Lecture 13 gave labelled particles a matrix S-matrix and the Yang–Baxter equation. So the question is

The answer: the ring condition is an eigenvalue problem for an inhomogeneous Heisenberg chain whose sites are the particles (Theorem C.2 and Proposition C.4), and its algebraic Bethe ansatz gives two coupled families of Bethe equations (Theorem C.6). The worked example is two particles (Example C.7).

The model is Lecture 11’s delta gas with one addition: each particle carries a spin-$\tfrac12$ label, and the particles are fermions. On a ring of circumference $L$, $$H=-\sum_{j=1}^{N}\frac{\partial^{2}}{\partial x_j^{2}} +2c\sum_{i<j}\delta(x_i-x_j),\qquad c>0,$$(C.1) acting on wavefunctions $\psi(x_1\sigma_1,\dots,x_N\sigma_N)$, $\sigma_j\in\{\uparrow,\downarrow\}$, that are totally antisymmetric under exchanging a position and its label together. This is the Yang–Gaudin model (Yang 1967, Gaudin 1967). Write $M$ for the number of down spins, so $S^{z}=\tfrac N2-M$, and take $M\le N/2$.

Lecture 13 did the two-body problem for labelled particles. On $x_1<x_2$ write $$\psi(x_1,x_2)=A\,e^{\ii(k_1x_1+k_2x_2)}+A'\,e^{\ii(k_2x_1+k_1x_2)}, \qquad A,A'\in\C^{2}\otimes\C^{2},$$(C.2) with slot $j$ of the tensor holding the label of the particle at position $j$, and extend by antisymmetry. Proposition 13.1 gives $A'=Y_{\mathrm F}(k_1-k_2)A$ with $$Y_{\mathrm F}(k)=-\frac{k\perm+\ii c}{k+\ii c} =S(k)\,\Pi_--\Pi_+,\qquad S(k)=\frac{k-\ii c}{k+\ii c},$$(C.3) $\perm$ the permutation of the two label factors (Definition 9.4) and $\Pi_\pm=\tfrac12(\Id\pm\perm)$ the projectors on labels symmetric and antisymmetric checked.

(C.3) is the whole reason this lecture exists. On the singlet — labels antisymmetric, positions symmetric — the delta acts and the amplitude picks up Lecture 11’s scalar $S(k)$. On the triplet the wavefunction vanishes where the particles meet, the delta never sees them, and the amplitude picks up $-1$ (Figure C.1). One two-body problem, two answers; a scalar Bethe equation cannot hold both.

Lecture 13 also fixed the object to work with. $Y$ is written in the order of positions; the natural bookkeeping for a scattering problem follows the momenta, and there the two-body matrix is $X=\perm Y$ (Equation (13.3)), for fermions $$X(u)=-\frac{u+\ii\perm}{u+\ii}=\frac{\Rmat(-u)}{u+\ii}=-\frac{\Lop{}(u)}{u+\ii}, \qquad u=\frac{k_1-k_2}{c},$$(C.4) with $\Rmat(u)=u-\ii\perm$ the course’s rational R-matrix and $\Lop{}(u)=u+\ii\perm$ the Heisenberg Lax operator of Lecture 14. We drop the subscript $\mathrm F$: in this lecture $X$ always means the fermionic one. Three facts, all checked symbolically, carry the lecture: $$X(0)=-\perm,\qquad X(u)X(-u)=\Id,\qquad X_{12}(u-v)X_{13}(u)X_{23}(v)=X_{23}(v)X_{13}(u)X_{12}(u-v).$$(C.5) The last is the Yang–Baxter equation (§13.3); the first is the only place a sign will come from in §C.3. Note that (C.4) uses $\Rmat(-u)$, not $\Rmat$: in the course’s $RTT$ ordering (Theorem 14.2) the Lax operator paired with $\Rmat(\bethe)=\bethe-\ii\perm$ is $\bethe+\ii\perm$.

