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The correlation length gives (approximately) the distance over which a spin flip has an effect. For systems with ordered phases, at low temperatures the correlation length is then small (since a single spin flip will have little affect due to the large energy needed to rotate all the spins).

But the one-dimensional Ising model (although it doesn't have an ordered phase) has a correlation length that diverges at low temperature. Given what I have said above and given at $T=0$ for the 1D Ising we do get an ordered phase, why is the correlation length diverging instead of going to zero?

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I assume that the question is not about how one shows that the correlation length diverges, as this is a simple computation when $d=1$, but rather about intuition behind the result.

First, note that the same also happens when you consider the $d$-dimensional Ising model, with $d\geq 2$, as the temperature approaches $T_c$ from above. (Actually, also from below, but let's stick to the situation you are interested in.)

In both cases, the reason is the same (although the mechanism is more subtle in higher dimensions). The correlation length is the relevant length-scale in the system. So, above $T_c$, it measures, for example, the typical size of clusters of spins taking the same values. The size of these clusters diverges as the system gets closer and closer to the critical temperature.

Rather than the kind of dynamical interpretation you seem to be using, you should rather interpret the correlation length in the following statistical manner: suppose you only observe a single spin of your system (say, the spin at $0$, $\sigma_0$) and you discover that it takes the value $+1$. What can you say about the value of the spin at $i\neq 0$? Well, if $i$ is "close enough" to $0$, then knowing that you have a $+1$ at $0$ makes it more likely to observe a $+1$ also at $i$. The correlation length quantifies what "close enough" means. Namely it is the typical distance up to which the probability of observing a $+1$, given that $\sigma_0=+1$, differs significantly from $1/2$.

Now, for the one-dimensional model at low temperature, observing that $\sigma_0=+1$ makes it very likely that you'll see only $+1$ up to very large distances, precisely because the energetic cost is huge to flip spin. This will occur (the cost being finite), but the density of pairs of neighbors with spins taking different values goes to zero as $T\downarrow 0$. The average distance between two consecutive such pairs will be of the order of the correlation length, so it diverges.

Addendum

Let me stress the difference with what happens as $T\downarrow 0$ in higher dimensions. In this case, the system is in an ordered phase. For definiteness, let's assume that it is in the $+$ state $\mu^+_T$. In this case, the correlation length measures the typical distance over which $$ \mu_T^+(\sigma_i=s \,|\, \sigma_0=s) \text{ differs significantly from } \mu_T^+(\sigma_i=s). $$ And this distance decays to $0$ as $T\downarrow 0$ , for reasons explained in this answer.

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  • $\begingroup$ Ah OK, your 'addendum' makes a lot sense to me. I am assuming that we can generalize this to the unordered case by just replacing $\mu_T^+$ with simply an average without ergodicity breaking. So in general the correlation length measure the typical difference over which $P(\sigma_i=s \mid \sigma_0=s)$ differs from $P(\sigma_i=s)$ where the probability $P$ is taken w.r.t the full equilibrium distribution in the unordered phase and the ergodicity broken distribution in the ordered. $\endgroup$ Jun 9, 2018 at 8:33
  • $\begingroup$ @Quantumspaghettification : exactly :) . $\endgroup$ Jun 9, 2018 at 8:34
  • $\begingroup$ Sorry to come back to this. There is one point I am worrying about with this question related to this recent question I asked. After symmetry breaking in something like the Heisenberg model as $T \rightarrow 0$ as far as I can tell we will not get one definite state but our partition function will contain all ground states which where not lost under ergodicity breaking. In such a case this would bring us back to the 1d case and thus indicate that $\xi \rightarrow \infty$. What am I missing? $\endgroup$ Jun 9, 2018 at 15:14
  • $\begingroup$ @Quantumspaghettification : I am not sure I understand what you mean by "we will not get one definite state". In dimension $3$ and more, below the critical temperature, the Heisenberg model has a continuum of pure states, indexed by a unit vector in $\mathbb{R}^3$ (corresponding to the possible orientations of the spontaneous magnetization). From this point of view, there is no difference with what happens in the Ising model. [...] $\endgroup$ Jun 9, 2018 at 15:42
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    $\begingroup$ @Quantumspaghettification: Yes, this should remain true in the quantum case. Some aspects are proved in Simon's book (The Statistical Mechanics of Lattice gases). Of course, things are not as transparent in the quantum case as they are in the classical case... $\endgroup$ Jun 10, 2018 at 17:17

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