New answers tagged

0

Effective mass is calculated from the dispersion relation in a material. I can give a simplified explanation assuming your minimum is at $k=0$ and the dispersion relation is isotropic. The concept is that free electrons have a dispersion of $$E = \frac{\hbar^2 k^2}{2m}$$ Now in a solid, at the bottom of the band (small $E$) the dependence of $E$ on $k$ will ...


2

In my view, the simplest way to see SSB (spontandeuos symmetry breaking) in superconductor is to start from path integral for superconductor. We can approximate attractive interaction between electrons (which comes from phonon exchange) to contact constant interaction with coupling $g$. Obtained theory consists 4-fermion interaction. This interaction can be ...


2

If your angular brackets denote expecatation in BCS bround state in which $\beta|gnd\rangle=0$ your result should be immediate--- no algebra required. The diagonalized Hamiltonian is a sum $\sum_i E_i \beta_i^\dagger\beta_i$ and commutes with the quasiparticle number. But $\beta^\dagger_i\beta^\dagger_j|gnd\rangle$ is an energy eigenstate that differs by 2 ...


-2

If you simulate something with gain or losses use the scattering matrix, otherwise the transfer matrix


2

On applying L'Hôpital's rule, we get $ y = \frac{2Msin(Mx)cos(Mx)}{2sin(x)cos(x)}$. Again applying L'Hôpital's rule $\frac{sin(Mx)}{sin(x)} = M$, giving $y=M^{2}$. Just here, it is proved that $\frac{sin^{2}(Mx)}{sin^{2}x}$ has maxima at $ x = n\pi$. Here n is any integer and not just even integers. In $\frac{sin^{2}(\frac{1}{2}M\vec{a_{1}}.\vec{\Delta k)...


0

At the top of page 455 in Ashcroft and Mermin, they note that one can expand the Bose-Einstein distribution in powers of $1/T$ to get the high-temperature expansion. The first term yields the Law of Dulong and Petit, and the rest of the terms decay as a function of $T$. Due to the natural high-frequency cut-off caused by the finite distance between atoms, ...


0

First of all, you indeed have the periodic potential of the lattice, formed by the positive ions. Remarkably enough, it is not this periodic structure that is the main origin of the resistance. Bloch's wave function given by $$\Psi (r) = u(r)e^{ikr}$$ takes into account this periodic structure of the material. As you can see, this is still a stationary ...


2

So what is illustrated in this figure is the energy band structure, these represent the dispersion, i.e. the energy $E$ in function of the wave vector $k$. Now, you can find in standard textbooks on solid state physics that the effective mass is defined as $$(m^*)^{-1} =\frac{1}{\hbar^2}\frac{\partial^2 E}{\partial k^2}$$ So the curvature of these (...


0

Fast answer: you can have these centered lattices... they are just not useful. Either they are incompatible with symmetry or they can be reduced to smaller cells. Imagine/draw a hexagonal C face-centered cell. The resulting lattice is incompatible with the 6-fold symmetry. (For the same reason, you cannot have cubic face-centered. Only cubic all-faced ...


0

Classically, we picture electrons being drawn towards the ions and, on impact, dispersing energy throughout the lattice. The higher the electromagnetic attraction of the individual ion the more likely that an electron is going to be drawn to and collide with it. Quantum mechanically, we picture it as electron waves being dispersed due to defects in the ...


1

I do not know if I can add much, but I can try to explain how these matrix elements arise. The integral should indeed be over all space, but the lattice structure manifests itself in that the momentum $K$ only takes discrete values: the reciprocal lattice vectors. If you make a Fourier transform of the Coulomb potential, you indeed encounter a problem of the ...


1

It's the energy gap of the single-particle spectrum. If one tries to probe the system with a STM, for example, you will get no tunneling below this gap. It can be read from the retarded Green function which in matrix form is $$ g^r(\omega) = -i\frac{\pi\rho_0}{\sqrt{(\omega^2-\Delta^2)}}\left(\begin{array}{cc}\omega & \Delta \\ \Delta & \omega \end{...


