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3

A quote from http://en.wikipedia.org/wiki/Symmetry_protected_topological_order : The SPT order (for both frermionic and bosonic systems) has the following defining properties: Distinct SPT states with a given symmetry cannot be smoothly deformed into each other without a phase transition, if the deformation preserves the symmetry. However, they all can ...


1

You may start from the density matrix $\rho$ of the grand canonical ensemble (assuming $k_BT=1$ as the unit of energy), $$\rho=Z^{-1}e^{-H+\mu n},$$ where $Z=\text{Tr}\,e^{-H+\mu n}$ is the partition function. Then $\langle n\rangle$ and $\langle n^2\rangle$ are defined by tracing with the density matrix $$\langle n\rangle=\text{Tr}\,n\rho\text{, and ...


1

Yes, this is a known problem in the condensed matter community. There are two different definitions of short range entangled (SRE) states. The key difference is whether the fermion symmetry protected topological (SPT) state belongs to the SRE state. Strictly speaking, the fermion state is not SRE, because fermion statistics is already a phenomenon of long ...


4

This is a very good question. The same operator algebra does not imply that $H(J,h)$ and $H(h,J)$ have the same spectrum. As has been mentioned in Dominic's answer, even the ground state degeneracy is different under the interchange of $J$ and $h$ ($J\gg h$: symmetry-broken two-fold degeneracy, and $J\ll h$ unique ground state), therefore it is impossible to ...


7

There is quite a big controversy these days about the correct definition of the entropy in the microcanonical ensemble (the debate between the Gibbs and Boltzmann entropy), which is closely related to the question. Everyone agrees that the correct definition of the density matrix is given by $$\rho(E)=\frac{\delta(E-H)}{\omega(E)},$$ where $H$ is the ...


6

A quick answer The bottom line is the spontaneous breakdown of a global U(1) symmetry and the concomitant rigidity of this U(1) phase. By rigidity, I mean something reminiscent of a restoring force felt when one tries bending or distorting a solid stick, which, fundamentally, originates from the translational symmetry breaking in a typical crystalline ...


0

I've tracked a variety of quasiparticle issues for over 20 years now, including a couple of deep literature dives on fractionally charged topological solitons in materials such as polyacetylene. The term quasiparticle has always dominated in the articles I've found. That's true even for integer spin excitations, which after all are quite particle-like in ...


1

Taking a step back, I'd suggest a look at Ashcroft and Mermin's "Solid State Physics" where they treat the harmonic crystal modes (chapter 22 in my edition). Nowhere do they suggest that LO-TO splitting occurs only in ionic solids. Instead, they make it clear that a Bravais lattice with a mono-atomic basis has acoustic modes only. Once a poly-atomic ...


1

1) In general, an algebra can have many representations. In this case, however, if you assume that there is a unique joint +1 eigenstate of the $\sigma_i$'s, that determines the representation uniquely. [All the other states can be found from this state by applying products of $\sigma_i^x$to it. And from the anti-commtation of $\sigma_i^x$ and ...


5

In all theories (London, Ginzburg-Landau, Bardeen-Cooper-Schrieffer, Bogoliubov-Gennes) the Meissner effect is accounted for as the constitutive relation $j\propto A$ : the current is proportional to the vector-potential. (The proportionality factor depends on the system of units, so let us forget about that here.) For the London's phenomenological theory, ...


4

Here is an algebraic approach to understand the edge state. Let us start from a generic Dirac Hamiltonian for the bulk fermions in the $d$-dimensional space. $$H=\sum_{i=1:d}\mathrm{i}\partial_i\alpha^i+m(x_i)\beta,$$ where $\alpha^i$ and $\beta$ are anti-commuting gamma matrices ($\{\alpha^i,\alpha^j\}=2\delta^{ij}$, $\{\alpha^i,\beta\}=0$, $\beta\beta=1$), ...


3

Because $|\phi|^2=\rho$ is the boson density, and you expect to have finite boson density in the Mott phase (like one boson per site), so you can not just let the boson field $\phi$ goes to zero. In fact, there are two ways to disorder a superfluid and restore the U(1) symmetry, either by the amplitude fluctuation or by the phase fluctuation. The ...


3

The topological insulator has the charge $U(1)$ symmetry, while the topological superconductor does not. In fact, this difference is general, and applies to all insulators and superconductors. In condensed matter physics, insulators and superconductors are both gapped fermion systems. Depending on whether the charge $U(1)$ symmetry is presented or not, they ...


4

For the sake of simplicity, let us limit this discussion to non-interacting fermionic phases: e.g. band insulators with electron quasiparticles and fully gapped superconductors with Bogoliubov quasiparticle excitations. Classification of interacting and/or non-fermionic phases is still work in progress. Before outlining basic differences between topological ...


0

(I assume this is a light or neutron scattering question.) In the Guinier regime, the intensity curve follows an exponential decay: $I(q)\propto \exp[-(qR_g)^2/3]$. If you plot the (natural) log of this vs. $q^2$, you can get the radius of gyration from the slope.


1

Since the superconductor is not dissipative (at least for very low frequencies), it will not generate thermal noise the same way a resistor does. However, every superconductor of finite length forming a loop of non-zero area has an inductivity L. Just like capacitors held at a finite temperature will generate charge fluctuations for a charge measurement of ...


