Past exam of the quantum Condensed Matter Field Theory course of the University of Cambridge / 2016 / Paper 1 / 1

For the XXZ spin chain, write $\Delta =1+\alpha$ and replace each spin by a classical vector of length $S$. A bond is $J{S}^2{\mathbf{n}}_m^{\mathsf{T}}\mathrm{diag}(1,1,\Delta ){\mathbf{n}}_{m+1}$, whose minimum is $-J{S}^2\max (1,|\Delta |)$. For an even periodic chain all bonds can attain that bound simultaneously. The classical ground states of an XXZ spin chain therefore give $\alpha >0$ gives the two easy-axis Néel states ${\mathbf{S}}_m=(-1{)}^{m}S\hat{\mathbf{z}}$ and its reversal, with energy $-JN{S}^2(1+\alpha )$.

For $-2<\alpha <0$ the ground states are planar Néel configurations, ${\mathbf{S}}_m=(-1{)}^{m}S(\cos \phi ,\sin \phi ,0)$, with arbitrary $\phi$: their degeneracy is a circle and their energy is $-JN{S}^2$. This includes the easy-plane interval used later. Part (a), however, does not restrict negative $\alpha$ to that interval. For $\alpha <-2$, $\Delta <-1$ and the lowest states are instead the two uniform $z$-polarized ferromagnetic configurations, with energy $JN\Delta S^2$. At $\alpha =-2$, a rotation by $\pi$ about $z$ on alternating sites maps the classical energy to an isotropic ferromagnet: ${\mathbf{S}}_m=S((-1{)}^m{n}_x,(-1{)}^m{n}_y,n_z)$ for any unit vector $\mathbf{n}$ is a ground state. This special degeneracy is a sphere. At the other boundary, $\alpha =0$, the ordinary Néel direction is arbitrary on a sphere. These statements concern classical minimizing configurations, not the degeneracy of the exact finite-chain quantum ground state.

Let $[a,a^{†}]=1$, $n=a^{†}a$, and take $S^{+}=(S^{-}{)}^{†}=\sqrt{2S-n}a$. The physical Holstein–Primakoff occupation constraint limits the Hilbert space, which contains only $n=0,1,\dots ,2S$. Acting in this number basis givesThe upper and lower endpoints are annihilated as required. ConsequentlyThe change in $n$ also gives $[S^z,S^{\pm }]=\pm S^{\pm }$. These matrix elements prove the spin algebra on the full physical spin-$S$ space. Extending the oscillator space to arbitrary occupation would not give a physical finite spin: the occupation constraint and operator ordering in the square root matter. bosons on distinct sites have $[a_m,a_n^{†}]={\delta }_{mn}$ and all other site commutators zero.

Rotate the spin on one sublattice through $\pi$ about its $x$ axis, using the unitary product of the corresponding spin-rotation operators. On a rotated site the old components obey $S_{\mathrm{old}}^x=S^x$, $S_{\mathrm{old}}^y=-S^y$, $S_{\mathrm{old}}^z=-S^z$. Every bond joins opposite sublattices, soThe longitudinal product changes sign. Hence the rotated bond is $-J\Delta S_m^z{S}_{m+1}^z+\frac{J}{2}(S_m^{+}{S}_{m+1}^{+}+S_m^{-}{S}_{m+1}^{-})$, giving the requested form. The canonical spin algebra is unchanged because the transformation is unitary. A staggered easy-axis reference configuration is now the uniform all-up state. At $\alpha =0$ one first chooses a Néel direction, here $z$, before performing the same rotation.

For occupations small compared with $S$, use $S^z=S-n$, $S^{-}\simeq \sqrt{2S}{a}^{†}$ and $S^{+}\simeq \sqrt{2S}a$. The longitudinal bond contributes $-J\Delta S^2+JS\Delta (n_m+n_{m+1})+O(S^0)$; the transverse pair terms contribute $JS(a_m{a}_{m+1}+a_m^{†}{a}_{m+1}^{†})+O(S^0)$. Thus the direct real-space expansion isTake $a_m=N^{-1/2}{\sum }_k{e}^{ikm}{a}_k$, with $k=2\pi j/N$ over one full Brillouin zone and the lattice spacing set to one in $k$. Symmetrizing $k$ and $-k$ changes the phase in the pair sum to $\cos k$, givingnormal ordering the second Nambu spinor component then yieldsThe printed matrix has a factor-of-two error on its diagonal. With a full-zone sum its printed $2\Delta$ would produce $4\Delta a_k^{†}{a}_k$, and even its vacuum expectation would differ from the direct expansion by $J\Delta NS$. A half-zone convention cannot repair both coefficients: one must then group each $k,-k$ pair and double its off-diagonal pair coefficient as well. The formula above follows from the specified Hamiltonian and has the expected gapless isotropic limit; subsequent physical results use this consistent expansion. Terms quartic in bosons first enter at order $S^0$.

