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I am studying a system for which I observe a power-law decay in the correlation function: $\left\langle s\!\left(0\right)\cdot s\!\left(r\right) \right\rangle \propto r^{-\alpha}$

I am interested in computing the correlation length ($\xi$) for this system, but I don't know how. Typically one assumes $\left\langle s\!\left(0\right)\cdot s\!\left(r\right) \right\rangle \propto \exp(-r/\xi)$ and then it is straightforward to extract $\xi$ as the negative of the slope of a plot of $\log(\left\langle s\!\left(0\right)\cdot s\!\left(r\right) \right\rangle)$ vs. $r$. However, in my system, such a plot is nonlinear, but I find that $\log(\left\langle s\!\left(0\right)\cdot s\!\left(r\right) \right\rangle)$ vs. $\log(r)$ is linear. For different states of the system the exponent $\alpha$ of the power-law decay ranges from about $0.29$ to about $0.69$.

1. How can I compute the correlation length in such a system?

And...

2. What does this power-law decay in the correlation function mean?

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2 Answers 2

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Let $C(r)$ be the correlation function. The correlation length is defined as $$ \xi=-\lim_{r\to\infty}\frac{r}{\log C(r)} $$

If $C(r)$ decays slower than an exponential, the limit above diverges, and $\xi=\infty$. If it decays faster, $\xi=0$.

A power law behaviour is typical of scale-invariant systems, such as a CFT.

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  • $\begingroup$ This formula is super useful. Could you please provide a reference for the same? $\endgroup$ Commented Oct 31, 2017 at 13:16
  • $\begingroup$ Hi @SayanMandal. Unfortunately, I don't really have a reference. The "proof" is straightforward though: just let $C(r)\sim \mathrm e^{-r/\xi}$ and solve for $\xi$. Cheers! $\endgroup$ Commented Oct 31, 2017 at 17:58
  • $\begingroup$ Hi @AccidentalFourierTransform, thanks a lot. I was wondering about the $r\rightarrow\infty$ limit. I guess this has to do with looking at large scale effects. $\endgroup$ Commented Oct 31, 2017 at 21:22
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If the correlation function decays as a power law, then the correlations are scale invariant (that is, $C(\lambda r)=\lambda^{-\alpha}C(r)$), which is another way to say that there isn't any (macroscopic) length scale. In another word, there isn't any correlation length, and one typically says that it is infinite (as scale invariant correlations can be obtained from the limit $\xi\to\infty$ of a non-scale invariant function).

In simulations, if the system has a finite correlation length, one can assume that for small momentum $q$, $$ C(q)=\frac{C(0)}{1+q^2\xi^{2}} .$$ Since the simulations are done on finite system size $L$, the smallest non-zero momentum is $q=2\pi/L$, and an estimate of the correlation length is $$\xi_{\rm sim} = \frac{L}{2\pi}\sqrt{\frac{C(0)}{C(2\pi/L)}-1}.$$ In the limit of large system sizes, this will converge to the true correlation length, whereas it will diverge if the system is scale invariant.

Concerning the meaning of a power law decay, there are typically two possibilities. One is that the system is at the critical point of a (second order) phase transition, where the correlation length diverges. Another possibility is that the system is in a broken symmetry phase (of a continuous symmetry), where Goldstone modes dominate the physics at long distance, since they are by definition gapless.

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  • $\begingroup$ Where does the expression $C(q)=C(0)/(1+q^2\xi^2)$ come from ? $\endgroup$
    – J.A
    Commented May 3 at 12:57
  • $\begingroup$ @J.A from the Fourier transform of an exponentially decaying correlation function. $\endgroup$
    – Adam
    Commented May 4 at 6:09

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