Does a photon have any measurable size as a "particle"? If we determine its exact location and take away all motion in an isolated static environment, how large is it? Is it sub-Planck in size? If there is no motion in this static environment, is the photon also sub-Planck in terms of time?
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4$\begingroup$ As far as we know, there is no meaningful notion of size for fundamental particles. $\endgroup$– AccidentalFourierTransformCommented Feb 17, 2018 at 3:50
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$\begingroup$ How about time? If there is no wavelength, but simply a static particle or bundle of unexpressed potential energy, would this not also indicate that there is no meaningful notion of time (other than the present) in such a situation? $\endgroup$– Vincent NovoCommented Feb 17, 2018 at 3:55
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$\begingroup$ have you done any research? please add any to the body of your post and explain why it does not answer your question $\endgroup$– pentaneCommented Feb 17, 2018 at 4:21
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$\begingroup$ Personally, I only have published research by others and personally performed thought experiments based on well known research, such as the varying wavelengths of light. My question has more to do with whether or not there is any measurable size and length of duration of a photon in a "fixed" state within the realm we are operating. $\endgroup$– Vincent NovoCommented Feb 17, 2018 at 4:29
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$\begingroup$ I think we can't find its exact position beacause we know its exact velocity. So heisenbergs uncertainity principle says that uncertainty in position is infinite. $\endgroup$– Gurbir SinghCommented Feb 24, 2018 at 8:19
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2 Answers
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According to superstring theory, the physical size of any elementary particle (including photons) is that of one superstring, or the Planck length. As a wave though, we can think of the "size" of a photon as being that of its wavelength. Note however that the Heisenberg uncertainty principle means that trying to measure the amount of space that a photon occupies will proportionally increase the uncertainty in its momentum and energy. The Planck length would represent the smallest size that is physically possible to measure in theory (at one Planck energy).
answered Feb 17, 2018 at 9:07
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In the standard model of particle physics which is continuously validated by experiments, the photon is a zero mass,spin one, point ,elementary particle in the table of particles that build up all that we know. Its energy is $h*ν$ where $ν$ is the frequency of the classical electromagnetic wave that it can build up in superposition with zillions of other photons.
If we determine its exact location and take away all motion in an isolated static environment, how large is it?
Zero mass means it is moving with speed c in all Lorentz frames as all zero mass particles, so this scenario can never happen.
Does a photon have any measurable size as a “particle”?
The size of the footprint of the photon will depend on the measuring instrument. The footprint will be bounded by the Heisenberg uncertainty principle .
Here one observes the footprints of single photons, in a double slit experiment.
The footprint is a dot within the measurement errors, and certainly bounded by the uncertainty principle, since h, the Planck constant, is very small, macroscopic measurements always obey it.
Thus , a photon, as described by current quantum mechanical theories is a point particle, and it demonstrates an effective size according to the boundary conditions of the measurement.
