The Bethe Ansatz

As was mentioned in a_h, the monodromy matrix \(A,B,C,D\) operator entries are not simply related by the usual complex conjugation. This section gives the explicit relations.

Let us start from expression h.lj for the \(L\) operator on site \(j\):

\begin{equation*} L_j (\lambda) = \frac{1}{\lambda -i/2} \left( \begin{array}{cc} \lambda {\boldsymbol 1}_j -i S^z_j & -i S^-_j \\ -i S^+_j & \lambda {\boldsymbol 1}_j + i S^z_j \end{array} \right). \end{equation*}

Consider a single site so \(N = 1\). Then (using a subscript to remind us of the number of sites),

\begin{equation*} T_1(\lambda) = L_1 (\lambda) = \left( \begin{array}{cc} A_1 (\lambda) & B_1 (\lambda) \\ C_1 (\lambda) & D_1 (\lambda) \end{array} \right) \end{equation*}

so

\begin{align*} A_1(\lambda) &= \frac{1}{\lambda-i/2} \left( \lambda {\boldsymbol 1}_1 - i S^z_1 \right), \hspace{5mm} B_1(\lambda) = \frac{-i}{\lambda-i/2} S^-_1, \\ C_1(\lambda) &= \frac{-i}{\lambda-i/2} S^+_1, \hspace{5mm} D_1(\lambda) = \frac{1}{\lambda-i/2} \left( \lambda {\boldsymbol 1}_1 + i S^z_1 \right). \end{align*}

This allows us to write

\[ A_1^\dagger(\lambda) \equiv \left[ A_1 (\lambda^*) \right]^\dagger = \frac{1}{\lambda + i/2} \left( \lambda {\boldsymbol 1}_1 + i S^z_1 \right) \]

so considering all operators, we get

\begin{align*} A_1^\dagger(\lambda) &= \frac{\lambda-i/2}{\lambda+i/2} \,D_1(\lambda), \hspace{5mm} B_1^\dagger(\lambda) = -\frac{\lambda-i/2}{\lambda+i/2} \,C_1(\lambda), \\ C_1^\dagger(\lambda) &= -\frac{\lambda-i/2}{\lambda+i/2} \,B_1(\lambda), \hspace{5mm} D_1^\dagger(\lambda) = \frac{\lambda-i/2}{\lambda+i/2} \,A_1(\lambda), \end{align*}

which we can succinctly write as

\[ T_1^\dagger (\lambda) \equiv \left[T_1(\lambda^*)\right]^\dagger = \frac{\lambda-i/2}{\lambda+i/2}\,\sigma^z_a \,T_1(\lambda) \,\sigma^z_a \]

with the Pauli matrix acting in auxiliary space.

For general \(N\), we proceed by induction. We first hypothesize that

\[ T_N^\dagger (\lambda) \equiv \left[T_1(\lambda^*)\right]^\dagger = d^{-1}_N (\lambda)\,\sigma^z_a \,T_N(\lambda) \,\sigma^z_a \]

in which \(d_N (\lambda) = \left[ \frac{\lambda + i/2}{\lambda -i/2} \right]^N\) is h.ad with subscript specifying the number of sites. Namely, we hypothesize that

\begin{align*} A_N^\dagger(\lambda) &= \frac{D_N(\lambda)}{d_N (\lambda)}, \hspace{5mm} B_N^\dagger(\lambda) = -\frac{C_N(\lambda)}{d_N (\lambda)}, \\ C_N^\dagger(\lambda) &= -\frac{B_N(\lambda)}{d_N (\lambda)}, \hspace{5mm} D_N^\dagger(\lambda) = \frac{A_N(\lambda)}{d_N (\lambda)}. \end{align*}

In view of TprodL, let us now consider

\[ T_{N+1} (\lambda) = L_{N+1} (\lambda) T_N (\lambda) \]

which we can write explicitly as

\[ \left( \begin{array}{cc} A_{N+1} (\lambda) & B_{N+1}(\lambda) \\ C_{N+1} (\lambda) & D_{N+1}(\lambda) \end{array} \right) = \frac{1}{\lambda - i/2} \left(\begin{array}{cc} \lambda {\boldsymbol 1}_{N+1} -i S^z_{N+1} & -i S^-_{N+1} \\ -i S^+_{N+1} & \lambda {\boldsymbol 1}_{N+1} + i S^z_{N+1} \end{array} \right) \left( \begin{array}{cc} A_N (\lambda) & B_N(\lambda) \\ C_N (\lambda) & D_N(\lambda) \end{array} \right). \]

Elementwise, the conjugacy thus takes the form

\begin{align*} A_{N+1}^\dagger (\lambda) &= \frac{1}{\lambda + i/2} \left( \left( \lambda {\boldsymbol 1}_{N+1} + i S^z_{N+1} \right) A_N^\dagger (\lambda) + i S^+_{N+1} C_N^\dagger (\lambda) \right), \\ B_{N+1}^\dagger (\lambda) &= \frac{1}{\lambda + i/2} \left( \left( \lambda {\boldsymbol 1}_{N+1} + i S^z_{N+1} \right) B_N^\dagger (\lambda) + i S^+_{N+1} D_N^\dagger (\lambda) \right) \end{align*}

with similar-looking equations for \(C, D\). Using our hypothesis for the conjugacy of \(A_N, \ldots D_N\) then recursively proves that our general conjugacy relations for any \(N\) are

\begin{align*} A^\dagger(\lambda) &= \frac{D(\lambda)}{d (\lambda)}, \hspace{5mm} B^\dagger(\lambda) = -\frac{C(\lambda)}{d (\lambda)}, \\ C^\dagger(\lambda) &= -\frac{B(\lambda)}{d (\lambda)}, \hspace{5mm} D^\dagger(\lambda) = \frac{A(\lambda)}{d (\lambda)} \end{align*}

or more succinctly

\[ T^\dagger (\lambda) = d^{-1} (\lambda) \,\sigma^z_a \,T(\lambda) \,\sigma^z_a. \]