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In insulators, the valence band and lower bands comprise just enough quantum states for the number of electrons in uncharged material. If there's no thermal energy to promote electrons into higher bands, the valence band will, at equilibrium, be full. And since it is an insulator, there's no electrons in the next higher (conduction) band. If the material ...


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Yes, there is a method known as McCoy’s method. You can find a detailed description in section 2.2 of the paper http://arxiv.org/abs/hep-th/0307095 . The idea is to use the fact that the eigenvalues of the transfer matrix are (Laurent) polynomials together with the Baxter TQ equation. Knowing a priory the eigenvalues allows you to use the TQ equation to find ...


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"Because it does not have a crystal structure, it is hard to find physical similarities with a solid" simply proves that you're using the wrong definition of "solid". Relatively few solids have crystalline structure, so a good definition of solid simply can't depend on that. Let me see if I can offer a better definition. Rigid body (Idealization) The ...


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I fear your question is not very precise. Unfortunately my reputation is not high enough to allow me to comment instead of answer. What exactly do you mean by the term "exciton"? The way it is normaly understood, is a pair of one electron and one hole. So there are no multiple electrons whose spins could line up parallel oder antiparallel. In a biexciton ...


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Glass is a typical amorphous solid. Amorphous materials typically show no melting point but do have a Glass Transition Point ($T_g$). Below it, the material behaves like a solid, with a glass-like fracture surface when fractured. Typical amorphous materials include several types of elastomer (rubber) like natural rubber (NR), with a $T_g$ of around ...


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Sputtering deposition is not normally preformed at ultra high vacuum pressures, thus the films tend to be polycrystalline while e-beam evaporated metal films could be done at much lower pressures resulting in a more uniform film, even single crystalline depending on other conditions like the substrate, lattice mismatch and so on. This is just one difference. ...


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Mathematically the reason why we typically only consider the first Brillouin zone is that it is the only area necessary for completely defining both the wavefunction and energy of the system. From Bloch's theorem it follows that $$ \psi_{n,\vec{k} + \vec{K}}(\vec{r}) = \psi_{n,\vec{k}}(\vec{r}) $$ and $$ E_{n,\vec{k} + \vec{K}} = E_{n,\vec{k}} $$ Thus, ...


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opinion based question, so it may be closed. The author Vincent (family name) has a very good introduction to group theory for molecules. I like this book as it has questions for you to answer as you go along so you really learn it as you read. If you are interested in solid state then you will have to go further to space groups with another text - this ...


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deriving the desired equation for the density of states we are reminded that the surface which we are integrating over is a surface of constant energy in the reciprocal space which is denoted by $S(e)$, where $e$ is the 3-dimensional dispersion relation we know that the number of states between the surfaces $S(e)$ and $S(e + de)$ is given by the integral. ...


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$b_1, b_2$ and $b_3$ are reciprocal primitive vectors. $G$ is the set of all vectors that are in the reciprocal lattice and, as you said, is given by the linear combination of the reciprocal primitive vectors. The set of points, $G$, just define the lattice vectors or the locations of the origin of each Brillouin zone. Now we need to look within each ...


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deriving the desired expression for $\boldsymbol k$ and figuring out what $\boldsymbol k$ really is we will begin with the definition of bloch's theorem. $$\psi_{\boldsymbol k}(\boldsymbol r + \boldsymbol t_n) = e^{i \boldsymbol k \cdot \boldsymbol t_n} \psi_{\boldsymbol k} (\boldsymbol r)$$ here $\boldsymbol k$ is a wave vector defined in the primitive ...


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It's a conventions issue. When you define a system (a cell, a car tire, an atom), there is the system, and its surroundings. If you're interested in describing the system, you will discuss the properties of the system relative to the surroundings. Saying any given description of the system is positive is saying that the system has higher pressure / greater ...


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To construct a crystal you need a lattice and a basis. The lattice represents the translational symmetry of the system. Namely, graphene has a hexagonal lattice, meaning the two lattice vectors are 60 degress apart. Since the brillouin zone is constructed by inverting the lattice vectors, the brillouin zone is shaped based upon the lattice, but not the ...


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For a transition like electron-hole recombination, its probability is linear with time with some characteristic time scale that depends on the system: the probability of having a transition between $0$ and $dt$ is $\frac{dt}{\tau}$. So there is a non-vanishing transition probability at arbitrarily small times. In momentum space, the evolution of a ...


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The "only" thing you need to do is to establish a mapping. You have a basis function at $$ \vec{R} = a\vec{X} + b\vec{Y}$$ with index i. In other words, your basis is $\phi_{abi}$. Since Matlab only understands (well) vectors and matrices, you need to map this to a continuous index n. For example, a square with sides $N_a$, and $N_b$ and $N_i$ basis ...


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$|\psi(+)|^2$ has its peaks at the ions, while $|\psi(-)|^2$ has its maxima between the ions. Since those represent the chance of finding the electron at a certain place we expect $|\psi(+)|^2$ to have a lower potential energy as $|\psi(-)|^2$, lower energy because the coulomb potential between the electron and the ions is attractive. What he is calculating ...


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If you carry your example forward, then the total power emitted by a 1 m long copper wire that is carrying 1 A is just the sum of the power emitted by each electron. How many charge carriers are there in a 1 m copper wire with a resistance of 100 Ohm (since you had a field of 100 V/m)? Resistivity of copper is $1.7\cdot 10^{-8}~ \Omega\cdot m$, so a 1 meter ...


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For an infinite chain with periodic boundary conditions, you have $n$ translation symmetry. This means you can look for a basis of solutions whose $n$ dependence is $e^{ikn}$. The boundary conditions constrain you to $k=m\frac{2\pi}N$ with $m=0$..$N-1$


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If you're talking about a solid, the integral is bounded to the first Brillouin zone. This gives you a finite number of states. Otherwise, you have for each energy an infinite set of possible $k_z$ so the DOS diverges.


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Because crystal momentum and the vector potential appear together, introducing the vector potential changes the conserved quantity from just crystal momentum to crystal momentum + electromagnetic momentum.


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It might help to just look at the vacuum level, and think of how the structure reaches equilibrium. In your first picture, electrons will start to flow from right to left, towards the lower fermi level. This will charge the left side negatively, and leave positive charge on the right side. Therefore the vacuum level will curve on both sides, with a U ...


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Regarding your question on sandwiching GaAs between two AlGaAs barriers: If you do this for a narrow quantum well (like you sketched above), the electron wavefunction protrudes into the barrier quite a bit. As the barrier material is a ternary alloy, the electrons are exposed to alloy scattering. This is simply due to the fact that Ga and Al atoms are ...


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The first case you mention where you just sandwich GaAs inside AlGaAs is worked out in page 66 of Semiconductor Nanostructures. By adding a doping layer you control the carrier density and with that the resistivity. Like you say, it's necessary to draw the electrons away from the doping layer using attraction to the smaller-gapped GaAs. This creates bound ...



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