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Application to the case N=2

The matrix L_{ik} defined in Eq. (16) is then \begin{equation*} \hat{L}=\left(\begin{array} {cc} 1 & -\frac{1}{\sqrt{3}}\\ 1 & \frac{1}{\sqrt{3}}\end{array}\right), \end{equation*} with an inverse given by \begin{equation*} \hat{L}^{-1}=\frac{\sqrt{3}}{2}\left(\begin{array} {cc} \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}}\\ -1 & 1\end{array}\right). \end{equation*} The weights are given by the matrix elements 2(L_{0k})^{-1} . We have thence \omega_0=1 and \omega_1=1 .