Is the actual energy of a photon ever measured? How is it done?
I read that a photon is usually identified by diffraction, that means its wavelength is measured, is that right? In this way we determine that a red-light photon makes a full oscillation in roughly $700 nm$Then, by theoretical reasons, we deduce its energy is about $400 THz$. I wonder if, when this relation was established, ther was an instrument capable of precisely counting $4*10^{14}$ oscillations in an exact second.
What I am trying to understand is what exacly is a photon, when we can talk about the existence of a photon. I'll try to explain what is my problem:
we usually receive light (or radio waves) from a continuous source: the Sun, a flame, a bulb etc…, right? how/when can we isolate a single photon? Is a photon the whole set of oscillations during the span of a second? can we consider a photon of red light a single oscillation of the EM field lasting only $\frac{1}{4*10^{14}}$ second, even if no instrument would be able to detect it? Yet, that photo would propagate all the same at $c$ in vacuum for a second or more or forever, but, though its wavelength its still $700 nm$, its energy would probably not be in the region of THz anymore, or would it? To frame it differently, does energy change if the unit of time is halved or doubled?
I hope you can understand my questions even though my exposition is confused.
Edit
The energy of anything is a definite issue, it is the result of a meaurement, and does not depend on theoretical considerations, QM, classical or other models, or onthe fact whether it is a wave or not.
I'll try to clarify my main concern with a concrete example: consider something you can control and manipulate. You can produce low frequency EMR (short radio wave about 10m wavelength) making a charge oscillate up and down 30 million times a second, right?
Now, suppose you make that charge oscillate only for 1/1000 th of a second. all the same That wave will propagate at C and will oscillate 3*10^7 times a second and will be diffracted revealing a wavelength of 10m, is that right or will it oscillate only 30 000 times a second?
Furthermore, whenit hits something or you determine its frequency, will it still have the same energy of a wave that has been produced making a charge oscillate for one whole second or for ten seconds?
Best Answer
The classical light beam, an electromagnetic wave, emerges from zillions of photons which travel with velocity $c$ and build it up.
The energy of a photon is $E=h\times \nu$, where $h$ the Planck constant, and $\nu$ is the frequency which will appear in a classical wave built up by this energy photons. The way this happens is explained mathematically here, but is not simple to understand without quantum mechanics and field theory. The photon itself is not oscillating in (x,y,z,t). Only travelling with velocity $c$.
The energy of the classical wave is given by the average intensity, for example , for a plane wave it can be written
$$S=\frac{1}{c\mu_0}E^2\overline{\sin^2(kx-\omega t)}=\frac{1}{c\mu_0}\frac{E_m^2}{2} $$
where here $E$ is the electric field of the classical light wave.
The individual energy of $h\nu$ of photons will add up to the energy transferred by the collective electromagnetic wave.
The velocity of the photon is fixed and does not change unless there is an interaction, as in Compton scattering,. It is an elementary particle of the standard model .
The theoretical model, called quantum electrodynamics, is so well validated with experimental data that one can identify the energy of the photon with the frequency of the classical light beam, and use classical interference set ups. Individual photons from known frequency light beams have been observed through the double slit experiment, as dots on a screen. The existence of photons and their frequency with energy association is well validated.
When energies become large , as in X rays and gamma rays, different laboratory techniques can identify the energy of a single photon, as with the photoelectric effect, and with the electromagnetic calorimeters in particle experiments identifying single gamma rays of great energy.
For example this Higgs to gamma gamma event:
The green lines display the energy deposited in the calorimeters by each gamma. It is known as a photon because it does not interact in the tracking chambers, and deposits the energy in the electromagnetic ones.
so to your questions
Yes, as seen above.
No. as said the photon is an elementary particle and the classical light beam rides on zillions of photons , each contributing in synergy a tiny part to the electric and magnetic fields of the electromagnetic field.
No the energy of the photon is always equal to $h\nu$, for the whole spectrum. $ν$ is just there for a photon only as a handle to inform what type of light beam a zillion of such energy photons will generate, see the table.
There is a mathematical quantum mechanical connection between the classical electrodynamic solutions and the quantum mechanical solution, because maxwell's equations are quantized and give rise to the wavefunction of the photon. In the wavefunction, which is complex, i.e not measurable, there are the same E nd B and $ν$ that will be built up by zillions of photons. This is the frequency appearing in the double slit experiment , one photon at a time. It is in the probability of detection of the photon that the frequency plays a role.
You ask in a comment:
The units are not understandable, but no, an oscillating charge will not give one photon, it will depend on the boundary values and will give a spectrum of classical frequencies which will be composed of innumerable photons.