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After a bipartite spin rotation, a leading Holstein–Primakoff transformation expansion gives $H_2=-J\Delta N{S}^2+JS{\sum }_k[2\Delta n_k+\cos k(a_k{a}_{-k}+a_k^{†}{a}_{-k}^{†})]$. The bosonic Bogoliubov diagonalization gives the displayed spectrum and vacuum energy $-J\Delta N{S}^2+JS{\sum }_k[\sqrt{{\Delta }^2-{\cos }^{2}k}-\Delta ]$. The anisotropy gap is $2JS\sqrt{{\Delta }^2-1}$. At $\Delta =1$ the finite-angle transformation is singular at its zero modes, and quantum depletion of Néel order is infrared divergent in one dimension.

For the XY model with equal ferromagnetic exchange $J$ in its plane and field term $-hS{\sum }_j{S}_j^x$, rotate the local quantization axis along the Classical spin. The Holstein–Primakoff transformation gives the number coefficient ${\alpha }_k=S(|h|+2J-J\cos k)$ and anomalous coefficient ${\Delta }_k=JS\cos k$ with one transverse-axis convention. Bosonic Bogoliubov diagonalization gives ${ϵ}_k=\sqrt{{\alpha }_k^2-{\Delta }_k^2}$ and field gap $S\sqrt{|h|(|h|+2J)}$. A nonzero field stabilizes every oscillator mode.

Use the unitary discrete Fourier transformThe transformed operators obey $[a_k,a_q^{†}]={\delta }_{kq}$. The hopping terms contribute $2A\cos k$ to the number coefficient. In the pairing sum, the odd sine part cancels between $k$ and $-k$. ThuswhereThe factor one half in the full momentum sum avoids counting an anomalous pair twice. For $H_f>0$, ${\alpha }_k>|{\Delta }_k|$ for every allowed momentum. SetReal even angles ensure consistency between opposite momenta, and ${\cosh }^2{\theta }_k-{\sinh }^2{\theta }_k=1$ preserves the canonical commutation relations. The anomalous coefficient becomes ${\Delta }_k\cosh 2{\theta }_k-{\alpha }_k\sinh 2{\theta }_k=0$. Normal ordering then giveswithThe field-polarized planar ferromagnetic spin wave has energy gap $ϵ(0)=S\sqrt{H_f(H_f+2J)}$. The momenta run over the whole Brillouin zone once, including the self-inverse momenta $0$ and, for even $L$, $\pi$; these use single-mode squeezing rather than introducing an extra independent partner.

The zero-field case has a zero-mode obstruction to bosonic Bogoliubov diagonalization. Its nonzero-momentum modes have $ϵ(k)=2\sqrt{2}JS|\sin (k/2)|$, but ${\alpha }_0={\Delta }_0=JS$, so no finite squeezing angle exists for the uniform mode. With $Q_0=(a_0+a_0^{†})/\sqrt{2}$, its Hamiltonian is $JS{Q}_0^2-JS/2$, not an identically zero oscillator. For example, its matrix element between the vacuum and the two-boson number state is $JS/\sqrt{2}$, whereas $ϵ(0)b_0^{†}{b}_0$ would vanish. Consequently the requested oscillator diagonalization is literal for nonzero field; at zero field treat the collective orientation separately or introduce a field regulator before taking a limit.

For nonzero field, the spin-wave ground state is the vacuum of all $b_k$. The inverse Bogoliubov transformation gives $⟨a_k^{†}{a}_q⟩={\delta }_{kq}{\sinh }^2{\theta }_k$ and vanishing one-point oscillator averages. Translational invariance therefore gives the depletionRotating back to the physical axes yields, to the retained spin-wave order,The zero transverse components also follow from rotation by $\pi$ about the physical $x$ axis in a symmetry-preserving ground state. The nonzero depletion describes zero-point quantum fluctuations: the original boson vacuum is not the ground state because the anomalous terms create pairs. It is nonnegative and must satisfy $\delta S\ll S$ for the expansion to be reliable. At strong field it tends to zero, with $\delta S\sim J^2/(8H_f^2)$ in the infinite-chain limit.

For fixed $H_f>0$, the momentum sum has a finite thermodynamic limit obtained by $L^{-1}{\sum }_k\to (2\pi {)}^{-1}{\int }_{-\pi }^{\pi }dk$. At zero field the uniform mode already has no oscillator vacuum, as explained in part d. Even after removing that mode, the nonzero modes behave asThus assigning a finite ordered moment by simply setting $h=0$ in the infinite momentum sum is invalid. The divergence indicates failure of this expansion about a fixed planar orientation; it is not a negative physical magnetization. At finite size an exact zero-field state chosen to have definite total $S^z$ has zero planar one-point magnetization by the continuous rotation symmetry. Establishing its full correlations requires a treatment beyond the divergent ordered-state approximation.

