Optional lecture C

Nesting and the Hubbard model

What this lecture does

Off the path: nothing on it uses this lecture. Give the particles of a gas a spin, and carrying one round the ring acts on a vector of labels instead of multiplying by a number. The one idea of this lecture is that this eigenvalue problem is itself an integrable model — an inhomogeneous Heisenberg chain whose sites are the gas’s own particles — solved by Lecture 15’s algebraic Bethe ansatz, with a second set of rapidities. Then the Hubbard model, whose equations have the same shape with kk replaced by sin⁡k\sin k.

Reading and tutorial. Material marked Reading is studied on your own; material marked Tutorial is worked in the tutorial. Everything else is the lectured core.

C.1 The question

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

how is a gas of particles with a spin quantised on a ring, when the two-body S-matrix is a matrix?

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).

C.2 Fermions with a spin, and a matrix S-matrix

The model is Lecture 11’s delta gas with one addition: each particle carries a spin-12\tfrac12 label, and the particles are fermions. On a ring of circumference LL, H=−∑j=1N∂2∂xj2+2c∑i<jδ(xi−xj),c>0, 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 ψ(x1σ1,…,xNσN)\psi(x_1\sigma_1,\dots,x_N\sigma_N), σj∈{↑,↓}\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 MM for the number of down spins, so Sz=N2−MS^{z}=\tfrac N2-M, and take M≤N/2M\le N/2.

Remark C.1 (Units, and two names used twice). Units are Lecture 11’s, ℏ=2m=1\hbar=2m=1. The Hubbard hopping tt is a number, the transfer matrix t(λ)\transfer(\bethe) always carries its argument; the Hubbard Λα\Lambda_\alpha is a spin rapidity, Lecture 15’s eigenvalue Λ(λ)\Lambda(\bethe) always carries its argument.

The two-body problem is two problems

Lecture 13 did the two-body problem for labelled particles. On x1<x2x_1<x_2 write ψ(x1,x2)=A ei(k1x1+k2x2)+A′ ei(k2x1+k1x2),A,A′∈C2⊗C2, \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 jj of the tensor holding the label of the particle at position jj, and extend by antisymmetry. Proposition 13.1 gives A′=YF(k1−k2)AA'=Y_{\mathrm F}(k_1-k_2)A with YF(k)=−kP+ick+ic=S(k) Π−−Π+,S(k)=k−ick+ic, 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) P\perm the permutation of the two label factors (Definition 9.4) and Π±=12(1±P)\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)S(k). On the triplet the wavefunction vanishes where the particles meet, the delta never sees them, and the amplitude picks up −1-1 (Figure C.1). One two-body problem, two answers; a scalar Bethe equation cannot hold both.

Figure C.1. The two channels of (C.3) are spin states of the colliding pair: antisymmetric labels force symmetric positions and the delta bites (left); symmetric labels give a wavefunction vanishing where the particles meet (right). ∣↑↓⟩=(∣S⟩+∣T0⟩)/2\ket{\uparrow\downarrow}=(\ket S+\ket{T_0})/\sqrt2 feels both.

Following the momenta

Lecture 13 also fixed the object to work with. YY 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=PYX=\perm Y (Equation (13.3)), for fermions X(u)=−u+iPu+i=R(−u)u+i=−L(u)u+i,u=k1−k2c, 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 R(u)=u−iP\Rmat(u)=u-\ii\perm the course’s rational R-matrix and L(u)=u+iP\Lop{}(u)=u+\ii\perm the Heisenberg Lax operator of Lecture 14. We drop the subscript F\mathrm F: in this lecture XX always means the fermionic one. Three facts, all checked symbolically, carry the lecture: X(0)=−P,X(u)X(−u)=1,X12(u−v)X13(u)X23(v)=X23(v)X13(u)X12(u−v). 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 R(−u)\Rmat(-u), not R\Rmat: in the course’s RTTRTT ordering (Theorem 14.2) the Lax operator paired with R(λ)=λ−iP\Rmat(\bethe)=\bethe-\ii\perm is λ+iP\bethe+\ii\perm.

C.3 Periodicity is an eigenvalue problem

Inside the region x1<x2<⋯<xNx_1<x_2<\dots<x_N the Bethe ansatz reads ψ∣x1<⋯<xN=∑R∈SNAR ei∑jkR(j)xj,AR∈(C2)⊗N, \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 jj of ARA_R holding the label of the particle at position jj; total antisymmetry then determines ψ\psi everywhere. The contact conditions at xj=xj+1x_j=x_{j+1} are Lecture 13’s walls, ARsj=Yj,j+1(kR(j)−kR(j+1))ARA_{Rs_j}=Y_{j,j+1}(k_{R(j)}-k_{R(j+1)})A_R (Equation (13.5)), and in the momentum-ordered amplitudes BR=ΠRARB_R=\Pi_RA_R of Lemma 13.2 they become BRsj=Xab(uab)BR,a=R(j),b=R(j+1),uab=ka−kbc. 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 aa and bb multiplies by XabX_{ab} (Figure C.2). Nothing is left to choose: the interaction has been used up. What remains is the ring.

Figure C.2. Left: a crossing is the matrix XabX_{ab} on the two labels — the momenta are permuted, as classically (Lecture 12), the labels carried across by an operator. Right: particle aa once round the ring crosses every other once and moves past N−1N-1 others; both counts appear in Theorem C.2.

Theorem C.2 (Periodicity). Let k1,…,kNk_1,\dots,k_N be pairwise distinct, no two differing by ±ic\pm\ii c (so that every Xab(uab)X_{ab}(u_{ab}) is defined), and let ψ\psi be the totally antisymmetric extension of (C.6) with amplitudes satisfying (C.7). Then ψ\psi is periodic in every variable if and only if, for every R∈SNR\in S_N and a=R(1)a=R(1), (−1)N−1eikaL Xa,R(N)(ua,R(N))⋯Xa,R(2)(ua,R(2)) BR=BR. (-1)^{N-1}e^{\ii k_aL}\,X_{a,R(N)}\big(u_{a,R(N)}\big)\cdots X_{a,R(2)}\big(u_{a,R(2)}\big)\,B_R=B_R . (C.8) These N!N! conditions are equivalent to NN conditions on a single vector B=BidB=B_{\mathrm{id}} sketched here; proved in the reading below, (−1)N−1eikaL XaB=B,Xa=Xa,a−1⋯Xa,1 Xa,N⋯Xa,a+1, (-1)^{N-1}e^{\ii k_aL}\,\mathcal X_aB=B,\qquad \mathcal X_a=X_{a,a-1}\cdots X_{a,1}\,X_{a,N}\cdots X_{a,a+1}, (C.9) every factor at its own argument uab=(ka−kb)/cu_{ab}=(k_a-k_b)/c.

