The link is: https://www.feynmanlectures.caltech.edu/I_26.html#Ch26-F3

Hi, I'm reading Feynman's lecture on Optics: The principle of least time, and I'm wondering if I got what he's saying right. For context

That is the feature which is, of course, not known in geometrical optics, and which is involved in the idea of wavelength; the wavelength tells us approximately how far away the light must “smell” the path in order to check it. It is hard to demonstrate this fact on a large scale with light, because the wavelengths are so terribly short. But with radiowaves, say 33-cm waves, the distances over which the radiowaves are checking are larger. If we have a source of radiowaves, a detector, and a slit, as in Fig. 26–13, the rays of course go from S to D because it is a straight line, and if we close down the slit it is all right—they still go. But now if we move the detector aside to D′, the waves will not go through the wide slit from S to D′, because they check several paths nearby, and say, “No, my friend, those all correspond to different times.” On the other hand, if we prevent the radiation from checking the paths by closing the slit down to a very narrow crack, then there is but one path available, and the radiation takes it! With a narrow slit, more radiation reaches D′ than reaches it with a wide slit!

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To my understanding, he is saying that the ray "probes" the path ahead (the extent of this probing depends on its wavelength) to determine which path it takes. This probing can be blocked by a narrow slit. If the probing is blocked, then the ray becomes "undetermined" and "may" bend for another path. Can somebody validate and supplement my take on this?


1 Answer 1


Path-integral-based explanations are quite different from typical dynamical physics explanations, and switching back and forth between them is quite tricky.

A dynamical physics explanation starts with initial conditions and then uses equations which evolve those initial conditions into a final answer. The final end point is then explained by the initial conditions and the dynamical equations.

But a path-integral explanation works very differently. In the case of a photon, the "inputs" to the calculation are the initial location and the final location. Then, instead of dynamical equations, there's a calculation which takes into account all possible paths the photon might have taken (between those two endpoints) and uses all those paths to calculate a probability. The output of the calculation is the probability that the photon is emitted from the initial location and is detected at the final location. There is no implication about what actually happened in the middle, no dynamical equations, just a probability which is calculated "all at once".

The problem then is switching back and forth. When Feynman talks about the photon in some anthropomorphic way, "smelling" or "checking" or acting like something which makes decisions, he's pushing back into the typical dynamical explanation, treating photons as if they are following some dynamical rules. But there are no such rules; certainly he would never say that the photon is taking some single particle-like path between points A and B. Perhaps he's just recognizing that most people don't think in this "all at once" perspective and trying to translate it back to the dynamical story. Evidently, given your reasonable confusion about this, it was likely a mistake to try to switch modes of explanation in midstream.

In a nutshell, the path integral viewpoint is not consistent with a photon "probing" or "sniffing" or trying to follow any dynamical rules whatsoever. It's a very different way of solving the problem that doesn't translate back to dynamical explanations.

Here's an essay which discusses these different explanatory 'Schemas', and another paper which shows that you can't interpret the path integral in terms of a "one real path" account.

  • $\begingroup$ Ok, thank you! I think I'm starting to get a hang of this now. So when the slit is wide, the probability of a photon going to D' is about 0, whereas when it is narrow enough, it could be significant? $\endgroup$
    – frank guo
    Commented Aug 24, 2023 at 17:43
  • $\begingroup$ Yes, that's the point he's making. Ignore all that anthropomorphic-photon language. :-) $\endgroup$ Commented Aug 24, 2023 at 17:50

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