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I Feynman path integral

  1. Nov 14, 2017 #1
    I'm reading "Teaching Feynman’s sum-over-paths quantum theory" by Taylor et al.
    http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.374.4480&rep=rep1&type=pdf,

    I'd like to confirm whether my understanding is correct, so a couple of questions.

    1. We need to try and think of all kinds of different classical, not-quite-classical, and frankly weird trajectories that are going to contribute to the final "arrow" (propagator?). But have to restrict ourselves to trajectories that complete the trip from A to B in the same time as that of a straight line flight from A to B. -- Is this correct?

    2. Once we select a path, we integrate the difference between KE and PE along that path. How do we calculate this for a photon if we consider a segment where it is flying, say, at 2 x c ?

    3. Consider an interferometer in a laboratory. There is a detector where we see a complete null due to destructive interference. Now on another workbench is a mirror that has nothing do do with the experiment at all. But based on the nature of diffraction and gaussian beams (which have an inevitable "tail" that never goes to zero) we know that if we move that mirror, there will be a tiny change in the position of the null because some non-zero part of the energy bounces off that unused mirror. Now my question is, when we consider 'non-classical' Feynman paths that bounce off that stray mirror, is this merely a way of including weak diffraction contributions, or is there something way deeper going on?
     
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  3. Nov 14, 2017 #2

    PeterDonis

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    It depends on what you mean by "in the same time". The coordinate time will be the same, yes, because the coordinate time of A and B is part of the definition of A and B--they are events in spacetime, not points in space. But the "elapsed time" for the particle traveling along a particular path--what the paper you linked to would call the number of turns of the particle's stopwatch hand--is not the same for all paths: if it were, all paths would have the same phase and there would never be any quantum interference between paths.

    For an electron, yes. See below.

    You don't. The KE - PE version applies to an electron, not a photon. For a photon, you calculate the rotation rate of the stopwatch hand by assuming it is the same as the frequency of the corresponding classical wave (item 3 on p. 191). The reason you have to bring in the KE - PE thing for an electron is that there is no classical wave corresponding to an electron (item 6 on p. 191).

    In principle, yes. But you might want to try running some numbers to see how small this effect actually is for a typical laboratory experiment.

    What's the difference?
     
  4. Nov 14, 2017 #3
    Thank you!

    Exactly. I'm like, "Is that all it is, or is there something cool that I'm missing?"
     
  5. Nov 15, 2017 #4
    If we want to find the clock rotation for a small path segment dx, dt where say v = dx/dt = ##\alpha c## , and the frequency is ##\omega## , what is the formula for the clock rotation over that interval? The paper doesn't provide that, and other works I could find are far beyond my current level. Would we just take the Lorentz-transformed dt based on ##\alpha##, and multiply by ##\omega##? And we don't apply any Doppler shift to the omega, just keep it constant?
     
    Last edited: Nov 15, 2017
  6. Nov 15, 2017 #5

    vanhees71

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    Already reading the abstract nearly made me faint. Don't teach the students about trajectories of photons. There are none!
     
  7. Nov 15, 2017 #6

    PeterDonis

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    Where does this "v" come from?

    Go read what you quoted from my post again. It already contains the answer to this question.
     
  8. Nov 15, 2017 #7
    It comes from any single assumed path (in spacetime) that we happened to choose, as a candidate that is going to contribute to the final total arrow. Say I decided to find the arrow rotation for a path where the photon traveled from event A to event B in spacetime. But let's say it is a straight line in terms of the x, y, z space but it traveled faster than c for some time and slower than c for some time. So this path is a valid member of all paths that start at event A and end at event B. And the "v" is the velocity in that path at some point along it.

    Hope I've not gone wrong up to this point. If it's ok, then how do we get the arrow rotation for a small segment dt, dx. As you pointed out, we can't assume that the "t" is the same as that for a classical path. So I'm guessing the dt's have to be the Lorentz transformed dts. But if that is correct, does the ##\omega## have to be transformed, too?
     
  9. Nov 15, 2017 #8

    PeterDonis

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    How does a path give you a ##v##?

    That question has already been answered. Go back and read what you quoted from me, again.
     
  10. Nov 15, 2017 #9
    I think part of the problem is, I have been using the same word "path" to talk about both "history" and "curve in x,y,z space". I'll think through it again and get back. Thank you.
     
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