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How does Fig. 26-3 (shown below) correspond to the following paragraph from this Feynman lecture?

Finally, we give a very crude view of what actually happens, how the whole thing really works, from what we now believe is the correct, quantum-dynamically accurate viewpoint, but of course only qualitatively described. In following the light from A to B in Fig. 26–3, we find that the light does not seem to be in the form of waves at all. Instead the rays seem to be made up of photons, and they actually produce clicks in a photon counter, if we are using one. The brightness of the light is proportional to the average number of photons that come in per second, and what we calculate is the chance that a photon gets from A to B, say by hitting the mirror. The law for that chance is the following very strange one. Take any path and find the time for that path; then make a complex number, or draw a little complex vector, ρeiθ, whose angle θ is proportional to the time. The number of turns per second is the frequency of the light. Now take another path; it has, for instance, a different time, so the vector for it is turned through a different angle—the angle being always proportional to the time. Take all the available paths and add on a little vector for each one; then the answer is that the chance of arrival of the photon is proportional to the square of the length of the final vector, from the beginning to the end!

Any insight greatly appreciated!! Thanks in advance!!

EDIT: Here is a lecture where Feynman talks about this

enter image description here

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The picture is a way to compute the probability of going from an initial to a final state. In other words, we are

following the light from A to B.

The little arrows are the result of following this algorithm:

Take any path and find the time for that path; then make a complex number, or draw a little complex vector, ρeiθ, whose angle θ is proportional to the time. The number of turns per second is the frequency of the light. Now take another path; it has, for instance, a different time, so the vector for it is turned through a different angle—the angle being always proportional to the time. Take all the available paths and add on a little vector for each one

In interpreting the above paragraph, note that the space that the picture is being drawn in is an abstract space (the complex plane), and not real space. Real life space is not present in this picture, so for each arrow in the picture, you have to imagine what the path of the photon looks like in real life in your head.

The big arrow is the "final answer" and is the result of adding the little arrows using vector addition. The big arrow is related to the probability of going from initial state to final state, as Feynman says:

then the answer is that the chance of arrival of the photon is proportional to the square of the length of the final vector, from the beginning to the end!

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    $\begingroup$ Hi @eball. There is no such thing as "the length of the final path", and the arrows in the picture don't tell you very much about the motion of the photon. It's best not to think of the photon taking one path from A to B. All we can measure is that the photon starts and A and ends up at B, with some probability. This picture is a way to compute the probability. The way we do it is to take every possible path from A to B, assign it an arrow, and put it on this picture. The arrow doesn't have any direct interpretation in terms of the path (eg it's not like the arrow is the direction that the... $\endgroup$
    – Andrew
    Nov 20, 2020 at 8:06
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    $\begingroup$ ...photon travelled, or something). There is just some rule that assigns an arrow to each path: the angle the arrow for a given path makes with the x axis is proportional to the time it takes for the photon to go along that path. In terms of the final, big arrow, the interpretation is that the length of this arrow squared is the probability that the photon went from A to B. It has nothing to do with the "final path the photon took," since there is no "final path." The path is not measurable anyway, the only thin we can say about the photon is the probability of it going from A to B. $\endgroup$
    – Andrew
    Nov 20, 2020 at 8:09
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    $\begingroup$ Hi @eball : Yes, this is a good point! If you drew this diagram honestly, there would be an infinite number of small arrows, one for each of the infinite number of paths. Of course, it isn't possible to draw an infinite number of arrows, so this is really just a schematic drawing, giving you the idea of a more complete calculation. If you haven't read Feynman's "Mayan bean counter" analogy (bradford-delong.com/2019/06/…), you should. It gives some context for what he is trying to do here. $\endgroup$
    – Andrew
    Nov 24, 2020 at 1:51
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    $\begingroup$ "What do all the little arrows have to do with computing the final probability of the photon going from A to B?" The answer is that the square of the length of the arrow you get by summing all the little arrows is equal to the probability of going from A to B. If your question is "why should that be the case?"... sadly, there is not really a deeper answer. This is a way of expressing a solution to the Schrodinger equation, which is simply a postulate (assumption) of quantum mechanics, with no deeper explanation. (A relevant Feynman video: youtube.com/watch?v=36GT2zI8lVA) $\endgroup$
    – Andrew
    Nov 24, 2020 at 1:53
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    $\begingroup$ @eball Exactly! If you want the more detailed explanation, here is an example of lecture notes with all the gory details: tpi.uni-jena.de/~wipf/lectures/pfad/pfad2.pdf Eq 2.34 (or maybe 2.32) is the mathematical equation Feynman is representing with his arrows. Each little arrow corresponds to a phase factor $e^{iS[w]/\hbar}$, where $S[w]$ is the classical action associated with path $w$, and the deceptively simple looking $\int \mathcal{D} w$ is a stand-in for the limiting procedure needed to sum over all possible paths. $\endgroup$
    – Andrew
    Nov 24, 2020 at 18:01

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