14

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 ...


12

That is not a definition of correlation length. (It is a definition of the critical exponent.) The correlation length is defined in terms of the 2-point correlation function of spin observables. Pick points $x$ and $y$ on the lattice, and consider the expectation value $\langle s(x) s(y) \rangle$ of the product of the spin observable at $x$ and the spin ...


11

After days of thinking, searching, discussions, and testing, I can finally answer my own question now. The answer is much more involved than I expected from such a "simple" XY model (even just for the Ising model)! All "correct solutions/spectrum" stated below are checked against results from exact diagonalization. Simply put, there's nothing wrong with ...


10

I'm not a 100% this will address the question, so this might be more of a long comment than an answer. My main point is, perhaps, that given a Hamiltonian, you'll still want to specify the dynamics to do simulations. Even the Ising model can be simulated in several different ways, all satisfying detailed balance, where the relaxation towards equilibrium is ...


10

The correlation length of the 2d Ising model has been computed explicitly. You can find the expression in the famous book by McCoy and Wu. Here's a plot of the inverse correlation length (i.e., $1/\xi$) at various temperatures, taken from this recent review paper: This is only to show the directional dependence, as the radial scale is not the same for all ...


9

The reason that the systems energy is lowered when spins are aligned comes from the coulomb (electrostatic) potential, not magnetism. The details are non-trivial, but basically if you combine the Pauli Exclusion Principle with the Coulomb Potential you find that the ground state occurs when spins are aligned. The repulsion with two macroscopic magnets is a ...


9

Even though the explicit commutator you wrote is wrong--you should not have conjugated $\Pi$ in the second term-- your conclusion is sound that you cannot possibly satisfy the Born-Heisenberg commutation relation with 2x2 matrices. In fact, there is a general Theorem: The Heisenberg algebra does not admit faithful finite-dimensional (matrix) representations. ...


9

As this is a list-like question, let me list a few things (without much discussion -- feel free to ask specific questions about individual points). Each item mentions what the Heisenberg model (HM) has as opposed to the Ising model (IM). continuous symmetry vs. discrete symmetry as a consequence: gapless excitations whenever the symmetry is broken (i.e. in ...


8

Just a small addition to what user1504 said: the correlation length can be defined for $T<T_c$ as well, so that $\big\langle\big(s(x)-\langle s(x) \rangle \big)\big(s(y)-\langle s(y) \rangle \big)\big\rangle=e^{-\frac{|x-y|}{\xi}}$


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 first thing to realize is that there are no "true" phase transitions (in the sense of non-analytic behaviour of thermodynamic potentials) in finite systems. This is the main difficulty one faces when analysing phase transitions using (most) computer simulation schemes. In particular, such simulations are only reliable as long as the observed correlation ...


8

Basically, the answer is yes: $H$ is TRI because it is real. Reality condition really means that the Hamiltonian obeys a certain anti-unitary symmetry. In this case, the time-reversal operation is simply $T=K$ where $K$ is the complex conjugation. It is not the usual one($T=K\prod_i i\sigma^y_i$), and in particular $T^2=1$, so there is no Kramers' theorem ...


8

The critical point is not the same thing as the RG fixed point. Let $\mathcal{T}$ denote "theory space" meaning the set of all possible probability measures on real valued fields on the fixed unit lattice $\mathbb{Z}^2$. Block-spin or decimation etc. give you a map $R:\mathcal{T}\rightarrow \mathcal{T}$, namely, a renormalization group transformation. The ...


8

Another commonly used notion of "solving a theory" is to find a procedure to compute, at least in principle, all local observables. Sometimes one might also add non-local observables to the mix, it depends on what people are most interested about. Local observables are correlation functions $\langle O_1(x_1)\cdots O_n(x_n)\rangle$, where $O_i$ may ...


8

Let me write the Hamiltonian $$ H = -J \sum_i S_i^z S_{i+1}^z. $$ This choice will avoid some annoying (and irrelevant) signs. One way to formulate the statement in the OP precisely is as follows. Consider the variables $\delta_i=S_i^zS_{i+1}^z$. Since $\delta_i=1$ when the spins at $i$ and $i+1$ agree and $\delta_i=-1$ when the spins at $i$ and $i+1$ ...


7

A universality class is an equivalence class of physical models – field theories, quantum field theories, or models of classical or quantum statistical physics – where the equivalence is defined by two or several models' having the same mathematical description of the behavior at very long time scales and distance scales. So if two models' behavior at very ...


7

I don't think that there is a direct, natural physical interpretation (one can of course always cook up something ex post facto). There are however close relations with other topics. Here, I'll try to explain some close links with Markov chains. I'll stick to the case of the one-dimensional Ising model, to keep things concrete, but it should be clear from ...


