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Im not so familliar with particle physics and QFT and my logic gets muddy when I am transfering from classical field theory to particle interactions.

Suppose we have an antenna that is hit by an EM plane wave. The plane wave has an oscillating electrical field that excites the electrons in the antenna. The electrons inside the antenna start to oscillate up and down along the antenna according to the direction of the oscillating electric field. Simple enough.

On the otherhand, in particle physics, an EM plane wave consist of photons, as in energy quanta. The individual photons hit the electrons inside the antenna and transfer their energy thus exciting them, according to Compton scattering.

Here is when my logic gets muddy.

The electrons are supposed to oscillate up and down along the antenna according to the electrical field, but on the otherhand the individual photons, hitting the electrons, should move the electrons according to Compton scattering.

How does the photon energy transfer to the electrons determine the direction (along the antenna, up and down ) that the electrons move? How is it determined when a photon hits an electron inside the antenna does it move up or down?

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The electrons are supposed to oscillate up and down along the antenna according to the electrical field, but on the otherhand the individual photons, hitting the electrons, should move the electrons according to Compton scattering.

It can be shown in field theory that the individual photons build up the classical electromagnetic wave. This means that the classical framework of electric and magnetic fields changing in space to produce the classical wave is a good mathematical description. The individual scattering photon is forced by the lattice of the antenna to Compton scatter in accordance with the classical formulas too, producing the current in the antenna collectively.

As an analogy think of the flow of water down a pipe, classically: the individual particles scatter elastically or give up some energy to the walls, but the direction of the flow is down the pipe because of the constraints of its walls. When one has to do with grand ensembles of particles in boundary conditions one has to think through more than how the individuals in the ensemble behave in scattering.

The individual photon interactions, depend also on the ensemble (lattice electrons) boundary conditions.

How does the photon energy transfer to the electrons determine the direction (along the antenna, up and down ) that the electrons move? How is it determined when a photon hits an electron inside the antenna does it move up or down?

It depends on the beam direction and the boundary conditions of the antenna lattice, and of course in energy and momentum conservation. The photons in a beam travel in the direction of the beam and when scattering individually will have the highest probability of giving off momentum in the direction of the beam. Do not forget they have low energy, and travel with the velocity of light, so each individual electron gets very many scatters, in "tune" with the field theoretical build up of the beam into a classical formula.

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The photons are polarised, in exactly the same way that the electromagnetic field is polarised. So the atoms know which direction they are being pushed in by the polarisation of the photons.


When people talk about quantum particles like photons they most certainly do NOT mean little billiard balls. In a quantum sense light behaves exactly like it does classically (it has polarisation, wavelength, coherence length frequency) with only one major alteration:

Whenever the amount of energy that has been absorbed/radiated into the light field is measured, one will find that it is an exact multiple of $\hbar \omega$ (with $\omega$ the photon angular frequency). So if you measure that the energy absorbed is $101 \hbar \omega$ you would say "101 photons were absorbed".

$\hbar$ is a tiny number, so in most cases the "steps" are so small we don't notice (or care) so that the transfer of energy between light and matter appears to be continuous. It is only in experiments with low intensities of light and sensitive detectors that the quantisation of the energy matters.

Aside from this discretisation of the energy the light field behaves just like the classical one you are used to, so it hits the antenna with some frequency, and polarisation, and drives the electric current in exactly the same way you already understand.

(Indeed the Hamiltonian describing the interaction is derived from Maxwell's equations and the classical electromagnetic Greens function: https://arxiv.org/abs/quant-ph/0006121. Section 3.2 of the paper may help clarify my answer (if you like mathematical answers), they start with the classical equations of the light and then they "quantise" them, essentially build-in the rule that the energy transfered will always be measured to be a fixed number of $\hbar\omega$'s.)

As far as I am aware their is no Compton scattering at all in an antenna, Compton scattering is a different effect where photons down-convert to a lower frequency. That is very different to photon absorption in an antenna.

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  • $\begingroup$ Thanks for the answer! Although, I dont know do did I still understand. I am under the impression that when ever a photon hits a free charged particle inelstic scattering (Compton scattering?) happens. As in some of the momentum of the photon is transferred to the free charged particle. As an antenna is simply a conductor with abundace of free electrons, I assumed that this would be the method of energy transfer inside the antenna? $\endgroup$
    – mmm
    Oct 22 '19 at 16:59
  • $\begingroup$ The wavelengths of light that are resonant with the antenna are determined by the antenna's length and shape. This arises (I believe) because the electrons in the antenna are only free up to movements of this size (they cannot leave the antenna), so I think the energy transfer method in the antenna is actually the resonant motion of the electrons, IE they are not free electrons at that wavelength scale. Compton scattering gives you a lower frequency photon, that certainly doesn't normally happen in a radio antenna. $\endgroup$
    – Dast
    Oct 23 '19 at 8:56
  • $\begingroup$ Also, I was thinking about this more and have edited my answer to (hopefully) improve it. $\endgroup$
    – Dast
    Oct 23 '19 at 8:58

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