# Heisenberg ferromagnet in continuum limit

I consider the case of the simple, say 2D, Heisenberg ferromagnet with exchange interaction between the nearest neighbors. The Hamiltonian is: $$H = -J \sum_{<ij>} \mathbf S_i \mathbf S_j,$$ where $\mathbf S_{i,j}$ are classical spin vectors. I often encounter the following chain of arguments while transiting to the continuum limit. First, we express the scalar product of spins in terms of the small angle between them: $$\mathbf S_i \mathbf S_j \approx S_0^2 \cos \theta_{ij} \approx S_0^2 -\frac{1}{2}S_0^2 \theta_{ij}^2,$$ where $S_0=|\mathbf S|$ and $\theta_{ij}$ is the angle between $i$th and $j$th spins. The Hamiltonian then turns into: $$H =const +\frac{J}{2}S_0^2 \sum_{<ij>}\theta_{ij}^2 \rightarrow_{cont. \ lim.} \rightarrow \frac{J}{2}S_0^2 \int d^2 r (\nabla \theta(\mathbf r))^2.$$

The question is about the last transition: what is the correct way to perform the transition to the continuum limit rigorously?

Here I provide my attempt to obtain the continuum-case expression for a square lattice, which obviously has a flaw and leads to an incorrect answer. However, I don't see where the problem is.
First, I substitute an integral instead of the sum: $\sum_{<ij>} \rightarrow \int \frac{dx dy}{a^2}$, where $a$ is the distance between the neighbors. The expression for the angle may approximately be rewritten as $$\theta_{ij} \rightarrow \theta(\mathbf r+\Delta \mathbf r) - \theta(\mathbf r) \approx (\Delta\mathbf r, \nabla )\theta(\mathbf r).$$ Next, I need to write the square of it in the Hamiltonian: $$H \rightarrow \frac{J}{2} \int \frac{d^2 r}{a^2} \left [ (\Delta \mathbf r, \nabla )\theta( \mathbf r)\right ]^2 = \frac{J}{2} \int \frac{d^2 r}{a^2} \left [ a_x \frac{\partial \theta }{\partial x} + a_y \frac{\partial \theta }{\partial x}\right ]^2 = \\ =\frac{J}{2}\int \frac{d^2 r}{a^2} \left ( a_x^2 \left(\frac{\partial \theta }{\partial x} \right )^2+ a_y^2 \left(\frac{\partial \theta }{\partial y} \right )^2 + 2a_x a_y \frac{\partial \theta }{\partial x} \frac{\partial \theta }{\partial y}\right ),$$ where $a_{x,y}=\Delta r_{x,y}$. The last term in the round brackets clearly is absent in the correct expression. Where am I wrong in my calculations?

Remember that the Hamiltonian involves a sum over all pairs of neighbouring sites. Assume that the sites are located on a square lattice so that their positions are given by $\vec r=a(n_x\vec e_x+n_y\vec e_y)$ where the n's are integer. The Hamiltonian reads \eqalign{ H&=-J\sum_{\vec r} \big(\vec S_{\vec r+a\vec e_x}.\vec S_{\vec r} +\vec S_{\vec r+a\vec e_y}.\vec S_{\vec r}\big)\cr &=-JS^2\sum_{\vec r} \Big[\cos\big(\theta_{\vec r+a\vec e_x} -\theta_{\vec r}\big)+\cos\big(\theta_{\vec r+a\vec e_y} -\theta_{\vec r}\big)\Big]\cr &\simeq -JS^2\sum_{\vec r} \Big[\cos\big(a\partial_x\theta\big) +\cos\big(a\partial_y\theta\big)\Big]\cr &\simeq -{JS^2\over 2}\sum_{\vec r} \Big[1-a^2 \big(\partial_x\theta\big)^2-a^2 \big(\partial_y\theta\big)^2\Big]\cr &={\rm Cst}+{JS^2a^2\over 2}\sum_{\vec r} ||\vec\nabla\theta||^2 } On an hexagonal lattice, one needs to distinguish two sublattices. On one of them, the three neighbours are at $\vec a=\vec e_x$, $\vec b=-{1\over 2}\vec e_x+{\sqrt 3\over 2}\vec e_y$ and $\vec c=-{1\over 2}\vec e_x-{\sqrt 3\over 2}\vec e_y$. The energy on site $\vec r$ then reads \eqalign{ &-J\big(\vec S_{\vec r+a\vec a}.\vec S_{\vec r} +\vec S_{\vec r+a\vec b}.\vec S_{\vec r} +\vec S_{\vec r+a\vec c}.\vec S_{\vec r}\big)\cr &=-JS^2\Big[\cos\big(\theta_{\vec r+a\vec a} -\theta_{\vec r}\big)+\cos\big(\theta_{\vec r+a\vec b} -\theta_{\vec r}\big)+\cos\big(\theta_{\vec r+a\vec c} -\theta_{\vec r}\big)\Big]\cr &\simeq -JS^2\Big[\cos\big(a\partial_x\theta\big) +\cos\Big(-{a\over 2}\partial_x\theta+a{\sqrt 3\over 2} \partial_y\theta\Big)+\cos\Big(-{a\over 2}\partial_x\theta -a{\sqrt 3\over 2}\partial_y\theta\Big)\Big]\cr &\simeq -{JS^2\over 2}\sum_{\vec r} \Big[1-a^2 \big(a\partial_x\theta\big)^2-a^2\Big(-{1\over 2}\partial_x\theta +{\sqrt 3\over 2}\partial_y\theta\Big)^2-a^2\Big(-{1\over 2} \partial_x\theta-{\sqrt 3\over 2}\partial_y\theta\Big)^2\Big]\cr &={\rm Cst}+{JS^2a^2\over 2}\times {3\over 2}||\vec\nabla\theta||^2 } The only difference with the square lattice is a geometric factor $3/2$.
• Thanks for the input. However, it seems that I didn't make an assumption that $\nabla \theta$ is divergence. What more, I did not obtain $\nabla \theta$ solely, but rather $(\Delta \mathbf r, \nabla) \theta = a_x \frac{\partial \theta}{\partial x }+a_y \frac{\partial \theta}{\partial y }$ in my derivation, where $a_{x,y}=\Delta r_{x,y}$. – Gretchen Oct 23 '17 at 12:46
• I do not know any general proof. I added the case of an hexagonal lattice. I guess that the same kind of result is obtained in 2D as long as the neighbours are uniformly distributed over the unit circle around the point $\vec r$. – Christophe Oct 24 '17 at 8:41