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13

The Ising model is a model, originally developed to describe ferromagnetism, but subsequently extended to more problems. Basically, it is an interaction model for spins. Imagine you have a system which is a collection of $N$ spins. Each spin $S_i$ has two possible states $+1$ or $-1$. Here you can imagine already a possible extension to more states. You can ...


10

There is a result I only heard about recently: it has been proven that computing partition functions for the Ising-model in dimensions > 2 is NP-complete. (The paper can be found at http://www.cs.brown.edu/people/sorin/pdfs/Ising-paper.pdf; a more readable one is here http://www.siam.org/pdf/news/654.pdf - both can be found on the Wikipedia on the Ising ...


8

There are two problems at low temperatures. One is intrinsic to the Ising model, the other to the Metropolis algorithm. The Ising one is a bit more serious. Problem with the Ising Model at low temperatures The Ising model of ferromagnetism posits that the electrons in a model can be modeled as little magnetic dipoles, confined to only two quantized spin ...


8

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 ...


6

Maybe I didn't get the question, but the whole point of the discussion of Ising model is not getting zero magnetization. This is a topic of the spontaneous symmetry breaking (further "SSB"). The symmetry you are talking about is broken spontaneously leading to nonzero magnetization. I must admit that I never got into details of Onsager solution, but I know ...


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) ...


5

Off the top of my head, the example I can think of is the whole work that Boltzmann did. He based his entire theory of statistical mechanics on the concept of indivisible particles (i.e. that all matter is made up out of atoms). Doing this, his theory (using theoretical mathematical methods as you said) was able to predict how the atoms determine the visible ...


5

It seems pretty clear that if you take a very diluted subset of, say, the horizontal line through $0$, then you'll be able to make a Peierls argument. For example, put $h=+\infty$ (worst possible case, amounting to fixing the corresponding spins to $+1$) at all vertices with coordinates (10^k,0), with $k\geq 1$. Then, when removing a contour surrounding a ...


5

Yes mean-field theory is wrong for the one-dimensional case (and wrong for the two and three dimensional cases as well, where the transition exists but the mean-field approximation gets the wrong critical temperature and exponents). In fact it's a typical first year exercise to solve the 1D Ising model exactly using transfer matrices, and I suggest you look ...


5

I am about halfway the most important part of Onsager's paper, so I'll try to summarize what I've understood so far, I'll edit later when I have more to say. Onsager starts by using the 1D model to illustrate his methodology and fix some notations, so I'm gonna follow him but I'll use some more "modern" notations. In the 1D Ising model, only neighbouring ...


5

1) The Metropolis algorithm is more general than just the sampling scheme used: the sampling scheme that you describe is a "local update," however there exist more general sampling schemes ("cluster updates" like Wolff algorithm, Swendsen-Wang, etc.). The point is that local update sampling isn't inefficient at low temperatures, per se, it's inefficient near ...


5

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 ...


5

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 ...


5

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 ...


4

I think that you are really interested in the $q$-state clock model, which is similar to the Potts model, and is defined as follows. Fix an integer $q\geq2$. For each $i\in\mathbb{Z}^d$, let $$ \theta_i \in \bigl\{\frac{2\pi}{q} k\,:\, k\in\{0,1,\ldots,q-1\}\bigr\}, $$ and define the spin at site $i$ by $$ \mathbf{S}_i = (\cos\theta_i,\sin\theta_i) . $$ The ...


4

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 ...


4

Onsager found 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 can be found here (you will need institutional access) L. Onsager, "Crystal Statistics. I. A Two-Dimensional Model with an Order-Disorder Transition", ...


4

This doesn't seem quite like an appropriate question for a math site, so I guess it will be transferred shortly, but anyway, the answer to your question is that the dipolar interaction is by no means neglected in theories of ferromagnetism, it is the long-range interaction that governs the appearance of domain walls. The short-range exchange interaction, on ...


4

This is an assumption, but in the case of the of the Ising model it is an obviously correct one. The statement that $m$ is the same at all sites is the assumption that the state of the system respects translational invariance. In general a system may spontaneously break translational symmetry, just as it breaks the $s\rightarrow-s$ symmetry. So when you do ...


4

Seeing as no one is trying to give an answer, I'll take a stab at it myself. Shortly after writing this question I learned (in this cute answer of Raskolnikov's) about Baxter's wonderful book on exact solutions in statistical mechanics. Slowly but surely I realized that Ising model has been solved so many times by some many different methods by virtually ...


4

Let me ask a seemingly naive question first: what is actually a statistical ensemble? Mathematically, it's a prescription of an assignment of probability to every microstate. So this is a probability measure. But we have to be careful with measures defined on infinite systems. You certainly can't blindly work with $\exp(-\beta E)$ factors as energy will ...


4

Probably the best book in the field of spin models is the classic book by Baxter "Exactly Solved Models in Statistical Mechanics", it is very well written and contains both the basics and some more advanced topics.


4

For nearest-neighbor interactions in 1D and 2D, the free energy of the system can be computed analytically. We can then check that this free energy is at its global minimum for a certain state. In 3D, we do not know the free energy analytically, so we have to resort to some kind of simulation (Monte-Carlo probably). If you reach a final state of your ...


4

This is mostly a question of definitions: Spontaneous symmetry breaking occurs when the underlying laws of a physical system have a symmetry, but the ground state does not. For an Ising system with $B=0$, $$H = \sum_{i,j} J_{ij} s_i s_j$$ we can see explicitly that the energy of a state $\{s_i\}$ is precisely the same as the energy of the state with every ...


3

I wish I could do your question justice, but I will content myself with a remark on the connection between two of the solution methods mentioned in Barry McCoy's article, namely Baxter's commuting transfer matrix method, and Onsager's original algebraic approach. In a certain sense, these methods have to be considered distinct since Baxter's method is ...


3

The 3d Ising model can be "solved" in a certain sense, it can be recast as the problem of a discrete lattice Fermionic string. This method is explained in detail in the last chapter of Polyakov's "Gauge Fields and Strings", and it is the natural generalization of Onsager's method to 3d. This method does not produce analytical computable critical exponents ...


3

If I understand you correctly, you seem to be asking the fundamental question of how on earth broken symmetries are possible. After all, there is an apparent contradiction: on one hand, the thermal average $\langle M \rangle= \sum_{\{s_i\}} M \exp(-\beta E)$ must vanish due to symmetry, while on the other hand, physicists claim that the 2D Ising model gives ...


3

The simplest model is the spin-1/2 chain with Majumdar–Ghosh interaction: $$H=\sum_i P_{3/2}(i-1,i,i+1),$$ where $P_{3/2}(i,j,k)$ is the projection operator that projects a state onto the subspace with total spin-3/2 on sites $i,j,k$. The ground states are two dimer states (see the figure on wikipedia Majumdar–Ghosh model): ...


3

It is a direct consequence of the usual (weak) spatial Markov property enjoyed by the Ising model. Namely, if $\Lambda$ is a (deterministic) finite subset of $\mathbb{Z}^d$, and $f$ a local function with support inside $\Lambda$, then the expected value of $f$ in the box $\Lambda$, with a given frozen configuration $\omega$ outside $\Lambda$ only depends on ...


3

It is only exactly at the critical temperature that this CFT result works. You haven't mentioned if you have used the critical temperature when you did the monte-carlo. At/near critical point, autocorrelation time becomes huge. (If I am not mistaken, autocorrelation time must blow up exactly at critical temperature, however it is cut-off due to finiteness ...



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