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B Observing a Wave function

  1. Mar 7, 2017 #1
    What exactly do you mean by observing a state/ collapsing wave function. What is observing? Is it seeing the particle? Hearing?
    Also how cautious do you have to be near a quantum computer so that you don't collapse its wave function?
     
  2. jcsd
  3. Mar 7, 2017 #2

    Strilanc

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    You don't have to be any more cautious around a quantum computer than around any other lab equipment. As long as you don't trip into the dilution refrigerator and pull all of the wires out in some kind of zany fall, you'll be fine.

    The definition you should have in mind for "the state collapsed" is "thermodynamically irreversible effect that depended on the state of the system". It has nothing at all to do with humans.
     
  4. Mar 8, 2017 #3
    Observing must be equivalent to applying some operator to the wave function which changes the state. Is there a measurement operator?
     
  5. Mar 8, 2017 #4
    Hermitian operator.
     
  6. Mar 8, 2017 #5

    bhobba

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    Collapsing is not part of the QM formalism - only of some interpretations. In those interpretations its the instantaneous change in state after an observation. The usual interpretation that has it is Copenhagen where the instantaneous thing isn't worried about because it's simply a state of knowledge that appears in a theorists calculations - its like the Bayesian view of probability - who cares if your knowledge of something suddenly changes.

    An observation is trickier. That's because at the beginner and intermediate level its sort of left up in the air as some kind of 'mark' or something that happens here in the common sense classical world. That is a bit of a blemish because QM is supposed to explain that classical world - how can it explain whats assumed in the first place.

    In more advanced treatments an observation is a purely quantum process that happens once decoherence occurs without detailing what that is. That way the issue is resolved, but, again without going into details, some problems still remain.

    Thanks
    Bill
     
  7. Mar 11, 2017 #6
    This is a common misconception. You do not apply a Hermitian operator that corresponds to a measurement to the state to get the state after the measurement. That will be correct (except the normalization) only if the state is an eigenstate of the operator (so it is unchanged). The correct operator to apply, assuming that repeating the measurement immediately gives the same outcome, is the projection operator corresponding to the observed outcome.
     
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