10.1 The question

Part I found a classical many-body system’s conserved quantities as the traces of one matrix, commuting because of one local identity. Part II builds the quantum counterpart: Lecture 14 constructs commuting operators as the traces of one matrix again, and Lecture 15 obtains the chain’s Hamiltonian from their logarithmic derivative and diagonalises them. The classical count does not carry over, since for a finite quantum chain counting commuting operators decides nothing (Lecture 0’s caveat). It starts, as the subject did, with an exact solution found by a guess. So the question of this lecture is

The answer is Bethe’s ansatz: plane waves of flipped spins, whose only interaction is a two-body phase (Proposition 10.1), quantised by the Bethe equations (Theorem 10.2). The rule for passing from Part I to Part II comes first.

The rule of translation

Quantum mechanics keeps the shape of the classical equation of motion and changes its ingredients. States are vectors in a Hilbert space, observables are Hermitian operators, and in the Heisenberg picture an observable evolves by $$\frac{\dd A}{\dd t}=\frac{1}{\ii\hbar}\comm{A}{H}.$$(10.1) Set against $\dot f=\pb{f}{H}$, this is the substitution $\pb{\cdot}{\cdot}\to\comm{\cdot}{\cdot}/\ii\hbar$ — checked on the one bracket that fixes it, $\pb{q}{p}=1$ against $\comm{q}{p}=\ii\hbar$. An operator $A$ is conserved exactly when $\comm{A}{H}=0$, and two conserved operators are “in involution” when they commute. From here on $\hbar=1$, except in Lecture 14, where the classical limit $\hbar\to0$ is taken and $\hbar$ is written.

10.2 The model

The spin-$\tfrac12$ Heisenberg chain on a ring of $N$ sites is $$H=J\sum_{n=1}^{N}\Big(\mathbf S_n\cdot\mathbf S_{n+1}-\tfrac14\Big) =\frac J2\sum_{n=1}^{N}\big(P_{n,n+1}-1\big),\qquad \mathbf S_{N+1}=\mathbf S_1,$$(10.2) with $P_{n,n+1}$ the permutation of the spins on sites $n$ and $n+1$ (Definition 9.4). On the basis of up and down spins the bond term is simple: $\tfrac J2(P-1)$ gives zero on a parallel pair, and on an antiparallel pair it returns $\tfrac J2$ times (the swapped pair minus the pair itself).

Symmetries. The total spin $\mathbf S=\sum_n\mathbf S_n$ commutes with every $P_{n,n+1}$, since exchanging two spins does not change their sum; so $H$ commutes with $S^{x},S^{y},S^{z}$ — the $\SU(2)$ symmetry — and with the shift $U$ that moves every spin one site on. $S^{z}=\tfrac N2-M$, where $M$ is the number of down spins, so $H$ does not mix different $M$.

The reference state. In $\ket{\Uparrow}=\ket{\uparrow\uparrow\cdots\uparrow}$ every bond is parallel, so $H\ket{\Uparrow}=0$. Since $P$ has eigenvalues $\pm1$, each bond term has eigenvalues $0$ and $-J$. The sign of $J$ decides what $\ket\Uparrow$ is. For $J<0$ every bond term is $\ge0$, so $H\ge0$ and $\ket\Uparrow$ is a ground state: the ferromagnet, and the intuition of this lecture, where flipped spins are excitations that cost energy. For $J>0$, the antiferromagnet and the course’s frozen convention, $\ket\Uparrow$ is the highest level; for even $N$ the ground state lies at $M=N/2$ (Lecture 16); for odd $N$ the lowest level is made of spin-$\tfrac12$ multiplets — two of them for $N=3,5,7,9$ (the check script) — with highest weights at $M=(N-1)/2$. Every formula below holds for either sign.

What must be found. Because $H$ commutes with $\SU(2)$, its eigenspaces are sums of $\SU(2)$ multiplets, and each multiplet of total spin $s$ contains exactly one highest-weight vector, killed by $S^{+}=\sum_nS^{+}_n$, with $S^{z}=s$; the rest of the multiplet is got from it by applying $S^{-}$. Decomposing $(\C^{2})^{\otimes N}$ (a prerequisite), the multiplets of spin $\tfrac N2-M$ number $$\binom NM-\binom N{M-1},\qquad 0\le M\le\tfrac N2 .$$(10.3) So the task is: find the highest-weight eigenvectors with $M$ down spins, for each $M\le N/2$. The method looks for them as waves of flipped spins.

