The Bethe Ansatz

Eigenstates of the transfer matrix a.R.s.t

Remember that the monodromy matrix entries \(A(\lambda), B(\lambda), C(\lambda), D(\lambda)\) are taken to be functions taking operator values in some Hilbert space which we haven't specified yet. These obey commutation relations (a quadratic algebra) specified by ABCDcr, once the \(R\)-matrix is specified. We expect that these operator-valued functions are sufficient to navigate our whole Hilbert space, however we choose the latter to be defined. Let us try to develop some intuition of their meaning.

First of all, our generating function for conserved charges, the transfer matrix tau, takes from tautrT, TeABCD a simple form in terms of these operators:

\begin{equation} \tau (\lambda) = \mbox{Tr}_{{\cal A}} T(\lambda) = A(\lambda) + D(\lambda). \tag{tauAD}\label{tauAD} \end{equation}

We are looking for states simultaneously diagonalizing all conserved charges: these are obtained automatically by seeking eigenstates of the transfer matrix itself, so we will be looking for eigenstates of \(A(\lambda) + D(\lambda)\). On the other hand, the form of the monodromy matrix TeABCD suggests to interpret \(B\) as a raising operator and \(C\) as a lowering operator within our Hilbert space.

To make a start building actual eigenstates, we assume the existence of a pseudovacuum \(|0\rangle\), which will be interpreted as the highest-weight state of the monodromy operators algebra. For definiteness, we also assume that it simultaneously diagonalizes the \(A\) and \(D\) operators summing up to transfer matrix, and that the lowering operator \(C\) annihilates it:

\begin{equation} A(\lambda) | 0 \rangle = a(\lambda) | 0 \rangle, \hspace{1cm} D(\lambda) | 0 \rangle = d(\lambda) | 0 \rangle, \hspace{1cm} C(\lambda) | 0 \rangle = 0. \tag{pv}\label{pv} \end{equation}

where the functions \(a(\lambda)\) and \(d(\lambda)\) can be chosen arbitrarily.

The pseudovacuum is therefore by construction an eigenstate of the transfer matrix:

\begin{equation} \tau(\lambda) | 0 \rangle = \tau (\lambda | \emptyset ) | 0 \rangle, \hspace{1cm} \tau (\lambda | \emptyset ) = a(\lambda) + d(\lambda). \tag{taupv}\label{taupv} \end{equation}

To construct further eigenstates of the transfer matrix, we consider applying our raising operator \(B\) repeatedly on the pseudovacuum. In generality, let us consider the state

\begin{equation} | \{ \lambda_j \}_M \rangle \equiv \prod_{j=1}^M B (\lambda_j) | 0 \rangle \tag{prodBpv}\label{prodBpv} \end{equation}

for generic \(M\) and \(\{ \lambda_j \}_M\). Note the very important fact that the order in the product is immaterial, in view of the commutation relation ABCDcr \((14)\).

To check how the transfer matrix acts on such a state, we start by computing the action of \(A\). The result is

\begin{align} A(\lambda) \prod_{j=1}^M B (\lambda_j) | 0 \rangle = \Lambda^A \prod_{j=1}^M B (\lambda_j) | 0 \rangle + \sum_{l=1}^M \Lambda^A_l B (\lambda) \prod_{j\neq l}^M B (\lambda_j) | 0 \rangle, \nonumber \\ \Lambda^A = a(\lambda) \prod_{j=1}^M b^{-1} (\lambda_j, \lambda), \hspace{1cm} \Lambda^A_l = - a(\lambda_l) \frac{c(\lambda_l, \lambda)}{b(\lambda_l, \lambda)} \prod_{j\neq l}^M b^{-1} (\lambda_j, \lambda_l). \tag{AprodBpv}\label{AprodBpv} \end{align}

The form of the action of this operator is easy to establish intuitively. When commuting \(A(\lambda)\) through a \(B(\lambda_j)\) using ABCDcr \((\bar{13})\), two types of terms are produced: one maintaining the function arguments, the other exchanging them. The only possible arguments of the \(M+1\) operators \(A, B\) are thus \(\lambda, \{ \lambda_j \}\). Assuming that all \(\lambda_j\) are distinct, once we have commuted \(A\) all the way through, if \(A\) has argument \(\lambda\), the first term with coefficient \(\lambda\) is obtained (and we have always used the first term on the right-hand side of ABCDcr \((\bar{13})\)). Otherwise if the argument \(\lambda\) is exchanged to an operator \(B\), the second term is obtained.

