The Bethe Ansatz

General considerations and the Yang-Baxter equation a.g

Let us start from the naive definition of quantum integrability, namely: \({\bf QI:N}\) A system is quantum integrable if:

  • there is a complete set of quantum operators \(\hat{\boldsymbol Q} = (\hat{Q}_1, \hat{Q}_2, ...)\) which are constants of motion (conserved charges), \(\left[ \hat{H}, \hat{\boldsymbol Q} \right] = 0\).
  • all conserved charges are in involution (commute with each other), \(\left[ \hat{Q}_n, \hat{Q}_m \right] = 0 ~~ \forall n, m\).

Let us use this as inspiration, and try to construct integrable models from scratch. First of all, we'd like to do the whole job in one go. Instead of constructing the charges one by one, we'd thus like to construct a generating function for them. Let's define a function \(\tau\) (which we'll end up calling the transfer matrix), taking operator values in the Hilbert space and taking as argument a spectral parameter \(\lambda\):

\begin{equation} \tau (\lambda) = \mbox{exp}~ \sum_{n=0}^\infty \frac{i_n}{n!} Q_n (\lambda - \xi)^n \tag{tau}\label{tau} \end{equation}

(in which \(i_n\) are arbitrary numerical constants). Then,

\begin{equation} Q_n = i_n^{-1}\frac{d^n}{d\lambda^n} \ln \tau (\lambda) |_{\lambda = \xi}. \tag{Qn}\label{Qn} \end{equation}

The statement that all conserved charges are in involution then becomes equivalent to the statement that the transfer matrix commutes for all values of the spectral parameter,

\begin{equation} \left[ \tau(\lambda), \tau(\mu) \right] = 0, ~~ \forall \lambda, \mu. \tag{tauc}\label{tauc} \end{equation}

The central idea of the Algebraic Bethe Ansatz is to provide a formalism for generating such commuting transfer matrices.

To find representations of tauc, we can proceed as follows. First, we introduce the notion of an auxiliary space: that is, we let the transfer matrix be the trace over a space \({\cal A}\) of a new object \(T(\lambda)\) which we'll call the monodromy matrix:

\begin{equation} \tau(\lambda) = \mbox{Tr}_{\cal A} T(\lambda) \tag{tautrT}\label{tautrT} \end{equation}

The auxiliary space is any space we want. For simplicity, we start here by taking it to be a two-dimensional vector space isomorphic to \(\mathbb{C}^2\). The monodromy matrix therefore acts in the tensor product space \({\cal A} \otimes {\cal H}\) of auxiliary and Hilbert spaces.

Condition tauc is fulfilled provided

\begin{equation*} \left[ (\mbox{Tr}_{{\cal A}} T(\lambda)), (\mbox{Tr}_{{\cal A}} T(\mu)) \right] = 0. \end{equation*}

Let's now consider the tensor product of two separate auxiliary spaces \({\cal A}_{1,2}\) (the tensored auxiliary space). Define the operators \(T_{1,2}(\lambda)\) acting in \({\cal A}_1 \otimes {\cal A}_2 \otimes {\cal H}\) as

\begin{equation*} T_1 (\lambda) = T(\lambda) \otimes {\bf 1}_2, \hspace{1cm} T_2 (\lambda) = {\bf 1}_1 \otimes T(\lambda). \end{equation*}

Since the product of traces in a tensor product is the trace in the tensor product space, we have

\begin{equation} \mbox{Tr}_{{\cal A}_1 \otimes {\cal A}_2} \left[ T_1 (\lambda), T_2 (\mu) \right] = 0, \tag{trT1T20}\label{trT1T20} \end{equation}

or in other words

\begin{equation} \mbox{Tr}_{{\cal A}_1 \otimes {\cal A}_2} T_1 (\lambda) T_2 (\mu) = \mbox{Tr}_{{\cal A}_1 \otimes {\cal A}_2} T_2 (\mu) T_1 (\lambda). \tag{trT1T2c}\label{trT1T2c} \end{equation}

(this relation, since we have taken the trace in both auxiliary spaces, is now really a commutation relation in Hilbert space).

