Rational Yang–Baxter transfer matrices and Heisenberg charges

In [F2] take $x=u$, $y=v$, $z=0$ and label the three factors $1=a$, $2=b$, $3=n$. For each site $n$ this gives $R_{ab}(u-v)R_{an}(u)R_{bn}(v)=R_{bn}(v)R_{an}(u)R_{ab}(u-v)\mathrm{.}$ This identity acts on $V_a\otimes V_b\otimes V_n$ and as the identity on every other site.

Write $R_{ab}=R_{ab}(u-v)$, $R_{an}=R_{an}(u)$ and $R_{bn}=R_{bn}(v)$. For $1\le n\le N$ set $T_a^{(n)}(u):=R_{an}(u)\cdots R_{a1}(u)$ and $T_b^{(n)}(v):=R_{bn}(v)\cdots R_{b1}(v)$. The case $n=1$ of the desired identity is step 1.1. Suppose it holds for $n-1$. Operators $T_a^{(n-1)}$ commute with $R_{bn}(v)$, and $R_{an}(u)$ commutes with $T_b^{(n-1)}$, because their auxiliary and physical tensor factors are disjoint. Therefore $\begin{array}{rl}{\displaystyle R_{ab}{T}_a^{(n)}{T}_b^{(n)}}& {\displaystyle =R_{ab}{R}_{an}{R}_{bn}{T}_a^{(n-1)}{T}_b^{(n-1)}}\\ {\displaystyle }& {\displaystyle =R_{bn}{R}_{an}{R}_{ab}{T}_a^{(n-1)}{T}_b^{(n-1)}}\\ {\displaystyle }& {\displaystyle =R_{bn}{R}_{an}{T}_b^{(n-1)}{T}_a^{(n-1)}{R}_{ab}}\\ {\displaystyle }& {\displaystyle =T_b^{(n)}{T}_a^{(n)}{R}_{ab},}\end{array}$ where the second equality is step 1.1 at site $n$, the third is the induction hypothesis, and the final equality uses the other disjoint-site commutation. The argument holds for each $n$ up to $N$, proving the claimed RTT identity.

Taking $n=N$ in step 2.1 gives the stated monodromy relation with $T_a^{(N)}=T_a$ and $T_b^{(N)}=T_b$. No invertibility assumption on $R_{ab}(u-v)$ is used, so the identity also holds at its singular spectral differences.

At $u=0$, every factor is $R_{aj}(0)=i{P}_{aj}$, so $t(0)=i^N{\mathrm{tr}}_a(P_{aN}\cdots P_{a1})$. On an elementary chain tensor $e_{j_1}\otimes \cdots \otimes e_{j_N}$, start the auxiliary factor at each basis vector $e_a$ in turn. The swaps pass $e_{j_1},\dots ,e_{j_N}$ successively to the auxiliary factor and shift the preceding entries one site right; the auxiliary trace retains exactly the term with $e_a=e_{j_N}$. The output is therefore $e_{j_N}\otimes e_{j_1}\otimes \cdots \otimes e_{j_{N-1}}$. These tensors form a basis, hence ${\mathrm{tr}}_a(P_{aN}\cdots P_{a1})=U$. As $U$ permutes a basis, it is invertible, proving the first identity and invertibility of $t(0)$.

Differentiate the finite product defining $T_a(u)$ and use $R_{aj}^{\mathrm{\prime }}(0)=I_{aj}$. For each $m$, let $D_m={\mathrm{tr}}_a(P_{aN}\cdots P_{a,m+1}{P}_{a,m-1}\cdots P_{a1})$, with the factor $P_{am}$ omitted. Following the auxiliary basis vector through the remaining swaps and taking its trace gives, for $2\le m<N$, $D_m(e_{j_1}\otimes \cdots \otimes e_{j_N})=e_{j_N}\otimes \left({⨂}_{r=1}^{m-2}{e}_{j_r}\right)\otimes e_{j_m}\otimes e_{j_{m-1}}\otimes \left({⨂}_{r=m+1}^{N-1}{e}_{j_r}\right)\mathrm{.}$ An empty tensor block in this display is omitted. For $m=N$ the same calculation gives $e_{j_{N-1}}\otimes e_{j_1}\otimes \cdots \otimes e_{j_{N-2}}\otimes e_{j_N}$; for $m=1$ it gives $e_{j_1}\otimes e_{j_N}\otimes e_{j_2}\otimes \cdots \otimes e_{j_{N-1}}$. In each case this is exactly $U{P}_{m-1,m}$ applied to the input, with $P_{0,1}:=P_{N,1}$. Since the elementary tensors form a basis, $D_m=U{P}_{m-1,m}$ as operators. Thus $t^{\mathrm{\prime }}(0)=i^{N-1}{\sum }_{m=1}^{N}U{P}_{m-1,m}$.

