Past exam of the quantum Condensed Matter Field Theory course of the University of Cambridge / 2014 / Paper 1 / 1

Use the Holstein–Primakoff transformation only on the physical occupation sector $0\le n\le 2S$, with $n=a^{†}a$ and $S^{-}=a^{†}\sqrt{2S-n}$. Acting on number states givesConsequently $S^{+}{S}^{-}=(n+1)(2S-n)$ and $S^{-}{S}^{+}=n(2S-n+1)$, soThe endpoint factors keep the spin operators inside the physical sector. The square-root ordering matters; replacing it by an unrestricted linear oscillator would not give the exact finite-spin algebra.

For the even periodic chain, every classical Néel state ${\mathbf{S}}_m=(-1{)}^{m}S\mathbf{n}$ has energy $-JN{S}^2$, with arbitrary unit vector $\mathbf{n}$. Thus there is a continuous sphere of classical orientations, not merely two Ising-like states. Opposite staggered directions are connected by a global spin rotation.

A quantum Néel product is not an eigenstate: transverse exchange flips opposite neighbouring spins. The ground state therefore contains quantum superpositions and zero-point spin waves, reducing the classical staggered moment. A finite isotropic balanced chain has a symmetry-invariant ground state; a symmetry-broken reference is an approximation for analysing fluctuations, not its exact finite-system eigenstate.

Rotate alternate sites by $\pi$ about a transverse axis so all local reference spins point up. Choosing the transverse axes to give positive anomalous coupling, the linear spin-wave approximation yieldsUse lattice spacing one and $a_m=N^{-1/2}{\sum }_k{e}^{ikm}{a}_k$, with all $N$ momenta $k=2\pi j/N$, $j=-N/2,\dots ,N/2-1$, in $[-\pi ,\pi )$. Then ${\gamma }_k=\cos k$ andReordering $a_{-k}{a}_{-k}^{†}=a_{-k}^{†}{a}_{-k}+1$ gives the requested Nambu form with constant $-JNS(S+1)$. Its extra $JSN$ restores $-JN{S}^2$ before diagonalization. This full-zone convention has no additional sublattice or half-zone factor.

For real even ${\theta }_k$, the Bogoliubov transformation givesEquivalently, its Nambu matrix obeys $T_k{\sigma }_z{T}_k^{†}={\sigma }_z$. The bosonic canonical metric is preserved; $T$ need not be unitary in the ordinary positive-definite metric.

The inverse transformation has negative off-diagonal entries. Choosing $\tanh 2{\theta }_k={\gamma }_k$ removes anomalous terms and givesThere are gapless points at $k=0$ and the zone boundary; near zero, ${\omega }_k\simeq 2JS|k|$, a linear antiferromagnetic spin wave with velocity $2JS$ in the stated units. The zero modes are collective rotations: their divergent squeezing cannot be treated as ordinary normalized independent oscillator vacua. A finite-size cutoff or infinitesimal symmetry-breaking field regulates them.

As a constant-term check, the continuum spin-wave vacuum energy per site is $-JS(S+1)+(2/\pi )JS$. This is a spin-wave prediction, not an exact spectrum for every finite-$S$ one-dimensional chain.

Choose the reference staggered direction so its classical moment is positive. The Bogoliubov vacuum has $⟨a_k^{†}{a}_k⟩={\sinh }^2{\theta }_k$, henceNear each soft point the integrand is proportional to $1/|k|$, so the one-dimensional thermodynamic integral diverges logarithmically. The ordered-reference expansion fails; it does not predict a physical negative magnetization. Long-range Néel order is destroyed in this isotropic one-dimensional setting. The exact finite symmetric ground state also has zero staggered expectation without selecting a direction.

For linearly dispersing modes in $d$ dimensions, zero-temperature depletion behaves as ${\int }_{0}dk{k}^{d-2}$ and is infrared finite for $d>1$, allowing order in suitable two- and three-dimensional antiferromagnets. Finite depletion is not a guarantee for every frustrated quantum magnet. At nonzero temperature, the extra Bose occupation changes the singularity to $\int dk{k}^{d-3}$: short-range isotropic systems cannot sustain continuous-symmetry long-range order for $d\le 2$. This thermal conclusion is distinct from the one-dimensional zero-point argument.