Numbers commute, so for Lecture 11’s gas and Lecture 12’s chain the two routes of Figure 12.6 could not disagree. Give each particle a spin, and a collision can exchange the spins as well as the momenta. So the question is

The answer: the Yang–Baxter equation (13.6), which the three-body contact conditions force (Theorem 13.3), and which Yang’s matrix satisfies (Proposition 13.4). The worked example is two labelled particles (Proposition 13.1).

Let each particle carry a label in $\C^{d}$ — a spin component, a hyperfine state — and let the delta be blind to it. A two-particle wavefunction takes values in $\C^{d}\otimes\C^{d}$, the first factor for particle $1$’s label, and is symmetric (bosons) or antisymmetric (fermions) under exchanging positions and labels together: $$\psi(x_1,x_2)=\pm P\,\psi(x_2,x_1),$$(13.1) with $P$ the permutation of the two factors (Definition 9.4). On $x_1<x_2$ write $$\psi(x_1,x_2)=A\,e^{\ii(k_1x_1+k_2x_2)}+A'\,e^{\ii(k_2x_1+k_1x_2)},\qquad A,A'\in\C^{d}\otimes\C^{d},$$ and let (13.1) fix $\psi$ on $x_2<x_1$.

For spin-$\tfrac12$ fermions, then, the interacting phase lives in the spin singlet, and the triplet scatters as free fermions do: two fermions with parallel spins never meet.

In terms of the $t$ and $r$ of §11.3’s reading, $Y_{\mathrm B}=t(k)\,P+r(k)$: transmission carries the labels across the collision, so the particle now on the left has the label that was on the right — the $P$ — and reflection leaves them where they were. For fermions $Y_{\mathrm F}=-t(k)\,P+r(k)$.

Following the momentum. $Y$ is written in the order of positions (slot $1$ is the particle on the left). Figure 11.5 follows the momenta, and there the natural object is $$X(u)=PY,\qquad u=\frac{k_1-k_2}c:\qquad X_{\mathrm B}(u)=\frac{\Rmat(u)}{u+\ii},\quad X_{\mathrm F}(u)=\frac{\Rmat(-u)}{u+\ii}=-\frac{L(u)}{u+\ii},$$(13.3) where $\Rmat(u)=u-\ii P$ is the course’s rational R-matrix and $L(u)=u+\ii P$ the Lax operator of the Heisenberg chain (Lecture 14). Both obey unitarity, $X(u)X(-u)=\Id$. $X_{\mathrm B}$ is the R-matrix of Lecture 14; $X_{\mathrm F}$, which Yang found for spin-$\tfrac12$ fermions, is the $u+\ii P$ of the reading at the end of this lecture up to a scalar, and it is why the algebraic Bethe ansatz of Lecture 15 solves the labelled gases of optional lecture C.

Three labelled particles, on $x_1<x_2<x_3$, with the factorised ansatz $$\psi=\sum_{\sigma\in S_3}A_\sigma\,e^{\ii(k_{\sigma(1)}x_1+k_{\sigma(2)}x_2+k_{\sigma(3)}x_3)}, \qquad A_\sigma\in(\C^{d})^{\otimes3},\quad k_1>k_2>k_3 .$$(13.4) The term $\sigma=\mathrm{id}$, fastest on the left, is the incoming wave; $\sigma=w_0$, the order reversed, is the outgoing one; the other four come after one or two collisions. The ansatz is non-diffraction written down; the question is whether the contact conditions allow it.

The walls. On the wall $x_{j+1}=x_j$ the terms pair up as before, $\sigma$ with $\sigma s_j$, $s_j=(j\;j{+}1)$, and the third particle is a spectator. So Proposition 13.1, applied to the labels in slots $j,j+1$, says $$A_{\sigma s_j}=Y_{j,j+1}\big(k_{\sigma(j)}-k_{\sigma(j+1)}\big)A_\sigma$$(13.5) for every $\sigma$ and $j$: all the conditions. Each is invertible, and read from $\sigma s_j$ it says the same thing: it is one contact condition.

