Two qubit states lie on the equator of the Bloch sphere: (azimuthal angle ) and…

PHYSICS · Multi-part · Fermi University Physics · Quantum Information & Technology · Quantum states, measurement and entanglement · Quantum states, measurement and entanglement

Quantum Information and Technology, Ch Qubits, registers and the Bloch sphere

Problem

Two qubit states lie on the equator of the Bloch sphere: ∣ψ1⟩=∣+⟩|\psi_1\rangle = |+\rangle (azimuthal angle ϕ1=0\phi_1=0) and ∣ψ2⟩=12(∣0⟩+ei2π/3∣1⟩)|\psi_2\rangle = \dfrac{1}{\sqrt2}\big(|0\rangle + e^{i2\pi/3}|1\rangle\big) (azimuthal angle ϕ2=120∘\phi_2 = 120^\circ). ((a)) Write down the Bloch vectors r⃗1\vec r_1 and r⃗2\vec r_2. ((b)) Use ∣⟨ψ1∣ψ2⟩∣2=(1+r⃗1⋅r⃗2)/2|\langle\psi_1|\psi_2\rangle|^2 = (1+\vec r_1\cdot\vec r_2)/2 to find the probability that a measurement testing for ∣ψ1⟩|\psi_1\rangle (i.e. in a basis containing ∣ψ1⟩|\psi_1\rangle) finds the qubit in ∣ψ1⟩|\psi_1\rangle, given it was prepared in ∣ψ2⟩|\psi_2\rangle. ((c)) Check the result by computing ⟨ψ1∣ψ2⟩\langle\psi_1|\psi_2\rangle directly.

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