0

The answer to my question is Yes. Embarrassingly, one of the simplest examples is given by the fermionic quasistring topological order I described in my paper http://arxiv.org/abs/1404.4385 . The magic is that the 5th oriented bordism group is generated by the mapping torus of complex conjugation on CP^2. Thus, if we consider the fermionic quasistring top ...


1

As the size of the lattice gets large, the superposition $$\vert \psi \rangle = \frac{1}{\sqrt{2}}\left(\vert \uparrow \uparrow \uparrow \cdots \rangle + \vert \downarrow \downarrow \downarrow \cdots \rangle\right)$$ in the Ising model becomes unstable to environmental perturbation. Just like Schrodinger's cat (a superposition $\vert \mathrm{live} ...


0

I thinking a lot of time, then I came in this assertion: The kinetic part of the Anderson's hamiltonian have a continuous spectrum. We want to work numerically with this part, so we need to find a some basis that diagonalizes approximately (the terms off-diagonal be small by controlling some discretization parameter) in such a way that now the "spectrum" is ...


1

Your question is deep and to the best of my knowledge only partially resolved, see Sec. II of 2007 article in Review of Modern Phsyics. The trick is called "the Wilson chain" and is essential for NRG to work on strongly correlated models with Kondo-like behavior. The Wilson chain construction makes NRG a computationally efficient procedure. My ...


2

Unfortunately, the premise is wrong. In general, lower dimension DOS can't converge to higher dimension DOS. That's one primary reason why 2D devices are interesting. For example, the van Hove singularities of the DOS can diverge in 2D, but only their derivatives can diverge in 3D.


2

[As requested, I convert my comment into an answer, as it might also be useful for other people.] There is a very interesting series of works by Lieb and Yngvason on entropy and the second law of thermodynamics, based on the kind of axiomatic approach you seem to be interested in. You can start with this introductory paper, or this, this or this more ...


3

The constant $C$ is not a part of the single-particle Hamiltonian. It is what is called the vacuum energy, and has no observable effect unless we look at gravitation. To be specific, adding this constant to the Hamiltonian makes $e^{-\beta H}\,\rightarrow e^{-\beta C}e^{-\beta H}$ and $Z=\rm{Tr}(e^{-\beta H})\, \rightarrow\, e^{-\beta C} Z$. That is, the ...


4

Ultimate physical motivation Strictly in the sense of physics, the entropy is less free than it might seem. It always has to provide a measure of energy released from a system not graspable by macroscopic parameters. I.e. it has to be subject to the relation $${\rm d}U = {\rm d}E_{macro} + T {\rm d} S$$ It has to carry all the forms of energy that cannot be ...


1

Thermodynamically entropy is defined by \begin{equation} \mathrm{d}S = \frac{\mathrm{d}Q_{rev}}{T} \, ,\end{equation} where $\mathrm{d}Q_{rev}$ is the heat, transferred reversibly. As you point out it can be shown that this quantity is a function of state. This implies that the entropy of any thermodynamic system has, up to a constant, a well defined ...


1

As you have in the commutation relations, $\sigma_i \sigma_j= \sigma_j\sigma_i$ e.g. spin operators on different sites commute, so there is no minus sign to pick up.


0

The first thing to realise here, is that in your Hamiltonian formulation, the lattice constant is implicitly stated as being $b=1$ (I'm calling it b, not to confuse it with the lattice operators). Next you can identify, if you're working in two dimensions, $x=am$ and $y=bn$ yielding $$\hat{H}_0=-J(\sum_{(x/b),(y/b)} ...


1

Well, when did you want it to be written down? As it is, it was pretty much simultaneously and independently arrived at by Hubbard, Kanamori, and Gutzwiller an awful long time ago. Probably others too. The point is, it was written down when there were experimental phenomena that justified including interactions in the model. It wasn't some great conceptual ...


1

min Zhang, The Hubbard model (offen reffered to as the U and J terms in ab-initio DFT or tight-binding models) is a little bit more complicated than it looked like. It is indeed an additional energy that you add locally to some states (d or f bands usually) to locallized them. Usually you want to do this because DFT tends to spouriously delocallize ...


1

Absorption isn't an instant event. At the level of simple quantum mechanics, this system can be described as follows. Evolution of electron in crystal is governed by Schrödinger's equation. External electromagnetic field, namely the light which we shine on the crystal, is a periodic addition to the Hamiltonian. When you start shining light at the crystal, ...


0

I would say in the same way a thick beam in a structure "knows" that you are trying to tear it off: it is temporarily displaced somewhat due to your effort, but then returns to its stable state. So the electrons are not totally immune to the photon (for example, the crystal acquires some minuscule momentum due to light pressure, if the photon is reflected ...


0

If you have the distance between atoms in a cubic lattice, then the (1,1,1) planes are the planes you get when you move one atom in X, one in Y, and one in Z. There is a nice presentation on this at https://nanohub.org/resources/5488/download/ch3_p3.pdf The following picture is taken from that presentation, and answers your question exactly:


0

Well I had an extensive discussion with experimentalist friend and he tells me that its the unfilled electrons i.e d-orbitals...... yea it can be more than spin half but we are discussing hopping in a lattice then it is a single electron with spin half and in condensed matter(since we cannot have more than one electron to hop from the same site) it is the ...



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