Following the momenta. Reorder the slots of $A_\sigma$ so that slot $a$ holds the label of the particle that carries $k_a$: $B_\sigma=\Pi_\sigma A_\sigma$, where $\Pi_\sigma$ moves the factor in slot $j$ to slot $\sigma(j)$.

So each crossing of momentum lines $a$ and $b$ multiplies by $X_{ab}(u_{ab})$: Figure 11.5 with matrices. From the incoming wave to the outgoing one there are two routes. Along $\mathrm{id}\to s_1\to s_1s_2\to w_0$ the crossings are $12$, $13$, $23$; along $\mathrm{id}\to s_2\to s_2s_1\to w_0$ they are $23$, $13$, $12$ (Figure 13.1).

This is “derived, not postulated”: nothing was assumed but that the three-body wave is a sum of rearranged plane waves, and the equation is what that costs. The logic runs one way: the equation is necessary for a consistent factorised ansatz; it does not by itself prove that an interaction is non-diffractive.

The proof used only the form $u-\ii P$ and the additivity of $u$; the variant $u-\ii(1+\epsilon u)P$, just as harmless-looking, fails (13.6) for every $\epsilon\ne0$ (the check script; the reading at the end).

Scalar $S$: automatic. Without labels, $d=1$, $P=1$ and $X=(u-\ii)/(u+\ii)=S$: both sides are $S_{12}S_{13}S_{23}$, and the order of the crossings does not matter. The Toda chain is in the same position: no labels, numbers (displacements) at the crossings, and its braid relation (Figure 12.6) is the scalar, automatic case. The equation earns its keep only with labels.

$N$ particles sketched. The amplitudes sit on the permutations of $S_N$, the conditions on the edges $\sigma\to\sigma s_j$, and they are consistent if every closed walk multiplies to $\Id$. By Coxeter’s presentation of $S_N$ (cited from Gaudin §10.1) the closed walks are generated by an edge and back (unitarity, $X_{21}(-u)X_{12}(u)=\Id$), the squares $s_js_l=s_ls_j$, $\abs{j-l}\ge2$ (automatic: four different lines), and the hexagons $s_js_{j+1}s_j=s_{j+1}s_js_{j+1}$ (the Yang–Baxter equation). So the Yang–Baxter equation and unitarity make the $N$-body factorised ansatz consistent; the check script confirms it for $N=4$.

The $N$-body S-matrix is the ordered product of two-body matrices, one per pair of momentum lines, in the order the lines cross, $$S_N=\prod_{\text{crossings, in time order}}X_{ab}(u_{ab}),$$(13.7) and by the Yang–Baxter equation every order of crossings gives the same product: the many-body problem is determined by the two-body one. On a ring, carrying one particle round gives, as in Theorem 11.3, a condition — now on a vector of labels: an eigenvalue problem for a product of $X$’s, a transfer matrix, solved by the nested algebraic Bethe ansatz (optional lecture C).

• The consistency condition of Part II, derived rather than postulated: a factorised three-body wave exists for every incoming amplitude exactly when the two orders of Figure 13.1 agree.

• Yang’s matrix, $\Rmat(u)=u-\ii P$ up to a scalar, the R-matrix from which Lecture 14 builds the Heisenberg chain’s commuting transfer matrices.

• Factorised scattering for $N$ particles, sketched: with unitarity, the equation makes every order of crossings give the same product (13.7).

• Status. Proposition 13.1, Theorem 13.3 and Proposition 13.4 are proved. The $N$-body statement is sketched, with Coxeter’s presentation of $S_N$ cited, and checked for $N=4$. The Zamolodchikov–Faddeev algebra of the reading is cited.

• Necessary, not sufficient. The equation is what a consistent factorised ansatz costs. It does not by itself prove that an interaction is non-diffractive.

• Unitarity is a second condition, $X(u)X(-u)=\Id$ for matrices that commute with $P$; Yang’s satisfy it.

• Without labels the equation is automatic, so Lecture 11’s gas and Lecture 12’s chain never tested it.

Suppose three-body scattering is a sequence of two-body events. There are two ways for three lines to cross pairwise. Draw each particle’s path in space and time (a world line), with time running up. Three world lines with three different speeds meet pairwise in three crossings, and the middle line can pass either below or above the crossing of the other two (Figure 13.2). The two orders are different sequences of two-body events, and they describe the same three-body collision. They must agree.

