Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 75 / 1

Use spin one-half states $|\uparrow ⟩,|\downarrow ⟩$ with $S^z=\pm 1/2$, so ${\sigma }^i=2S^i$ and $H=-J{\sum }_n({\sigma }_n^x{\sigma }_{n+1}^x+g{\sigma }_n^z)$. At $g=0$, every ferromagnetic bond has its lowest energy $-J$ when adjacent $x$ spins agree. On the connected periodic chain the normalized ground states arewhere $|\pm x⟩=(|\uparrow ⟩\pm |\downarrow ⟩)/\sqrt{2}$. Their span is the two-dimensional ground-state subspace; any normalized superposition is another ground state. They are exchanged by the global spin-flip discrete symmetry generated by ${\prod }_n{\sigma }_n^z$.

As $g\to +\infty$, the field dominates and selectsThe limiting polarized state is unique. At large finite $g$, exchange admixes virtual spin flips, so the product state is the limiting wavefunction rather than an exact finite-field eigenstate. The bulk expansion below gives $E_0/N=-J[g+1/(4g)+O(g^{-3})]$. The exact twofold degeneracy stated at $g=0$ should be distinguished from thermodynamic spontaneous symmetry breaking in the ordered phase: a finite chain at nonzero field can have a split pair of symmetry eigenstates.

Put $n_n=c_n^{†}{c}_n$ and $P_n={\prod }_{m<n}(1-2n_m)$. The Jordan–Wigner transformation is $S_n^{+}=P_n{c}_n$, $S_n^{-}=P_n{c}_n^{†}$ and $S_n^z=1/2-n_n$. Since ${\sigma }_n^x=P_n(c_n+c_n^{†})$, $P_{n+1}=P_n(1-2n_n)$ and $(c_n+c_n^{†})(1-2n_n)=c_n^{†}-c_n$, adjacent bonds becomeThus, apart from the end bond,The end bond contains the global fermion parity and sets the sector-dependent periodic or antiperiodic fermion modes. It contributes an order-one boundary term, which is negligible for the thermodynamic energy density; it is not identically zero for the finite spin chain.

Choose the discrete Fourier transform convention $c_n=N^{-1/2}{\sum }_k{e}^{-ikn}{c}_k$. Hopping gives $-2J\cos k{c}_k^{†}{c}_k$. Opposite-momentum pairing gives $iJ\sin k(c_k^{†}{c}_{-k}^{†}+c_k{c}_{-k})$, using the canonical anticommutation relations to antisymmetrize the coefficient. ThereforeFor the Ising-chain Nambu spinor ${\Psi }_k=(c_k,-i{c}_{-k}^{†}{)}^T$, expansion ofrecovers every term: the diagonal contributes $J(g-\cos k)(n_k+n_{-k}-1)$, and the two off-diagonal entries supply the pairing. Since ${\sum }_k\cos k=0$, the constant is $-JgN$. Reversing the Fourier sign changes the pairing convention; the specified sign makes the displayed matrix agree directly.

Write $a_k=g-\cos k$, $s_k=\sin k$ and $r_k=\sqrt{a_k^2+s_k^2}$. The Ising-chain Bogoliubov diagonalization uses the eigenvalues $\pm r_k$ of the two-by-two Nambu matrix. For distinct paired modes $k\ne -k$, choose $\cos {\theta }_k=a_k/r_k$, $\sin {\theta }_k=-s_k/r_k$ and ${\theta }_{-k}=-{\theta }_k$. The rotation $U_k=e^{-i{\theta }_k{\sigma }_y/2}$ satisfies $U_k^{†}{M}_k{U}_k=r_k{\sigma }_z$, and its first transformed component isThe opposite signs of the paired angles preserve the canonical anticommutation relations. Self-paired spinless fermion modes have zero pairing and are treated directly as occupied or empty number levels; they do not require this paired-angle formula.

Both $k$ and $-k$ occur in the sum, so the physical quasiparticle coefficient is $2J{r}_k$, not $J{r}_k$. ConsequentlyThis is the bulk quadratic result under the stipulated boundary simplification. Restoring the finite-chain parity sectors adjusts allowed modes and global excitation constraints. In the thermodynamic limit, the ground-state energy per site isFor $g\ge 0$, the bulk quasiparticle gap is $2J|g-1|$. The requested three regimes are shown in the original sketch and discussed separately below.

At zero field, $r_k=1$ and ${ϵ}_k=2J$ for every $k$. The Ising-chain quasiparticle dispersion is flat: the bulk quadratic excitation costs no momentum-dependent energy. In spin language these are domain-wall excitations. A periodic finite spin configuration has an even number of domain walls, so the single bulk quasiparticle energy $2J$ does not imply an allowed isolated one-wall spin state; the first such closed-chain spin excitation costs $4J$.

Expanding $r_k$ at large positive $g$ gives $r_k=g-\cos k+{\sin }^{2}k/(2g)+O(g^{-2})$, soThe branch is strongly gapped and relatively flat, with width approaching $4J$ against a mean energy of order $2Jg$. These are field-polarized spin-flip quasiparticles with a small relative exchange modulation. Averaging the expansion over momentum also gives $E_0/N=-J[g+1/(4g)+O(g^{-3})]$.

At the critical field,The quasiparticle gap closes and the low-energy branch is linear, unlike a massive excitation. This is the Ising-chain quantum critical point of the Transverse-field Ising model; the cusp in the nonnegative dispersion represents the two opposite propagation directions.

The Hellmann–Feynman theorem gives ${\partial }_g{E}_0=⟨{\partial }_{g}H⟩=-2J{\sum }_n⟨S_n^z⟩$. Differentiating the Ising-chain ground-state energy yields ${\partial }_g{E}_0=-J{\sum }_k(g-\cos k)/r_k$. HenceThis is also obtained from the occupations of the Bogoliubov quasiparticle vacuum: $⟨c_k^{†}{c}_k⟩=[1-(g-\cos k)/r_k]/2$ and $S_n^z=1/2-n_n$. At an exact zero mode the finite-sector limiting ground state specifies the occupation. In the bulk limits, the magnetization per site is zero at $g=0$, tends to $1/2$ as $g\to \infty$, and equals $1/\pi$ at $g=1$: there the integrand $(1-\cos k)/r_k=|\sin (k/2)|$.