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The short answer is, in the thermodynamic limit, k could always be defined as a good quantum number for bulk states despite of different boundary conditions. This can be shown by considering a finite potential of several equally distributed wells and infinite high walls as boundary. The solution of wave function is then planewave-like and has some phase ...

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The interaction is needed for the validity of the mean field approximation, which assumes that the gap remains intact. It is clear that without interaction the argument fails for reasons you have mentioned (no gap). It is unfortunately not possible to derive analytically for which interaction the gap will survive, so the argument only gives a possible way ...

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In transition-edge sensors (TES), the superconducting detector is held very close to its critical temperature, so that it's resistance is a very steep function of the temperature. The figure below, from a review paper on TES, demonstrates the resistance vs. temperature in the right panel. In operation, the sensor temperature will be kept somewhere on the ...

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See the following link for a beautiful explanation: https://stanford.edu/~jeffjar/statmech/lec4.html#solving

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Here is a theoretical understanding of polarization and polarizability using functionals. Let us denote the external electric field generated by free charges (which are all outside of the dielectric medium) as $\textbf{E}_0$ . Here we treat this field and the polarization field on equal footing. The electrostatic potential energy is denoted as $U[\textbf{E}... 2 Any Taylor series expansion in$q$will work, but it's probably easier to write \begin{equation} \ln(1\pm x) = \pm x -\frac{1}{2}x^2 \pm \frac{1}{3}x^3 -... \end{equation} so that \begin{equation} \ln \left (\frac{1+x}{1-x}\right ) = \ln(1+x)-\ln(1-x) = 2x+\frac{2}{3}x^3 +... \end{equation} and then write \begin{equation} \ln \left (\frac{i\omega_m+v_F q}{i\... 2 [Note: There are already some nice answers here, but they focus on the case of non-interacting fermions. While that is certainly an interesting case, it is quite a small fraction of the landscape of topological phases of matter, and some of its key features (in particular, the topological space being a vector bundle defined by the energy eigenstates) do not ... 1 I can see two big differences between a hypothetical perfect conductor and a superconductor: 1. A "perfect" conductor will keep its magnetic flux condition. A superconductor will always expel the magnetic flux (Meissner effect). What does this mean? Let's say that you have a regular conductor in an external magnetic field, which, of course, is also ... 1 Yes, it looks like$\delta^{ab}$is effectively the Kronecker delta. Your expression of$G^{ab}_p$cannot be correct because it is of the form$[(A\delta^{ab}+B)^{-1}]^{ab}$. In fact the expression given for$G^{ab}_p$is just$((p^2+r)\delta^{ab}-\Sigma_p \delta^{ab})^{-1}. 1 Using teh resolution of identity: $$1=\sum_m|m\rangle\langle m|,$$ we can write $$\langle n|H'(E_n^0-H^0)^{-1}H'|n\rangle = \sum_{m,m'}\langle n|H'|m\rangle\langle m|(E_n^0-H^0)^{-1}|m'\rangle\langle m'|H'|n\rangle = \sum_{m,m'}\frac{\langle n|H'|m\rangle\langle m'|H'|n\rangle}{\langle m|E_n^0-H^0|m'\rangle}=\\ \sum_{m,m'}\frac{\langle n|H'|m\rangle\langle ... 6 The misunderstanding many have is that topology just the study of topological spaces. It is really also about continuous functions between two topological spaces. If one has an infinite lattice model, where the Hilbert spaces is more-or-less \ell^2(\mathbb{Z}^d), and if the Hamiltonian is periodic (a huge limitation) then one has momentum space that is ... 8 Based on OP's elaboration in the comments on their background, I will try to make this answer pedagogical and self-contained, erring ont he side of simplicity in the beginning; however, since OP mentioned some familiarity with topological manifolds and vector bundles, the concluding section will become more sophisticated. Before beginning, however, I'll do ... 3 I suspect the authors are referring to the general principle that "Quantum physics reduces to classical physics when everything commutes". The usage of "noncommutative" here is fairly loose, after all it makes no sense to say "\sigma^z doesn't commute" without specifying what it fails to commute with. I believe the mental ... 6 We want to calculate the expectation value \langle H - E\rangle on the excited state we are constructing. This is going to be much easier if the state we construct is correctly normalized. We could explicitly normalize the state but calculating the normalization constant for many-body states can be painful. Applying a unitary operator to an already ... 