Tag Info

New answers tagged

0

What exactly makes them different from ordinary insulators is the number of edge states. In a 2D topological insulator, you are guaranteed to have the Fermi energy at the edge of the sample cross the edge bands an odd number of times in half the edge Brillouin zone while in a trivial band insulator, if there are edge states, the Fermi energy will cross the ...


2

To grasp the relevant physics at a sloppy level, perhaps you simply need a few examples. You know a concept is commonly constructed by the manner you refer to it together with other concepts. Symmetry breaking usually results in ground state degeneracy and long range order. Order parameter field aids you in identifying degenerate sectors with the symmetries ...


19

Because the electrical force on an electron is around 10^39 times that of gravity. Given the equivalence between gravitational and acceleration forces, you would have to shake it quite hard. Before you got to the point where an electron would drop out the entire material would disintegrate and all kinds of other phenomena would take precedence over you ...


12

Because there is an energy barrier between the metal and vacuum. Consider the ions in the metal as a uniformly distributed positive charge. Near the metal surface, the free electron wave function spread out a little into the vacuum, thus near the surface of metal the electric dipole forms with the electric field points to vacuum. Thus a gradual potential ...


2

In spin liquids, the ordered state is broken by zero-point fluctuations even at $T=0$. Even though it is common for spin liquids to be frustrated, it is not necessarily so. The $S=1$ Heisenberg spin chain (AFM), for example, is a spin liquid without being frustrated. The name spin liquid comes (I believe), from the exponentially decaying correlation (like ...


0

A highly subjective answer: I've always imagined "collective excitation" more as a subset under "quasiparticle" than as an alternative. The classification I usually do is that a quasiparticle is either a "dressed state" or a "collective excitation." The difference between the two is that a dressed state is adiabatically connected to some particle in a ...


0

There is a deeper meaning of the magnetic length $l_B$. It is the lattice constant of a two-dimensional (2D) artificial structure. Let's consider a particle moving in 2D plane under a perpendicular uniform magnetic field of strength $B$. In classical mechanics, the system is obviously translationally invariant under any translation. However, in quantum ...


0

There are a few ways to extract transport properties from your single-particle temperature Green's functions. By analytically continuing it to real time $t$, one gets information about how a particle propagates in the medium. More exactly, you get the probability of the particle traveling a distance $x = |x_1-x_2|$ in the interval of $t$. From this, you ...


1

By inhomogeneous I assume you mean disordered, i.e., a system with a noisy/random potential landscape. I'm not sure which Mahan book you are referring to, however I found Akkermans and Montambaux' Mesoscopic Physics of Electrons and Photons to give a good discussion of the problem of wave propagation in disordered media. Essentially the problem is "solved" ...


0

It depends on context. Isotropy means "the same in all directions". In the spin context, this can sometimes mean actual spatial directions, but usually it means directions in "spin-space". That is, an isotropic exchange interaction is one where the x-components of the two spins interact the same as the y-components, and both interact the same as the ...


0

For what it's worth, it is claimed that the critical exponents differ above and below the critical point for some exactly solvable 2-dimensional model: http://www.ujp.bitp.kiev.ua/files/journals/49/11/491114p.pdf (Ukr. J. Phys., v.49, #11, p.1122 (2004)).


2

Critical exponents are properties of the RG fixed point that drives the phase transition. They are computed by linearising the RG flow equations close to the fixed point. The exponents are the derivatives of the beta functions evaluated at the fixed point. They know nothing of the way you approach the fixed point. In particular if you are flowing slightly ...


0

From the wikipedia article linked in your question, we see that Wannier functions can also be defined as $$ \underbrace{\phi_{\textbf{R}}(\textbf{x})}_{\textbf{R} \text{ Bravais lattice vector}} \equiv \frac{1}{\sqrt{N}} \underbrace{\sum_{\textbf{k}}}_{\substack{\text{summed over first}\\\text{Brillouin zone}}} e^{-i\textbf{k}\cdot\textbf{R}} ...


2

I wanted to post this as a comment, but it grew too long. The final question that was a bit hidden, but that several other users seemed to be also interested in, would be about why the blue LED was maybe so much harder to construct than the red one. Reading through Steve B's link to the nobel scientific background provides me with enough information that I ...


2

Additionally, blue was the last of the primary colours so its invention made the production of white LEDs possible. Ordinary lamps could then be replaced with extremely energy efficient LED alternatives.


1

The statement is not true, because there are counter examples. A U(1) spin liquid is gapless, but it is insulating. An $s$-wave superconductor is fully gapped, but it is (super)conducting.


39

The Nobel website scientific background is good. Basically, when you try to make gallium nitride, you usually end up with a material that is (1) chock-full of defects, and (2) n-doped (even when you were trying to p-dope it). So blue LEDs required The invention of MOCVD technology for growing crystals (early 1970s); Finding the right recipe to grow good ...


4

The "critical" part was in finding and producing a structure with a large enough bandgap to produce blue photons. The first LEDs produced relatively longwave infrared (IR) photons, which have far less energy than the green or blue photons now available from LEDs. In general, the larger the desired bandgap, the harder it is to manufacture a suitable ...


7

The book The Blue Laser Diode: The Complete Story deals with the issues of p-type doping of GaN. The difficulty of growing high quality GaN crystalline films lies in the problem of finding a suitable substrate material. (...) The link above points to the chapter you may be interested in.


