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I can follow the equitations from John Rennie's answer in the above thread, but considering real 3D space, I don't understand how wavelength would make the location of a photon less precise. Can anyone explain that more?

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- Thread starter Chris Frisella
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- #1

- 76

- 6

I can follow the equitations from John Rennie's answer in the above thread, but considering real 3D space, I don't understand how wavelength would make the location of a photon less precise. Can anyone explain that more?

- #2

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What's described by QFT are the detection probability densities of photons for given single-photon Fock states. As it turns out this probability is defined by the energy density of the electromagnetic field. Now such a state is most easily intuitively depicted as a wave packet. If you want a photon state with a well-determined wavelength of this field, this means that the field must be very narrow in momentum space. In position space, which is the Fourier transform of the wave packet in momentum space, the wave packet is pretty broad, i.e., the photon is not well "localized" in the sense that the probability distribution for detecting it (given by the normalized energy density of the field) is pretty broad. It's an example of a kind of uncertainty relation, but this has to be taken with a grain of salt since strictly speaking it's not an uncertainty relation in the precise sense, because (as stressed above) there's no position operator for photons, but you can think about it along the lines of detection probability, and you are closer to the right picture when using the wave-point of view than the particle-point of view.

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Follow up question: Why would a photon ever be detected at a lower "energy density" area in the probability wave? Like, if you fired off one single photon, do you know why it would ever appear at an area of low probability? What is the underlying mechanic that creates this?

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