n-vector model

The continuum limit can be understood to be the sigma model. This can be easily obtained by writing the Hamiltonian in terms of the product

${\displaystyle -{\tfrac {1}{2}}(\mathbf {s} _{i}-\mathbf {s} _{j})\cdot (\mathbf {s} _{i}-\mathbf {s} _{j})=\mathbf {s} _{i}\cdot \mathbf {s} _{j}-1}$

where ${\displaystyle \mathbf {s} _{i}\cdot \mathbf {s} _{i}=1}$ is the "bulk magnetization" term. Dropping this term as an overall constant factor added to the energy, the limit is obtained by defining the Newton finite difference as

${\displaystyle \delta _{h}[\mathbf {s} ](i,j)={\frac {\mathbf {s} _{i}-\mathbf {s} _{j}}{h}}}$

on neighboring lattice locations ${\displaystyle i,j.}$ Then ${\displaystyle \delta _{h}[\mathbf {s} ]\to \nabla _{\mu }\mathbf {s} }$ in the limit ${\displaystyle h\to 0}$, where ${\displaystyle \nabla _{\mu }}$ is the gradient in the ${\displaystyle (i,j)\to \mu }$ direction. Thus, in the limit,

${\displaystyle -\mathbf {s} _{i}\cdot \mathbf {s} _{j}\to {\tfrac {1}{2}}\nabla _{\mu }\mathbf {s} \cdot \nabla _{\mu }\mathbf {s} }$

which can be recognized as the kinetic energy of the field ${\displaystyle \mathbf {s} }$ in the sigma model. One still has two possibilities for the spin ${\displaystyle \mathbf {s} }$: it is either taken from a discrete set of spins (the Potts model) or it is taken as a point on the sphere ${\displaystyle S^{n-1}}$; that is, ${\displaystyle \mathbf {s} }$ is a continuously-valued vector of unit length. In the later case, this is referred to as the ${\displaystyle O(n)}$ non-linear sigma model, as the rotation group ${\displaystyle O(n)}$ is group of isometries of ${\displaystyle S^{n-1}}$, and obviously, ${\displaystyle S^{n-1}}$ isn't "flat", i.e. isn't a linear field.