Proof. Because ψ\psi is antisymmetric, periodicity in one argument at every configuration gives it in all of them; take the first. Let 0≤x1<⋯<xN<L0\le x_1<\dots<x_N<L and shift x1x_1 by LL. The shifted point has its arguments in the order x2,…,xN,x1+Lx_2,\dots,x_N,x_1+L, so moving the first argument past the other N−1N-1 costs (−1)N−1(-1)^{N-1}: ψ(x1+L,σ1;… )=(−1)N−1 ψ(x2σ2,…,xNσN,x1+L,σ1),\psi(x_1{+}L,\sigma_1;\dots) =(-1)^{N-1}\,\psi(x_2\sigma_2,\dots,x_N\sigma_N,x_1{+}L,\sigma_1), and both sides of ψ(x1+L,… )=ψ(x1,… )\psi(x_1{+}L,\dots)=\psi(x_1,\dots) are now values of (C.6). Matching the coefficient of each plane wave: the term RR on the right contributes eikR(N)Le^{\ii k_{R(N)}L} and the momentum word (R(N),R(1),…,R(N−1))(R(N),R(1),\dots,R(N-1)), while the slots of ARA_R are read in the cyclic order (σ2,…,σN,σ1)(\sigma_2,\dots,\sigma_N,\sigma_1). Writing cc for the cycle 1↦N1\mapsto N, j↦j−1j\mapsto j-1, and C=Πc−1\mathcal C=\Pi_{c^{-1}} for the operator that permutes slots that way, the condition is AR=(−1)N−1eikaL C ARc−1,a=R(1). A_R=(-1)^{N-1}e^{\ii k_aL}\,\mathcal C\,A_{Rc^{-1}},\qquad a=R(1). (C.10) Now pass to BR=ΠRARB_R=\Pi_RA_R. Since Π\Pi is a homomorphism, ΠRc−1−1=ΠcΠR−1=C−1ΠR−1\Pi_{Rc^{-1}}^{-1}=\Pi_{c}\Pi_R^{-1}=\mathcal C^{-1}\Pi_R^{-1}, so CARc−1=ΠR−1BRc−1\mathcal C A_{Rc^{-1}}=\Pi_R^{-1}B_{Rc^{-1}} and the relabelling operator cancels: (C.10) is BR=(−1)N−1eikaLBRc−1B_R=(-1)^{N-1}e^{\ii k_aL}B_{Rc^{-1}}. The permutation Rc−1Rc^{-1} is RR with its momentum word shifted one place left, reached from RR by s1s2⋯sN−1s_1s_2\cdots s_{N-1}, and at each step (C.7) applies with first leg aa throughout: BRc−1=Xa,R(N)⋯Xa,R(2)BRB_{Rc^{-1}}=X_{a,R(N)}\cdots X_{a,R(2)}B_R. That is (C.8).

The conditions for different RR reduce to (C.9): those with the same aa are conjugate by the Yang–Baxter equation, and a chain of the X\mathcal X’s carries the others onto BidB_{\mathrm{id}} (below). 

Proof of the reduction, and the two places it can go wrong (★). Same aa. If R′=RsjR'=Rs_j with j≥2j\ge2, so that RR and R′R' share a=R(1)a=R(1) and differ by exchanging two letters b,b′b,b' of the word, then BR′=Xbb′BRB_{R'}=X_{bb'}B_R, and the Yang–Baxter equation (C.5) in the form XabXab′=Xbb′ Xab′Xab Xbb′−1 X_{ab}X_{ab'}=X_{bb'}\,X_{ab'}X_{ab}\,X_{bb'}^{-1} (C.11) conjugates the whole product: the operator in (C.8) for R′R' is Xbb′X_{bb'} times the one for RR times Xbb′−1X_{bb'}^{-1}, because the remaining factors Xa,lX_{a,l} with l≠b,b′l\ne b,b' have legs disjoint from b,b′b,b'. So the two conditions are the same condition. The order in (C.11) matters: with the two Xbb′X_{bb'} exchanged the identity is false, and at N=3N=3 with (ξa,ξb,ξb′)=(13,−57,25)(\xi_a,\xi_b,\xi_{b'})=(\tfrac13,-\tfrac57,\tfrac25) the two sides differ by an exact rational 0.430.43 in the largest entry checked.

Different aa. Every RR with R(1)=aR(1)=a is therefore equivalent to the cyclic word Ca=(a,a+1,…,N,1,…,a−1)C_a=(a,a+1,\dots,N,1,\dots,a-1), with C1=idC_1=\mathrm{id}; but that is a condition on BCaB_{C_a}, not on BidB_{\mathrm{id}}. They are the same condition, by an induction on aa. The word Ca+1C_{a+1} is CaC_a with its first letter moved to the back, reached by s1⋯sN−1s_1\cdots s_{N-1}, and the crossings accumulated on the way are exactly Xa,Ca(N)⋯Xa,Ca(2)=XaX_{a,C_a(N)}\cdots X_{a,C_a(2)}=\mathcal X_a, so BCa+1=Xa BCa. B_{C_{a+1}}=\mathcal X_a\,B_{C_a}. (C.12) If the condition for aa holds, Xa\mathcal X_a acts on BCaB_{C_a} as the non-zero scalar qa=(−1)N−1e−ikaLq_a=(-1)^{N-1}e^{-\ii k_aL}, so BCa+1B_{C_{a+1}} is a non-zero multiple of BCaB_{C_a}. Starting from C1=idC_1=\mathrm{id}, every BCaB_{C_a} is a multiple of BB and the NN conditions are (C.9); conversely (C.12) runs the same induction backwards. All N!N! conditions, and the chain, are verified at N=3N=3 and N=4N=4 checked. 

The fermionic sign in (C.9) is not cosmetic: at N=2N=2 dropping it changes eikL=(u+i)/(u−i)e^{\ii kL}=(u+\ii)/(u-\ii) to eikL=−(u+i)/(u−i)e^{\ii kL}=-(u+\ii)/(u-\ii), and moves the root by a factor 2.0072.007 at L=10L=10, c=3c=3 checked.

