The Bethe Ansatz

The Hamiltonian of the Heisenberg magnet which we will use throughout is

\begin{equation} H = J \sum_{j=1}^N \left[ S^x_j S^x_{j+1} + S^y_j S^y_{j+1} + \Delta (S^z_j S^z_{j+1} - 1/4) \right]. \tag{xxz.h}\label{xxz.h} \end{equation}

\(J\) is the exchange coupling, with \(J > 0\) (resp. \(J < 0\)) being the antiferromagnetic (resp. ferromagnetic) case. The parameter \(\Delta \in {\mathbb R}\) is called the anisotropy of the model. The spin-\(1/2\) operators \(S^{\alpha}_j\) are equipped with indices \(\alpha = x, y, z\) and \(j\) which labels the lattice site. We consider a closed periodic chain, so that

\begin{equation} S^{\alpha}_{j+N} \equiv S^{\alpha}_j. \tag{xxz.pbc}\label{xxz.pbc} \end{equation}

These operators obey canonical \(su(2)\) commutation relations

\begin{equation*} \left[ S^{\alpha}_j, S^{\beta}_k \right] = i \hbar \delta_{jk} \epsilon^{\alpha \beta \gamma} S^{\gamma}_j \end{equation*}

where \(\epsilon^{\alpha \beta \gamma}\) is the completely antisymmetric tensor, and the Kronecker symbol \(\delta_{jk}\) ensures commutation of operators on different sites. More convenient operators for calculations are the spin raising and lowering operators

\begin{equation*} S^{\pm}_j = S^x_j \pm i S^y_j \end{equation*}

with commutation relations

\begin{equation*} \left[S^z_j, S^{\pm}_k \right] %= \left[ S^z_j, S^x_k \pm i S^y_k \right] = \pm \hbar \delta_{jk} S^{\pm}_j, \hspace{1cm} \left[ S^+_j, S^-_k \right] = 2\hbar \delta_{jk} S^z_j. \end{equation*}

These provide an equivalent form of xxz.h,

\begin{equation} H = J \sum_{j = 1}^N \left[ \frac{1}{2} \left(S^+_j S^-_{j+1} + S^-_j S^+_{j+1}\right) + \Delta \left(S^z_j S^z_{j+1} - 1/4\right) \right]. \tag{xxz.hp}\label{xxz.hp} \end{equation}

Most of our review will concern the spin-\(1/2\) chain, for which the spin operators can be represented using Pauli spin matrices,

\begin{equation*} S^{\alpha}_j = \frac{\hbar}{2} \sigma^{\alpha}_j, \end{equation*}

with the standard definitions used for each site

\begin{equation*} \sigma^x = \left( \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array} \right), \hspace{1cm} \sigma^y = \left( \begin{array}{cc} 0 & -i \\ i & 0 \end{array} \right), \hspace{1cm} \sigma^z = \left( \begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array} \right). \end{equation*}

The Pauli ladder operators are defined as

\begin{equation*} \sigma^+ = \frac{\sigma^x + i \sigma^y}{2} = \left(\begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array} \right) \equiv S^+, \hspace{1cm} \sigma^- = \frac{\sigma^x - i \sigma^y}{2} = \left(\begin{array}{cc} 0 & 0 \\ 1 & 0 \end{array} \right) \equiv S^-. \end{equation*}

The Hilbert space \({\cal H}_j\) on each lattice site \(j\) is then spanned by the two states \(| \pm \rangle_j\), chosen as eigenstates of the \(S^z_j\) operator (N.B.: from now on, we take \(\hbar = 1\)):

\begin{equation*} S^z_j | \pm \rangle_j = \pm \frac{1}{2} | \pm \rangle_j, \hspace{1cm} S^{\pm}_j | \mp \rangle_j = | \pm \rangle_j, \hspace{1cm} S^{\pm}_j | \pm \rangle_j = 0. \end{equation*}

The full Hilbert space is obtained by tensoring all the on-site spaces, \({\cal H} = \oplus_{j=1}^N {\cal H}_j\). It is spanned by the set of \(2^N\) basis states \(\{ | \epsilon_1, ..., \epsilon_N \rangle \}\) with \(\epsilon_j = \{ +, -\} ~\forall j\). One particular member of this set will be of importance later on: the state with all spins pointing up along \(\hat{z}\),

\begin{equation} | 0 \rangle = \otimes_{j = 1}^N | + \rangle_j. \tag{xxz.r}\label{xxz.r} \end{equation}

We will refer to this state as the reference state.

The \(XXZ\) Hamiltonian xxz.h commutes with the \(\hat{z}\)-projection of the total spin operator, \(S^z_{\rm tot} = \sum_{j=1}^N S^z_j\), \[ \left[ H, S^z_{\rm tot} \right] = 0, \] so that the Hilbert space separates into subspaces of fixed magnetization along the \(\hat{z}\) axis. We label each of these subspaces \({\cal H}_M\) by the integer \(M \in \{0, 1, ..., N \}\) representing the number of down spins, i.e. \(S^z_{\rm tot} = \frac{N}{2} - M\). The dimensionality of each subspace is then given by the binomial coefficient \(\mbox{dim} ({\cal H}_M) = \left( \begin{array}{c} N \\ M \end{array} \right)\), fulfilling the requirement \(\sum_{M = 0}^N \mbox{dim} ({\cal H}_M) = 2^N\).