Relationship of Light's Wavelengthvs and Probability Wave?

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SUMMARY

The discussion focuses on the relationship between a photon's wavelength and its localization in three-dimensional space, emphasizing that photons are described as quanta of the electromagnetic field rather than classical particles. It is established that due to the massless spin-1 nature of photons, a position observable cannot be defined in the traditional sense. The detection probability densities of photons are linked to the energy density of the electromagnetic field, illustrating that a well-defined wavelength corresponds to a broad wave packet in position space, leading to uncertainty in localization. This highlights the importance of understanding photons through quantum field theory (QFT) rather than classical particle concepts.

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  • Quantum Field Theory (QFT)
  • Understanding of massless spin-1 fields
  • Fourier transforms in quantum mechanics
  • Concept of wave packets in quantum physics
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  • Study the implications of massless spin-1 fields in quantum mechanics
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Chris Frisella
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http://physics.stackexchange.com/questions/103904/size-of-a-photon

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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Photons are neither particles nor fields (wave) in the classical sense but the quanta (i.e., single-"particle" Fock states) of the electromagnetic field. One should note that the particle picture is very problematic particularly for photons since the quantum field that's used to describe them is a massless spin-1 field, and massless fields with spin ##\geq 1## do not allow the definition of a position observable in the usual sense. Thus you cannot even define a position as an observable.

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.
 
I get some of that :)

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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