Inside the region $x_1<x_2<\dots<x_N$ the Bethe ansatz reads $$\psi\big|_{x_1<\dots<x_N} =\sum_{R\in S_N}A_R\,e^{\ii\sum_{j}k_{R(j)}x_j}, \qquad A_R\in(\C^{2})^{\otimes N},$$(C.6) slot $j$ of $A_R$ holding the label of the particle at position $j$; total antisymmetry then determines $\psi$ everywhere. The contact conditions at $x_j=x_{j+1}$ are Lecture 13’s walls, $A_{Rs_j}=Y_{j,j+1}(k_{R(j)}-k_{R(j+1)})A_R$ (Equation (13.5)), and in the momentum-ordered amplitudes $B_R=\Pi_RA_R$ of Lemma 13.2 they become $$B_{Rs_j}=X_{ab}(u_{ab})B_R,\qquad a=R(j),\quad b=R(j+1),\quad u_{ab}=\frac{k_a-k_b}{c}.$$(C.7) Each crossing of momentum lines $a$ and $b$ multiplies by $X_{ab}$ (Figure C.2). Nothing is left to choose: the interaction has been used up. What remains is the ring.

The fermionic sign in (C.9) is not cosmetic: at $N=2$ dropping it changes $e^{\ii kL}=(u+\ii)/(u-\ii)$ to $e^{\ii kL}=-(u+\ii)/(u-\ii)$, and moves the root by a factor $2.007$ at $L=10$, $c=3$ checked.

(C.9) is $N$ eigenvalue problems for $N$ different operators, which looks hopeless. It is not: the $N$ operators are values of one commuting family, and it is a family we have already solved.

Its quantum spaces are the labels of the particles, its $N$ “sites” the particles themselves, at inhomogeneities equal to their own momenta: no lattice, no spin Hamiltonian, a bookkeeping device that happens to be integrable.

So all $N$ conditions can be imposed at once: ask for a common eigenvector of the family $\transfer(\bethe)$ and read off the $N$ numbers $\transfer(\xi_a)$. This is the whole of nesting. The first Bethe ansatz reduced the gas to a finite linear-algebra problem in the labels; the second solves that problem.

Lecture 15 solved this chain for equal inhomogeneities, and nothing in the solution used that they are equal: with $\Lop{an}(\bethe-\xi_n)$ at site $n$ the $RTT$ relation holds as before (Theorem 14.2 needs only a common $\Rmat$), the exchange relations are unchanged, and the reference-state eigenvalues (15.4) become $\prod_n(\bethe-\xi_n+\ii)$ and $\prod_n(\bethe-\xi_n)$ checked, $N=4$. Work with the unnormalised monodromy of Proposition C.5, $$\widetilde T(\bethe)=\Lop{aN}(\bethe-\xi_N)\cdots\Lop{a1}(\bethe-\xi_1) =\begin{pmatrix}A&B\\ C&D\end{pmatrix},$$(C.14) whose entries are the $A,B,C,D$ below — not those of (C.13), which differ from them by the scalar and would give $A\ket0=(-1)^{N}\ket0$. Its reference state is all labels up, $\ket0$, on which $\widetilde T$ is triangular with $A(\bethe)\ket0=\prod_n(\bethe-\xi_n+\ii)\ket0$ and $D(\bethe)\ket0=\prod_n(\bethe-\xi_n)\ket0$; $B$ creates; and for pairwise distinct $v_1,\dots,v_M$, none equal to $0$ or $-\ii$, the vector $\Psi=B(v_1)\cdots B(v_M)\ket0$ is an eigenvector of the whole family, with $$\Lambda(\bethe)=\prod_n(\bethe-\xi_n+\ii)\prod_{\alpha}\frac{\bethe-v_\alpha-\ii}{\bethe-v_\alpha} +\prod_n(\bethe-\xi_n)\prod_{\alpha}\frac{\bethe-v_\alpha+\ii}{\bethe-v_\alpha},$$(C.15) provided the residues at $\bethe=v_\alpha$ cancel (Corollary 15.9).

