Regular Bethe data give transfer-family eigenpairs

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.