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Two simultaneous measures of complementary properties

  1. Mar 26, 2012 #1

    dpa

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    hi all,
    what happens if we employ two devices such that one measures position and another measures momentum but that they measure at the same time.
     
  2. jcsd
  3. Mar 26, 2012 #2

    DrChinese

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    Are these on the same particle or on different particles?

    At any rate it doesn't matter. The Heisenberg Uncertainty Principle (HUP) applies, and the results will be consistent with that. If you measure a particles q, measure its p, and finally its q again, you will find that q1 is different than q2. Each observation places the measured property into an eigenstate, and its complementary property(-ies) will be placed in a completely uncertain (read random) state.
     
  4. Mar 26, 2012 #3

    dpa

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    yes! I know that but i am not asking first this then that. I mean simultaneously.
     
  5. Mar 26, 2012 #4

    dpa

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    yes! I know that but i am not asking first this then that. I mean simultaneously.
    And thats for singular particle.
     
  6. Mar 26, 2012 #5

    Cthugha

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    To paraphrase what DrChinese said: It is completely impossible to perform simultaneous measurements of two complementary properties. You can either perform measurement A and then measurement B or the other way round, but a consequence of qm and the uncertainty principle is that there are no simultaneous measurements of such kind.
     
  7. Mar 26, 2012 #6

    Demystifier

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    It is not really that you can't try to measure both. Indeed, if you have an apparatus that measures p and another apparatus that measures x, nobody can stop you to turn on both apparatuses at the same time. And if you do that, the apparatuses will certainly show some values. But can the result of such a procedure be interpreted as a reliable measurement of x, p, or both? The only way to find out is to repeat the measurement of x, p, or both at a slightly later time. And if you do that, the new value of x, p, or both will typically be TOTALLY DIFFERENT from the first value. Quantum mechanically, that means that your procedure did not really collapse the wave function to an eigenstate of either x or p. Practically, that means that you cannot really interpret the result of such a procedure as a reliable measurement of either x or p.

    By the way, if you measure ONLY x, or ONLY p, and repeat the measurement at a slightly later time, you will obtain the same value. Therefore, you can reliably measure only x or only p. But not both at once.
     
  8. Mar 26, 2012 #7

    DrChinese

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    To expand on what Demystifier is saying:

    It is possible to have a particles that are effectively clones of each other (because they are entangled). A measure of p on one allows you to know accurately the p of the other. Or a measure of q on one allows you to know the q of the other.

    And so you can measure p on one and q on the other. So by inference, it would seem as if you now know p and q for both, in violation of the HUP. But as Demystifier points out, you have a simultaneous value for both but this doesn't really mean anything. Each will show a different value or p or q than predicted in a subsequent measurement.

    In other words, you do not have a particle with well defined values of p and q when you are done. Which is just a restatement of the HUP.
     
  9. Mar 26, 2012 #8

    Cthugha

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    Indeed I should have posed that more carefully. I just wanted to emphasize that 'switching some detectors on and getting some values' does not really give you what one would naively think about as 'performing a simultaneous measurement'. I hope we all agree about that.
     
  10. Mar 27, 2012 #9

    Demystifier

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    One additional note. As I said, if you try to measure both x and p, the state will collapse neither to an x-eigenstate nor to a p-eigenstate. However, it does not mean that it will not collapse to anything interesting. Under certain optimal conditions, it can be shown that it will collapse into a COHERENT STATE, characterized by certain average values of x and p, and a minimal uncertainty product Delta x Delta p = hbar/2. So by attempting to measure both x and p you actually prepare coherent states, which may be very useful in practice.

    Indeed, coherent states are known to be the closest quantum analogue of classical states with both p and x being well defined. Quantum coherent states are often denoted as |x,p>, where x and p are average values of position and momentum. They have a Gaussian distribution of both position and momentum around their average values.
     
    Last edited: Mar 27, 2012
  11. Mar 27, 2012 #10

    Cthugha

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    Does this work for an arbitrary initial state or are there some prerequisites?
     
  12. Mar 27, 2012 #11

    Demystifier

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    It works for arbitrary initial state. But of course, you cannot predict into WHICH coherent state it will collapse. As usual, the collapse is effectively random, with probabilities depending on the initial state.
     
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