# The Bethe Ansatz

#### Definitionsc.h.d

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 $$\sum_{M = 0}^N \mbox{dim} ({\cal H}_M) = 2^N$$. Except where otherwise noted, all content is licensed under a Creative Commons Attribution 4.0 International License.

Created: 2023-06-07 Wed 16:02