2.1.1 Exchange energy

The Heisenberg Hamiltonian of the exchange interaction is usually written in the form

$$\mathcal H_\mathrm{exch}=-\sum_{i,j=1}^M J_{ij}\mathbf{S}_i\cdot\mathbf{S}_j \quad,$$

where $J_{ij}$ is the exchange integral, which can be calculated using quantum mechanics [11,12]. It decreases rapidly with increasing distance between the atoms, and so the sum has to be taken only for nearest neighbours and we can write $J$ for $J_{ij}$. $\mathbf{S}_{i,j}$ stands for the spin operators. If we replace them by classical vectors and rewrite the dot product, we obtain for the exchange energy

$$E_\mathrm{exch}= -J S^2 \sum_{i,j\vert i \not= j} \cos \phi_{i,j} \quad.$$

Next, we assume, that the angles $\phi_{i,j}$ are small and develop the cosine into its Taylor series expansion. We also take the sum for each pair of nearest neighbours only once and redefine the zero level of the exchange energy by removing the constant term.

$$E_\mathrm{exch}= J S^2 \sum_{NN} \phi_{i,j}^2$$

If we use the continuous variable $\mathbf{m}=\mathbf{M} / M_\mathrm{s}$ for the magnetization, we get for small angles

$$\vert\phi_{i,j}\vert \approx \vert\mathbf{m}_i - \mathbf{m}... ...prox \vert(\mathbf{r}_{i} \cdot \nabla) \mathbf{m}\vert \quad,$$

where $\mathbf{r}_{i}$ is the position vector from lattice point $i$ to $j$. Then, the exchange energy is given by

$$E_\mathrm{exch}= J S^2 \sum_{i}\sum_{\mathbf{r}_i} [(\mathbf{r}_{i} \cdot \nabla) \mathbf{m}]^2 \quad.$$

Changing the summation over $i$ to an integral over the ferromagnetic body, we get

$$E_\mathrm{exch}= \int_V A \left[ (\nabla m_x)^2+(\nabla m_y)^2+ (\nabla m_z)^2\right] \,d{^3r}\, \quad.$$ (2.5)

The exchange constant $A$ is given by

$$A=\frac{J S^2 c}{a} \quad,$$

where $a$ is the distance between nearest neighbours and $c \in \{1, 2, 4\}$ for a simple cubic, body centred cubic, and face centred cubic crystal structure, respectively.