The product of crossings is a transfer matrix

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

Definition C.3 (The auxiliary chain). For momenta as in Theorem C.2, put ξj=kj/c\xi_j=k_j/c, adjoin an auxiliary C2\C^{2} labelled aa, and set Ta(λ)=XaN(λ−ξN)⋯Xa1(λ−ξ1),t(λ)=tra⁡Ta(λ). T_a(\bethe)=X_{aN}(\bethe-\xi_N)\cdots X_{a1}(\bethe-\xi_1), \qquad \transfer(\bethe)=\tra T_a(\bethe). (C.13)

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

Proposition C.4 (The crossings are the transfer matrix). For every jj, t(ξj)=−Xj\transfer(\xi_j)=-\mathcal X_j.

Proof. At λ=ξj\bethe=\xi_j the jjth factor of (C.13) is Xaj(0)=−PajX_{aj}(0)=-\perm_{aj}: Ta(ξj)=−XaN⋯Xa,j+1 Paj Xa,j−1⋯Xa1.T_a(\xi_j)=-X_{aN}\cdots X_{a,j+1}\,\perm_{aj}\,X_{a,j-1}\cdots X_{a1}. Push Paj\perm_{aj} to the right using PajXal=XjlPaj\perm_{aj}X_{al}=X_{jl}\perm_{aj} for l≠jl\ne j; the factors to the right of it become operators on the quantum spaces alone, M=Xj,j−1⋯Xj1\mathcal M=X_{j,j-1}\cdots X_{j1}, which involves only the spaces 1,…,j1,\dots,j and so commutes with XaN⋯Xa,j+1X_{aN}\cdots X_{a,j+1}. Hence Ta(ξj)=−M XaN⋯Xa,j+1PajT_a(\xi_j)=-\mathcal M\,X_{aN}\cdots X_{a,j+1}\perm_{aj}, and tra⁡(FaPaj)=Fj\tra(F_a\perm_{aj})=F_j, valid for any FF not acting on space jj, gives t(ξj)=−M XjN⋯Xj,j+1=−Xj\transfer(\xi_j)=-\mathcal M\,X_{jN}\cdots X_{j,j+1}=-\mathcal X_j. 

Proposition C.5 (The chain is Heisenberg). Ta(λ)=(−1)N[∏n(λ−ξn+i)]−1 LaN(λ−ξN)⋯La1(λ−ξ1)T_a(\bethe)=(-1)^{N}\big[\prod_n(\bethe-\xi_n+\ii)\big]^{-1}\, \Lop{aN}(\bethe-\xi_N)\cdots\Lop{a1}(\bethe-\xi_1): the auxiliary chain is an inhomogeneous Heisenberg chain, up to one scalar. Consequently [t(λ),t(μ)]=0[\transfer(\bethe),\transfer(\mu)]=0 and [t(λ),Sz]=0[\transfer(\bethe),S^{z}]=0 for all λ,μ\bethe,\mu away from the poles ξn−i\xi_n-\ii.

Proof. The displayed identity is (C.4) factor by factor. The RTTRTT relation (Theorem 14.2) needs only that every site operator obeys RLLR\Lop{}\Lop{} with one common R\Rmat, and the shifts cancel in the argument of R\Rmat, (λ−ξn)−(μ−ξn)=λ−μ(\bethe-\xi_n)-(\mu-\xi_n)=\bethe-\mu, so commutativity follows as in Theorem 14.3; scalars commute with everything checked. 

So all NN conditions can be imposed at once: ask for a common eigenvector of the family t(λ)\transfer(\bethe) and read off the NN numbers t(ξa)\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.

C.4 The nested Bethe equations

Lecture 15 solved this chain for equal inhomogeneities, and nothing in the solution used that they are equal: with Lan(λ−ξn)\Lop{an}(\bethe-\xi_n) at site nn the RTTRTT relation holds as before (Theorem 14.2 needs only a common R\Rmat), the exchange relations are unchanged, and the reference-state eigenvalues (15.4) become ∏n(λ−ξn+i)\prod_n(\bethe-\xi_n+\ii) and ∏n(λ−ξn)\prod_n(\bethe-\xi_n) checked, N=4N=4. Work with the unnormalised monodromy of Proposition C.5, T~(λ)=LaN(λ−ξN)⋯La1(λ−ξ1)=(ABCD), \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,DA,B,C,D below — not those of (C.13), which differ from them by the scalar and would give A∣0⟩=(−1)N∣0⟩A\ket0=(-1)^{N}\ket0. Its reference state is all labels up, ∣0⟩\ket0, on which T~\widetilde T is triangular with A(λ)∣0⟩=∏n(λ−ξn+i)∣0⟩A(\bethe)\ket0=\prod_n(\bethe-\xi_n+\ii)\ket0 and D(λ)∣0⟩=∏n(λ−ξn)∣0⟩D(\bethe)\ket0=\prod_n(\bethe-\xi_n)\ket0; BB creates; and for pairwise distinct v1,…,vMv_1,\dots,v_M, none equal to 00 or −i-\ii, the vector Ψ=B(v1)⋯B(vM)∣0⟩\Psi=B(v_1)\cdots B(v_M)\ket0 is an eigenvector of the whole family, with Λ(λ)=∏n(λ−ξn+i)∏αλ−vα−iλ−vα+∏n(λ−ξn)∏αλ−vα+iλ−vα, \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 λ=vα\bethe=v_\alpha cancel (Corollary 15.9).

The spin rapidities below are λα=vα+i2\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.

