Past exam of the theoretical physics major option course of the University of Oxford / 2016 / 2

For a uniform classical spin $S\mathbf{n}$ with ${\mathbf{n}}^2=1$, the energy per bond isThe strict inequalities among the couplings make the $x$ direction the unique easy axis. ThusThese uniform configurations are also global classical minima: the largest singular value of the diagonal exchange matrix is $J^x$, so each bond has energy at least $-J^x{S}^2$, and both configurations saturate that bound.

A global rotation by $\pi$ about the $z$ axis sends $(S^x,S^y,S^z)$ to $(-S^x,-S^y,S^z)$, preserves every exchange product, and exchanges the two ground states. A $\pi$ rotation about $y$ does the same. The broken discrete symmetry can be represented by either operation.

The Hamiltonian has the discrete symmetry described in solution, whereas either selected easy-axis ground state has a nonzero order parameter $⟨S^x⟩$ and is not invariant under it. This is spontaneous symmetry breaking: the equations preserve the symmetry while the selected macroscopic quantum state does not.

An infinitesimal longitudinal magnetic field selects one orientation. The defining order of limits isin the ordered zero-temperature phase, with the opposite sign for $h\to 0^{-}$. A symmetric finite-volume quantum state can instead have zero magnetization. Quantum ordered states are generally dressed by fluctuations rather than exact fully polarized product states, because unequal transverse exchange can create pairs of spin waves. Discrete easy-axis symmetry does not imply a gapless Goldstone mode; indeed solution finds a nonzero spin wave gap. Nor does this argument assert positive-temperature order in a one-dimensional short-range chain.^[ref]

The Holstein–Primakoff transformation replaces each finite-dimensional spin algebra by a bosonic creation operator and annihilation operator, while restricting the local occupation number to $0\le n_j\le 2S$. The number $n_j=a_j^{†}{a}_j$ counts deviations from the fully polarized reference quantum state. Its zero-boson Fock vacuum has ${\tilde{S}}^z=S$.

The representation is exact on this physical subspace: the ordered square root gives ${\tilde{S}}^{+}|n⟩=\sqrt{n(2S-n+1)}|n-1⟩$ and ${\tilde{S}}^{-}|n⟩=\sqrt{(n+1)(2S-n)}|n+1⟩$, reproducing the spin commutation relations. For $n_j\ll S$,so the leading Hamiltonian is quadratic in bosons and describes spin waves.

The square root generates boson interactions beyond leading order and its operator ordering matters. Truncation also enlarges the Hilbert space to unphysical occupations unless the low-density approximation remains self-consistent. Large fluctuations, particularly near a gap closing or in low dimensions, can invalidate the linear spin-wave approximation even when its algebraic manipulation is straightforward.

Choose the positive-$x$ reference ground state and use the cyclic rotationThis is a right-handed permutation, so it preserves the spin commutation relations. It puts the ordered direction along the Holstein–Primakoff transformation's quantization axis. To the required order,The longitudinal exchange contributes $-L{J}^x{S}^2+2S{J}^x{\sum }_j{n}_j+O(S^0)$. The transverse terms contributeDropping the constant and all terms of order $S^0$ or smaller gives the stated linear spin-wave approximation, withThe anomalous pair terms are present because the two transverse couplings differ. A different right-handed choice of transverse axes can change the sign of $B$ by a boson phase convention, while leaving the dispersion relation unchanged.

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.