Timeline for Does a classical computer really require $2^n$ complex numbers to represent the state of $n$ qubit quantum computer?
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|Apr 18 '20 at 17:45||comment||added||tparker||@QC-Novice I don't know what you mean by "restated a confusion" - confusion is a state of mind, not a statement. Are you claiming that any of the claims in my answer are incorrect? I also don't know what you mean by "specification complexity" - do you mean Kolmogorov complexity? If so, then your claim is wrong: the computational complexity of a quantum state (or more precisely, the problem of producing a quantum state from the all-0 state) must be greater than or equal to its Kolmogorov complexity.|
|Apr 18 '20 at 14:00||comment||added||QC-Novice||Thank you, but I actually feel that you have restated the confusion that I think exists in the literature. There is a tendency in the literature to claim that the need to specify coefficients of every computational basis vector in a quantum computer is what makes classical simulation that reaches this state computationally complex. In a way, I am simply pointing out that computational complexity of a quantum state can be much lower than its specification complexity. My question is in essence this: Why should that be different for a classical simulation?|
|Apr 18 '20 at 1:39||history||answered||tparker||CC BY-SA 4.0|