Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 81 / 1

Each bond of the classical Heisenberg antiferromagnet is minimized when its two spins are antiparallel. Since the periodic chain has an even number of sites, all bonds can be minimized simultaneously:The ground-state degeneracy is a continuous sphere of orientations $\hat{\mathbf{n}}\in S^2$. Choose the Néel state with even sites in $|S,S⟩$ and odd sites in $|S,-S⟩$.

This product state is not an energy eigenstate. The spin ladder operators inproduce states with both neighboring magnetic quantum numbers changed. On an up-down bond, $S_n^{-}{S}_{n+1}^{+}$ has a nonzero matrix element $2S$, hence coefficient $JS$ in the Hamiltonian. These orthogonal spin configurations show explicitly why the classical minimum is not an exact quantum eigenstate for $S>0$.

Make a bipartite spin rotation: rotate every odd site's spin through $\pi$ about the $x$ axis, using $U={\prod }_{n\mathrm{odd}}{e}^{-i\pi S_n^x}$. This unitary conjugation preserves the spin commutation relations. At an odd site it sends $(S^x,S^y,S^z)$ to $(S^x,-S^y,-S^z)$ and exchanges the spin raising operator and spin lowering operator.

Each bond joins one rotated and one unrotated site. In the transformed operators,ThusThe selected Néel state becomes an all-up reference state, so a single Holstein–Primakoff transformation convention works on both sublattices.

Use the Holstein–Primakoff transformation on the physical occupation number subspace $0\le n\le 2S$. With the canonical commutation relation $[a,a^{†}]=1$, the ordered square root givesBoth endpoints are respected: $S^{+}|0⟩=0$ and $S^{-}|2S⟩=0$. Consequently,Since these Fock states form a basis of the physical spin space, $[S^{+},S^{-}]=2S^z$ there. Similarly $[S^z,S^{\pm }]=\pm S^{\pm }$. The Holstein–Primakoff occupation constraint is essential: unrestricted bosonic occupation would not represent a spin-$S$ Hilbert space.

The linear spin-wave approximation expands about the rotated all-up state, with ${\hat{n}}_n=a_n^{†}{a}_n$:In the longitudinal product, $S_n^z{S}_{n+1}^z=S^2-S({\hat{n}}_n+{\hat{n}}_{n+1})+O(S^0)$. Each site belongs to two bonds, so the resulting quadratic Hamiltonian isThe omitted terms are of order $S^0$ at fixed small occupation number. The pair terms describe the quantum fluctuations that were missing from the classical Néel state.

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.

Under the bipartite spin rotation, the staggered magnetization becomes the uniform transformed $S^z$. In the Bogoliubov transformation vacuum ${\alpha }_k|0_{\alpha }⟩=0$,The quantum depletion of Néel order therefore gives, with the zero modes regulated,Near each zero of $\sin k$, the integrand behaves as $1/(2|k-k_0|)$, producing a logarithmic divergence. With a finite-size cutoff of order $N^{-1}$, the depletion grows as ${\pi }^{-1}\log N+O(1)$. Thus the large-$S$ expansion about a state with finite Néel order is not self-consistent in the infinite one-dimensional chain. The divergent expression is not a negative physical magnetization; it signals breakdown of that ordered approximation. It does not determine whether the exact excitation spectrum is gapped.