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B Meaning of the word 'instantaneous'

  1. Jul 2, 2016 #1
    " an electron emits a photon instantaneously" or consider other instantaneous reactions.

    What is meant by instantaneous here?

    Does it mean there is no time lag between the emission? Does it mean that the emission takes place at a speed greater than the speed of light?
     
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  3. Jul 2, 2016 #2

    vanhees71

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    I've no clue. Of course, in nature nothing happens instantaneously, and a photon needs a (however very short time) to form. Where does this statement come from? I hope, it's not from a serious textbook but from a popular-science book. Note that there are almost no good popular-science books on physics. There are some exceptions: Feynman's books like QED, Weinberg, The first three minutes, Ledermann and Teresi, The God Particle.
     
  4. Jul 2, 2016 #3

    Nugatory

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    Questions like this are why we have to be cautious about natural language descriptions of phenomena that are more precisely described mathematically - words like "instantaneous" may not be as precise as the speaker had hoped.

    You haven't provided the source of the quotation so we have no context and can only guess at what was intended. However, there's a fair chance that they were trying to say that we start in a state with no photon and end up in a state with a photon - but that there are no observable in-between states in which the photon is only partly emitted.
     
  5. Jul 2, 2016 #4
    This is from a debate/discussion on the nature of reality on you tube in which 9-10 scientists participated.

    The actual point is 'is there really no time lag when an electron emits a photon.'
     
  6. Jul 3, 2016 #5
    I think people are not viewing this thread (especially the expert ones)

    OK. I ask in a different way.

    What does instantaneous mean?

    1. No time lag.

    2. A very-very-very small time lag.

    This would be easy, hopefully...
     
  7. Jul 3, 2016 #6

    vanhees71

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    Again, experts are not too interested in pseudo-science. You also have not given your source, where this at best inaccurate statement comes from. It starts with the fact that for itself a single electron cannot emit a photon. You either need it to scatter, leading to bremsstrahlung, or it's bound in an atom and changes from an excited energy level within the atom to a lower-lying energy level (either by stimulated or spontaneous emission).

    As I already stated according to quantum theory there is no instanteneous and also no jumps. However, within the here applicable quantum field theory, it is impossible to interpret the transient states during the time evolution where interactions are relevant, in a particle-like fashion. All that's possible to calculate from QFT (in this case QED) are S-matrix elements, which describe the transition-probability rates from an asymptotic free initial (here an electron plus some other particle it scatters from or an electron bound within an atom) to another asymptotic free final state (here an electron + other particles + a photon, or an electron bound in another lower-energy atomic state + a photon).
     
  8. Jul 3, 2016 #7

    ftr

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    What about EPR?
     
  9. Jul 3, 2016 #8

    vanhees71

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    By construction of relativistic QFT there are no instantaneous interactions at a distance either. This has been discussed countless times in this forum.
     
  10. Jul 3, 2016 #9

    ftr

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    You seem to have made a sweeping statement. EPR is considered to be an instantaneous phenomenon(not interaction).
     
  11. Jul 3, 2016 #10

    Paul Colby

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    One could construct a classical version of EPR using two counter rotating gyroscopes. A small explosive device separates the two. Sometime later you measure one gyroscope and, as if by magic, in an instant, you predict the direction of the unmeasured one. Clearly, no non-local interaction is needed to account for this astounding fact. What sticks in peoples craw is the quantum nature of the observables not the correlation.

    Now, in regard to the photon emission, how exactly is the time of emission to be known?
     
  12. Jul 3, 2016 #11

    ftr

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    The explanation you allude to is not a universal one(a minority). As for emission, that is exactly the point, the formalism imply it is instantaneous.
     
  13. Jul 3, 2016 #12

    Paul Colby

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    Which formalism is that? Last I checked the formalism provides one with an absorption or detection rate. How is this a statement about emission time?
     
  14. Jul 3, 2016 #13

    ftr

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    Because there is no statement, that implies it is instantaneous.
     
  15. Jul 3, 2016 #14

    ftr

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    Moreover, I don't know why you find it strange since with superposition an electron has undefined state before measurement. That is even more stronger than "instantaneous".
     
  16. Jul 3, 2016 #15

    Paul Colby

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    Not following your logic here. My point, in the limit one finds this interesting, is how exactly is one to frame this question from an experimental or observational point of view? It's unclear to me one may even define the emission time for an individual decay.
     
  17. Jul 3, 2016 #16

    ftr

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    The emission did happen, right?
     
  18. Jul 3, 2016 #17

    Paul Colby

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    One may measure with a finite accuracy or time interval a time of detection. One may then infer again with some finite time interval the emission time based on the distance to the emitter. Neither of these may be confused with instantaneous. Nor do either of these numbers relate directly to the time it took to emit. For this one needs the time of excitation. How do you propose to determine that, and to what accuracy?
     
  19. Jul 3, 2016 #18

    ftr

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    The first quantization of interaction of light and matter does not go into mechanism. I think we are going in a circle.
     
  20. Jul 3, 2016 #19

    Paul Colby

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    I was trained as an experimentalist. If you think this is a circle that's likely because you frame questions based on an incomplete view of the theory. What is or is not an observable is a non-trivial question. Not everything in field theory is observable. For example the "blue" component of the quark field can't be observed directly because this would violate color symmetry. So, are you asking a question about something that is in principle observable? Framing even an idealized experiment helps give one insight into this type of question.

    Okay, so what does the second quantization treatment tell us? Why limit your discussion to a theory known to be incomplete.
     
  21. Jul 3, 2016 #20

    ftr

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    Last edited by a moderator: May 8, 2017
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