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This question has nothing in common with the question it is proposed to be duplicate e.g about virtual photons. The question is about a real radiation field where is known that no virtual photons exist.

The electric field E moving up and down (in its value) as a sinusoidal wave, which has been regularly shown in EM books should move a probe electron up and down (in space). How does a photon picture make this pattern, as the photon is moving forward (not up and down) its momentum is perpendicular to E? Of course there is also magnetic momentum but it is too faints. If one uses Feynman diagrams (as is proposed in the answer of Lubos in the proposed duplicate question) for electron photon scattering the momentum of the electron should also be in the photon's direction and not perpendicular as the EM picture suggests.

Mercury
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  • The em wave diagrams you see, are oscillating electric and magnetic fields , on a single line, these are not spacial oscillations , they are the oscillations of the EM field associated with a particular POINT in space. – jensen paull Mar 13 '22 at 17:39
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    Does this answer your question? Virtual photon description of $B$ and $E$ fields – John Rennie Mar 13 '22 at 20:00
  • @jensenpaull i didn't say that E is oscillating in time and space. The oscillation are of the electron in the E field. The electron is oscillating because the force is F=eE. – Mercury Mar 13 '22 at 20:44
  • @JohnRennie well I never thought that the E filed is comprised of virtual photons. Is the E field also virtual? I will look at your link definitely. – Mercury Mar 13 '22 at 20:47
  • You are talking about an oscillating charge, this oscillating charge causes electromagnetic waves, which are oscillating Electric and magnetic fields. The "wave diagram" of an EM wave are field oscillations, not spacial ones. – jensen paull Mar 13 '22 at 21:09
  • @jensenpaull well this is true but one can easily neglect them for the picture of the EM wave to be more clear. By the way the electron could not be possible to follow the field. I don't know why you repeat that the osc. of the field are not spacial. I told you that only the electron is supposed to move, but its hard for it too. Anyway one can consider a EM in a medium with slower speed. There are medium where the light can be slowed to almost 0. – Mercury Mar 14 '22 at 12:47
  • how could you neglect field oscillations when talking about an EM wave? The field oscillations IS what we call an "electromagnetic wave". If an incoming electromagnetic wave hits an electron, the electron will follow the force described by the electromagnetic field in the particles location. If the incoming electromagnetic wave is sinusoidal, as you have described, then the Electron will osscilate sinusoidally as when an EM wave passes a location, it could e.g follow $\vec{E_{0}}cos(wt)$ so as it passes through space, the associated electric field of the wave will cause ossilations – jensen paull Mar 14 '22 at 13:02
  • The Momentum of the E and B field is in the direction of the wave vector, and has zero components in the perpendicular originally. However the reason momentum is conserved is due to the maxwell stress tensor. I've posted another question to clear up my doubts – jensen paull Mar 14 '22 at 16:39
  • The E field is always created by an electron in an atom, the photon does not start the electric field, you have it backwards. The EM field carries the energy to another atom/electron and this is where the E field is finished. The transmission of the energy is called the photon, there is no photon without a starting and ending atom/electron. We can never measure the E field of the photon unless we destroy it. This is also why 2 lasers impacting a surface DO NOT cancel each other, all photons have their own path and they never cancel each other. – PhysicsDave Mar 14 '22 at 19:46
  • @PhysicsDave What you say is trivial. And does not answer how the photon momentum which is perpendicular to E (and B) makes the e- oscillate with E. Maybe this is where I am wrong (I doubt that e- will oscillate with E as I taught about). Possible solution is that e- because of its mass can not follow E but stays in place, nevertheless producing some current j (which must be a true QM magic) and the j x B gives a force forward. So e- will move only forward and not along E. – Mercury Mar 16 '22 at 13:51
  • @PhysicsDave As it goes to the Dirac dictum I read a note by Ray Glauber (Nobel) around 1990. He says that Dirac is wrong totally (there was an experiment with two lasers interference also).. Unfortunately I can not read the article it is not free. – Mercury Mar 16 '22 at 14:21
  • https://aapt.scitation.org/doi/10.1119/1.17790 here is the comment from Glauber on Dirac interference. Glauber states that "2 photon states" are the reason to discount Dirac .... Glauber's explanation is based on statistics and formulas and lacks the weight of Dirac's statement. The "photon interfering with itself" is interpreted today as "the photon determines its own path' (Feynman). – PhysicsDave Mar 21 '22 at 00:16
  • the e- in the atom is not at rest, it has many momentums and directions that enable it to interact with the photon on a probabilistic basis. It is not realistic to say the e- is standing still waiting for the photon to come. – PhysicsDave Mar 21 '22 at 00:19
  • @PhysicsDave This is just the 1 page not the article. There are the Magyar Mandel, and Louradour exp. on 2 lasers interference. The observable effect of interference is possible only by very coherent lasers (e.g. there must be indistinguishability of the photons). This is also seen in Hong Ou Mandel exp. When there is not good coherence there are beats of the frequencies observed by many. – Mercury Mar 21 '22 at 08:48
  • Here's a good paper: The Twiss-Hanbury Brown Controversy : A 40-Years Perspective by P. Goodman, W . S. Langer and S. P . Brumby. A quote states (3). " since the frequency difference between the two beams is greater than the band-width of each beam, each detected photon is 'Dichromatic': its localisation within an interference fringe results in our inability to ascribe it to either (laser) separately". – PhysicsDave Mar 24 '22 at 00:30
  • @PhysicsDave I took a ride through some threads about interference here and found this one - Does a photon interfere only with itself? The answer of Lubos Motl is categorical. Quote 1: "A single photon can only interfere with "itself". However, "itself" is ill-defined because all photons are identical in quantum mechanics. Because of their Bose-Einstein statistics, the wave function of all photons is symmetric - invariant under all permutations of the individual photons. So the states in which some photons are permuted actually do interfere with each other - the symmetry must be preserved." – Mercury Mar 30 '22 at 10:07
  • @PhysicsDave Quote 2: "If those variables include positions, the interference can only occur in between states whose all other quantum numbers take identical values." So if two photons have same wavelength and polarization they will interfere. I think if the lasers are tuned exactly they will be interference. Where does a certain photon start from (laser 1 or 2) seems irrelevant to me. – Mercury Mar 30 '22 at 10:08
  • Lubos answer is here https://physics.stackexchange.com/questions/6234/does-a-photon-interfere-only-with-itself and your original question is here https://physics.stackexchange.com/questions/696469/can-the-spot-of-a-laser-on-a-plane-be-extinguished-by-another-laser. Your question was about destructive interference. So we have to be very careful about implying that photons can erase each others energy .. that's not possible. – PhysicsDave Mar 31 '22 at 19:07
  • The simplest meaning of interference in the DSE implies cancellation of energy which is not possible. For the DSE Dirac and others stated the photon interferes with itself .... the modern meaning is that each photon determines its own path ... this interpretation explained single photon "interference". – PhysicsDave Mar 31 '22 at 19:17
  • There is Hong Ou Mandel "interference", there is 2 photon Glauber "interference" and probably a few others. All of these modes involve electrons creating/detecting EM waves in the EM field. In the HOM effect the photons take the same path .... interference is that one photon followed the other based on favourable (least action) conditions in the EM field ... – PhysicsDave Mar 31 '22 at 19:19

