Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 81 / 1

Use dimensionless spin angular momentum operators, so their commutators have no additional $\hslash$. A fully polarized state $|F⟩={⨂}_n|S,S{⟩}_n$ is an exact ground state of the Heisenberg ferromagnet, with $E_0=-NJ{S}^2$. 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 $(S_{\mathrm{tot}}^{-}{)}^r$, $r=0,\dots ,2NS$, 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 $⟨[S_{\mathrm{tot}}^x,S_{\mathrm{tot}}^y]⟩=i⟨S_{\mathrm{tot}}^z⟩\ne 0$, 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.

The Holstein–Primakoff transformation acts on the physical number-state subspace $0\le n\le 2S$. Its operator ordering matters: $S^{+}=\sqrt{2S-\hat{n}}a$ lowers the number first. Thereforewhile $S^z|n⟩=(S-n)|n⟩$. The square roots vanish at the appropriate ends, so the spin operators preserve the Holstein–Primakoff occupation constraint. The bosonic Fock space outside this subspace is not part of the spin-$S$ representation.

These matrix elements give $[S^z,S^{\pm }]=\pm S^{\pm }$. The two products on $|n⟩$ are $(n+1)(2S-n)$ and $n(2S-n+1)$, respectively, so their difference is $2(S-n)$. ThusAlso $(S^z{)}^2+\frac{1}{2}(S^{+}{S}^{-}+S^{-}{S}^{+})=S(S+1)I$. The transformation therefore has both the correct algebra and the correct spin Casimir operator, not just its small-occupation approximation.

For occupation much smaller than $2S$, expand the square root in the Holstein–Primakoff transformation. To leading order $S_n^{+}=\sqrt{2S}{a}_n+O(S^{-1/2})$, $S_n^{-}=\sqrt{2S}{a}_n^{†}+O(S^{-1/2})$ and $S_n^z=S-a_n^{†}{a}_n$. ConsequentlyPeriodic boundaries count each occupation in two bonds. The linear spin-wave approximation is thereforeThe displayed fluctuation Hamiltonian is bilinear and conserves boson number. The omitted terms are magnon interactions, quartic at the first omitted order. With occupation held of order one, they are smaller by $1/S$ than the order-$S$ quadratic terms. The polarized vacuum has no zero-point depletion in this ferromagnetic expansion.

Use $a_n=N^{-1/2}{\sum }_k{e}^{ikn}{a}_k$ with $k=2\pi j/N$, choosing $N$ distinct values modulo $2\pi$ in the Brillouin zone. The lattice spacing is one; for spacing $a$ the dimensionless variable here would be $ka$. Orthogonality of the discrete Fourier modes diagonalizes the bilinear Hamiltonian:This is the nearest-neighbour ferromagnetic magnon dispersion. The printed expression is missing a factor of four for the stated exchange Hamiltonian. If ${\omega }_k$ denotes a physical frequency, $\hslash {\omega }_k={\epsilon }_k$; setting $\hslash =1$ does not remove that factor. The printed coefficient would require an exchange constant $J/4$ in the original Hamiltonian.

An independent check uses one spin lowering at site $n$. The exact one-magnon Hamiltonian relative to $E_0$ has diagonal element $2JS$ and off-diagonal elements $-JS$ at $n\pm 1$, giving the same Fourier eigenvalue. At small $k$, ${\epsilon }_k\simeq JS{k}^2$, and the $k=0$ state belongs to the degenerate ground multiplet. The quadratic low-energy branch is the ferromagnetic type-B Goldstone boson.

Repeat the linear spin-wave approximation for bonds of lengths one and two. The polarized energy is $E_0=-N{S}^2(J_1+J_2)$ and each bond length contributes its own lattice Laplacian. ThusThe next-nearest-neighbour ferromagnetic magnon dispersion remains nonnegative for the stated positive exchanges and has its only zero at $k=0$ modulo $2\pi$ when $J_1>0$. At small $k$, ${\epsilon }_k\simeq S(J_1+4J_2)k^2$: longer-range ferromagnetic exchange increases the stiffness.

At this harmonic order, a bosonic occupation configuration has energy $E_0+{\sum }_k{\epsilon }_k{n}_k$. The one-magnon dispersion is also exact. General multi-magnon energies acquire the interactions omitted in the large-$S$ expansion; the formula does not claim an exact diagonalization of the full interacting many-spin Hamiltonian.