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The correlation length $\xi$ is related to critical temperature $T_c$ as $$ \xi\sim|T-T_{c}|{}^{-\nu}, $$

where $\nu$ is the critical exponent.

  1. Is this the formal definition of correlation length? If not, what is the formal definition of correlation length (for phase transition in the Ising model)?
  2. Can you give a physical understanding of correlation length?
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That is not a definition of correlation length. (It is a definition of the critical exponent.)

The correlation length is defined in terms of the 2-point correlation function of spin observables. Pick points $x$ and $y$ on the lattice, and consider the expectation value $\langle s(x) s(y) \rangle$ of the product of the spin observable at $x$ and the spin observable at $y$. This quantity tells you how strongly correlated the spin at $x$ and the spin at $y$ are, as a function of the temperature, coupling constant, and the distance between $x$ and $y$. If $T > T_c$, then the correlation function dies off exponentially fast in $|x-y|$.

$\langle s(x) s(y) \rangle \sim e^{-\frac{|x-y|}{\xi(T)}}$

The correlation length is, by definition, the constant (in $x$ and $y$, but not in $T$) which tells you how fast the correlation function vanishes.

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    $\begingroup$ So how will you give a formal definition of the correlation length (for Ising model)? $\endgroup$ – cosmicraga Apr 1 '13 at 14:05
  • $\begingroup$ What I wrote above is the formal definition: the correlation length is the exponential decay rate of the 2-point correlation function. $\endgroup$ – user1504 Apr 1 '13 at 14:19
  • $\begingroup$ Correlation length is found out by plotting $\langle s(x) s(y) \rangle $ vs. r and it is the length where the curve first changes its sign by crossing the r axis. $\langle s(x) s(y) \rangle \sim e^{-\frac{|x-y|}{\xi(T)}}=0$. \\ $-\frac{|x-y|}{\xi(T)} = 1$ \\ $|x-y| = -\xi(T)$ is it correct ? $\endgroup$ – cosmicraga Apr 1 '13 at 14:44
  • $\begingroup$ No, that's wrong. $\langle s(x) s(y)\rangle$ never changes sign. You can extract $\xi$ by looking at the ratio $\frac{\langle s(0) s(y)\rangle}{\langle s(0) s(2y)\rangle}$. $\endgroup$ – user1504 Apr 1 '13 at 15:02
  • $\begingroup$ Also, you've botched the math: $e^a = 0$ does not imply $a=1$. $\endgroup$ – user1504 Apr 1 '13 at 15:20
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Just a small addition to what user1504 said: the correlation length can be defined for $T<T_c$ as well, so that $\big\langle\big(s(x)-\langle s(x) \rangle \big)\big(s(y)-\langle s(y) \rangle \big)\big\rangle=e^{-\frac{|x-y|}{\xi}}$

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  • $\begingroup$ Yes, +1, thanks for adding this. (I was being lazy, so I only discussed the case $T > T_c$, where $\langle s(x) \rangle = 0$. ) $\endgroup$ – user1504 Apr 1 '13 at 19:36

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