Write $u_k=\cosh {\theta }_k$, $v_k=\sinh {\theta }_k$, taking real even ${\theta }_k$. The transformation givesThe second equality uses $u_{-k}=u_k$ and $v_{-k}=v_k$, while the first uses $u_k^2-v_k^2=1$. The creation-creation commutator vanishes by conjugation. Therefore the $b$ modes are canonical bosons. In the paired basis,Its determinant is one and its inverse reverses the signs of the off-diagonal entries. Hence $\boxed{a_k=u_k{b}_k-v_k{b}_{-k}^{†}}$. At self-inverse momenta $k=0,\pi$ this is the corresponding one-mode squeezing transformation, still canonical for finite ${\theta }_k$.

For the easy-axis XXZ antiferromagnetic spin-wave spectrum, the consistent quadratic Hamiltonian in part (d) has $A=\Delta$ and $B_k=\cos k$. The inverse transformation acts on the Nambu matrix by a congruence. Its diagonal and off-diagonal entries become $A\cosh 2{\theta }_k-B_k\sinh 2{\theta }_k$ and $B_k\cosh 2{\theta }_k-A\sinh 2{\theta }_k$. The pair term vanishes whenFor $\alpha >0$ the transformation is real and finite for every $k$. Summing the two Nambu number operators over the full zone givesHere $E_0$ names the constant in the requested oscillator form; the actual quadratic ground-state energy is $E_{\mathrm{gs}}^{(2)}=E_0+\frac{1}{2}{\sum }_k{ϵ}_k=-J\Delta N{S}^2+JS{\sum }_k({\Omega }_k-\Delta )$. This distinguishes the normal-ordering constant from the total zero-point contribution. bosonic Bogoliubov diagonalization is a canonical, not an ordinary unitary, diagonalization of the Nambu matrix.

At $\alpha =0$ and $k\ne 0,\pi$, the same formula holds with ${\Omega }_k=|\sin k|$. The exact zero modes have $|\tanh 2{\theta }_k|=1$, so no finite squeezing angle diagonalizes them as positive-frequency oscillators. Treat collective rotations separately or retain a small anisotropy regulator before taking the limit. Finally, if one instead diagonalizes the literal erroneous printed matrix, it gives $\tanh 2{\theta }_k=\cos k/(2\Delta )$ and ${ϵ}_k^{\mathrm{print}}=2JS\sqrt{4{\Delta }^2-{\cos }^{2}k}$. That result remains gapped at $\alpha =0$ and is not the spin-wave spectrum of the Hamiltonian stated at the start.

For $\alpha >0$, the dispersion is periodic and has minima at $k=0,\pi$ in the rotated full-zone convention, withThe maximum is $2JS(1+\alpha )$ at $k=\pm \pi /2$. The easy-axis state has no continuously broken symmetry, so a gap is allowed. Near a minimum the energy behaves as $2JS\sqrt{\alpha (2+\alpha )+(\delta k{)}^2}$. At $\alpha =0$ it becomes ${ϵ}_k=2JS|\sin k|$, linear near its zeros, with small-$k$ slope $2JS$ in the chosen units. Restoring lattice spacing $a$ and $\hslash$ gives spin-wave speed $2JSa/\hslash$. Zone folding reorganizes the same modes into the magnetic Brillouin zone.

These are linear spin-wave predictions, not a statement that the exact one-dimensional quantum chain always has long-range Néel order. At the isotropic point the zero-point spin reduction contains $N^{-1}{\sum }_k(|\sin k{|}^{-1}-1)/2$, which diverges logarithmically as the infrared cutoff vanishes. This failure of the assumed ordered reference must not be mistaken for a calculation of the exact interacting spectrum.

In this part $0<\Delta =1+\alpha <1$, so the ordered reference lies in the $xy$ plane. Choosing one direction breaks the continuous rotation symmetry within that plane. A gapless, linearly dispersing phase mode is therefore expected in the spin-wave approximation, while tilting the Néel axis out of the plane costs anisotropy energy. It is incorrect to continue the easy-axis expansion about an unstable $z$-ordered reference and interpret its imaginary frequencies as physical modes.

For the easy-plane XXZ antiferromagnetic spin-wave spectrum, take the Néel direction along old $x$, then rotate alternating local spins so that all longitudinal components point up. One such local frame gives the full-zone quadratic coefficientsand thereforeNear $k=0$, ${ϵ}_k\simeq JS\sqrt{2(1+\Delta )}|k|$. The zone-edge mode has gap ${ϵ}_{\pi }=2JS\sqrt{2(1-\Delta )}$. A different sublattice phase convention shifts the zero by $\pi$; folding into the magnetic zone displays an acoustic and a gapped polarization. The figure uses this explicitly chosen frame. One-dimensional long-wavelength fluctuations again limit the ordered-reference interpretation.