Past exam of the computer science course of the University of Oxford / 2019 / Quantum information / 2 / d

Consider one normalized measurement vector $|u⟩=a|0⟩+b|1⟩$ on C. The unnormalized partial inner product on AB isIts tensor-product coefficient matrix and determinant areThe two-qubit product-state determinant criterion therefore says that this branch is entangled whenever $a\ne 0$. Its probability is $(|b{|}^2+2|a{|}^2)/3>0$, so such a branch is physically possible and can be normalized without changing whether its determinant vanishes.

For an orthonormal basis $\{|u_0⟩,|u_1⟩\}$, completeness gives $|⟨0|u_0⟩{|}^2+|⟨0|u_1⟩{|}^2=1$. At least one vector has $a\ne 0$. Hence at least one nonzero-probability outcome leaves AB entangled, proving conditional entanglement of the three-qubit W state. Some bases give two entangled branches; the computational basis gives one entangled and one product branch. The claim is not that every outcome must entangle AB.

Suppose $|W⟩=(U_A\otimes U_B\otimes U_C)|GHZ⟩$. Measure C in the orthonormal basis $\{U_C|0⟩,U_C|1⟩\}$. Its two AB conditional vectors would be proportional torespectively. Both are product states, because measuring the original GHZ state's C factor in the computational basis selects $|00⟩$ or $|11⟩$. This contradicts part (i), which applies to every basic measurement basis of the W state's C factor. Therefore no local unitary transformation maps GHZ to W.

An additional invariant makes the distinction visible without measurement records. Across the partition AB versus C, the GHZ reduced density matrix on C has eigenvalues $1/2,1/2$, whereas the W state's are $2/3,1/3$. By local-unitary invariance of reduced-state spectra, these spectra cannot be changed by local unitaries. This supplies the same impossibility and emphasizes that the proposed transformation is a product of unitaries, not arbitrary postselected local processing.