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Use dimensionless <spin angular momentum> operators, so their commutators have no additional $\hbar$. A fully polarized state $|F\rangle=\bigotimes_n|S,S\rangle_n$ is an exact <ground state> of the <Heisenberg ferromagnet>, with \b[$E_0=-NJS^2$]. On each bond the largest eigenvalue of $\mathbf S_n\cdot\mathbf S_{n+1}$ is $S^2$, obtained in the two-spin total-spin-$2S$ sector. Hence the bondwise lower bound on the Hamiltonian is attained by this state.
The <ferromagnetic ground-state multiplet> has total spin $NS$. Applying the global lowering operator $(S^-_{\rm tot})^r$, $r=0,\ldots,2NS$, generates its $2NS+1$ magnetic states; their linear combinations are also <ground states>. Global spin rotations generate polarized <ground states> pointing in arbitrary directions and span the same multiplet. Uniform reorientation therefore costs no energy. Slowly varying orientations give gapless <ferromagnetic magnons>, with quadratic rather than linear long-wavelength dispersion. In the ordered thermodynamic <ground state>, the two broken spin generators are canonically paired because $\langle[S^x_{\rm tot},S^y_{\rm tot}]\rangle=i\langle S^z_{\rm tot}\rangle\ne0$, producing one <type-B Goldstone boson>. For a finite chain the zero-wavevector motion remains inside the exact ground multiplet; nonzero wavevectors have a finite-size spacing that vanishes as $N^{-2}$. This is a ground-state statement, not a claim of finite-temperature long-range order in one dimension.
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