The chains and gases of Lecture 10, Lecture 11 and Lecture 15–Lecture 17 needed the Bethe ansatz because their particles scatter. Some chains only look as if they do. So the question is

The answer, for the transverse-field Ising chain: the Jordan–Wigner map (Theorem 18.3) makes it quadratic in fermions, sector by sector, and Fourier and Bogoliubov give its whole spectrum in both parity sectors (Theorem 18.6). The worked example is the chain itself, against exact diagonalisation (Figure 18.3).

On a ring of $N$ sites, each carrying a spin $\tfrac12$ with Pauli matrices $\sigma^{x},\sigma^{y},\sigma^{z}$, the transverse-field Ising model (TFIM) is $$H=-J\sum_{n=1}^{N}\sigma^{x}_n\sigma^{x}_{n+1}-h\sum_{n=1}^{N}\sigma^{z}_n, \qquad J,h\ge0,\qquad \sigma_{N+1}=\sigma_1 .$$(18.1) The first term wants neighbouring spins aligned along $x$; the second, the transverse field, wants every spin up along $z$. The two do not commute, and that competition is the whole model. $N$ is even throughout. (For odd $N$ the momentum $k=\pi$ moves from one of the grids of §18.5 to the other; nothing changes in principle, but the bookkeeping does, and it is not treated.)

Two limits by inspection. At $J=0$ the ground state is $\ket{\uparrow\uparrow\cdots\uparrow}$ with energy $-hN$, and flipping one spin costs $2h$. At $h=0$ there are two ground states, all spins along $+x$ or all along $-x$, energy $-JN$; the cheapest excitation of a ring is a pair of domain walls, each breaking one bond at cost $2J$, so $4J$ in all. Between the limits something must happen, and the reading at the end finds it at $h=J$.

The symmetry. Let $$\Pi=\prod_{n=1}^{N}\sigma^{z}_n .$$(18.2) Each $\sigma^{z}_n$ commutes with the field term and anticommutes with $\sigma^{x}_n$, so $\Pi$ commutes with every bond $\sigma^{x}_n\sigma^{x}_{n+1}$ (two anticommutations) and with $H$. It is the spin flip $\sigma^{x}\to-\sigma^{x}$, it squares to $1$, and it splits the Hilbert space into an even sector $\Pi=+1$ and an odd sector $\Pi=-1$. It exchanges the two $h=0$ ground states, so the combinations with a definite parity are their sum (even) and difference (odd).

Everything about these operators follows from (18.3). Taking $m=n$ in the second relation gives $c_j^{2}=0$ and $(c_j^{\dagger})^{2}=0$: no mode can be created twice — Pauli’s principle. Then $n_j^{2}=c_j^{\dagger}(1-c_j^{\dagger}c_j)c_j=n_j$, so each $n_j$ has eigenvalues $0$ and $1$, and the $n_j$ commute with each other.

Fock space. Suppose a unit vector $\ket0$, the vacuum, has $c_j\ket0=0$ for all $j$. The $2^{N}$ vectors $$c^{\dagger}_{j_1}c^{\dagger}_{j_2}\cdots c^{\dagger}_{j_M}\ket0,\qquad j_1<j_2<\dots<j_M,$$ are eigenvectors of every $n_j$, with $n_j=1$ exactly for the $j$ listed; they are orthonormal, their span is mapped to itself by every $c_j$ and $c_j^{\dagger}$, and the action there is fixed by the CAR alone. The order of the $j$’s matters only through a sign: $c_1^{\dagger}c_2^{\dagger}\ket0=-c_2^{\dagger}c_1^{\dagger}\ket0$. That sign is what makes fermions fermions, and it is where spins differ.

On one site, $\sigma^{+}=\ket\uparrow\!\bra\downarrow$ and $\sigma^{-}=\ket\downarrow\!\bra\uparrow$ satisfy $\{\sigma^{-},\sigma^{+}\}=1$ and $(\sigma^{-})^{2}=0$: one spin is one fermionic mode, with $\ket\uparrow$ empty and $\ket\downarrow$ occupied. But on two different sites, $\sigma^{-}_1$ and $\sigma^{-}_2$ act on different tensor factors, so they commute; their anticommutator is $2\sigma^{-}_1\sigma^{-}_2\ne0$. Spin flips on different sites are hard-core bosons: they exclude each other on a site and commute between sites. The fix has to supply a minus sign whenever two flips are exchanged, and it cannot be local.

