# Photons of RF energy

So I start with the understanding that light is photons of electromagnetic energy, and that as the color gets more red the wavelength increases, and that an optical photon has both an electric and a magnetic component. So far so good.

If I were to run an LED at low enough current, I could get one photon per time interval on average at random times. Those photons would radiate away, each in some more or less random direction, at the speed of light, not decreasing in energy until they are absorbed somewhere.

So keep moving down in frequency. It seems that somewhere the photons start acting differently. Assume I have a device that can radiate a photon of RF at some frequency (GHz? kHz?).

According to everything I've read, the energy has to make it out past the radiansphere where it detaches from the source, and then it radiates away in all directions (unlike the optical photon) and while the total energy may remain constant, an antenna or other device of a real size can then only intercept a fraction of the photon, unlike optical photons where as Feynman famously said (paraphrasing from Six Easy Pieces) "They're magic bullets. You can't have half a bullet. You get one whole bullet or nothing) If we had an encompassing sphere around the optical emitter, we'd see the photon strike one tiny point at some location on the sphere, and if we could see RF, we'd see the entire sphere glow slightly as the RF hit it.

• So obviously the photons and the radio waves are doing the right things, what am I misunderstanding here?
• Is it simply that RF isn't Photons? Why? Where and how does this transition happen as we decrease in frequency?
• Jan 4, 2020 at 19:10
• Here's a related question on the similarities & differences of optical vs RF photons: physics.stackexchange.com/q/482541/123208 Jan 4, 2020 at 22:40

Those photons [from an LED] would radiate away, each in some more or less random direction, at the speed of light, not decreasing in energy until they are absorbed somewhere.

This isn't quite right. Photons don't move like bullets or billiard balls. They propagate according to the wave equation. They diffract, just like classic electromagnetic waves. So we can't say a photon moves in a particular direction.

And when emitted from an LED, their state is uncertain so that they in fact propagate in all directions, not in some well-defined direction.

then it [an RF photon] radiates away in all directions (unlike the optical photon) and while the total energy may remain constant

The RF photon radiates in all directions just like an optical photon, not unlike an optical photon.

It might diffract more strongly because it has a longer wavelength than the optical photon. But that's entirely expected and predicted as much by classical electromagnetic theory as by quantum theory.

Is it simply that RF isn't Photons? Why? Where and how does this transition happen as we decrease in frequency?

It's mainly that the RF photon energy is vastly smaller than an optical photon energy. If you compare an optical photon at 500 THz to an RF photon at 500 kHz, the optical photon has 1 billion times the energy of the RF photon. That means if you have, for example, a 1 mW source of RF photons, its power is divided up into 1 billion times as many "magic bullets" as a 1 mW source of optical photons.

There are also differences in the size of structures that cause noticeable diffraction of RF sources vs optical sources due to the vastly different wavelengths.

And the fact that thermal sources at everyday temperatures are producing much more RF and far-IR radiation than near-IR or visible radiation, so that RF receivers and optical receivers have different background noise considerations.

And the fact that RF wavelengths are similar to the size of antenna structures we can realistically build, whereas building a 100-1000 nm antenna is generally impractical, so we use atomic receivers (photodiodes, for example) rather than things like dipole antennas, to receive optical signals.

So when a photon leaves an LED or other source, and arrives at exactly one spot on a target

I disagree. A photon leaving an LED has entirely uncertain direction of travel. It propagates according to the same mathematics as the classical EM wave equation.

If we receive it with a photodiode, then we think it was moving from the LED in the direction of the photodiode, but this isn't any more true for the optical photon than it would be for a radio photon.

If we emit a radio photon from one antenna and receive it with another one, we could make the same argument that the photons involved must have travelled from the source in the direction of the receiver to be received, but this would be equally incorrect. In both cases the photons were emitted from the source in all directions and only the measurement at the receiver re-localized them at the receiver location.