1

In a given element of the disk, with a distance $r$ from the centre, length $\Delta r$, and cross section $r \Delta \theta$, the balance of forces are: $$r \Delta \theta ( \sigma_{r+\Delta r} - \sigma_{r}) = \mu r \Delta \theta \Delta r \omega^2 r$$ Dividing by $r \Delta \theta \Delta r$ and taking the limit when $\Delta r->0$: $\frac{d\sigma}{dr} = \...


0

There are waves in the wire, and they can be reflected back from the closure. As an example: when I was working with preamplifiers to increase the signal measured in the oscilloscope, I had to use a 50 ohm closure at the end of every wire, to get a good enough reflection otherwise the signal kind of "leaked" , got noisy. I guess some electrodynamics ...


1

We always index the wavefunctions with the wavevector $k$. Sometimes this is even written as $$\psi(k,x)=\mathrm{e}^{\mathrm{i}kx}u(k,x)$$ which makes it looks like the function depends on real space and reciprocal space simultanously. That would be weird, but that's not what the notation is saying. For example, consider the eigenstates $|n \...


1

This arises from the discrete translational symmetry of the lattice (Bloch's theorem). Since the potential is periodic, the Schrodinger's equation is invariant to discrete translations by lattice vectors. Suppose, $\left|0\right>$ is the original state we start with, and $\left|a\right>$ represent the state after a translation by a. $ H(0)\left|0\right&...


0

I think it is better to really think physically about how energy bandstructures determine the electronic transport. First, let's answer the question which electrons participate in the electronic transport of the material? Well, transport requires some empty available energy levels throughout the material in which electrons can move "freely". The electrons ...


1

No, it is impossible. The reason is very simple if one recalls the meaning of phonon dispersion curves. They give the set of possible frequencies, $\omega_s({\bf k})$, for the eigenmodes of the $s-$th dispersion branch. The energy density of a harmonic crystal, at a temperature $T$ is: $$ u = u_{static} + \frac{1}{V}\sum_{{\bf k},s} \frac{1}{2} \hbar \...


1

Oh my I think you made some mess of most of the concepts and terms... In metals there are a lot of free electrons already. Increasing temperature increases the number of free (unbound) electrons, but it also increases lattice vibrations (phonons) which make the electrons scatter, and this effect is much stronger. Therefore, increasing the temperature makes ...


1

For metals: You don't have formation of more free electrons, there is decrease of resistivity as temperature falls. You can express conductivity as inverse of resistivity so you have $\sigma= 1/ \rho = \frac{1}{\rho_0 (1+\alpha(T - T_0))}$. As you can see conductivity depends on temperature as $\frac{1}{T}$. For semiconductors: Formation of Cooper pairs ...


1

That depends strongly on many factors. The reason for the decrease in thermal conductivity at high temperature is that the mean free path goes down because of phonon-phonon scattering (interaction with the lattice). At some point the phonons do not propagate much further than a lattice period and the lattice melts. So those are the shortest mean free paths. ...


3

Phonons are lattice vibrations. The distance between two consecutive phonons is of the order of 1/N where N is the number of atoms in the lattice. At room temperature a phonon travels approx 10 to 100 lattice constants before scattering. In this article https://www.nature.com/articles/srep17131 they say that a phonon travels $< 1 \mu m$ before ...


2

In monolayer TMDs, the inversion symmetry is broken, which results in a non-zero Berry curvature of the electron bands in the vicinity of the K and K' valleys in the Brillouin zone. In addition, these compounds have strong spin-orbit coupling which results in the spin-splitting of the valence bands (and to a much lesser extent in the CB). Time reversal ...


1

We have the Bloch's theorem to get the ansatz for the wavefunction, $ψ_{k}(x)=e^{ikx}u_{k}(x)$ to solve the Schrodinger's equation. We know that $u_{k}(x)$ is a periodic function in x with a period equal to the lattice constant, but we are still left with the task of finding what exactly $u_{k}(x)$ is. Substituting our ansatz into the Schrodinger equation ...


Top 50 recent answers are included