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If a photon has lower Energy (b.e.) it must pass a high energy filter unaltered. Let it be a red photon through a blue filter. As a photon is in fact a package of many waves with different wavelengths (if it has Poisson distribution there would be also blue wavelengths) and its energy cannot be divided - How does it pass a filter unaltered? The filter can cut its higher frequencies but they posses energy. So 'part' of the photon would be cut off by the filter and it would be very much different both in shape and energy.

Mercury
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    Why is "a photon" a package of many waves with different wavelengths? – Jon Custer Oct 25 '22 at 17:25
  • Photons are amounts of energy, momentum and angular momentum. They are not objects and they are not wave packets. If you want to have "light with many different wavelengths", then you need a large number of photons. – FlatterMann Oct 25 '22 at 18:16
  • @FlatterMann No matter are they objects they interact with almost localized electrons and then they are localized too. In order to account for this property you need a wavepacket of almost all f - Fourier theorem. If one has only one f in the photon it is an infinite wide spatial distribution. How is this spatially distributed energy squeezed in almost point in no time? They (the photons) have also been born by localized electrons. – Mercury Oct 25 '22 at 20:35
  • I have to disappoint you, but that is a semi-classical picture that "is just not so". Photons do not have frequencies at all. They only have energy, momentum and angular momentum. Waves with wave lengths and frequencies are emergent phenomena that are made up by the statistical properties of a large number of photons. – FlatterMann Oct 26 '22 at 04:11
  • @FlatterMann This is according to what theory? – Mercury Oct 26 '22 at 05:12
  • This is according to every experiment that has ever been done at the photon level. – FlatterMann Oct 26 '22 at 05:31
  • @FlatterMann IMO you are denying wave particle duality!? Are you denying that in DSE made with one photon at a time there is not the wave behavior for an individual photon revealed? How will the statistics made by itself in many repetition when the particle has no wave? Famously when the wavelength goes to zero (greater mass) the wave behaviour is supressed and statistics is different. How does statistics know it in order to change? – Mercury Oct 26 '22 at 16:10
  • Let us continue this discussion in chat. – Mercury Oct 26 '22 at 16:22
  • I am simply saying that your understanding of quantum mechanical phenomenology is wrong. – FlatterMann Oct 26 '22 at 18:28
  • Ok maybe I am wrong. But why? The wave - particle duality WPD is an established fact at least in mainstream QM. And it goes about single particles. WPD is not emerging but fundamental feature of almost all in the interpretations. Let me know in what interpretation WPD is emergent. – Mercury Oct 26 '22 at 19:22

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In the mainstream theory of physics, a photon is an elementary point particle , of energy $hν$ where $h$ is Planck's constant $ν$ is the that frequency a large number of photons of that energy,will have as classical electromagnetic radiation. A photon is not a bundle of classical electromagnetic waves, classical waves develop from the underlying quantum field theory of particles.

So depending on the type of filter a photon will either go through unscathed or interact with the fields if the filter changing its energy accordingly.

Harris Spivack
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anna v
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  • If it is a point particle its wavefunction has to be a superposition of all frequencies by Fourier Theorem. And the wavefunction for photons cannot be other than electromagnetic wave of Maxwell. Do you know other wavefunction for photons? – Mercury Oct 25 '22 at 21:06
  • It can, though for the quantum level solution using maxwell' equations https://arxiv.org/abs/quant-ph/0604169 – anna v Oct 26 '22 at 03:07
  • @Mercury A wave function is an ensemble description. It only makes statements about the statistical averages of many repetitions of a system. This is especially hard to "grok" in linear optics where a wave made out of many photons behaves exactly like the ensemble would because in linear optics photons don't interact with each other, so ensemble and multi-particle states behave identical to each other for CLASSICAL states of light. This is not so in general, though. You can find an endless number of questions online about "the double slit with single photons", all of which are nonsensical. – FlatterMann Oct 26 '22 at 04:16
  • @annav You write "ν is the that frequency a large number of photons of that energy will have as classical electromagnetic radiation." So if the photon is single what does ν means? Does not a single photon has wave properties? – Mercury Oct 26 '22 at 16:26
  • @Mercury I state it in my answer. It is the frequency of the classical electromagnetic radiation that will exist once the number of photons becomes very large, as seen in the experiment with the double slits. When the photons are few, there is no classical interference. The classical electromagnetic interference pattermn appears in the accumulation of a large number of photons. – anna v Oct 26 '22 at 17:21
  • @Mercury For a single photon v is meaningless. – FlatterMann Oct 26 '22 at 18:29
  • @Mercury my anser to this question may clear things up , I confused your comment question with a similar answer I gave here https://physics.stackexchange.com/questions/733703/shorter-wavelength-of-photon-in-detecting-a-particle/733865?noredirect=1#comment1641763_733865 – anna v Oct 26 '22 at 18:36
  • @annav I see you state that single photon has no wavelength. This is very strange I haven't seen such statement in any QM book. In fact there the wave particle duality is fundamental. I didn't read that photons make exceptions nevertheless m=0 only for them. I cannot see (also did the establishers of QM) how the DSE and similar can be accounted for without a wave. You know a single photon can almost not land in the dark areas. How does it know that when it has no wavelength? – Mercury Oct 26 '22 at 21:08
  • I cannot see how a multitude of photons will express wave behavior when the constituent 1 photon has them not and there isn't any interaction between them moreover they come one at a time in taylored DSE. – Mercury Oct 26 '22 at 21:09
  • That is why quantum mechanics and its probability wavefunction was proposed and the theory fitted the data almost perfectly. It can be shown mathematically that tha wave behavior in probability space builds up builds up the classical electromagnetic radiation in space time. It needs the quantum field theory mathematics. see answers https://physics.stackexchange.com/questions/93430/how-is-the-classical-em-field-modeled-in-quantum-mechanics – anna v Oct 27 '22 at 03:25
  • @annav Surely wave behaviour of single photons (evident in one at a time photon DSE) builds up an EM wave when N is great. So the photon has wave behaviour and hence WL. It is a probability WL for single photons and rise to EM WL (no surpringly). But single photon has to have many frequencies around a central one which is dominant because is photon localized with the distance between slits of DSE. Also the Ei Ej levels in atoms where the transition Ei-Ej build a photon are somewhat smeared hence uncertainty in E of photon. – Mercury Oct 27 '22 at 17:14
  • @Mercury this is your own theory tryng to fit the classical behavior on quantum entities. It is not mainstream. – anna v Oct 27 '22 at 17:43
  • This is what I read in QM books - wavepackets for 1 particle state b.e. Peskin eq. 4.65. Where does it stand that photon has no wave properties and no WL if I understood it correctly? You refer to the blog of Motl but it doesn't exist anymore so I cannot understand the meaning of all this. – Mercury Oct 27 '22 at 19:36
  • @Mercury I do not have the book, but the quantum waves are in probability, that displays wave behavior in space. probability , not individual point particle. – anna v Oct 28 '22 at 04:12
  • @annav I see you are proponent of Ensemble Interpretation. "The ensemble interpretation of quantum mechanics considers the quantum state description to apply only to an ensemble of similarly prepared systems, rather than supposing that it exhaustively represents an individual physical system". Maybe you have Ballentine in view. – Mercury Oct 28 '22 at 10:32