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As far as I can understand, the pseudopotential should be used in connection with the nearly free electron (NFE) approximation. A preliminar warning is that, while NFE approximation works well with the alkali metals and reasonably well with most of the s-p bonded metallic systems, application to $d-$electrons metals somewhat transcends the reasonable limits ...

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One uses simple perturbation theory, but the parts of the Hamiltonian chosen as the ground state and the perturbation are different for $g\rightarrow 0$ and for $1/g\rightarrow 0$. In the context of Ising model and similar spin systems these approximations are referred to as high-temperature expansion and low-temperature expansion, one can check, e.g., this ...

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Yes. You need a field $\lambda(x)$ in @Alexandro Nikolaenko's answer above. One lambda for each site or point $x$. $$\int d[\lambda(x)]e^{\int dx \lambda(x)(b^\dagger(x) b(x)+...)}$$

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The $\theta$ variable is simply a representation of the ignorance that you have on determining a wave-function because what actually is physically observable is the state (or rather expectation values of the state). When you have a Hilbert space, your (normalized) wave-function is a vector $|\psi\rangle$. Multiplying by a phase factor $|\psi\rangle \to e^{i\... 0 The author here maybe concentrate on how to construct a tensor by a vector, so the average over direction means the average over direction of vector,namely $$\langle T_{ij}\rangle=\frac{1}{4\pi}\int T_{ij}d\Omega$$ where$d\Omega$is the solid angle The argument a symmetric traceless tensor will yield zero when averaged over all directions has some ... 1 Can the "well-known properties" be developed using free particles? (I think that the answer should be "yes".) Because the two equations are the same for free particles. For free particles: $$\epsilon\left(\vec{k}\right) = \frac{\hbar^2k^2}{2m}$$ and $$\psi\left(\vec{x}, t\right) = e^{i\left(\vec{k}\cdot\vec{x}-\omega t\right)}$$ (If it ... 1 The solution to the problem goes as follows: Observation one: The plane partition intersects the wall of any of the aforementioned containers along a Young tableaux, as the following figure illustrate: Let's write that tableaux as$\mu^{T}= (\mu^{T}_{1},...,\mu^{T}_{n})$with the sequence$\mu^{T}_{1}...,\mu^{T}_{n}$non-increasing (by definition). ... 1 If referring to electrical (not thermal) conductivities, then yes they are the same thing. Though the term Hall usually pops up when dealing with the Hall effect (classical or quantum) meaning the transverse conductivity is dominated by the effect of an external magnetic field. 3 Yes, water can form a "glassy" solid states, called amorphous ice. Depending on conditions and formation process it can have quite different properties. In particular it is not necessary for it to be compressed to very high density: it is actually possible to obtain amorphous ice that is less dense than liquid water, by vapour deposition (see Wiki).... 1 A thermal state describes a system at non-zero temperature. For "temperature" to have a thermodynamic meaning, as usual, you assume your system in in thermal equilibrium with a large reservoir at temperature$T$. So thermal states are the states of a system in thermal equilibrium with a reservoir. I find the best way to introduce people to the ... 2 Consider that using a k quantum number means that there is no definite position of your particle (it is a consequence of Heisenberg uncertainty principle). However, electrons in solids are described as wave packets narrowly centered around k in momentum distribution. Said so, in Aschroft, N. W., and N. D. Mermin. "Solid State Physics (Brooks-Cole, ... 2 The diagonalization of general quadratic bosonic Hamiltonians is discussed in J.H.P. Colpa, Physica A: Statistical Mechanics and its Applications 93, 327 (1978). It's a problem commonly encountered in spin-wave theories for noncollinear magnets with a higher number of sublattices, for example. The central results are as follows. If you can write your ... 2 For the sake of this explanation, let's concentrate on systems that have a spectral gap (not the most general scenario but it shall do). Let$P$be the Fermi projection of some topological material$H$such that its Fermi energy is placed inside of a spectral gap of$H$. We have the Riesz formula $$P = -\frac{1}{2\pi\mathrm{i}}\oint(H-zI)^{-1}\mathrm{d}z$$ ... 3 Topological means large scale. If you can detect a property by looking at a small patch of your space, then that property is not topological, by definition. Something is topological when you need to look at points separated by a finite distance. A local order parameter is an operator situated at a point. As such, it can only probe that point, and a small ... 