# Tag Info

6

Electricity needs charges particles (or quasi-particles) to conduct. Heat can be conducted with almost any quasi-particle. Diamond is one of the best conductors of heat in existence, and it's because of phonons, ie quasi-particles of lattice vibrations, which are strong because the diamond lattice is strong.

3

If you have a very low frequency, the material behaves as a conductor: the electrons respond instantaneously to the excitation, and therefore the metal becomes a reflector (the boundary condition of "no E field parallel to the surface" is met). If you have a very high frequency, the electrons don't have "time to react" at all - so the amplitude of their ...

3

Let's see if this explanation works for you. It is more intuitive than rigorous - but maybe it will help. In a "normal" simple harmonic oscillator, the potential well is a parabola, and the restoring force is proportional to the displacement. We know that the equation of motion for such a well is a sinusoid. Now if we make the potential well "more than ...

3

Yes, para hydrogen, in a limited manner. Recent work at Göttingen has revealed convincing evidence for superfluidity in liquid hydrogen, the only liquid other than helium to exhibit this quantum behaviour. From a spectroscopic experiment on droplets of parahydrogen, it has been discovered that properties of superfluid are observed in a system of ...

2

Question: The above analysis seems to treat the oscillators as distinguishable. But aren't they indistinguishable? In Einstein's model, the oscillators are supposed to sit at (oscillate around) definite place in space. So you could say they are distinguishable. For example, by their cartesian coordinates with respect to lab frame. 1)That the Gibbs ...

2

I'm not quite sure what you mean by "crossing at $[E_F,0.5]$" but the Fermi level doesn't change with temperature by definition. The Fermi level is defined as the energy of the highest energy electrons at zero temperature when the system is in its ground state. It is a property of the system that is only dependent on the quantum mechanical eigenfunctions and ...

1

The lack of the $e^{-i \omega t}$ term is just because we're using complex wave notation. If you've ever taken an electrical engineering course, it's the same sort of thing that is used there: We're using $A e^{i \omega t}$ to stand for $A \cos (\omega t - \delta)$, with the implicit assumption that we're only interested in the real part of the quantities ...

1

A single photon with 3-momentum $p=\hbar k$ incident on the system is scattered to an outgoing state with momentum $\hbar k'$. If the interaction operator between the photon and the sample is $U$, and the initial and final states are labelled by $|k\rangle$ and $|k'\rangle$ respectively (we ignore photon polarization/spin), then the amplitude for this ...

1

Crystal lattices are classified in a way that is not necessarily the most natural one. A first classification is by their lattice class, which is determined by its associated unit cell, which is not necessarily the same as a fundamental domain for the lattice. There can be different Bravais lattices in each lattice class, determined by possible additional ...

1

Just to be clear, the two or more atoms do not have to be of different type. Optical phonons are related to the relative vibrations of atoms within the unit cell, while acoustic phonons describe the relative vibration of different unit cells. Optical phonons arise whenever the unit cell has at least one such degree of freedom, meaning at least two atoms in ...

1

There are essentially two ways of generating coherent radiation from solid-state devices: Classical electronic oscillators, in which charge is made to oscillate back and forth within a device... the frequency of radiation corresponds to the frequency of charge oscillation. Solid-state lasers, in which charge-carriers undergo a transition between two ...

1

Sorry I was too lazy to retype everything, but here is a set of notes I made for myself a little while ago. Note that below $\epsilon(q,0) = 1+V(q)\Pi(q)$, where $V(q) = e^2/\epsilon_0q^2$. If my notes aren't clear, let me know.

Only top voted, non community-wiki answers of a minimum length are eligible