Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 51 / 4

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.

Use the first copy to determine computational-basis parity and the second to determine phase parity. On the first pair, Alice and Bob each measure the Pauli operator $Z$ and communicate their signs $z_A,z_B$. On the second, each measures $X$ in the $|+⟩,|-⟩$ basis and communicates the signs $x_A,x_B$.

Every Bell state is an eigenstate of both $Z\otimes Z$ and $X\otimes X$. Direct application to its two basis terms gives the tableFor example, $Z\otimes Z$ acts positively on $|00⟩,|11⟩$ and negatively on $|01⟩,|10⟩$; $X\otimes X$ exchanges the two terms in each pair, detecting their relative sign. Thus individual outcomes are random, but the products in the table are certain. The Bell parity and phase observables label the state uniquely.

The two communicated parities identify all four states with probability one. Each measurement is local and all remaining operations are classical, so the protocol is LOCC discrimination of four Bell states using two copies. Alice can simply send her two signs to Bob, who combines them with his own. Measuring $Z$ on the first copy destroys its phase information, but the second identical copy still supplies it; this is why the two-copy assumption enables this protocol.

Charlie measures his qubit in the $X$ basis, $|\pm ⟩=(|0⟩\pm |1⟩)/\sqrt{2}$. Rewriting the GHZ state in that basis givesEach outcome occurs with probability $1/2$. Charlie records $c=0$ for $+$ and $c=1$ for $-$ and sends that bit to Bob. Alice and Bob's conditional Bell state is $(I\otimes Z^c)|{\Phi }^{+}⟩$. Bob applies $X{Z}^c$, with $Z^c$ applied first. ThenThus every measurement branch gives the requested state after a known local correction: success is deterministic. All quantum operations act on only one party, and the only nonlocal resource used during the protocol is classical communication. This is localizing a Bell pair from a GHZ state by LOCC.

Ignoring Charlie's outcome instead gives Alice and Bob the average state $\frac{1}{2}(|{\Phi }^{+}⟩⟨{\Phi }^{+}|+|{\Phi }^{-}⟩⟨{\Phi }^{-}|)=\frac{1}{2}(|00⟩⟨00|+|11⟩⟨11|)$, which is separable. The classical record supplies the correction needed to use the conditional entanglement; no communication-free preparation of a known entangled state is being claimed.