1

It is not generally true that a gapped system is insulating. Or more precisely, this statement is not detailed enough to be said true or false generically. One case where this is true is for non-interacting particles (say, free electron in a lattice). For interacting particles, it is much more subtle. In particular, just stating "gapped system" is not ...


1

Your understanding is basically correct. Some materials, intrinsic semiconductors, have small band gaps so that at room temperature (for example) there is enough thermal energy around to promote some electrons to the conduction band, and some holes to the valence band. These materials have the kind of gap that you describe, but are not insulators. Another ...


0

You can use either potential for either purpose; it's just that some potentials are better for the different purposes. The reason is that these are empirical potentials; their constants are tweaked to work for a certain purpose. For example, if you're looking at the phonon band structure, you want $\omega\left(\vec{q}\right)$ to be as accurate as possible. ...


0

When you Fourier transform the tight-binding Hamiltonian, $$c_{j,s}=\sum_{k,s} a_{k,s} e^{i R_j k},$$ with periodic boundary conditions, you will be left with a diagonalized Hamiltonian in the desired form. For details see section 2.3 of these lectures notes, starting on p.18: http://manybody.skku.edu/Lecture%20notes/Solid%20State%20Physics.pdf


0

Phonons are collective modes in solids and a general derivation is needed independent of particular lattice constants to first order. Lattice constants define individual solids. Are you aware of the harmonic oscillator approximation? All symmetric potentials have as a first term in their expansion the quadratic, thus the harmonic oscillator ...


0

Any increasing sequence $(\Lambda_n)_{n\geq 1}$ of finite subsets of $\mathbb{Z}^d$, $d\geq 2$, such that $\bigcup_{n\geq 1} \Lambda_n =\mathbb{Z}^d$ will do. All sequences $(\mu_{\Lambda_n}^+)_{n\geq 1}$ of finite-volume Gibbs measures in $\Lambda_n$ with $+$-boundary condition converge to the same infinite-volume Gibbs measure $\mu^+$, under which there is ...


1

My previous comments are almost ok, your actual problem seems to be that you do not average over the magnetization. You are measuring <cos(theta)>, which is the average of m_x. So just change m = magnetization_cossin(); to magnetization_cossin(&mx, &my); where you define void magnetization_cossin(double* mx, double* my) { int x, y, z; ...


1

For a metal, the permittivity can is typically described by the Drude model with a permittivity given by, \begin{equation} \epsilon = \epsilon' - i\epsilon'' = \epsilon_\infty - \frac{\omega_p^2}{\omega(\omega - i\gamma)} = \epsilon_\infty - \frac{\omega_p^2}{\omega^2 + \gamma^2} + i\gamma\omega\frac{\omega_p^2}{\omega^2 + \gamma^2} \end{equation} where ...


5

First let me make two comments before answering the question. The difference between metal and insulator rest in the existence of the itinerant electron Fermi surface or not. Ising (or Heisenberg) model is just an effective theory of local moments (localized electrons in the atoms), which contains no information of the itinerant electron, so there is no ...


3

The main justification for considering the Ising model is that it's exactly solvable in one & two dimensions (and that it shows critical behavior which is universal in some sense). It is not particularly meaningful as an approximation to a real physical system. The Heisenberg model does a much better job, but it is also a lattice model. If you really ...


4

I think that you are really interested in the $q$-state clock model, which is similar to the Potts model, and is defined as follows. Fix an integer $q\geq2$. For each $i\in\mathbb{Z}^d$, let $$ \theta_i \in \bigl\{\frac{2\pi}{q} k\,:\, k\in\{0,1,\ldots,q-1\}\bigr\}, $$ and define the spin at site $i$ by $$ \mathbf{S}_i = (\cos\theta_i,\sin\theta_i) . $$ The ...


0

(I will write $\ell$ instead of $l$, because it confuses me less frequently.) Assuming that $\delta$ definition is a typo, and it is supposed to be $$ \delta=(\ell+1)[2(\ell-1)j_{\ell-1}(kR)-kRj_\ell(kR)] $$ Your matrix's determinant is nonzero... \begin{align} \alpha\delta-\beta\gamma &= \ell \left[R j_{\ell}(kR) k - 2 j_{\ell+1}(kR) \left(\ell + ...


1

There is a very nice property that works, in practice, for most systems that is that of equivalence between the Gibbs' ensembles in the thermodynamic limit. The prototypical example is that of the equivalence between the canonical ensemble and the microcanonical ensemble. One way to state it is to say that the free energy $F(T,N,V) \equiv -k_BT \ln Q(T,N,V) ...


1

The statistical ensembles differ in the constraints imposed to them. In the canonical ensemble, the number of particles $N$, the volume $V$ and the temperature $T$ are fixed. In the grand-canonical ensemble, the number of particles is not fixed, it is determined by the chemical potential $\mu$, which plays the same role on $N$ as temperature on energy or ...


3

You are right, these terms are related. Metastability usually comes about in systems which are described by a Landau free energy which contains a cubic or power 6 term on top of the usual $\phi^4$-theory. E.g. the Landau free energy with a cubic term is shown below. There are three special temperatures: $T^{**}$ at which an additional local minimum forms ...


0

Note that this is a two-site problem, so the many-body Hilbert space is finite (and actually just four). Therefore one can simply express all the operators in terms of $4\times 4$ matrices. Let us order the many-body basis states as $|s_1\rangle=a_{1\uparrow}^\dagger a_{2\downarrow}^\dagger|\Omega\rangle$, $|s_2\rangle=a_{2\uparrow}^\dagger ...



Top 50 recent answers are included