From step 1.1, $t(0{)}^{-1}=i^{-N}{U}^{-1}$. The definition of $U$ shows $U{P}_{m-1,m}{U}^{-1}=P_{m,m+1}$ for $2\le m\le N$, while $U{P}_{N,1}{U}^{-1}=P_{1,2}$. Therefore $t^{\mathrm{\prime }}(0)t(0{)}^{-1}=i^{-1}{\sum }_{m=1}^N{P}_{m,m+1}=-i{\sum }_{m=1}^N{P}_{m,m+1}$. Multiplication by $i$ proves the claimed logarithmic-derivative identity.

Put $s=u-v$. Since $P_{ab}^2=I$ and the mixed terms cancel, $R_{ab}(s)R_{ab}(-s)=-(s^2+1)I$; hence $R_{ab}(s)$ is invertible for $s\mathrm{\ne }i,-i$. By [F3], $T_b(v)T_a(u)=R_{ab}(s)T_a(u)T_b(v)R_{ab}(s{)}^{-1}$. The factor $R_{ab}$ acts only on the finite auxiliary space $V_a\otimes V_b$, so the ordinary finite trace over that space is cyclic even with operator-valued coefficients on $V_N$. Also ${\mathrm{tr}}_{ab}(T_a(u)T_b(v))=t(u)t(v)$ and ${\mathrm{tr}}_{ab}(T_b(v)T_a(u))=t(v)t(u)$, by expanding each auxiliary matrix in matrix units. Taking ${\mathrm{tr}}_{ab}$ therefore proves $[t(u),t(v)]=0$ whenever $u-v\mathrm{\ne }\pm i$.

Each entry of $t(u)$ is a polynomial in $u$, because it is the trace of a finite product of matrices whose entries are linear polynomials. Thus every entry of $[t(u),t(v)]$ is a polynomial in $(u,v)$. For each fixed $v$, it vanishes for every $u$ except at most $v+i$ and $v-i$; a one-variable polynomial with infinitely many zeros is zero. This holds for every $v$, so $[t(u),t(v)]=0$ for all complex $u,v$, including the two singular differences.

By [F2], $t(0)$ is invertible; hence $M(u)=t(0{)}^{-1}t(u)$ is analytic near $0$ and $M(0)=I$. Choose a submultiplicative matrix norm. Continuity of $M$ gives a neighborhood of $0$ where $\mathrm{\parallel }M(u)-I\mathrm{\parallel }<1$, and there define $\log M(u)={\sum }_{n\ge 1}(-1{)}^{n+1}(M(u)-I{)}^n\mathrm{/}n$, a convergent matrix power series. Step 2.1 implies $t(u)$ commutes with $t(w)$ and every $t(v)$; since $t(0)$ is invertible, $t(0{)}^{-1}$ commutes with them as well. Thus $M(u)$, $M(w)$, and each $t(v)$ commute. Their logarithmic power series preserve these commutations. Comparing the convergent two-variable Taylor coefficients proves the coefficient matrices commute pairwise and with every $t(v)$.

Fix $u\in \mathbb{C}$. Differentiate $[t(u),t(v)]=0$ from [F3] with respect to $v$ at $v=0$; the entries are polynomials, so this gives $[t(u),t^{\mathrm{\prime }}(0)]=0$. Also $[t(u),t(0)]=0$. Since $t(0)$ is invertible by [F2], multiplying $t(u)t(0)=t(0)t(u)$ on the left and right by $t(0{)}^{-1}$ shows $[t(u),t(0{)}^{-1}]=0$. Therefore $t(u)$ commutes with $t^{\mathrm{\prime }}(0)t(0{)}^{-1}$.

Substitute [F2] into step 1.1 and multiply by the scalar $iJ$. The sum ${\sum }_j{P}_{j,j+1}$ therefore commutes with every $t(u)$, and so does $H_N=J{\sum }_j{P}_{j,j+1}+cI$, since scalar multiples of the identity commute with all endomorphisms.

