Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Qed and probability

  1. Jun 25, 2013 #1
    I'm not sure whether to post this here or not, but since the book is about quantum i figured it was as good a place as any. I don't know too much about this, so I am just trying to understand exactly what the heck Feynman was talking about. I understood most of it but I have a few questions I was hoping some of you guys here could help me out with. So here are my questions:

    Is Feynman saying that the reason why we use quantum mechanics is to be able to talk about the possibility of an action rather than the cause of it? E.g. that a photon will interact with an electron with some probability, but not why the photon is reacting in the first place?

    Second, are the photons which bind electrons to nuclei 'virtual' and if they are, what does he mean by this? If they don't exchange actual photons, why is he using this as an explanation for what is occurring?

    Finally, Why did he choose to have a rotating stopwatch hand as the means of choosing the angle at which the probability amplitude points?

    I think this stuff is really interesting, so i'd like to understand it as well as possible. Thanks for the help =)
     
  2. jcsd
  3. Jun 25, 2013 #2

    Fredrik

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Something like that, yes. QM assigns probabilities to possible results of measurements, but it doesn't really paint a picture that can be thought of as a description of what's "really happening".

    It's possible that someone may slap me for this, because I've seen some different views about this subject expressed here before, and I'm not particularly good at quantum field theory. But I would say that all internal lines in a Feynman diagram represent virtual particles. The diagrams themselves probably shouldn't be thought of as descriptions of what's actually happening (we don't know if that picture is accurate or not). So I suggest that you think of them as representations of the terms of a series that shows up when we do the calculation.

    One thing that distinguishes the virtual particles from the real ones is that when we do the calculation, we have to consider virtual particles with all possible velocities, not just velocities for which the speed is less than c.

    A complex number is something that can be written as x+iy, where x and y are real numbers, and ##i## is a weird mathematical object that has the property ##i^2=-1##. Since a complex number is uniquely identified by an ordered pair (x,y) of real numbers, they can be visualized as points in a plane. A point (x,y) can also be represented by an arrow drawn from (0,0) to (x,y). The number ##\sqrt{x^2+y^2}## is called the modulus or absolute value of the complex number x+iy. Every complex number can be written in the form ##re^{i\theta}##, where r is the absolute value, and ##\theta## is an angle between 0 and ##2\pi##. A complex number with absolute value 1 can therefore be written as ##e^{i\theta}##, and represented as a point on the unit circle (the circle of radius 1 centered at (0,0)), or as an arrow from (0,0) to a point on the unit circle. This is how Wikipedia illustrates it:

    http://upload.wikimedia.org/wikipedia/commons/7/7a/Complex_number_illustration_modarg.svg

    (Their ##\varphi## is my ##\theta##). Feynman is talking about complex numbers with absolute value 1. Specifically, he's talking about numbers ##e^{iSt}##, where t is the time and S is a different number for each path. Hm, OK now I have to think. I think that the argument goes that paths that are very different from the one that minimizes S will not contribute much to the final result, and can therefore be ignored. So we can choose to only consider paths that are very close to the path that minimizes S. This means that all our S's will have values that are close to each other, so we can assume that they're all equal, without introducing a big error. So now you can think of S as just a number, and t as the time, which can also be interpreted as an angle when you represent the complex number ##e^{iSt}## as an arrow.
     
  4. Jun 25, 2013 #3
    Ah I see. So you are saying that the only terms of the series that contribute greatly to the probability are the ones that minimize S, right? And that we ought to only seriously consider these ones because all the others just add insignificant bits, and even though they all say something, they aren't big enough to affect the general idea.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Qed and probability
  1. Feynman's QED (Replies: 28)

  2. Symmetry in QED (Replies: 1)

  3. QED questions (Replies: 1)

  4. Photons and QED (Replies: 26)

  5. QED Theory (Replies: 8)

Loading...