Theorem C.6 (The nested Bethe equations, Yang and Gaudin 1967 proved, given the reduction of Theorem C.2, sketched here). Let k1,…,kNk_1,\dots,k_N be pairwise distinct with no two differing by ±ic\pm\ii c (so that no XX in (C.9) is singular), let λ1,…,λM\bethe_1,\dots,\bethe_M be pairwise distinct and ≠±i2\ne\pm\tfrac\ii2 (Corollary 15.9’s regular roots, vα≠0,−iv_\alpha\ne0,-\ii), put ξj=kj/c\xi_j=k_j/c, and let eikjL=∏α=1Mξj−λα+i2ξj−λα−i2,j=1,…,N,∏j=1Nλα−ξj+i2λα−ξj−i2=∏β≠αλα−λβ+iλα−λβ−i,α=1,…,M.\begin{align} e^{\ii k_jL}&=\prod_{\alpha=1}^{M} \frac{\xi_j-\bethe_\alpha+\tfrac\ii2}{\xi_j-\bethe_\alpha-\tfrac\ii2}, &&j=1,\dots,N,\tag{C.16}\\ \prod_{j=1}^{N}\frac{\bethe_\alpha-\xi_j+\tfrac\ii2}{\bethe_\alpha-\xi_j-\tfrac\ii2} &=\prod_{\beta\ne\alpha}\frac{\bethe_\alpha-\bethe_\beta+\ii}{\bethe_\alpha-\bethe_\beta-\ii}, &&\alpha=1,\dots,M . \tag{C.17}\end{align} Then Ψ=B(λ1−i2)⋯B(λM−i2)∣0⟩\Psi=B(\bethe_1-\tfrac\ii2)\cdots B(\bethe_M-\tfrac\ii2)\ket0, if it is not zero, satisfies every periodicity condition (C.9), and the resulting eigenfunction of (C.1) has E=∑j=1Nkj2,K=∑j=1Nkj,Sz=N2−M. E=\sum_{j=1}^{N}k_j^{2},\qquad K=\sum_{j=1}^{N}k_j,\qquad S^{z}=\frac N2-M . (C.18)

Proof. (C.17) is Corollary 15.9’s system (15.17) for the inhomogeneous chain, written in λα=vα+i2\bethe_\alpha=v_\alpha+\tfrac\ii2: the residue of (C.15) at vαv_\alpha vanishes exactly when ∏n(vα−ξn+i)/∏n(vα−ξn)=∏β≠α(vα−vβ+i)/(vα−vβ−i)\prod_n(v_\alpha-\xi_n+\ii)/\prod_n(v_\alpha-\xi_n) =\prod_{\beta\ne\alpha}(v_\alpha-v_\beta+\ii)/(v_\alpha-v_\beta-\ii). So Ψ\Psi is a common eigenvector and Proposition C.4 turns (C.9) into NN scalar equations, (−1)N−1eikaL⋅(−ΛX(ξa))=1(-1)^{N-1}e^{\ii k_aL}\cdot(-\Lambda^{X}(\xi_a))=1, with ΛX\Lambda^{X} the eigenvalue of (C.13). Evaluate it. By Proposition C.5, ΛX(λ)=(−1)NΛ(λ)/∏n(λ−ξn+i)\Lambda^{X}(\bethe)=(-1)^{N}\Lambda(\bethe)\big/\prod_n(\bethe-\xi_n+\ii); at λ=ξa\bethe=\xi_a the second term of (C.15) carries the factor ∏n(ξa−ξn)=0\prod_n(\xi_a-\xi_n)=0 and dies, and the first term’s prefactor cancels the denominator exactly, leaving ΛX(ξa)=(−1)N∏αξa−λα−i2ξa−λα+i2. \Lambda^{X}(\xi_a)=(-1)^{N}\prod_{\alpha} \frac{\xi_a-\bethe_\alpha-\tfrac\ii2}{\xi_a-\bethe_\alpha+\tfrac\ii2}. (C.19) Then (−1)N−1⋅(−1)⋅(−1)N=1(-1)^{N-1}\cdot(-1)\cdot(-1)^{N}=1 for every NN, and what is left is (C.16). Inside the sectors every term of (C.6) is a plane wave of energy ∑kj2\sum k_j^{2} and momentum ∑kj\sum k_j, as in Theorem 11.3. 

The two sets do different jobs. The kjk_j are momenta and carry the energy (C.18). The λα\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, kjL+2∑αarctan⁡ ⁣(2(ξj−λα))=2πIj,∑jarctan⁡ ⁣(2(λα−ξj))−∑β≠αarctan⁡(λα−λβ)=πJα, 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 Ij∈Z+M2I_j\in\Z+\tfrac M2 and Jα∈Z+N−M+12J_\alpha\in\Z+\tfrac{N-M+1}{2}, and K=(2π/L)(∑jIj+∑αJα)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.

Example C.7 (N=2N=2: the two channels, again). Take N=2N=2, M=1M=1. (C.17) reads (λ−ξ1+i2)(λ−ξ2+i2)=(λ−ξ1−i2)(λ−ξ2−i2)(\bethe-\xi_1+\tfrac\ii2)(\bethe-\xi_2+\tfrac\ii2) =(\bethe-\xi_1-\tfrac\ii2)(\bethe-\xi_2-\tfrac\ii2), whose only root is λ=12(ξ1+ξ2)\bethe=\tfrac12(\xi_1+\xi_2); then ξ1−λ=12(ξ1−ξ2)\xi_1-\bethe=\tfrac12(\xi_1-\xi_2) and (C.16) becomes eik1L=k1−k2+ick1−k2−ic,e^{\ii k_1L}=\frac{k_1-k_2+\ii c}{k_1-k_2-\ii c}, exactly Lecture 11’s Lieb–Liniger equation (11.9) for two particles. The other M=1M=1 state is the descendant λ→∞\bethe\to\infty, where the right-hand side of (C.16) tends to 11: free fermions. The two channels of Figure C.1, recovered from the nesting 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.

C.5 The Hubbard model

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 LL sites, LL even, with cjσ†c^{\dagger}_{j\sigma} creating an electron of spin σ\sigma at site jj and njσ=cjσ†cjσn_{j\sigma}=c^{\dagger}_{j\sigma}c_{j\sigma}, H=−t∑j=1L∑σ(cjσ†cj+1,σ+cj+1,σ†cjσ)+U∑j=1Lnj↑nj↓, 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=1t=1: the Yang–Gaudin gas put on a lattice, with the dispersion −2cos⁡k-2\cos k, which is not quadratic, so there is no Galilean invariance.

