New answers tagged

1

To clarify, the Bloch result that you wish to show is that $\psi(x) = e^{i k x}u_k(x)$ where $k = 2\pi s / Na$ and $u_k(x)$ is a function which has the same periodicity as the potential $U(x)$, that is, $u_k(x) = u_k(x+a)$. We can directly verify that (2) fulfills these conditions. From (2), we have $u_k(x) = e^{-ikx}\psi(x)$. To verify that this has the ...


2

Does Bose-Einstein condensation depend on boundary conditions? No. This can be shown rigorously in 'On the Bose-Einstein condensation of an ideal gas' by L. J. Landau and I. F. Wilde (1979). The proof lies in computing the fugacity $z = \mathrm{e}^{\beta \mu}$ (called activity in the paper) and showing that it exhibits a non-analytic behaviour at some $T = ...


-1

Usually the periodic boundary conditions and the hard wall boundary conditions refer to different situations: Periodic boundary conditions are a mathematical trick to deal with the continuum of states - one usually takes in the end $L\longrightarrow\infty$, recovering the continuous spectrum. Hard wall boundary conditions imply actual walls, such as ...


0

The holes itself do not possess any charge it is the site at which there used to be an electron present but the electron leaves creating a potential at that position and when the near by electron comes in its place the hole moves to new position, they keep flowing on path discontinues unlike electron, the holes are said to be positive as they move in the ...


2

Yes, the correlation matrix of the subsystem is the submatrix of the correlation matrix for the entire system. This is since the correlation matrix is defined as $$ C_{ij} = a \,\mathrm{tr}[c_i c_j] $$ with $c_i$, $c_j$ all fermionic operators on the whole system (what $a$ and $c_i$ are exactly depends on the convention chosen by the respective paper), and ...


3

Assuming a cubic sample of side $L$ and periodic boundary conditions. The wavevector $\mathrm{k}$ can assume any value on the grid \begin{equation} \mathrm{k} = (k_x, k_y, k_z) = \left(\frac{2\pi}{L}n_x, \frac{2\pi}{L}n_y, \frac{2\pi}{L}n_z\right) \end{equation} where $n_x, n_y, n_z$ are integers. The Fermi momentum is the maximum momentum allowed for ...


2

Let's first be clear about some basics then we go on to questions: The periodic boundary condition requires that any wave in the sample $e^{ikr}$ have the same value for a position $r$ as it has for $r+L$ (after one round around the circle). This imposes quantization of $k$ $$k=\frac{2\pi n}{L} \ \ \ n\ \text{is integer}.$$ In three dimensions, $$\mathbf{k}=\...


2

It's a derivative: $\frac{dQ}{dT} = kT^{3}$. You have to integrate from the initial temperature $T_0$ to the double: $2T_0$. $$ Q = k\int_{T_0}^{2T_0}T^{3}dT. $$ And you have the correct result. Don't confuse heat capacity with specific heat capacity.


1

I think we should speak here not of semiclassical model, but of semiclassical approximation, which preserves some of the quantum features in favor making things more tractable. Indeed, if we were to demand that the position and momentum can be measured simulatenously, we could not have quantum effects. However, if we demand that they are measured ...


1

First of all, it's important to remember that we are dealing with models here, which are mathematical expressions that are derived from certain assumptions. Generally speaking, in materials systems we need to work with very large numbers of particles that can interact in non-trivial ways, so there is a tendency to simplify things to make the mathematics ...


0

This question seems to be based on a false premise, namely, that systems that are nonlinear classically are linear when quantized. Really, the opposite tends to be the case. E.g., Maxwell's equations in vacuum are exactly linear but in QED there is nonlinearity due to interactions mediated by electron loops. Quantum mechanics is linear at the level of the ...


0

In the canonical ensemble, the density matrix of the system is given by $$\rho={1\over Z}e^{-H/k_BT}$$ By choosing to define the density matrix as $\rho=|G.S.\rangle\langle G.S.|$, you implicitly considered the case of a temperature $T=0$. As you have shown, the reduced density matrix is then $$\rho_A=|1\rangle_A\langle 1|_A$$ In your example, the two ...


2

Graphene Much of the basic theory underlying graphene was known long ago - it has served as the zeroth order approximation for describing graphite, whose electronic, emchanical and other properties are well described as those of weakly coupled graphene sheets. Heterostructures Heterostructures are somewhat different - here the properties of a bulk region are ...


3

When faced with a repeating pattern, the conceptual issue is to identify the underlying Bravais lattice and a possible basis to describe the crystalline structure. Of course, there are many (infinite) possibilities for these elements. Still, the constraint of having the minimal basis and the most symmetric primitive unit cell for the Bravais lattice are good ...


1

The energy $E$ is related to the moment $p$ in the following way (with $c=1$) $$E=\sqrt{p^2+m^2}\rightarrow p=\sqrt{E^2-m^2}\tag{1}.$$ $d^3p$ is a infinitesimal volume in the $p$-space. Since $f(p)$ depends only in the magnitude of $\vec{p}$, we can use spherical coordinates, so $$d^3p=p^2\sin\theta \,dp\,d\theta\,d\phi$$ $$n=\frac{g}{2\pi^3}\int f(p)d^3p=\...


