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If photons have frequency then they must vibrate. So i was wondering, what kind of vibration do photons have? But according to physicists, photons do not vibrate. Can someone explain me, how is that possible?

Qmechanic
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Look at this experiment, which shows in one page what is the meaning of frequency for light, electromagnetic wave, and what is the meaning of fequency for photons, elementary particls.

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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.

The frame on the far right shows the classical interference pattern for electromagnetic waves of frequency $ν$ of the experiment.

The leftmost frame shows the footprints of the individual photons, a particle hitting a small area on the screen.

On the left it looks random. On the right one gets the distribution of the probability for photons of energy $hν$ to be at the (x,y) of the screen. It is the probability that waves for the individual photon,, the individual photon leaves a small footprint consistent with the standard model hypothesis that it is a point elementary particle with a wavefunction whose $Ψ^*Ψ$ gives the probability of finding the photon at a particular (x,y,z,t).

It can be shown that classical electromagnetic waves and their E and B fields that classically carry the energy of the electromagnetic wave, emerge from the superposition of zillion of photons with energy $hν$.

Similar double slit experiments exist for electrons and other particles and the same reasoning is behind the plots. It is the probability of finding the particle that waves, not the particle.

anna v
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  • This single photon experiment proves that, “A single photon behaves like a particle while many photons together behaves like an EM wave”, am i right? If i am right then, hν < hc/λ, hν ≠ hc/λ. –  Dec 21 '19 at 03:33
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    @user248881 the sentence is correct, I do not understand what you are saying with the equations. As far as the photons are concerned , it is the energy that defines them. The $ν$ is just there by dividing the energy with the constant found by Planck to make the black body radiation formula consistent with data.. Wavelength is also a collectve effect seen in classical EM. In a medium the speed of light may be less than in vacuum, classically. The photons always go with c, only the superpostion of the wave functions changes the velocity fir EM waves. – anna v Dec 21 '19 at 05:44
  • In reference to my sentence, i wanted to say that, energy of a photon is less than energy of an EM wave (hν < hc/λ); and also the energy of a photon is not equal to energy of an a EM wave (hν ≠ hc/λ). –  Dec 21 '19 at 06:42
  • The energy of the electromagnetic wave is given by the average E in the wave http://hyperphysics.phy-astr.gsu.edu/hbase/Waves/emwv.html It is an emergent value summed (cosnsevation of energy) from the zillion of photons $h*ν$. so what you say is not correct. In the picture you can see the addition of energy, one photon hits a few pixels on the screen, no EM wave. %00.000 photons show EM wave srtucture and you could sum the energy in the right most frame, if you took the trouble to multiply $hν$ by the number of photons – anna v Dec 21 '19 at 07:05
  • Is this (E = hν), not the formula to measure energy of a single photon of certain frequency? Or this equation (E = hν) is used to measure energy of zillions of photons? –  Dec 21 '19 at 09:18
  • E=hν comes from the black body formula, the data show that the EM field is quantiised, each quantum' energy equal to h*ν http://hyperphysics.phy-astr.gsu.edu/hbase/mod6.html . It is the classical wave that is measured so ν ιs known only classically. The photons energy can be measured, and then we know that a great number of such energy photons will ake anEM wave of frequency E/h – anna v Dec 21 '19 at 11:24
  • I am sorry but i asked you another question and you gave me another answer. If you want to continue further discussion, please reply my last question (which is above your last comment). –  Dec 23 '19 at 02:47
  • I am answering above. The formula is for connection between quantum and classical. The h is the proportionality constant for this. Only the energy can be measured for a photon, not frequency. . Frequency appears collectively for the probability of detecting a photon, can only be seen in classical light, the accumulation of photons. But I think we should stop the discussion here. – anna v Dec 23 '19 at 05:30
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You are confusing to different concepts. The photon is a particle whereas the classical description of light is a wave, vibrating or oscillating electric and magnetic field vectors. The dual description is the hallmark of quantum mechanics and quantum field theory. The same could be said of material particles like the electron. They have an associated frequency but they do not "vibrate" in the traditional sense.

Material vibrations occur in acoustics and are oscillations in a material like air, water, or elastic solids (there are other examples). Our experience leads us to expect "vibrations" need a medium. For a long time people thought this was true of light and postulated an ether for light waves to travel in. In the late 1800s to early 1900s we discovered that there was no need for this and that we could describe "light" as oscillations in electric and magnetic field vectors. This is a classical mechanics description of light.

Experiments seem to indicate that light (which we thought of as a wave) has particle like properties and that matter (electrons, protons, etc which we thought of as particles) could behave like waves. This led to the development of quantum mechanics in which particle-wave duality is an underlying paradigm.

Any particle has an associated frequency and wavelength related to its energy and momentum. Similarly any quantized field (like light) has associated particles related to its irreducible quantum states.

In the context of QM and QFT any "particle" has a frequency even thought that particle is not necessarily vibrating in the classic sense of acoustics.

  • So according to your view on different particles, an electron should have wavelength, do they? –  Dec 21 '19 at 03:53
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    It is not my view, but part of the foundation of qm. –  Dec 21 '19 at 11:44
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Start by having a read through the answers to Do photons truly exist in a physical sense or are they just a useful concept like $i = \sqrt{-1}$? As explained there, while it's tempting to imagine the photon as a little ball of light this is not the case. Unless the photon is being created or destroyed it is usually in a delocalised state and behaves like an electromagnetic wave rather than a particle. So the vibration is not like a particle bouncing to and fro. It is the oscillation of the electromagnetic field associated with the particle. Note that this oscillation is of the values of a field not a physical motion in space like the wave moving along a string.

All particles have oscillations like this when they are propagating as waves, and the oscillation is in the value of the wavefunction that describes the particle. For particles like electrons the physical meaning of the wavefunction is somewhat elusive, and we cannot simply relate a physical property to the oscillation. So while we can observe the wave like properties in diffraction experiments we cannot simply measure the oscillation of electrons as they pass. Light is something of a special case since the wavefunction is effectively just the complex field described by Maxwell's equations. So when we measure the oscillation of the electric and magnetic fields we are effectively measuring the oscillations of the photon.

John Rennie
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