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41

Arnold Neumaier's comment about statistical mechanics is correct, but here's how you can prove it using just thermodynamics. Let's imagine two bodies at different temperatures in contact with one another. Let's say that body 1 transfers a small amount of heat $Q$ to body 2. Body 1's entropy changes by $-Q/T_1$, and body 2's entropy changes by $Q/T_2$, so ...

27

From a fundamental (i.e., statistical mechanics) point of view, the physically relevant parameter is coldness = inverse temperature $\beta=1/k_BT$. This changes continuously. If it passes from a positive value through zero to a negative value, the temperature changes from very large positive to infinite (with indefinite sign) to very large negative. ...

13

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 ...

10

Let me try to answer my own questions to thank those who gave voice and support for undeleting this question. My main reference is the chapter 3 Basic quantum statistical mechanics of spin systems of the unfinished book "Modern Statistical Mechanics" by Paul Fendley. The spin-1 valence-bond-solid state (VBS) in Fig. 2 can be imagined as the following: ...

8

After thinking about it I must say it is not as simple as I thought it would be. The JW transformation on the transverse Ising model contains quite a few subtleties. So to proceed, 1) Take your ground state for ANY $h$ expressed in the spinless fermion language. I stress ANY because this condition is true always - it's not just for $h<1$. Now this is ...

8

The name DMRG is somewhat of a historical accident, and its modern day incarnations are not directly linked to the renormalization group or phase transitions. Instead, it is better understood as a variational technique based on matrix-product states (MPS) ansatzes. Still, there is a historical link between the two tools, and it can be useful to know about it....

7

Take a hydrogen gas in a magnetic field. The nuclei can be aligned with the field, low energy, or against it, high energy. At low temperature most of the nuclei are aligned with the field and no matter how much I heat the gas I can never make the population of the higher energy state exceed the lower energy state. All I can do is make them almost equal, as ...

6

The spin of a quasiparticle can be determined from a number of ways: If the quasi-particle is a "compound" object, you just add the individual spins according to the appropriate rules for adding angular momenta. An example would be the polaron, which is an electron dressed with a bunch of phonons. The electron has spin $1/2$, the phonons have spin $0$, so ...

4

Never mind, I got it figured out. For those interested: Applying $H$ to $|\psi>$, the result can be written as $\sum_{1\leq n_1< n_2\leq L}^L \alpha(n_1,n_2) |n_1,n_2>+\sum_{n_1=1}^L\beta(n_1) |n_1,n_1+1>$, where $\alpha$ and $\beta$ are functions containing "illegal" terms like $f(n_1,n_1)$. The next step would be demanding $\alpha(n_1,n_2)=... 4 It is correct that if you proceed the way you describe it, you obtain a 4-local parent Hamiltonian$H=\sum h$, where$h\ge0$, and$h\vert\Psi\rangle=0$, where$\vert\Psi\rangle$is the AKLT state. For a parent Hamiltonian constructed this way, one can show (for an arbitrary injective MPS) that the ground state is unique with a gap above. However, if you ... 3 Ah, but who says that negative absolute temperatures exist at all? This is not without its controversies. There's a nature paper here which challenges the very existence of negative absolute temperatures, arguing that negative temperatures come about due to a poor method of defining the entropy, which in turn is used to calculate the temperature. Other ... 3 The canonical example for MPS (in fact, the first MPS ever) is the AKLT model (http://journals.aps.org/prl/abstract/10.1103/PhysRevLett.59.799, https://projecteuclid.org/euclid.cmp/1104161001). The 2nd reference also discusses the 2D (=PEPS) version of the state. Another example of an exact MPS/PEPS model are (nearest-neighbor) RVB states (https://arxiv.org/... 3 Haldane chain is an one-dimensional spin-1 XXZ chain, with Hamiltonian:$H = J \sum_{i} (S_i^+ S_{i+1}^- + \mathrm{H.c.}) + V\sum_{i} S_i^z S_{i+1}^z + U \sum_i (S_i^z)^2$With certain values of the parameters($J,\,U,\, V$), it is in a symmetry-protected topological (SPT) phase, with novel properties, e.g. symmetry fractionalization, etc. It is in the ... 3 Usually, the argument relies on the fact that there is a band of spinons (i.e., a continuum of modes) above the ground states, so I don't see how your argument above works. One then argues that those modes are occupied according to the Bose-Einstein distribution, and one then computes the correction to the magnetization from these modes, and finds that the ... 3 To expand on LubošMotl's comment, see the following classic paper by Lieb, Schultz and Mattis. For one-dimensional systems and nearest neighbor interactions, the spin chain that you mention as an example in the comment can be converted into a free fermionic model. See section II in the above paper for details. 3 In one dimension, MERA naturally capture critical systems (i.e., systems with power-law decaying correlations and a log-divergence in the entanglement entropy). MPS (i.e., one-dimensional PEPS), one the other hand, have exponentially decaying correlations and a constant entanglement entropy. (Note: This is for a constant bond dimension and does not preclude ... 3 The gap of the parent Hamiltonian does not depend on the number$D$of blocks (at least not directly). The spectral gap of the parent Hamiltonian in the block-injective case is analyzed and a lower bound is given in B. Nachtergaele, Commun. Math. Phys. 175, 565 (1996), arXiv:cond-mat/9410110. 3 Your mistake is in claiming that that is 'the' string order. There is no such unique string order; rather, it depends on the phase of matter one wants to probe. Let me illustrate the point with an example that is more familiar: suppose someone tells you that 'the' symmetry-breaking order parameter is $$\lim_{|i-j| \to \infty}\langle S^x_i S^x_j \rangle.$$ ... 3 For completeness I'll summarize the answer here. After a fun conversation in the comments, we saw that it will be more illuminating to write $$H=-\sum_ {i=1}^{N-1} \sigma_i^x \sigma_ {i+1} ^x + h \sum_ {i=1} ^N \sigma_i^z$$ as $$H=-\sum_ {i=1}^{N-1} \sigma_i^x \sigma_ {i+1} ^x + h\left( \sum_ {i=1} ^{N-1} 1_1 \otimes \cdots 1_{i-1} \otimes \sigma_i^... 3 Take a product state |\psi\rangle=|+,+,\dots,+\rangle, on N spins. Then, to write it in the first form takes a tensor$$ \psi_{i,j,\dots} $$with 2^N non-zero elements. For N=100, there is no way this fits in your computer. On the other hand, as an MPS, this can be written with$$ A_i^{[s]}=\frac{1}{\sqrt{2}} $$for all s, so you only need 2N ... 3 I'll go ahead and say that one can roughly classify systems into Systems with a unique ground state. Systems that have multiple (possibly infinitely many) degenerate ground states, but that will tend to select a unique one by some mechanism. Systems with topological ground state degeneracy. (I won't rule out there existing additional classes, and I'd ... 3 Now I believe, that it's hard to assign some physical meaning to this value, but it can be used to calulate the correlation (as pointed by @lcv), that is:$$ Corr(S_i^z, S_j^z) = \frac{\langle S_i^z \cdot S_j^z \rangle - \langle S_i^z \rangle \cdot \langle S_j^z \rangle}{\sqrt{\langle {(S_i^z)}^2 \rangle - \langle S_i^z \rangle ^2}\sqrt{\langle {(S_j^z)}^2 \... 2 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$\...

