Qubits 1 and 2 are prepared in the Bell state (Chapter 4)

PHYSICS · Multi-part · Fermi University Physics · Quantum Information & Technology · Quantum states, measurement and entanglement · The quantum circuit model and quantum algorithms

Quantum Information and Technology, Ch The quantum circuit model

Problem

Qubits 1 and 2 are prepared in the Bell state ∣Φ+⟩12=12(∣00⟩+∣11⟩)|\Phi^+\rangle_{12} = \frac{1}{\sqrt2}(|00\rangle + |11\rangle) (Chapter 4). An ancilla qubit 3 is prepared in ∣+⟩|+\rangle (Chapter 1). Recall that Rn^(θ)=e−iθn^⋅σ⃗/2R_{\hat n}(\theta) = e^{-i\theta\hat n\cdot\vec\sigma/2} for any axis n^\hat n, so Rn^(2π)=−IR_{\hat n}(2\pi) = -I on a single qubit. The gate C(Rz^(2π))3→1C(R_{\hat z}(2\pi))_{3\to1}, with qubit 3 as control and qubit 1 as target, is applied. ((a)) Write the 4×44\times4 matrix of C(Rz^(2π))C(R_{\hat z}(2\pi)), control first, qubit 3, and target second, qubit 1, in the computational basis ∣00⟩,∣01⟩,∣10⟩,∣11⟩|00\rangle,|01\rangle,|10\rangle,|11\rangle. ((b)) Because Rz^(2π)=−IR_{\hat z}(2\pi)=-I, every state of qubit 1, indeed every state of the pair of qubits 1 and 2 treating that pair as the target, is an eigenstate of it with eigenvalue −1-1, whether or not qubit 1 is entangled with qubit 2. Use phase kickback to find the exact state of the ancilla and the Bell pair after C(Rz^(2π))3→1C(R_{\hat z}(2\pi))_{3\to1} is applied to ∣+⟩3⊗∣Φ+⟩12|+\rangle_3\otimes|\Phi^+\rangle_{12}, and state which Bloch-sphere fact from Chapter 1 this result makes observable. ((c)) The protocol now measures qubit 3 in the XX basis, an outcome certain from part (b), and uses the classical outcome to decide whether to apply a ZZ gate to qubit 2 before Bob measures qubit 2 in the computational basis. Use the deferred measurement principle to redraw this as a circuit in which qubit 3 is measured only at the end, and explain why Bob's measurement statistics on qubit 2 are the same either way.

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