# Argument on why spin correlation functions in Ising model decay exponentially with a correlation length?

I'm reading Quantum Field Theory in Strongly Correlated Electronic Systems, Nagaosa.

Consider 1D Ising model, $$H=J_z\sum_i S^z_iS^z_{i+1}.$$ on page 3, it says

The groud stae is 2-fold degenerate because the Hamiltonian is invariant under the transformation $$S^i_z \rightarrow -S^i_z$$, performed at all sites $$i$$. Calling these two ground states $$A$$ and $$B$$ and assuming that the system at the right-hand side is in state $$A$$, and at the left-hand side in state $$B$$, then somewhere there must exist a boundary between region $$A$$ and region $$B$$. This boundary is called a kink or soliton. Because at finite temperature this excitation occurs with a finite density, the spin correlation function $$F(r) =\langle S^z_iS^z_{i+r}\rangle$$ will decay exponentially with a correlation length $$\xi$$.

I know how to directly calculate the correlation function, but I wonder how the argument for exponential decay of correlation function is made here and how to understand it.

Any help would be highly appriciated!!

Let me write the Hamiltonian $$H = -J \sum_i S_i^z S_{i+1}^z.$$ This choice will avoid some annoying (and irrelevant) signs.

One way to formulate the statement in the OP precisely is as follows.

Consider the variables $$\delta_i=S_i^zS_{i+1}^z$$. Since $$\delta_i=1$$ when the spins at $$i$$ and $$i+1$$ agree and $$\delta_i=-1$$ when the spins at $$i$$ and $$i+1$$ disagree, you can identify them with the kinks in your question (that is, there is a kink between $$i$$ and $$i+1$$ when $$\delta_i=-1$$).

Introducing the variables $$\delta_i=S_i^zS_{i+1}^z$$, the Hamiltonian becomes $$H = J^z \sum_i \delta_i.$$ It follows that the random variables $$\delta_i$$ are independent and identically distributed. One can easily compute their expectation: since $$P(\delta_i = 1) = \frac{e^{\beta J^z}}{e^{\beta J^z} + e^{-\beta J^z}},$$ one has $$\langle \delta_i \rangle = \frac{e^{\beta J^z} - e^{-\beta J^z}}{e^{\beta J^z} + e^{-\beta J^z}} = \tanh(\beta J^z).$$ Finally, noting that $$S_i^zS_{i+r}^z = \delta_i\delta_{i+1}\cdots\delta_{i+r-1}$$, we obtain $$\langle{S_i^zS_{i+r}^z}\rangle = \langle\delta_i\delta_{i+1}\cdots\delta_{i+r-1}\rangle = \langle \delta_i \rangle^r = (\tanh(\beta J^z))^r.$$

In words, the fact that kinks proliferate in the system (at each $$i$$, there is a positive probability that a kink is present, so there will be a positive density of them in the system) prevents the ordering of the spins.

• Thanks! What you post is what I considered "directly calculate the correlation function", but I wonder would it be fair to say that in systems that can have finite excitation density (a gapped system?), some "correlation function" will decay exponentially with a correlation length? Nov 3 '20 at 6:36
• The 1d Ising model being so simple, once you formulate the problem precisely, you can usually just compute everything explicitly... But the qualitative answer is the one I give at the end: the probability of having an interval of length L in which all spins are equal decays exponentially since between each pair of neighbors, you have a positive probability of having a kink. Nov 3 '20 at 11:11
• Why must it decay "exponentially" then? Nov 3 '20 at 12:02
• For the same reason that the probability that you obtain n heads in a row when flipping a coin decays exponentially with n... Nov 3 '20 at 12:04
• Another explanation : start with the spin at the left end of the interval. Each time you encounter a kink, the spin changes sign. So, the spin at the right end will have the same sign as the spin at the left end if and only if you have an even number of kinks. But the probability of this happening converges to 1/2 exponentially fast with the length of the interval. Nov 3 '20 at 12:08