# Tag Info

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I'm afraid you heard wrong: there is no exact solution for the Bose-Hubbard model (BHM) in 1D. There are approximate solutions in the superfluid ($J\gg U$) and Mott insulator ($J\ll U$) limits, but neither works in the intermediate regime $J\sim U$. For $D>1$, mean-field theory is often used to interpolate between these limits, but is obviously an ...

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Someone might give a more precise answer in terms of group theory but I'll give it a go anyway ; feel free to edit my post. Instead of considering the case of an odd number of fermions, one can consider just a single spin $1/2$ - fermion to discuss $2 \pi$ rotations. Spin $1/2$ is a representation of dimension 2 of the rotation group, which is called the ...

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About the limit: $\frac{\sin[\frac{\pi}{d}(1/N+1)x]}{\sin[\frac{\pi x}{dN}]}= \sin[\frac{\pi}{d}(1/N+1)x]\times \frac{\frac{\pi x}{dN}}{\sin[\frac{\pi x}{dN}]}\times \frac{dN}{\pi x}\approx \sin[\frac{\pi}{d}x] \times 1 \times \frac{Nd}{\pi x}$ in the last step I used $\lim_{x\to 0}\frac{\sin x}{x}=1$ and $1/N+1\approx 1$. after rearranging the terms it ...

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I am a little unsure what you are asking. You seem to think that two Majorana fermions can interact in such a way that one gets to Dirac fermions. The Majorana fermion is its own antiparticle. The charge conjugation of $\psi$ is $C\psi~=~i\psi^*$. The appearance of $\psi$ and $C\psi$ in the Lagrangian means that the Majorana field must be electrically ...

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Charge density waves and Wigner crystals are two ways of understanding the same broken symmetry. In the Wigner crystal picture we imagine the electrons sitting on lattice sites; the electronic charge density thus has broken translation and rotation symmetry. In the CDW picture we imagine an essentially uniform distribution of electrons that develops a ...

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Interactions have a crucial role in the process of making a BEC. Bose-Einstein Condensation without interactions/collisions In theory, it is possible to produce a BEC starting from an ideal gas (i.e. a gas of non-interacting particles) of bosons and by cooling it down. Note that most of time, the way the gas is cooled down is never specified, the ...

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So spin 0 particles can pass freely through fermions and other particles if there is lack of EM repulsion between them? There is never lack of EM repulsion because baryonic matter is made of charged particles. The superfluid will stay into the container.

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First of all, I have not directly worked on BEC or laser cooling per se, but what I am writing is my understanding after discussing this subject with a person who is directly involved in this activity. Hence if anything is wrong or inconsistent please let me know. As rightly said by @Lagrangian that it is not required that the particles should be at same co-...

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The general misconception is that the bosons in BEC and superfluids are in the same quantum state including the same spatial coordinates. This would result in stacking of each particles' wave function and unlimited reduction in volume of such a substance. The truth is that the particles in BEC or superfluids do not necessarily crowd, but have the capability ...

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This question seems to have been posted for quite a while... I don't know if you ve already found the answer.. Anyway, there are at least two methods to derive it (as far as I know). One popular method is the equation of motion'' method which formulates the contour Dyson equation first and then uses Langreth rules to obtain similar recursive equations for ...

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When you discuss polarization of matter due to the propagation of electromagnetic waves you usually consider low-energy photons, meaning photons of long wavelenghts. Given that the inter-lattice spacing $a$ is a few angströms, if you consider for instance visible light of $\lambda \sim$ a few $100$ nm, you have $\lambda >> a$. As the wavevector goes ...