10.3 One magnon

Let $\ket x$ be the state with one down spin, at site $x$. Of the $N$ bonds only the two touching $x$ are antiparallel, so $$H\ket x=\frac J2\big(\ket{x-1}+\ket{x+1}-2\ket x\big),$$(10.4) a particle hopping on a ring. The plane wave $\ket k=\sum_xe^{\ii kx}\ket x$ is an eigenvector with $$E(k)=J(\cos k-1),\qquad e^{\ii kN}=1,\quad k=\frac{2\pi I}N,\ I=0,\dots,N-1.$$(10.5) This is a magnon, or spin wave: one flipped spin spread over the ring with momentum $k$. For the ferromagnet, $E=\abs J(1-\cos k)\approx\tfrac12\abs Jk^{2}$ at small $k$: a quasiparticle with a quadratic dispersion.

Which are highest weight? $S^{+}\ket k=\big(\sum_xe^{\ii kx}\big)\ket\Uparrow$, which vanishes for $k\ne0$ and not for $k=0$. And $\ket{k=0}=\sum_x\ket x$ is $S^{-}\ket\Uparrow$: not a new state, but the ferromagnetic multiplet one step down. So the $N-1$ magnons with $k\ne0$ are the highest-weight states with $M=1$, as (10.3) says.

The rapidity

From now on the momentum is traded for a rapidity $\lambda$, $$e^{\ii k}=\frac{\lambda+\ii/2}{\lambda-\ii/2},\qquad\text{i.e.}\qquad \lambda=\tfrac12\cot\tfrac k2,\qquad E=-\frac{2J}{4\lambda^{2}+1}.$$(10.6) It maps $0<k<2\pi$ onto the real line, $k=\pi$ to $\lambda=0$, and $k\to0$ to $\lambda\to\pm\infty$. The reason for the change appears in the next section: the two-magnon phase depends on $k_1$ and $k_2$ in a complicated way, but on $\lambda_1$ and $\lambda_2$ only through their difference. Three names for one number — $k$, $e^{\ii k}$ and $\lambda$ — are a standard place to get lost; the lecture fixes $\lambda$ here and keeps it.

10.4 Two magnons

The equations

Let $\ket{x_1,x_2}$, $x_1<x_2$, have down spins at $x_1$ and $x_2$, and look for an eigenvector whose amplitude on $\ket{x_1,x_2}$ is $\psi(x_1,x_2)$. For the moment think of a long open line and ignore the ends; the ring enters at the end of this section. There are two cases: $$E\psi(x_1,x_2)=\tfrac J2\Big[\sum_{\pm}\big(\psi(x_1\pm1,x_2)+\psi(x_1,x_2\pm1)\big) -4\psi(x_1,x_2)\Big]\qquad(x_2>x_1+1),$$(10.7) $$E\psi(x,x+1)=\tfrac J2\big[\psi(x-1,x+1)+\psi(x,x+2)-2\psi(x,x+1)\big] \qquad(x_2=x_1+1).$$(10.8) In the bulk each down spin hops independently, with four antiparallel bonds. When the two are neighbours the bond between them is parallel: neither can hop onto the other, and there are only two antiparallel bonds (Figure 10.1).

Bethe’s ansatz and the contact condition

Two far-apart magnons should move as two plane waves, so try $$\psi(x_1,x_2)=A_{12}\,e^{\ii(k_1x_1+k_2x_2)}+A_{21}\,e^{\ii(k_2x_1+k_1x_2)} :$$(10.9) the two momenta, carried in either order. Each term solves the bulk equation (10.7) with the same energy $$E=E(k_1)+E(k_2),$$(10.10) because in the bulk the two coordinates hop independently. The formula (10.9) defines $\psi$ on the whole plane, the diagonal included, and satisfies the right-hand side of (10.7) at every point. Compare (10.8) with (10.7) evaluated at $x_2=x_1+1$: the two differ by exactly the two missing hops, and the extra diagonal term, $$\text{\text{(10.7)} at }(x,x+1)\;-\;\text{\text{(10.8)}} =\tfrac J2\big[\psi(x+1,x+1)+\psi(x,x)-2\psi(x,x+1)\big].$$ So the ansatz solves the true equations if and only if the contact condition $$\psi(x,x)+\psi(x+1,x+1)-2\psi(x,x+1)=0\qquad\text{for all }x$$(10.11) holds. It concerns values of $\psi$ at configurations that do not exist; that is the point of the trick.

