The Bethe Ansatz

Now that the structure of our R-matrix is set, let's look more closely at our monodromy matrix. We want to get rid of auxiliary space altogether, and write all our algebraic relations in terms of operators in the Hilbert space. For the particular choice of R-matrix given by R1, we thus get

\begin{align*} R (\lambda, \mu) T_1 (\lambda) T_2 (\mu) = \left( \begin{array}{cc} A(\lambda) A(\mu) & A(\lambda) B(\mu) \\ b(\lambda, \mu) A(\lambda) C(\mu) + c(\lambda, \mu) C(\lambda) A(\mu) & b(\lambda, \mu) A(\lambda) D(\mu) + c(\lambda, \mu) C(\lambda) B(\mu) \\ c(\lambda, \mu) A(\lambda) C(\mu) + b(\lambda, \mu) C(\lambda) A(\mu) & c(\lambda, \mu) A(\lambda) D(\mu) + b(\lambda, \mu) C(\lambda) B(\mu) \\ C(\lambda) C(\mu) & C(\lambda) D(\mu) \end{array} \right. \hspace{1mm}\nonumber\\ \left. \begin{array}{cc} B(\lambda) A(\mu) & B(\lambda) B(\mu) \\ b(\lambda, \mu) B(\lambda) C(\mu) + c(\lambda,\mu) D(\lambda) A(\mu) & b (\lambda, \mu) B(\lambda) D(\mu) + c(\lambda, \mu) D(\lambda) B(\mu) \\ c(\lambda, \mu) B(\lambda) C(\mu) + b(\lambda,\mu) D(\lambda) A(\mu) & c(\lambda, \mu) B(\lambda) D(\mu) + b(\lambda, \mu) D(\lambda) B(\mu) \\ D(\lambda) C(\mu) & D(\lambda) D(\mu) \end{array} \right) \end{align*}

and

\begin{align*} T_2(\mu) T_1(\lambda) R (\lambda, \mu) = \left( \begin{array}{cc} A(\mu) A(\lambda) & b(\lambda, \mu) B(\mu) A(\lambda) + c(\lambda,\mu) A(\mu) B(\lambda) \\ C(\mu) A(\lambda) & b(\lambda, \mu) D(\mu) A(\lambda) + c(\lambda,\mu) C(\mu) B(\lambda) \\ A(\mu) C(\lambda) & b(\lambda, \mu) B(\mu) C(\lambda) + c(\lambda,\mu) A(\mu) D(\lambda) \\ C(\mu) C(\lambda) & b(\lambda, \mu) D(\mu) C(\lambda) + c(\lambda,\mu) C(\mu) D(\lambda) \end{array} \right. \hspace{1mm}\nonumber\\ \left. \begin{array}{cc} c(\lambda, \mu) B(\mu) A(\lambda) + b(\lambda,\mu) A(\mu) B(\lambda) & B(\mu) B(\lambda) \\ c(\lambda, \mu) D(\mu) A(\lambda) + b(\lambda,\mu) C(\mu) B(\lambda) & D(\mu) B(\lambda) \\ c(\lambda, \mu) B(\mu) C(\lambda) + b(\lambda,\mu) A(\mu) D(\lambda) & B(\mu) D(\lambda) \\ c(\lambda, \mu) D(\mu) C(\lambda) + b(\lambda,\mu) C(\mu) D(\lambda) & D(\mu) D(\lambda) \end{array} \right) \end{align*}

This represents a set of nontrivial commutation/product relations between the \(A,B,C,D\) operator functions (for practical reasons, these are often written such that the order of the two operators is changed from one side of the equality to the other):

