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This question already has an answer here:

My question refers to Photon flux spectrum diagrams.

The diagram shows the number of photons at different wavelengths. My question is whether the graph is granular or continuous. Do photons exist at all possible wavelengths, or are there some "forbidden" wavelengths?

--- I want to get a PFD data set but not sure how to treat the 0.5 nm increments in the https://www.nrel.gov/grid/solar-resource/spectra.html (G173-03) Spectral data.

Below my calcuations are in green. I am afraid I am double counting photons the way ive done it so far.

(small note the coloumn D should say PFD and not PPFD*)

enter image description here

enter image description here

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marked as duplicate by ahemmetter, Martin, Jon Custer, GiorgioP, ZeroTheHero Mar 22 at 3:03

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ It is possible to have a single photon with a non-well-defined wavelength whose spectrum spans an interval, you know. $\endgroup$ – Emilio Pisanty Mar 21 at 7:33
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    $\begingroup$ That's an interesting question! For a given source there are of course wavelengths that are forbidden (e.g. only certain transitions exist in an atom, etc), and you always have the option to make a pulse shorter (thereby "stretching" it across wavelengths), or move the source. Hence I don't immediately see a reason why it shouldn't be possible to have photons of arbitrary wavelength. It would be interesting to see if there is an upper or lower limit. $\endgroup$ – ahemmetter Mar 21 at 8:00
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    $\begingroup$ However, it is possibly a duplicate of How many possible electromagnetic wavelengths are possible? $\endgroup$ – ahemmetter Mar 21 at 8:02
  • $\begingroup$ @ahemmetter please see amendments. $\endgroup$ – dlight Mar 21 at 8:41
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    $\begingroup$ Are you interested in all photons from the Sun, or all photons in general? With regards to your recent edit, the wavelength doesn't need to be a multiple of a nanometer; 373.0nm and 373.5nm are different photons with different energies. In fact there are photons with a wavelength of only a fraction of a nanometer, e.g. gamma rays have wavelength in the picometer range. It just depends if your spectrometer can resolve such small differences in wavelength. $\endgroup$ – ahemmetter Mar 21 at 10:47
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Do photons exist at all possible wavelengths?

Yes. You can essentially treat the background illumination as a full continuum over wavelength, such that any arbitrary wavelength interval, no matter how small, will contain a nonzero amount of energy.

For data sets like the one that underlies the plot in your image, if it is reported at a regular grid in wavelength (i.e. values of photon spectral flux at each wavelength in the list $\lambda_0$, $\lambda_0+\Delta\lambda$, $\lambda_0+2\Delta\lambda$, $\lambda_0+3\Delta\lambda$, $\ldots$), then the standard assumption will be that this represents the integrated spectral photon flux over an interval of length $\Delta\lambda$ around the reported wavelenght, i.e. the reported spectral photon flux at $\lambda_0+n\Delta \lambda$ is the integral of the spectral photon flux over the interval $[\lambda_0+(n-\tfrac12\Delta \lambda),\lambda_0+(n+\tfrac12\Delta \lambda)]$, divided by $\Delta \lambda$.

However, it's always possible that the dataset you're working with is doing something fancier - in which case this will be specified in detail in the dataset's associated documentation.

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  • $\begingroup$ photons do not have wavelengths, they have frequencies. The wavelength is hidden in the complex wavefunction and is not measurable except in probability distributions. The energy is measurable. $\endgroup$ – anna v Mar 21 at 12:50
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    $\begingroup$ @annav As far as the formalism here is concerned, $\lambda$ is by definition $c/\nu$, regardless of how you choose to interpret it. If it has a frequency, then it has a wavelength for all practical purposes regarding the measurements OP asks about. $\endgroup$ – Emilio Pisanty Mar 21 at 14:50
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Photons are quantum entities, they are elementary point particles of energy= $hν$ where h is the Planck constant and $ν$ the frequency of the wave that will be built up when photons of this frequency are accumulated, the emergent classical electromagnetic wave. For single photon behavior see this double slit single photon experiment.. The accumulated interference pattern is a measure of probability distribution of the photons impinging on the double slits.

The wavelength corresponding to the $ν$ is mathematically within the complex valued wave function of the photon, and can be seen only in probability distributions. As energy has only the limit of conservation of energy, values of frequencies can go from zero to infinity, bounded by the possibility of having an appropriate source, and so do the wavelengths of the emergent classical light beams.

As said in the comments black body radiation has no limit, except on how probable it is for a very low or very high energy photon to be emitted .

The figure you are describing is a transmission (label on the right) figure too, so the material interposed will limit by absorption the photon frequencies that can be transmitted.The spectra seen are absorptions from atomic and molecular levels of the transmitter.

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  • $\begingroup$ Single photons need not have a well-defined frequency, in exactly the same way that single electrons need not have a well-defined energy. $\endgroup$ – Emilio Pisanty Mar 21 at 12:23
  • $\begingroup$ Or, to lay out the reasoning more expansively: in QFT, photons and electrons have exactly the same status (i.e. excitations of a quantum field), and electrons can have non-well-defined energy. In order to claim that all photons need to have well-defined frequencies, you need to be claiming that (i) photons and electrons are not equivalent in QFT (and if so, how?) or (ii) that QFT somehow does not apply to photons. $\endgroup$ – Emilio Pisanty Mar 21 at 12:36
  • $\begingroup$ @EmilioPisanty I am evidently talking about measurable photons, not about virtual or other theoretical photons, because the plot in the question is measured photons. Dragging in the formalism of quantum field theory on such an elementary question is not useful, imo of course. Basic quantum mechanics, which always holds no matter higher level mathematical complications, should answer measurement questions. $\endgroup$ – anna v Mar 21 at 12:48
  • $\begingroup$ Quantum field theory can be applied to everything, I have been taught of a QFT of nuclear physics, it is a formalism. When photons do not have a well defined frequency, they are off mass shell in the corresponding QFT feynman diagrams, i.e.virtual. $\endgroup$ – anna v Mar 21 at 12:48
  • $\begingroup$ It is perfectly possible to have an on-shell, fully physical single-photon state without a well-defined frequency - simply take a superposition of multiple states with different frequencies. $\endgroup$ – Emilio Pisanty Mar 21 at 14:49

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