Past exam of the mathematics course of the University of Cambridge / 2026 / ii / Paper 3 / 15E (Quantum Information and Computation)

Across the cut $A|BC$, the two conditional $BC$ vectors $\alpha |00⟩$ and $\gamma |00⟩+\beta |11⟩$ are linearly independent when $\alpha \beta \ne 0$, so the Schmidt rank exceeds one. Hence the state cannot be a three-qubit product for any $\gamma$.

Charlie measures in the $X$ basis and sends one bit to Alice or Bob; the remaining pair becomes $|{\Phi }^{+}⟩$ or $|{\Phi }^{-}⟩$, corrected by a local $Z$. Alice then performs standard teleportation: CNOT from $|\xi ⟩$ to her GHZ qubit, Hadamard, two computational measurements, and sends the two outcomes to Bob, who applies the corresponding Pauli correction.

No. Without classical communication Bob's reduced state is independent of Alice's unknown input by no-signalling, whereas successful transfer would make it equal to that input.

Reordering as $A_1{A}_2{B}_1{B}_2$ gives ${\alpha }^2|0000⟩+\alpha \beta (|0011⟩+|1100⟩)+{\beta }^2|1111⟩$. Each party applies CNOT $1\to 2$ and measures qubit 2. When both obtain 1, qubits $A_1{B}_1$ are proportional to $|00⟩+|11⟩$. The success probability is $2|\alpha \beta {|}^2$.

After local phases, take $\alpha ,\beta \ge 0$. Alice measures with Kraus operators $M_1=\mathrm{diag}(\alpha ,\beta )$ and $M_2=\mathrm{diag}(\beta ,\alpha )$, which satisfy $M_1^{†}{M}_1+M_2^{†}{M}_2=I$. Outcome 1 directly gives $\alpha |00⟩+\beta |11⟩$. Outcome 2 gives $\beta |00⟩+\alpha |11⟩$; Alice tells Bob the outcome and both apply $X$, swapping the two coefficients. Thus the conversion succeeds with certainty, in agreement with Nielsen majorization.