# Tag Info

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The question made me google "water nano structure" and sure enough: Water: Nanostructure and fluctuations S. D. Zakharov Recently a model of local organization of water was experimentally justified, in which tetrahedrally coordinated water clusters of 1–2 nanometers arise and disappear in liquid composed of H2O molecules with partially broken ...

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The quantity of electrons traveling across this interesting capacitor is necessarily hindered by the formula you used. Although there would be this consideration for electron emission, the potential through the dielectric would be chiefly concerned with the polarization and I would therefore consider that. Polarization is the net shift in dipole moments of ...

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Wood ash is most mineral. Opinions seem to differ about the composition, but it's things like calcium and potassium carbonates, phosphates and oxides. The minerals in wood are distributed throughout the wood, so as the organic material burns away you are left with a very fine network of aggregated particles of the minerals. Googling will find you lots of ...

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Consider the Ising antiferromagnet $$H=\sum Z_i Z_{i+1}$$ on a ring. It even has product ground states $|0101\cdots\rangle$ and $|1010\cdots\rangle$ (and only those product ground states). Thus, the lowest energy product states are not translational invariant.

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There are no easy to verify criteria which will work in general. One way to see it is by noting that for every classical StatMech model, we can define a PEPS with the same correlation functions (for which the tensors can be easily constructed from the StatMech model), see http://arxiv.org/abs/quant-ph/0601075. On the other hand, for StatMech models it is ...

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You can only distinguish the sublattices in this case because you've tagged them A,B. The process of inversion only exchanges identical carbon with carbon, leaving the crystal physically unchanged. If you gave me a crystal with one orientation and I then returned it to you without telling you whether or not it's been inverted, you'd have no way of knowing. ...

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Yes, pairing is possible from a repulsive interaction. The reason behind this is that pairing has to occur in a certain angular momentum channel : $l=0$ for s-wave superconductivity, $l=1$ for p-wave, and so on. To see this, you can expand the repulsive $k$-dependent interaction on Legendre polynomials. Check this review that deals with the Kohn-Luttinger ...

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It is indeed possible to change between these phases adiabatically. Since, as you noted, the ground state changes between being a superfluid and a Mott insulator, starting in the ground state and making an adiabatic change means that you track that change in state by definition. Note that this diagram is only formally true for the Grand Canonical ensemble, ...

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The way mobility depends on average scattering time of the carriers is given here: A simple model gives the approximate relation between scattering time (average time between scattering events) and mobility. It is assumed that after each scattering event, the carrier's motion is randomized, so it has zero average velocity. After that, it accelerates ...

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I have no experience with either CVD or PLD, but it was interesting to think about this question. In a humble attempt to build on Peter Diehr's answer, here is some theory (at a very heuristic and simplified level). The deposition of each new layer can be thought of as being governed at large scales by some mixture of Laplacian and Eden growth in two ...

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I have a lot of experience with CVD and sputtering, but limited experience in PLD; however, several of my colleagues did this all the time in our shared a laser lab. When they were attempting to reproduce a specific result all of the parameters had to be systematically varied, from the laser fluence to the substrate conditioning and temperature, and more. ...

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You cannot distinguish this, they are all symmetric. The hopping that is clockwise with respect to one cell is anticlockwise with respect to the neighboring cell. If you add this "with respect to", then orientation is meaningfull. Well, any graph can be oriented. Just arbitrarily chose for each edge a positive direction and call the other one negative. ...

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A compositional superlattice is a periodic layer structure of different materials. These typically have different bandgaps, effective masses, refractive indices etc. There are limitations on which materials can be stacked. They need to have the same crystal structure and lattice constant or at least negligible strain. The model system would be $GaAs$/$AlAs$ ...

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I think the first helpful fact to clarify is that there are two different kinds of topological phases: there are so-called Symmetry Protected Topological (SPT) Phases (displaying 'symmetry protected topological order') and there are (intrinsic) Topological Phases (displaying '(intrinsic) topological order'). As some quick examples: topological insulators and ...

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In a perfect lattice, the resistance would be zero. The reason for resistance is electrons being scattered by departures from a perfect lattice, for example lattice vibrations or imperfections in the lattice due to, for example, the presence of impurities or irregularities.

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Metals are in solid state. So conductors are basically solids. (Don't take it too much. There are no solids in quantum physics). A conductor is hence characterized by tightly packed atoms with plenty of availability of valence electrons that are ready to get off from what that holds them. Also a good conductor like silver, iron etc. have very good atomic ...