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Any free fermion Hamiltonian where the Fermi energy is chosen such that it is exactly at the bottom (or top) of the band (in the case of a single band) is of this form: It is obviously gapless, and its ground state is the vacuum, i.e., short-range correlated (or rather uncorrelated). One such example would be the 1D XX model, $$H=-\tfrac12\sum (\sigma_x^i\... 2 The canonical example for MPS (in fact, the first MPS ever) is the AKLT model (http://journals.aps.org/prl/abstract/10.1103/PhysRevLett.59.799, https://projecteuclid.org/euclid.cmp/1104161001). The 2nd reference also discusses the 2D (=PEPS) version of the state. Another example of an exact MPS/PEPS model are (nearest-neighbor) RVB states (https://arxiv.org/... 8 Yes. Bose-Einstein condensation was experimentally achieved in systems of: Rubidium atoms (first experimental realization, 2001 Nobel prize) Potassium atoms Cesium atoms Lithium atoms Sodium atoms Exciton polaritons Photons Phonons (?) If you are using atoms, they must behave like bosons (see also here). 0 Let me work on this part of the matrix equation$$H=\sum_{\bf{k}} \begin{pmatrix}a_{\bf{k}}^\dagger & b_{\bf{k}}\end{pmatrix} \begin{pmatrix}1 &\gamma_{\bf{k}}\\\gamma_{\bf{k}} & 1\end{pmatrix} \begin{pmatrix}a_{\bf{k}} \\ b_{\bf{k}}^\dagger\end{pmatrix}=\sum_{\bf{k}} \begin{pmatrix}\alpha_{\bf{k}}^\dagger & \beta_{\bf{k}}\end{pmatrix}\...

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No, it's unitary transformation, but only when you consider the Hamiltonian's electron & hole together.