Three properties of $S$ carry the rest of the lecture. It depends on the rapidities only through $\lambda_1-\lambda_2$, which is why rapidities were introduced. For real $\lambda$ it is a phase, $\abs{S(\lambda)}=1$, and $S(\lambda)S(-\lambda)=1$. And $S(0)=-1$: with equal rapidities $A_{21}=-A_{12}$ and $k_1=k_2$, so the Bethe vector is zero — a Pauli principle that was not put in. (Whether a rescaled limit could still give a state belongs to the completeness question, a reading; roots are taken pairwise distinct from now on.) Physically, the two magnons meet, and the outgoing wave carries the same two momenta, exchanged, times a phase: no reflection into new momenta, and $S$ is the entire effect of the collision. For two classical particles of equal mass that is forced by energy and momentum conservation (Proposition 0.1); for magnons it is not, at least at total momentum $K=\pi$, where $E(k_1)+E(k_2)=2J\big(\cos\tfrac K2\cos\tfrac{k_1-k_2}2-1\big)$ is $-2J$ however $K$ is split. Non-diffraction here is found by the ansatz, not supplied by kinematics.

The ring

On the ring, $\ket{x_1,x_2}$ with $1\le x_1<x_2\le N$ is the same state as the down spins at $x_2$ and $x_1+N$. Extend $\psi$ by requiring $$\psi(x_1,x_2)=\psi(x_2,x_1+N);$$(10.13) then the equations of the ring are the equations of the line for $x_1<x_2<x_1+N$, including the pair adjacent across the seam, $x_1=1$, $x_2=N$. Substituting (10.9) and matching the coefficients of the two plane waves gives $A_{12}=A_{21}e^{\ii k_1N}$ and $A_{21}=A_{12}e^{\ii k_2N}$. With Proposition 10.1 these are the Bethe equations for two magnons; the reading at the end of the lecture states them as a theorem, and §10.5 does $M$ magnons at once.

10.5 $M$ magnons

For $M$ down spins at $x_1<\dots<x_M$, Bethe’s ansatz is a sum over all $M!$ ways of handing out the momenta: $$\psi(x_1,\dots,x_M)=\sum_{\sigma\in S_M}A_\sigma\, \exp\Big(\ii\sum_{j=1}^{M}k_{\sigma(j)}x_j\Big),\qquad A_\sigma=\prod_{\substack{a<b\\ \sigma\text{ puts }b\text{ before }a}}S(\lambda_a-\lambda_b),$$(10.14) where “$\sigma$ puts $b$ before $a$” means $\sigma^{-1}(b)<\sigma^{-1}(a)$; $A_{\mathrm{id}}=1$, and for $M=2$ this is (10.12). The amplitude collects one factor of $S$ for each pair of magnons whose momenta have been exchanged.

Reading the Bethe equations. Write (10.15) as $$e^{\ii k_jN}\prod_{l\ne j}S(\lambda_j-\lambda_l)=1 .$$(10.17) Carry magnon $j$ once round the ring: it picks up $e^{\ii k_jN}$ from the distance travelled and one factor $S$ from each other magnon it passes, and it must come back to the same wavefunction (Figure 10.2). The many-body problem has become $M$ coupled equations for $M$ numbers.

