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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?

Quantum spaghettification
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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.

Yvan Velenik
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  • 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. – Quantum spaghettification Jun 09 '18 at 08:33
  • @Quantumspaghettification : exactly :) . – Yvan Velenik Jun 09 '18 at 08:34
  • 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? – Quantum spaghettification Jun 09 '18 at 15:14
  • @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. [...] – Yvan Velenik Jun 09 '18 at 15:42
  • [...] From the point of view of decay of correlations, things are more complicated. Indeed, I'd expect that correlations decay as a power law in this case, so that the correlation length is infinite for all $T<T_c$. Such a power law decay was proved in the $3d$ classical XY model (see this paper). – Yvan Velenik Jun 09 '18 at 15:43
  • My point is: For the Ising model at $T<T_c$ we know that spontaneous symmetry breaking will give us as $T\rightarrow 0$ all spins up or all spins down i.e. a pure state. For the Heisenberg model as $T<T_c$ I think all ground states with a sub-extensive number of spin flips (looking at the $h\rightarrow 0$ interpretation of SSB) will contribute equally to the ergodicity broken partition function. Thus as $T\rightarrow 0$ we will get a mixed state. Do you agree with this? – Quantum spaghettification Jun 09 '18 at 15:55
  • Are you talking about the quantum Heisenberg model? I am talking about the classical one (my knowledge of quantum spin systems is very superficial). In the classical Heisenberg model, rotation invariance is broken and you get a continuum of extremal Gibbs measures, one for each possible directions the spontaneous magnetization might point to. This is in complete analogy with what happens in the Ising model. In this case, as $T\downarrow 0$, each of these measures concentrates on the corresponding ground state (with all spins pointing in the same direction). – Yvan Velenik Jun 09 '18 at 16:06
  • Ok I was talking about the quantum case, but I think the arguments you make in the comments to your answer here https://physics.stackexchange.com/a/396512/70392 should hold in the quantum case (what do you think?) which would then clear up my confusion – Quantum spaghettification Jun 10 '18 at 13:08
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    @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... – Yvan Velenik Jun 10 '18 at 17:17