 Solution

Rotate the spin on one sublattice through $\pi$ about its $x$ axis, using the unitary product of the corresponding spin-rotation operators. On a rotated site the old components obey $S_{\mathrm{old}}^x=S^x$, $S_{\mathrm{old}}^y=-S^y$, $S_{\mathrm{old}}^z=-S^z$. Every bond joins opposite sublattices, soThe longitudinal product changes sign. Hence the rotated bond is $-J\Delta S_m^z{S}_{m+1}^z+\frac{J}{2}(S_m^{+}{S}_{m+1}^{+}+S_m^{-}{S}_{m+1}^{-})$, giving the requested form. The canonical spin algebra is unchanged because the transformation is unitary. A staggered easy-axis reference configuration is now the uniform all-up state. At $\alpha =0$ one first chooses a Néel direction, here $z$, before performing the same rotation.