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No-cloning and uncertainty principle

  1. Apr 13, 2014 #1
    In an explanation to distinguish the Heisenberg Uncertainty Principle from the Observer Effect, on p. 89 of "Quantum Computation and Quantum Information", Nielsen and Chuang start by writing:
    "The correct interpretation of the uncertainty principle is that if we prepare a large number of quantum systems in identical states....."
    However, according to the no-cloning theorem, we cannot perfectly copy an arbitrarily chosen unknown quantum state. So, do the authors mean "...if we were able to prepare (although in reality we cannot) a large number...." or "....if it happened that there were a large number of quantum systems in identical states and we knew that without measurement (which would collapse the states)...." Seems almost metaphysical to me, but maybe I am misinterpreting what Nielsen and Chuang meant, or maybe I am misinterpreting the no-cloning theorem, or maybe I am misapplying.
    Any help would be appreciated.
     
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  3. Apr 13, 2014 #2

    mfb

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    Sure, but we can perfectly copy a known quantum state, and apply the same transformations to all systems afterwards. That's all you need.
     
  4. Apr 13, 2014 #3
    Thanks, mfb. That makes sense. To get to a known quantum state without measurement, I presume we are talking about states that are known because we have produced them, e.g. by sending them through a polarized filter or some such.
     
  5. Apr 13, 2014 #4

    mfb

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    That is a good example, right.
     
  6. Apr 13, 2014 #5

    Fredrik

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    The state is the mathematical thing that represents the real-world preparation procedure. The state is uniquely determined by the preparation procedure. If you subject many identical systems to identical preparation procedures, their states will be identical.
     
  7. Apr 13, 2014 #6
    Thanks, Fredrik.

    That one can thus prepare identical states has become clear, and so gives a sense to the original quote I was asking about; my question was that I was originally (and, as it turns out, unnecessarily) thinking about states of particles which appear in nature which were not prepared by humans, and hence with an unknown state. So, even though my original question was answered by the idea that we are dealing with known states, I nonetheless wonder whether you include these unknown states in the "real-world preparation procedure".
     
  8. Apr 13, 2014 #7

    Fredrik

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    Yes, I would say that doing nothing (or doing nothing other than to separate a specimen from its environment) is a preparation procedure.

    It's tempting to think of physical systems as always being in a pure state, so that we only need to use mixed states when we don't know which pure state we're dealing with. But to me this looks like an additional assumption that doesn't change any of the theory's predictions. So if you're thinking of particles as always being in some state (prepared by the environment), you should probably think in terms of state operators (density matrices) rather than state vectors.
     
  9. Apr 13, 2014 #8
    Thanks again, Fredrik. Makes sense.
     
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