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About Feynman's 'QED:The strange theory of light and matter'

  1. Aug 19, 2009 #1
    I've just read Feynman's book:'QED:The strange theory of light and matter' and, in order to measure the probability that a photon will reflect or transmit through a partially reflective piece of glass, Feynman introduces an 'imaginary stopwatch'. Can you explain me:
    -what do the hands of this stopwatch represent;
    -why does Feynman oppose the sense of the first hand in the case of a reflection by two surfaces;
    -how does he know the lenght of the hands when he measures the probability?

  2. jcsd
  3. Aug 19, 2009 #2
    I read this book about a year ago (and loved it) so i might be wrong on a couple of things, but here goes. The 'watch hands' represent the direction in which photons will scatter. Then, using tip to tail vector addition method you can add up all of these tiny arrows and get a resultant vector, therefore showing you the 'average (if you will)' way in which photons will reflect.

    I'm pretty sure, although he didn't explain this, that he determines the length of each arrow based on probabilities. For example, the uses a 'unit arrow' say in gradients of 5%. That way, if a photon has a 5% chance of scattering one way the arrow's length will be 'x' where as when it has a 20% chance of going one way the arrow's length will be '4x'.

    Hope this helps a bit! And hopefully someone corrects me if i'm wrong :)
  4. Aug 19, 2009 #3


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    mg0shtisha, no I do not think that is correct. Feynman is attempting to explain his path integral theory for QED. In path integrals, each path contributes the same amount to the probability density for the wave function. So each path that he calculates always has the same magnitude. What changes is the phase of the path. This phase is represented by his stopwatch, progressing through the 360 degrees of phase. In a path integral, the phase component is directly proportional to the action of the desired path, that is it is the time integral over time of the path's Lagrangian.

    In classical physics, the path of motion for a particle in a given system is given by the stationary path of the action. About the stationary point, there will be very little change in the action. This means that the phase around the classical path of Feynman's path integral will vary slowly. This is shown by his stopwatch moving slowly near the classical path. It also means that the paths around the classical path contribute the most to the path integral since the phases do not cancel out.

    Away from the classical path, the action is highly oscillatory. This means that the phases of paths that are very close to eachother will cancel out. Feynman talks about his stopwatch being at 12 o'clock and then suddenly going to 6 o'clock resulting in a net contribution of zero. So the adding of the vectors from the stopwatch, going from tip to tail, would be another way of adding up the phase contributions of various paths. If the phase contributions cancel out, then the addition of all the stopwatch vectors would end up back at the tail of the first phase vector. This is just a vectoral representation of the addition of phase contributions.

    This is very much like using the techniques of steepest descent or stationary phase when integrating an integral. If we have an integral that is highly oscillatory in phase, but constant in magnitude, then the points where the oscillation in the phase is the slowest will have the greatest contributions to the final result. We can get a good estimate of the integral by simply integrating around the stationary points as opposed to the entire length of the integral. With a path integral, we need to consider every path that can be made from point A to point B. Obviously, this is a monumental task to integrate over for even the simplest of situations. However, having knowledge about the stationary points, the classical path that is, allows us to truncate the paths that we need to consider since we know paths that deviate far from the classical path will be highly oscillatory and thus will not contribute greatly to the final path integral.

    In regards to the OP:

    The path for the reflection is the classical path of that problem. If we said that light goes through the minimal path length, then we would not get any reflected path, light would only go from point A to point B directly. A reflected path comes about due to Fermat's Principle: http://en.wikipedia.org/wiki/Fermat's_principle . This principle allows for more paths than just the direct path to be valid and, as expected, is the same as Feynman's path integral in the classical limit. Fermat's principle will give you the classical paths that I have mentioned, so Feynman's path integral will naturally pick out the same classical paths as the main paths that contribute to the wave function. The proof of this would be to work out the Lagrangian and find the stationary paths of the action from the Lagrangian. This is purely a mathematical exercise.

    It has been a while since I read QED but I do not recall any real variation in the length of the hands. I would expect them to always be the same length since the contribution from each path is the same, only the phase (represented by the direction of the hands) is different between the paths.

    EDIT: I think this would obviously be more appropriately situated in the Quantum forum.
    Last edited: Aug 19, 2009
  5. Aug 19, 2009 #4


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    I don't have a copy of QED with me out here but I think I may remember the arrow length. I think that Feynman represents the integral as the vector sum of the arrows generated by his stopwatch. This would be true since each path contributes the same magnitude but different phase (represented by the length and direction of the arrow respectively). The addition of all these vector arrows is analagous to the integration, adding up all the contributions of the path. The final vector sum is representative of the probability density squared. This would be \psi squared in Schroedinger quantum mechanics. Is this the arrow length that you are talking about or does he vary the length of the arrows used in the vector sum? It's just been so long since I read the book and I cannot find any representative pages online.
  6. Aug 19, 2009 #5
    Born2bwire, thanks a lot for that. As i said i hadn't read it in a while, but i was still way off. I haven't delved that deep into the material yet as i'm just an incoming freshman and a state university, but i was just trying to see if i remembered right! Your correction was extremely interesting and informative though, thanks again.
  7. Aug 20, 2009 #6


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    The length of the hand is always the same. The hand represents a complex number of magnitude 1. Such a number can always be expressed in the form [itex]e^{it}[/itex], where t is a real number. Complex numbers can be represented by vectors ("arrows") in a plane. t is the angle (measured in radians) by which you would have to rotate an arrow pointing to the right counterclockwise to get it to point in the same direction as the arrow representing your complex number.

