1. Jan 13, 2010

### tiredryan

Hello. I'm trying to understand electromagnetic radiation. Any help would be appreciated.

From my understanding a photon is a basic "unit" of electromagnetic radiation with an energy corresponding to hv. Additionally radios transmit electromagnetic waves by sending an AC through an antenna. The amount of EM waves transmitted is a function of the power supplied by the radio. As a thought experiment, if we continuously reduce the power supplied to the antenna, will we reach a point where there would be only one photon emitted (corresponding to when E=hv; P=hv/t)? Then will no radio waves be emitted below that power? What will happen if the antenna receives 1.5 times the power (when E=1.5hv; P=1.5hv/t); will there be only 1 photon emitted and the remaining energy lost as heat?

Thanks.

2. Jan 13, 2010

### Born2bwire

Yes, if you reduce the power enough you will experience granularity with the signal due to the fact that the radiation must be emitted using discrete quanta of energy.

In answer to your other questions, one thing to note is that the antenna is actually excited by an electromagnetic wave. Any AC signal in an electrical circuit is propagated by electromagnetic waves. These waves excite the currents in the circuit (or the voltages that cause the build up of charges). So for you to send any amount of power to the antenna you will have to send at least a photon's worth of quanta otherwise there will be no wave to excite the currents on the antenna. The antenna itself will dissipate energy from conductive losses. In addition, efficiency will be reduced by reflection of energy off of the antenna due to impedance mismatch and further reduction in radiated power is due to the fact that there will invariably be energy trapped in the near field of the antenna as opposed to propagating out into space.

These efficiencies, on a quantum level, will probably be modifiers to the statistical behavior of the transmitted signal. It is really no good to talk of the photons on an individual basis as any meaningful analysis and measurement will be done on a statistical basis, a large number of samplings. So in some instances the photon that is sent to the antenna will not be transmitted, it's energy will be lost as heat due to conductive losses. Or it's energy may be trapped in the near field. But on a statistical set of photons sent I would imagine that you would start to see measurements that would reflect the effective losses and inefficiencies of the antenna. Still, this is a bit of mixing with classical theory and when it comes to just singular photon behavior we need to think purely in the quantum regime for true accuracy.

3. Jan 13, 2010

### Bob S

It is probably useful to consider the 1420-MHz (L-band) microwave signal as an electromagnetic wave, and also as the quantum hyperfine transition in hydrogen (like in interstellar gas). On one hand, antennas can easily transmit and receive these electromagnetic waves. On the other hand, hydrogen atoms can receive and emit the signal only as discrete quanta (photons). As the voltage on a 1420-MHz transmitting antenna is reduced, eventually the antenna will transmit a few photons per second.
Bob S

4. Jan 13, 2010

### f95toli

It is also worth pointing out that one common way of testing single photon detectors is to use a conventional RF source but then attenuate it (a lot), if the source is attenuated enough it will eventually start to behaving as something that is almost like a "true" single photon source (which it can never really truly be since no conventional source is capable of generating a number state, but this is a technicality) and the received signal will start following Poisson statistics.

5. Jan 13, 2010

6. Sep 7, 2011

### particle7

I know this is an old thread, but this question is really bothering me.

Recently I thought I'd see if I could do some simple quantum experiments at home. You can buy single photon detector tubes (ye olde worlde- style vacuum tube type things) quite cheaply, and lasers these days go for less than a good book.

But after thinking about it a bit, I realised to my horror that neither the version of Young's double slit experiment where one photon at a time is used to create an interference pattern, nor the type of photoelectric experiment originally explained by Einstein, would prove whatsoever that photons exist (whatever "exist" means). Both experiments merely prove that light is emitted and absorbed in quanta.

Now I understand why Einstein never accepted quantum mechanics fully. Looking into it further, I see that other experiments have been done ("anti-bunching" experiments, entanglement, etc) that purport to prove the existence of the photon.

However, leaving that whole kettle of worms aside, I'm wondering if any experiments have actually been done that show the quantum nature of radio waves, since experiments demonstrating the same thing with visible light are already tricky to do and often dubious in nature.

[By "dubious", I don't mean to be provocative here --- it's just that there are two types of photon experiments; ones that are suggestive of photons existing but do not constitute logical proof, such as Young's Double Slit with faint light, and ones that are claimed to constitute proof such as the anti-bunching experiments, and I want to ask if either type can be done with radio waves]

As I understand it, visible light is typically produced or absorbed by electrons undergoing quantum transitions; however, radio waves are typically produced by effectively delocalised electrons in metal wires being shifted about. Isn't this the case? So I imagine that quantum-type effects would be far more difficult to show with radio waves, if radio waves lack sufficient energy to trigger electron shell transitions.

Anyone care to comment? Are we just taking it on faith that radio waves are quantized simply because higher-energy EM waves are quantized?

Last edited: Sep 7, 2011
7. Sep 7, 2011

### Bob S

Intergalactic hydrogen atoms (density about one per cubic meter) can absorb and emit single radio-frequency (1.4 GHz) photons. This is a quantum level transition, due to the hyperfine structure in neutral hydrogen atoms.

The best single photon detector I know of is a vacuum photomultiplier, with a quantum efficiency of less than 50%. This is for visible photons.

Bob S