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New answers tagged condensed-matter

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 ...

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In graphene, there are symmetry elements which connect sublattices. For instance, there are few 180 degrees rotations which exchange $A$ and $B$. In principle, applying this transform you should get a connection between $\Delta_A$ and $\Delta_B$. Naively, they should be equivalent. Probably, up to the sign, but the sign also depends on a particular choice of ...

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The Hamiltonian can be written as $\sum_k \psi^\dagger M \psi$ where $\psi=\begin{pmatrix}a_k \\ b_k\end{pmatrix}$ and $M=\begin{pmatrix}\omega_0 & \Omega f_k^* \\ \Omega f_k & \omega_0\end{pmatrix}$. We introduce a new set of operators $\phi=\begin{pmatrix}c_k \\ d_k\end{pmatrix}$, via $\psi=U \phi$ where $U$ is neccesarily a 2x2 matrix. This ...

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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 ...

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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 ...

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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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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 ...

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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 ...

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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 ...

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The solution of this problem is in principle very easy. Since $$z^{4} + 4\left(\mu + 1\right)z^{2} + 8\lambda\Delta z + 4\left(\mu^{2} + \Delta^{2} - V^{2}\right) = 0\text{,}$$ is a fourth order equation I can solved this equation by the hand or with Mathematica/Maple and get 4 solutions. The interesting case is that with two purely real and two complex ...

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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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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 ...

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The oxygen rich condition will normally lead to a change in stoichiometry up to a certain degree. Thus, you will expect to have interstitial oxygen atoms possibly compensated by cation vacancies.

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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 ...

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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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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 ...

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Symmetry is the key to distinguish charge density wave (CDW) and other charge modulations. CDW is not just a wavy pattern in the charge density as literally indicated by its name, it is an order that spontaneously breaks the translation symmetry. Note that the symmetry must be spontaneously broken but not be broken by hand. This means that the Hamiltonian ...

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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 ...

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The mapping between the quantum and the classical system is formal, but as you say, we can usually interpret a quantum phase transition of a $d$ dimensional quantum system (that is, a phase transition at zero temperature) as a (classical) phase transition in a $d+1$ dimensional classical system. The temperature of the quantum system maps to the size of the ...

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Losing translation invariance is one of the big problem when studying disordered system (for example). In that case, one usually averages over the disorder to render the system translation invariant (on average), with its own technical difficulties. The applications for transport properties are explained in the Mahan or most good text book. If the ...

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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: $$\epsilon(k) = ... 4 What are those qubit in essence? Are they some kind of ultimate thing that build up our world? Yes. In the string-net picture of elementary particles, the qubits are the ultimate things that build up our world. We live inside a quantum qubit world (ie a quantum information world) (see http://blog.sciencenet.cn/blog-1116346-736093.html ) Such an emergence ... 2 The dispersion relation gives you information regarding the relation between momentum of electrons, and energy of such electron. Heisenberg's uncertainty principle relates uncertainty in the position versus uncertainty in momentum, which is a very different issue. If you consider a single massive free particle, it also possesses a dispersion relation in the ... 0 Conventionally that what a unit cell is supposed to do, to capture as much symmetry found in the lattice. Otherwise, a primitive cell will be sufficient to describe a crystal. So yes it is possible to define a unit cell with all point group symmetries. 0 Can we say that if the mass gap of the anyon fractional exicitations \Delta are larger than the thermal energy E_H=k_b T (temperature at T),$$ \Delta >> E_H=k_b T  then we still can observe topological order at small but finite temperature.(?)

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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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I think this is in nature a problem of relativity theory. Since we(observers) take the metal as a reference frame to perform experimental measurements on electrons, then according to the equivalence principle of general relativity, the effects of 'suddenly move' is physically equivalent to gravity. But as we know, the electrons are very light, and usual ...

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hmm this is usually the standard notation with \begin{align*} \text{NN} &= \text{NN}\\ \text{NNN} &= \text{2NN}\\ &\ldots \end{align*} The only thing I could imagine is some kind of numbering of the nearest neighbors: If you have 5 nearest neighbors like in your example with 1NN, 2NN, 3NN, etc. you could mean the first atom which is NN, the ...

1

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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I think we can't compare a hollow pipe (filled with gas) to that of solid conductor. Gas molecules/atoms are free to move. In solids kernels aren't free to move, though conduction electrons are free to move. Drift velocity of free electrons will be very less (even during current flow). So, we can't compare motion of gaseous atom or electrons with that 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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I believe it will be disturbed (though it will probably be very small). I remember reading that in thermodynamics, all the formulas assume a fixed inertial reference frame. In fact this problem can be equivalently thought of as the application of an external time-dependent potential on the metal $V(t)$ (just like the uniform gravitational potential) so we ...

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