2

I should say that you have 3 related questions, namely 1) To what extent can we trust the approximations based on HP and Jw transformations, 2) The nature of the low excitation spectrum and 3) The relation with Goldstone modes. We shall look first at the Holstein-Primakoff method. The spin ladder operators for at a site $j$ are given by $S^-_j = \sqrt{2S}... 2 I don't see how you get mixed terms. Your$Uacts on every other site. For instance (as operators acting on different sites commute) $$\ldots+ U \sigma^{2i}_x \sigma^{2i+1}_x U^{-1} +\ldots= \ldots \sigma^{2i}_z \sigma^{2i}_x \sigma^{2i}_z \sigma^{2i+1}_x \ldots=\ldots i \sigma^{2i}_y \sigma^{2i}_z \sigma^{2i+1}_x \ldots=\ldots-\sigma^{2i}_x \sigma^{2i+1}... 2 Here is one example. The book Quantum Inverse Scattering Method and Correlation Functions by Korepin, Bogoliubov and Izergin introduces the coordinate Bethe ansatz first for the 1d Bose gas (Chapter I.1), with wave function governed by the non-linear Schrödinger equation$$i \ \partial_t \Psi = -\partial_x^2 \Psi + 2\ c\ |\Psi|^2 \ \Psi \ ,$$which is very ... 2 How about a two-particle problem? See the model in this paper Tamm-Hubbard surface states in the continuum. The model is very simple but not so trivial. It is also a beautiful model as it can be solved exactly by Bethe ansatz---Bethe ansatz in the baby form. I think it can make a very good exercise in solid state physics. It is a variant of the model in ... 2 Remember that the Hamiltonian involves a sum over all pairs of neighbouring sites. Assume that the sites are located on a square lattice so that their positions are given by \vec r=a(n_x\vec e_x+n_y\vec e_y) where the n's are integer. The Hamiltonian reads$$\eqalign{ H&=-J\sum_{\vec r} \big(\vec S_{\vec r+a\vec e_x}.\vec S_{\vec r} +\vec S_{\vec ... 2 Yes and no. The quantum Ising model in 1+1 dimensions has a phase transition as you vary the coupling in the Hamiltonian. This model is directly related to the statistical mechanics of the classical Ising model in 2 spatial dimensions via the usual correspondence between quantum field theory and classical statistical mechanics. But this is only a phase ... 2 This is essentially an answer to your questions R-matrix for spin chains, Elliptic R-matrix and Yang Baxter solution for XYZ model,R$matrix for XYZ spin chain, Algebraic Bethe Ansatz and$R\$-matrices, which all basically ask the same question anyway. In short: to the best of my knowlegde, coming up with an R-matrix is an art, not a derivation. (Cf. the ...

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