I am studying the exchange interaction, which can be described with the Heisenberg Hamiltonian:
$\hat{H} = -\sum_{i,j}J_{ij}\hat{\mathbf{S_i}}\cdot \hat{\mathbf{S_j}}$
In the framework of constant exchange energy J and by calling $\phi_{ij}$ the angle between the spins, which are therefore treated classically, the energy becomes:
$E = -JS^2\sum_{\langle ij \rangle} \cos{\phi_{ij}} \simeq const + \frac{JS^2}{2} \sum_{\langle ij \rangle}\phi_{ij}^2$
for sufficiently small $\phi$ values.
If I define the unitary vector of the magnetic dipole and call their distance $r_{ij}$ by exploiting the Taylor expansion one gets:
$|\phi_{ij}| \simeq |m_i-m_j| \simeq \left |\left(\mathbf{r_{ij}}\cdot \nabla \right)\mathbf{m} \right| $
Keeping this guy inside the energy expression:
$E = \frac{JS^2}{2}\sum_{\langle ij \rangle} \left|\left(\mathbf{r_{ij}}\cdot \nabla \right)\mathbf{m} \right|^2 = \frac{A}{2}\int\left[\left(\nabla m_x\right)^2 + \left(\nabla m_y\right)^2 \left(\nabla m_z\right)^2\right] $
where A is the exchange stiffness. I tried to compute it for a simple cube, and I managed. I have three equivalent direction. What about HCP, do I have to choose c along the third direction such as I have 12 nearest neighbors?
The expected value is $A_{hcp} = \frac{4\sqrt{2}JS^2}{a}$
