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What will happen to a single photon when it goes through a prism?

Will it just be deflected in the direction related to its frequency?

I made this drawing for better understanding, here if I place a set of detectors after the prism, only one of them would detect the photon, contrary to a polychromatic wave light where all of them would trigger. Does this actually happen? enter image description here

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  • $\begingroup$ Do you mind telling me how you created the graphic? $\endgroup$ – garyp Dec 3 '18 at 20:48
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A good answer to this question would require knowing the source of the single photon you're asking about. If the beam from a laser emitting at, say, 532 nm and having a bandwidth of, say, 0.1 nm, is passed through a prism, the angular spread of the beam will be extremely small -- let's say an angle of $\alpha$. Any individual photon in that beam will contain a mix of wavelengths and will land somewhere within the angular spread $\alpha$. If the beam comes from a continuum laser which has a bandwidth that covers the visible spectrum from ~650 nm to ~400nm, then the beam will spread into a rainbow upon passing through the prism, and will cover a much wider angle $\beta$. And, any single photon in the beam will land somewhere within the angular spread $\beta$.

A photon can only be detected once. Before detection, its frequency (and polarization) are indeterminate. That means those properties don't have a value until they are detected/measured. It's not that the properties are not known; the properties don't have a value. The wave packet constituting a photon before detection is just a packet of probability densities, specifying the likelihood that a measurement will yield any particular value. So the wave packet of any photon in the beam is spread by a prism just as the beam is spread, but as soon as the photon is detected somewhere, we assign it a wavelength that it didn't have just before detection. If instead of detecting the photon at that point, we let it pass through a slit in an aperture, we are constraining the wave packet so that the photon, if detected downstream from the slit, is sure to have a wavelength within the range of wavelengths passed by the slit.

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  • $\begingroup$ Re, "It's not just that the properties are not known; the properties don't have a value." Why is it more satisfying to say it that way instead of saying that, even though the theory makes accurate predictions about the statistical population of photons in the experiment, it says nothing about individual photons? $\endgroup$ – Solomon Slow Dec 3 '18 at 17:16
  • $\begingroup$ The point is that a photon (or any other quantum particle) does not have a quantum state until the state is measured. The alternate view ("hidden variables"), that a particle has a state but we don't know what it is until it's measured, turns out to be incorrect, as is revealed by the statistics of the results of measurements on single particles (e.g., "single-photon" double-slit interferometry). In the context of this question, the quantum state that's being measured is wavelength, so I'm saying that a photon does not have a definite wavelength until the wavelength is measured. $\endgroup$ – S. McGrew Dec 3 '18 at 20:37
  • $\begingroup$ You say, "a photon...until the state is measured." But what photon are you talking about? You can't even know that there was a photon until its state is measured. You can predict how many photons will be detected in an experiment, and you can predict the spatial distribution and the energy spectrum of the detection events, but a theory that predicts where and when the next one will strike simply does not exist. Maybe I'm confused about what "photon" means. I'm thinking of something that can be counted by a detector, but maybe you're thinking of it as the wave function that predicts... $\endgroup$ – Solomon Slow Dec 3 '18 at 21:10
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    $\begingroup$ Some folks argue that the very idea of "photon" is misleading, and that the term should be abandoned. I think they are right, but I don't know what term to use instead. This is getting a bit off-topic. Would you like to move it to a chat room? $\endgroup$ – S. McGrew Dec 3 '18 at 21:23
  • $\begingroup$ @S.McGrew There’s no reason to get rid of the term photon and there’s no justification to assume they don’t exist. $\endgroup$ – Bill Alsept Dec 5 '18 at 2:02

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