Past exam of the mathematics course of the University of Cambridge / 2024 / ii / Paper 3 / 10E (Quantum Information and Computation)

If the Schmidt rank is one, the decomposition is $|\psi ⟩=|\alpha ⟩|\beta ⟩$ after absorbing its sole coefficient, so the state is a product. Conversely, a product state has coefficient matrix $X=a{b}^T$, which has rank one. Therefore a bipartite pure state is entangled exactly when its Schmidt rank is at least two.

The state is alreadyIt therefore has Schmidt rank one across each of $A|BC$, $B|AC$, and $C|AB$.

Across $A|BC$ the state is the product $|0{⟩}_A\otimes |{\varphi }^{+}{⟩}_{BC}$, so its Schmidt rank is one. Across $B|AC$ it isand the two states on each side are orthonormal, so the Schmidt rank is two. The same argument with $B$ and $C$ exchanged handles $C|AB$. This is precisely the entanglement of the Bell state shared by $B$ and $C$.

Across $A|BC$, the GHZ state has the Schmidt decompositionof rank two. Symmetry gives the same rank-two decomposition across $B|AC$ and $C|AB$, so the state is entangled across every bipartition.

Projecting the GHZ state onto Charlie's outcome $0$ leaves the unnormalised state $|00{⟩}_{AB}/\sqrt{2}$. After normalization Alice and Bob have $|00⟩$, a product state.

Projecting the three-qubit $W$ state onto Charlie's outcome $0$ leavesAfter normalization this is $(|01⟩+|10⟩)/\sqrt{2}$, a Bell state and hence an entangled state of Alice and Bob.