= Solution
The <Holstein–Primakoff transformation> converts a <spin> algebra into bosonic operators while retaining the finite physical <spin> <Hilbert space>. With $[a_n,a_m^\dagger]=\delta_{nm}$, write
$$
S_n^z=S-a_n^\dagger a_n,\qquad
S_n^+=\sqrt{2S-a_n^\dagger a_n}\,a_n,\qquad
S_n^-=a_n^\dagger\sqrt{2S-a_n^\dagger a_n}.
$$
The <Holstein–Primakoff occupation constraint> is $0\le n_n\le2S$. Acting on $|n\rangle$, the raising <matrix> element is $\sqrt{n(2S-n+1)}$ and the lowering element is $\sqrt{(n+1)(2S-n)}$. Their squared difference is $2(S-n)$, reproducing $[S^+,S^-]=2S^z$. Operator ordering inside the square roots is essential. For low <boson> occupation relative to $S$, expand them to obtain $S^+\simeq\sqrt{2S}\,a$ and $S^-\simeq\sqrt{2S}\,a^\dagger$. Keeping quadratic terms gives the <linear spin-wave approximation>; higher terms describe <magnon> interactions. Large $S$ helps control the expansion, but a low-dimensional <infrared divergence> can invalidate it even at large finite $S$.
For the <ferromagnet> choose $H_F=-J\sum_n\mathbf S_n\cdot\mathbf S_{n+1}$, $J>0$, with unit lattice spacing and $\hbar=1$. Expand around the all-up state. A longitudinal bond contributes $-JS^2+JS(n_n+n_{n+1})$, and the transverse exchange contributes $-JS(a_n^\dagger a_{n+1}+a_{n+1}^\dagger a_n)$. After a <Fourier transform>,
$$
H_F^{(2)}=-JNS^2+\sum_k\omega_F(k)a_k^\dagger a_k,
\qquad\boxed{\omega_F(k)=2JS(1-\cos k)\simeq JS k^2.}
$$
These are <ferromagnetic magnons>. The polarized vacuum is an exact <ground state> of the full ferromagnetic <Hamiltonian>: every bond is in its maximum <total-spin sector>, minimizing that bond <energy>. It therefore has no quantum zero-point reduction of its <magnetization>. The one-magnon dispersion is also exact, although interacting many-magnon states need not be free <bosons>. Rotating the <polarization> generates the maximum-total-spin ground-state multiplet. Selecting a definite direction in a thermodynamic symmetry-breaking description breaks $SU(2)$ to $U(1)$; the two transverse <spin> components form a canonical pair, producing a single quadratic <type-B Goldstone boson>. A small aligning field gaps it by the Zeeman <energy>.
At nonzero <temperature> the one-dimensional <ferromagnet> is different: its thermal <magnon> number contains $\int dk\,[e^{\beta JS k^2}-1]^{-1}$, with infrared behaviour $\int dk\,T/k^2$. The divergent <magnetization> reduction signals the absence of finite-temperature <long-range order> in an infinite short-range isotropic chain, consistent with the <Mermin-Wagner theorem>. It does not disallow the polarized zero-temperature <ground states> just constructed.<Absence of Ferromagnetism or Antiferromagnetism in One- or Two-Dimensional Isotropic Heisenberg Models by N. D. Mermin and H. Wagner>[abstract and finite-temperature theorem]{ref}
For the <antiferromagnet> use $H_{AF}=+J\sum_n\mathbf S_n\cdot\mathbf S_{n+1}$, still $J>0$, on an even bipartite chain. The classical reference state alternates up and down. Rotate one sublattice through $\pi$ so that both local quantization axes point along their reference <spins>. In unrotated physical components, to leading order $S_A^z=S-a^\dagger a$, $S_B^z=-S+b^\dagger b$, $S_A^+\simeq\sqrt{2S}\,a$ and $S_B^+\simeq\sqrt{2S}\,b^\dagger$. Choosing Fourier phases using the actual site <positions> gives
$$
H_{AF}^{(2)}=-JNS^2+2JS\sum_{k\in\mathrm{RBZ}}
\left[a_k^\dagger a_k+b_k^\dagger b_k+
\cos k\,(a_kb_{-k}+a_k^\dagger b_{-k}^\dagger)\right].
$$
The reduced <Brillouin zone> has $M=N/2$ momenta and <length> $\pi$. The pair creation and annihilation terms express the fact that an antiparallel reference product state is not an <eigenstate>. A bosonic <Bogoliubov transformation> with $u_k=\cosh\theta_k$, $v_k=\sinh\theta_k$, $\tanh2\theta_k=\cos k$ diagonalizes each two-mode block. With the sign convention $a_k=u_k\alpha_k-v_k\beta_{-k}^\dagger$ and $b_{-k}=u_k\beta_{-k}-v_k\alpha_k^\dagger$, the result is
$$
H_{AF}^{(2)}=E_{AF}^{(2)}+\sum_{k\in\mathrm{RBZ}}\omega_{AF}(k)
(\alpha_k^\dagger\alpha_k+\beta_k^\dagger\beta_k),
$$
$$
\boxed{\omega_{AF}(k)=2JS\sqrt{1-\cos^2 k}=2JS|\sin k|\simeq2JS|k|,}
\qquad E_{AF}^{(2)}=-JNS^2+\sum_{k\in\mathrm{RBZ}}[\omega_{AF}(k)-2JS].
$$
Thus the harmonic antiferromagnetic modes have linear, rather than quadratic, long-wave dispersion. In an unfolded Brillouin-zone description the same result is represented by zeros at both zero and the antiferromagnetic <wavevector>; these are not extra independent degrees of freedom beyond the two-sublattice accounting.
The Bogoliubov vacuum has no $\alpha$ or $\beta$ excitations but contains original <spin> <bosons>, with $v_k^2=\tfrac12(|\sin k|^{-1}-1)$. Its <staggered magnetization> is $m_s=S-M^{-1}\sum_{k\in\mathrm{RBZ}}v_k^2$. This <quantum depletion of Néel order> diverges logarithmically at small $k$ in one dimension. \b[Linear spin-wave theory therefore does not establish a stable ordered antiferromagnetic <ground state> for this chain.] Its nominal gapless dispersion is a result of expanding around an invalid ordered reference, not a general proof of the exact excitation spectrum. For the nearest-neighbour isotropic chain, integer-spin systems have a Haldane gap, whereas half-integer-spin systems are gapless; the spin-one-half chain in particular has fractional <spin> excitations rather than a simple dilute gas of ordered-state <magnons>. This distinction requires physics beyond the harmonic expansion, including the topological term in the continuum theory.<Nonlinear Field Theory of Large-Spin Heisenberg Antiferromagnets: Semiclassically Quantized Solitons of the One-Dimensional Easy-Axis Néel State by F. Duncan M. Haldane>[pp.1153–1156, integer versus half-integer spin]{ref}
\Image[/past-exam-of-the-quantum-condensed-matter-field-theory-course-of-the-university-of-cambridge/2015/paper-1-spin-wave.png]
{title=Original harmonic chain dispersions; the antiferromagnetic curve is a spin-wave approximation, not the exact spectrum for every spin}
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