A qubit is prepared in the state and used as the target register of phase estimation…

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

Quantum Information and Technology, Ch The quantum Fourier transform and phase estimation

Problem

A qubit is prepared in the state ∣ψ⟩=cos⁡(π/6)∣0⟩+sin⁡(π/6)∣1⟩|\psi\rangle = \cos(\pi/6)|0\rangle + \sin(\pi/6)|1\rangle and used as the target register of phase estimation for U=ZU = Z with t=2t = 2 control qubits. Z∣0⟩=∣0⟩Z|0\rangle = |0\rangle (eigenphase φ=0\varphi = 0) and Z∣1⟩=−∣1⟩Z|1\rangle = -|1\rangle (eigenphase φ=1/2\varphi = 1/2). Someone argues: "since the eigenphases are 00 and 1/21/2 with weights cos⁡2(π/6)=0.75\cos^2(\pi/6) = 0.75 and sin⁡2(π/6)=0.25\sin^2(\pi/6) = 0.25, phase estimation must output the weighted-average phase 0.75×0+0.25×0.5=0.1250.75\times0 + 0.25\times0.5 = 0.125, giving a single sharp peak at register reading 0.125×4=0.50.125\times4 = 0.5, which rounds to 0101." ((a)) Find the error in this argument. ((b)) Give the correct output distribution over the four possible two-bit register readings, with their probabilities.

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