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A States in relativistic quantum field theory

  1. Sep 12, 2016 #1

    A. Neumaier

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    No. This is a noncovariant, observer-specific view.


    In the covariant, observer-independent view of fields, states are labeled instead by the causal classical solutions of hyperbolic field equations. On the collection of these the Peierls bracket is defined, which is the covariant version of the Poisson bracket. Each observer picks out a particular frame and with it at each time a Cauchy surface that intersects a causal classical solution exactly once - giving the instantaneous field labels of the observer's state satisfying a functional Schroedinger equation that reduces after space discretization to the nonrelativistic treatments.


    Thus the same covariant state appears different to each observer, just as in classical relativistic physics the same covariant length appears different to different observers.


    In classical relativistic physics, one can directly compare only events modeled by a single observer; models of different observers have no connection unless they agree on the information encoded in it and use the rules of relativity to translate their models into mathematically equivalent things. The same holds even more so in quantum relativistic physics.


    The collapse is a sudden change of the model used by an observer to reinterpret the situation when new information comes in, hence depends on when and whether the observer cares to take notice of a physical fact. This clearly cannot affect other observers and their models of the physical situation. Hence there is no nonlocal effect. Nonlocal correlations appear only when a single observer compares records of other (distant) observer's measurements. At that time the past light cone of this observer contains all the previously nonlocal information, so that locality is again not violated
     
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  3. Sep 12, 2016 #2

    zonde

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    Realism requires that observations of different observers can be unified into single consistent system. If it's impossible to form that consistent system without nonlocal effects then they are real.
     
  4. Sep 12, 2016 #3

    A. Neumaier

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    Each observer sees a consistent picture given its information. The final observer sees the complete picture with his complete information. Thus there is nothing inconsistent in the whole setting although there are nonlocal correlations without nonlocal action.
     
  5. Sep 12, 2016 #4

    atyy

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    For some interpretation of all your words - I agree with all your words except one. The only word I don't agree with is the first word, which should be "Yes" instead of "No".
     
    Last edited: Sep 12, 2016
  6. Sep 13, 2016 #5

    Demystifier

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    If only @vanhees71 could accept that definition of the collapse. :smile:
     
  7. Sep 13, 2016 #6

    vanhees71

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    But I do accept this view. That's what I've been emphasizing the whole time! Only in #4 again it is very clear that atyy doesn't accept it, because of course if you accept it, then the first word in #1 indeed must be clearly "no"!
     
  8. Sep 13, 2016 #7

    Demystifier

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    I see. Then the problem must be with @atyy who, at least in some posts, seems to suggest that collapse could be something more than a mere update.
     
  9. Sep 13, 2016 #8

    vanhees71

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    Yes!
     
  10. Sep 13, 2016 #9

    atyy

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    If the observer pretends that the collapse is nonlocal, he makes no mistake.

    Hence the answer is yes - reality is a tool to calculate the results of experiments.

    BTW, in general, I am agnostic to whether collapse is only an update - I leave open the option that collapse is something more than an update. Similarly, although I don't know whether the reality in which the wave function collapses is real, I leave open the possibility that it is. Incidentally, Cohen-Tannoudji, Diu and Laloe are careful about this point in their text - they say that collapse is an update - but they do not say that collapse is only an update.
     
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