Ensemble average in quantum computing

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SUMMARY

The discussion centers on the concept of ensemble averages in quantum computing, specifically regarding the measurement process that collapses quantum states. It highlights the use of Nuclear Magnetic Resonance (NMR) to measure billions of spins and questions the minimum number of ensembles required for accurate results in quantum systems. The participants debate whether the number of ensembles needed scales as O(n) or O(2^n) for n qubits, emphasizing that there is no fundamental minimum number of ensembles, drawing parallels to coin toss experiments.

PREREQUISITES
  • Quantum state measurement techniques
  • Nuclear Magnetic Resonance (NMR) in quantum computing
  • Understanding of ensemble averages
  • Basic principles of quantum mechanics
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Quantum computing researchers, physicists, and anyone interested in the measurement processes and accuracy challenges in quantum systems.

morphemera
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In the last step to get result from a quantum computer, measurement is required to collapse the quantum state. This can be done by the measuring a large ensemble average of same computed result. One realization I know is to use NMR to measure billions of spin. So, what is the minimum number of ensembles are required to get a 'good' result?

Furthermore, suppose the number of one qubit requires O(1) ensemble, how many ensembles are required for n qubits system to have same accuracy? I suppose the answer should be O(n), but if it is O(2^n), then the realization of quantum computer still seems useless to me. Any explanation of the answer?
 
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Is there a minimum number as such?

Let me ask you this: What's a good enough number of coin flips to see almost half the flips are heads?

You are talking about an accuracy in your second question, how much accuracy are you aiming for?

My point is there's no fundamental minimum. It's exactly like a coin toss experiment, nothing peculiar to quantum computing here, in this context.
 

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