\begin{align} \tag{ABCDcr} \\ &(11) \hspace{1cm} & \left[ A(\lambda), A(\mu) \right] &= 0, \nonumber \\ &(\bar{13}) & A(\lambda) B(\mu) &= \frac{1}{b(\mu, \lambda)} B(\mu) A(\lambda) - \frac{c(\mu, \lambda)}{b(\mu, \lambda)} B(\lambda) A(\mu), \nonumber \\ &(\bar{12}) & B(\lambda) A(\mu) &= \frac{1}{b(\mu, \lambda)} A(\mu) B(\lambda) - \frac{c(\mu,\lambda)}{b(\mu,\lambda)} A(\lambda) B(\mu), \nonumber \\ &(21) & A(\lambda) C(\mu) &= \frac{1}{b(\lambda, \mu)} C(\mu) A(\lambda) - \frac{c(\lambda, \mu)}{b(\lambda, \mu)} C(\lambda) A(\mu), \nonumber \\ &(31) & C(\lambda) A(\mu) &= \frac{1}{b(\lambda, \mu)} A(\mu) C(\lambda) - \frac{c(\lambda, \mu)}{b(\lambda, \mu)} A(\lambda) C(\mu), \nonumber \\ &(22) & \left[ A(\lambda), D(\mu) \right] &= \frac{c(\lambda,\mu)}{b(\lambda,\mu)} ( C(\mu) B(\lambda) - C(\lambda) B(\mu) ), \nonumber \\ &(14) & \left[ B(\lambda), B(\mu) \right] &= 0, \nonumber \\ &(23) & \left[ B(\lambda), C(\mu) \right] &= \frac{c(\lambda,\mu)}{b(\lambda,\mu)} ( D(\mu) A(\lambda) - D(\lambda) A(\mu)), \nonumber \\ &(24) & B(\lambda) D(\mu) &= \frac{1}{b(\lambda,\mu)} D(\mu) B(\lambda) - \frac{c(\lambda,\mu)}{b(\lambda,\mu)} D(\lambda) B(\mu), \nonumber \\ &(34) & D(\lambda) B(\mu) &= \frac{1}{b(\lambda,\mu)} B(\mu) D(\lambda) - \frac{c(\lambda,\mu)}{b(\lambda,\mu)} B(\lambda) D(\mu), \nonumber \\ &(41) & \left[ C(\lambda), C(\mu) \right] &= 0, \label{eq:CC}\nonumber \\ &(\bar{43}) & C(\lambda) D(\mu) &= \frac{1}{b(\mu,\lambda)} D(\mu) C(\lambda) - \frac{c(\mu,\lambda)}{b(\mu,\lambda)} D(\lambda) C(\mu), \nonumber \\ &(\bar{42}) & D(\lambda) C(\mu) &= \frac{1}{b(\mu,\lambda)} C(\mu) D(\lambda) - \frac{c(\mu,\lambda)}{b(\mu,\lambda)} C(\lambda) D(\mu), \nonumber \\ &(44) & \left[ D(\lambda), D(\mu) \right] &= 0. \label{eq:DD} \nonumber \end{align}

(in which the notation \((ab)\) to the left of each equation means we have used the \(a,b\) matrix element directly, and \((\bar{cd})\) the \(cd\) one with \(\lambda \leftrightarrow \mu\)). Note that bc12 ensure that (\bar{13}) and (\bar{12}) are consistent with each other, and similarly for other pairs of equations.

For notational convenience, we define the following functions (pay attention to the order of the arguments)

\begin{equation} f(\mu, \lambda) \equiv \frac{1}{b(\lambda, \mu)}, \hspace{10mm} g(\mu, \lambda) \equiv \frac{c(\lambda, \mu)}{b(\lambda, \mu)} \tag{fg}\label{fg} \end{equation}

(note that \(g(\mu, \lambda) = -g(\lambda, \mu)\) due to bc12), which allows us to rewrite ABCDcr into the slightly shorter form

\begin{align} \tag{ABCDcrs} \\ &(11) \hspace{1cm} & \left[ A(\lambda), A(\mu) \right] &= 0, \nonumber \\ &(\bar{13}) & A(\lambda) B(\mu) &= f(\lambda, \mu) B(\mu) A(\lambda) + g(\mu, \lambda) B(\lambda) A(\mu), \nonumber \\ &(\bar{12}) & B(\lambda) A(\mu) &= f(\lambda, \mu) A(\mu) B(\lambda) + g(\mu, \lambda) A(\lambda) B(\mu), \nonumber \\ &(21) & A(\lambda) C(\mu) &=f(\mu, \lambda) C(\mu) A(\lambda) + g(\lambda, \mu) C(\lambda) A(\mu), \nonumber \\ &(31) & C(\lambda) A(\mu) &=f(\mu, \lambda) A(\mu) C(\lambda) + g(\lambda, \mu) A(\lambda) C(\mu), \nonumber \\ &(22) & \left[ A(\lambda), D(\mu) \right] &= g(\mu, \lambda) ( C(\mu) B(\lambda) - C(\lambda) B(\mu) ), \nonumber \\ &(14) & \left[ B(\lambda), B(\mu) \right] &= 0, \nonumber \\ &(23) & \left[ B(\lambda), C(\mu) \right] &= g(\mu, \lambda) ( D(\mu) A(\lambda) - D(\lambda) A(\mu) ), \nonumber \\ &(24) & B(\lambda) D(\mu) &= f(\mu, \lambda) D(\mu) B(\lambda) + g(\lambda, \mu) D(\lambda) B(\mu), \nonumber \\ &(34) & D(\lambda) B(\mu) &= f(\mu, \lambda) B(\mu) D(\lambda) + g(\lambda, \mu) B(\lambda) D(\mu), \nonumber \\ &(41) & \left[ C(\lambda), C(\mu) \right] &= 0, \nonumber \\ &(\bar{43}) & C(\lambda) D(\mu) &= f(\lambda, \mu) D(\mu) C(\lambda) + g(\mu, \lambda) D(\lambda) C(\mu), \nonumber \\ &(\bar{42}) & D(\lambda) C(\mu) &= f(\lambda, \mu) C(\mu) D(\lambda) + g(\mu, \lambda) C(\lambda) D(\mu), \nonumber \\ &(44) & \left[ D(\lambda), D(\mu) \right] &= 0. \nonumber \end{align}