# Tag Info

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One of the biggest failure of theoretical condensed matter and/or material sciences is that up to now, nobody has ever been able to predict what compounds will be a good superconductors. Of course, since we don't really understand High-Tc superconductivity, we cannot predict which ceramic will or will not be a nice superconductor. But even in the case of ...

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In principle yes, but the electrons will respond at around their natural frequency of oscillation. This is the plasma frequency and for most metals is around the frequency of visible light or about $10^{14}$ Hz. So the electrons will only be displaced for a few fractions of a picosecond. The analogy with sound is that the motion creates a sound wave that ...

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An atom in isolation offers a potential well, and electrons form bound states in the well. The energy of those bound states can be calculated exactly in the case of a single-electron (hydrogen-like) atoms or by variational computational methods for more complicated cases. Now when you put several atoms together in a tight and regular array, they offer a ...

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What are those qubit in essence? Are they some kind of ultimate thing that build up our world? Yes. In the string-net picture of elementary particles, the qubits are the ultimate things that build up our world. We live inside a quantum qubit world (ie a quantum information world) (see http://blog.sciencenet.cn/blog-1116346-736093.html ) Such an emergence ...

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$k$ is just a quantum number. $\hbar k$ gets its name "crystal impulse" from the fact, that the formula for a band structure without interaction (free electrons) coincides with the formula you get with the definition of classical impulse in terms of $k$, but it is NOT an actual impulse. For a free electron we have the energy dispersion: $$\epsilon(k) = ... 2 The dispersion relation gives you information regarding the relation between momentum of electrons, and energy of such electron. Heisenberg's uncertainty principle relates uncertainty in the position versus uncertainty in momentum, which is a very different issue. If you consider a single massive free particle, it also possesses a dispersion relation in the ... 2 There are two different kinds of symmetry breaking involved in your question. The first would be spontaneous symmetry breaking. In this case we are dealing with a theory that is invariant under a certain symmetry group, but its vacuum is not. The breaking of the symmetry corresponds to a specific choice of the vacuum, the freedom of choosing a vacuum results ... 2 I have to admit that I have no idea about the model you are working on, but the standard way to determine whether a gauge theory is confining or not is to calculate the vacuum expectation value expectation value of Wilson loops. The latter are gauge invariant operators that describe parallel transport around a closed loop in spacetime. If the vacuum ... 2 In a single free atom, electrons have well defined energy levels and are somewhat bound to atom. Consider the following quantum mechanical model of atom to get an idea about an isolated atom. When all this isolated atoms come together to form the crystal, the atoms do not have well defined energy levels. There will be molecular orbitals. When the atoms ... 2 I would say they are not entirely the same, but it depends on the context. First the definitions: the Wigner transform of an operator \hat{A} is defined as$$\tilde{W}\left[\hat{A}\right]=\int dz\left[e^{\mathbf{i}pz/\hbar}\left\langle x-z/2\right|\hat{A}\left|x+z/2\right\rangle \right]$$and this is a strange function. You see that on the left, the ... 1 This is an eigenvalue problem. Let's assume your Bogoliubov transformation is of the form: (a_k,b_k)^T=X(c_k,d_k)^T. What this transformation do is let your Hamiltonian become: H_k=w_1c_k^\dagger c_k+w_2 d_k^\dagger d_k, with the anti-commute relation holds for new field operators c_k and d_k. Now you can check that X is just the matrix where its ... 1 Hamiltonian is already diagonalized by momentum. You need to define new Bose-operators c_k = u_k a_k + v_k b_k \\ d_k = w_k a_k+x_k b_k  This is general form, with some complex constants u_k, v_k, w_k, x_k for each k independently. There are also c^+_k and d^+_k, conjugated with previous one. Now you need c_k and d_k correspond to some ... 1 The intuition is that the valence electrons are so far away from their nucleus that when they combine to form metals, they feel the attraction of all the other nuclei as strongly as from theirs. In a more rigorous description, the orbitals for the valence electrons fully overlap with their neighbouring atoms, so their "play field" extends all over the ... 1 First, a somewhat minor point is that x = 0,0.01a,0.02a,...a,1.01a,....2a....100a actually gives a list of 10001 points, not 10000 points. I will assume that you actually meant to say x = 0,0.01a,...a,1.01a,....2a....99.99a. Second, you say that$$V(x)=\sum_{K}e^{iKx}V_{K}$$where K =\frac{2\pi n}a and n=0,1,2,3, but this gives a non-Hermitian ... 1 To simplify the problem, we may neglect the potential energy term V(r), as it is simply irrelevant to our derivation. So we write the Hamiltonian as$$H=\frac{1}{2}(-i\partial_x-A)^2.$$The ground state is given by minimization of the energy. As the Hamiltonian is a square of (-i\partial_x-A), so it is minimized when (-i\partial_x-A)=0. Which means on ... 1 First, I will set e=1 for simplicity. Let \psi_0 denote the wave function that satisfies the free Schrodinger equation: $$i \frac{\partial \psi_0}{\partial t} = -\frac{1}{2m}\mathbf{\nabla}^2 \psi_0 + V \psi_0 \tag{1}$$ Furthermore, let \psi be the wave function that obeys the Schrodinger equation for a non-vanishing vector ... 1 A simple method to judge the chirality (or in your words "orientation") of the Hamiltonian is to evaluate the following quantity$$f=\frac{i}{2}\mathrm{Tr}\frac{\partial h}{\partial{q_x}}\frac{\partial h}{\partial{q_y}}\frac{\partial h}{\partial{m}}. The sign of this quantity $f$ gives the chirality of the Hamiltonian. Example: Given the two ...

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The electron motion does feed back to the oscillator, but that is another diagram, known as the bubble diagram, in which you calculate the self-energy correction of the oscillator. That self-energy presumably contains imaginary part, which is then interpreted as the damping of the oscillator. You can either calculate the self-energy corrections ...

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I am assuming that those are bare Green's function in the diagram, and the dotted line is the harmonic oscillator $<a^\dagger a>$ ? 1) The diagram you drew does not "know" about the damping of the harmonic oscillator, (although it is closely related by the optical theorem and such to the diagrams that would calculate the damping, I suppose that is ...

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You misunderstood the classification I believe. Let's take an example. In class D and 1D, the classification tells you there are two possible vacua (you understood this apparently). This is the famous $\mathbb{Z}_{2}$ ensemble in the classification. Next the classification tells you also that: at the boundary between the two gapped vacua, a Majorana mode ...

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