The Bethe Ansatz

Determinant representation for matrix elements

We now compute the matrix elements of the local spin operators in the basis of Bethe eigenstates. Let's do the simplest example in some detail, the matrix element of the \(\sigma^-_j\) operator in the basis of the (as yet unnormalized) Bethe eigenstates:

\begin{equation*} F_j^- (\{ \mu \}_{M+1}, \{ \lambda \}_M) = \langle 0 | \prod_{k=1}^{M+1} C(\mu_k) ~\sigma^-_j \prod_{l=1}^M B(\lambda_l) | 0 \rangle. \end{equation*}

Here, we consider that both sets \(\{ \mu \}_{M+1}\) and \(\{ \lambda \}_M\) are solution to Bethe equations. Using sigABCD, and the fact that Bethe eigenstates are eigenstates of the transfer matrix with eigenvalues

\begin{equation*} (A(\xi_j) + D(\xi_j)) \prod_{l=1}^M B(\lambda_l) | 0 \rangle = \prod_{k=1}^M b^{-1} (\lambda_k, \xi_j) \prod_{l=1}^M B(\lambda_l) | 0 \rangle, \end{equation*}

we can therefore rewrite the matrix element as

\begin{align*} F_j^- (\{ \mu \}_{M+1}, \{ \lambda \}_M) &= \left[ \prod_{l=1}^{M+1} \prod_{k=1}^{j-1} b^{-1} (\mu_l, \xi_k)\right] \left[ \prod_{l=1}^{M} \prod_{k=j+1}^{N} b^{-1} (\lambda_l, \xi_k)\right] \langle 0 | \prod_{k=1}^{M+1} C(\mu_k) B(\xi_j) \prod_{l=1}^M B(\lambda_l) | 0 \rangle \nonumber \\ &= \phi_j^{-1} (\{ \lambda \}) \phi_{j-1} (\{ \mu \}) S_{M+1} (\{ \mu \}_{M+1}, \{ \xi_j, \{ \lambda \}_{M}) \end{align*}

in which we have defined

\begin{equation*} \phi_j (\{ \lambda \}) = \prod_{l=1}^M \prod_{k=1}^j b^{-1} (\lambda_l, \xi_k) \end{equation*}

and used the fact that \(\phi_N (\{ \lambda \}) = 1\), as can be seen from the Bethe equations.

Our job is done, since we can now make use of Slavnov's theorem ssp to obtain the scalar product \(S_{M+1}\) as a determinant. The final answer for the desired matrix element is then

\begin{equation} F_j^- (\{ \mu \}_{M+1}, \{ \lambda \}_M) = \frac{\phi_{j-1} (\{ \mu \})}{\phi_{j-1} (\{ \lambda \})} \frac{\prod_{l=1}^{M+1} \varphi(\mu_l - \xi_j + \eta)}{\prod_{l=1}^M \varphi(\lambda_l -\xi_j + \eta)} \times \nonumber \\ \times \frac{\det_{M+1} H^- (\{ \mu \}, \{ \lambda \})}{\prod_{M+1 \geq l > m \geq 1} \varphi(\mu_l - \mu_m) \prod_{1 \leq l < m \leq M} \varphi(\lambda_l - \lambda_m)} \tag{sigminME}\label{sigminME} \end{equation}

in which the \(H^{-}\) matrix entries are

\begin{align} \left. H^{-}_{ab} \right|_{b < M+1} &= \frac{\varphi(\eta)}{\varphi(\mu_a - \lambda_b)} \left( a(\lambda_b) \prod_{l=1, \neq a}^{M+1} \varphi(\mu_l - \lambda_b + \eta) - d(\lambda_b) \prod_{l=1, \neq a}^{M+1} \varphi (\mu_l - \lambda_b -\eta) \right), \nonumber \\ H^{-}_{a M+1} &= \frac{\varphi(\eta)}{\varphi(\mu_a - \xi_j + \eta) \varphi(\mu_a - \xi_j)}. \tag{Hmin}\label{Hmin} \end{align}

A similar but slightly longer calculation gives the matrix element for the \(\sigma^z\) operator,

\begin{equation} F_j^z (\{ \mu \}_{M}, \{ \lambda \}_M) = \frac{\phi_{j-1} (\{ \mu \})}{\phi_{j-1} (\{ \lambda \})} \frac{\prod_{l=1}^{M} \varphi(\mu_l - \xi_j + \eta)}{\prod_{l=1}^M \varphi(\lambda_l -\xi_j + \eta)} \times \frac{\det_{M} (H (\{ \mu \}, \{ \lambda \}) - 2P(\{\mu\}, \{\lambda \}))} {\prod_{M+1 \geq l > m \geq 1} \varphi(\mu_l - \mu_m) \prod_{1 \leq l < m \leq M} \varphi(\lambda_l - \lambda_m)} \tag{sigzME}\label{sigzME} \end{equation}

in which the \(H\) matrix entries are

\begin{equation} H_{ab} = \frac{\varphi(\eta)}{\varphi(\mu_a - \lambda_b)} \left( a(\lambda_b) \prod_{l=1, \neq a}^{M} \varphi(\mu_l - \lambda_b + \eta) - d(\lambda_b) \prod_{l=1, \neq a}^{M} \varphi (\mu_l - \lambda_b -\eta) \right), \tag{sigzH}\label{sigzH} \end{equation}

and \(P\) is a rank one matrix with entries

\begin{equation} P_{a b} = \varphi(\eta)\frac{\prod_{l=1}^M \varphi(\lambda_l - \lambda_b + \eta)}{\varphi(\mu_a - \xi_j + \eta) \varphi(\mu_a - \xi_j)}. \tag{sigzP}\label{sigzP} \end{equation}

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Author: Jean-Sébastien Caux

Created: 2023-06-07 Wed 16:02