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

A normalized bipartite pure state is entangled when it cannot be written as a product state $|a{⟩}_A\otimes |b{⟩}_B$. Equivalently, its Schmidt rank exceeds one. This definition concerns the specified partition into systems A and B; the mixed-state definition of separability requires a different convex-mixture condition.

A bipartite unitary operator is an entangling gate if it maps at least one product state to an entangled state. It need not entangle every product input. For example, the controlled-NOT gate entangles $|+⟩|0⟩$ but leaves $|00⟩$ as a product. A gate that preserves every product input is not entangling, even if it exchanges the two subsystems.

The joint Born rule uses the tensor product of the two outcome bras. Its amplitude isThe other source component vanishes on contraction with $⟨0|$ on A. ThereforeThis is a joint probability: $\mathrm{Pr}(A=0)=16/25$, while $\mathrm{Pr}(B={\theta }_0\mid A=0)={\cos }^2\theta$. Dropping the first factor would give only the conditional probability.

Here $⟨+y|=(⟨0|-i⟨1|)/\sqrt{2}$. Keeping that complex conjugation, the joint amplitude isHence $\mathrm{Pr}(A=+,B=+y)=0$. The cancellation is destructive quantum interference between the two source components, not the absence of either local outcome. Each local outcome individually has probability $1/2$, since each reduced density matrix is $I/2$. Omitting the conjugation in the second bra would turn the cancellation into reinforcement and give the wrong answer.

For deterministic classical responses, denote the two possible local answers by $x_0,x_1$, $y_0,y_1$ and $z_0,z_1$. Certainty in the GHZ game would requireTaking the exclusive or of all four equations gives zero on the left, since every answer occurs twice, but one on the right. This contradiction rules out deterministic certainty. Pre-shared randomness cannot cure it: conditioning on the random seed fixes a deterministic response table, and none succeeds on every allowed input. Thus noncommunicating classical players cannot win with certainty. No uniform distribution on the referee's questions is needed for this worst-case claim.

For the quantum strategy, share the GHZ state and measure the Pauli X gate observable when the local question is 0, or the Pauli Y gate observable when it is 1. Return bit 0 for measurement eigenvalue $+1$ and bit 1 for eigenvalue $-1$. The four relevant operator identities areFor the first identity, the three bit flips exchange $|000⟩$ and $|111⟩$. For each remaining identity, the two Y actions contribute $i^2=-1$ on $|000⟩$ and $(-i{)}^2=-1$ on $|111⟩$, so they exchange the terms with a minus sign. Local measurements on separate systems commute, and the product of their observed eigenvalues is therefore fixed by the displayed joint eigenvalue. That product equals $(-1{)}^{x+y+z}$, giving exactly the required output parity for each promised input. Hence the quantum strategy wins with certainty without communication. This is the GHZ theorem in nonlocal-game form; it does not let a player control another player's individual outcome.

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.

For every nontrivial dimension $d\ge 2$, no such pure state exists. The literal claim needs this qualification: for $d=1$, the only measurement outcome is necessarily the same for both parties. The contradiction below requires allowing all complex orthonormal bases, as quantum theory does.

If equal outcomes were certain in the computational basis, all amplitudes with different labels would vanish, so the normalized joint state would have the formChoose distinct labels $j,k$ and replace their computational vectors in both local bases by $|+{⟩}_{jk}=(|j⟩+|k⟩)/\sqrt{2}$ and $|-{⟩}_{jk}=(|j⟩-|k⟩)/\sqrt{2}$, leaving all other vectors fixed. The joint amplitude for the different outcomes $+,-$ is $(c_j-c_k)/2$. Certainty of equal outcomes would force $c_j=c_k$.

Now use the complex pair $|+i{⟩}_{jk}=(|j⟩+i|k⟩)/\sqrt{2}$ and $|-i{⟩}_{jk}=(|j⟩-i|k⟩)/\sqrt{2}$ in the same way on both systems. The amplitude for different outcomes $+i,-i$ is $(c_j+c_k)/2$, because the conjugated bra factors on the $|kk⟩$ term multiply to $(-i)(i)=1$. Thus equal outcomes would also force $c_j=-c_k$. Combining the two requirements gives $c_j=c_k=0$. Every label belongs to such a pair when $d\ge 2$, forcing the zero vector and contradicting normalization. This proves impossibility of universal identical-basis perfect correlations.

The maximally entangled state $d^{-1/2}{\sum }_j|jj⟩$ is not a counterexample: it correlates matching labels when one party uses a basis and the other its coordinatewise complex conjugate basis. Requiring the two complex bases to be identical is different. This distinction also follows from conjugate-state steering from a maximally entangled pair; restricting attention to real bases would miss the contradiction.