Love thy neighbor, and everyone else too

Once again, the ground state is given by the arrangement

$\ldots \uparrow \uparrow \uparrow \uparrow \uparrow \uparrow \uparrow \ldots$

which has the energy

$\int\limits_{-N}^{+N}di\int\limits_{-N}^{+N} \frac{dj}{\lvert i - j \rvert^{\gamma_r}} \approx 2N\sum\limits_{i=1}^N i^{-1}$

and the entropy

$S = k_B \log 2$

The lowest excitation from the ground state is the formation of a single domain wall, which is a hard boundary between up and down spins:

$\ldots \uparrow \uparrow \uparrow \uparrow \downarrow \downarrow \downarrow \downarrow \ldots$

This state has entropy $S = k_B \log N$ because the domain wall can be placed at any of the $N$ positions in the 1d line.

The energy of this state, relative to the ground state, is found by summing the spin interactions between each pair of unlike spin states. For convenience, we place the boundary at between $i=n$ and $i=n+1$ .

First, we integrate over each interaction of a given up spin, then we integrate over all up spins. In a bit more detail, the steps are

• Integrate over all interactions of "up" spin $n$ to get an expression for the energy due to a given spin variable $n$
• Take the limit $N\rightarrow\infty$ , to get kill off terms that die in the infinite limit.
• Integrate over all of the "up" spins to get an expression for the energy of all up-down interactions as a function of the number of particles in the system $N$
• Consider the behavior of $E(\text{domain wall})$ in the infinite limit for different value of $sigma$

The integral we face is given by

$E(\text{domain wall})=\lim\limits_{N\rightarrow\infty}\int\limits_1^n di \int\limits_{n+1}^N dj \frac{\Delta E}{\lvert i-j \rvert^{\gamma_r}}$

To ease calculation, let's make the change of variables

$\begin{aligned} l&=i-j \\ dl&=-dj \end{aligned}$

so that $l$ goes from $n-1$ to $N-i$ , and $i$ goes from $1$ to $N$ .

Proceeding with the integral over $l$

$\begin{aligned} E(\textrm{domain wall}) &= \Delta E \int\limits_1^n di \int\limits_{n-i}^{N-i} \frac{dl}{l^{\gamma_r}} \\ &= \frac{- \Delta E}{\gamma_r-1}\int\limits_1^n di \left[\frac{1}{\left(N-i\right)^{{\gamma_r}-1}} - \frac{1}{\left(n-i\right)^{\gamma_r-1}}\right] \\ \end{aligned}$

which becomes

$E(\text{domain wall}) \approx \frac{\Delta E}{{\gamma_r}-1}\int\limits_1^n di\frac{1}{\left(n-i\right)^{{\gamma_r}-1}}$

in the limit $N\rightarrow\infty$ . Performing the integral over $i$ , we have

$E(\textrm{domain wall}) \approx \Delta E \frac{ (n+1)^{2-\gamma_r}}{\left(\gamma_r-1\right)\left(\gamma_r-2\right)}$

which goes as $\sim N^{2-{\gamma_r}}$ .

Recall, the entropy of the domain wall is given by $\sim k_B\log N$ since we can place the wall at any of the $N$ spin variables. $N^{2-{\gamma_r}}$ clearly outpaces $\log N$ for all $\sigma < 2$ and so, entropy will always win for low enough temperatures.

In other words, there will always be a finite $T$ below which $N^{2-{\gamma_r}} > k_B T\log N$ , i.e. in which the ground state is favored over the domain wall, in the infinite limit. This implies that for $\gamma_r<2$ , the infinite 1d spin lattice will find long range order at finite temperatures and, so, the upper bound for $\gamma_r$ is 2.