A fermion is a down spin, and the string counts how many fermions stand to the left (Figure 18.1): since $\sigma^{z}=1$ on $\ket\uparrow$ and $-1$ on $\ket\downarrow$, $$n_n=c_n^{\dagger}c_n=\sigma^{-}_n\sigma^{+}_n=\tfrac12(1-\sigma^{z}_n),\qquad S_n=\prod_{m<n}(1-2n_m),\qquad \Pi=(-1)^{N_f}.$$(18.5) The strings cancel in $n_n$ because $S_n^{2}=1$ and $S_n$ acts on other sites than $\sigma^{\pm}_n$. The vacuum is $\ket{\uparrow\cdots\uparrow}$, and the spin parity (18.2) is the fermion parity.

The field term is immediate: $-h\sigma^{z}_n=2hn_n-h$. A bond between neighbours is almost as easy, because the strings of $n$ and $n+1$ differ by one factor.

The boundary bond is the one term that is not a quadratic form in the $c$’s: it carries the parity of the whole chain. But $\Pi$ commutes with $H$, so on each sector it is a number, and there the boundary term is quadratic after all. Define $c_{N+1}:=-c_1$ in the even sector and $c_{N+1}:=+c_1$ in the odd sector. Then in both, (18.8) reads $(c_N^{\dagger}-c_N)(c_{N+1}^{\dagger}+c_{N+1})$, the same form as the bulk, and $$H=H_{+}\,\frac{1+\Pi}{2}+H_{-}\,\frac{1-\Pi}{2},\qquad H_{\pm}=-J\sum_{n=1}^{N}(c_n^{\dagger}-c_n)(c_{n+1}^{\dagger}+c_{n+1}) +2h\sum_{n=1}^{N}c_n^{\dagger}c_n-hN,$$(18.9) with $c_{N+1}=\mp c_1$: the fermions are antiperiodic in the even sector and periodic in the odd one. Each $H_{\pm}$ is quadratic, commutes with $\Pi$ (every term changes $N_f$ by $0$ or $\pm2$), and acts on all of Fock space. So the spectrum of $H$ is found in two steps: diagonalise $H_{+}$ and $H_{-}$ completely, then keep the even states of the first and the odd states of the second. The half thrown away is not part of the spin chain.

In each sector the fermions obey $c_{n+N}=\mp c_n$, so the plane waves that respect this are $e^{\ii kn}$ with $e^{\ii kN}=\mp1$: $$\text{even sector (NS):}\ k\in\Big\{\tfrac{\pi(2m+1)}{N}\Big\},\qquad \text{odd sector (R):}\ k\in\Big\{\tfrac{2\pi m}{N}\Big\},\qquad m=0,\dots,N-1,$$(18.10) with $k$ taken mod $2\pi$. (The names are borrowed from string theory, Neveu–Schwarz and Ramond, where the same two boundary conditions occur.) Both grids are symmetric under $k\to-k$. The R grid contains $k=0$ and $k=\pi$, which are their own negatives; the NS grid, for even $N$, contains neither (Figure 18.2). In the sector’s grid set $$c_n=\frac1{\sqrt N}\sum_ke^{\ii kn}c_k,\qquad c_k=\frac1{\sqrt N}\sum_{n=1}^{N}e^{-\ii kn}c_n .$$(18.11) The matrix $(e^{-\ii kn}/\sqrt N)$ is unitary, so the $c_k$ satisfy the CAR, and $c_{N+1}$ computed from (18.11) is $\mp c_1$ automatically.

