Lecture 10 solved a lattice model. Lecture 0 asked why a one-dimensional gas might not thermalise, and pointed at the gas of the cradle. So the question is

Yes, for bosons with a repulsive contact interaction (Theorem 11.3), with real roots (Proposition 11.4) and one state per set of quantum numbers (Proposition 11.5); and only in one dimension, because only there do two-body collisions keep the set of momenta (§11.5).

$N$ identical bosons move on a ring of length $L$ and interact only when they touch: $$H=-\sum_{j=1}^{N}\frac{\partial^{2}}{\partial x_j^{2}}+2c\sum_{i<j}\delta(x_i-x_j), \qquad c>0 .$$(11.1) This is the Lieb–Liniger model (1963), repulsive for $c>0$, in units $\hbar=2m=1$: a free particle of momentum $k$ has energy $k^{2}$ and velocity $2k$.

It is the gas §0.2 promised. In the thin tubes of the quantum Newton’s cradle (Figure 2) the transverse motion is frozen, and the atoms’ short-range interaction becomes, along the tube, a delta of a strength fixed by the scattering length and the tube width (Olshanii 1998) cited. Between collisions the particles are free; the delta imposes a condition wherever two of them meet.

For two particles use the centre of mass $X=\tfrac12(x_1+x_2)$ and the separation $r=x_2-x_1$. Then $$-\partial_1^{2}-\partial_2^{2}=-\tfrac12\partial_X^{2}-2\partial_r^{2}, \qquad H=-\tfrac12\partial_X^{2}-2\partial_r^{2}+2c\,\delta(r).$$(11.2) Integrate $H\psi=E\psi$ over $-\epsilon<r<\epsilon$ and let $\epsilon\to0$: $-2\big[\partial_r\psi\big]_{r=0^-}^{r=0^+}+2c\,\psi\big|_{r=0}=0$. ($\psi$ itself must be continuous: a jump would put a $\delta'$ into $\partial_r^{2}\psi$ that nothing cancels.) So the $r$-derivative jumps by $c\psi$. For bosons $\psi$ is even in $r$, its derivative odd, and the jump is twice the one-sided value. With $\partial_r=\tfrac12(\partial_2-\partial_1)$ this is the contact condition $$\big(\partial_2-\partial_1-c\big)\psi=0\qquad\text{at }x_2=x_1,\ \text{approached from }x_1<x_2 .$$(11.3) A symmetric $\psi$ is fixed by its values on $x_1<x_2$, so the problem is: solve the free equation on $x_1<x_2$, subject to (11.3) on its edge (Figure 11.1).

On $x_1<x_2$ try, as in Lecture 10, $$\psi(x_1,x_2)=A_{12}\,e^{\ii(k_1x_1+k_2x_2)}+A_{21}\,e^{\ii(k_2x_1+k_1x_2)}, \qquad E=k_1^{2}+k_2^{2}.$$(11.4) Both waves solve the free equation with the same energy; only the contact condition is left.

$S$ is Proposition 10.1’s formula at $\lambda=k/c$. No rapidity is needed: by Galilean invariance, which the lattice lacks, the phase depends on the momenta only through their difference. Its properties:

• $\abs{S(k)}=1$ for real $k$, and $S(k)S(-k)=1$. Write $$S(k)=-e^{\ii\theta(k)},\qquad \theta(k)=2\arctan\frac kc,\qquad \theta'(k)=\frac{2c}{k^{2}+c^{2}}>0$$(11.6) (Figure 11.2, left).

• $S(0)=-1$: with equal momenta $A_{21}=-A_{12}$ and the wave vanishes — a Pauli principle for bosons, which nobody put in.

• As $c\to\infty$ at fixed $k$, $S\to-1$: the wavefunction vanishes where two particles meet, and these hard-core bosons have the energies of free fermions (Girardeau 1960), the Tonks–Girardeau gas. On the ring the fermions obey $\psi(\dots,x_j+L,\dots)=(-1)^{N-1}\psi(\dots,x_j,\dots)$, periodic for odd $N$ and antiperiodic for even $N$, and this is what “free fermions” means below. As $c\to0$ at fixed $k\ne0$, $S\to1$: free bosons. (At $k=0$, $S=-1$ for every $c$, so this limit is not uniform in $k$.)

Reflection and transmission. Were the two particles distinguishable, with the same interaction, the relative wave coming in from $r>0$ would be partly transmitted to $r<0$ and partly reflected, with amplitudes (the same integration) $$t(k)=\frac{k}{k+\ii c},\qquad r(k)=\frac{-\ii c}{k+\ii c},\qquad \abs t^{2}+\abs r^{2}=1 .$$ For identical bosons a transmitted pair and a reflected pair are the same state, both amplitudes feed the one outgoing wave, and $t+r=S$.