• Comments are not for extended discussion; this conversation has been moved to chat. Jan 6, 2020 at 17:28
• If a photon is emitted by stimulated emission (laser) could we say it is going in a particular direction. Yes there is a very small chance that diffraction at the laser opening will cause a few photons to appear elsewhere but for the most part they seem to have direction. Jan 18, 2020 at 1:10

All electromagnetic radiation - from VLF to extreme X-ray and beyond, exhibit both wave and particle behavior. What you perceive a lot depends on your set up/test equipment and what you're looking for.

For an interesting read on this, do an on-line search for the 2-slit (or double slit) experiment.

• Ok, I've read that one from multiple sources including Feynman's Six Easy Pieces lectures, which I love. I understand that light and photons both diffract and produce interference patterns, but the photons "shot" at the target don't illuminate the wall behind the photon source. RF can of course be focused into a beam, but that's not the natural state. At low frequencies (100's kHz) a ferrite rod antenna radiates in a donut shaped pattern once the energy clears the radiansphere, and if you enclosed it in the equivalent of an integrating sphere, the whole inner surface would light when hit. Jan 4, 2020 at 20:56
• @user103218 A ferrite rod antenna emits a donut shaped pattern when emitting monochromatically because a coherent source cannot emit isotropically. Common light sources (light bulbs, LEDs, etc) are typically not coherent and so can emit isotropically. Neither is a "natural state", rather different types of sources are more practical at different frequencies. Jan 5, 2020 at 4:45
• @user103218 : Clearly, you need to use a bigger waveguide for your RF (so that aperture diffraction doesn't fool you into believing you had a non-directional source). These are common for microwave radio communications and for all sorts of radar, starting with Barrow's work in the 1930s using 40 cm waves. Jan 5, 2020 at 11:26

To get a radio wave one accelerate electrons in the antenna rod. If this will be done only once, the electrons get accumulated inside the rod and especially its end. During the acceleration each involved electron emits photons, somewhere between X-rays and infrared radiation. This depends from the tuning of the antenna system and the power of the antenna generator.
All the emitted photons are polarized; for a vertically mounted antenna rod the electric field component of the involved photons are oriented vertically and the magnetic field component are aligned horizontally.

Such burst of photons repeats with every phase turn of the antenna generator. What we get is a modulated EM radiation. If you are measuring such radiation at a far enough distance, it is obvious that you will receive only single photons, still with a sinusoidal distribution of the incoming number of photons.

I always wondered, why the representatives of the quantum mechanical model vehemently oppose this view. Let them treat modulated EM radiation as a whole wave. But allow please a more detailed view to this process.

This is a very interesting theoretical issue. At high energies, say above 100GHz electromagnetism has a more distinctly particle nature, explainable in terms of QED. Below 100GHz it takes on more of a wave characteristic more easily understood in terms of Maxwell's formulation. There can be no doubt though as Feynman is reported to have said, it is particles (tiny bullets) all the way down.

The hydrogen line at around 1420 MHz tells us that quantum effects involving photons continue down to low energy levels.the problem is that individual photons below about 100GHz have too low energy to be individually discernable.

For example there are claims the BICEPS receiver at the south pole can detect individual photons to around 100gHz and ensembles of photons exhibiting quantisation down to about 50GHz. Below that energy level individual photons are indiscernible and quantum effects difficult to see. However, provided the photon ensemble is of sufficient magnitude the wave energy can be detected as radio waves.

Whether photons at radio frequencies are point like or have a large diffuse nature is a moot point. I have seen both points being argued by apparently well qualified people. If they are point like particles it is conceptually easy to see how they could impact an antenna and excite electrons in a circuit. The situation with diffuse particles is harder to discern in terms of QED.

Interesting issues are what happens if we increase the diameter of a radio receiver's antenna conductor, would this allow more photons, if they are point like, to strike the antenna? Another interesting conundrum is the high frequency miniwhip active antenna comprising a small plate about 12 cm long by about 5cm wide that provides quite good shortwave reception in the 3mHz-30mHz range, is it intercepting waves or particles? An example of this antenna is used on the University of Twentes high performance on line SDR receiver in the Netherlands. By way of contrast how exactly do we explain the operation of a Faraday loop antenna normally explained by Maxwell's equations, in terms of QED.? No easy answers that I can see.