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7

The most immediate answer would seem to be that a great variety of different crystal phases can exist because their long-range order makes it possible to classify them based on the different symmetries of their lattice structure. Since the liquid (or amorphous solid) phase only has short-range order and the gaseous phase doesn't even have that, it seems ...


5

This is essentially a result of the equipartition theorem where each degree of freedom contributes $k_B T/2$ to the energy. Given that the specific heat in this context is just ${\partial E}/{\partial T}$ then each degree of freedom contributes $k_B/2$ to the specific heat. For the classical model of lattice vibrations in solids this leads to the Dulong-...


5

While the trace is invariant under a transform to another basis, you need to take into account here that the coherent state basis is not an orthogonal basis and it is overcomplete. We can evaluate the trace of an operator $A$ by inserting identity operators in front of and after the operator and then using resolution of identity in terms of the coherent ...


3

I tend to agree with Sanya in that I am not sure about the universality of this. There might of course be instances where this is the case. A pure metal has a periodic lattice of ions. There is then a conduction band of electrons that fills the space between the ions. These electrons have wave vectors in the reciprocal space. In space the occurrence of ...


2

The first term (sum) in $\bar H$ obviously commutes with all $\sigma_x$ variables because it's a function of $\sigma_x$ only and they commute with each other. The second term (sum) in $\bar H$ also commutes with the product of all $\sigma_x$ because the first term in the summand is a $c$-number and the second term $\sigma_z^j \sigma_z^{j+1}$ anticommutes ...


2

The given ansatz includes two assumptions. (1) The approximate ground states is seeked on a subspace of the Hilbert space consisting of rotated versions of single constant vector. (This subspace is a $2$-sphere parametrized by a unit vector in $\mathbb{R}^3$ (2) The value of the spin projection in the direction of the unit vector is half of the number of ...


2

Necessary and sufficient condition for conjugacy: see Sylvester's law of inertia. In your notation, the two symmetric matrices $K$ and $K'$ are conjugate by a transformation $W^TKW = K'$ if and only if they have the same number of positive and negative eigenvalues (no null eigenvalues since both are nonsingular). The eigenvalues themselves need not be ...


2

No, you don't need to work in the basis where the Hamiltonian is diagonal. It's a fact of linear algebra that the sum of the diagonal elements of a matrix is the same no matter what basis you're in, so you can easily evaluate the trace in whichever basis is convenient.


2

Mostly kinetic energy. The kinetic energy of a free particle is not quantized. It becomes so when the particle is closed in a box. But even in this case the energy levels are often so closely spaced that the spectrum is almost continuous. In fact, if you solve the Schroedinger equation for a particle in a 1D infinite square well you will find the following ...


2

Let $[G,H]=0$, and consider an eigenstate $|\psi\rangle$ of $H$, $$ H|\psi\rangle = E\psi\rangle\ . $$ Then, $|\psi'\rangle := G|\psi\rangle$ is also an eigenstate of $H$ with the same eigenvalue, since $$ H|\psi'\rangle = HG|\psi\rangle = GH|\psi\rangle = EG|\psi\rangle = E|\psi'\rangle\ . $$ Thus, also $|\chi\rangle = \tfrac{1}{\sqrt{2}}(|\psi\rangle + |\...


2

1) The reduction they are referring to is explain in the middle of page 13: "We now write (4.5) in the invariant sector as a sum of $n/2$ commuting $2×2$ Hamiltonians that we can diagonalize." 2) They are "reducing" the difficult problem of diagonalizing a very large matrix to the much easier problem of separately diagonalizing $n/2$ different $2 \times 2$ ...


2

\begin{equation} b_{j} = \frac{1}{\sqrt{n}}\sum_{p}e^{-i\pi pj/n}\beta_{p}\qquad b_{j+1} = \frac{1}{\sqrt{n}}\sum_{p}e^{-i\pi q(j+1)/n}\beta_{q} \end{equation} Then \begin{equation} b_{j}^{\dagger}b_{j+1}^{\dagger} = \frac{1}{n}\sum_{p,q}e^{\pi i(p+q)j/n}e^{\pi iq/n}\beta_{q}^{\dagger}\beta_{p}^{\dagger} = \sum_{p} e^{-\pi ip/n}\beta_{-p}^{\dagger}\beta_{p}^{...