The spin rapidities below are $\bethe_\alpha=v_\alpha+\tfrac\ii2$, minus Lecture 10’s convention; for generic inhomogeneities the mirrored roots do not solve the spin equations, and the reading at the end of this lecture says why the sign is not free.

The two sets do different jobs. The $k_j$ are momenta and carry the energy (C.18). The $\bethe_\alpha$ are dimensionless, as many as there are down spins, and carry none of it — they enter (C.16) only as phase shifts. That is spin–charge separation in its rawest form (a reading draws two states).

The counting. In logarithms, $$k_jL+2\sum_\alpha\arctan\!\big(2(\xi_j-\bethe_\alpha)\big)=2\pi I_j,\qquad \sum_j\arctan\!\big(2(\bethe_\alpha-\xi_j)\big) -\sum_{\beta\ne\alpha}\arctan(\bethe_\alpha-\bethe_\beta)=\pi J_\alpha,$$(C.20) with $I_j\in\Z+\tfrac M2$ and $J_\alpha\in\Z+\tfrac{N-M+1}{2}$, and $K=(2\pi/L)(\sum_jI_j+\sum_\alpha J_\alpha)$. The parities are part of the equations: a wrong one solves (C.16) with an extra minus sign, invisible in (C.20) checked.

Whether these solutions exhaust the spectrum is a separate question, and the answer is Lecture 10’s reading’s for the Heisenberg chain: they do not, without the complex ones. The reading at the end says what is known cited: Takahashi ch. 12; Essler et al. ch. 4.

The Hubbard model’s Bethe equations are stated here, and their derivation cited; the tutorial solves the lattice two-body problem, which fixes the correspondence with §C.4, sign included.

On a ring of $L$ sites, $L$ even, with $c^{\dagger}_{j\sigma}$ creating an electron of spin $\sigma$ at site $j$ and $n_{j\sigma}=c^{\dagger}_{j\sigma}c_{j\sigma}$, $$H=-t\sum_{j=1}^{L}\sum_{\sigma} \big(c^{\dagger}_{j\sigma}c_{j+1,\sigma}+c^{\dagger}_{j+1,\sigma}c_{j\sigma}\big) +U\sum_{j=1}^{L}n_{j\uparrow}n_{j\downarrow},$$(C.21) and we set $t=1$: the Yang–Gaudin gas put on a lattice, with the dispersion $-2\cos k$, which is not quadratic, so there is no Galilean invariance.

With $\Lambda_\alpha=c\bethe_\alpha$ they are (C.16)–(C.17) under $$k_j\longmapsto\sin k_j,\qquad c\longmapsto \frac{U}{2t},$$(C.24) with that sign: with $c\mapsto-U/2t$ none of the eighteen singlet states of the two-electron $L=6$ ring satisfies (C.22); with it, all do checked. Shastry’s $\Rmat$-matrix behind them is not of difference form, which Lecture 14 and Lecture 15 used (the reading). For $\kappa$ species the construction nests $\kappa-1$ times, one level per node of the Dynkin diagram $A_{\kappa-1}$ (the reading).

• Nesting: a matrix S-matrix on a ring is an eigenvalue problem, and the eigenvalue problem is an inhomogeneous Heisenberg chain whose sites are the particles (Propositions C.4 and C.5), so the algebraic Bethe ansatz solves it.

• Two families of rapidities: the momenta carry all of the energy, the spin rapidities none of it (Theorem C.6).

• The Hubbard model’s equations, the same system under $k\mapsto\sin k$, $c\mapsto U/2t$ (Theorem C.8, cited).

• Status. Theorem C.2 is proved, except that its reduction of the $N!$ conditions to $N$ on one vector is sketched here and proved in a reading; Propositions C.4 and C.5 are proved; Theorem C.6 is proved given that reduction and the algebraic Bethe ansatz of Lecture 15 with unequal inhomogeneities, checked here at $N=4$. The Lieb–Wu equations are cited; their $N=2$ case is derived in the tutorial.

• Completeness is not claimed, for the gas or for the lattice: the nested equations need complex spin roots, and the regular Lieb–Wu solutions miss states (the readings).