Each $P_{j,j+1}$ is a real permutation matrix with $P_{j,j+1}^2=I$; therefore $P_{j,j+1}^{\ast }=P_{j,j+1}^{-1}=P_{j,j+1}$, so the real linear combination $H_N$ is Hermitian. On the ordered basis $\mathrm{\mid }\uparrow \uparrow ⟩,\mathrm{\mid }\uparrow \downarrow ⟩,\mathrm{\mid }\downarrow \uparrow ⟩,\mathrm{\mid }\downarrow \downarrow ⟩$ of $V\otimes V$, the swap $P$ fixes the first and fourth vectors and exchanges the middle two. Direct multiplication of the displayed Pauli matrices gives $\begin{array}{rl}{\displaystyle \frac{1}{2}(I+X\otimes X+Y\otimes Y+Z\otimes Z)\mathrm{\mid }\uparrow \uparrow ⟩}& {\displaystyle =\mathrm{\mid }\uparrow \uparrow ⟩,}\\ {\displaystyle \frac{1}{2}(I+X\otimes X+Y\otimes Y+Z\otimes Z)\mathrm{\mid }\downarrow \downarrow ⟩}& {\displaystyle =\mathrm{\mid }\downarrow \downarrow ⟩,}\\ {\displaystyle \frac{1}{2}(I+X\otimes X+Y\otimes Y+Z\otimes Z)\mathrm{\mid }\uparrow \downarrow ⟩}& {\displaystyle =\mathrm{\mid }\downarrow \uparrow ⟩,}\\ {\displaystyle \frac{1}{2}(I+X\otimes X+Y\otimes Y+Z\otimes Z)\mathrm{\mid }\downarrow \uparrow ⟩}& {\displaystyle =\mathrm{\mid }\uparrow \downarrow ⟩\mathrm{.}}\end{array}$ These four vectors form a basis, so the displayed Pauli operator equals $P$.

Apply [F3] at $x=u-i\mathrm{/}2$ and $y=v-i\mathrm{/}2$. Since the two shifts cancel in $x-y$, the shifted monodromies obey $R_{ab}(u-v)T_a^L(u)T_b^L(v)=T_b^L(v)T_a^L(u)R_{ab}(u-v)\mathrm{.}$ All operators act on finite-dimensional auxiliary and chain tensor factors.

On a single site with the site vector $e_{\uparrow }$, direct application of $L_{aj}(u)=(u-i\mathrm{/}2)I+i{P}_{aj}$ gives the auxiliary matrix $L_{aj}(u)e_{\uparrow }={\left(\begin{array}{cc}(u+i\mathrm{/}2)e_{\uparrow }& i{e}_{\downarrow }\\ 0& (u-i\mathrm{/}2)e_{\uparrow }\end{array}\right)}_a\mathrm{.}$ It is upper triangular; multiplying these local matrices in the prescribed site order keeps the product upper triangular on $\mathrm{\Omega }=e_{\uparrow }^{\otimes N}$. Its diagonal entries multiply, so $C(u)\mathrm{\Omega }=0,A(u)\mathrm{\Omega }=a(u)\mathrm{\Omega },D(u)\mathrm{\Omega }=d(u)\mathrm{\Omega },a(u)=(u+i\mathrm{/}2{)}^N,d(u)=(u-i\mathrm{/}2{)}^N\mathrm{.}$

In the ordered auxiliary basis $\mathrm{\mid }\uparrow \uparrow ⟩,\mathrm{\mid }\uparrow \downarrow ⟩,\mathrm{\mid }\downarrow \uparrow ⟩,\mathrm{\mid }\downarrow \downarrow ⟩$, $R_{ab}(x)$ has diagonal entries $x+i,x,x,x+i$ and the two off-diagonal middle entries $i$. Comparing the row $\mathrm{\mid }\uparrow \uparrow ⟩$ and column $\mathrm{\mid }\downarrow \uparrow ⟩$ entries in step 1.1 gives $(x+i)B(u)A(v)=iB(v)A(u)+xA(v)B(u),x=u-v\mathrm{.}$ Exchange $u,v$ and solve for $A(u)B(v)$ to obtain $A(u)B(v)=f(x)B(v)A(u)+g(x)B(u)A(v)$ for $x\mathrm{\ne }0$, where $f(x):=(x-i)\mathrm{/}x$ and $g(x):=i\mathrm{/}x$. The row $\mathrm{\mid }\uparrow \uparrow ⟩$, column $\mathrm{\mid }\downarrow \downarrow ⟩$ entry gives $(x+i)B(u)B(v)=(x+i)B(v)B(u)$, hence the $B$ operators commute for $x\mathrm{\ne }-i$; both sides are polynomial matrices, so this holds also at $x=-i$. The row $\mathrm{\mid }\downarrow \uparrow ⟩$, column $\mathrm{\mid }\downarrow \downarrow ⟩$ entry gives $xD(u)B(v)+iB(u)D(v)=(x+i)B(v)D(u)$. Thus for $x\mathrm{\ne }0$, $D(u)B(v)=h(x)B(v)D(u)+k(x)B(u)D(v)$, where $h(x):=(x+i)\mathrm{/}x$ and $k(x):=-i\mathrm{/}x$.