Use bosonic Bogoliubov diagonalization with $A_k=2JS$, $B_k=2JS{\gamma }_k$. Away from the zero modes, letHere $u_k^2-v_k^2=1$ preserves the canonical commutation relations, while the sign of $v_k$ cancels anomalous pairing. In hyperbolic notation $u_k=\cosh {\theta }_k$, $v_k=\sinh {\theta }_k$, this condition is $\tanh (2{\theta }_k)=-B_k/A_k$. Combining the normal ordering constants givesThe dispersion vanishes linearly near $k=0$ and $\pi$, with spin-wave velocity $2JS$ in unit lattice spacing. The exact zero modes make the Bogoliubov coefficients singular; use a small infrared regulator or a symmetry-selected reference and treat the global rotations separately. A finite transformation at ${\omega }_k=0$ is not asserted.

On a hypercubic lattice of coordination $z=2d$, the same hypercubic antiferromagnetic spin-wave dispersion usesThe corresponding constant is $-dNJS(S+1)$, and ${\omega }_{\mathbf{k}}\sim 2JS\sqrt{d}|\mathbf{k}|$ near a Goldstone point. Restoring lattice spacing $a$ multiplies this velocity by $a$.

Apply a discrete Fourier transform, with lattice spacing one:The wavevectors lie in a Brillouin zone, which we choose as $[-\pi ,\pi )$. Pairing $k$ with $-k$ makes the quadratic HamiltonianA bosonic Bogoliubov diagonalization uses $a_k=\cosh {\theta }_k{\alpha }_k-\sinh {\theta }_k{\alpha }_{-k}^{†}$. The canonical commutation relations hold because ${\cosh }^2{\theta }_k-{\sinh }^2{\theta }_k=1$. The anomalous terms vanish when $\tanh 2{\theta }_k=B_k/A=\cos k$, givingNear $k=0$ and $k=\pi$, the dispersion relation is linear: ${\omega }_k\sim 2JS|k-k_0|$. These are the low-energy antiferromagnetic spin waves; the two zeroes are related by the two-sublattice description. The transformation is singular at the exact zero modes, which require an infrared regulator or separate treatment of the collective rotation.

For the easy-axis XXZ antiferromagnetic spin-wave spectrum, the consistent quadratic Hamiltonian in part (d) has $A=\Delta$ and $B_k=\cos k$. The inverse transformation acts on the Nambu matrix by a congruence. Its diagonal and off-diagonal entries become $A\cosh 2{\theta }_k-B_k\sinh 2{\theta }_k$ and $B_k\cosh 2{\theta }_k-A\sinh 2{\theta }_k$. The pair term vanishes whenFor $\alpha >0$ the transformation is real and finite for every $k$. Summing the two Nambu number operators over the full zone givesHere $E_0$ names the constant in the requested oscillator form; the actual quadratic ground-state energy is $E_{\mathrm{gs}}^{(2)}=E_0+\frac{1}{2}{\sum }_k{ϵ}_k=-J\Delta N{S}^2+JS{\sum }_k({\Omega }_k-\Delta )$. This distinguishes the normal-ordering constant from the total zero-point contribution. bosonic Bogoliubov diagonalization is a canonical, not an ordinary unitary, diagonalization of the Nambu matrix.

At $\alpha =0$ and $k\ne 0,\pi$, the same formula holds with ${\Omega }_k=|\sin k|$. The exact zero modes have $|\tanh 2{\theta }_k|=1$, so no finite squeezing angle diagonalizes them as positive-frequency oscillators. Treat collective rotations separately or retain a small anisotropy regulator before taking the limit. Finally, if one instead diagonalizes the literal erroneous printed matrix, it gives $\tanh 2{\theta }_k=\cos k/(2\Delta )$ and ${ϵ}_k^{\mathrm{print}}=2JS\sqrt{4{\Delta }^2-{\cos }^{2}k}$. That result remains gapped at $\alpha =0$ and is not the spin-wave spectrum of the Hamiltonian stated at the start.

Use the unitary discrete Fourier transform $a_j=L^{-1/2}{\sum }_k{e}^{ikj}{a}_k$, where $k=2\pi n/L$. Thenwith ${\alpha }_k=S[2J^x-(J^y+J^z)\cos k]$ and ${\beta }_k=-S(J^y-J^z)\cos k$. The factor $1/2$ accounts for the duplicated $k,-k$ pair in the full momentum sum.

