 Solution

Diagonalize the density operator in the Hadamard gate basis $|\pm ⟩=(|0⟩\pm |1⟩)/\sqrt{2}$. Its eigenvalues are $2/3,1/3$, soChoose orthonormal records on a second qubit to construct the first purification of a density operator:It is normalized. In its outer product the cross terms have partial trace zero because $⟨0|1{⟩}_B=0$, while the diagonal terms give precisely ${\rho }_A$.

A different purification follows from unitary freedom of purification. Apply the Hadamard gate to $B$:The same partial-trace calculation uses $⟨+|-{⟩}_B=0$ and gives ${\rho }_A$. The two global states are distinct, not related by a global phase, although their reduced density operators agree. Explicitly the second isIts reduction has diagonal entries $a^2+b^2=1/2$ and off-diagonal entries $2ab=1/6$, checking all the required matrix entries.