I've found nothing on this topic. Everyone says a photon is one wavelength of whatever beam of energy it belongs to, but no one says why this needs to be the case. If anyone has an answer, I'd be glad to read it. Thank you in advance for your help.
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3The statement isn't true. A single photon can (and indeed, always does) contain a superposition of many frequencies/wavelengths. – Mark Mitchison Jul 18 '18 at 14:12
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2Possible duplicate of Do photons truly exist in a physical sense or are they just a useful concept like $i=\sqrt{-1}$? – John Rennie Jul 18 '18 at 14:18
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1See also What is the relation between electromagnetic wave and photon?. These two questions go some way to explaining what a photon is and how it relates to light rays. – John Rennie Jul 18 '18 at 14:20
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Your answer, although it answers my previous question, sparks a new one in my mind. My new question is why do photons have to be in a superposition. – Anthony Ducharme Jul 18 '18 at 14:29
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1They don't "have" to be, but they mostly are because 1) quantum mechanics allows superposition, and 2) most processes that emit photons have some finite bandwidth (you don't expect a delta-function like sharpness in nature). – wcc Jul 18 '18 at 14:57
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"Everyone says ..." Could you be more specific? You have to be careful in these days of tweets and facebook. – my2cts Jul 18 '18 at 19:50
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1Possible duplicate of What is the relation between electromagnetic wave and photon? – Jon Custer Jul 19 '18 at 15:25
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To put things plainly: photons do not have to have a well-defined wavelength.
The states of the field which have a single photon present (i.e. states with well-defined photon number equal to one) and which have a well-defined energy (and therefore frequency, and therefore wavelength) are a convenient basis with which to work with.
Moreover, in many situations (particularly those available to the technology of the early 20th century, during which many of the prevailing attitudes about QM were forged) it can be a great approximation to just take those states as representative of your physical situation.
However, it is important to note that photons with a well-defined energy are unphysical states, in that the Heisenberg uncertainty principle $\Delta E \,\Delta t \gtrsim \hbar$ dictates that those states must be completely stationary. This means that they cannot move or change in any way whatsoever, and if they're travelling (like plane waves) then they must fill all of the space they're travelling to and from.
The useful, physical states of the EM field within quantum mechanics are superpositions of those states over a finite bandwidth, which then lets the Heisenberg uncertainty principle allow for processes that take finite time. They might still be single-photon states, but the photon has some quantum-mechanical "fuzziness" of indefinition as to its frequency and its energy, which is precisely what allows it to evolve in time.
For introductory applications, that will very often not matter, so it's often OK to pretend that those considerations don't exist and that each photon must have a well-defined frequency, but that's a strict limitation to the sorts of phenomena that those presentations can cover.
If you want more details, the best reference is probably David Tannor's Introduction to Quantum Mechanics: A Time-Dependent Perspective, at the level of a mid- to advanced undergraduate course in QM.
Emilio Pisanty
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