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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, ...

4

A mode expansion is an expansion in terms of a complete set of orthonormal modes. The orthonormality ensures that the set of modes is linearly indepenedent and so the expansion coefficients are uniquely determined. Coherent states are an overcomplete set of modes that are not orthonormal. You can expand things out in terms of them, but the expansion ...

3

You can use the general formula for denisty-of-states $$N(E)=2\sum_\alpha\delta(E-E_\alpha)$$ Now using our energy dispersion relation we get \begin{align*} N(E) &= 2\sum_{\alpha=n,k_x,k_y}\delta(E-E_\alpha)\\ &\approx 2\frac{L_xL_y}{(2\pi)^2}\int_{-\infty}^\infty dk_x \int_{-\infty}^\infty dk_y\sum_n\delta(E-E_n(k))\\ &= 2\frac{L_xL_y}{(2\pi)^2}\...

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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 ...

1

Condensation is the phenomena in which a macroscopic number of bosons occupy the same microscopic quantum state. Usually, the number of bosons occupying a state with energy $E$ in a system held at a fixed temperature $T$ is given by the Bose-Einstein function $$n(E) = \frac{1}{e^{E/T}-1}$$ However, another restriction is of course that all bosons in the ...

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I found the problem can be solved if one looks at the definition as following: \begin{align} f(\vec{q}+\vec{G},\vec{q}'+\vec{G}';\omega) & = \int d\vec{r}d\vec{r}' e^{-i(\vec{q}+\vec{G})\cdot\vec{r}} f(\vec{r},\vec{r}';\omega) e^{-i(\vec{q}'+\vec{G}')\cdot\vec{r}} = f_{\vec{G},\vec{G}'}(\vec{q},\vec{q}';\omega) \\ & \Rightarrow f_{\vec{G},\vec{G}'}(...

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There is no need to symmetrize or antisymmetrize the tensor product of the spin spaces. Each spin is tied to a particlular site in the lattice, so different spins are distinguishable. It would be different if the objects with spin were able to hop/tunnel from site to site. Then their statistics would matter.

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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 ...

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Let's start from the basic physical reasoning. The idea is that we want to find the general form of the wavefunction that has the symmetry of the problem. Let's say that we have a one-dimensional lattice with lattice spacing $a$. So, a wavefunction that shares the symmetry of the lattice would be invariant under translations by $na$ where $n\in\mathbb{Z}$. ...

1

This is actually the fourier transformation. Normally you see $$f(\vec{x})=N\sum_{\vec{k}}e^{i\vec{k}\vec{x}}f(\vec{k})$$ where N is a normalization parameter. This in terms of linear algebra changes your basis from eigenvalues of the position operator to eigenvalues of the momentum operator. The same thing happens here, but instead of functions you ...

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Answer to your question has been hinted in the comment.The spin-1/2 operator for the itinerant (or conduction) electron, $\mathbf{s}$ has expectation $\langle \uparrow | s_{z} | \uparrow\rangle = 1/2$ and $\langle \uparrow | s_{x} | \uparrow\rangle = \langle \uparrow | s_{y} | \uparrow\rangle = 0$, where $| \uparrow\rangle$ is the eigenstate of the $z$-...

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To see how this can happen mathematically, consider the matrix $$A=\left[\begin{array}{cc} 0 & 1\\ 1 & 0 \end{array}\right]$$ which is Hermitian and real (so in the standard basis). Changing basis means conjugating by a unitary, such as $Q$ where $$Q=\left[\begin{array}{cc} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}}\\ \frac{-i}{\sqrt{2}} & \... 4 Normally the index at an element tells me "how many atoms of that element there are", but in this case we are discussing values of x = 0.04 etc. so this does not make sense. That is still what it means. You just need to beat it with a rubber hammer until it starts making sense. In this particular instance, for example, it means that four atoms out of ... 2 If 100% of the medium is composed of the two elements A and B, than a easy ways to write the percentages p_A and$$x=p_B=1-p_A$$is A_{p_A}B_{p_B}. Thus, in your case it becomes A_{1-x}B_x. 1 if you have A an operator which is linear in bosonic operators, and you want to calculate its expectation value when the bosons are non-interacting, then$$ \langle e^A \rangle = \exp\left[\langle A \rangle + \frac{1}{2}\left(\langle A^2 \rangle-\langle A \rangle^2\right) \right]$$which is a consequence of Wick's theorem. 4 Bloch's theorem states that the eigenvalues and eigenvectors of a periodic one-electron Hamiltonian can be labeled by Bloch's wavevectors \bf k lying in the first Brillouin zone. For each \bf k there is an infinite set of solutions of the eigenvalue problem, labeled by another quantum number, n, the band index. The eigenvalues E_n({\bf k}) depend on ... 1 That depends strongly on many factors. The reason for the decrease in thermal conductivity at high temperature is that the mean free path goes down because of phonon-phonon scattering (interaction with the lattice). At some point the phonons do not propagate much further than a lattice period and the lattice melts. So those are the shortest mean free paths. ... 0 I cannot give you a comprehensive answer as I am no expert in the field. But in the effective monopole description that you talk about, the monopole is not a physical monopole that resides in real space, but it resides in the space of the parameter that we use to define the Berry connection. In the case of the band structure of Weyl semimetals, this should ... 1 Well, for what it's worth, OP's heuristic^{\dagger} zeta function evaluation works in principle:$$ \begin{align}{\rm Det}(i\frac{d}{dt}) ~=~&\prod_{n\in\mathbb{Z}}\lambda_n ~=~\prod_{n\in\mathbb{Z}}(-(n+1/2))\cr ~=~&\left[ \prod_{n\in\mathbb{Z}}(-1/2)\right]\left[ \prod_{n\in\mathbb{Z}}(2n+1)\right]\cr~\stackrel{(2)}{=}~& \prod_{n\in\mathbb{Z}...

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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 ...

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"Is it true that smooth surfaces can not be created without a harder material?" - no. Look up how diamonds are polished - the polishing surface is another diamond or diamond dust.

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It is true that in the context of pseudopotential theory (which has a much broader scope than DFT) people use the term "atomic configuration" in a slightly ambiguous way. Strictly speaking what they are speaking about is the electronic configuration ot the atom. For instance, the ground state of a neutral sodium atom is $1s^22s^22p^63s^1$ or, ore briefly $[... 2 About the second question of yours just note that the following worldline action $$S=\int d\tau \left(\dot{x}^{\mu}p_{\mu}+l^{\mu}p_{\mu}\right)$$ with$\mu=0$to$2$reproduces Chern-Simons perturbativelly. (For more details see this). About the first question of yours a good way to start is to realize that imposing the supersymmetry to be local leads ... 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 ... 3 Say you know$\psi_n(x)$. I will show that$\psi_n(x)$can be written as$e^{ikx}u_n(x)$for an infinite number of different values of$k$and$u_n$. Let's assume you know one decomposition$\psi_n(x)=e^{ikx}u_n(x)$. Bloch's theorem guarantees one such decomposition exists. Then we can also write$\$ \psi_n(x) = e^{i(k+\frac{2\pi N}{a})x} e^{-\frac{2\pi i N}{...

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