Feynman's QED; 36,000 revolutions per inch for red light.

In summary, Feynman explains in his book QED that the significance of 36,000 is the wavelength of red light, which is about 1/36000 inches (700 nm). He suggests that the angle of the "little arrows" in his model corresponds to both the phase angle of the EM oscillations in the photon and the probability amplitudes of virtual photons. However, he does not elaborate on this further.
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anorlunda
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In Feynman's book, QED, The Strange Theory of Light and Matter, page 27 he says, "As long as the photon moves, the stopwatch hand turns (about 36,000 times per inch for red light); when the photon end up at the photomultiplier, we stop the watch."

I've been puzzling about the significance of 36,000. The wavelength of red light is about 1/36000 inches (700 nm). So it sounds like he is saying that the angle of the "little arrows" is the same as the phase angle of the EM oscillations in the photon. But my reading of the book is that the angle of the arrows refers to probability amplitudes of virtual photons, not the phase angle of a real photon. Feynman did not elaborate.

So I'm left confused. What is the significance of 36,000? Hopefully, an I-level answer is possible.
 
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anorlunda said:
I've been puzzling about the significance of 36,000. The wavelength of red light is about 1/36000 inches (700 nm). So it sounds like he is saying that the angle of the "little arrows" is the same as the phase angle of the EM oscillations in the photon. But my reading of the book is that the angle of the arrows refers to probability amplitudes of virtual photons, not the phase angle of a real photon. Feynman did not elaborate.
In Feynman's heuristic model it's both. If we take the double-slit, for example. The QED model must largely match the classical EM wave model. The phase of the probability amplitude must match the frequency/wavelength of the classical wave.
 
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