The pairing term creates or destroys fermions two at a time, at $k$ and $-k$: the fermion number is not conserved, only its parity. So group the modes in pairs. For $0<k<\pi$ on the grid write $\Psi_k=(c_k,\,c_{-k}^{\dagger})^{T}$, $$a_k=2(h-J\cos k),\qquad b_k=2J\sin k,\qquad h_k=\begin{pmatrix}a_k&-\ii b_k\\ \ii b_k&-a_k\end{pmatrix}.$$ Collecting the terms of $k$ and $-k$ in (18.12), and reordering $c_{-k}^{\dagger}c_{-k}=1-c_{-k}c_{-k}^{\dagger}$, $$H_{\pm}=\sum_{0<k<\pi}\Psi_k^{\dagger}h_k\Psi_k +\sum_{k\in\{0,\pi\}}a_k\big(c_k^{\dagger}c_k-\tfrac12\big),$$(18.13) where the second sum is present only in the odd sector. The constants cancel: the reordering produces $\sum_{0<k<\pi}a_k$, and $\sum_k\cos k=0$ on either grid, so $\sum_{0<k<\pi}a_k-hN=-\tfrac12\sum_{k\in\{0,\pi\}}a_k$.

A $2\times2$ Hermitian matrix is diagonalised by a rotation, and $h_k$ has eigenvalues $\pm\sqrt{a_k^{2}+b_k^{2}}=\pm\varepsilon(k)$, where $$\varepsilon(k)=2\sqrt{J^{2}+h^{2}-2Jh\cos k}\ \ge0 .$$(18.14) Let $\theta_k$ be the angle with $\cos2\theta_k=a_k/\varepsilon(k)$, $\sin2\theta_k=b_k/\varepsilon(k)$ (for $J=h=0$ the whole block vanishes and any $\theta_k$ will do; otherwise $\varepsilon(k)>0$ at every paired $k$, since it vanishes only at $k=0$ when $h=J$), and $$\gamma_k=\cos\theta_k\,c_k-\ii\sin\theta_k\,c_{-k}^{\dagger},\qquad \gamma_{-k}=\cos\theta_k\,c_{-k}+\ii\sin\theta_k\,c_k^{\dagger}.$$(18.15) In matrix form $\Psi_k=U_k\Gamma_k$ with $\Gamma_k=(\gamma_k,\gamma_{-k}^{\dagger})^{T}$ and $U_k=\big(\begin{smallmatrix}\cos\theta_k&\ii\sin\theta_k\\ \ii\sin\theta_k&\cos\theta_k\end{smallmatrix}\big)$, and a two-line computation gives $$U_k^{\dagger}h_kU_k=\begin{pmatrix}a_k\cos2\theta_k+b_k\sin2\theta_k&\ii(a_k\sin2\theta_k-b_k\cos2\theta_k)\\ -\ii(a_k\sin2\theta_k-b_k\cos2\theta_k)&-(a_k\cos2\theta_k+b_k\sin2\theta_k)\end{pmatrix} =\begin{pmatrix}\varepsilon(k)&0\\0&-\varepsilon(k)\end{pmatrix}.$$ Three things make (18.15) a legitimate change of fermions. The $\gamma$’s satisfy the CAR — $\{\gamma_k,\gamma_k^{\dagger}\}=\cos^{2}+\sin^{2}=1$, $\{\gamma_k,\gamma_{-k}\}=\ii\sin\theta_k\cos\theta_k-\ii\sin\theta_k\cos\theta_k=0$, and the rest are immediate. The formula for $\gamma_{-k}$ is the formula for $\gamma_k$ at $-k$, since $\theta_{-k}=-\theta_k$ ($a$ is even in $k$, $b$ odd). And the state $\ket{\Omega_k}=(\cos\theta_k+\ii\sin\theta_k\,c_k^{\dagger}c_{-k}^{\dagger})\ket0$ is killed by both $\gamma_k$ and $\gamma_{-k}$, as one checks in a line: it is the pair’s new vacuum, and it contains an even number of fermions. Then $$\Psi_k^{\dagger}h_k\Psi_k=\varepsilon(k)\big(\gamma_k^{\dagger}\gamma_k-\gamma_{-k}\gamma_{-k}^{\dagger}\big) =\varepsilon(k)\big(\gamma_k^{\dagger}\gamma_k+\gamma_{-k}^{\dagger}\gamma_{-k}-1\big).$$

The board check. Take $h=0$. Then $\varepsilon(k)=2J$ for every $k$, and the even sector’s lowest level is $-\tfrac12\cdot N\cdot2J=-JN$. In the odd sector $e_0=-2J$ is negative: occupying $k=0$ lowers the energy. Its lowest level has $n_0=1$ and everything else empty — one fermion, odd, allowed — and equals $-\tfrac12\big(-2J+2J+(N-2)2J\big)-2J=-JN$. The two $h=0$ ground states of §18.2 are reproduced, one in each sector. Had the odd sector been given NS momenta, or had $e_0$ been replaced by $\abs{e_0}$, its lowest level would have come out at $-JN+2J$ and the degeneracy would have been lost. Figure 18.3 compares the theorem with exact diagonalisation for $N=8$ at twenty-one fields.

