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Quantum Phase Estimation

Anitya Gupta
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⚡ Quantum Brief
A 2025 Q&A thread clarifies the Quantum Phase Estimation (QPE) algorithm implementation using a unitary operator S and eigenstate |1⟩ as the target qubit, with 3 counting qubits and 1 target qubit. The goal is estimating an unknown phase φ ∈ [0,1) with 3-bit precision by measuring the counting qubits, where φ ≈ j/2³ and j is the measured integer output. Confusion arose over the target qubit’s role, which must remain an eigenstate of S throughout the process to ensure accurate phase estimation via controlled-S operations. The 12.5% output probability per state reflects uniform superposition in the 3-qubit system, yielding 8 possible measurement outcomes (2³) for phase approximation. Experts emphasized proper MathJax formatting for equations and noted QPE’s precision scales exponentially with additional counting qubits, highlighting its efficiency for quantum algorithms.
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Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Stack Overflow for Teams is now called Stack Internal. Bring the best of human thought and AI automation together at your work. Bring the best of human thought and AI automation together at your work. Learn more Stack InternalKnowledge at workBring the best of human thought and AI automation together at your work.Implement the Quantum Phase Estimation (QPE) algorithm using the unitary operator S and the eigenstate ∣1⟩ as the target qubit. Use 3 counting qubits and 1 target qubit. Assumptions / notation: Assume the unitary S satisfies S∣1⟩ = exp (2πiφ) ∣1⟩ for some unknown phase φ∈[0,1). Your QPE circuit must estimate φ to 3-bit precision (i.e., output the best 3-bit binary approximation of φ).I am confused with the target Qubit and when I am taking the 3 Qubit then it is having 12.5% Probability for each as an output? Please any suggestion is helpfulTarget qubit is eigenstate of the unitary S. In QPE you can estimate $\theta$ by $\theta=\frac{j}{2^m}$ where $j \in \{0, 1, 2, .., 2^m-1\}$. Here $j$ is a measured value from your counting 3 qubits, so $m=3$. Hence it has 3-bit precision.Thanks for contributing an answer to Quantum Computing Stack Exchange!But avoid …Use MathJax to format equations. MathJax reference.To learn more, see our tips on writing great answers.Required, but never shown By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy. Start asking to get answersFind the answer to your question by asking.Explore related questionsSee similar questions with these tags.To subscribe to this RSS feed, copy and paste this URL into your RSS reader. Site design / logo © 2025 Stack Exchange Inc; user contributions licensed under CC BY-SA . rev 2025.11.27.37399

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