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Feynman Path Integral's Meaning

  1. Apr 15, 2010 #1

    jaketodd

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    Does the math of the Feynman path integral dictate particles taking many paths or is that just an english language gloss-over?

    Thanks,

    Jake
     
  2. jcsd
  3. Apr 15, 2010 #2
    all paths contribute, we dont know and can not know which path the particle took
     
  4. Apr 15, 2010 #3

    jaketodd

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    But some paths have a greater probability than others, correct?
     
  5. Apr 15, 2010 #4
    yes, each path is weighted with exp(-i/hbar * S ) where S is the classical action

    but since we can't tell which path the particle took, we must add these probabilites coherently
     
  6. Apr 15, 2010 #5
    Consistent Histories provides a way of assigning probabilities to non-interfering sets of paths though. So we can assign probabilities, but not to individual paths.
     
  7. Apr 15, 2010 #6
    Why don't you get the answer from the father of the path integral himself?

    In QED: The Strange Theory of Light and Matter, Feynman explains, in a highly accessible way, what path integrals are and how they relate to QED. You can find that book in just about any library.

    If you want nuts-and-bolts explanations there is no easy way other than learning about quantum field theory.
     
  8. Apr 15, 2010 #7

    jaketodd

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    Why not? Because I think even him dismissed the interpretation that the particle takes infinitely many paths and those paths cancel to probabilities for paths, yet, to my humble knowledge, that is what the math of the path integral states happens.
     
  9. Apr 15, 2010 #8
    In physics, when one has a crazy idea that works, it may be necessary to state that it is not the only way to look at it so people dont simply dismiss it because it seems crazy.

    The essence is that the classical path is a stationary point of the action, i.e. a maximum/minimum/ or saddle point. However in QM, we know that uncertainty is built in and for short times we can have large energy fluctuations and short distances large momentum fluctuations. The leading quantum corrections to the classical physics comes from taking into account these quantum fluctuations on the classical path, hence many paths are now accessible and "feeling" these paths leads to nontrivial quantum physics.
     
  10. Apr 15, 2010 #9
    Does the original "Sum Over All Histories" for a particle make the notion of a PI any clearer?

    @jaketodd: He did, but that doesn't mean it isn't useful or worth learning.
     
  11. Apr 15, 2010 #10
    My understanding is that Feynman did not care much for interpretation anyway. "Shut up and calculate!"
     
  12. Apr 16, 2010 #11
    Incidentally, this is a nice video:
     
    Last edited by a moderator: Sep 25, 2014
  13. Apr 16, 2010 #12
    Yup.
     
  14. Apr 16, 2010 #13
    As first said by Mermin :)
     
  15. Apr 17, 2010 #14

    jaketodd

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    I didn't say learning about the path integral and related things isn't useful or worth learning. I was saying that I think Feynman didn't publicly support the implications of the math that his path integral stated, and therefore he is not the person to get the proper interpretation from.

    Doesn't it all come down to the math in the end? If the math says a particle takes infinitely many paths, then that's what is happening if it's correct regardless of the shrink wrap you put around it. However, a good question would be: "Is there a different way to mathematically reproduce the same results (and therefore the same accuracy) without the infinitely many paths being built into the math?" Or, am I wrong and that interpretation is not built into the math?

    Thanks,

    Jake
     
  16. Apr 17, 2010 #15
    I would say that makes him uniquely qualified, being able to use a tool without believing that it describes a complete physical reality.

    You're right. You're wrong. What you just said is why it's a useful mathematical tool, but NOT a physical description. In other words, the math is just fine, but if it has no physical reality it is still an approximation, or tool. That's not a bad thing, but it doesn't mean that until we can determine what occurs at that scale, our description reflects reality.
     
  17. Apr 17, 2010 #16

    jaketodd

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    Isn't the path integral the best we have in terms of what happens on that scale? Could it be the best we will ever have given the inherent uncertainty in quantum theory? If not, is it not reasonable to think what would be an improvement on it would contain the ideas of the math of the path integral and then the something else?

    Thank you,

    Jake
     
  18. Apr 17, 2010 #17
    The first, I would agree with, but the second... how can we tell? I would say an understanding of the Planck (and sub-Planck) scale would help to eliminate these approximations.
     
  19. Apr 17, 2010 #18

    RUTA

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    “There was a time when the newspapers said that only twelve men understood the theory of relativity. I do not believe there ever was such a time. There might have been a time when only one man did, because he was the only guy who caught on, before he wrote his paper. But after people read the paper a lot of people understood the theory of relativity in some way or other, certainly more than twelve. On the other hand, I think I can safely say that nobody understands quantum mechanics. … Do not keep saying to yourself, ‘But how can it be like that?’ because you will get ‘down the drain,’ into a blind alley from which nobody has yet escaped. Nobody knows how it can be like that.” Richard Feynman, The Age of Entanglement, Lousia Gilder, Vintage Books, New York (2008), pp 228-229. She cites “November 1964…Lectures”: Feynman, Character of Physical Law, chapter 6.
     
  20. Apr 17, 2010 #19

    RUTA

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    The phasor exp(-iS/hbar) is unit length for any S. What S changes for the phasor of each path is its angle, not its length, so the phasors add more coherently about the extremum of S. Of course, the extremum of S gives the classical equations of motion via the least action principle.
     
  21. Apr 17, 2010 #20

    RUTA

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    It does not even require the existence of particles, a fortiori paths. Thus, the answer to the first part of your question is "no."
     
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