1D Ising model: exact result

In the 1D Ising model with fixed $J_{ij} = J$, without magnetic field, the density of states (dos) can be calculated exactly. There is a caveat in the case of periodic boundaries (which I don't expect to change the result considerably), but let us consider a linear spin chain.

The Hamiltonian (taking $J=1$) $$ H = - \sum\limits_{\langle i,j \rangle} \sigma_i \sigma_j $$ takes values $-N+1, -N+3, \ldots, N-3, N-1$ for a system with $N$ spins (and periodic boundaries). For $z=2$ this has the special property that the number of terms in the Hamiltonian is $\frac{(N-1)z}{2} = N-1$, one less than the number of spins because the two ends have no interaction. Note that the terms $H_{i,i+1} \equiv -\sigma_i \sigma_{i+1}$ are linearly independent. In fact, there is a 2-to-1 correspondence between the set of spins $\{\sigma_i\}$ and the set of interaction terms $\{H_{ij}\}$. Each set of interactions is invariant under $\sigma_i \rightarrow -\sigma_i$ (for all spins). Therefore, the number of states with given energy $E(k) = -N+1+2k$ with $k \in \{0, 1, \ldots, N-1\}$ is given by $$ g(E(k)) \equiv g(k) = 2 \binom{N-1}{k} $$ Indeed, summing over all $k$ gives $$ \text{Total number of states} = \sum\limits_{k=0}^{N-1} 2 \binom{N-1}{k} = 2^N. $$

Higher dimensions: are there analytical approximations?

Now I've been looking into ways to extend such an argument to higher dimensions, but so far found nothing. The problem is that for $d>1$ there is no neat correspondence between the spins and the interaction terms. For instance, given two spins on a 2D lattice, there are multiple paths between the two, so the interaction terms are limited by the constraint that the spin values are consistent over each of those paths. If one path predicts spin 1 should have an opposite sign from spin 2 and the other that the signs should be equal, then such a set of interactions is not permitted.

I know that there is a neat algorithm called the Wang-Landau algorithm for calculating the density of states. In fact, the results for the 2D Ising model look similar to the binomial distribution, though appearing somewhat steeper (see the original article). However, I was not able to find any analytical work on calculating the dos. Are there any methods that can be used to obtain an (approximate) analytical result?


1 Answer 1


If you look for an exact density of state (DOS) for a 2D Ising model I recommend the Paul Beale's article PhysRevLett.76.78.

In the article the author shows how to correctly obtain the density of states starting from the definition of the partition function


where n is the number of columns, m is the number of rows, $K=J/k_BT$ is the coupling, $k_B$ is Boltzmann’s constant, $x=e^{-2K}$ is the low-temperature expansion variable, and the coefficient $g_k$ is the number of config- urations with energy $4kJ$ above the two ground states (all spins +1 or all spins -1).

The result of the partition function has a binomial term that represents the density of state. The solution has many other terms and your calculation is very hard work. So the author give a link for download a notebook from Wolfram Matematica with the implementation of the calculations of the density of states and other thermodynamics quantities.

In the article the link is broken, but I found another in the author’s personal website. I hope to have help with this question.

Link for download the notebook

  • $\begingroup$ This is not an answer to the posted question. The poster is not asking for a reference to an article but rather for a method to solve its problem. You should at least provide a brief overview of the content of the article and how this could solve the poster's problem. $\endgroup$
    – ZaellixA
    Commented Apr 21, 2022 at 15:35
  • $\begingroup$ While this link may answer the question, it is better to include the essential parts of the answer here and provide the link for reference. Link-only answers can become invalid if the linked page changes. - From Review $\endgroup$
    – ZaellixA
    Commented Apr 21, 2022 at 15:35
  • $\begingroup$ I understand your careful with the validity of the answer, but the link for the article is permanent. It is placed in a well known journal and your change is very difficult. $\endgroup$ Commented Apr 22, 2022 at 17:48
  • $\begingroup$ I understand the fact that this is a well established journal, but it is a matter of the regulations of the site to not provide "link-only" answers. $\endgroup$
    – ZaellixA
    Commented Apr 22, 2022 at 17:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.