# Why does a divergent correlation function (not length) imply lack of the order in the system?

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?

• 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. May 15 '20 at 15:13
• @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.
– SRS
May 15 '20 at 15:18
• don't refer but say what you mean yourself. May 15 '20 at 15:19
• Correlation function has logarithmic divergence in 2D, goes as $\log (|{\bf x}-{\bf x^\prime}|)$.
– SRS
May 15 '20 at 15:21
• 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. May 15 '20 at 15:31

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.

• 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... May 15 '20 at 14:36
• 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
– SRS
May 15 '20 at 15:44
• 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. May 15 '20 at 15:55
• 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. May 15 '20 at 16:00
• 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). May 15 '20 at 16:18