A}_j = {|{\tilde
\Psi}_j\rangle} {\langle {\tilde \Psi}_j|} \in {\mathbb C}^{n \times n}$ are given by ${|{\tilde \Psi}_j\rangle} = {|\Psi_j\rangle} \oplus {|\Phi_j\rangle} \in
{\mathbb C}^n$. The extended vectors ${|{\tilde \Psi}_j\rangle}$ are the columns of the matrix $${\tilde M} = \left( \begin{array}{ccc} {|\Psi_1\rangle} & \ldots &
{|\Psi_n\rangle} \\ {|\Phi_1\rangle} & \ldots & {|\Phi_n\rangle} \end{array}
\right) \in {\mathbb C}^{n \times n},$$ that is an arbitrary unitary matrix containing $M$ as upper part of size $d \times n$. The extension of $M$ to a unitary matrix ${\tilde M}$ is always possible since the rows of $M$ are orthonormal. This is guaranteed by the fact that each POVM $(A_j)$ satisfies $\sum_j A_j =I_d$. The probability distribution ${\tilde p}_j = {\rm tr}({\tilde \rho} {\tilde A}_j)$ equals the distribution $p_j$ of the original POVM since $${\tilde p}_j = {\rm
tr}\left({\tilde \rho} {\tilde A}_j \right) = {\rm tr}\left(
\left(\rho \oplus 0_{n-d} \right)  \left(\begin{array}{cc}
{|\Psi_j\rangle}{\langle \Psi_j|} & {|\Psi_j\rangle} {\langle \Phi_j|} \\ {|\Phi_j\rangle}
{\langle \Psi_j|} & {|\Phi_j\rangle} {\langle \Phi_j|}
\end{array} \right) \right) = {\rm tr}\left( \rho A_j \right) = p_j.$$ In our example, we have $$M= \sqrt{\frac{1}{3}} \left( \begin{array}{cccc} 1 &
1 & 1 & 0 \\ 1 & \omega & \omega^2 & 0 \end{array} \right) \in
{\mathbb C}^{2 \times 