    The book is based on his notes for a series of lectures that are available online. Link.
  8. Aug 20, 2009 #7
    First, thanks a lot for all these precious details.

    About the arrow lenght, what I understood is that it is always the same and that Feynman found this lenght experimentally:Since the probability for a photon to be reflected by a piece of glass is of 4% and since this probability is given by the area of the square which side has the same lenght as the arrow, then this arrow must be 0.2 centimeter long (0.2^2=0.04). But I may be wrong?

    There are still one or two points that are not clear for me:
    -I read that the stopwatch was linked to the photon sent. The watch hand will accomplish one turn each time the photon will have covered the distance of one wave-lenght (ie 26000turns per inch for a red photon for instance). Can you explain me what exactly is the link between the stopwatch and the photons?
    -I still can't understand why does Feynman oppose the sense of the first arrow in the case of a reflection by two surfaces.
  9. Aug 20, 2009 #8


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    Ok, here's the thing. The path integral only tells you the probability density of a particle (or state) starting at point A and time T_A being at the point B at time T_B. The path integral does not say anything about how that particle got from point A to point B. If you want to specify a specific path, you need to take into account the probability of the particle going to and from each desired point of that path. So in the reflection case, we need to specify that the particle hit the mirror at say point D, time T_D. So we first calculate the path integral from A to D and then use that to scale the path integral from D to B. Now, let's say we want to the particle to just reflect off of any point along the mirror for times between T_A and T_B. That's a troublesome problem because we have to account for a lot of paths (well, infinite paths technically). So Feynman probably just simplified it by saying, or finding, what that probability turns out to be anyway. I am not sure if this is what you mean because again I do not have a copy of the book and I cannot gleam much from what you have written. I am not sure if you are saying he calculated the probability of reflection and then scaled it according to experiment, or if he was calculating another probability that involved a reflection and then scaled that probability further by the probability of reflection.

    In general, there is no way to scale the arrow. Technically, what you would do is scale it by the number of arrows you have. Essentially, he is replacing his integral with a finite summation represented by adding the arrows together. If I do a simple Riemann sum of a function for the integral, I sample N points that have a distance of delta x between them. I estimate the integral as the sum of the function values times delta x,

    [tex] \int f(x)dx \approx \sum_{n=1}^Nf(x_n)\Delta x[/tex]

    However, Feynman's sum here (the path integral), must be a the square root of the probability. So, if all the arrows lined up, then that would maximize the integral and the maximum probability that you can have is 1. So if N arrows all add up to a length of N units, then we would have to scale by 1/N. This is just a guess but it could be a rough idea of how to scale the arrows.

    In the end though, the length of the arrows is meaningless because it is a graphical analogy of the path integral. He is just trying to give you a brief idea of the method but this is not how it is really done.

    2. Ok, I finally found a picture of his mirror spiel, huzzah. http://lesswrong.com/lw/pk/feynman_paths/

    My impression is that Feynman is cheating. With the path integral, you have to specify not only the starting and ending locations in space but also in time. However, Feynman is only specifying space in his problem and is assuming that the photon is moving at the speed of light and is measuring the time it takes the photon to go along the path. Technically, you should be able to make this relationship. Fermat's principle states that light will travel the path of least time, which is what he is replicating here.

    The actual Lagrangian for a photon in free space is rather complicated. However, since we know that Fermat's principle can be distilled from QED, there should be a way to express the photon path integral in a manner that he is using.
  10. Aug 20, 2009 #9


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  11. Aug 21, 2009 #10
    Thanks again for the precious explanation you added Born2bwire.
    " I am not sure if you are saying he calculated the probability of reflection and then scaled it according to experiment, or if he was calculating another probability that involved a reflection and then scaled that probability further by the probability of reflection."
    I apologize for not having the language to express it better but in fact, when I posted this message, I didn’t know how Feynman did to compute this length.

    I read again this book and I found answers to some of my questions (This book is based on lectures given by Feynman; you can find these lectures by following the link introduced by Fredrick). Actually, I wouldn’t say that Feynman cheats, but he simplifies considerably his calculation. Indeed, in order to get the probability for a photon to be reflected or not, you need to consider the probability that a photon in absorbed by an electron on the surface, and then sent towards the good direction (I strongly recommend you the third lecture for this).
    As you said, the length of the arrows has no real importance, but it is the "relative length" of one arrow in relation with the other that is important.
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