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The poles of the spectral density give you the particles of the system, whether they are fundamental or bound states. There is no fundamental difference between these two types: they behave exactly the same, phenomenologically, and they are indistinguishable theoretically. This becomes even clearer in light of the concept of duality: the same theory often ...


4

There's actually no contradiction between what you say and what the two linked documents say. In fact, the two documents describe the solution of the wave equation in the form of a tight-binding solution, which is indeed a Bloch wavefunction, but it's constructed as the sum of wavefunctions with terms centered in the two sublattices. Recall that the idea of ...


4

You may want to look for a discussion in Hakens "Quantum field theory of solids". In principle, one can quantize any field, by expanding it in its eigenmodes and imposing the commutation relations. The quantization procedures are thus the same in the QFT and the solid state, where one has such effective particles as "free" electrons, ...


4

They overlap at least, and they may even be synonymous, depending on who you ask. Specialists in different fields often end up borrowing each other's ideas and methods, and the lines between them often become blurred or found to be nonexistent in the first place. Refs 1, 2, 3, and 4 are just a few of the many examples. And by the way, the author of ref 1 is ...


4

For most semiconductor, e.g. GaAs, a exciton has binding energy about 6 meV ($0.006 eV$) and with a binding length about $200 \dot{A}$. That is a very loosely binded state, can be broken into free electron and hole, easier to propagate out of the surface region. On the contrary, an organic exciton is usually located within the molecule and has a binding ...


3

The gist of such self-consistent is the (implied) assumption of a unique solution. If we indeed have a unique solution then the self-consistent solution we find under certain assumptions is indeed the correct one. In the case you discuss here it is indeed the case, and you can show this by writing the equation for the total number of particles as a function ...


3

There is no direct correlation between the band structure of a covalent material and the directional character of the corresponding Wannier functions. The reason is quite general. There is no direct relation between the spectrum of the eigenvalues and the spatial properties of the eigenfunctions. The only thing which is possible to establish easily is how ...


3

At $k=0$, acoustic phonons correspond to macroscopic translation of the entire crystal, which naturally are completely in phase and cost zero energy. This behavior is essentially by definition and is always true as acoustic phonons are the Goldstone modes of translation. In other words, the energy of the crystal should be the same if it is located in London ...


3

Bound vs. extended states In condensed matter physics bound states are the states with the wave functions decaying towards infinity, as opposed to the extended states, as, e.g., plane waves. In this sense the definitive answer to the question can be given only by the exact diagonalization of the Hamiltonian and studying the behavior of the wave functions. ...


2

The assumption behind the question is that materials have finite conductivity due to electron scattering from impurities and the imperfections of the crystal lattice. This is not quite the case. Firstly, the scattering from impurities and crystal imperfections is coherent scattering, so, in principle, it doesn't cause any dissipation, unless combined with an ...


2

you can use a Gruneisen second order approximation for the zero-pressure equation of state [Vocadlo L, Knight K S, Price G D and Wood I G 2002 Phys. Chem. Miner. 29 132]. In order to apply this equation you must have at least some unit cell volume values at different temperatures, you must know the Debye temperature and the bulk modulus value of your ...


2

The atomic mass is either (i) the mass per atom, as measured in amu, or (ii) the mass per mole, as measured in grams. As a result, since $\rho_m$ is specified to be in grams per cubic cm (rather than amu per cubic cm), $\rho_m/A$ is the number of moles per cubic cm.


2

In your second approach, you correctly solve the hardcore boson problem (in essence, hardcore bosons are exactly two-level spin systems). In your first approach, on the other hand, you diagonalize the single-particle problem. As an approach to solve the many-particle problem, where you subsequently fill all the negative energy modes, this only works for non-...


2

After reading the comments and searching a bit more, I think I've managed to find the solution. Considering the two displacements $r_1$ and $r_2$: applying the projection formalism should give, for $r_1$ or $r_2$, $$P r_1 = \frac{1}{4}\left(D(E)+D(C2)+D(\sigma_v)+D(\sigma'_v)\right) r_1$$ $$\Leftrightarrow P r_1 = \frac{1}{2}(r_1 + r_2)$$ which indicates us ...