2

Ordinary silica glass really is amorphous. You can study the short range order of a material by measuring its radial distribution function. That is, take any atom and plot the density as a function of distance from that atom. This is easily measured using X-ray or neutron scattering. In a crystal the RDF is just a function of the crystal structure and is ...


2

First of all, I want to see an (experimental) proof that any metal has a higher resistance than any alloy (at any pressure, temperature and volume). What I presume your teacher might have wanted to hear is something along the following lines: a perfect, perfectly static crystal would be, if I remember correctly, perfectly transparent to an electron, so there ...


2

The fractional quantum Hall effect (FQHE) is a physical phenomenon in which the Hall conductance of 2D electrons shows precisely quantised plateaus at fractional values of e^2/h . It is a property of a collective state in which electrons bind magnetic flux lines to make new quasiparticles, and excitations have a fractional elementary charge and possibly ...


2

When you take the limit $q \rightarrow 0$, your formula becomes $$\chi( 0,0) \sim - \sum\limits_{\bf{k}} \frac{d f(\varepsilon_k)}{d \varepsilon_k}$$ where $f$ is the occupancy number of the electronic states. This susceptibility represents the response of the system to some external perturbation. Usually, this perturbation tends to displace electrons to ...


1

The susceptibility quantifies the response of a material to an external electric field due to the redistribution of electronic charge within the material. The formula you give for the charge in a particular electronic band (in practice the total susceptibility comes from the contributions from all bands). It is known as the Lindhard model and is derived from ...


1

It looks like the sum appears because of the completeness relation $$\mathbf{1} = \sum_{\alpha} \left| \alpha \right> \left< \alpha \right|$$ taken over all the $\{ \alpha_j \}$ to give $$\mathbf{1} = \sum_{\{ \alpha_j \}} \left| \alpha_j \right> \left< \alpha_j \right|$$ Starting with the first expression, which I believe you corrected in ...


1

Your question is really broad. As correctly pointed out by @John Rennie there can be so many signals that can pass from the solids. It must be noted that even if the signal can pass through solid it will experience certain losses. Mechanical waves such as sound took advantage of elasticity of the material. The sound oscillations are transferred through ...


1

The edge states of a topological insulator are not superconducting, because the current is carried by ordinary electrons, not by a supercurrent of condensed Cooper pairs. The electrons in the edge state of a topological insulator are indeed prohibited by time-reversal symmetry from scattering off a stationary impurity (elastic backscattering). On the other ...


1

For $\epsilon=0$, $W$ may contain true eigenvectors of $A$, which can be arranged to be real (and only these matter for boundary modes). But in general the rows of $W$ concatenate real and imaginary parts of the complex eigenvectors of $A$. Say $$ Au_j = i\epsilon_j u_j\\ Au^*_j = -i \epsilon_j u^*_j\\ u^\dagger_j u_k = u^\dagger_j u^*_k = 0\;\;\text{for}...


1

To see that there are two sector, corresponding to the eigenvalues of $G$ note that $G^2=1$ since $$ G^2 = (\prod^n_{j=1}\sigma^{(j)}_x)^2 = \prod^n_{j=1}(\sigma^{(j)}_x)^2 = 1$$ Thus there are two eigenvalues to $G$ that are $\pm1$. These sectors need not be of the same size. Consider just 2 spins in theire singlet and triplet configurations. The singlet ...


1

The energy spectra of Equation (13) is the bulk energy spectra as a function of the momentum $q$. It describes the energy levels of the chain where the first and the last sites 1 and $N$ are connected in a ring. The spectra can be obtained in three steps: 1) first, Fourier transform Eq. 4, so that you will obtain a Hamiltonian in momentum space, whose ...


1

Luttinger liquids are a case of bosonization. Two fermions $\psi$ and $\psi'$ are written according to a boson field $\phi$ $$ \psi^\dagger\psi' = exp(i\phi), \psi'^\dagger\psi = exp(-i\phi^\dagger), $$ where the "field" $\phi$ serves as a phase and is real valued. For the fermion operator $c_j$ the number operator $n_j = c_j^\dagger c_j$ permits us to ...



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