# The Bethe Ansatz

#### Solution of the quantum inverse problema.me.sqip

Until now, we have thus managed the first two steps needed to obtain correlation functions within the framework of the Algebraic Bethe Ansatz, namely: we can construct our eigenstates, and normalize them properly. However, the relationship between the operator entries of the monodromy matrix and local spin operators of the spin chain remains obscure at this stage. This problem is known as the quantum inverse problem, and we here present its solution.

The crucial remark is that, since $$b(0) = 0$$ and $$c(0) = 1$$, the $$L_n$$ operator becomes a permutation matrix exchanging the auxiliary space with the $$n$$-th quantum space when the spectral parameter is put equal to the on-site inhomogeneity:

$$L_{n} (\xi_n, \xi_n) = R_{a n} (0) = P_{a n} = \left( \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{array} \right). \tag{LasP}\label{LasP}$$

We consider the homogeneous case $$\xi_j = \xi$$, $$j = 1, ..., N$$ where things are slightly simpler. The monodromy matrix evaluated at $$\lambda = \xi$$ takes the simple form

\begin{equation*} T(\xi) = P_{a N} P_{a N-1} ... P_{a 1} = P_{a 1} P_{1N} P_{1N-1} ... P_{12} = P_{a 1} U \end{equation*}

where $$U = P_{1N} P_{1N-1} ... P_{12}$$ is the cyclic shift operator in quantum space, and we have used $$P_{jk}P_{jl} = P_{kl}P_{jk}$$. We can write this last equation in matrix form:

\begin{equation*} \left( \begin{array}{cc} A(\xi) & B(\xi) \\ C(\xi) & D(\xi) \end{array} \right) = \left( \begin{array}{cc} \frac{1 + \sigma^z_1}{2} & \sigma^-_1 \\ \sigma^+_1 & \frac{1- \sigma^z_1}{2} \end{array} \right) U \end{equation*}

where $$\sigma^z_1$$, $$\sigma^{\pm}_1 = \frac{\sigma^x_1 \pm i \sigma^y_1}{2}$$ are the Pauli matrices associated to site $$1$$. Solving for these gives

\begin{equation*} A(\xi) + D(\xi) = U, \hspace{0.3cm} A(\xi) - D(\xi) = \sigma^z_1 U, \nonumber \\ B(\xi) = \sigma^-_1 U, \hspace{1cm} C(\xi) = \sigma^+_1 U. \end{equation*}

These are the required formulas for the first site. For the other sites, it's simply necessary to use the translation operator $$\sigma_j = U^{j-1} \sigma^a_1 U^{1 - j}$$ and $$U^N = {\bf 1}$$ to obtain the final formulas

\begin{align*} \sigma^z_j &= \left[ A(\xi) + D(\xi) \right]^{j-1} (A(\xi) - D(\xi)) \left[ A(\xi) + D(\xi) \right]^{N - j}, \nonumber \\ \sigma^-_j &= \left[ A(\xi) + D(\xi) \right]^{j-1} B(\xi) \left[ A(\xi) + D(\xi) \right]^{N - j}, \nonumber \\ \sigma^+_j &= \left[ A(\xi) + D(\xi) \right]^{j-1} C(\xi) \left[ A(\xi) + D(\xi) \right]^{N - j}. \end{align*}

In the inhomogeneous case, a similar proof gives

\begin{align} \sigma^z_j &= \prod_{k=1}^{j-1} \left[ A(\xi_k) + D(\xi_k) \right] (A(\xi_j) - D(\xi_j)) \prod_{k=j+1}^N \left[ A(\xi_k) + D(\xi_k) \right], \nonumber \\ \sigma^-_j &= \prod_{k=1}^{j-1} \left[ A(\xi_k) + D(\xi_k) \right] B(\xi_j) \prod_{k=j+1}^N \left[ A(\xi_k) + D(\xi_k) \right], \nonumber \\ \sigma^+_j &= \prod_{k=1}^{j-1} \left[ A(\xi_k) + D(\xi_k) \right] C(\xi_j) \prod_{k=j+1}^N \left[ A(\xi_k) + D(\xi_k) \right]. \tag{sigABCD}\label{sigABCD} \end{align}

This therefore provides the third ingredient needed for the computation of the correlation functions.