetry. In ...

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Because without magnetic field, the chemical potential is at the Dirac point. In other words all the valence bands are occupied. The system is particle-hole symmetric. With an external magnetic field added, the energy levels are quantized into Landau levels(LLs). Hlaf of the zero-th LL comes from the original valence band and the other half from the original ...

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