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?

  • $\begingroup$ 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

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.

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.