Quantum Fourier transform

jimmycricket
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When using the Quantum Fourier transform to find the period of the function f(x)\equiv a^x\mod N why is it that the input register is 2n qubits in size and the output register is n qubits?
 
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The overall period finding circuit probably uses ancilla bits, but turns them into measured garbage along the way. The Quantum Fourier Transform doesn't need ancilla bits, or measurement, so it's not the cause for adding them.

Personally, I wouldn't count ancilla bits against the input size. The circuit takes n input bits and n ancilla bits, burns the neg-entropy in the ancilla bits, and produces n outputs.
 
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. Towards the end of the first lecture for the Qiskit Global Summer School 2025, Foundations of Quantum Mechanics, Olivia Lanes (Global Lead, Content and Education IBM) stated... Source: https://www.physicsforums.com/insights/quantum-entanglement-is-a-kinematic-fact-not-a-dynamical-effect/ by @RUTA
If we release an electron around a positively charged sphere, the initial state of electron is a linear combination of Hydrogen-like states. According to quantum mechanics, evolution of time would not change this initial state because the potential is time independent. However, classically we expect the electron to collide with the sphere. So, it seems that the quantum and classics predict different behaviours!

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