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The time reversal and chiral symmetry are special because they are antiunitary symmetries, in contrast to the other unitary symmetries like translation and rotation symmetries. Antiunitary symmetry operation involves complex conjugation of the wave function of the system, which is a non-trivial operation beyond unitary transforms. Unitary symmetries are ...

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It is not true that all superconductors are gapped. For example, d-wave superconductors in cuprates are gapless. The energy gap in the superconductor arises from the fact that breaking the Cooper pair requires finite energy. The low-lying quasi-particle excitations are all pair breaking excitations, so they are gapped from the ground state by the amount of ...

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You find particle hole symmetry (PHS) for example in superconductors, where you can take for the Hamiltonian the Bogoliubov-de Gennes (BdG) Hamiltonian as a mean-field approximation. This is the only experimental example I know. There are maybe other systems with particle hole symmetry that I don't know about, but I will use superconductors as an example in ...

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Collected Papers of L.D. Landau edited by D. Ter Haar The Discovery of Super Fluidity These above books might help you. Start doing your own research, this is how we study.

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The key is: Landau theory doesn't assume the order parameter is small. All it assumes is that the free energy is analytic in the order parameter. One then usually expands this free energy up to some order (which is possibly by definition of 'analytic'). It is key to realize that expanding a function in a variable to some order does not mean this variable has ...

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The spin orbit coupling can be derived from the nonrelativistic limit of the dirac equation and is given by $$H_{\text{s-p}} = \frac{\varepsilon_0}{2m_e^2c^2}\mathbf{\hat{s}}\cdot\left(\mathbf{E}\times\mathbf{\hat{p}} \right)$$ $\mathbf{E}$ is the total electric field acting on an electron, which consist of a microscopic electric field ...

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When we say that a lattice has a particular symmetry we mean that the lattice is mapped onto itself by the symmetry. So if I have a (2d) material which has inversion symmetry in the bulk and which has an atom at a point $(x,y)$ then inversion symmetry tells me that there is another, identical atom at $(-x, -y)$. At the surface, however, this is no longer ...

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What bothers me is that if (ferro)magnetism stems from the arrangement of electrons in various orbitals (the imbalance in total electron spin), why don't a lot more materials, including basic elements, have a Curie point? What makes the very few ferromagnetic materials so unique? Not all materials display magnetism or paramagnetism. Materials may ...

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The problem is that you are looking at the gap at the $\Gamma$ point ($k=0$) and not at the real gap at the Brillouin zone boundary. Here is the plot for the phonon dispersion relation of a diatomic linear chain. You can see that the real gap at the $k=\pi/a$ point is vanishing for $m_1=m_2$. This plot looks different as typical plots for a monoatomic ...

