0
$\begingroup$

In the book of Statistical Field Theory by Giuseppe Mussardo, on page 51, it is given while talking about Ising model that

One arrives to the same conclusion by analysing the possibility of a non-zero expectation value of the spin, i.e. a non-vanishing limit $$ |\langle\sigma\rangle|^{2}=\lim _{r \rightarrow \infty} G^{(2)}(r) $$

where $G^2$ is the two-point correlation function and $\sigma_i$ would denote the spin at location $i$ (not exact sure what to understand is $\sigma$ doesn't contain an index).

How does the author derives/arrives this equality? I don't get it.

$\endgroup$
15
  • 1
    $\begingroup$ At large distances (compared with the correlation length), fluctuations in spins are uncorrelated, so $\langle \sigma(0)\sigma(r)\rangle - (\langle\sigma\rangle)^2$ approaches zero as $r\rightarrow\infty$. The rigorous proof (that a correlation length exists) is somewhat nontrivial however. $\endgroup$
    – TLDR
    Apr 19 at 17:28
  • $\begingroup$ @TLDR but what the equation in the question states is that the square of the mean of $\sigma$ is related to the correlation function. That is my question: how? $\endgroup$
    – Our
    Apr 19 at 17:38
  • $\begingroup$ @TLDR The equation isn't talking about fluctuations, etc. $\endgroup$
    – Our
    Apr 19 at 17:38
  • $\begingroup$ Do you see how statistical fluctuations are relevant to the definition of $G^{(2)}$? $\endgroup$
    – TLDR
    Apr 19 at 17:40
  • 1
    $\begingroup$ Neither is "=". $\endgroup$
    – TLDR
    Apr 19 at 17:42
1
$\begingroup$

A general statement.

Let us first state a general result, valid in any dimension. Let $f$ and $g$ be two local functions (that is, functions depending only on finitely many spins). Then $$ \lim_{|i|\to\infty} \langle f \cdot (g\circ\theta_i) \rangle^+ = \langle f \rangle^+ \langle g \rangle^+ , \tag{1} $$ where I used $\theta_i$ to denote the translation by $i\in\mathbb{Z}^d$ and $\langle\cdot\rangle^+$ denotes the $+$ state (that is, the state obtained using $+$ boundary condition). Identity (1) is an easy consequence of the FKG inequality, see Exercise 3.15 in this book (note that its solution can be found in Appendix C).

Application to the 2-point function.

In particular, setting $f=g=\sigma_0$, Identity (1) reduces to $$ \lim_{|i|\to\infty} \langle \sigma_0 \sigma_i \rangle^+ = \bigl( \langle \sigma_0 \rangle^+ \bigr)^2. \tag{2} $$ It can, in addition, be shown that $\langle \sigma_0 \rangle^+=m^*$, where $m^*$ denotes the spontaneous magnetization density (defined as the limit, as $h\downarrow 0$, of the magnetization density in the presence of a magnetic field $h$); see Remark 3.30 in the book.

Extension to the free and periodic states

Note that (2) (but not Identity (1) above) also holds for the state $\langle\cdot\rangle$ obtained using free or periodic boundary conditions. This follows from the decomposition $\langle\cdot\rangle = \frac12 \langle\cdot\rangle^+ + \frac12 \langle\cdot\rangle^-$, combined with the identity $\langle\sigma_0\sigma_i\rangle^+ = \langle\sigma_0\sigma_i\rangle^-$: $$ \lim_{|i|\to\infty} \langle \sigma_0 \sigma_i \rangle = \lim_{|i|\to\infty} \bigl( \tfrac12 \langle\sigma_0 \sigma_i\rangle^+ + \tfrac12 \langle\sigma_0 \sigma_i\rangle^-\bigr) = \lim_{|i|\to\infty} \langle \sigma_0 \sigma_i \rangle^+ = \bigl( \langle \sigma_0 \rangle^+ \bigr)^2. $$ Note that it is still $m^*=\langle \sigma_0 \rangle^+$ that appears in the right-hand side (in fact, $\langle\sigma_0\rangle = 0$ by symmetry).


Additional remarks:

  • Identity (1) is actually true for any extremal state in any model, but the proof is much more abstract (see Theorem 6.58 in the same book).
  • The decomposition $\langle\cdot\rangle = \frac12 \langle\cdot\rangle^+ + \frac12 \langle\cdot\rangle^-$ is not a trivial fact (it is, for instance, not always true for the Ising model on a tree). It follows from the fact that the state $\langle\cdot\rangle$ is translation invariant and that all translation-invariant states are convex combinations of the states $\langle\cdot\rangle^+$ and $\langle\cdot\rangle^-$. Proofs can be found here or here.
$\endgroup$
0

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.