Past exam of the computer science course of the University of Oxford / 2021 / Quantum information / 1

In an orthonormal computational basis, the squared norm is the sum of the squared moduli of the coefficients:Thus the vector is normalized and represents a pure state of a qubit. The imaginary part contributes positively to the squared modulus; squaring the coefficient without complex conjugation would give the wrong norm.

For the Hadamard basis, $|\pm ⟩=(|0⟩\pm |1⟩)/\sqrt{2}$. The Born rule givesWriting the two computational basis amplitudes as $a,b$, normalization gives $\mathrm{Pr}(\pm )=\frac{1}{2}\pm \mathrm{Re}(a^{\ast }b)$, and $\mathrm{Re}(a^{\ast }b)=\sqrt{3}/8$. HenceThese probabilities sum to one. The cross term is interference between probability amplitudes, so the probabilities cannot be obtained by treating the input as an incoherent mixture of the computational basis alternatives.

No. Perfect discrimination of pure states requires orthogonality, whereasTo see the obstruction directly for a basic measurement, let $\{|b_j⟩\}$ be its orthonormal basis. Certain identification would require disjoint sets of possible outcome labels for two different preparations. For every $j$, at least one of $⟨b_j|{\varphi }_0⟩$ and $⟨b_j|{\varphi }_1⟩$ would therefore vanish. Completeness would then givecontradicting the computed overlap. Failure to distinguish just this pair already rules out certain identification of the three-state family.

Identification is stronger than quantum state exclusion. Here an outcome only needs to rule out its associated preparation. Use the following orthonormal basis for a basic measurementIts defining overlaps satisfy $⟨b_m|{\varphi }_m⟩=0$ for $m=0,1,2$. By the Born rule, observing $b_m$ makes preparation ${\varphi }_m$ impossible. Measure in this basis and send its outcome label $m$; the prize is guaranteed.

For completeness, the conditional probability matrix, with rows $m$ and columns the prepared-state label $j$, isEvery column sums to one and the diagonal vanishes. Each result leaves two possible preparations, so this projective state exclusion does not contradict the impossibility of perfect identification. The family consists of antidistinguishable quantum states.

Prepare Bob in $|0{⟩}_B$ and encode the bit using Alice's preparation $|m{⟩}_A$. The controlled-NOT gate givesBob measures in the computational basis and reads $m$ with probability one. Alice's final measurement is unnecessary. The interaction transmits a classical bit; it does not clone an arbitrary unknown qubit.

Reverse communication is also possible. Prepare Alice in $|+{⟩}_A$. Bob encodes $m=0$ as $|+{⟩}_B$ and $m=1$ as $|-{⟩}_B$. Since $X|\pm ⟩=\pm |\pm ⟩$, phase kickback giveswhere $|(-{)}^0⟩=|+⟩$ and $|(-{)}^1⟩=|-⟩$. Alice measures in the Hadamard basis: $+$ means zero and $-$ means one, deterministically.

The target's Pauli X gate eigenvalue becomes a relative phase on the control. This is the operational content of CNOT control reversal in the Hadamard basis; the computational basis control arrow alone does not determine every possible direction of communication.

Charlie cannot send a bit to Alice with the stated preparation, fixed interaction and final local measurements. Write $V={\mathrm{CNOT}}_{B\to A}$ and $W={\mathrm{CNOT}}_{B\to C}$, so the complete gate is $U=WV$ with the rightmost gate acting first. Alice has finished interacting before the gate involving Charlie acts.

For arbitrary product input density matrices, put ${\tau }_{AB}=V({\rho }_A\otimes {\rho }_B)V^{†}$. The final reduced density matrix of Alice isThe equality follows because a unitary operator on the traced-out systems leaves the partial trace unchanged. The right-hand side is independent of Charlie's preparation. Therefore every outcome probability of every local measurement by Alice is independent of Charlie's message. Quantum no-signalling applies after Alice's last interaction, even though phase kickback can allow communication between the two qubits of an individual CNOT gate.

Alice cannot send a bit to Charlie either. An argument based only on the apparent chronological chain would be insufficient: Alice interacts with Bob before Bob interacts with Charlie. The extra fact is that these particular gates commute.

In Bob's computational basis the complete gate isBoth controlled-NOT gates preserve Bob's control label, and their target operations act on different systems. Thus $WV=VW$. We may consequently regard Bob–Charlie as the first interaction and Bob–Alice as the last one. The final operation on $AB$ cannot change Charlie's reduced density matrix, and Charlie's earlier state cannot depend on Alice's initially separate preparation.

The target no-signalling in a common-control unitary formula makes the independence explicit:Alice's density matrix does not appear. The formula includes arbitrary complex initial amplitudes and every final measurement on Charlie. This conclusion uses the specified pair of gates; a general intervening operation on Bob could change the communication possibilities.