2 Answers2

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A photon is not a particle in the classical sense of the word.

A photon is a quanta of the radiation field.

Suppose you've got your sinusoidal Electric field, Suppose as you fix the energy say, hv, then the phase is maximally unknown.(Quantum Mechanics)

In a way a photon is much more similar to a field than a particle.

The emergence of the classical Electric field from photons can be understood in the "coherent state" representation of the Hilbert space.(The point is that, the field becomes classical in the limit where the average number of photons of a favored mode goes to infinity and the field becomes similar to a classical sinusoidal wave in the favored mode)

Bastam Tajik
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  • It is well known that a photon has momentum forward. Where it is is also known at the moment of the interaction (electron accepts it). The question is why the electron should move up-down when the momentum is forward. Mathematics just makes the things obscured. – Mercury Mar 14 '22 at 08:29
  • Well it's a good question. For wavelengths below the Compton wavelength of the target, the wave like behavior of the photon is dominant, and when the Compton wavelength of the target is greater than the wavelength of the incoming photon then the particle like nature of the photon is dominant. That's the very cocept of particle-wave duality that can be unified in the context of any QFT including QED. – Bastam Tajik Mar 14 '22 at 08:44
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Photons in mainstream physics, are quantum mechanical entities which in great numbers build up the classical electromagnetic radiation, which is what you are describing with the changing fields.

This picture for circularly polarized light gives an intuition how the classical wave is built up from a quantum mechanical substrate of photons.

polphot

One needs the mathematics of quantum mechanics in order to really understand this.

Also this double slit experiment can give an intuition how photons, when detected one by one leave a footprint of a particle on the screen,

enter image description here

Figure 1. Single-photon camera recording of photons from a double slit illuminated by very weak laser light. Left to right: single frame, superposition of 200, 1’000, and 500’000 frames.

but in a large accumulation the classical interference pattern of the wave nature appears on the right, connected with the probabilistic nature of quantum mechanics.

One needs to study quantum mechanics to master this.

anna v
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  • Do you know why the photon which is hitting forward is causing an electron to move up-down e.g. perpendicular to the push F=eE. Do really the field jiggle up-down as represented (because of the mass of the electron it is quite impossible to follow the speed c of EM wave)? Maybe the picture of EM is illusion. Electron can not follow it. – Mercury Mar 14 '22 at 08:38
  • You need quantum mechanics for electron photon interactions . for beam of electrons and laser beam see https://uspas.fnal.gov/materials/08UMD/Lecture5.pdf – anna v Mar 14 '22 at 09:41
  • In QM the electron is not afaik free but in atom and there are transitions li>-> lj>. I think you mean QED but it is also not useful because there only Feynman diagrams and the momentum is forward, which is not what EM is pictured. I'm not interested in dense fields and electron jets, but a single electron + monochromatic asap EM wave. – Mercury Mar 14 '22 at 12:18
  • google "feynman diagram electron photon scattering" and you will see how many QED diagrams exist depending on the energies involve between free electrons and free photons. – anna v Mar 14 '22 at 12:44
  • "perpendicular to the push F=eE." at the level of photons there is no push, as the photon has no electric field, just photon electron scattering quantum mechanically, only the probability of scattering is predicted, and is seen in the data of the double slits on the right. – anna v Mar 16 '22 at 05:51
  • At the level of photons there are their Energy and momentum. QM or QED (look at Feynman diagrams especially) obey the both conservation laws among others. So when I say push I mean the momentum conservation law. In Feynman diagrams the photon is scattered by the e- and e- receives a subsequent momentum (e.g. push in popular speech). – Mercury Mar 16 '22 at 13:35
  • Of course there are probabilities for all angles of scatter but I think that the mean value is very predominantly in the direction of the photon's momentum (e.g. perpendicular to E). – Mercury Mar 16 '22 at 13:38