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In quantum error correction, we start with the fact that no-cloning theorem doesn't allow cloning of unknown states, hence we need to come up with other strategies.

But often we know the circuit that produces the state that we want to protect, so can't we just prepare the same state multiple times to in-effect clone it?

I realize that the circuits used for preparation would themselves incur error. But wouldn't there be a way to go around it by using more copies?

Is there a precise way to see that this pursuit is completely useless - perhaps one could show that there is no way to do this without using an exponential number of copies?

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The problem with this approach is that there's no mechanism to pump noise out of the system, so noise just accumulates until everything is broken. You need some kind of repeatable crosscheck that can detect noise, and inform you of what to do to remove it.

Consider that, without a way to pump noise out, even having a trillion copies barely extends the lifetime of the system. If the expected time for one system to die is T, then the expected time until the last good copy out of a trillion dies is less than 30T. Paying a trillion times more resources for 30 times more fidelity is clearly not scaling well. You're paying exponential cost to get linear improvement. With normal error correction things are the other way around: pay polynomial cost to get exponential benefit. For example, using the surface code, paying a thousand times more resources can give a trillion times more lifetime.

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  • $\begingroup$ thank you for the nice answer. :) Just one question - how did you arrive at the 30T estimate? $\endgroup$
    – user1752323
    Sep 1, 2022 at 18:32
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    $\begingroup$ @user1752323 Honestly I just wildly assumed it'd be $\ln(n)$ times bigger because of pattern matching to other problems like coupon collector. I did a quick simulation using numpy and it seems to be right. It's maybe much clearer using an argument based on half lives, where it's 40 times longer because after 40 half lives you've got 1 in $2^{40} \approx$ 1 in a trillion of the copies left. $\endgroup$
    – Craig Gidney
    Sep 1, 2022 at 19:49

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