For pairwise distinct parameters $v_1,\dots ,v_M$ and a variable $u$ distinct from them, the exchange relations of step 2.1 imply $\begin{array}{rl}{\displaystyle A(u)\prod _{k=1}^{M}B(v_k)\mathrm{\Omega }=}& {\displaystyle a(u)\prod _{k=1}^{M}f(u-v_k)\prod _{k=1}^{M}B(v_k)\mathrm{\Omega }}\\ {\displaystyle }& {\displaystyle +\sum _{j=1}^{M}a(v_j)g(u-v_j)\prod _{k\mathrm{\ne }j}f(v_j-v_k)B(u)\prod _{k\mathrm{\ne }j}B(v_k)\mathrm{\Omega },}\\ {\displaystyle D(u)\prod _{k=1}^{M}B(v_k)\mathrm{\Omega }=}& {\displaystyle d(u)\prod _{k=1}^{M}h(u-v_k)\prod _{k=1}^{M}B(v_k)\mathrm{\Omega }}\\ {\displaystyle }& {\displaystyle +\sum _{j=1}^{M}d(v_j)k(u-v_j)\prod _{k\mathrm{\ne }j}h(v_j-v_k)B(u)\prod _{k\mathrm{\ne }j}B(v_k)\mathrm{\Omega }\mathrm{.}}\end{array}$ For $M=0$ these follow from step 1.2. Suppose they hold for $M-1$ and move $A(u)$ past the first factor $B(v_1)$ using step 2.1, then apply the induction formulas to the remaining $M-1$ factors. The unwanted coefficient for a fixed $j\ge 2$ is the common factor $a(v_j){\prod }_{k\mathrm{\ne }1,j}f(v_j-v_k)$ times $f(x)g(y)+g(x)g(y-x)$, with $x=u-v_1$ and $y=u-v_j$. Direct substitution of $f(z)=(z-i)\mathrm{/}z$ and $g(z)=i\mathrm{/}z$ gives $f(x)g(y)+g(x)g(y-x)=g(y)f(x-y)$; here $x-y=v_j-v_1$, the missing factor in the target product. For $j=1$, the second term of the first exchange gives the displayed coefficient directly. For $D$, the same induction uses $h,k$ and the identity $h(x)k(y)+k(x)k(y-x)=k(y)h(x-y)$, verified by substituting $h(z)=(z+i)\mathrm{/}z$ and $k(z)=-i\mathrm{/}z$. This proves both formulas for all $M$.

Apply step 3.1 with $v_j={\lambda }_j$ and add its $A$ and $D$ equations. The coefficient of ${\prod }_{k}B({\lambda }_k)\mathrm{\Omega }$ is exactly $\mathrm{\Lambda }(u)$ in the statement. For each $j$, the combined coefficient of $B(u){\prod }_{k\mathrm{\ne }j}B({\lambda }_k)\mathrm{\Omega }$ is $g(u-{\lambda }_j)[a({\lambda }_j){\prod }_{k\mathrm{\ne }j}f({\lambda }_j-{\lambda }_k)-d({\lambda }_j){\prod }_{k\mathrm{\ne }j}h({\lambda }_j-{\lambda }_k)],$ because $k(u-{\lambda }_j)=-g(u-{\lambda }_j)$. The bracket is zero: after substituting $a,d,f,h$, its equality is precisely the Bethe equation in [F2], whose denominators are nonzero by the regularity hypotheses. Thus $\tau (u)\mathrm{\Psi }=\mathrm{\Lambda }(u)\mathrm{\Psi }$ for every $u$ outside the finite root set.

Put $Q(u)={\prod }_{j=1}^M(u-{\lambda }_j)$. The numerator of $\mathrm{\Lambda }(u)$ is $P(u)=a(u){\prod }_{j=1}^M(u-{\lambda }_j-i)+d(u){\prod }_{j=1}^M(u-{\lambda }_j+i),$ a polynomial. At $u={\lambda }_j$, its value is $P({\lambda }_j)=i[d({\lambda }_j){\prod }_{k\mathrm{\ne }j}({\lambda }_j-{\lambda }_k+i)-a({\lambda }_j){\prod }_{k\mathrm{\ne }j}({\lambda }_j-{\lambda }_k-i)]=0$ by the same Bethe equation. Since the ${\lambda }_j$ are distinct, each is a simple root of $Q$, so the factor theorem gives $Q\mid P$ and $\mathrm{\Lambda }=P\mathrm{/}Q$ is a polynomial. Both $\tau (u)\mathrm{\Psi }$ and $\mathrm{\Lambda }(u)\mathrm{\Psi }$ are then polynomial vector-valued functions agreeing away from finitely many points by step 4.1; their difference has infinitely many zeros and is identically zero. As $\mathrm{\Psi }\mathrm{\ne }0$ by [F2], this is an eigenvector identity for every $u$, with no claim that other eigenvectors arise this way.