Figure 18.4 shows $\varepsilon(k)$ for three fields.

• A complete solution: every eigenvalue and eigenvector of the chain, in both parity sectors (Theorem 18.6).

• A conserved occupation $n_k$ for every mode, the data Lecture 19 needs to say what a relaxed state remembers.

• The same free fermions elsewhere: at infinite repulsion Lecture 16’s gas has the free-fermion ground-state energy (Proposition 16.6), and Girardeau’s map makes it free fermions for every state, with a boundary condition that depends on $N$ (the reading at the end).

• Status. Theorem 18.3 and Theorem 18.6 are proved; the spectrum is also checked against exact diagonalisation at $N=8$.

• $N$ is even throughout; odd $N$ changes the bookkeeping of the grids and is not treated.

• The two sectors are not optional. One boundary condition for both, or none, gives a quadratic model that is simple and wrong.

• Free is special. Jordan–Wigner turns the Heisenberg chain’s $\sigma^{z}\sigma^{z}$ term into a quartic one: its fermions interact, and it needed the Bethe ansatz.

Three sites by hand. On three sites, $c_1=\sigma^{+}_1$, $c_2=\sigma^{z}_1\sigma^{+}_2$, $c_3=\sigma^{z}_1\sigma^{z}_2\sigma^{+}_3$. Then $\{c_1,c_2\}=(\sigma^{+}\sigma^{z}+\sigma^{z}\sigma^{+})_1\,\sigma^{+}_2=0$, where $\sigma^{+}_1$ and $\sigma^{+}_2$ alone commuted. On states, $$c_2^{\dagger}c_3^{\dagger}\ket{\uparrow\uparrow\uparrow} =\sigma^{-}_2\sigma^{z}_2\,\sigma^{-}_3\ket{\uparrow\uparrow\uparrow} =\ket{\uparrow\downarrow\downarrow},\qquad c_3^{\dagger}c_2^{\dagger}\ket{\uparrow\uparrow\uparrow} =\sigma^{z}_2\sigma^{-}_2\,\sigma^{-}_3\ket{\uparrow\uparrow\uparrow} =-\ket{\uparrow\downarrow\downarrow}:$$ the string at site $3$ sees the fermion at site $2$ or not, according to the order of creation.

The construction of local charges. Every occupation $\gamma_k^{\dagger}\gamma_k$ commutes with $H_{\pm}$; the local charges come from weighting whole blocks: for an even function $f$, weight the term of $k$ in (18.13) by $f(k)$. The result commutes with $H_{\pm}$ block by block, and if $f(k)=\cos mk$ the weighted coefficients are still trigonometric polynomials in $k$, so the operator is a sum of fermion bilinears of range at most $m+2$. The odd combinations $\sum_k\sin mk\,c_k^{\dagger}c_k$ also commute, because pairing creates $k$ and $-k$ together and leaves $n_k-n_{-k}$ alone. In spin language the first few are these.

At $N=8$ and $h=0.7J$ a computer search (the check script) finds that, modulo multiples of the identity, the conserved local charges of range at most three are spanned by $H$, $Q_{+}$, $Q_{-}$, $Q_{-}'$ — at that size and coupling only. Spin forms follow from (18.7) and its analogues: a fermion bilinear on sites $n$ and $n+d$ carries the string of the $d-1$ sites between them, and on the ring the terms that wrap round pick up $-\Pi$ exactly as the boundary bond did, so one spin operator serves both sectors.