7

First note that, as you say, the 2-point function $\langle \sigma_i \sigma_j\rangle$ does not tend to zero as $\|j-i\|\to\infty$ when $T<T_{\rm c}$; namely, $$ \lim_{\|j-i\|\to\infty} \langle \sigma_i \sigma_j\rangle^+ = (m^*(T))^2, $$ where $m^*(T)=\langle\sigma_0\rangle^+$ is the spontaneous magnetization (the $+$ superscript indicates that the ...


7

Your argument only applies to finite systems (otherwise the energy is ill-defined) and there are no phase transitions in finite systems. So, there is no contradiction there. Moreover, your argument only applies when both $h=0$ (no magnetic field) and you use free or periodic boundary conditions. Indeed, were it not the case, then you would not have symmetry ...


7

In fact, the two-dimensional Ising model was first solved by Lars Onsager on a cylinder/torus, which, while lacking intrinsic curvature, has a 'nontrivial' topology and moduli space (you can play with the solution by changing how the cylinder/torus closes on itself.) More generally, given a graph $G = (V, E)$, it is possible to define an Ising model on $G$ ...


6

I think that the most prominent example of "prediction before observation" in statistical physics is the Bose-Einstein condensate. It was predicted in ~1925 by, well, Bose and Einstein, obviously. Then after more than ten years it was proposed as an explanation for superfluidity and superconductivity. And the actual BEC of atoms (as a new state of matter) ...


6

Onsager computed the partition function of the 2D periodic square lattice (toroidal boundaries) Ising model. It is arguably one of the most elegant proof of modern statistical mechanics. The original paper is available on the APS website below: (you will need institutional access) L. Onsager, "Crystal Statistics. I. A Two-Dimensional Model with an Order-...


6

I would argue that this maybe due to the way you calculate your autocorrelation. An autocorrelation like that straight line is the result of a large square signal. The Ising model has a phase transition at the critical temperature. Above it, it's disordered; below it, it becomes ordered, which means that the magnetization stops flipping back and forth. This ...


6

Here is how Chandler does his counting: Take the (square) lattice to be of infinite extent or a finite lattice with periodic boundary conditions in both directions. The total number of edges is equal to $4N/2=2N$ if there are $N$ sites. The ground state corresponds to all spins being in the same state (all up or all down). The ground state energy is $-J$ ...


6

I will explain how I measured the spin-spin correlation function for the 2d Ising model. Generalization to more than 2 dimensions should be straightforward as long as you have hypercubic lattices. Just to get the notation straight: Let's use the name $\sigma_{(i,j)}$ for the spin at position $\vec r_{(i,j)} = \begin{pmatrix} i \\ j \end{pmatrix}$. Let's ...


6

wsc's answer (i.e., Onsager's computation of the free energy) provides one alternative road to a proof of a phase transition in the Ising model. It implies the existence of a phase transition in dimension 2 (for the nearest-neighbor model). Combined with correlation inequalities, this implies existence of a phase transition in any dimension d≥2, and ...


6

Eq. 2.A.30 is a somewhat non-trivial identity for the ground-state $|\psi_0\rangle$ which only uses the ground-state property that $\eta_q|\psi_0\rangle =0$. (Of course, as the OP has noted, $c_i|\psi_0\rangle \neq 0$.) What we need to show is that $$I=\langle \psi_0 | (c_j+c_j^\dagger)(c_i+c_i^\dagger)|\psi_0\rangle =\delta_{ij}.$$ Using Eq. 2.A.37a, we ...


6

You are correct that for $h=0$ the quantum Ising model reduces to the classical model. Assuming a 2D square lattice this model has been solved exactly by Onsager. It undergoes a phase transition at a certain critical temperature which is signaled by the order parameter $M_2 = \left( \frac{1}{N} \sum_i S_i^z \right)^2$. Below this temperatur the system breaks ...


6

This is an extremely deep question that still isn't fully understood. Such a "macroscopic superposition" $|\uparrow \uparrow \uparrow \uparrow \dots \rangle + |\downarrow \downarrow \downarrow \downarrow \dots \rangle$ is a perfectly valid state in the Hilbert space, and yet we never see it experimentally (at least not without a lot of careful work to ...


6

This is a great question. Pages 15 and 16 of these notes argues that no nontrivial spin Hamiltonian can ever be a fixed point under spin decimation, but I don't understand why their argument doesn't hold in the 1D case. The notes end with the cryptic comment there are many RG’s. The goal is not to see how many don’t work, but rather to find one that ...


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