Theorem C.8 (Lieb and Wu 1968). The regular eigenstates of (C.21) with NN electrons and MM down spins — those with finite, pairwise distinct roots, and 2M≤N≤L2M\le N\le L — are given by charge momenta k1,…,kNk_1,\dots,k_N and spin rapidities Λ1,…,ΛM\Lambda_1,\dots,\Lambda_M solving eikjL=∏α=1Msin⁡kj−Λα+iU/4sin⁡kj−Λα−iU/4,∏j=1NΛα−sin⁡kj+iU/4Λα−sin⁡kj−iU/4=∏β≠αΛα−Λβ+iU/2Λα−Λβ−iU/2,\begin{align} e^{\ii k_jL}&=\prod_{\alpha=1}^{M} \frac{\sin k_j-\Lambda_\alpha+\ii U/4}{\sin k_j-\Lambda_\alpha-\ii U/4},\tag{C.22}\\ \prod_{j=1}^{N}\frac{\Lambda_\alpha-\sin k_j+\ii U/4}{\Lambda_\alpha-\sin k_j-\ii U/4} &=\prod_{\beta\ne\alpha}\frac{\Lambda_\alpha-\Lambda_\beta+\ii U/2}{\Lambda_\alpha-\Lambda_\beta-\ii U/2}, \tag{C.23}\end{align} with E=−2∑jcos⁡kjE=-2\sum_j\cos k_j cited: Lieb and Wu (1968); Essler et al. ch. 3. They are not all the states, and not all the rest are symmetry descendants either (the reading gives one of each on a six-site ring).

With Λα=cλα\Lambda_\alpha=c\bethe_\alpha they are (C.16)–(C.17) under kj⟼sin⁡kj,c⟼U2t, k_j\longmapsto\sin k_j,\qquad c\longmapsto \frac{U}{2t}, (C.24) with that sign: with c↦−U/2tc\mapsto-U/2t none of the eighteen singlet states of the two-electron L=6L=6 ring satisfies (C.22); with it, all do checked. Shastry’s R\Rmat-matrix behind them is not of difference form, which Lecture 14 and Lecture 15 used (the reading). For κ\kappa species the construction nests κ−1\kappa-1 times, one level per node of the Dynkin diagram Aκ−1A_{\kappa-1} (the reading).

Dictionary

Elastic non-diffractive scattering: the asymptotic momenta are only permuted, and a collision leaves a displacement (Theorem 12.2, cited)⟶the momenta are still only permuted ((C.18)) and a collision still leaves a phase, but the labels are transported by XX (Figure C.2), so the ring condition is an eigenvalue problem and not a number (type A; the Part II side proved, the Part I side cited)

C.6 What it buys

  • 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↦sin⁡kk\mapsto\sin k, c↦U/2tc\mapsto U/2t (Theorem C.8, cited).

C.7 Caveats, in one place

  • Status. Theorem C.2 is proved, except that its reduction of the N!N! conditions to NN 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=4N=4. The Lieb–Wu equations are cited; their N=2N=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.

Reading

C.8 The auxiliary chain drawn; which shift; two states drawn

Figure C.3. The auxiliary chain of Definition C.3, drawn as Lecture 14 drew the Heisenberg chain — with the one difference that is the commonest place to stall: its sites are not lattice sites but the particles of a gas on a continuous ring, each at its own rapidity. Setting λ\bethe equal to one of them collapses the horizontal line onto a vertical one (Proposition C.4), which is why the commuting family contains the NN periodicity operators.

Remark C.9 (Which shift, and why it is not free). Write the spin rapidities as λα=vα+i2\bethe_\alpha=v_\alpha+\tfrac\ii2. Lecture 15’s homogeneous chain names the same roots vα=−λα−i2v_\alpha=-\bethe_\alpha-\tfrac\ii2 ((15.6)), so this lecture’s λα\bethe_\alpha is minus Lecture 10’s. There the sign was a labelling: the homogeneous system is invariant under λ→−λ\bethe\to-\bethe and the two choices name a mirror doublet, which only the vector tells apart. Here the ambiguity generically does not arise — for generic inhomogeneities the spin equations are not invariant and the mirrored roots do not solve them, although a symmetric set such as ξ=(−1,0,1)\xi=(-1,0,1) is invariant (at M=1M=1 both roots ±15/6\pm\sqrt{15}/6 solve it) — but the check script builds B(v1)⋯B(vM)∣0⟩B(v_1)\cdots B(v_M)\ket0 anyway and confirms it is the eigenvector used checked.

Figure C.4. Two spin-singlet Bethe states at N=4N=4, M=2M=2, L=9L=9, c=3c=3, total momentum zero. Left: both have four real momenta, so this panel is one axis and the two rows are only a device for keeping the states apart. Each is drawn against its own 2πIj/L2\pi I_j/L (open marks, arrow the shift), and the shift is inward for both. The two states do not share their charge quantum numbers: (C.20) gives (−2,−1,1,2)(-2,-1,1,2) for one and (−3,−2,2,3)(-3,-2,2,3) for the other, which no residual of (C.16) can see. Right: one state has a real spin pair, the other a 22-string on the imaginary axis — a bound state of two spin waves — with energy 7.177.17 against 3.443.44. From scripts/figures/nested_roots.py.

Reading

C.9 The Hubbard model’s symmetry, and a check

The equations of Theorem C.8 are checked against exact diagonalisation of a six-site ring, where 2020 of the 2121 states of the singlet sector are regular Bethe states and the twenty-first is the η\eta-pair of Remark C.10. Eighteen of the twenty — those with K≠πK\ne\pi, complex roots included — are checked to 10−1410^{-14} by reconstructing the roots from (K,E)(K,E); that reconstruction degenerates at K=πK=\pi, so the two states there are checked directly against the equations instead checked. The SO(4)SO(4) commutators below are computed exactly, on one even ring; they are why LL is even.

Remark C.10 (SO(4)SO(4)). (C.21) has more symmetry than the spin su(2)su(2). Let η†=∑j(−1)jcj↑†cj↓†\eta^{\dagger}=\sum_j(-1)^{j}c^{\dagger}_{j\uparrow}c^{\dagger}_{j\downarrow} create a pair on a site, with the staggering. Then [H,η†]=Uη†,[H,S+]=0,[η†,S+]=0,[H,\eta^{\dagger}]=U\eta^{\dagger},\qquad [H,S^{+}]=0,\qquad [\eta^{\dagger},S^{+}]=0, all three for even LL: the hopping terms cancel in pairs along the ring, and the closing bond cancels only if (−1)L=1(-1)^{L}=1. At L=3L=3, U=7U=7 the commutator misses Uη†U\eta^{\dagger} by 22 checked. So H−U2N^H-\tfrac U2\hat N carries a second su(2)su(2) commuting with the first. As Lie algebras su(2)⊕su(2)≅so(4)su(2)\oplus su(2)\cong so(4); the Z2\Z_2 of the familiar statement (SU(2)×SU(2))/Z2≅SO(4)(SU(2)\times SU(2))/\Z_2\cong SO(4) lives at the group level, and no commutator can see it.

