# If quantum computing requires hundreds of digits of accuracy, how will it be possible?

Leonid Levin said, "Exponential summations used in QC require hundreds if not millions of decimal places accuracy. I wonder who would expect any physical theory to make sense in this realm." See https://groups.google.com/forum/m/#!msg/sci.physics.research/GE5cz3xefCc/e0eh34MZGdwJ

Given that no machine has ever been designed to be sensitive to physical quantities to hundreds of digits of accuracy, how will quantum computing ever be possible in the real world?

To explain what I mean, in a QC model, the state vector has exponential size dimension in which the squares of the entries add to one. So if all of the entries are equal, when they are rounded to say the billionth digit, they will all be zero on a 100 qubit machine, contradicting the fact that they all must add to one. This is a big problem.

EDIT: I think the question is this: How can we possibly perform sensible measurements on quantum computers, given their extreme sensitivity?

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Playing devil's advocate on your wording: I am 100% certain that I have made a code that is accurate to 302 digits (strict Fortran 90), easily extensible to thousands of digits of accuracy. –  Kyle Kanos Mar 30 '14 at 1:23
But those 302 digits are merely the contents of thousands of memory locations on the computer. These digits don't represent anything physical. –  Craig Feinstein Mar 30 '14 at 1:32
Oh, I see. By physical machine you actually mean a measuring device like a ruler, scale, thermometer, etc, and not a computer (which is how I interpreted it). –  Kyle Kanos Mar 30 '14 at 1:50
TL;DR, give me (and other viewers) the short version. –  Kyle Kanos Mar 30 '14 at 1:58
The question should be at least reasonably self-contained. Give us the abstracted version. –  dmckee Mar 30 '14 at 2:06

If you believe the fault-tolerant threshold theorem for quantum computers, you do not require hundreds of digits of accuracy.

Levin does not believe this theorem. More precisely, he believes that the hypotheses required for the theorem to work do not apply to the actual universe.

I believe his mental model of quantum mechanics resembles the idea that the physics of the universe is being simulated on a classical machine which has floating point errors. I don't believe this is true.

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Ok, now I see the point of disagreement. –  Craig Feinstein Mar 30 '14 at 15:07
But what about the fact that QM is not exactly correct and has to be corrected by QED, which is only known to be only valid to 10 or so decimals? –  Craig Feinstein Mar 30 '14 at 18:12
The real question is whether the rules of the universe are exact unitary evolution or something else. If they're exact unitary evolution and you have locality of action (quantum field theories, including QED, satisfy these) then the fault-tolerant threshold theorem holds. If the universe has extra levels of weirdness under the quantum field theory, then it's not clear the hypotheses are satisfied. –  Peter Shor Mar 30 '14 at 19:09
Where is the evidence that the unitary evolution is exactly unitary? The experiments that have been done to confirm QM could be interpretted as having satisfied approximate unitary evolution (to say 10 decimal places) as well as exact unitary evolution. –  Craig Feinstein Mar 30 '14 at 20:22
Beyond a certain number of decimal places, we don't know whether the evolution is unitary or not. If it's not, the hypotheses of our fault-tolerant threshold theorems are certainly not satisfied. But this doesn't mean the fault-tolerance protocols wouldn't work. I don't believe anybody has a concrete proposal for a kind of non-unitarity which would have to cause quantum computing to fail, and not cause any other observable changes in physics. –  Peter Shor Mar 31 '14 at 21:18

Craig, I think you're confusing two important things. First, your original question was something along these lines: Given the fact that we can expand a state ket $|\psi\rangle$ in term of basis kets $$|\psi\rangle = \sum_n c_n |\psi_n \rangle$$ and that there can be infinitely many $n$, let's consider a state which has equal probability to be in any of these (infinitely many) basis states. The question is, what happens when you measure this state? In QM, when you measure, you're supposed to get one of the basis states as the outcome for your measurement.

Now here comes your point: You presume that, if the chance to be in any of the basis states is tiny enough, if you measure some sort of rounding error (?) will cause you to always NOT measure it in that basis state. You then reason that this will happen for each and every basis state, so that you can never measure $|\psi\rangle$ in any basis state, and we have a contradiction with the original statement that $|\psi\rangle$ could be expanded like we did.

Fortunately for quantum mechanics, your reasoning is flawed. If you measure a state, YOU WILL ALWAYS GET SOME OUTCOME. There is no such thing as a rounding error in that sense in nature, this would contradict all sorts of continuity theorems and the likes.

Now, there is a more interesting, and relevant, question hidden in here. The fact is that quantum computers are extremely sensitive, and making controlled measurements without disturbing the system too much to ruin the computation is a legitimate problem in quantum computation. I added this to your question in an edit, for I think it's a question worthwhile posing. I hope this clears up the confusion for everyone involved.

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In your edit, you got what I was saying correctly. My point was that QC is extremely sensitive, so how will it work in the real world? –  Craig Feinstein Mar 30 '14 at 15:04

Actually, it's possible, and all thanks to the use of more powerful processors, but not common processors, that kind of technology requires cold (or less heat) like about a few nanodegrees over 0 Kelvin. With the cold the particles are not so thick, so the electrons can flow easily trough the computer. Look for a video called NOVA MAKING STUFF COLDER Also the size matters, if it's small is faster. LOOK TOO FOR NOVA MAKING STUFF SMALLER

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While this link may answer the question, it is better to include the essential parts of the answer here and provide the link for reference. Link-only answers can become invalid if the linked page changes. –  Neuneck Mar 30 '14 at 11:25