7 OK, another proof according to the Remark by Emilio that made clear the interpretation of the question. We have to average the tensor (as a bilinear map) in the space of the rotations. The only way to do it, as Emilio wrote, is to take an integral with respect to the normalized (left-invariant) Haar measure of$SO(3)$: $$\langle T\rangle_{SO(3)}(u,v) := \... 10 The order parameter is a meso- or macroscopic quantity which is formed by adding together the tensor contributions from a bunch of different molecules: if each molecule \alpha has a tensor T^{(\alpha)}, then the total order parameter is the sum of the T^{(\alpha)} from all the molecules in the relevant sample of material, i.e.,$$ T = \sum_\alpha T^{(\... 5 Symmetry requirement is not necessary. Let us take an orthonormal basis$e_1,e_2,e_3$and consider all the unit vectors$n \in S^2$. $$\int_{S^2} T(n,n) d n = \sum_{i,j=1}^3T(e_i,e_j) \int_{S^2} n^i n^j dn\:.$$$dn$is the standard rotation-invariant measure on$S^2$with total value$4\pi$, i.e., referring to standard spherical coordinates $$dn = \sin \... 1 One possible approach is writing the sum of logs as a log of product and using the formulas for infinite products in Gradshtein and Ryzhik. 1 S=\sum_n\ln(-i\omega_n+\epsilon)=\sum_n\ln(i\omega_n-\epsilon)+C (usually this is an action and the constant is irrelevant. This transform is unneccessary just for convenient) =\mathrm{Res}\left\{\ln(z-\epsilon)g(z)\right\} when g(z) is what you've mentioned. Then the problem is to evaluate this integral. We could select the branch cut as (-\infty,\... 3 In this context, "intrinsic" means that the Hall conductivity comes from the Berry curvature. That is, it's a contribution intrinsic to the band structure. Disorder can produce "extrinsic" contributions to the anomalous Hall effect through so-called side-jump and skew-scattering mechanisms. If you're interested, you can see this answer of ... 1 For this kind of stuff you usually integrate by parts. First change your sum to an integral:$$\sum_n \log\left(-i\omega_n +\frac{k^2}{2m}+\mu\right) \Rightarrow \int \mathrm{d}\omega \, \log(f - \mathrm{i}\omega), $$where f here is k^2/2m+\mu which I am assuming are not functions of \omega. Then integrate by parts:$$\int \mathrm{d}\omega \, \... 0 While SCF one-particle wavefunctions for electron in a molecule are molecular orbitals, it is not always true that molecular orbitals are always solutions of SCF equations. An example of SCF molecular orbitals are solutions of Kohn-Sham Schrödinger-like equations for electronic orbitals in a molecule. An example of electronic orbitals which are not connected ... 0 These are sound modes. As the fluid moves backward and forward (motion of the order parameter$\langle \psi\rangle$along the potential minumum$\theta$direction) the density also changes (radial mtion in$\rho$direction -- the "Higgs" mode) so that moving forward corresponds to higher density and moving backwards to lower. Thus, in a superfluid,... 0 The "bound current" are not real currents but are "volume current density"$J_m$equivalent to the magnetization vector field$M$they represent$J_m=\nabla\times M$. So to answer you questions: \ "When we apply a voltage bias to this sample, do the bound currents affect in any way the flow of the transport current?" : they ... 3 So I assume you know in the BCS theory, fermions form Cooper Pairs (bound electrons) at very low temperatures. Since their paired state has a lower energy than the Fermi energy, they are bound. So the pair is now a composite boson with spin 0 or 1 instead of the fermions spin 1/2. This allows them to "condense" into the same quantum state which is ... 0 The off-diagonal terms can be reorganized into the form:$H_{od}=e^{i\vec{k}\cdot\vec{\delta_1}}+e^{i\vec{k}\cdot\vec{\delta_2}}+e^{i\vec{k}\cdot\vec{\delta_3}}$with$\vec{\delta_1}=(-a,0)$,$\vec{\delta_2}=(\frac{1}{2}a,\frac{\sqrt{3}}{2}a)$, and$\vec{\delta_3}=(\frac{1}{2}a,-\frac{\sqrt{3}}{2}a)$Now let$\vec{k}=\vec{K}+\vec{q}$, we can easily expand ... 0 Solution The solution is actually quite simple. Let us for simplicity consider two dimensions$(x,y)$. The electron and hole dispersions are \begin{equation} \mathbf{k}^2 = \frac{2m}{\hbar}\left(\mu + E\right)\, \mathrm{and}\, \mathbf{q}^2 = \frac{2m}{\hbar}\left(\mu - E\right) \end{equation} respectively. Therefore, the group velocities are \begin{equation} ... 1 The equilibrium properties at low-enough temperatures (for metals at room temperatures,$k_BT\ll E_F$where$E_F$is the Fermi energy) can be determined by knowing the properties of the ground state. At any temperature,$\mathrm{T}$, the equilibrium state of a system is dictated by the minimization of its Helmholtz free energy: $$F=U-T S\tag{1}$$ where $$U=\... 