3 In a general space-time dimension, helicity is the eigenvalue of a rotation that leaves the direction of motion unchanged. The group of such rotations is called the “little group”. For a massless particle, the little group plays an especially important role, since it is not possible to bring this particle to a rest frame where all rotations have an ... 0 Edit: this was written before J. Murray's excellent answer Saying solid state physics isn't my expertise is an understatement, so I'm not sure my answer is correct, but I'll do my best: According to my understanding, the topological space involved is the electron band structure (see for example this gif from Wikipedia which shows the structure for some ... 1 I think you did not correctly understand what density of states (DOS) mean. DOS is a probability density function (PDF). As Andrew pointed out, it takes energy as input and returns the number of states for a given energy. You cannot discretize E as they are not eigenvalues of any observable. It is the input parameter, and discretizing it simply does not ... 1 From equation (22.1) \vec{r}(\vec{R}) = \vec{R} + \vec{u}(\vec{R}) so that r_\mu = R_\mu + u_\mu and hence,$$\tag{1} \frac{\partial\phi}{\partial u_\mu} = \frac{\partial\phi}{\partial r_\mu} \frac{\partial r_\mu}{\partial u_\mu} = \frac{\partial\phi}{\partial r_\mu} $$so that the two definitions of \phi_{\mu\nu} are identical. \vec{R} being a ... 1 I am not sure is this is all you need, but the answer to your question is yes, the operators are the same, just expressed in real vs k-space:$$ c({\bf{r}})=\frac{1}{\sqrt{V}}\sum_k c_k e^{-i \bf{k r}},\quad c({\bf k})=\frac{1}{\sqrt{2\pi}}\int c(r) e^{i \bf{k r}}d{\mathbf r} $$(the sign convention was picked up from (2.24)) As of the "formal correct ... 1 In particular, for the first equation we need small |q|, |p_1|≃p_F, and |p_2|≃p_F. Don't we just need for the second equation small |q_1|, small |q_2|, and |p|≃p_F? This is correct. But notice, while small |q| requires fine-tuning in all 3 directions in k-space, the condition |p_1|\simeq p_f requires fine-tuning only in one direction - ... 0 The only change you need to make is to write your last equation as \begin{equation} \phi_{\mu \nu}(R-R') = \left . \frac{\partial^2 \phi(R-R'+u(R)-u(R'))} {\partial u_\mu(R) \partial u_\nu(R')}\right |_{u=0} \end{equation} which you can see is the contributing term from U. Note that U is the full potential, not the potential at the equilibrium points in ... 3 I think you need to be careful about the terminology. None of the papers or quotes you posted uses the word "qubit". "Qubit" is a term you often find in Quantum Information Theory literature and refers to any two-state quantum system. These could be bosonic, fermionic or even (in the case of topological quantum computation) anyonic in ... 13 Qubits are neither fermionic nor bosonic, but you can use either fermionic or bosonic degrees of freedom to store qubits. The notion of a qubit has nothing to do with exchange symmetry nor with the commutators or anticommutators that apply to raising and lowering operators for quantum fields. That's why a qubit, qua qubit, is neither fermionic nor bosonic. ... 6 Qubits in the quantum-circuit model are assumed to be distinguishable, so it makes no difference whether they're implemented using bosons or fermions. "A qubit is a spin-1/2 system" is true only in the sense that an abstract qubit is isomorphic to an abstract spin-1/2 system, such as an electron confined so that it has no spatial degrees of freedom.... 1 I evaluated the partial derivative, both by hand and in Mathematica. I got the same result as you got. I don't possess any knowledge regarding condensed matter but I can assure you that your calculations are correct. 0 In your example the forces in question that have to be in balance are the rigidity of the box walls and the ever increasing pressure of the gas (assuming a totally ideal gas that would never condense). The reason a black hole eventually forms is that matter is in so small a space that the local geometry warps enough so that "outward" is no longer ... 2 Astronomical objects made of gas/plasma in which outward pressure forces are in balance with inward gravitational attraction are called stars. 0 Reminder: semiconductors are just insulators with a narrow bandgap (that is the bandgap typiclaly comparable to the room temperature). Thus, in an intrinsic semiconductor at zero temperature all the electrons are in the valence band and none of them is excited above the gap to teh conduction band, thus an intrinsic semiconductor at zero temperature has no ... 