Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 2 / 10D (Quantum Information and Computation)

Expanding either Bell state and using $⟨0|1{⟩}_B=0$ removes the cross terms:Thus the two expectation values agree. Equivalently, both states have the same reduced density matrix $I/2$ on qubit $A$; the positivity of $Q$ makes it a possible measurement effect but is not needed for the algebraic equality.

Yes. The states $|{\Phi }^{+}⟩$ and $|{\Phi }^{-}⟩$ are orthogonal, sinceA joint projective measurement in the Bell basis therefore distinguishes them with certainty. Concretely, a controlled-NOT followed by a Hadamard gate converts the Bell basis into the computational basis before measurement.

No. Taking the partial trace over $B$ givesEvery measurement on $A$ alone therefore has identical outcome probabilities in the two cases. This is the local indistinguishability of Bell-state phase.

Yes. Alice and Bob both measure in the Hadamard basis $|+⟩,|-⟩$. The identitiesshow that their outcomes agree for $|{\Phi }^{+}⟩$ and differ for $|{\Phi }^{-}⟩$. After exchanging one classical bit they know which state they had. This is the LOCC discrimination of two Bell states.

No. In quantum dense coding, Alice's four encodings produce four different joint Bell states, but the qubit she transmits has reduced statein every case. Charlie intercepts only that qubit, so every measurement he can perform has the same probability distribution for messages $00,01,10,11$. He gains no information about the two-bit string.