1 I'm going to give a few examples to add to the other answers. Spin liquids are low-temperature magnetic phases of matter which do not spontaneously break any symmetries. Generally, some type of frustration prevents the system from adopting any particular ground state, the origin of frustration could be competing energetic interactions or can be caused by ... 0 Apart from the answers already provided, one reason why ground states are studied is because they are quite robust. Leave a system, coupled to a bath (a system with much larger degrees of freedom), then the state of the system tends to be the one that minimises the overall energy, ie, the ground state. This is because the ground state cannot be de-excited ... 22 To add to Vadim's answer, the ground state is interesting because it tells us what the system will do at low temperature, where the quantum effects are usually strongest (which is why you're bothering with QM in the first place). OR it is interesting because the finite temperature behavior can be treated as a perturbation above the ground state. For example, ... -1 Actually, the main reason is that because HgTe is a zero-gap semiconductor. Of course, Graphene is also a zero-gap semiconductor but which is made out of carbon elements—a light element with a weak spin-orbit interaction. At this point, better to look at other materials with strong spin-orbit interactions, made from heavy elements. 14 Ground state contains information about most thermodynamic properties of the system at zero temperature. In fact, it can be thought of as a limiting case of the partition function at zero temperature. In many respects many physical systems never depart far from their ground state (although this notably not the case when dealing with phase transitions). ... 0 The connection between the electrostatic (Hartree in QM problems) potential and chemical potential (or node voltage in electronics) is one that always causes a great deal of confusion. The short answer is that, to get it right, one has to solve a nonlinear electrostatic screening problem, which has no helpful symmetries, and in cases like the one under ... 0 You’re going wrong by considering the optical conductivity to be real only. Conductivity is complex just as permittivity is! It can be related to epsilon through Ampere’s Law:$$\nabla\times H = J + \frac{dD}{dt}$$For time-harmonic fields, we get$$\nabla\times H = (\sigma - i \omega \epsilon)E.$$So the complex relative permittivity is related to the ... 1 partial answer: \pi is the vacuum polarization tensor since its kubo you want only one loop calculation assume we have feynman rules let this=e\hat{\mathbb{v}}_a now let this =G_{ab} finally the vacuum polarization tensor is now assume periodic imaginary time we have$$\pi_{ab}(\textbf{q},\omega)=\frac{ie^2}{\beta}\sum_{\textbf{k},\omega_n}\mathrm{... 0 First, 'ordered' does not mean there's more symmetry. Ordered phase means order parameter select a specific nonzero value. Think about Ising$Z_2$case. If total magnetization is zero, system is$Z_2$symmetric(i.e. if you flip all the spins your magnetization is still zero). However, if nonzero magnetization appears, then$Z_2$is broken. Therefore, ordered ... 2 If we move to Fourier space, we have $$\mathbf H(\mathbf r,t) = \int_{-\infty}^\infty \hat{\mathbf H}(\mathbf r,\omega) e^{-i\omega t} d\omega$$ $$\mathbf E(\mathbf r,t) = \int_{-\infty}^\infty \hat{\mathbf E}(\mathbf r,\omega) e^{-i\omega t} d\omega$$ $$\mathbf D(\mathbf r,t)= \int_{-\infty}^\infty \epsilon(\omega)\hat{\mathbf E}(\mathbf r,\omega) e^{-i\... 1 I see two possible problems in your consideration. You've investigated perturbations of ferromagnetic ground state. When spin variations \delta m are zeros, spins on three sublattices are the same:$$ S_i = (0, 0, 1),\quad \forall i.$$The Landau-Lifshitz equation is a nonlinear one. Effective field${\cal H}_{i,{\rm eff}}\$ depends on neighboring spins. ...

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I suggest checking the Mahan's book - the notation seems resembling his derivation. Generally, when the Kubo formula is written in terms of one-particle Green's functions, it means that we are either dealing with a non-interacting case or one of the vertices has been "dressed" by the interactions. Non-interacting case in this context is much more ...

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Whenever one classifies the topology of a system by means of a Chern number, which is the case for systems breaking time-reversal symmetry, a non-trivial Chern number indicates the given material realizes the quantum anomalous Hall effect. In other words, without underlying external magnetic fields, a material with a non-trivial Chern number, a so-called ...

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