For the printed HH the η\eta’s are spectrum-generating, not symmetries: [H,η†]=Uη†[H,\eta^{\dagger}]=U\eta^{\dagger} moves the energy rather than fixing it, and it is H−U2N^H-\tfrac U2\hat N whose degeneracies SO(4)SO(4) organises cited: Yang (1989); Essler et al. ch. 2. The staggering makes η†\eta^{\dagger} anticommute with the translation, so it raises the momentum by π\pi, and η†∣0⟩\eta^{\dagger}\ket0 is an exact eigenstate of (C.21) of energy UU at K=πK=\pi checked — and outside Theorem C.8, because at K=πK=\pi every finite Bethe pair has cos⁡k1+cos⁡k2=0\cos k_1+\cos k_2=0, so E=0E=0, while exact diagonalisation at L=6L=6 returns 0,00,0 and UU.

Reading

C.10 Higher rank: where the Dynkin diagram comes in

Nothing in §C.3 used that the label space is C2\C^{2}. For κ\kappa species the two-body matrix is still X(u)=R(−u)/(u+i)X(u)=\Rmat(-u)/(u+\ii) with R(u)=u−iP\Rmat(u)=u-\ii\perm on Cκ⊗Cκ\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 κ×κ\kappa\times\kappa in the auxiliary index — the transfer matrix t=tra⁡Ta\transfer=\tra T_a still acts on the κN\kappa^{N}-dimensional label space — its reference state breaks slκsl_\kappa to slκ−1sl_{\kappa-1}, and its first row carries κ−1\kappa-1 creation operators B2(λ),…,Bκ(λ)B_2(\bethe),\dots,B_\kappa(\bethe) in place of Lecture 15’s single BB. A Bethe vector Bi1(λ1)⋯BiM(λM)∣0⟩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 κ−1\kappa-1, whose auxiliary problem needs another, and so on — κ−1\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. R(u)=u−iP\Rmat(u)=u-\ii\perm commutes with A⊗1+1⊗AA\otimes\Id+\Id\otimes A for every A∈slκA\in sl_\kappa checked, so the hierarchy is slκsl_\kappa-invariant, and the κ−1\kappa-1 levels are the nodes of the Dynkin diagram Aκ−1A_{\kappa-1}: level aa carries MaM_a rapidities λ(a)\bethe^{(a)}, and the equations couple level aa to level bb through the Cartan matrix, ∏j=1Nλα(a)−ξj+i2δa1λα(a)−ξj−i2δa1=∏b=1κ−1 ∏β(b,β)≠(a,α)λα(a)−λβ(b)+iCab/2λα(a)−λβ(b)−iCab/2,C=(2−1−12) for A2, \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 11 at every other node, δa1=0\delta_{a1}=0 — and the term (b,β)=(a,α)(b,\beta)=(a,\alpha) excluded: with Caa=2C_{aa}=2 it would contribute i/(−i)=−1\ii/(-\ii)=-1 and turn the κ=2\kappa=2 case into 1=−11=-1 checked. Equation (C.17) is that κ=2\kappa=2 case (Figure C.5); the A2A_2 pair, whose second equation contains no ξ\xi at all, is in the reading cited: Sutherland (1968), (16)–(17); Babelon, de Vega and Viallet (1982).

Figure C.5. The levels of the nesting are the nodes of the Dynkin diagram and the gas hangs off the first. Each node carries its own rapidities λ(a)\bethe^{(a)}; nodes joined by an edge have Cab=−1C_{ab}=-1 and appear in each other’s equations, unjoined ones have Cab=0C_{ab}=0 and do not. For κ=2\kappa=2 there is one node, (C.17). Only that first node sees the ξj\xi_j.

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 sl3sl_3’s fundamental from it.

In the reading of optional lecture B the root systems AN−1A_{N-1}, BNB_N, CNC_N, DND_N index classical Calogero–Moser models, one interaction per positive root; here Aκ−1A_{\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.

Dictionary

The root system of slNsl_N, indexing the classical Calogero–Moser models — one pair interaction per positive root (optional lecture B, reading)⟶the Dynkin diagram of Aκ−1A_{\kappa-1}, indexing the levels of the nested Bethe ansatz for κ\kappa-component particles — one set of rapidities per node, coupled by the Cartan matrix (C.25) (type A, cited)

Reading

C.11 Shastry’s R-matrix, and two transcription traps

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↑C↓/2)\exp(hC^{\uparrow}C^{\downarrow}/2) whose parameter is tied to the spectral one by sinh⁡2h=usin⁡2z\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 t\transfer at a special point from being the shift operator: that is regularity, R(z,z)∝P\Rmat(z,z)\propto\perm, which Shastry’s matrix has. What difference form buys is one R\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 A2A_2 the second node of (C.25) reads 1=∏β≠αλα(2)−λβ(2)+iλα(2)−λβ(2)−i∏γλα(2)−λγ(1)−i2λα(2)−λγ(1)+i2,1=\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 ξ\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 −1-1 is taken across cited: Sutherland (1968), (16)–(17).

The two statistics. The bosonic XB=R(u)/(u+i)X_{\mathrm B}=\Rmat(u)/(u+\ii) of Lecture 13 also has XB(0)=−PX_{\mathrm B}(0)=-\perm, because R(0)=−iP\Rmat(0)=-\ii\perm: the two statistics are told apart by u→−uu\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=6L=6, N=2N=2, U=23/3U=2\sqrt3/3 the symmetric wavefunction eik0(x+y)(3−∣x−y∣)e^{\ii k_0(x+y)}(3-\abs{x-y}) with k0=7π/6k_0=7\pi/6 is an exact eigenstate of energy 232\sqrt3 at K=π/3K=\pi/3. Any two-root representation of it must satisfy E=−4cos⁡(K/2)cos⁡12(k1−k2)E=-4\cos(K/2)\cos\tfrac12(k_1-k_2), which here forces k1=k2k_1=k_2: at coincident roots the cusp condition gives A′=−AA'=-A and the ordinary ansatz vanishes, so this state is a regularised limit of a colliding pair, reached as q→0q\to0 along k=k0±iqk=k_0\pm\ii q. Nor is it a descendant: η\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∑(n↑−12)(n↓−12)U\sum(n_\uparrow-\tfrac12)(n_\downarrow-\tfrac12), so its energy carries an extra U4(L−2N)\tfrac U4(L-2N); its coupling is u=U/4tu=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 Δμ\Delta_\mu is the chemical-potential discontinuity. The same book’s Δ\Delta is the single-holon threshold, exactly half of it, and is printed in Bessel form with u=U/4tu=U/4t. Which gap is meant is not a detail: the two differ by a factor of two at every coupling checked.