1 At 0K, all valence electrons are indeed bonded in covalent bonds and thus are localized. The semiconductor behaves like an insulator (in the absence of crystallographic defects). At temperatures above 0K, the electrons can break away from their bonds due to the thermal energy supplied to them. They then can move around the lattice. If a voltage is applied, ... 3 This is because of a global U(1) phase symmetry$$\phi^{\prime}~=~e^{i\theta}\phi, \qquad \bar{\phi}^{\prime}~=~e^{-i\theta}\bar{\phi}, $$for the complex fields. Only correlators$$\langle F[\phi,\bar{\phi}] \rangle_0~=~\langle F[\phi^{\prime},\bar{\phi}^{\prime}] \rangle_0~=~e^{iq\theta}\langle F[\phi,\bar{\phi}] \rangle_0$$with no net U(1) charge q ... 2 Berry phase is equal to the surface integration of Berry Curvature. In the first case, Berry curvatures are located in the tube, so you move your particle around the tube will collect all the Berry phase and it get quantization result. By the way - if you move the particle into the tube, you also get non-quantized Berry phase. In the spin model, the source ... 1 Superfluidity is a macroscopic quantum effect, and for the quantum nature of the single particles to emerge and dominate, their "effective range" has to spatially extend and become comparable to the interparticle spacing. Only in this way will the particles be "aware" of each other, and the constraints imposed by quantum statistics be ... 3 One can always think of an operator as a matrix and refer to the corresponding matrix methods. Since the operator in question is non-hermitian, its right and left eigenvectors are generally not the same, and the diagonalizing transformation is generally not a unitary one, i.e., one cannot restrict it to matrices satisfying S^\dagger=S. The more general ... 0 By using distributional derivatives (as suggested by @Jakob in comments), I was able to solve above integrals. According to Wikipedia, we have identity:$$ \int [\nabla_x \delta(x)] \phi(x) dx = -\int \delta(x) [\nabla_x \phi (x)]dx $$So, I get:$$ \nabla\cdot J = \frac{-i\hbar}{2m}\bigg(+\int dr \delta(s-r) \nabla_r (c_s^+ (\nabla_rc_r)) - \int dr \delta(... 0 I think that it's probably because the spatial order due to the vortex lattice emerges at a scale (i.e. the typical distance between vortices) that is greater than the scale of ODLRO. At this scale the system is practically in a hydrodynamic regime, in the sense that QM is not needed to describe it. In fact, the Tkachenko waves (that are basically elastic ... 0 I have recently noticed the reference given by @akhmeteli, which greatly helps me understand MFA. I shall do some math here (with respect to the original problem): $$\newcommand{\expect}{\langle #1 \rangle} \newcommand{\Tr}{\text{Tr}} H = \sum \epsilon_k c^\dagger_k c_k + \sum_{k k'} V_{kk^{\prime}}c_{k\uparrow}^{\dagger}c_{-k\downarrow}^{\dagger}c_{-k^{\... 0 Only you are not doing the math right in your Edit 1:$$ \begin{aligned} \nabla\cdot \vec J =& \frac{-i\hbar e}{2m}\int d\vec r \{\psi_r^+ (\nabla_r[\psi_r^+,\psi_{r'}])\nabla \psi_r +\nabla \psi_r^+ (\nabla[\psi_{r},\psi_{r'}^+])\psi_r\}\\ =& \frac{-i\hbar e}{2m}\int d\vec r \{ \color{red}{ - \psi_r^+ \nabla_r\delta(r-r')\nabla \psi_r} +\nabla \... 0 I'm thinking of exactly the same question two weeks ago. And I was thinking this symmetry seemed to be a U(1) symmetry or something, because the system is invariant under continuous variation of that phase theta, just as you said. Later I find this is actually the chiral symmetry of the left-moving and right-moving fields, and if you write the order ... 0 I figured this out, it's just a change of basis. $$\langle f_\mu^\dagger f_\nu \rangle = \langle\, \vert c_{a_N}\dotsc_{a_1}\, f_{\mu}^\dagger f_\nu \, c_{a_1}^\dagger\dots c_{a_N}^{\dagger}\vert\, \rangle\\ = \sum_{i,j} \langle \mu\vert a_i\rangle\langle a_j\vert\nu\rangle \langle\,\vert c_{a_N}\dots c_{a_1}\, c_{a_i}^{\dagger} c_{a_j}\, c_{... 3 These are called Discrete Fourier Transforms. We define discrete Fourier transform g_p (p=0, 1, 2, \dots, N-1) of a function f is$$ g_p = \frac{1}{\sqrt{N}} \sum_{k=0}^{N-1} f_k \, e^{2\pi ikp/N} $$Inverse Fourier transform is defined in similar way,$$ f_j = \frac{1}{\sqrt{N}} \sum_{p=0}^{N-1} g_p \, e^{-2\pi ijp/N}\$ Refer Arfken, 7th Edition, 20.6, ...

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I see this post is old but I hope I can help those who read it in the future: In the string model that (I assume) is being explained, one is only interested in the vertical component of the spring's force (it is easy to see that the horizontal one must be, at least on average, always zero), which is obtained by multiplying the net force times the sine of the ...

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