• The fermionic sign and the Hubbard sign are not conventions; each wrong choice gives wrong numbers.

The equations of Theorem C.8 are checked against exact diagonalisation of a six-site ring, where $20$ of the $21$ states of the singlet sector are regular Bethe states and the twenty-first is the $\eta$-pair of Remark C.10. Eighteen of the twenty — those with $K\ne\pi$, complex roots included — are checked to $10^{-14}$ by reconstructing the roots from $(K,E)$; that reconstruction degenerates at $K=\pi$, so the two states there are checked directly against the equations instead checked. The $SO(4)$ commutators below are computed exactly, on one even ring; they are why $L$ is even.

Nothing in §C.3 used that the label space is $\C^{2}$. For $\kappa$ species the two-body matrix is still $X(u)=\Rmat(-u)/(u+\ii)$ with $\Rmat(u)=u-\ii\perm$ on $\C^{\kappa}\otimes\C^{\kappa}$, it still obeys the Yang–Baxter equation checked, and Theorem C.2 and Proposition C.4 go through word for word. What changes is §C.4. The auxiliary monodromy is now $\kappa\times\kappa$ in the auxiliary index — the transfer matrix $\transfer=\tra T_a$ still acts on the $\kappa^{N}$-dimensional label space — its reference state breaks $sl_\kappa$ to $sl_{\kappa-1}$, and its first row carries $\kappa-1$ creation operators $B_2(\bethe),\dots,B_\kappa(\bethe)$ in place of Lecture 15’s single $B$. A Bethe vector $B_{i_1}(\bethe_1)\cdots B_{i_M}(\bethe_M)\ket0$ therefore carries a vector of indices, and choosing that vector is itself a Bethe-ansatz problem of rank $\kappa-1$, whose auxiliary problem needs another, and so on — $\kappa-1$ levels, the last of them scalar. That is why the procedure is called nesting, and the extra creation operators are what Lecture 15’s reading pointed at.

The bookkeeping is the Lie algebra’s. $\Rmat(u)=u-\ii\perm$ commutes with $A\otimes\Id+\Id\otimes A$ for every $A\in sl_\kappa$ checked, so the hierarchy is $sl_\kappa$-invariant, and the $\kappa-1$ levels are the nodes of the Dynkin diagram $A_{\kappa-1}$: level $a$ carries $M_a$ rapidities $\bethe^{(a)}$, and the equations couple level $a$ to level $b$ through the Cartan matrix, $$\prod_{j=1}^{N} \frac{\bethe^{(a)}_\alpha-\xi_j+\tfrac\ii2\delta_{a1}} {\bethe^{(a)}_\alpha-\xi_j-\tfrac\ii2\delta_{a1}} =\prod_{b=1}^{\kappa-1}\ \prod_{\substack{\beta\\(b,\beta)\ne(a,\alpha)}} \frac{\bethe^{(a)}_\alpha-\bethe^{(b)}_\beta+\ii C_{ab}/2} {\bethe^{(a)}_\alpha-\bethe^{(b)}_\beta-\ii C_{ab}/2}, \qquad C=\begin{pmatrix}2&-1\\-1&2\end{pmatrix}\ \text{for }A_2,$$(C.25) the source appearing only at the node the particles attach to — the left-hand side is $1$ at every other node, $\delta_{a1}=0$ — and the term $(b,\beta)=(a,\alpha)$ excluded: with $C_{aa}=2$ it would contribute $\ii/(-\ii)=-1$ and turn the $\kappa=2$ case into $1=-1$ checked. Equation (C.17) is that $\kappa=2$ case (Figure C.5); the $A_2$ pair, whose second equation contains no $\xi$ at all, is in the reading cited: Sutherland (1968), (16)–(17); Babelon, de Vega and Viallet (1982).

The same diagram indexes the other series once the label space carries a different representation, and then the representation data — weights, matrix elements of the generators — is what the computation needs; the rep-engine package supplies it exactly, and the check script takes $sl_3$’s fundamental from it.