Reading

C.12 Half filling, the Mott gap, and spin–charge separation

At half filling, N=LN=L, the Hubbard model is an insulator for every U>0U>0. Lieb and Wu’s own paper established it from (C.22)–(C.23) in the thermodynamic limit: the charge gap, meaning the discontinuity μ+−μ−\mu_+-\mu_- of the chemical potential, is Δμ=16t2U∫1∞y2−1sinh⁡(2πty/U) dy,\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>0U>0, tending to U−4tU-4t from above and to (8/π)Ut e−2πt/U(8/\pi)\sqrt{Ut}\,e^{-2\pi t/U} as U→0U\to0 cited: Lieb and Wu (1968), (21)–(23). Say which gap. The reference book’s Δ\Delta (its (7.18)) is the single-holon threshold, exactly half of this, and is written in Bessel form with u=U/4tu=U/4t; the two agree to every digit once converted checked. There is no metal–insulator transition at finite UU in one dimension; the transition is at U=0+U=0^{+}. The spin sector meanwhile stays gapless, and at large UU the half-filled model reduces to the Heisenberg antiferromagnet of Lecture 10 with J=4t2/UJ=4t^{2}/U: second-order hopping of an electron onto an occupied site and back costs UU and gains t2t^{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-3J=-12/U, the two agree to 0.62%0.62\% at U=40U=40 and 0.039%0.039\% at U=160U=160 checked.

That is spin–charge separation, and it is visible already in §C.4: the energy (C.18) does not contain λα\bethe_\alpha. In the thermodynamic limit the two sectors acquire different sound velocities — vcv_{\mathrm c} from the kk’s and vsv_{\mathrm s} from the λ\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.

Reading

C.13 What is not claimed about the gas

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 λ\bethe’s, one of which is drawn in Figure C.4 — while for c>0c>0 the momenta kjk_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.

Reading

C.14 Bosons with a label, and the attractive gas

Two variants cost one line each and are worth knowing.

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

Attraction. For c<0c<0 the kjk_j need no longer be real (Proposition 11.4 assumed c>0c>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=cλ±ic/2k=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=2N=2, K=0K=0, c=−g<0c=-g<0 the even bound wave is cosh⁡q(x−L/2)\cosh q(x-L/2) with 2qtanh⁡(qL/2)=g2q\tanh(qL/2)=g, so q>g/2q>g/2; at g=1g=1, L=2L=2 the true qq is 0.77170.7717 against the string’s 0.50.5 checked. Nothing in §C.3 used the sign of cc; everything in §C.4 about real solutions did.

Tutorial

C.15 Two electrons on a lattice, from scratch

The Lieb–Wu equations were stated, not derived. Here is the N=2N=2 case in full, on a lattice of LL sites, and it is enough to fix (C.24) including its sign.

Take one up electron and one down electron; the spin singlet has a symmetric spatial wavefunction ψ(x1,x2)\psi(x_1,x_2). For x1≤x2x_1\le x_2 put f(x1,x2)=A ei(k1x1+k2x2)+A′ ei(k2x1+k1x2),f(x_1,x_2)=A\,e^{\ii(k_1x_1+k_2x_2)}+A'\,e^{\ii(k_2x_1+k_1x_2)}, set ψ=f\psi=f there and ψ(x1,x2)=f(x2,x1)\psi(x_1,x_2)=f(x_2,x_1) elsewhere. Away from the diagonal the lattice Schrödinger equation is free and gives E=−2(cos⁡k1+cos⁡k2)E=-2(\cos k_1+\cos k_2) for any A,A′A,A'.

The diagonal. At x1=x2=xx_1=x_2=x the equation reads −[ψ(x−1,x)+ψ(x+1,x)+ψ(x,x−1)+ψ(x,x+1)]+Uψ(x,x)=E ψ(x,x),-\big[\psi(x{-}1,x)+\psi(x{+}1,x)+\psi(x,x{-}1)+\psi(x,x{+}1)\big] +U\psi(x,x)=E\,\psi(x,x), and by symmetry ψ(x+1,x)=f(x,x+1)\psi(x{+}1,x)=f(x,x{+}1) and ψ(x,x−1)=f(x−1,x)\psi(x,x{-}1)=f(x{-}1,x), while the free equation for ff continued to the diagonal is the same expression with every ψ\psi replaced by ff and no UU. Subtracting, [f(x+1,x)+f(x,x−1)−f(x−1,x)−f(x,x+1)]+Uf(x,x)=0.\big[f(x{+}1,x)+f(x,x{-}1)-f(x{-}1,x)-f(x,x{+}1)\big]+U f(x,x)=0 . Every term carries ei(k1+k2)xe^{\ii(k_1+k_2)x}; cancelling it leaves a linear relation whose coefficients are, with sj=sin⁡kjs_j=\sin k_j, [2i(s1−s2)+U]A+[−2i(s1−s2)+U]A′=0,soA′A=s1−s2−iU/2s1−s2+iU/2\big[2\ii(s_1-s_2)+U\big]A+\big[-2\ii(s_1-s_2)+U\big]A'=0, \qquad\text{so}\qquad \frac{A'}{A}=\frac{s_1-s_2-\ii U/2}{s_1-s_2+\ii U/2} checked. That is Lecture 11’s S(k)=(k−ic)/(k+ic)S(k)=(k-\ii c)/(k+\ii c) with k↦s1−s2k\mapsto s_1-s_2 and c↦U/2c\mapsto U/2 — (C.24), and its sign, derived rather than transcribed.

