indicating that classically, the staggered magnetization vector $\mathbf{N}$ rotates around the (constant) total magnetization vector $\mathbf{M}$. Let $\mathbf{S}_{A(B)}=S_{A(B)}\hat{r}_{A(B)}$, where $S_{A(B)}$ are the magnitudes of the spins $\mathbf{S}_{A(B)}$ and $\hat{r}_{A(B)}$ are unit vectors indicating the directions of $\mathbf{S}_{A(B)}$; then, the classical ground state has $\hat{r}_A=-\hat{r}_B$ with $\mathbf{M}=0$, i.e., the two spins are antiferromagnetically aligned. Note that the equation of motion given in Eq.  implies that ${\partial(\mathbf{N}^2)\over\partial t}=0$, i.e., the magnitude of $\mathbf{N}$ remains unchanged during its motion. Therefore, if we write $\mathbf{N}=N\hat{n}$, where $N$ is the magnitude of $\mathbf{N}$ and $\hat{n}$ is the unit vector denoting its direction, we find that only $\hat{n}$ changes under the equation of motion given in Eq. .

The effects of quantum mechanics can be seen most easily by observing that the equation of motion given in Eq.  describes the dynamics of a free rotor (a rigid rod with one end fixed such that the rod can rotate freely around the fixed end). A free rotor can be represented by a vector $\mathbf{r}=r_0\hat{r}$, where $r_0=$ constant is the length of the rod and $\hat{r}$ is the unit radial vector describing the orientation of the rod. The rod has an angular momentum of $$\label{rotor1}
   \mathbf{L}=\mathbf{r}\times\mathbf{p}=r_0\hat{r}\times\mathbf{p},$$ where $\mathbf{p}=mr_0\dot{\hat{r}}$ 