Final energy difference

Since the spin to be flipped takes only two values, $s_l^1=\pm 1$ and $s_l^2=\pm 1$, it means that if $s_l^1= 1$, then $s_l^2=-1$ and if $s_l^1= -1$, then $s_l^2=1$. The other spins keep their values, meaning that $s_k^1=s_k^2$. If $s_l^1= 1$ we must have $s_l^1-s_{l}^2=2$, and if $s_l^1= -1$ we must have $s_l^1-s_{l}^2=-2$. From these results we see that the energy difference can be coded efficiently as $$\begin{equation} \Delta E = 2Js_l^1\sum_{< k>}^{N}s_k, \tag{3} \end{equation}$$ where the sum runs only over the nearest neighbors $k$ of spin $l$. We can compute the change in magnetisation by flipping one spin as well. Since only spin $l$ is flipped, all the surrounding spins remain unchanged.

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