The Bethe Ansatz

Computing the norm of Bethe eigenstates seems difficult, since the wavefunctions are made up of factorially many terms. In fact, this illustrates the general difficulties encountered when attempting to calculate anything starting from the explicit Bethe Ansatz expression for the wavefunctions. Let's go back to the two-particle case, and consider the wavefunction

\begin{equation*} \Psi_2 (x_1, x_2| \lambda_1, \lambda_2)|_{0 \leq x_1 < x_2 \leq L} = e^{i\lambda_1 x_1 + i\lambda_2 x_2 -\frac{i}{2} \phi(\lambda_1, \lambda_2)} - e^{i\lambda_2 x_1 + i\lambda_1 x_2 +\frac{i}{2} \phi(\lambda_1, \lambda_2)}. \end{equation*}

Let's compute the norm of this in the domain \(0 \leq x_1 < x_2 \leq L\):

\begin{equation*} {\mathbb N}_2 = \int_0^L dx_1 \int_{x_1}^L dx_2 | \Psi_2|^2 = I_1 - 2 \Re I_2 \end{equation*}

where

\begin{equation*} I_1 = 2 \int_0^L dx_1 \int_{x_1}^L dx_2 = L^2 \end{equation*}

and (using the notation \(\lambda_{12} \equiv \lambda_1 - \lambda_2\))

\begin{equation*} I_2 = \int_0^L dx_1 \int_{x_1}^L dx_2 e^{i\lambda_{12} x_{12} -i\phi_{12}} = \frac{e^{-i\phi_{12}}}{\lambda_{12}} \left[ \frac{1 - e^{-i\lambda_{12}L}}{\lambda_{12}} - iL\right]. \end{equation*}

From the Bethe equations, we get that \(e^{-i\lambda_{12}L} = e^{2i\phi_{12}}\) so (making use of the definition of \(\phi_{12}\)) we get

\begin{equation*} 2\Re I_2 = -\frac{L}{\lambda_{12}} 2\sin \phi_{12} = -\frac{L}{\lambda_{12}} \frac{4c \lambda_{12}}{\lambda_{12}^2 + c^2} = -\frac{4cL}{\lambda_{12}^2 + c^2} \end{equation*}

Therefore, the square norm of the two-particle state is

\begin{equation*} {\mathbb N}_2 = L^2 + \frac{4cL}{\lambda_{12}^2 + c^2} \end{equation*}

This coincides with the following determinant:

\begin{equation*} \left| \begin{array}{cc} L + \frac{2c}{\lambda_{12}^2 + c^2} & \frac{-2c}{\lambda_{12}^2 + c^2} \\ \frac{-2c}{\lambda_{12}^2 + c^2} & L + \frac{2c}{\lambda_{12}^2 + c^2} \end{array} \right| = ~\mbox{det} \left(\frac{\partial^2 S}{\partial_{\lambda_j} \partial_{\lambda_l}}\right). \end{equation*}

That is, for two particles in the Bose gas, the square norm of the wavefunction coincides with the determinant of the Hessian matrix of the Yang-Yang action (we will call this matrix the Gaudin matrix from now on).

Amazingly, this results also holds for higher particle numbers,

\begin{equation*} {\mathbb N}_{N} = ~\mbox{det}_N \left(\frac{\partial^2 S}{\partial_{\lambda_j} \partial_{\lambda_l}}\right). \end{equation*}

This result, first conjectured by M. Gaudin b-Gaudin based on the corresponding infinite-size formulas 1971.Gaudin.JMP.12.I 1971.Gaudin.JMP.12.II, was proved by V. Korepin using the Algebraic Bethe Ansatz in 1982.Korepin.CMP.86. This determinant is also the Jacobian of the transformation of variables from the rapidities \(\lambda\) to the quantum numbers \(I\), \(\frac{D(\lambda_1, ..., \lambda_N)}{D(I_1, ..., I_N)}\). Since the measure for state summation in quantum number space is completely flat, it is not surprising that the norm of the state should correspond to such a Jacobian.

Since it really only relies on the form of the Bethe Ansatz wavefunction, Gaudin's formula also holds true for other integrable models. In particular, for spin chains, the square norm of Bethe eigenstates is given by the determinant of the appropriate Gaudin matrix.