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Maybe you would like to trace over other base. For example, your interest eigenstate is $N-$ body state $|\psi\rangle$, then you have density matrix $\rho=|\psi\rangle\langle\psi|$, and to derive a single site's information you only have to trace over other state, i.e., $\rho_j=\prod_{i\neq j}\sum_{\sigma_i}\langle \sigma_i|\rho|\sigma_i\rangle$, and get a $... 0 Fourier transformation of the creation/annihilation operators is no different than any other Fourier transform; the spin index just comes along for the ride. For instance, $$c^\dagger_{i \alpha} = \frac{1}{\sqrt{2\pi}} \int_\text{BZ} d^2k\,c^\dagger_{\mathbf{k} \alpha}e^{i\mathbf{k}\cdot\mathbf{x}_i}$$ With that operator in hand, your second term (ignoring ... 0 The short answer would be the empirical Matthiessen's rule: the total resistivity of a crystalline metallic specimen is the sum of the resistivity due to thermal agitation of the metal ions of the lattice and the resistivity due to the presence of imperfections in the crystal (scattering). There are of course deviations from that rule: it assumes that ... 2 As you pointed out, the phonon-mediated BCS-type superconductors exhibit a gap$\Delta_0$which is isotropic in$k$-space, we call it an s-wave gap. As @leongz pointed out, it comes from the fact that the electron-phonon interaction used in the BCS model does not depend a momentum ; inserting it into the gap equation gives a s-wave gap. The precise ... 0 There is no spontaneous symmetry breaking at the water-gas phase transition, because it's a first-order transition and symmetry breaking typically happens at second-order phase transitions. Physicists usually think of a phase as a region of parameter space that's connected by paths that don't cross any phase transitions, so a physicist would indeed say that ... 0 One way to see this is using the last equality of Eq.(48), i.e.$K_F^{(1)}[t_0](t) = K_\mathrm{eff}^{(1)}(t) - K_\mathrm{eff}^{(1)}(t_0)$and then applying Eq.(47). A bit confusing given that Eq.(44) appears in the previous subsection. Also, note that the weight function$f(t-t') = (1+2\frac{t-t'}{T})$appearing in the integrand of Eq.(47) [and thus ... 0 There is an interesting experiment to construct a Black Hole in ultracold atom systems. The idea is to artificially create a velocity speed field greater than the superfluid speed, which indicates somehow a particle cannot escape the boundary. Ref: Phys. Rev. Lett. 85, 4643 arXiv:0803.0669 0 Following the Wikipedia article on the sign function,$d\mathrm{sgn}(x)/dx=2\delta(x)$. So you could use that to replace the last line of your derivation. You could also switch the order of the arguments at the end to eliminate the minus sign. 1 A topological invariants is a continuous map$n$: $$\mathfrak{H}\ni H\mapsto n\left(H\right)\in S$$ where$H$is the Hamiltonian of your system and$S$is some topological space.$\mathfrak{H}$is the space of all admissible Hamiltonians. Unfortunately the exact definition of$\mathfrak{H}$is still a matter of current research. To keep things short, we ... 0 Fermi level is an energy in which the electron distribution probability is 1/2. Due to Pauli Exclusion, the electrons will pile up so the Fermi level for electrons will move up. Chemical potential is the same with Fermi level. To calculate the value, you have to integral the Fermi-Dirac distribution function times the density of states. For a rough ... 2 I have asked a professor about this and he gave me the answer. After replacing$\mathbf{k}$by$-i\nabla$in$H(h\mathbf{k})$, we are actually getting a new Hamiltonian that acts on envelop of wave functions. To make this answer relatively complete, I will briefly introduce the main steps focusing on only one band. Suppose the band we are interested in ... 0 As I know there is a notion of "bulk boundary correspondence" if you terminate the infinite slab of graphene along an axis in an arbitarily direction then you get an edge and surface states. It will support chiral movement of electrons provided that its bulk(I.e unterminated graphene) be in a topological phase characterized by some topological quantity. This ... 0 You should use parity analysis in wien2k by group theoretical considerations. You must first converge your material then run "x irrep -so" for spin orbit inclusion or without it. You should just notice to use case.vector file from the k-path of the band to have a comparison between case.band.agr and case.outputir[so] or case.irrep. don't forget to use ... 5 Nothing in the laws of thermodynamics forbids multiple liquid phases for a single substance. The only limit is the simultaneous coexistence of at most three phases (at triple points). Water has a solid-liquid-gas triple point and several soid-solid-liquid and solid-solid-solid triple points; see the phase diagram of water and ice. In addition, although not ... 1 There is actually only one disordered phase - from a physicist's perspective, the liquid and the gas are actually the same phase because one can continuously vary the external parameters (temperature and pressure, in this case) to get from the liquid to the gas without passing through any phase transition, because the phase transition line terminates within ... 1 Technically speaking all solids have crystalline structure. Anything that is truly amorphous is known as supercooled liquid (such as glass). Depending on how well crystals are oriented they can be divided into two categories, as pointed out by @lemon, single crystals and polycrystals. Single crystals have nearly perfect orientation of sizes few mm long. ... 2 Solids are rigid and resist a change in volume. There are three general types of solid: As you can see from the picture, a crystal is a solid for which the atoms/molecules are highly ordered to form a periodic lattice. Examples of (poly)crystalline solids include diamond and table salt. Examples of non-crystalline solids include glass and amorphous ... 2 When you take the limit$q \rightarrow 0$, your formula becomes $$\chi( 0,0) \sim - \sum\limits_{\bf{k}} \frac{d f(\varepsilon_k)}{d \varepsilon_k}$$ where$f$is the occupancy number of the electronic states. This susceptibility represents the response of the system to some external perturbation. Usually, this perturbation tends to displace electrons to ... 14 The most immediate answer would seem to be that a great variety of different crystal phases can exist because their long-range order makes it possible to classify them based on the different symmetries of their lattice structure. Since the liquid (or amorphous solid) phase only has short-range order and the gaseous phase doesn't even have that, it seems ... 1 The susceptibility quantifies the response of a material to an external electric field due to the redistribution of electronic charge within the material. The formula you give for the charge in a particular electronic band (in practice the total susceptibility comes from the contributions from all bands). It is known as the Lindhard model and is derived from ... 4 A eigenstate of a crystal hamiltonian can be written as a Bloch function in space representation $$\psi(\mathbf{r}) = e^{i\mathbf{k}\mathbf{r}} u_\mathbf{k}(\mathbf{r})$$$u$is periodic with respect to the unit cell. The momentum is now given by$\$ \langle \psi|\hat{\mathbf{p}} |\psi\rangle = -i\hbar \int e^{-i\mathbf{k}\mathbf{r}}u_\mathbf{k}^*(\mathbf{...

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