1

If we take, assuming isotropy on p: $$d^{3}p = 4\pi p^{2}dp, $$ and taking $c = 1$ in the energy-momentum relation: $$E^{2} = p^{2} + m^{2}. $$ We have: $$|\frac{dp}{dE}| = \frac{E}{\sqrt{E^{2} - m^{2}}}. $$ Then: $$d^{3}p = 4\pi p^{2}dp = 4\pi \frac{(E^{2} - m^{2})}{\sqrt{E^{2} - m^{2}}}EdE.$$ So, finally: $$d^{3}p = 4\pi E\sqrt{E^{2} - m^{2}}dE.$$


2

Essentially, there is nothing special about the lattice of a semiconductor that would prevent it to "host" Cooper pairs. The process that you describe of "pairing all electrons before entering [your preferred material here]" is called proximity effect. You just juxtapose a superconductor with another material (generally a non-...


1

A gas to liquid transition involves latent heat and is hence a first-order transition (apart from the critical point where transition happens with no coexistence). The isotherms for this transition are shown below: (image from Lumen Learning | Physics | Phase Changes) On the liquid side, the isotherms are very steep - that means that the "phase" ...


2

If you insist on not breaking one type of terms in the Hamiltonian - say, the plaquette constraints - then the remaining theory is a $Z_2$ gauge theory with the plaquette constraints as the gauge constraints. Thus, within this subspace where the plaquette constraints are not violated, the theory behaves just as a gauge theory. But this is no longer true when ...


0

I completely understand your confusion, because that was exactly mine too. The logic is that, generally by the Schrodinger equation $\hat{H} \psi(x) = \epsilon \psi(x)$, the energy $\epsilon$ is determined by the full wave function $\psi(x)$. Similarly, if the potential is not periodic, in Fourier space we will need the full set of $c_\mathbf{q}$ (with $\...


0

The Pauli exclusion principle only refers to particles in exactly the same state. If they are local wavepackets localised at different locations then they aren't in exactly the same state. So it doesn't apply here. In fact, since Pauli is a (somewhat weak) corollary to the fact that fermion wavefunctions must be anti-symmetric under exchange, that you wrote ...


1

As @Quantumwhisp has correctly pointed out, the semiconductor junctions described in the classical textbooks are usually treated within the semiclassical approximation. However, nowadays we are able to fabricate devices where the wave function is coherent across the junction is everyday reality - these range from tunnel diodes (one tunnel junction) to ...


2

The magic words are "semiclassical model" (you can read about it for example in Ashcroft / Mermin, chapter 12), https://www.fzu.cz/~knizek/literatura/Ashcroft_Mermin.pdf). You are right in that the solution of an electron with crystal momentum $\vec{k}$ for a lattice is itself periodic. You can however superimpose wave functions with differing $\...


0

For it to be hermitian, the Hamiltonian should read \begin{equation} H_{k} = -\sum_{k}( 2t \cos(k) + \mu) c^{\dagger}_{k} c_{k} + \Delta e^{-ik}c^{\dagger}_{k} c^{\dagger}_{-k} + \Delta^{*} e^{ik} c_{-k}c_{k}) \ , \end{equation} or something the like. This is what your derivations gives: Substitute $k\to-k$ and use $\sum_k=\sum_{-k}$, and you get $$ \sum k ...


1

A formal form of writing: $$ \rho(E) = \sum_n \delta(E - E_n). $$ where $$ E_n = \frac{n^2\hbar^2\pi^2}{2 m a^2}. $$


2

You are correct. There is a $c$-number energy shift: $$ \hat H_{\rm Bogoliubov}= a^\dagger_i H_{ij}a_j +\frac 12 \Delta_{ij} a^\dagger_i a^\dagger_j +\frac 12 \Delta^{\dagger}_{ij} a_i a_j\\ = \frac12 \left(\matrix{ a^\dagger _i &a_i}\right)\left(\matrix{ H_{ij}& \phantom {-}\Delta_{ij}\cr ...


0

For most systems, the derived macroscopic thermal quantities from anyone of the formulations should have a same expression. Lets examine the simplest two-level system, this example appeared in many statistical text books, being treated in details for both microcanonical and canonical ensembles. The two-level non-interaction system assumes particle can exists ...


2

Actually, downfolding (see section 5.2) on its own does not require any approximations. It also does what you are looking for. One defines two projectors $P, Q$ onto sub-Hilbert spaces, where we are interested in the subspace corresponding to $P$. Together $P+Q = 1$, i.e., they together project onto the full Hilbert space. One then gets an effective ...


3

The name "fracton" was coined by Vijay, Haah, and Fu (https://arxiv.org/abs/1505.02576) precisely because of the phenomenon you have described: individual fractons are immobile, but composites of multiple fractons may not be. Composites of these fundamental excitations, however, are topological excitations that are free to move within sub-...