A plane wave has no position, so superpose: a narrow spread of $k_1$ around $\bar k_1$ and of $k_2$ around $\bar k_2<\bar k_1$, the faster particle on the left. A packet sits where its phase is stationary in the momentum. In the incoming term the $k_1$ phase is $k_1x_1-k_1^{2}t$, stationary at $x_1=2\bar k_1t$. In the outgoing term it is $k_1x_2-k_1^{2}t+\arg S(k_1-k_2)$, and $\arg S=\pi+\theta$, so after the collision sketched $$x_2=2\bar k_1t-\theta'(\bar k_1-\bar k_2),\qquad x_1=2\bar k_2t+\theta'(\bar k_1-\bar k_2)$$(11.7) Each momentum comes out on a line parallel to its incoming one: the line carrying the larger momentum is shifted by $-\theta'$, the other by $+\theta'$ — in the centre-of-mass frame, both particles are delayed (Figure 11.2, right, checks this against numerically superposed exact solutions).

Cut the ring at an origin, so that $0\le x_j<L$. The $\binom N2$ planes $x_i=x_j$ cut this cube into $N!$ sectors, one for each order of the particles; periodicity glues cyclically related sectors, which is where the Bethe equations will come from. A symmetric $\psi$ is fixed by its values on the fundamental domain $x_1<\dots<x_N$, whose images under permutations are the other sectors (Figure 11.3); it has $N-1$ walls, $x_j=x_{j+1}$.

The wall condition is §11.3’s, wall by wall. Rigorously $H$ is defined as a self-adjoint operator through its quadratic form, in which the places where three particles meet have measure zero and impose nothing (the KLMN theorem; Reed and Simon, vol. II, §X.2) cited.

On the fundamental domain Bethe’s ansatz is Equation (10.14) with momenta for rapidities: $$\psi(x_1,\dots,x_N)=\sum_{\sigma\in S_N}A_\sigma \exp\Big(\ii\sum_{j=1}^{N}k_{\sigma(j)}x_j\Big),\qquad A_\sigma=\prod_{\substack{a<b\\ \sigma\text{ puts }b\text{ before }a}}S(k_a-k_b),$$(11.8) one factor $S$ for each pair of momenta that $\sigma$ has exchanged.

The Heisenberg chain has complex roots, strings that make bound states (a reading of Lecture 10). Repulsion forbids them.

With real roots take logarithms, using $(k+\ii c)/(k-\ii c)=-e^{-\ii\theta(k)}$: $$k_jL+\sum_{l=1}^{N}\theta(k_j-k_l)=2\pi I_j,\qquad I_j\in\Z+\tfrac{N-1}2,$$(11.11) the $N-1$ signs going into $I_j$ ($\theta(0)=0$, so the term $l=j$ is harmless). Summing over $j$, the $\theta$’s cancel in pairs and $K=(2\pi/L)\sum_jI_j$.

The reason: (11.11) is the gradient of a strictly convex function of $k_1,\dots,k_N$, which has exactly one minimum.

So for $c>0$ the Bethe states are labelled exactly by sets of $N$ distinct $I_j\in\Z+\tfrac{N-1}2$, one real solution each; coincident roots come only from coincident $I$’s, and then $S(0)=-1$ kills the vector. These are the quantum numbers of $N$ free fermions, the $c\to\infty$ limit $k_j=2\pi I_j/L$. That the Bethe states are all the eigenstates is here a theorem (Dorlas 1993, who also proves them orthogonal) cited, unlike for the Heisenberg chain (a reading of Lecture 10). The ground state is the most compact set, $I_j=-\tfrac{N-1}2,\dots,\tfrac{N-1}2$, at every $c>0$ (Yang and Yang 1969) cited; Figure 11.4 follows its roots.

Rapidities. On a ring the particles never become free, yet every eigenstate carries a set $\{k_j\}$, the rapidities, which enter $E$ and $K$ as if they were. The Bethe states form an orthogonal basis (Dorlas 1993) cited, so a state’s weight on each set $\{k_j\}$ is constant in time. That alone decides nothing about relaxation: the weights on the eigenstates of any Hamiltonian are constant. The distribution of §0.2 is a coarser object. Lecture 16 turns a large set $\{k_j\}$ into a density of rapidities. Lecture 19 shows, for the Ising chain, how conserved mode occupations fix a non-thermal steady value of a local observable, and Lecture 20 how a complete family of conserved charges fixes the rapidity density of a uniform gas. In the trap of the cradle that density is not conserved: Lecture 20 follows its approximate motion, a shape that keeps returning.

