 Solution

For the XXZ spin chain, write $\Delta =1+\alpha$ and replace each spin by a classical vector of length $S$. A bond is $J{S}^2{\mathbf{n}}_m^{\mathsf{T}}\mathrm{diag}(1,1,\Delta ){\mathbf{n}}_{m+1}$, whose minimum is $-J{S}^2\max (1,|\Delta |)$. For an even periodic chain all bonds can attain that bound simultaneously. The classical ground states of an XXZ spin chain therefore give $\alpha >0$ gives the two easy-axis Néel states ${\mathbf{S}}_m=(-1{)}^{m}S\hat{\mathbf{z}}$ and its reversal, with energy $-JN{S}^2(1+\alpha )$.

For $-2<\alpha <0$ the ground states are planar Néel configurations, ${\mathbf{S}}_m=(-1{)}^{m}S(\cos \phi ,\sin \phi ,0)$, with arbitrary $\phi$: their degeneracy is a circle and their energy is $-JN{S}^2$. This includes the easy-plane interval used later. Part (a), however, does not restrict negative $\alpha$ to that interval. For $\alpha <-2$, $\Delta <-1$ and the lowest states are instead the two uniform $z$-polarized ferromagnetic configurations, with energy $JN\Delta S^2$. At $\alpha =-2$, a rotation by $\pi$ about $z$ on alternating sites maps the classical energy to an isotropic ferromagnet: ${\mathbf{S}}_m=S((-1{)}^m{n}_x,(-1{)}^m{n}_y,n_z)$ for any unit vector $\mathbf{n}$ is a ground state. This special degeneracy is a sphere. At the other boundary, $\alpha =0$, the ordinary Néel direction is arbitrary on a sphere. These statements concern classical minimizing configurations, not the degeneracy of the exact finite-chain quantum ground state.