You need to build up a certain number of photons to get what classically is an electromagnetic wave. For a single photon the photon number and phase dont commute, that is to say if you have a single photon, the uncertainty on the phase is 2pi, ie it's not defined. Classsic texts on quantum describe how classically you arrive at an electromagnetic wave.

Neil

A photon is another way of looking at an electromagnetic wave and it is a quantum-mechanical particle. However, the photon has zero mass and no electric charge.

The relationship between the photon energy, wavelength and frequency is:

p = h/λ, where p is the momentum and h being the Planck constant.

The momentum of a photon (zero mass) is given by p = E/c, and the wavelength by λ = c/f, where c is the speed of light in vacuum.

--> E = hc/λ = hf

You need to build up a certain number of photons to get what classically is an electromagnetic wave. For a single photon the photon number and phase dont commute, that is to say if you have a single photon, the uncertainty on the phase is 2pi, ie it's not defined. Classsic texts on quantum describe how classically you arrive at an electromagnetic wave.

Neil

the photon is an approach to represent and understand the propagation of energy. The photon is invisible but we see its effects. In the visible electromagnetic waves we see the effect of the photon because its energy is relatively high, on the other hand in the hertizian waves the photon is always present but its effect is not visible because its energy is relatively weak.

Dear Richard,

Thank you for all the equations. Do you have any published paper about this?

Thanks

Roland

@ Dear Richard,

I don't think the equations given by Hage are wrong. Just a discontinuity about frequency (sometimes f in λ = c/f and sometimes v in E = hv ). Your equations are also correct.

@ Dear Roland,

If you are looking for any reference. You may refer to Wikipedia first.

https://en.wikipedia.org/wiki/Photon

Thank you.

I edited them just now :)

Thanks to Richard for pointing out the errors!

@ Morten and Richard,

Thank you both. :)

Same as you relate light and photons. In fact, you can think of "antenna aperture" as being analogous to optical lens aperture, when dealing with light. The bigger the aperture, the more photons you can gather. You see this reflected in antenna gain equations and in path loss equations.

So for instance, use the example of RF path loss through free space.

Path loss (dB) = 32.45*logbase10(freq in MHz) +20*logbase10(dist in km)

Why does this equation claim that path loss increases with frequency, through free space? Because for equal gain antennas, say unity gain, the unity gain antenna at higher frequencies is physically SMALLER than the unity gain antenna at low frequencies. Antenna gain depends on antenna design, but in general, is a function of wavelength. So short wavelength, at equal gain, makes for a smaller antenna aperture. The smaller antenna gathers fewer photons. Therefore, path loss appears bigger at the higher frequencies.

(Note that I am using free space here, so there's no excuse of obstacles being in the way, for the higher frequencies to attentuate faster, as there might be on the earth's surface. If obstacles are in the way, sure, higher frequencies won't diffract around the obstacle as well. But in free space, you have no obstacles. And yet, higher frequencies attenuate faster. It's actually a matter of antenna aperture differences, for equal gain.)

The energy and frequency relationships have been given by our other colleagues, and all I can add is, the concept of antenna aperture is best explained if you can relate photons to RF waves, as you do with light.

Albert,

I have been thinking about The Spectrum, and what remains consistent moving up and down the frequency. One thing that is consistent is that there are photons involved from one end to the other. Your "... think of "antenna aperture" as being analogous to optical lens aperture, when dealing with light." brings a new perspective.

This, in turn, has taken me to the land of the single pixel camera, which I am now exploring for inspiration.

Thank you,

Patrick

the obvious question: why do optical waves reflect on material obstacles and this is not the case for the Hertezian waves?

Thanks for the education colleagues

It may be instructive to also consider the physical size limits of a spherically convergent RF/EM field in a dielectric media. The circumference of such an inwardly convergent/focused EM field is limited to the media wavelength. Such a concept may help one consider a different kind of relationship between an RF/EM field and a photon, both with a physical minimum limited size related to the wavelength in the media.

Photons are not smaller than the antenna aperture (unless the beam is). Photons appear to be as wide as the transmitted coherent beam - a double slot causes interference fringes even with one photon and even when the slits are very far apart, provided that both slots are in the coherent beam. A wide antenna aperture detects more photons because the larger aperture increases the probability of the place the photon is detected being within the aperture - each photon is very spread out but is only detected at one place in its spread.

Thank you, but my comment was related to a convergence of an aperture or an array that surrounds the target region causing RF/MW energy to converge where the convergence focus is not able to be any smaller than the size that I had mentioned. One could ask why can't such a focal zone converge to a smaller size than a circumference that is a wavelength of the media? The question I ask is could this relate to a concept of a photo where there is a limit in such a convergent focus inversely related to frequency.

I do not know the wave mechanics answer, but my experience in physical optics indicates that spot sizes less than a wavelength circumference might be achievable by using superdirectivity. This requires a lot of stored energy in the aperture and the generation of (in your case) spherical modes with high mode numbers that are evanescent at the focus, so require large amplitudes. Double negative, or negative refractive index materials achieve small spot sizes by reversing the "loss" of evanescent waves from a sub-wavelength circumference object so that they are still significant at the focus. If you can generate these evanescent waves with enough amplitude then you can get the small spot size you want.

Do a modal decomposition of the spot you want, to find out the spherical mode content (or any other set of modes). A large number of these modes will be evanescent at the focus. If you can get these modes to the focus with the right amplitude (by making them very big farther away) then they will add up to the small spot you want.

A model of photon as a wave packet in the luminiferous medium one can read here: Article On the Nature of Photon

The of " photon amplitude " concept is introduced, through which energy is represented similarly to quantum (phonon) energy of elastic mechanical wave. Photon volume and shape are discussed.

Roland asked the question: "How can we relate photon in electromagnetic waves? What is photon in Radio waves?" Answer to this question: A monochromatic Radio wave is a set of photons in the same quantum state. And the relationship of theirs energy to frequency and wavelength given above.

All the best.

Hi,

E = hf, where E is the energy of the photon, h is the Planck's constant and f is the frequency of the wave/radiation.

Regards,

Suhas. D.

De Broglie defined the characteristics that a localized electromagnetic photon must have to obey Maxwell's theory in all respects.

So, by very definition, if he had it right, it can be connected to frequency, energy and wavelength in the very same manner as any point on the expanding wavefront of a theoretical Maxwell spherically expanding electromagnetic wave:

https://www.omicsonline.org/open-access/on-de-broglies-doubleparticle-photon-hypothesis-2090-0902-1000153.pdf