In the reading of optional lecture B the root systems $A_{N-1}$, $B_N$, $C_N$, $D_N$ index classical Calogero–Moser models, one interaction per positive root; here $A_{\kappa-1}$ indexes the levels of a quantum Bethe ansatz, one set of rapidities per node. The same combinatorial object, two unrelated uses, and no theorem in this course connecting them.

The R-matrix. Shastry (1986) built a two-dimensional vertex model whose transfer matrix commutes with (C.21), by taking two six-vertex models at their free-fermion point and coupling them with a local diagonal factor $\exp(hC^{\uparrow}C^{\downarrow}/2)$ whose parameter is tied to the spectral one by $\sinh2h=u\sin2z$. Two things are worth separating. Jordan–Wigner is the transformation between the fermions and the spin variables, not the coupling. And the loss of difference form is not what stops $\transfer$ at a special point from being the shift operator: that is regularity, $\Rmat(z,z)\propto\perm$, which Shastry’s matrix has. What difference form buys is one $\Rmat$ generating a whole commuting family from a single argument, which is what Lecture 14’s construction used cited: Shastry (1986); Essler et al. ch. 12.

The other nodes. For $A_2$ the second node of (C.25) reads $$1=\prod_{\beta\ne\alpha} \frac{\bethe^{(2)}_\alpha-\bethe^{(2)}_\beta+\ii} {\bethe^{(2)}_\alpha-\bethe^{(2)}_\beta-\ii} \prod_{\gamma} \frac{\bethe^{(2)}_\alpha-\bethe^{(1)}_\gamma-\tfrac\ii2} {\bethe^{(2)}_\alpha-\bethe^{(1)}_\gamma+\tfrac\ii2},$$ with no $\xi$ in it at all: the second level sees the particles only through the first. Sutherland’s original writes both levels with an external minus and the self-root included, which is the same equation once the self-factor $-1$ is taken across cited: Sutherland (1968), (16)–(17).

The two statistics. The bosonic $X_{\mathrm B}=\Rmat(u)/(u+\ii)$ of Lecture 13 also has $X_{\mathrm B}(0)=-\perm$, because $\Rmat(0)=-\ii\perm$: the two statistics are told apart by $u\to-u$, not by the value at the origin, and Proposition C.4 holds for both. What they do not share is the periodicity condition checked.

An exceptional state. Theorem C.8 is restricted to regular solutions, and its escape clause is not decorative. At $L=6$, $N=2$, $U=2\sqrt3/3$ the symmetric wavefunction $e^{\ii k_0(x+y)}(3-\abs{x-y})$ with $k_0=7\pi/6$ is an exact eigenstate of energy $2\sqrt3$ at $K=\pi/3$. Any two-root representation of it must satisfy $E=-4\cos(K/2)\cos\tfrac12(k_1-k_2)$, which here forces $k_1=k_2$: at coincident roots the cusp condition gives $A'=-A$ and the ordinary ansatz vanishes, so this state is a regularised limit of a colliding pair, reached as $q\to0$ along $k=k_0\pm\ii q$. Nor is it a descendant: $\eta$ annihilates it and the spin generators preserve the total spin checked.

The Hamiltonian. The reference book writes (C.21) in its particle–hole symmetric form, $U\sum(n_\uparrow-\tfrac12)(n_\downarrow-\tfrac12)$, so its energy carries an extra $\tfrac U4(L-2N)$; its coupling is $u=U/4t$, ours is written out; and its spin equation is the reciprocal of (C.23). All three differences are invisible in a formula copied out of context checked.

The gap. §C.12’s $\Delta_\mu$ is the chemical-potential discontinuity. The same book’s $\Delta$ is the single-holon threshold, exactly half of it, and is printed in Bessel form with $u=U/4t$. Which gap is meant is not a detail: the two differ by a factor of two at every coupling checked.

