Past exam of the theoretical physics major option course of the University of Oxford / 2022 / 2

The continuous internal symmetry is a common rotation around $z$:Both the transverse exchange and the Dzyaloshinskii-Moriya interaction contain one raising and one lowering spin operator, so the phases cancel. Equivalently, $[H,{\sum }_j{S}_j^z]=0$ and the continuous group is U(1).

The discrete internal symmetry is time reversal, $\mathcal{T}{\mathbf{S}}_j{\mathcal{T}}^{-1}=-{\mathbf{S}}_j$, together with $\mathcal{T}i{\mathcal{T}}^{-1}=-i$. Each term in $H$ contains two spin operators, so it is invariant. The operation is antiunitary; treating the reversal of all three spin components as a unitary rotation would violate the spin commutation relations. It reverses the magnetization and interchanges the two easy-axis ordered orientations. A convenient equivalent representative, composed with a $z$ rotation, is $R_z(\pi )\mathcal{T}$, which reverses $S^z$ while preserving $S^x,S^y$ as operators.

The statement about one continuous and one discrete internal symmetry describes generic parameters. At $D=0$, additional unitary $\pi$ rotations about transverse axes are allowed; at $D=u=0$, the continuous internal symmetry enlarges to full spin rotation symmetry. Spatial translations of the uniform chain are separate symmetries of the sites.

The Holstein–Primakoff transformation maps the finite spin basis $|S,m⟩$ to bosonic Fock states $|n⟩$ with $n=S-m$. Thus $n=0$ is the fully polarized reference state, and creating a boson lowers $S^z$ by one. The physical subspace has the Holstein–Primakoff occupation constraint $0\le n\le 2S$, not the whole infinite oscillator Hilbert space.

The specified ordering of the square root givesThese are the exact spin ladder operator matrix elements; in particular, the lowering amplitude vanishes at $n=2S$. The underlying oscillator obeys its usual canonical commutation relation on the enlarged space, while the spin operators preserve the constrained physical subspace. Projecting the oscillator operators themselves to a finite space would change their commutator at its upper endpoint.

Expanding $\sqrt{2S-n}=\sqrt{2S}(1-n/(4S)+\cdots )$ converts an ordered spin problem to approximately free magnons, with interactions in higher orders of $1/S$. Two complications are the nonpolynomial, order-sensitive square roots, which generate interacting bosons, and loss of the occupation bound when a truncated expansion uses the full bosonic Fock space. Large depletion, high temperature, or soft modes near a phase transition can invalidate the assumption $⟨n⟩\ll S$ even when $S$ is large.

Expand around the $+z$ polarized ground state. With $n_j=a_j^{†}{a}_j$ fixed as $S\to \infty$, the Holstein–Primakoff transformation gives $S_j^{+}=\sqrt{2S}{a}_j+O(S^{-1/2})$, $S_j^{-}=\sqrt{2S}{a}_j^{†}+O(S^{-1/2})$ and $S_j^z=S-n_j$. The longitudinal exchange contributesUseThe transverse exchange then gives $-JS(a_{j+1}^{†}{a}_j+a_j^{†}{a}_{j+1})$, and the Dzyaloshinskii-Moriya interaction gives $-iDS(a_{j+1}^{†}{a}_j-a_j^{†}{a}_{j+1})$. Discarding the constant $-LJ(1+u)S^2$ and all $O(S^0)$ interactions yieldsHere periodic boundary conditions make each $n_j$ occur twice in the longitudinal sum. The approximation is the linear spin-wave approximation, controlled by small local magnon occupation compared with $S$.

Choose the unitary discrete Fourier transformThe finite-sum orthogonality identity gives $[b_k,b_q^{†}]={\delta }_{kq}$. It also gives ${\sum }_j{a}_{j+1}^{†}{a}_j={\sum }_k{e}^{-ik}{b}_k^{†}{b}_k$. Substituting $A,B$ from solution therefore gives the dispersion relationUsing the opposite sign in the Fourier transform reverses $k$ and the sine term together, without changing the spectrum.

For $u\gg 1$ every $ϵ(k)>0$, so the ground state of this quadratic Hamiltonian is the bosonic vacuum, $b_k|\mathrm{GS}⟩=0$ for every $k$. In the original spin variables it is the fully $+z$ polarized product state. Its energy is zero after the constant subtraction. Repeating the construction about the fully $-z$ polarized state describes the other easy-axis sector of the original Hamiltonian.

A one-magnon excitation $b_k^{†}|\mathrm{GS}⟩$ has energy $ϵ(k)$ above the reference ground state. A normalized many-magnon Fock state with occupations $n_k$ has excitation energy ${\sum }_k{n}_kϵ(k)$, within the dilute regime of the linear spin-wave approximation. The lowest excitation in this sector has the minimum allowed $ϵ(k)$.

Let $R=\sqrt{J^2+D^2}$ and $k_0=\arctan (D/J)$. Since $J\cos k+D\sin k=R\cos (k-k_0)$, the continuum minimum of the dispersion relation isFor $D>0$, the energy gap closes at a positive $u_c$, at wave vector $k_0$. Below it, the quadratic expansion about the polarized state has a negative mode and is no longer a stable ground state approximation; higher-order terms and a different ordered reference state are needed.

Two qualifications are required by the actual assumptions. First, $D=0$ is permitted, in which case $u_c=0$; it is a concrete counterexample to the printed strict inequality $u_c>0$. Second, a finite ring only has discrete momenta. Its finite-ring spin-wave threshold with antisymmetric exchange isThis approaches $u_c$ in the thermodynamic limit and equals it when $k_0$ is allowed. For example, $L=4$ and $0<D<J$ give $u_c(L)=0$, since the maximum of $1,D/J,-1,-D/J$ is one. Thus the continuum formula and its positive threshold require the large-ring interpretation. The two symmetry-related polarized ground states already have equal exact energy; the calculated energy gap refers to excitations about one chosen orientation.