The cyclic property of the trace in tensored auxiliary space and the property trT1T2c means that the products \(T_1 T_2\) and \(T_2 T_1\) must be related by a similarity transformation \(R_{12} (\lambda, \mu) \in {\cal A}_1 \otimes {\cal A}_2\) (the subindices indicate in which spaces the \(R\) matrix acts), in other words we can solve trT1T20 by requiring that \(T_1\) and \(T_2\) are intertwined by an invertible \(R\)-matrix,

\begin{equation*} R_{12} (\lambda, \mu) T_1 (\lambda) T_2 (\mu) (R_{12}(\lambda,\mu))^{-1} = T_2 (\mu) T_1 (\lambda) \end{equation*}

or in other words

\begin{equation} R_{12} (\lambda, \mu) T_1 (\lambda) T_2 (\mu) = T_2 (\mu) T_1 (\lambda) R_{12} (\lambda, \mu). \tag{RTTeTTR}\label{RTTeTTR} \end{equation}

Finding a representation of this relation (i.e. finding an \(R\) matrix and a monodromy matrix associated to it) thus constitutes the key to defining a set of commuting quantum operators, and therefore a class of \({\bf QI:N}\) integrable models.

We can investigate the restrictions on possible \(R\) matrices. First of all, any \(R\) matrix must be nonsigular,

\begin{equation*} \mbox{Det} ~R_{12} (\lambda,\mu) \neq 0, \end{equation*}

except at isolated points of \(\lambda,\mu\), which guarantees the existence of an inverse (which is both a left- and right-inverse since \(R\) is square by construction). Second, applying the permutation operator \(P_{12}\) (such that \(P_{12} M_{12} P_{12} = M_{21}\) for any matrix \(M\) in \({\cal A}_1 \otimes {\cal A}_2\)) in RTTeTTR leads to

\begin{equation*} (R_{21} (\mu,\lambda))^{-1} T_1 (\lambda) T_2 (\mu) = T_2 (\mu) T_1 (\lambda) (R_{21} (\mu,\lambda))^{-1}. \end{equation*}

We must thus have

\begin{equation*} R_{12} (\lambda,\mu) = f(\lambda,\mu) (R_{21}(\mu,\lambda))^{-1} \end{equation*}

where \(f(\lambda,\mu)\) is a \(\mathbb{C}\)-valued function. Since \(R_{12}(\lambda,\mu)\) is invertible from the left or right, we have that \(f(\lambda,\mu) = f(\mu,\lambda)\) and \(\left[R_{12} (\lambda,\mu), R_{21} (\mu,\lambda) \right] = 0\). We can rescale \(R\) by an arbitrary scaling function without modifying RTTeTTR, and therefore can without loss of generality impose the constraint

\begin{equation} R_{12} (\lambda,\mu) R_{21} (\mu,\lambda) = 1. \tag{RRe1}\label{:RRe1} \end{equation}

A further important constraint on possible \(R\) matrices comes from the compatibility conditions for RTTeTTR. Namely, if we were to consider the triple tensor product of auxiliary spaces and define \(T_1\), \(T_2\) and \(T_3\), we can change the product \(T_1(\lambda) T_2 (\mu) T_3 (\nu)\) to \(T_3(\nu) T_2(\mu) T_1(\lambda)\) in two different ways; a sufficient (but not necessary) condition for the result to be identical is for the \(R\) matrix to obey the Yang-Baxter relation

\begin{equation} R_{12} (\lambda,\mu) R_{13}(\lambda,\nu) R_{23}(\mu,\nu) = R_{23} (\mu,\nu) R_{13}(\lambda,\nu) R_{12}(\lambda,\mu). \tag{YB}\label{YB} \end{equation}

This automatically guarantees that all higher products are also consistently defined.

Note that in this whole story, we haven't mentioned anything about dimensionality of the physical space in which we'd like our model to live. This restriction will come later, as a convenience to define local Hamiltonian densities.

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Author: Jean-Sébastien Caux

Created: 2023-06-07 Wed 16:02