At half filling, $N=L$, the Hubbard model is an insulator for every $U>0$. Lieb and Wu’s own paper established it from (C.22)–(C.23) in the thermodynamic limit: the charge gap, meaning the discontinuity $\mu_+-\mu_-$ of the chemical potential, is $$\Delta_\mu=\frac{16t^{2}}{U}\int_1^{\infty} \frac{\sqrt{y^{2}-1}}{\sinh(2\pi ty/U)}\,\dd y ,$$ positive for every $U>0$, tending to $U-4t$ from above and to $(8/\pi)\sqrt{Ut}\,e^{-2\pi t/U}$ as $U\to0$ cited: Lieb and Wu (1968), (21)–(23). Say which gap. The reference book’s $\Delta$ (its (7.18)) is the single-holon threshold, exactly half of this, and is written in Bessel form with $u=U/4t$; the two agree to every digit once converted checked. There is no metal–insulator transition at finite $U$ in one dimension; the transition is at $U=0^{+}$. The spin sector meanwhile stays gapless, and at large $U$ the half-filled model reduces to the Heisenberg antiferromagnet of Lecture 10 with $J=4t^{2}/U$: second-order hopping of an electron onto an occupied site and back costs $U$ and gains $t^{2}$, and it can only happen if the two spins are antiparallel. On a four-site ring, where the Heisenberg ground energy is $-3J=-12/U$, the two agree to $0.62\%$ at $U=40$ and $0.039\%$ at $U=160$ checked.

That is spin–charge separation, and it is visible already in §C.4: the energy (C.18) does not contain $\bethe_\alpha$. In the thermodynamic limit the two sectors acquire different sound velocities — $v_{\mathrm c}$ from the $k$’s and $v_{\mathrm s}$ from the $\bethe$’s — and an added electron’s charge and spin propagate apart. The Luttinger-liquid description of the resulting low-energy theory, and the angle-resolved photoemission experiments on quasi-one-dimensional conductors that look for the two dispersing features, are the obvious continuation cited: Essler et al. ch. 8; Giamarchi, Quantum Physics in One Dimension.

That the solutions of (C.16)–(C.17) exhaust the spectrum is not proved here, and the situation is the one Lecture 10’s reading (§10.10) described for the Heisenberg chain: real roots do not suffice. Here the complex ones live in the spin sector — strings of $\bethe$’s, one of which is drawn in Figure C.4 — while for $c>0$ the momenta $k_j$ stay real. Proposition 11.4 proves that for the spinless gas; for the spinful one it is Yang’s and Gaudin’s statement, and it is not proved here. Completeness is a counting statement with known exceptions cited: Takahashi ch. 12; Essler et al. ch. 4. The thermodynamics built on that string hypothesis — the analogue of Lecture 17 for two coupled families — is a genuine subject and is not attempted.

Two variants cost one line each and are worth knowing.

Two-component bosons. (C.3) becomes $Y_{\mathrm B}=S\Pi_+-\Pi_-$, so $X_{\mathrm B}(u)=\Rmat(u)/(u+\ii)$ — the R-matrix itself rather than $\Rmat(-u)$, and the interacting channel is now the symmetric one. The auxiliary chain is again of Heisenberg type, the fermionic $(-1)^{N-1}$ of (C.9) is absent, and the nested equations come out with the shifts reflected. Substituting the bosonic $X$ into the fermionic condition without touching the statistics sign gives $e^{\ii kL}=-1$ in the singlet: antiperiodic quantisation, and a spectrum wrong by half a level checked.

Attraction. For $c<0$ the $k_j$ need no longer be real (Proposition 11.4 assumed $c>0$), and the gas forms bound pairs of one up and one down spin — the one-dimensional analogue of a Cooper pair. In the thermodynamic limit the pair is the string $k=c\bethe\pm\ii c/2$ cited: Gaudin, The Bethe Wavefunction, §11.4, where the string is presented as a large-system approximation. It is not exact on a finite ring: at $N=2$, $K=0$, $c=-g<0$ the even bound wave is $\cosh q(x-L/2)$ with $2q\tanh(qL/2)=g$, so $q>g/2$; at $g=1$, $L=2$ the true $q$ is $0.7717$ against the string’s $0.5$ checked. Nothing in §C.3 used the sign of $c$; everything in §C.4 about real solutions did.