I Relationship of Light's Wavelengthvs and Probability Wave?

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?
 
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. Towards the end of the first lecture for the Qiskit Global Summer School 2025, Foundations of Quantum Mechanics, Olivia Lanes (Global Lead, Content and Education IBM) stated... Source: https://www.physicsforums.com/insights/quantum-entanglement-is-a-kinematic-fact-not-a-dynamical-effect/ by @RUTA
If we release an electron around a positively charged sphere, the initial state of electron is a linear combination of Hydrogen-like states. According to quantum mechanics, evolution of time would not change this initial state because the potential is time independent. However, classically we expect the electron to collide with the sphere. So, it seems that the quantum and classics predict different behaviours!
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