4}.$$ A possible unitary extension ${\tilde M}$ of this matrix is given by $${\tilde M} = \sqrt{\frac{1}{3}} \left(
\begin{array}{cccc} 1 & 1 & 1 & 0 \\ 1 & \omega & \omega^2 & 0 \\ 1 &
\omega^2 & \omega & 0 \\ 0 & 0 & 0 & \sqrt{3} \end{array} \right) \in
{\mathbb C}^{4 \times 4}$$ leading to the vectors ${| {\tilde
\Psi}_1\rangle} = \sqrt{1/3}(1,1,1,0)^T$, ${| {\tilde \Psi}_2\rangle} =
\sqrt{1/3}(1,\omega ,\omega^2,0)^T$, ${| {\tilde \Psi}_3\rangle} =
\sqrt{1/3}(1,\omega^2,\omega,0)^T$, and ${| {\tilde \Psi}_4\rangle} =
(0,0,0,1)^T$. With the state ${\tilde \rho} = \rho \oplus
0_2 \in {\mathbb C}^{4 \times 4}$, for instance, we obtain the probability $${\tilde p}_2 = {\rm tr}\left( \left( \begin{array}{cc|cc} \rho_{11}
& \rho_{12} & 0 & 0 \\ \rho_{21} & \rho_{22} & 0 & 0 \\ \hline0 & 0 &
0 & 0 \\ 0 & 0 & 0& 0 \end{array} \right)  \left(
\begin{array}{cccc} 1 & \omega^2 & \omega & 0 \\ \omega & 1 & \omega^2
& 0 \\ \omega^2 & \omega & 1 & 0 \\ 0 & 0 & 0 & 0 \end{array} \right)
\right) = {\rm tr}\left( \rho \left( \begin{array}{cc} 1 &
\omega^2 \\ \omega & 1 \end{array} \right) \right) = p_2$$ for the second POVM 