From (18.14), $$\varepsilon(k)^{2}-4(J-h)^{2}=8Jh(1-\cos k)\ge0,$$(18.19) so the cheapest quasiparticle, at $k=0$, costs $2\abs{J-h}$: the quasiparticle gap. It closes at $h=J$ and nowhere else. The many-body gap — the distance from the ground state to the next level of the chain — is a different number, because each sector keeps only one parity of quasiparticle number. Riemann sums over the two grids are compared with one observation: for a $2\pi$-periodic function $f=\sum_m\hat f_me^{\ii mk}$, $\sum_ne^{\ii mk_n}$ over either grid vanishes unless $N$ divides $m$, so $$\sum_{k\in\mathrm{NS}}f(k)-\sum_{k\in\mathrm{R}}f(k)=-2N\sum_{j\ \mathrm{odd}}\hat f_{jN},$$(18.20) which tends to $0$ as $N\to\infty$ for smooth $f$, whose Fourier coefficients decay faster than any power.

For $N=8$ and $h=0.4J$ the numbers are: doublet splitting $2.6\cdot10^{-4}J$, against a quasiparticle gap of $1.2J$, and the next level $2.60J$ above the ground state, on its way to $4(J-h)=2.4J$ (check-l14-algebra.py). Each extra two sites cut the splitting by a factor approaching $(h/J)^{2}$. At $h=0$, the gap $4J$ above the doublet is the two domain walls of §18.2.

The doublet is spontaneous symmetry breaking seen through a finite window. In the ordered phase $h<J$ the even and odd ground states are the finite-$N$ versions of $\ket{\to\to\cdots}\pm\ket{\leftarrow\leftarrow\cdots}$; their splitting is the amplitude to tunnel between the two ordered states, which needs a domain wall to sweep round the whole ring — hence exponentially small. As $N\to\infty$ they become degenerate, and their combinations break the $\Pi$ symmetry, with $\langle\sigma^{x}\rangle\ne0$ (the reading computes it). In the disordered phase $h>J$ the ground state is unique and gapped, and a quasiparticle is a dressed flipped spin.

Between the phases the ground-state energy per site, $$e_0(h)=\lim_{N\to\infty}\frac{E_0^{+}}N=-\frac1{2\pi}\int_0^{\pi}\varepsilon(k)\,\dd k,$$(18.21) is continuous and has a continuous first derivative, but its second derivative diverges at $h=J$, like $\frac1{\pi J}\ln\abs{h-J}$ (Figure 18.5) sketched: near $k=0$ and $h=J$, $\varepsilon\approx2\sqrt{(h-J)^{2}+J^{2}k^{2}}$, whose second $h$-derivative integrates to a logarithm. A non-analyticity of the ground-state energy is what a quantum phase transition is, and the closing gap is its signature. At $h=J$ itself, $\varepsilon(k)=4J\abs{\sin(k/2)}$ and $e_0=-4J/\pi$.

The operator form of the duality, and the XY chain, are in the next reading.

The TFIM is one member of a family, $$H_{XY}=-J\sum_{n=1}^{N}\Big(\frac{1+\eta}{2}\,\sigma^{x}_n\sigma^{x}_{n+1} +\frac{1-\eta}{2}\,\sigma^{y}_n\sigma^{y}_{n+1}\Big)-h\sum_{n=1}^{N}\sigma^{z}_n,$$(18.22) with anisotropy $\eta$; $\eta=1$ is the TFIM. (The literature writes $\gamma$ for $\eta$; this course keeps $\gamma$ for Lecture 16’s coupling.) The XY chain goes through the same three steps as the TFIM, with the same sectors and grids, and gives Theorem 18.6 with $$\varepsilon(k)=2\sqrt{(h-J\cos k)^{2}+\eta^{2}J^{2}\sin^{2}k},$$(18.23) and signed $e_0$, $e_\pi$ unchanged (checked against exact diagonalisation at $N=8$ by the check script; the derivation is an exercise). The gap closes where both squares vanish: on the line $h=J$, and on the segment $\eta=0$, $h\le J$, where $\varepsilon$ vanishes at $\cos k=h/J$ (Figure 18.6). At $\eta=0$ the chain is the XX chain, which conserves the number of fermions; its momentum grids are compared below.

The XX chain ($\eta=0$) conserves the fermion number, and its $M$-fermion eigenstates are Slater determinants of plane waves. As spin flips, hard-core bosons, two of them pick up only the exchange sign, $S=-1$, whatever their momenta, and Lecture 10’s periodicity condition becomes $e^{\ii k_jN}=(-1)^{M-1}$: exactly the NS or R grid of (18.10) for the sector of parity $(-1)^{M}$. The Heisenberg chain differs from the XX chain exactly in that its $S$ depends on the rapidities.