2

The Wikipedia page links to the paper by Parkin and Mauri (1991) about the coupling of magnetic layers through a ruthenium spacer. The effect is seen up to 30 Å.


2

The band structure is the relation between energy and momentum. It will take the actual crystal orbitals to specify the actual wave function. From these you can control the Wannier orbitals. However, band structure calculations are based on translational symmetry and in this formalism it is not possible to include local correlations that are important in ...


2

The formula holds, because there is a sum over all ${\bf p}$. After that sum is performed, ${\bf v}\cdot{\bf v}'$ is essentially the only vector structure that is consistent with rotation symmetry. To prove the formula, first rewrite it so that the ${\bf v}$ and ${\bf v}$' are outside the sum: $$\sum_{{\bf p}}({\bf p}\cdot{\bf v})({\bf p}\cdot{\bf v}')F(p^{...


1

The Bohr-von-Leeuwen theorem applies to the absence of classical diamagnetism i.e to the magnetic effects on the classical theory of electron motion. On the other hand paramagnetism arsises from the field-induced alignment of the electrons instrinsic magnetic moment, an effect ignored in Bohr-von-Leeuwen. The electron's intrinsic magnetic moment is a ...


1

How about that: The left hand side of the equation has the dimension energy, while on the right hand side there is a $L^3$ (from the integral) times $e^2\,n^d$. The dimension of $n^d$ is $L^{-3d}$ and the dimension of $e^2$ is given by $M\,L^3\, T^{-2}$. Hence we obtain the dimension of $e^2\,n^d$ as $$ M\, L^3\, T^{-2}\, L^{-3d} \quad.$$ The right hand side ...


1

Roughly speaking, the Casimir effect occurs because the spacing between the plates is so small that it excludes electromagnetic waves from occupying that space. This exclusion process goes away for large spacings (EM waves have no difficulty getting in there) and gets stronger as the spacing gets progressively smaller (EM waves can't easily squeeze ...


1

These expressions are valid in finite systems (in any simply connected box), but the duality transformation maps a model with $+$ boundary condition to a model with free boundary condition, which affects the finite-volume free energy. Of course, as you say, in the thermodynamic limit, the boundary condition becomes irrelevant and one obtains a nontrivial ...


1

https://en.wikipedia.org/wiki/Bloch%27s_theorem are the eigenstates of an electron in a spatially periodic potential, such as that of a crystal lattice: $$ \psi(\mathbf{r}) = e^{i\mathbf{k}\mathbf{r}}u(\mathbf{r}). $$ That the eigenstates in a crystal have the form of the Bloch states can be shown by quite general symmetry arguments. However, thsi leaves ...


1

The bulk periodicity doesn't refer to the simplest structure on the surface, it refers to the lattice vectors within the bulk of the crystal where the effects of surface reconstruction are negligible (or alternatively, an ideally terminated surface before it relaxes into a reconstructed configuration). Consider this diagram from Wikipedia: In this instance,...


1

The Coulomb gauge condition $\nabla\cdot A$ does not uniquely specify a gauge. For any given field configuration, there are still infinitely many choices of $A$ which have that property. One says that there is still *residual gauge freedom * after imposing the Coulomb gauge condition. Normally this freedom goes away if you demand that the fields vanish at ...


1

As long as we are lacking any fundamental physical definition to a "states of matter", we will have arbitrary amount of these states of matter. This is well discussed in this question; What determines a state of matter? Condensate is practically a matter, where molecules holds their relative positions to each other in same order atleast over short ...


1

Let me give a completely mathematical description, since I think the accepted answer is more physically motivated. Let $U\in SU(2)$. Then its second-quantization $W_U$ on the $n$-particle Fock space would be just $U\otimes\cdots\otimes U$. In particular, if $U=e^{-iS}$ for some $S\in i \text{su}(2)$ (its Lie algebra), then $W_U = \exp(-ic^*S c)$ where $c^* S ...


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