This requirement is the Yang–Baxter equation, and it is a condition on the two-body scattering alone. For particles with nothing but a momentum it is automatic: each two-body event then contributes a number (classically, a shift in position; quantum mechanically, a phase), and numbers do not care about order. It bites when the particles carry an internal label — a spin, a colour — that a two-body event can exchange. Then each event is a matrix, matrices do not commute, and the equation is a strong constraint: a system whose two-body scattering fails it cannot scatter factorisably. Yang met it in 1967 as the consistency condition for exactly this kind of many-body scattering; Baxter met it in 1972 in the statistical mechanics of two-dimensional lattices.

The equation does not organise everything. The geometry of Part I — phase space, the Liouville–Arnold theorem, action-angle variables, the failure of integrability — owes it nothing. Nor does all of the algebra: the Lax pairs of Lecture 8 and the reduction behind Calogero–Moser (optional lecture B) do not come from it. What rests on it is one construction: commuting transfer matrices built from an $\rmat$-matrix. There the classical equation is the model-independent condition that makes the bracket consistent — sufficient, not forced, as Lecture 14’s reading explains — and it is realised by the periodic chain of Lecture 9; the quantum equation plays the same part from Lecture 14 on. (The open chain of Lecture 9’s readings has an $\rmat$-matrix bracket too, but of another kind, and its involution does not need the equation.) That construction is the course’s spine from Part I into Part II.

For readers who have met spin: let two particles each carry a spin-$\frac12$, and let them be fermions with a contact interaction. Their two-body scattering acts on the four-dimensional space $\C^{2}\otimes\C^{2}$ of the pair by $X_{\mathrm F}(u)=-L(u)/(u+\ii)$ of (13.3), with numerator $$L(u)=u\,\Id+\ii\,\perm,$$ where $\perm$ swaps the two spins and $u$ is the difference of the two particles’ rapidities, a parametrisation of their momenta. The scalar $-1/(u+\ii)$ makes $X_{\mathrm F}$ unitary, and it is the same three factors on both sides of the equation below, so the equation can be written with $L$. For three particles, write $L_{12}(u)$ for $L(u)$ acting on the spins of particles $1$ and $2$ and as the identity on the third, and so on. The two orders of Figure 13.2 are the two sides of $$L_{12}(u-v)\,L_{13}(u)\,L_{23}(v) =L_{23}(v)\,L_{13}(u)\,L_{12}(u-v),$$ an identity between $8\times8$ matrices. For this $L$ it holds for all $u$ and $v$; for the apparently harmless variant $u\,\Id+\ii(1+cu)\perm$ it fails for every $c\ne0$ (scripts/check-l0-algebra.py, group 6). This is Yang’s solution (1967), the one behind the Heisenberg chain of Lecture 10. This lecture derives the equation from three-body consistency, and Lecture 14 builds the rest of Part II on it.

Factorised scattering has an algebraic shorthand. Give each asymptotic particle an operator $Z(k)$ that creates it, with one component per label, and write $Z_a(k)$ for $Z(k)$ with its label in tensor slot $a$ (Lecture 9’s leg notation). Postulate the exchange relation $$Z_1(k_1)\,Z_2(k_2)=X_{12}(u_{12})\,Z_2(k_2)\,Z_1(k_1):$$ reordering two neighbours costs one two-body matrix. A product of three can be reversed in two ways, giving $X_{12}X_{13}X_{23}\,Z_3Z_2Z_1$ and $X_{23}X_{13}X_{12}\,Z_3Z_2Z_1$: the two sides of (13.6). Exchanging twice returns the product to itself only if $X(u)X(-u)=\Id$ (for matrices that commute with $P$, as Yang’s do): unitarity. This is the Zamolodchikov–Faddeev algebra (Zamolodchikov and Zamolodchikov 1979; Faddeev 1980) cited. In relativistic field theory, where it was born, the two-body matrix obeys a third condition, crossing symmetry, which relates a particle going forward in time to an antiparticle going backwards. It has no counterpart for the non-relativistic particles of this course, which is one reason field theory is outside it (Lecture 0).