Duality. Behind this is an exact map of operators (Kramers and Wannier 1941, for the classical two-dimensional Ising model) cited: the domain-wall variables $\tau^{z}=\sigma^{x}_n\sigma^{x}_{n+1}$ living on the bonds turn the bond term into a field and the field into a bond term, exchanging the ordered and disordered phases. It does not map the odd sector to itself — the signed mode $e_0=2(h-J)$ changes sign — which is why the proposition says “even”. Kramers and Wannier’s use of it survives intact: if there is one transition, it must be at the self-dual point.

The TFIM’s conserved charges are constructed directly, without a transfer matrix. Every occupation $\gamma_k^{\dagger}\gamma_k$ commutes with $H_{\pm}$, but a single one is not local. Local combinations exist: weighting the term of $k$ in (18.13) by $\cos mk$ gives an operator that still commutes with $H_{\pm}$, block by block, and whose coefficients are still trigonometric polynomials in $k$ — a sum of fermion bilinears of range at most $m+2$. The first beyond $H$ is, in spins, $$Q_{+}=\sum_{n=1}^{N}\Big[J\big(\sigma^{x}_n\sigma^{z}_{n+1}\sigma^{x}_{n+2}-\sigma^{z}_n\big) -h\big(\sigma^{x}_n\sigma^{x}_{n+1}+\sigma^{y}_n\sigma^{y}_{n+1}\big)\Big], \qquad \comm{H}{Q_{+}}=0,$$(18.24) a local charge of range $3$ (the tutorial, and checked exactly for $N=6,8,10$). Continuing in $m$ gives about $N$ local charges, as for XXX in Lecture 15’s tower; Lecture 19 uses them to say which combinations of the $n_k$ a relaxed state remembers.

Write $A_n=c_n^{\dagger}+c_n$ and $B_n=c_n^{\dagger}-c_n$, so that $A_n^{2}=1$, $B_n^{2}=-1$, and a bulk bond is $\sigma^{x}_n\sigma^{x}_{n+1}=B_nA_{n+1}$ by (18.7). Since $(\sigma^{x})^{2}=1$, $$\sigma^{x}_1\sigma^{x}_{1+r}=(\sigma^{x}_1\sigma^{x}_2)(\sigma^{x}_2\sigma^{x}_3)\cdots (\sigma^{x}_r\sigma^{x}_{r+1})=B_1A_2B_2A_3\cdots B_rA_{r+1},$$ a product of $2r$ fermion operators. In a Bogoliubov vacuum (a Gaussian state) such an expectation value is given by Wick’s theorem as a Pfaffian of two-point functions, and because $\langle A_lA_m\rangle=\delta_{lm}$ and $\langle B_lB_m\rangle=-\delta_{lm}$ it collapses to a determinant (Lieb, Schultz and Mattis 1961) cited: $$\langle\sigma^{x}_1\sigma^{x}_{1+r}\rangle=\det\big[g(m-l+1)\big]_{l,m=1}^{r},\qquad g(d)=\langle B_nA_{n+d}\rangle=\frac1{2\pi}\int_{-\pi}^{\pi}e^{-\ii kd}\, \frac{Je^{\ii k}-h}{\abs{Je^{\ii k}-h}}\,\dd k$$ in the thermodynamic limit (for finite $N$ the integral is the average over the NS grid). A matrix whose entries depend only on $m-l$ is a Toeplitz matrix, and the large-$r$ behaviour of Toeplitz determinants is a classical subject (Szegő’s theorem and its extensions). The results, due to Pfeuty (1970) and, for the XY chain, Barouch and McCoy (1971) cited: for $h<J$ the correlation tends to $(1-h^{2}/J^{2})^{1/4}$, so the spontaneous magnetisation is $\langle\sigma^{x}\rangle=(1-h^{2}/J^{2})^{1/8}$; at $h=J$ it decays like $r^{-1/4}$; for $h>J$ it decays exponentially. The check script confirms the determinant formula against exact diagonalisation at $N=12$, and the ordered-phase limit at $r=200$ to six digits. Franchini’s Chapter 1 and its appendices carry this out in full.