I am confused by something I have seen discussed but not with full clarity: it has to do with understanding antenna behavior at the quantum level for its own sake (yes, classical approach is easier but that is not what is intriguing me).

If photons are created only by electron energy transitions, which always result in very high frequency photon emissions, how can the atomic transitions excited by a radio source connected to an antenna cause the emission of “radio frequency photons”? Are those a popular mis-conception?

This is to say: if the antenna material is excited at 100 MHz by an AC source, the photons cannot be 100 MHz photons (E = hv), they would be in multi-THz range, dictated by the energy bands of the antenna material.

So this propagating radio wave must then be a train of high-frequency photons leaving the antenna material with an intensity modulated at 100 MHz? That modulated intensity train is received by photon absorption in the receiving antenna in kind, giving a 100 MHz signal current.

Correct? Then why is there a bunch of literature saying that radio waves can be seen as ‘radio frequency photons’? Is it all wrong?

And how exactly does QED explain this? I have not studied QED and QM closely, I come from electrical engineering and microwave engineering. Could you share a good lecture or reading source? I believe this is leading me to study how photons could be emitted in a modulated ‘packet’ so to speak.

  • $\begingroup$ Inter band transitions aren't the only way to produce photons. $\endgroup$ – The Photon Jun 22 '18 at 4:53
  • $\begingroup$ This is the main thing I want to understand. By what mechanism do electrons in a conductor emit photons with same frequency as the current excitation? And why is it that a continuous spectrum of photon frequencies is allowed (since obviously we have antenna working over broad continuous bandwidths) ? What is a good text describing this? $\endgroup$ – shawarma_king Jun 22 '18 at 14:44
  • $\begingroup$ See What are photons, EM radiation and radio waves $\endgroup$ – HolgerFiedler Jun 22 '18 at 15:13

The following is a simplified picture and as such I invite the experts and specialists here to offer their perspectives.

As pointed out by The Photon, electron transitions between orbitals within the electric field of a nucleus are not the only way of producing photons.

Any time an electrically charged object is accelerated, an electromagnetic field is propagated away from that object. A radio-frequency emitter (we'll call it an antenna) does not emit electromagnetic waves because of atomic-level electron energy transitions occurring within it. For practical purposes, an antenna at radio frequencies is conveniently approximated as an impedance-matching device which couples a source of high-frequency AC, high-voltage electrical current flow with the characteristic impedance of free space.

In the frequency range of DC to ~near-infrared microwave, the most convenient way of representing electromagnetic radiation, understanding its behavior, and calculating its effects is hence the wave model (via Maxwell's equations) and not the photon model. This does not mean you cannot in principle imagine radio waves as consisting of lots of photons with extremely low energies, it just means you don't have to in order to calculate.

Once you are in the infrared frequency region and above, the opposite is true: although you can certainly model IR radiation as consisting of waves, it is usually more convenient to deal with it as if it consists of photons (with high energies) and calculate according to the rules of QED instead.


Consider that you and I are standing facing each other and we both are holding electric charges of sufficient quantity and opposite polarity that they are attracted to each other due to the basic Coulomb force. We allow up-down and left-right movement, but no movement along the imaginary line that connects us. I raise my charge up and your charge follows, I lower it down and your charge follows. I move it to my left and your charge follows to your right.

If I wave it back and forth, your charge also wobbles back-and-forth. I am, effectively, a transmitting antenna and you are a receiving antenna. If I move my charge left-right a million times per second, your charge will also wiggle a million times per second and we should be able to tune this in with an AM radio. If I wag my charge back-and-forth 100 million times each second, you should be able to tune it in with an FM radio. If I move it back-and-forth 500 trillion times per second, you should see it as a blur of orange light.

That is what EM radiation is. If there was an astronomer equal distant from us both and this astronomer sees my charge move and your charge move later as a result, that time difference is the distance between us divided by the speed of light.

It's the same song and dance if you and I are as big as gods and we're both holding a planet that attracts the other. I can wave it back-and-forth and that would be emitting a gravitational wave that propagates at the same speed of $c$.

So, in the conductor of your antenna, there is charge in the the outer shell of free electrons of the metal atoms. All these outer shells with free electrons sorta comprise a "common" area where electrons and atoms change partners effortlessly. Hooking that up to a transmitter forces those electrons floating around in the "common" to slosh back and forth. This is why you normally need your antenna to be roughly $\frac12$ wavelength long, so that charge can slosh to one end of the driven antenna element and then start sloshing back, as driven by the transmitter connected to the antenna. same thing with the receiving antenna, to be efficient, the receiving antenna element should allow charge to slosh back-and-forth the length of the element in a single cycle. Then, similarly to us two people holding a "free" charge in our hands and one of us driving it back-and-forth and the other allowing it to respond, that is the mechanism of how antennas transmit an EM wave and detect or receive that EM wave.


The core problem with your treatment is this:

they would be in multi-THz range, dictated by the energy bands of the antenna material.

The movement of electrons that enacts the antenna current does not correspond to transitions between different bands in the conducting material of the antenna; instead, it corresponds to driven transitions between different states within the conduction band.

Thus, if you want to move this over into a QED formalism (a move which is generally speaking unnecessary, as real experiments will normally not be able to discriminate between the predictions of QED and classical electrodynamics, but it's still something we need to be able to do even if we don't use it) then the relevant transitions are those between the different states inside the band. These are very much different states, with orthogonal basis wavefunctions and all the necessary machinery, but (because they form a band) the energy differences between these states form a full continuum within the band, and there will be plenty of density of states at any central wavelength in the radio range that you care to name.

  • $\begingroup$ I think this is the key. How do I learn more about this continuum of states as compared to the radiation caused by transitions among orbitals of an atom? Is QED necessary to understanding it? $\endgroup$ – shawarma_king Jun 22 '18 at 14:54
  • $\begingroup$ No, QED is not necessary. Any textbook on solid-state physics should do. Ashcroft & Mermin is a solid starting point; this thread has more recommendations. $\endgroup$ – Emilio Pisanty Jun 22 '18 at 15:31

Antenna neither radiates/transmits nor transforms energy. It just directs energy. It is the whole system, Tx/Tx-Line/Antenna that transmits. At all times, the energy that is transferred onto an antenna follows the Poynting Vector's path, which is, at all times, tangential and along the Perfect-Emitting-Conductor's surface and thus DOES NOT radiate any energy at all. And as a matter of fact, antenna's elements themselves CONSUME that energy and heat-up. However, at points onto antenna dipole's elements where currents go from max to min and the other way around, skin electrons are -+ accelerated at all times. Photons generation follows as a consequence of that. According to the Principle of Least Action [Feynman, Lectures on Physics, Vol-II, Chapter-19], the photons generated ought to follow the lowest energy consuming direction into the surrounding space [photons lose energy due to the 120π resistance of empty space] Those trajectories are eventually loci of the values of the EM field emerging from the generated photons. And that is a classical Physics explanation.

However, space is not empty and as such quantum fluctuation happen at all time. Near an antenna, at the points where voltage goes from max to min e.t.c., a voltage gradient is generated between surface charges and quantum potential of the surrounding space of the antenna. That gives birth to virtual photons that interfere about the geometry of the antenna and according to its geometry form the respected fridges patterns that propagate into free space as plane waves. The so called near-field of an antenna is the space where virtual photons emerge and form the final radiation pattern. And this is the QED explanation of the radiation.

Hope this helps. John


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