§0.2 counted. For two equal masses, momentum and energy are two equations for two outgoing momenta, solved only by the incoming pair or its exchange (Proposition 0.1). Quantum mechanically the same holds for any short-range pair interaction, once the particles are free again: the outgoing wave of two particles contains only the momenta $k_1,k_2$, and scattering fixes nothing but amplitudes.

For three particles the count fails: two equations for three unknowns leave a circle of outgoing momenta, the six rearrangements being six points on it. A generic three-body collision sends waves out over the whole circle — diffraction — which is how a one-dimensional gas can thermalise (Sutherland, §2.2). Non-diffraction is not free. It is what the Bethe wavefunction asserts — in (11.8) only rearrangements of the incoming momenta ever appear — and Theorem 11.3 proves that the delta gas has it. One caution: the two-body count uses $E=k^{2}$; on a lattice even two magnons are not forced by kinematics (§10.4, at $K=\pi$).

In Figure 11.5 each crossing multiplies the amplitude by one $S$. Had the lines crossed in the other order, the same factors would come in another order, and nothing would change: for bosons without labels three-body consistency is automatic. It stops being automatic when a collision can do more than multiply by a number, which is Lecture 13’s subject.

• An interacting gas solved exactly, with a complete proof, and every state labelled by $N$ distinct quantum numbers, as for free fermions.

• A clean thermodynamic limit. Real roots and one state per set of quantum numbers are what Lecture 16 and Lecture 17 need: the roots become a density, with no strings to track.

• Rapidities that label every state, which Lecture 16 turns into a density: conserved in a uniform gas, where Lecture 20 shows what it fixes (Lecture 19 does the same for the Ising chain’s occupations), and moving in the cradle’s trap, where Lecture 20 follows it.

• Status. Theorem 11.3 is proved, wall by wall: unlike on the lattice, where three adjacent spins are a real configuration, the walls of Definition 11.2 never meet in the conditions. Proposition 11.4 and Proposition 11.5 are proved. Completeness and orthogonality of the Bethe states (Dorlas 1993), the ground state’s quantum numbers (Yang and Yang 1969) and the delta as a model of the tubes (Olshanii 1998) are cited.

• Repulsion is needed. For attractive $c<0$ the argument of Proposition 11.4 fails: complex roots appear, in strings, as bound states of bosons (McGuire 1964). That gas is not treated.

Two particles by hand. Put $k_{1,2}=K/2\pm q$ with $K=2\pi n/L$. Then $e^{\ii KL/2}=(-1)^{n}$, and the first equation of (11.9) becomes $qL+\theta(2q)=\pi(n+1)\pmod{2\pi}$, one equation for $q>0$ ($q=0$ gives the zero vector). Its left side increases from $0$, so each allowed right side gives one root, and $E=K^{2}/2+2q^{2}$. For $K=0$ the ground state has $qL+2\arctan(2q/c)=\pi$: $q\to\pi/L$ as $c\to\infty$ (free fermions, $k=\pm\pi/L$) and $q\approx\sqrt{c/L}$ as $c\to0$. The check script diagonalises two bosons on a ring of $M$ sites, $-\Delta/a^{2}+(2c/a)\delta$, with $a=L/M$, at $M=200,400,800$, extrapolates in $a$, and finds the twelve lowest levels to within $6\times10^{-9}$ of these.

Numerically, use damped Newton on (11.11): its Jacobian is the Hessian of the convex function $B$ of Proposition 11.5, positive definite, so every step is well defined; halve the step until the residual decreases. Start from the free-fermion values $2\pi I_j/L$.

Strong coupling. For $\abs{k}\ll c$, $\theta(k)\approx2k/c$, and with $\sum_lk_l=0$ for the symmetric ground state (11.11) becomes $k_j(L+2N/c)=2\pi I_j$, with error $O(c^{-3})$: the gas behaves as free fermions on a ring lengthened by $2N/c$.

Weak coupling. For $cL\ll1$ the roots of the ground state crowd together, but their differences, of order $\sqrt{c/L}$, stay large compared with $c$; then $\theta(x)\approx\pi\sgn x-2c/x$. The $\pi\sgn$ terms add up to exactly $2\pi I_j$ for the symmetric $I$’s and cancel, leaving $k_jL=\sum_{l\ne j}2c/(k_j-k_l)$. With $k_j=\sqrt{2c/L}\,h_j$ this is $h_j=\sum_{l\ne j}1/(h_j-h_l)$, Stieltjes’ characterisation of the zeros of the Hermite polynomial $H_N$ (Gaudin §4.3). Figure 11.4 shows both limits.