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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.

Uhlmann's theorem identifies quantum fidelity with the largest absolute overlap of purifications of a density operator. Choose a common reference Hilbert space $R$ of dimension at least that of the original system. Thenwhere $|u⟩,|v⟩\in {\mathcal{H}}_A\otimes {\mathcal{H}}_R$ have reduced density operators ${\rho }_A,{\sigma }_A$. One purification may be fixed in advance: the maximum is over the other, with the freedom to apply a unitary on the reference system. This is the unitary freedom of purification. Enlarging the reference by unused dimensions does not change the maximum.

Take maximizing purifications $|u⟩,|v⟩$ of ${\rho }_{AB},{\sigma }_{AB}$ on $ABR$. Their overlap has magnitude $F({\rho }_{AB},{\sigma }_{AB})$. Regard these same vectors as purifications of ${\rho }_A,{\sigma }_A$ with reference system $BR$. They are candidates in the larger optimization, so Uhlmann's theorem givesThis is monotonicity of quantum fidelity under partial trace: discarding a system cannot make two states more distinguishable according to their quantum fidelity.

Take Schmidt decompositions of the two purifications across $A:R$. Their squared Schmidt coefficients and their $A$-side eigenspaces are fixed by the same reduced state ${\rho }_A$. The reference-side Schmidt vectors are two orthonormal families, so a unitary $U_R$ maps one family to the other, including arbitrary choices inside degenerate subspaces. Hence the unitary freedom of purification givesIf the reference supports have different dimensions, the corresponding statement uses an isometry.

Construct ensemble purificationsThey have the same reduced state exactly when ${\rho }_1={\rho }_2$. By the unitary freedom of purification, this holds exactly when $|{\Phi }_2⟩=(I\otimes U)|{\Phi }_1⟩$ for a unitary $U$. Comparing reference-basis coefficients gives the Hughston–Jozsa–Wootters theorem relationConversely, substituting this relation and using $U^{†}U=I$ immediately gives ${\rho }_2={\rho }_1$.