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Consider a two-point correlation function defined as $$G_{ij}({\bf x},{\bf x}^\prime)\equiv \Big\langle\Big(\mathscr{O}_i({\bf x})-\big\langle\mathscr{O}_i({\bf x})\big\rangle\Big) \Big(\mathscr{O}_j({\bf x^\prime})-\big\langle\mathscr{O}_j({\bf x^\prime})\big\rangle\Big)\Big\rangle\tag{1}$$ $$~~~~~~~=\big\langle \mathscr{O}_i({\bf x})\mathscr{O}_j({\bf x}^\prime)\big\rangle-\big\langle \mathscr{O}_i({\bf x})\big\rangle\big\langle \mathscr{O}_i({\bf x}^\prime)\big\rangle\tag{2}$$ where $\mathscr{O}_i$ is the $i^{\rm th}$ component of a $n$-component order parameter ${\bf \mathscr{O}}$ and $\langle...\rangle$ denote thermal averages.

$1.$ If $G_{ij}({\bf x},{\bf x}^\prime)$ is identically zero, that implies $\mathscr{O}_i({\bf x})$ and $\mathscr{O}_i({\bf x}^\prime)$ are completely uncorrletaed or independent random variables, and vice-versa.

$2.$ If the function $G_{ij}({\bf x},{\bf x}^\prime)$ decays exponentially with increasing $|{\bf x}-{\bf x}^\prime|$ and thus quickly vanishes beyond a certain length-scale $\xi(T)$, that implies there is a correlation but on small length-scales, and that presumably can be referred to as a short-range order.

$3.$ At the phase transition point, $T=T_C$, where the function $G_{ij}({\bf x},{\bf x}^\prime)$ decays not exponentially but as some negative power of $|{\bf x}-{\bf x}^\prime|$, then that implies the divergence of the correlation length, i.e., $\xi(T_C)\to\infty$. At this point, perhaps, one would interpret this as the full system is trying to be correlated (?)

$4.$ If $G_{ij}({\bf x},{\bf x}^\prime)$ either remains constants or decays in such a way that even when $|{\bf x}-{\bf x}^\prime|\to\infty$, $G_{ij}({\bf x},{\bf x}^\prime)$ remains nonzero, finite, then there is the long-range order.


Question Apart from the four listed above, there is still one more important possibility. It is possible that the correlation function $G_{ij}({\bf x},{\bf x}^\prime)$ itself diverges at any temperature $T>0$. This is often the case for $2$-dimensional systems, for Hamiltonians with continuous symmetries and short-range interactions. This is at the heart of the Mermin-Wagner theorem.

How do we properly interpret the meaning of this divergence) preferably from the defining equations in Eq.$(1)$ or Eq.$(2)$). In particular, why does a divergent correlation function should mean a lack of order in the system?

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  • $\begingroup$ You should explain what you mean by the correlation diverging, especially if this is different from option 2 where $\xi$ diverges. Also, for Ising option 4 does not happen and it is not what one refers to as long range order. The latter is about $\langle\mathscr{O}(\mathbf{x})\mathscr{O}(\mathbf{x}')\rangle$ rather than the covariance $G$, having a finite nonzero limit as the points get infinitely far. $\endgroup$ May 15 '20 at 15:13
  • $\begingroup$ @AbdelmalekAbdesselam What about last few lines of this section en.wikipedia.org/wiki/… ? and the introduction section en.wikipedia.org/wiki/… I can also give textbook references. $\endgroup$
    – SRS
    May 15 '20 at 15:18
  • $\begingroup$ don't refer but say what you mean yourself. $\endgroup$ May 15 '20 at 15:19
  • $\begingroup$ Correlation function has logarithmic divergence in 2D, goes as $\log (|{\bf x}-{\bf x^\prime}|)$. $\endgroup$
    – SRS
    May 15 '20 at 15:21
  • $\begingroup$ Then that means the example you have in mind is the Gaussian free field on $\mathbb{Z}^2$ which does not quite make sense. See math.ucla.edu/~biskup/PIMS/PDFs/lecture1.pdf or the GFF chapter in the book by Yvan who answered below. $\endgroup$ May 15 '20 at 15:31
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Note that the 2-point function cannot diverge in models with bounded spins. So its divergence is clearly not at the heart of the Mermin-Wagner theorem. In fact, when the Mermin-Wagner theorem applies, the 2-point function usually still tends to 0 as the distance between the spins diverges. (See Chapter 9 of this book for more on this.)

