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-3

I had this same question. The proof and the missing steps are shown in the following book. You can follow it up. I'm providing a google book link and the name and page number of the book where the problem is solved. Book: Solid State and Quantum Theory for Optoelectronics Author: Michael A. Parker Google book link: ...


2

I have the feeling that there may not be a rigourous proof of the type you describe in your question. I remember a couple of years ago there was a proof that the close packed structures (ccp / hcp) gave the best 'space filling' characteristics. This proof by Hales was confirmed in August of 2014 The proof relies in part on computer checking that other ...


0

In classical physics, the canonical ensemble is defined over six-dimensional phase space $(\mathbf{x},\mathbf{p})$ only, i.e. three dimensions for coordinates and three dimensions for momentum. To answer your question of finding marginal energy density from full phase (state) space, it is sufficient to convert momentum density distribution $f(p)dp$ to the ...


0

http://en.m.wikipedia.org/wiki/File:HookesLawForSpring-English.png. I think this is also because of the spring constant which is I think is the gap present between the spring when it is coiled where the energy or the potential energy is stored and I don't think the atoms get affected


4

Firstly, you can deform material permanently..spring is no exception. On the atomic level, you are working against Coulomb forces that bind the material id est, that form the lattice. One primitive cell is well defined by the conditions of minimal energy. You can describe this potential as a quadratic, so you get harmonic forces, but it is not truly ...


8

You are asking two questions really 1) How is PE actually stored in a steel spring at the atomic level? The explanation for this lies in quantum mechanics 2) Could you explain in detail how/where potential energy is actually stored in a steel spring and why the material never surrenders to the bending forces taking a new shape? Replying to 1) ...


0

horizontal spring exerts a force $F = (−kx, 0, 0)$ that is proportional to its deflection in the $x$ direction. The work of this spring on a body moving along the space curve $s(t) = (x(t), y(t), z(t))$, is calculated using its velocity, $v = (vx, vy, vz)$, to obtain $$W=\int_0^t\mathbf{F}\cdot\mathbf{v}\mathrm\,{d}t =-\int_0^t kx v_x \mathrm\,{d}t = ...


1

You do not necessarily need Bloch states with the full translational symmetry to construct Wannier states. You can also do this in the presence of inhomogeneities and there is actually a well-defined approach as long as your band gap is well defined (and possibly beyond). The Wannier states at different sites may then have a different shape, but they remain ...


3

Strong correlation usually come with localized $d$ or $f$ orbitals in (or close to) a Mott insulating state, where the charge degrees of freedom is gapped by interaction, and the system become insulating even at half-filling. Silicon has no localized orbitals and it is a band insulator. Its valence band is fully filled, not half-filled, which means it can ...


1

Personally, I find it more intuitive to think in terms of the closely related quantity, the loss function, $-\text{Im}\frac{1}{\epsilon}$, rather than the optical conductivity. If one were to tune across a phase transition from a electronic liquid to an electronic solid, I suspect one would expect to see the softening of the free-carrier plasmon. Once the ...


1

Recall that the Fermi surface of the free electron gas, the total momentum is zero. How can electron gas be conductive? Well, when you apply the electric field, it will be non zero. The story is same for superconductivity.


3

Count: the central blue dot is shared by eight unit lattices.


0

Ok I think I got it: $$\langle \psi' | H| \psi\rangle =\sum_{n'=1}^N\frac{1}{\sqrt{N}}\langle n'|e^{-ikn' }(-\frac{J}{2}\sum_{i=1}^NS_i^+S_{i+1}^-)\frac{1}{\sqrt{N}}\sum_{n=1}^Ne^{ikn}|n\rangle+\sum_{n'=1}^N\frac{1}{\sqrt{N}}\langle n'|e^{-ikn' }(-\frac{J}{2}\sum_{i=1}^NS_i^-S_{i+1}^+)\frac{1}{\sqrt{N}}\sum_{n=1}^Ne^{ikn}|n\rangle ...


2

When a plane is parallel to two vectors, the dot product between its normal and that of the vectors is zero (they are perpendicular). So you are looking for a vector that is perpendicular to both $3\vec a + \vec c$ and $\vec b$. The answer, by inspection, is the vector $-\vec a + 0 \vec b + 3 \vec c$, from which the Miller indices are (if I remember this ...


1

To answer your question, one needs to understand a bit what is the Ginzburg-Landau (GL) formalism. Let us first recall the GL functional: $$F=\int dV\left[g\left|\left(\nabla-\dfrac{2\mathbf{i}e}{\hbar}A\right)\Psi\right|^{2}+a\left(T-T_{c}\right)\left|\Psi\right|^{2}+b\left|\Psi\right|^{4}+\dfrac{\left(\nabla\times A\right)^{2}}{2\mu}\right]$$ with ...


0

I think the “condensed matter physics” written by Michael Marder is a good one. The book is well writen and talks a lot about physics. He also have his lecture notes and syllabus onhiswebsite. You can start learning following his syllabus.


0

First of all I think we are thinking about the free electrons in the metal rather than the ones which are bound to individual atoms. Generally to apply a voltage we attach connections to each end of the metal and apply a potential difference. One way of thinking about what happens when we apply a voltage is that the fermi level gets slightly tipped from one ...


0

You should note that the index $j$ in the vector $\bf{r}_j$ denotes summation over nearest neighbors: vector can be decomposed: $$ \bf{r}_j = \bf{r}_i + \bf{r}_\delta $$ where $\delta$ denotes the nearest neighbors and change the summation from $\sum_{<i,k>}$ to $\sum_{i,\delta}$. You arrive at: $$ \sum_{k,k '} \sum_{i,\delta} e^{ik r_i - i k' r_i + i ...


1

Several sources give a higher (experimental) value of the Debye temperature for diamond - about 2220K: http://www.sbfisica.org.br/bjp/download/v03/v03a03.pdf , http://www.cvd-diamond.com/properties_en.htm , http://www.chm.bris.ac.uk/motm/diamond/diamprop.htm .


1

In answer to your first question: yes, as that figure shows. As for your second question, consider this: what would happen if you keep the diatomic basis but set the masses of the two atoms to be equal. The answer is that it'd look like the above figure but without a bandgap; it's the monoatomic case with the ends of the dispersion relation folded over to ...


1

My answer to a related question might help. The short version is that due to the periodic nature of the potential, a wave vector $k$ is identically the same as the wave vector $k+2\pi/a$. There are at least two ways of showing bands on a graph. One is the extended zone scheme which you have shown above. The other is the reduced zone scheme where we take ...



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