Tag Info

New answers tagged

0

Ok, I got the point: while classically I can (ideally) vary the energy of an oscillation mode by varying its amplitude from 0 to $\infty$, in QM I can vary the energy from $\frac{\hbar \omega}{2}$ to $\infty$ per steps of $\hbar\omega$.


0

In fact, the assumption of a uniform density neutralizing background is called the free electron model and is one of the first steps we take in the analysis of electronic properties of solids. Unfortunately, while the model makes multiple successful predictions (e.g., the Wiedemann-Franz law at high temperature), experimental evidence such as the temperature ...


1

There is no resistance is this scenario. You need a scattering mechanism like electron-phonon scattering or electron-defect scattering to obtain a non-infinite conductivity.


2

There is a book on the topic: B. Andrei Bernevig, Taylor L. Hughes: Topological Insulators and Topological Superconductors, Princeton University Press (2013).


2

I think two great reviews on the topic (cited by almost any paper on the matter) are: M. Z. Hasan and C. L. Kane, Colloquium: Topological insulators, Rev. Mod. Phys. 82, 3045 (2010); X.-L. Qi and S.-C. Zhang, Topological insulators and superconductors, Rev. Mod. Phys. 83, 1057 (2011);


0

I'm not too sure what you mean by interlayer and intralayer recombination. Photoluminescence occurs in bulk materials without any layers so this isn't an intrinsic requirement. In the case of LEDs the forward voltage drives an electron and hole current across the junction. With high design of electrons and holes in the same volume the recombination rate ...


0

A few less technical details than the ones you can find in Meng Cheng answer on this page are perhaps welcome. For practical applications, we are not interested in Majorana modes, since they are a simple mathematical rewriting of the fermionic creation and annihilation operators. Say differently, to any creation $c^{\dagger}$ and annihilation $c$ operators ...


2

You should forget about the name "Majorana fermions", although people have been using it (unfortunately) a lot in the literature on topological superconductivity. A much better terminology is "Majorana zero modes". "Majorana fermions" is more appropriate when you have propagating modes like chiral edge modes of a 2D topological superconductor, but not for ...


1

Yes, especially in research-level topics. There are several research groups that work with finding ways to apply differential geometry concepts to solid state systems (although condensed matter seems to be the preferred term nowadays). See for example the book by Altland and Simons, Condensed Matter Field Theory, Chapter 9 "Topology". This book is suitable ...


0

Differential Geometry has been useful in Mechanics of Crystalline Solids with Dislocations. See for example: http://arxiv.org/abs/1212.5125


1

I'll comment on the second issue that $E_F$ should always be greater than zero, which leads to a finite scattering rate. However, that's beside the point because in an ideal Fermi liquid at $T\rightarrow 0, |E_F-E|\rightarrow0$, the scattering rate becomes zero, which leads to an infinite lifetime. If in graphene, as you say, the independent particle picture ...


1

I think a classical example is electrical conductivity and resistivity (see Wikipedia), or any physical quantity which is described in the anisotropic case by a tensor (see also elasticity tensor as suggested in the comments by Jon Custer). Consider the Ohm's law in the anisotropic case $$ J_{i}=\sigma_{ij} E_j, \qquad E_i=\rho_{ij} J_j $$ The conductivity ...


2

Here's an example bandstructure (image from here). Each point on the horizontal axis is a point in k-space. Then the vertical axis shows the bands. k-space is three dimensional. In a perfect world we would make a four-dimensional graph with k-space on three axes versus band energy on the fourth axis. Alas, we live in a 3D universe and can most easily ...


0

LO-TO splitting is caused by the long-ranged nature of the Coulomb interaction (i.e. because the Fourier Transform of the Coulomb interaction,$4\pi e^2/q^2$, is not well-defined at $q=0$). Also, it occurs near the Brillouin zone center, but not at the exact Brillouin zone center because of retardation effects (i.e. the finite speed of light). At $q=0$, the ...


1

I will try to answer in a simple and intuitive way. If you need more details I suggest you, for example, Ashcroft and Mermin, Solid State Physics. The bands in the $E$-$k$ diagram are filled with electrons from the lowest energy state (the bottom of the dispersion curve) up to the Fermi energy. This is a consequence of the Pauli exclusion principle and of ...


0

It is not possible to get the dispersion in any direction using just the given dispersion. Some knowledge of the underlying crystal and electronic structure is required.


1

These are two different problems. What Jon was saying is correct. However, it does not explain LO-TO splitting. Like Jon said, because you can tell when you are on a Ga or As atom, the degeneracy of the optical modes are lifted at the Gamma point. This is in regards of 3 different optical modes separating. However, the phenomena Cardona is refers to involves ...



Top 50 recent answers are included