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Measurement effect upon wave function

  1. Jan 17, 2010 #1

    nomadreid

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    If I understand correctly, a measurement affects the wave function only at the moment of measurement. In other words, if one were able to compare two wave functions, one when there were measurements, and one in which there were no measurements, the only difference would be at the meager set of measurement points. (I am not starting a debate between MWI, Decoherence, Copenhagen, etc.: for many-world theorists, I am only considering our branch.) Is this correct?
     
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  3. Jan 17, 2010 #2
    If we take quantum mechanics at face value, then there is no way to answer your questions. There is no way to verify that the measurement process has any affect on the wave function.

    The wave function is defined in a linear function space and its explicit form depends on the preparation apparatus as well as the observable being measured. As far as we know, it does not propagate in 3-space and it does not interact with the measuring device. A collapsing wave function is not observable and there is no experimental evidence that such an event has ever occured.

    Although a quantum experiment consists of many parts it must be considered a single entity. (Bohr) The preparation apparatus, the particle, the particle detector and the measurement result are non-separable. The experiment must have a measurement result (Bohr), which completes the experiment; there is no experiment without an experimental result (Wheeler). In this sense, the experiment exists only at the moment the particle is detected.

    There is no wave function for a non-experiment that does not have a result. The wave function allows us to calculate the probability distribution of all possible results. But, if there are no results, how do we write the wave function? If we could, then what would we do with it?
     
  4. Jan 18, 2010 #3

    Matterwave

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    The most basic answer is no, this is not correct.

    If I have prepared 2 particles, and 1 goes "unobserved" while the other is "observed", wave-functions of the 2 particles are different.

    For example, if I have 2 particles confined to an infinite square well, and I know they are both in the first eigenstate, then their wave functions are the same atm (and will remain the same, absent perturbations, since the eigenstates are stationary states). If I measure one to be approximately at x=a/2 (where a is the length of the wall), then the wave function of the particle I just measured will no longer be in the first eigenstate. The particle I just measured will be confined to a small interval around x=a/2.

    The particle I didn't measure, will still be in the first eigenstate. If I measure the particle I DIDN'T measure, I have a probability of finding it according to the probability distribution of the first eigenstate. If I now measure the particle I DID measure (again), then it will still be confined to a small interval around x=a/2 as long as I measure it again fast enough that it hasn't had time to move (the probabilities between the two will not be the same).
     
  5. Jan 18, 2010 #4

    nomadreid

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    Thank you, eaglelake. That makes perfect sense, and I stand rebuked for stating a question with no experimental verifiability. Let me try to amend the question a little with a thought experiment. Suppose I take a number of measurements to determine a local wave function at time t1. Given a passage of time dt, with the system absorbing energy E, I do two things: I predict the probabilities using my proposed wave function, and I take another set of measurements to see if measurement and prediction correspond within a reasonable statistical range. Suppose they do, but I adjust my wave function to take into account the new measurements. I now do this again, after another lapse of time dt and energy E. I now have a new set of measurements. Unbeknownst to me, another experiment with the same initial conditions as my original set-up is being performed, but instead of taking three sets of measurements as I did, the other experimenter is taking only two sets of measurements, the same one that I did at the beginning, and the same one that I did at the end, except the other experiment was performed after waiting 2dt and having an absorption of 2E after the original set of measurements, without taking an intermediary set of measurements as I did. Would one expect the two sets of end measurements to be within statistically appropriate ranges of one another?
    I realize that I could have stated this better, but I hope you get the idea I am trying to get across.
    Thanks.
     
  6. Jan 18, 2010 #5

    nomadreid

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    Thanks, matterwave. You have given me something to contemplate. (I saw your reply after writing the reply to eaglelake.)
     
  7. Jan 18, 2010 #6

    nomadreid

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    To matterwave: a better response (I was in a hurry when I typed the last one): an excellent answer, thank you. The question was prompted by something I have often read that the wave function could be determined at all times in only knowing its initial conditions. But according to your response, this does not seem to be the case. Or am I wrong again? That's why I wrote that your answer gives me something to contemplate.
     
  8. Jan 18, 2010 #7

    Matterwave

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    If you know the initial condition of the wave function (and the potential the particle is in), then you can determined the way it evolves for all time, using the Schrodinger Equation, as long as it is undisturbed. Once you make a measurement, the particle wave function "collapses". This collapse does not happen according to the Schrodinger Equation, and is quite a mystery in QM (this is why certain formulations/interpretations of QM get rid of the wave function collapse).

    Wave-functions evolve in those 2 ways, either according to the SE, or collapsing. The evolution according to the SE is complicated but well studied (theoretically of course, you can't actually observe a wave function evolving). The collapse of a wave function...is not so clear.
     
  9. Jan 19, 2010 #8

    nomadreid

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    To Matterwave: thank you very much.
     
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