In fact, even with unbounded spins, things usually cannot go wrong (when the infinite system is well defined!). Consider a real-valued field $(\varphi_x)$. If your field is translation invariant and the spins have a finite variance $\sigma^2$, then the Cauchy-Schwarz inequality implies \begin{align} \lvert\langle\varphi_0\varphi_x\rangle - \langle\varphi_0\rangle \langle\varphi_x\rangle\rvert &= \lvert\bigl\langle (\varphi_0-\langle\varphi_0\rangle)(\varphi_x-\langle\varphi_x\rangle)\bigr\rangle\rvert \\ &\leq \sqrt{\bigl\langle(\varphi_0-\langle\varphi_0\rangle)^2\bigr\rangle \bigl\langle(\varphi_x-\langle\varphi_x\rangle)^2\bigr\rangle} \\ &= \sigma^2, \end{align} which shows that the two-point function again cannot diverge.

So, you would either need to consider non-translation invariant models or models in which the spins have an infinite variance.

As a simple example of a non-translation-invariant model with diverging 2-point function, you can consider the massless Gaussian free field on $\mathbb{Z}^2$, with boundary condition $\varphi_0=0$. Note that you need to fix the value of one spin, or do something similar, since otherwise the infinite-volume field does not exist (as a consequence of the Mermin-Wagner theorem, see Section 9.3 of the same book for more on this).


Update: I collect here the content of the comments made by Abdelmalek Abdesselam or myself, in case they disappear.

  1. Your definition of long-range order (point 4 in your list) is not the standard one. Indeed, you should rather use the non-truncated 2-point function for this, since the truncated one always tends to zero as $|x|\to\infty$ in pure states.
  2. Your question seems to find its roots in some confusions regarding the divergences in some versions of the physicists' "proof" of the Mermin-Wagner theorem.

    First, in many versions of the argument, the computation is carried by expressing the 2-point function $\langle S_0 \cdot S_x\rangle$ of the spin system (say an XY model) by a computation involving the two-dimensional massless Gaussian free field (after having approximated the cosine by a quadratic term and replaced the angle variables by real numbers). The correlation function $\bigl\langle(\varphi_0 - \varphi_x)^2\bigr\rangle$ does diverge (logarithmically) as $|x|\to\infty$ (this is closely related to the example I discuss above). This does not imply the divergence of the original 2-point function, which is necessarily finite, the spins being bounded. In fact, the 2-point function $\langle S_0 \cdot S_x\rangle$ actually tends to $0$ (as a power law) as $|x|\to\infty$.

    Second, in some versions of the argument, there is a second divergence of the correlation in the GFF as the distance between the point tends to $0$. This is (1) due to the (totally useless) replacement of the lattice GFF by a continuum GFF and has no physical relevance in the context of the Mermin-Wagner theorem.

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  • $\begingroup$ I am aware that this does not actually answer the question you asked. However it addresses a possible misconception underlying your question, by showing that the 2-point function generally does not blow up. (In particular, your discussion of the MW theorem makes me think that you might be confusing the divergence of the 2-point function with the divergence of an approximation of the latter used in some non-rigorous "proofs" of the MW theorem.) In any case, the question is: do you have any explicit example in mind? Otherwise, it makes answering your question particularly difficult... $\endgroup$ May 15 '20 at 14:36
  • $\begingroup$ In this note, the correlation functions are obtained for 2D systems. Expressions are given on page 3 and 4. weizmann.ac.il/condmat/oreg/sites/condmat.oreg/files/uploads/… Now I think, that it would not be meaningful to take $\delta r$ to zero due to finite lattice cut-off. So the logarithmic divergences talked about in this Wikipedia article are not strict divergences. en.wikipedia.org/wiki/Mermin%E2%80%93Wagner_theorem I am not completely sure. @YvanVelenik $\endgroup$
    – SRS
    May 15 '20 at 15:44
  • $\begingroup$ Yes, that's the type of approximations I thought you were relying on (computing, non-rigorously, correlations in the XY model using an approximating Gaussian free field). Note that the correlation function, in terms of the $\phi$ variables used in the approximation in the notes is $\langle e^{i(\phi(x)-\phi(0))}\rangle$ and is thus bounded by construction (it's the average of complex numbers of modulus $1$), as it should be. The divergence is only in the correlation of the $\phi$ variables, that is, the Gaussian free field I discuss in my answer. $\endgroup$ May 15 '20 at 15:55
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    $\begingroup$ To summarize, the 2-point function that is relevant in the notes you link is $\langle \mathbf{s}_0 \cdot \mathbf{s}_x\rangle$ and this is always bounded as I explain in my notes (the spins have norm $1$!), not $\langle(\phi(0)-\phi(x))^2\rangle$, which is just an intermediate quantity useful for the (heuristic) computation. $\endgroup$ May 15 '20 at 16:00
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    $\begingroup$ Note also that the logarithmic divergence at long distances is the only relevant one for the argument in the wikipedia article and in the notes you link; you should ignore the divergence at small distances, as it is completely artificial and actually comes from a second, totally useless approximation (from a lattice GFF to a continuum one). $\endgroup$ May 15 '20 at 16:18

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