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29

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 ...

21

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. ...

6

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 ...

6

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 ...

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

My wild idea would be to look for holomorphic linking of Wilson loops in the ABJM theory, working from its conjectured Grassmannian integral representation.

4

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 ...

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

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 ...

2

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 ...

2

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.

2

I don't see how you get mixed terms. Your $U$ acts 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 ...

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 ...

1

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 $S^-_j = ... 1 The answer to the apparent contradiction of these two transformations (the excitations seem to be either bosonic or fermionic) comes the fact that the spins are not equivalent to the fermions, because they have string attached to them, to respect the commutative nature of spins on different sites, see JW transformation on wiki. Therefore, even though the ... 1 1.) The easiest way to count the energy is as that of domain walls. Both of the configurations you have drawn have two domain walls. Each domain wall costs energy$E\Delta/2$, so both configurations have energy$E\Delta$above the ground state. I don't understand either how one would count the spin of a domain as 1. Perhaps that is the spin/site? 2.) To see ... 1 XX (and more generally XY) spin models became very popular lately, especially in connection to quantum-information theoretic problems. The XY model (without magnetic field though) has been introduced and properly solved in 1961 by Lieb, Schultz and Mattis (Ann. Phys. 16, 407, (1961)). By the way this is a great reference to start with, if you didn't read ... 1 Wick's theorem works for an arbitrary number of states equipped with their creation and annihilation oscillators. In the same way, Wick's theorem for spins applies to a collection of an arbitrary number of spins - or qubits, which is the same thing if their spin is$1/2\$. Wick's theorem is just a mathematical identity that works even for - and especially for ...

1

Some times it is not an efficient way to calculate the partition function using the functional integral. Even for a system of free fermions, we must have to evalute some complex Matsubara sums

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