1

In a conventional conductor, the current density and the electric field obey Ohm's Law, $\vec{J} = \sigma \vec{E}$. A perfect conductor, such as a superconductor, is the $\sigma \to \infty$ limit of this equation; this implies that in this limit, we must have $\vec{E} \to 0$ in order to have $\vec{J}$ approach a finite limit.


0

Why the current density j inside a superconductor has to be finite? Once the current density exceeds a certain critical current density the material stops superconducting. This critical current density is an important material property for many applications such as MRI. It dictates the amount of superconducting wire that is needed to achieve the desired ...


0

Why not use the Fermi surface in the tight-binding model? By convention, the Fermi surface separates filled from empty states, where these are infinitesimally close – that is, where the boundary falls in the middle of a band. If the boundary (the Fermi level) instead falls in a band gap, we say that a material has no Fermi surface. Thus one definition of a ...


0

First time using a turbo pump I presume. As one example, the Varian Turbo-V Vent Device tells you what you want to know: The Turbo-V vent device, consisting of a vent control unit and a vent valve, is a complete unit suitable for automatic venting of the Turbo-V pump when it is switched off or during a power failure switch... and The Turbo-V pump must be ...


7

From a mathematical perspective, ostensibly dodgy distributional issues often arise because the familiar formula for the Fourier transform is not the whole story. If a function (lets say of one variable) $f\in L^1(\mathbb R)$, then its Fourier transform $\hat f\in L^1(\mathbb R)$, and is given by $$ \hat f(k) = \int\mathrm dx \ e^{ikx} f(x)$$ If $f\in L^2(\...


3

I hope that I do not misunderstand something about your question, but $e^{-ik}$ goes once around the unit circle, so $$0=\sum_k e^{-ik} = \sum_k (a_k a_k^\dagger-a_k^\dagger a_k) e^{-ik}=\sum_k a_k a_k^\dagger e^{-ik}-\sum_ka_k^\dagger a_k e^{-ik}$$ and thus, $$\sum_k a_k a_k^\dagger e^{-ik}=\sum_ka_k^\dagger a_k e^{-ik}$$


1

Commensurate means that the ration of two parameters is an integer or, more generally, a ratio of integers. For example, if we consider a one-dimensional lattice with period $a$, and we impose on top of it a perturbation potential $$V(x)=V_0\cos(kx)$$ with spatial period $$\lambda=\frac{2\pi}{k}.$$ The sutuations where $\frac{\lambda}{a}=n$ or $\frac{\lambda}...


1

Term decay is rather rare in condensed matter physics, so I will assume that what you really mean is the finite lifetime. Condensed matter studies complex many-paryicle systems. For example, description of a crystal in terms of electron Bloch waves is valid only under assumption of a a rigid lattice and absence of Coulomb interaction. Once the interactions ...


1

At Weyl points ($K$, $q = 0$), we can approximate a tight-binding Hamiltonian as \begin{equation} H(\vec{K}+\vec{q})= v \vec{q} \cdot \vec{\sigma}. \end{equation} As this is a general $2x2$ Hamiltonian, we know that the eigenvalues will be of the form $ v|q|$. \ \end{definition} The corresponding eigenvectors are given as \begin{equation} |- \rangle = \...


1

You can have more bands than the size of the matrix. Think of Harper's equation: $$ \psi_{n+1}+\psi_{n-1} + (\lambda \cos q n)\psi_n = E\psi_n $$ This is a nearest-neighbour hopping problem with an position dependent on-site potential. The unit cell is not one site, but instead depends on the periodicity determined by $q$. If you have to go $N$ sites to get ...


1

After I posted my original answer, I realized that there's a better explanation, and it explains the name "continuum model": (for simplicity, I'll omit sublattice degree of freedom) Suppose that the eigenstate is: $$ \widetilde{\psi}(k_0)= \begin{pmatrix} \widetilde{\Psi}_1(k_0)\\ \widetilde{\Psi}_2(k_0) \end{pmatrix} = \begin{pmatrix} \sum_{k_1} \...


1

The Hamiltonian (2.131) is $$ H = \frac{1}{2m}({\bf p} + e{\bf A})^2 + V({\bf r}) + e {\bf E} \cdot {\bf r}. \tag{1}$$ The electric and magnetic fields are constant (homogeneous, so independent of position), and furthermore from section 2.1.6 we take the particular magnetic field ${\bf B} = (0,0,B)$ with choice of vector potential ${\bf A} = (0, x B,0)$. ...


1

I'm not fully satisfied with every step in this answer, I think I am missing a couple of details, but hopefully it is still helpful to you. Starting from (1), using $\nabla\cdot(\phi\mathbf{A})=\phi\nabla\cdot \mathbf{A}+\nabla\phi\cdot \mathbf{A}$, we get $$ n \nabla\cdot\frac{\partial \bar{r}}{\partial t}+ \nabla n\cdot\frac{\partial \bar{r}}{\partial t}=-\...


2

I suggest thinking about Fermi liquid theory as two successive levels of approximation (which are unfortunately not clearly delineated in the sentences you quote): In the limit of vanishing energy above the Fermi level, the eigenstates of the interacting theory have the same quantum numbers as free electrons, and the two are smoothly connected as the ...


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