The ring. Periodicity relates the two amplitudes once more, and here it is worth doing rather than quoting: the antisymmetry sign of Theorem C.2 does not appear, because we are working with the symmetric spatial function of the singlet. Since x2<x1+Lx_2<x_1+L, ψ(x1+L,x2)=f(x2,x1+L)\psi(x_1{+}L,x_2)=f(x_2,x_1{+}L), so f(x2,x1+L)=f(x1,x2)f(x_2,x_1{+}L)=f(x_1,x_2); matching the coefficients of ei(k2x1+k1x2)e^{\ii(k_2x_1+k_1x_2)} and ei(k1x1+k2x2)e^{\ii(k_1x_1+k_2x_2)} gives A eik2L=A′,A′eik1L=A,soeik1L=AA′=s1−s2+iU/2s1−s2−iU/2,A\,e^{\ii k_2L}=A',\qquad A'e^{\ii k_1L}=A, \qquad\text{so}\qquad e^{\ii k_1L}=\frac A{A'}=\frac{s_1-s_2+\ii U/2}{s_1-s_2-\ii U/2}, which is (C.22) at M=1M=1 once (C.23) has been used: at M=1M=1 its right-hand side is 11 and its left-hand side forces Λ=12(s1+s2)\Lambda=\tfrac12(s_1+s_2), so s1−Λ=12(s1−s2)s_1-\Lambda=\tfrac12(s_1-s_2) and the two displayed forms agree. Multiplying the two charge equations gives ei(k1+k2)L=1e^{\ii(k_1+k_2)L}=1: the total momentum is quantised, as it must be.

The test. Diagonalise (C.21) on L=6L=6 with one up and one down electron — a 36×3636\times36 matrix — and resolve it by total momentum and by spatial symmetry. There are 2121 symmetric states. Eighteen of them, all those with K≠πK\ne\pi, satisfy the displayed equation to 10−1410^{-14} at U=2U=2, 55 and 1212; some need complex k1,k2k_1,k_2 with k1+k2=Kk_1+k_2=K, which is what a bound kk–Λ\Lambda pair looks like checked. The other three sit at K=πK=\pi, where the energy −2(cos⁡k1+cos⁡k2)-2(\cos k_1+\cos k_2) vanishes for every finite pair with k1+k2=πk_1+k_2=\pi and so cannot fix k1−k2k_1-k_2: there the equation has to be read directly. It gives eik1L=−1e^{\ii k_1L}=-1, because k2=π−k1k_2=\pi-k_1 makes s1=s2s_1=s_2 and the ratio (iU/2)/(−iU/2)(\ii U/2)/(-\ii U/2), so kk lies on the antiperiodic grid (2m+1)π/6(2m+1)\pi/6 — the pairs {π/6,5π/6}\{\pi/6,5\pi/6\} with Λ=12\Lambda=\tfrac12 and {−π/6,7π/6}\{-\pi/6,7\pi/6\} with Λ=−12\Lambda=-\tfrac12, which are the two E=0E=0 states and are perfectly regular solutions checked; only the (K,E)(K,E) route to them fails. It is not true that any real pair summing to π\pi solves it: (0,π)(0,\pi) gives eik1L=1e^{\ii k_1L}=1 against −1-1. The third state, at E=UE=U, is the η\eta-pair of Remark C.10 and is outside Theorem C.8 altogether.

The continuum check, for comparison. Example C.7’s N=2N=2 gas can be solved without any Bethe ansatz. At zero total momentum, k1=−k2=κk_1=-k_2=\kappa, the relative coordinate obeys −2φ′′+2cδφ=Eφ-2\varphi''+2c\delta\varphi=E\varphi on a circle of circumference LL with φ\varphi even, so φ=cos⁡κ(x−L/2)\varphi=\cos\kappa(x-L/2) and the jump condition 2φ′(0+)=cφ(0)2\varphi'(0^{+})=c\varphi(0) gives tan⁡(κL/2)=c/2κ\tan(\kappa L/2)=c/2\kappa — which is the logarithm of Example C.7’s equation. At L=10L=10, c=3c=3 both give κ=0.2775645857\kappa=0.2775645857 checked. The factor 22 in the jump is the ℏ=2m=1\hbar=2m=1 units: ∂12+∂22=12∂X2+2∂x2\partial_1^{2}+\partial_2^{2}=\tfrac12\partial_X^{2}+2\partial_x^{2}.

Worth doing by hand. Set U→∞U\to\infty in the displayed equation along the scattering branch, where k1,k2k_1,k_2 stay real and EE stays bounded: the right-hand side tends to −1-1, so kjL∈π+2πZk_jL\in\pi+2\pi\Z and the two electrons quantise as free spinless fermions on a ring with antiperiodic boundary conditions. The spin has not disappeared — the state is still a singlet — but it has stopped costing energy. That is the lattice face of the impenetrable limit of Example C.7. The branches with E∼UE\sim U — the repulsively bound pair, k1,2=π±iqk_{1,2}=\pi\pm\ii q with q∼ln⁡(U/2)q\sim\ln(U/2), and the η\eta-pair — are not hard-core states and the limit does not describe them.

Summary

  • Two channels (C.3), and a crossing is X(u)=R(−u)/(u+i)X(u)=\Rmat(-u)/(u+\ii) (C.4).
  • Periodicity is a matrix condition (Theorem C.2).
  • The NN conditions are one commuting family (Propositions C.4 and C.5): Xa=−t(ξa)\mathcal X_a=-\transfer(\xi_a) for an inhomogeneous Heisenberg chain whose sites are the particles.
  • The nested Bethe equations (Theorem C.6): (C.16)–(C.17), with E=∑kj2E=\sum k_j^{2}; at N=2N=2, Lieb–Liniger in the singlet and free fermions in the triplet (Example C.7).
  • Hubbard: the Lieb–Wu equations are the same system under k↦sin⁡kk\mapsto\sin k, c↦U/2tc\mapsto U/2t (Theorem C.8, cited; its sign derived at N=2N=2 in the tutorial).

References

C. N. Yang, Phys. Rev. Lett. 19 (1967) 1312; M. Gaudin, Phys. Lett. A 24 (1967) 55; E. H. Lieb and F. Y. Wu, Phys. Rev. Lett. 20 (1968) 1445; Essler, Frahm, Göhmann, Klümper and Korepin, The One-Dimensional Hubbard Model (CUP 2005); B. S. Shastry, Phys. Rev. Lett. 56 (1986) 1529. The readings name their other sources.

The whole bibliography, lecture by lecture →