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4

Nothing in the laws of thermodynamics forbids multiple liquid phases for a single substance. The only limit is the simultaneous coexistence of at most three phases (at triple points). Water has a solid-liquid-gas triple point and several soid-solid-liquid and solid-solid-solid triple points; see the phase diagram of water and ice. In addition, although not ...


1

There is actually only one disordered phase - from a physicist's perspective, the liquid and the gas are actually the same phase because one can continuously vary the external parameters (temperature and pressure, in this case) to get from the liquid to the gas without passing through any phase transition, because the phase transition line terminates within ...


1

Technically speaking all solids have crystalline structure. Anything that is truly amorphous is known as supercooled liquid (such as glass). Depending on how well crystals are oriented they can be divided into two categories, as pointed out by @lemon, single crystals and polycrystals. Single crystals have nearly perfect orientation of sizes few mm long. ...


2

Solids are rigid and resist a change in volume. There are three general types of solid: As you can see from the picture, a crystal is a solid for which the atoms/molecules are highly ordered to form a periodic lattice. Examples of (poly)crystalline solids include diamond and table salt. Examples of non-crystalline solids include glass and amorphous ...


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 ...


12

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 ...


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 ...


0

A eigenstate of a crystal hamiltonian can be written as a Bloch function in space representation $$ \psi(\mathbf{r}) = e^{i\mathbf{k}\mathbf{r}} u_\mathbf{k}(\mathbf{r}) $$ $u$ is periodic with respect to the unit cell. The momentum is now given by $$ \langle \psi|\hat{\mathbf{p}} |\psi\rangle = -i\hbar \int e^{-i\mathbf{k}\mathbf{r}}u_\mathbf{k}^*(\mathbf{...


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

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 ...


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 ...


-2

Most objects are porous and allow the elector-magnetic signal through. Little electrons can pass through what appears to be solid material like photons through glass. Some are still trapped in the atoms. Denser metals like lead do not have enough space between the atoms to allow the electromagnetic waves through. Certain wavelengths can pass through certain ...


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 ...


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

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 ...


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-...


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}...


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 ...


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 ...


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

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 + |\...


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 ...


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

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

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 ...


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 ...


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

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 ...


0

There is a lot of experimental evidence showing that only a tiny fraction of the electrons close to the Fermi surface participate in the physical properties of a metal. One of them is the specific heat due to electronic degrees of freedom : at a certain temperature $T$, only electrons within a typical energy of $k T$ around the Fermi level can be excited by ...


0

the heat transfer coefficient (h) is equal to the thermal conductivity (k) divided by the thickness of the object. $$ h=\frac{k}{l} $$ the units for h are $[W/m^2 K]$ the units for k are $[W/mK]$


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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