Past exam of the computer science course of the University of Oxford / 2021 / Quantum information / 3

Prepare $|0⟩$, then apply $H$, then $T$. With $T=\mathrm{diag}(1,e^{i\pi /4})$,This uses only the allowed state preparation and elementary gates. The rightmost operator acts first, so $TH$ denotes the temporal order $H$ followed by $T$.

The required superposition is the Bell statePrepare $|00⟩$, apply $H$ to the first qubit, then apply CNOT from the first to the second. The intermediate state is $(|00⟩+|10⟩)/\sqrt{2}$, and the controlled-NOT gate gives the required Bell state. No final basis change is needed: the two displayed expressions represent exactly the same vector.

Write ${\rho }_0=|\varphi ⟩⟨\varphi |=(I+r_{x}X+r_{y}Y+r_{z}Z)/2$. Conjugating by a Pauli matrix keeps its own Bloch component and reverses the other two; for example,In the average of the four conjugations, the coefficient of every nonidentity Pauli matrix is zero. ConsequentlyThe Pauli operator basis gives the Bloch representation of a qubit density operator.^[ref]

This is the qubit Pauli twirl, the dimension-two case of the Heisenberg-Weyl twirling channel. The argument uses only trace one, so it also works for mixed qubit inputs. The preparation label must be forgotten: conditioning on the label would retain a known pure branch rather than the maximally mixed state obtained by the ensemble average.

Take the trine state ensembleAll three states are normalized. Since $1+\omega +{\omega }^2=0$, their average density matrix isGeometrically, their unit Bloch vectors form an equilateral triangle on the equator and sum to zero. Orthogonality is unnecessary for an ensemble decomposition of a mixed state: three distinct, nonorthogonal pure states can give the same maximally mixed average as two orthogonal states.

No test on Alice and Bob's systems gives any information about which global state was prepared, even with arbitrarily many independent copies. In the pure GHZ density matrix, the off-diagonal terms contain $|0⟩⟨1{|}_C$ or its adjoint. Their partial traces vanish. Hence both preparations have the same reduced density matrixFor any POVM effect $E$ acting jointly on $AB$, the outcome probability is $\mathrm{Tr}(E{\sigma }_{AB})$ in either case. Equal reduced states give identical local statistics, where local here means local to the entire accessible laboratory $AB$, not separate operations on $A$ and $B$.^[ref]

For $n$ independent copies, taking the partial trace of each inaccessible $C$ system yields ${\sigma }_{AB}^{\otimes n}$ under either hypothesis. Even an arbitrary collective quantum measurement on all $2n$ accessible qubits therefore has identical outcome probabilities. Adaptive local processing cannot change equality of the accessible density matrices. With equal prior probabilities the optimal guessing probability remains $1/2$; with other priors, no observation updates their ratio.

Charlie's Hadamard basis measurement, Alice's Bell-basis measurement and communicated correction bits give deterministic GHZ-assisted quantum teleportation. Let the unknown input be $|\chi {⟩}_Q=\alpha |0⟩+\beta |1⟩$. Charlie measures $C$ in the Hadamard basis and labels $+$ by $c=0$, $-$ by $c=1$. The identityshows that $AB$ is conditionally a Bell pair with phase label $c$. Charlie sends that one classical bit to Bob. The procedure for localizing a Bell pair from a GHZ state supplies precisely the missing conditional information; forgetting $c$ leaves the mixed reduced state of the preceding part.

Alice performs the usual Bell-basis measurement on $QA$. Operationally, she applies ${\mathrm{CNOT}}_{Q\to A}$, applies $H_Q$, and measures $Q$ and $A$ in the computational basis, obtaining bits $s,t$. She sends both bits to Bob. To track the correction without a convention ambiguity, conditional on $s,t$ the normalized state on $BC$ isProjecting Charlie's factor onto his recorded Hadamard basis state gives Bob, up to the global phase $(-1{)}^{ct}$,Consequently the inverse Pauli correction isBob applies $X^t$ first and $Z^{s\oplus c}$ second. The measurements on Charlie and on Alice commute because they act on different systems, so deriving the conditional state with Alice's measurement first gives the same protocol as Charlie measuring first. Each triple $(s,t,c)$ has probability $1/8$, independent of the unknown input, and every triple succeeds after correction.

This uses one previously shared GHZ state, two classical bits from Alice and one from Charlie, with no quantum transmission during the transfer. The standard Bell-measurement/Pauli-correction mechanism was established in the original quantum teleportation proposal.^[ref] Here Charlie's record supplies the Bell-pair phase, giving a controlled quantum teleportation protocol. Alice's input is consumed by her measurement; Bob reconstructs its state without learning $\alpha ,\beta$ and without making a second surviving copy.