10.6 The logarithmic form

For real $x$ let $\theta_n(x)=2\arctan(2x/n)$, so that $(x+\ii n/2)/(x-\ii n/2)=-e^{-\ii\theta_n(x)}$. Taking logarithms of (10.15) for real rapidities, $$N\theta_1(\lambda_j)-\sum_{l\ne j}\theta_2(\lambda_j-\lambda_l)=2\pi I_j,\qquad I_j\in\Z+\tfrac{N-M+1}2,$$(10.18) the signs $(-1)^{N}$ and $(-1)^{M-1}$ having gone into $I_j$: the Bethe quantum numbers are integers if $N-M$ is odd and half-odd integers if $N-M$ is even. Summing (10.18) over $j$ (the $\theta_2$’s cancel in pairs, $\theta_2$ being odd) and using $k=\pi-\theta_1(\lambda)$ gives the momentum $$K=\pi M-\frac{2\pi}N\sum_jI_j\pmod{2\pi}.$$(10.19) As $\lambda_j\to\pm\infty$ the left side of (10.18) tends to $\pm(N-M+1)\pi$, so $I_j=\pm\tfrac{N-M+1}2$ belongs to a root at infinity. The finite real roots of every solution met here have $\abs{I_j}\le\tfrac{N-M-1}2$, which leaves $N-M$ values (the limit explains the endpoint, not the bound). For $M=2$ at $N=6$ and $8$, where every solution is known, each choice of two distinct values gives exactly one real solution and there are no others — $\binom{N-M}M$ of them; at $N=8$, $M=3$, each of the $\binom53=10$ choices gives one (the check script; not proved in general). The $I_j$ label states the way occupation numbers label free fermions. The tutorial at the end of this lecture draws them as the crossings of a counting function, the picture Lecture 16 takes to $N\to\infty$.

Rewrite (10.17) with $S(\lambda)=-e^{\ii\theta_2(\lambda)}$: $$Nk_j+\sum_{l\ne j}\theta_2(\lambda_j-\lambda_l)=2\pi n_j+(M-1)\pi,\qquad n_j\in\Z .$$(10.20) The $(M-1)\pi$ collects the $M-1$ minus signs of $S=-e^{\ii\theta_2}$. With one magnon the sum is empty and (10.20) is $kN=2\pi n$: exactly the Bohr–Sommerfeld condition $\oint p\,\dd x=2\pi\hbar n$ of a particle going once round a ring of circumference $N$ (a reading of Lecture 6), and here it is not an approximation but the exact spectrum (10.5).

Highest weight

For $M=1$ this is the computation of §10.3: $S^{+}\ket k$ is $\sum_xe^{\ii kx}$, zero on the grid unless $k=0$, i.e. unless $\lambda=\infty$. In general $S^{+}\psi$ is a sum of $(M-1)$-magnon plane waves whose coefficients are, term by term, combinations of the Bethe equations, so vanish on solutions (Faddeev and Takhtajan 1981). The consequence follows because a highest-weight vector has $S^{z}=s\ge0$.

10.7 What it buys

• Exact eigenstates of an interacting chain, labelled by $M$ rapidities that solve $M$ coupled equations (Theorem 10.2), with energy and momentum read off them (10.16).

• Scattering that only exchanges momenta. The whole effect of a collision is the phase $S(\lambda_1-\lambda_2)$, and the $M$-magnon state is built from it alone: the second face of integrability that Lecture 0 described, now for a quantum system.

• Quantum numbers. In logarithmic form each real root carries a quantum number $I_j$ (10.18), as occupation numbers label free fermions; Lecture 16 takes them to $N\to\infty$ as a density.

10.8 Caveats, in one place

• Status. Proposition 10.1 is proved and Theorem 10.2 is sketched, its contact bookkeeping routine. That Bethe vectors are highest weight (Theorem 10.3) is cited.

• Completeness. Whether every highest-weight state is a Bethe state is not proved here. At $N=6$, $M=2$ all nine are found, six from real roots and three from complex or singular ones (the tutorial and the reading).

• Real quantum numbers. That each allowed set of distinct $I_j$ gives a real solution is checked here only in small cases, not proved in general. At $N=6$ and $8$ with $M=2$, where every solution is known, each set gives exactly one; at $N=8$ with $M=3$ each of the ten sets gives one, and uniqueness is not checked.

• The hypotheses of Theorem 10.2 — finite, pairwise distinct rapidities, none equal to $\pm\ii/2$, no two differing by $\pm\ii$ — exclude the singular solutions, which the reading treats.

• An analogy, not a limit. At $M=1$ the Bethe equation is Bohr–Sommerfeld; with $M$ magnons the resemblance is structural. No $\hbar\to0$ is taken, and magnons have no classical phase space.