= 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 $S^x_{\rm old}=S^x$, $S^y_{\rm old}=-S^y$, $S^z_{\rm old}=-S^z$. Every bond joins opposite sublattices, so
$$
S^x_{m,\rm old}S^x_{m+1,\rm old}+S^y_{m,\rm old}S^y_{m+1,\rm old}=S^x_mS^x_{m+1}-S^y_mS^y_{m+1}=\frac12(S_m^+S_{m+1}^++S_m^-S_{m+1}^-).
$$
The longitudinal product changes sign. Hence \b[the rotated bond is $-J\Delta S_m^zS_{m+1}^z+\tfrac J2(S_m^+S_{m+1}^++S_m^-S_{m+1}^-)$], 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.
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