Similarly, we obtain the action of \(D\) on states prodBpv:

\begin{align} D(\lambda) \prod_{j=1}^M B (\lambda_j) | 0 \rangle = \Lambda^D \prod_{j=1}^M B (\lambda_j) | 0 \rangle + \sum_{l=1}^M \Lambda^D_l B (\lambda) \prod_{j\neq l}^M B (\lambda_j) | 0 \rangle, \nonumber \\ \Lambda^D = d(\lambda) \prod_{j=1}^M b^{-1} (\lambda, \lambda_j), \hspace{1cm} \Lambda^D_l = - d(\lambda_l) \frac{c(\lambda, \lambda_l)}{b(\lambda, \lambda_l)} \prod_{j\neq l}^M b^{-1} (\lambda_l, \lambda_j). \tag{DprodBpv}\label{DprodBpv} \end{align}

A state prodBpv will thus be an eigenstate of the transfer matrix if \(\Lambda_j + \bar{\Lambda}_j = 0\), that is if

\begin{equation} \frac{a(\lambda_j)}{d(\lambda_j)} \prod_{l \neq j} \frac{b(\lambda_j, \lambda_l)}{b(\lambda_l, \lambda_j)} = 1 \tag{BER1}\label{BER1} \end{equation}

in which we have used bc12. These are the Bethe equations (constraints on the sets of rapidities \(\{ \lambda_j \}\)) guaranteeing that prodBpv are eigenstates of the transfer matrix tau

\begin{equation} \tau (\lambda) | \{ \lambda_j \}_M \rangle = \tau (\lambda | \{ \lambda_j \}_M ) | \{ \lambda_j \}_M \rangle, \tag{taue}\label{eq:taue} \end{equation}

with eigenvalues

\begin{equation} \tau (\lambda | \{ \lambda_j \}_M ) = a(\lambda) \prod_{j=1}^M b^{-1} (\lambda_j, \lambda) + d(\lambda) \prod_{j=1}^M b^{-1} (\lambda, \lambda_j). \tag{tauev}\label{tauev} \end{equation}

For later convenience, we also mention here the action of the operator \(C\) on a state, which can be derived according to a logic very similar to that leading to AprodBpv and DprodBpv:

\begin{equation} C(\lambda) \prod_{j=1}^M B (\lambda_j) | 0 \rangle = \sum_{k=1}^M \Lambda^C_k \prod_{j \neq k}^M B (\lambda_j) | 0 \rangle + \sum_{k_1 < k_2}^M \Lambda^C_{k_1 k_2} B (\lambda) \prod_{j \neq k_1, k_2}^M B (\lambda_j) | 0 \rangle, \tag{CprodBpv}\label{CprodBpv} \end{equation}


\begin{align} \Lambda^C_k &= a(\lambda_k) d(\lambda) \frac{c(\lambda, \lambda_k)}{b(\lambda, \lambda_k)} \prod_{j \neq k}^M b^{-1} (\lambda_j, \lambda_k) b^{-1}(\lambda, \lambda_j) + (\lambda_k \leftrightarrow \lambda), \nonumber \\ \Lambda^C_{k_1 k_2} &= a(\lambda_{k_1}) d(\lambda_{k_2}) \frac{c(\lambda, \lambda_{k_1})}{b(\lambda, \lambda_{k_1})} \frac{c(\lambda_{k_2}, \lambda)}{b(\lambda_{k_2}, \lambda)} b^{-1} (\lambda_{k_2}, \lambda_{k_1}) \prod_{j \neq k_1, k_2} b^{-1} (\lambda_j, \lambda_{k_1}) b^{-1} (\lambda_{k_2}, \lambda_j) + (\lambda_{k_1} \leftrightarrow \lambda_{k_2}) \tag{LC}\label{LC} \end{align}

which again looks more compact using the notations fg

\begin{align} \Lambda^C_k &= a(\lambda_k) d(\lambda) g(\lambda_k, \lambda) \prod_{j \neq k}^M f(\lambda_k, \lambda_j) f(\lambda_j, \lambda) + (\lambda_k \leftrightarrow \lambda), \nonumber \\ \Lambda^C_{k_1 k_2} &= a(\lambda_{k_1}) d(\lambda_{k_2}) g(\lambda_{k_1}, \lambda) g(\lambda, \lambda_{k_2}) f(\lambda_{k_1}, \lambda_{k_2}) \prod_{j \neq k_1, k_2} f (\lambda_{k_1}, \lambda_j) f (\lambda_j, \lambda_{k_2}) + (\lambda_{k_1} \leftrightarrow \lambda_{k_2}) \tag{LCfg}\label{LCfg} \end{align}

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Author: Jean-Sébastien Caux

Created: 2023-06-07 Wed 16:02