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Assuming the correctness of QM: Would the size of such a computer be smaller than the observable universe? If it were to represent all available information in the universe it seems that it's minimal necessary size would relate to a black hole of a certain mass. Would this lead us to believe this theoretical quantum computer (assuming it didn't represent itself) could be constructed smaller than the universe? Or if the information content of an object is truly expressed in relation to a surface area is the universe the object with the minimal surface required to represent the totality of it's information?

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The way you phrase your question is vague, and on points incorrect. 1) what makes you think the universe is less dense than a black hole? 2) if the universe is computationally irreducible, there is no shortcut and no prediction about exact future states is possible. In other words: it takes the full universe to 'play things out'. –  Johannes Feb 27 '11 at 18:26
    
@Johannes I will edit 1). But 2) sounds like an answer- although it seems that if the universe evolved from a smaller size then it is not computationally irreducible. On the vagueness? I don't know- any suggestions? –  jaskey13 Feb 27 '11 at 19:22
    
it will be difficult to render your question unambiguous. Firstly, if you assume QM to be valid, you can't introduce black holes (we have no consistent QM theory for black holes). Secondly, you need to specify out of which components you would build such a quantum computer. Thirdly, you can't assume the quantum computer "not to represent itself": if build it would be part of the universe. You can interpret 2) above as an answer. But it would logically not be possible to achieve what you are asking for (see e.g. Quantum non-xerox principle). –  Johannes Feb 27 '11 at 20:36
    
@Johannes I mention black holes to put a lower-bound on size in space. If this is problematic for you- ignore it- it is not the point. 2nd- this is a theoretical device/question, so as I can not specify how it would be built- I feel that I can still ask about the "computation reducibility/irreducibility" of the universe per your previous comment (which is exactly the content of my question) 3rd.1 is simply not a real hindrance to this question as it follows directly from the assumption of irreducibility which may or may not be the answer. –  jaskey13 Feb 27 '11 at 21:00
    
@Johannes 3rd.2- accurate representation of the early universe and it's subsequent evolution into what we see to today would not violate non-xerox- unless you contend that at each point each unique configuration of microstates of that system matched exactly one macrostate. –  jaskey13 Feb 27 '11 at 21:02
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2 Answers

If you are looking for the exact simulation of the whole universe, the answer is definitely no!

Because of this technical issue, that the computer itself is unfortunately a part of this universe, therefor it has to simulate itself as well alongside the rest of the universe! This means that the computer should produce results of its own calculations in the future (as well as the rest of the universe), which itself contains the whole simulation of the universe and the state of the computer, and again in the state of the computer the same thing is embedded again and again.

The other issue of a computer simulating itself is that it should produce its own results before they are even produced :)

In-fact larger grained simulations are possible(although not 100% accurate) and doable(even nowadays). Or if our universe had some fractal structure, then some smaller parts of it(like our computer) could represent its whole; which is unfortunately not the case.

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Assuming the correctness of QM, you can encode $2^{n}$ complex amplitudes with the computational basis of n entangled qubits. This suggests that the answer to the question is yes, much smaller than the universe in fact.

Unfortunately quantum computing technology isn't there yet to maintain useful amount of entangled qubits, or getting that much information out of the qubit system for that matter.

This is one of those questions that makes studying quantum computation very interesting.

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This doesn't make any sense. A quantum computer with n qubits has 2^n complex amplitudes, but the universe also has a number of quantum states that depends exponentially on the number of things in it. In other words, the universe contains m > n qubits, so it takes 2^m >> 2^n complex amplitudes to fully describe it. –  Keenan Pepper Feb 28 '11 at 6:57
    
The analysis is a bit naive, I agree. How many quantum states does it take to describe a macroscopical variable. What if you devise a quantum algorithm to allocate and compress the information, rather than bruteforcing the states as exponential degrees of freedom. –  Janne808 Feb 28 '11 at 12:45
    
I initially came to a similar conclusion as this answer, but soon ran into the problem noted by @Keenan. Since then I picked up a copy of "Quantum Computation and Quantum Information" by Micheal A. Nielsen and Isaac L. Chung (seems like a good place to start?) and from what I've digested so far: would not 2^m complex amplitudes represent more than the available amount of information as the m qubits are not all in completely mixed and unentangled states? This would allow for compression, as @Janne808 asks about above, with no loss of information. –  jaskey13 Feb 28 '11 at 15:05
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This answer is incorrect and tries to violate the Holevo bound. From an $n$-qubit system, you can at best extract $n$ bits. –  Joe Fitzsimons Sep 13 '11 at 3:29
    
Thanks for the correction. I haven't read up on Holevo's theorem yet. –  Janne808 Dec 30 '11 at 16:36
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