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To me, the Hubbard interaction per site is defined as $H_{int}=U n_{\uparrow} n_{\downarrow}$. (I suppressed the $i$, and also there is a factor of two because of your spin sum.) Then, the mean-field approximation is defined as $n_{\uparrow} n_{\downarrow}\to n_{\uparrow} \langle n_{\downarrow}\rangle+n_{\downarrow} \langle n_{\uparrow}\rangle-\langle ... 1 Within projected augmented wave (PAW) method it is incorrect to use the smooth pseudo wave function directly to obtain integrals. The same applies for ultra-soft pseudo potentials, but I will proceed discussing about PAW. The results can indeed be wrong. For example, for coin-metals, the d-band smooth wave functions$\tilde{\Psi}$can have norms of 0.2-0.4, ... 1 Index of refraction and the extinction coefficient are related to real and imaginary parts of dielectric function via a simple functions. Empirical real and imaginary parts of a dielectric function (or Lorentz oscillator fits to them) are usually used when simulating macroscopic Maxwell equations. So, what is missing (assuming you do not mean the trivial, ... 1 For a gas: Density Nuclear half-life of the components Viscosity Temperature of liquefaction Specific heat Speed of sound Rayleigh scattering coefficient 0 Take the fluorescence properties of a material for example. You cannot predict the time decay properties of the material subject to illumination just by knowing the refractive indexes. 0 I don't think this is true. Let us consider a 2 site hamiltonian:$H= - \mu_1^z \cdot \mu_2^z + \epsilon( \mu_1^x + \mu_2^x )$So the ground state manifold is nondegenerate and it spanned by vectors (-,+) and (+,-) for$\epsilon=0$for$\epsilon>0$but small, it is still nondegenerate and close a vector$\frac{1}{\sqrt{2}}((+,-) + (-,+))$. For ... 2 One way to estimate electron-phonon coupling is to take a look on the hot-electron relaxation rate, which can be more or less directly probed by pump-probe spectroscopy. The values for$\lambda$are around 0.2-0.5. There are few articles on this topic from C. Gadermaier. Here are the links http://dx.doi.org/10.1063/1.4726164 ... 2 You are probably aware that the underlying mechanism responsible for superconductivity in cuprates or Fe-based superconductors is still subject to intense debate and research. The usual BCS electron-phonon coupling (EPC) is too weak to account for superconductivity at temperatures of order$100$K, and produces a gap of s-wave symmetry which is incompatible ... 2 One can avoid the concept of symmetry breaking in this context, to avoid "non-conservation of the particle number". People have devised way to do that, see for example http://arxiv.org/pdf/cond-mat/0105058v1.pdf. However, all these approaches gives the same results than standard Bogoliubov-like methods in the thermodynamic limit. This is not too ... 2 The Landau criterion is not in itself a criterion for superfluidity, but a criterion for the breakdown of superfluidity. Indeed, if applied to insulators or ordinary fluids, it would tell you that all of these are also superfluid... What the Landau criterion tells you is the velocity of the superfluid flow at which excitations are created from the ... 0 The RKKY interaction is a generalization of the Kondo calculation in the case of two spins : it deals with finding the correct form of the interaction between two magnetic impurities via the Fermi sea. In this case you have a competition between two effects : the Kondo that describes the screening of each impurity bu the Fermi sea of electrons, and the RKKY ... 0 Ohm's law is a misnomer. It is not actually a true law, in the sense of Coulomb's of Ampère's; rather it is a 'rule of thumb' that applies pretty well in most circumstances. You will certainly not get a nobel price for finding an exception! A more general form of Ohm's law is $$\mathbf{J} = \sigma \mathbf{E},$$where$\mathbf{J}$is the current density, ... 0 You seem to be quite close to describing Andreev bound states. Recall that Andreev reflection involves an electron (hole) incident on an NS junction resulting in a Cooper pair in the superconductor, and a hole (electron) being reflected from the interface. In an SNS junction with a sufficiently narrow normal layer, where this process can occur at either ... 0 The only problem I have is that in most circuits there is more than one type of material, there are in fact many different materials that contribute to the overall current flow. It's not an ambiguity. Instead it's an unstated traditional over-simplification, as follows. A closed metal circuit may have many junctions between differing materials (e.g. ... 1 There's not enough information to say. It looks like your model has spring-like (harmonic) potentials between neighboring atoms, but phonon scattering requires anharmonic potentials. Harmonic potentials mean that the superposition principle still holds, so the phonons just pass right through each other without scattering. Since you have harmonic ... 0 Introduction: for metals, as you said, electrons can move to screen the fields. For low frequencies (less than the plasma frequency), the electrons are able to move fast enough to screen the fields; for high frequencies (greater than the plasma frequency), the electrons cannot move fast enough, and they do not "succeed" to screen the fields, which can ... 0 According to @Bercioux answer, if we choose the basis: $$\phi_1=(\Psi_{A+},\Psi_{B+},\Psi_{A-}，\Psi_{B-},\Psi_{A+}^\dagger,\Psi_{B+}^\dagger,\Psi_{A-}^\dagger,\Psi_{B-}^\dagger)$$ The BdG Hamiltonian should be written like this:$$H_{BdG}^1=\begin{pmatrix}H_+-E_F&0&0&\Delta_2\\0&H_--E_F&-\Delta_2&0\\ ... 0 Lets start from the beginning. I will drop$r$(to prove my lazyness). By definition$p_x \Psi = -i \sum_i c_i \frac{x}{a}(w(R_i+\hat{x})-w(R_i)) p_y \Psi = -i \sum_i c_i \frac{y}{a}(w(R_i+\hat{y})-w(R_i)) $Ignore the$\hbar$. I keep$x$and$y$to make the expression general. Now,$\Psi(\sigma_y p_x - \sigma_x p_y) \Psi\\ = \sum_{ij} c_i^\dagger ...

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