Make a bosonic Bogoliubov transformation $a_k=\cosh {\theta }_k{b}_k+\sinh {\theta }_k{b}_{-k}^{†}$. Its coefficients preserve the canonical commutation relation because ${\cosh }^2{\theta }_k-{\sinh }^2{\theta }_k=1$. The coefficient of the pair term vanishes when $\tanh 2{\theta }_k=-{\beta }_k/{\alpha }_k$. ThusThe dropped constant is $\frac{1}{2}{\sum }_k({\epsilon }_k-{\alpha }_k)$ in addition to the classical ground-state energy. Since $J^x>J^y>J^z>0$, ${\alpha }_k>|{\beta }_k|$ at every $k$, so the bosonic Bogoliubov diagonalization is stable. The minimum occurs at $k=0$ and gives the energy gap $2S\sqrt{(J^x-J^y)(J^x-J^z)}$. In the isotropic limit the pair coefficient vanishes and the familiar quadratic small-$k$ ferromagnetic dispersion relation is recovered.

Use the unitary discrete Fourier transformThe transformed operators obey $[a_k,a_q^{†}]={\delta }_{kq}$. The hopping terms contribute $2A\cos k$ to the number coefficient. In the pairing sum, the odd sine part cancels between $k$ and $-k$. ThuswhereThe factor one half in the full momentum sum avoids counting an anomalous pair twice. For $H_f>0$, ${\alpha }_k>|{\Delta }_k|$ for every allowed momentum. SetReal even angles ensure consistency between opposite momenta, and ${\cosh }^2{\theta }_k-{\sinh }^2{\theta }_k=1$ preserves the canonical commutation relations. The anomalous coefficient becomes ${\Delta }_k\cosh 2{\theta }_k-{\alpha }_k\sinh 2{\theta }_k=0$. Normal ordering then giveswithThe field-polarized planar ferromagnetic spin wave has energy gap $ϵ(0)=S\sqrt{H_f(H_f+2J)}$. The momenta run over the whole Brillouin zone once, including the self-inverse momenta $0$ and, for even $L$, $\pi$; these use single-mode squeezing rather than introducing an extra independent partner.

The zero-field case has a zero-mode obstruction to bosonic Bogoliubov diagonalization. Its nonzero-momentum modes have $ϵ(k)=2\sqrt{2}JS|\sin (k/2)|$, but ${\alpha }_0={\Delta }_0=JS$, so no finite squeezing angle exists for the uniform mode. With $Q_0=(a_0+a_0^{†})/\sqrt{2}$, its Hamiltonian is $JS{Q}_0^2-JS/2$, not an identically zero oscillator. For example, its matrix element between the vacuum and the two-boson number state is $JS/\sqrt{2}$, whereas $ϵ(0)b_0^{†}{b}_0$ would vanish. Consequently the requested oscillator diagonalization is literal for nonzero field; at zero field treat the collective orientation separately or introduce a field regulator before taking a limit.

For nonzero field, the spin-wave ground state is the vacuum of all $b_k$. The inverse Bogoliubov transformation gives $⟨a_k^{†}{a}_q⟩={\delta }_{kq}{\sinh }^2{\theta }_k$ and vanishing one-point oscillator averages. Translational invariance therefore gives the depletionRotating back to the physical axes yields, to the retained spin-wave order,The zero transverse components also follow from rotation by $\pi$ about the physical $x$ axis in a symmetry-preserving ground state. The nonzero depletion describes zero-point quantum fluctuations: the original boson vacuum is not the ground state because the anomalous terms create pairs. It is nonnegative and must satisfy $\delta S\ll S$ for the expansion to be reliable. At strong field it tends to zero, with $\delta S\sim J^2/(8H_f^2)$ in the infinite-chain limit.

For fixed $H_f>0$, the momentum sum has a finite thermodynamic limit obtained by $L^{-1}{\sum }_k\to (2\pi {)}^{-1}{\int }_{-\pi }^{\pi }dk$. At zero field the uniform mode already has no oscillator vacuum, as explained in part d. Even after removing that mode, the nonzero modes behave asThus assigning a finite ordered moment by simply setting $h=0$ in the infinite momentum sum is invalid. The divergence indicates failure of this expansion about a fixed planar orientation; it is not a negative physical magnetization. At finite size an exact zero-field state chosen to have definite total $S^z$ has zero planar one-point magnetization by the continuous rotation symmetry. Establishing its full correlations requires a treatment beyond the divergent ordered-state approximation.