# For spontaneous order, how does the correlation length $\xi(T)$ change for $T<T_C$?

The correlation length, $$\xi(T)$$, defined from the correlation function $$C(r)\sim \frac{1}{r^{d-2+\eta}}e^{-r/\xi(T)},~~T>T_C\tag{1}$$ is exponentially small above the transition temperature ($$T>T_C$$). The power law behaviour of the correlation function at $$T=T_C$$: $$C(r)\sim \frac{1}{r^{2-d+\eta}},~~T=T_C\tag{2}$$ tells that $$\xi$$ diverges at $$T=T_C$$. When the transition is associated with long-range order for $$T, the correlation function becomes homogeneous i.e., spatially uniform: $$C(r)\sim {\rm const.},~T

Question What is the correlation length for $$T and how does it change as $$T$$ is gradually decreased from $$T=T_C$$ to $$T\to 0$$? How can we see and understand that?

(I discuss the case of the Ising model, for concreteness, but most of what I say below holds much more generally.)

The correlation length below $$T_c$$ is defined through the truncated 2-point function, namely $$\langle \sigma_0\sigma_x\rangle^+ - \langle \sigma_0\rangle^+ \langle\sigma_x\rangle^+ \sim \frac{1}{|x|^{d-2+\eta}}e^{-|x|/\xi},$$ where $$\langle\cdot\rangle^+$$ denotes the expectation w.r.t. the $$+$$ state (one must precise the state below $$T_c$$ as there are several). I am cheating a bit (as you do), by doing as if $$\xi$$ was isotropic (it's not for lattice models, except asymptotically close to $$T_c$$).

Also, in the definition of $$\xi$$, one has to let $$|x|$$ go to infinity above. More precisely, one would define $$\frac{1}{\xi} = - \lim_{|x|\to\infty} \frac1{|x|} \log \bigl(\langle \sigma_0\sigma_x\rangle^+ - \langle \sigma_0\rangle^+ \langle\sigma_x\rangle^+ \bigr).$$ In particular, for $$T>T_c$$, you really have $$\langle \sigma_0\sigma_x\rangle \sim \frac{1}{|x|^{(d-1)/2}}e^{-|x|/\xi},$$ for all $$|x|$$ large compared to $$\xi$$. This exponent $$(d-1)/2$$ is typical of what is known as the Ornstein-Zernike behavior.

Similarly, when $$|x|\gg\xi$$, one has, for $$T, $$\langle \sigma_0\sigma_x\rangle^+ - \langle \sigma_0\rangle^+ \langle\sigma_x\rangle^+ \sim \frac{1}{|x|^{(d-1)/2}}e^{-|x|/\xi},$$ when $$d\geq 3$$, while $$\langle \sigma_0\sigma_x\rangle^+ - \langle \sigma_0\rangle^+ \langle\sigma_x\rangle^+ \sim \frac{1}{|x|^{2}}e^{-|x|/\xi},$$ when $$d=2$$ (this anomalous behavior when $$d=2$$ is the only claim that it really specific to the nearest-neighbor Ising model, and some relatives; for generic models, the behavior is also of Ornstein-Zernike type when $$d=2$$, that is, the exponent is $$1/2$$, not $$2$$).

As to the interpretation of the correlation length below $$T_c$$ and its behavior as $$T\downarrow 0$$, I already answered that elsewhere: see this answer (and also this one). Here, I'll only provide a graph of $$\xi$$ (along the horizontal direction) as a function of the temperature for the two-dimensional Ising model, where it can be computed explicitly (note that $$\xi$$ tends to $$0$$ both as $$T\downarrow 0$$ and as $$T\uparrow\infty$$, although the latter isn't visible in the picture below). • Is it a typo in the 3rd equation? Shouldn't it be $\frac{1}{|x|}e^{-x/\xi}$ instead of $\frac{1}{|x|^2}e^{-x/\xi}$?
– SRS
May 8 '20 at 5:59
• No, it's indeed $1/|x|^2$. This is a (well understood now) "pathology" of the planar Ising model. In the two-dimensional Ising model with non-nearest-neighbor interactions (thus non-planar), it is expected that the prefactor is again the usual $1/|x|^{(d-1)/2} = 1/\sqrt{|x|}$. May 8 '20 at 8:10
• I thought, for $T<T_C$, when a long-range order is established, the two-point correlation function should be nonzero no matter how large $x$ is. Is that true? If so, I can't see that from your expressions. @YvanVelenik
– SRS
May 15 '20 at 1:59
• The non-truncated 2-point function is bounded below. This follows from the FKG inequality: $\langle\sigma_0\sigma_x\rangle^+ \geq \langle\sigma_0\rangle^+\langle\sigma_x\rangle^+$, since the right-hand side is just the square of the magnetization density, which $>0$ in the $+$ state below $T_c$. The truncated 2-point function $\langle\sigma_0\sigma_x\rangle^+ - \langle\sigma_0\rangle^+\langle\sigma_x\rangle^+$ tends to $0$ (as $|x|\to\infty$) at all temperatures. It does so exponentially fast when $T<T_c$. May 15 '20 at 7:40
• As a complement to the previous comment: the fact that the truncated 2-point function tends to $0$ as $|x|\to\infty$ is true in any pure state, but can fail otherwise: for instance, if you consider the state obtained using free or periodic boundary conditions, then $\langle\sigma_0\sigma_x\rangle - \langle\sigma_0\rangle\langle\sigma_x\rangle = \langle\sigma_0\sigma_x\rangle$, which is still larger than the square of the magnetization density and thus $>0$ uniformly in $x$. May 15 '20 at 7:55