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Time Energy Uncertainty Question

  1. Dec 5, 2014 #1
    I would appreciate if someone could help figure out this thought experiment:

    Lets say I have two detectors named A and B.
    They both want to detect system C.
    For my naming convention I will say that C.B is the perturbed state of C after interacting with B

    Ok so both A and B decide to measure C at around the same time. If A measures C first then C becomes C.A and A becomes A.C. The instant after that, B will still be able to interact with C (as opposed to C.B) because time is an uncertainty, so now what is C? C.A or C.B? And is B now B.C or B.C.A?
  2. jcsd
  3. Dec 5, 2014 #2


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    Science Advisor

    There is no time uncertainty. If you choose your inertial frame, then whoever measures first will collapse the wave function first, and whoever measures second will be measuring on the collapsed wave function. An example is given in http://arxiv.org/abs/1007.3977 where the order of measurements depends on the choice of inertial frame, but the predicted probabilities are frame-independent.
  4. Dec 5, 2014 #3
    if its dependent on choice of frame then how can it be a property of the system?
  5. Dec 6, 2014 #4


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    Staff: Mentor

    The probability distribution of the results of the measurements is independent of the order and hence is a property of the system.
  6. Dec 6, 2014 #5


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    In quantum mechanics, only the measurement outcomes and the probabilities are real, partly because those are invariant events in the sense of classical special relativity. The wave function is not necessarily real, and is a tool for calculating the probabilities of measurement outcomes.
    Last edited: Dec 6, 2014
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