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Ising model is defiend as $$ \mathcal{H}=-H\sum_i S_i -\frac{1}{2}\sum_{i,j}J_{ij}S_i,S_j $$ In 1D we assume that indices $i,j$ are integers, $i,j\in\mathbb{Z}$, and that the coupling depends only on the distance between the spins: $$ J_{ij} = J(|i-j|). $$ Then (at least for the nearest neighbor coupling) the model does not exhibit a phase transition at finite temperature.

On the other hand, if we consider that $J_{ij}=J$ is a constant, we have the infinite-range Ising model, which can be solved by mean field theory and exhibits a phase transition at finite temperature. However, this formally falls under the 1D case described above.

Question: Is this contradiction appears only in infinite range limit (i.e., there is no phase transition in a finite range 1D Ising model) OR should finite range Ising model be considered as different from 1D?

Auxhiliary question: what really determines formally the dimensionality of the model? The number of nearest neighbors?

Background: what really interests me is a possibility of a phase transition in an Ising/Potts model with random couplings (one-dimensional in real space), but I would like to understand well the basics.

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Keeping with your notations, let us consider the Ising model on $\mathbb{Z}$ with coupling constants $J_{ij} = |i-j|^{-\alpha}$, with $\alpha > 1$ (otherwise the model is ill-defined; note that the mean-field model is pathological in that the interaction is dependent on the system size, $J_{ij} = J/N$ for a chain of length $N$).

Then, one can prove that this model displays long-range order at low temperatures if and only if $\alpha\leq 2$.

The fact that there is uniqueness in one dimension when the interaction decays faster than $r^{-2}$ is pretty general; see this answer for more information on the existence/absence of phase transitions in one-dimensional systems. In particular, any one-dimensional model with finite-range, translation-invariant interactions (and bounded spins) is disordered at all temperatures.

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  • $\begingroup$ By finite range you mean that $J_{ij}=0$ if $|i-j|$ is greater then some cutoff? $\endgroup$
    – Roger V.
    Commented May 6, 2021 at 11:08
  • $\begingroup$ Yes. Of course, as I said, you have disorder at all temperatures even for infinite-range interactions, provided they decay fast enough. $\endgroup$
    – Yvan Velenik
    Commented May 6, 2021 at 11:08
  • $\begingroup$ What I really have in mind is genetic sequence: it is physically linear, with every position taking 4 or 20 values (depending on whether we work on nucleotide or amino-acid level). There may be interactions between positions that are very far one from another - they do not encessarily decay with distance, but one could say that $J_{ij}$ is sparse. There is also random field at every position. $\endgroup$
    – Roger V.
    Commented May 6, 2021 at 11:28
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    $\begingroup$ That can, of course, be much harder than a homogeneous system. What I can say for sure is that the uniqueness result I state precisely in the other answer I mention applies essentially in complete generality including, presumably, to your model (this would provide a quenched result, so if you're interested in annealed results, it might not provide you with the best answer possible). Non-uniqueness is much harder and depends very strongly on the details of your model. So if the criterion I mention is not conclusive, then answering might be hard. $\endgroup$
    – Yvan Velenik
    Commented May 6, 2021 at 11:35

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