Hi Franz.

I don't know if you consider that photons could be permanently localized electromagnetic particles, but if you do, in the 3-spaces model, the photon energy is directly related to its frequency, and also to the transverse amplitude of the electromagnetic oscillation of its quantum of energy.

In all cases, this transverse amplitude can be calculated from the transverse wavelength of the photon's energy: (lamba alpha)/(2 pi).

The electromagnetic transverse wavelength can be used to calculate the energy of any permanently localized quantum of electromagnetic energy:

E= e2/(2 eps_0 alpha lambda)

or:

E= (mu_0 e2 c2) /(2 alpha lambda)

hi,

           can't we directly co-relate using the standard relation of the intensity of a light source (No. of photons per unit area per unit time) with the amplitude square ????

Regards

Amit

Hi Amit,

If your comment is about the 2 equations I gave, they are not for standard measurement of the number of photons crossing a unit area per unit time. They are for calculating the energy of individual light quanta.

The are equivalent to E=hf or E=hc/lambda, but without the need to use the Planck constant h, since they directly use the transverse amplitude of the electromagnetically oscillating quantum.

Franz, let me add a comment to the former answers:

Lets have a propagating wave. When it is detected, you can think of such detection either as a photon counting process or as an measurement of irradiance (flux per unit area), this last being proportional to the amplitude squared.

Then, if you reduce the amplitude of the wave to one half, you get one half or the photons, or you get an amplitude reduced to 1/sqrt(2) of the original. If you keep reducing and reducing the intensity of the beam -for the same frequency- then the measurement becomes a problem. Amplitude can be defined for arbitrarily low values -based on inverse square distance laws or other considerations- but detection of photons becomes an statistical issue and the actual measurement of the magnitude is no longer a straighforward task.

And, by the way, if you double the frequency of a propagating beam but keep the same amplitude of the wave, what you have is... less photons. ;-)

The wave and the photon are in no direct link. First, please note that photon is not a particle. It does not "exist" as something with properties. It is an EVENT. (And from what we know from quantum entanglement it is not even a localized event, but that is another story.)

Next, wave might be a single frequency or multiple frequency or time varying in any possible respect (amplitude, phase, frequency, polarization, OAM...), who cares. Now imagine an atom in the path of the wave. Let us say that is has a state of 1 eV which is not excited. If the wave decides to transfer the energy to the atom, and atom absorbs it by getting excited, the amount of energy transferred will be 1 eV. This event of energy transfer is called "photon". Some call it "click in the detector" but I'd say that the click is a consequence of a photon not a photon itself. Of course, to have a non-negligible probability for a photon of 1 eV the wave must contain a component of any of the frequencies (n + 1/2) times 1 eV. Now, n is related to the amplitude of that component, and the minimum required to make a 1 eV photon is n>=1.

Since we only have classical logic to describe our thoughts, or worse yet - to understand the Nature, and the Nature is quantum, I believe that the above explanation is the best you can get.  (If you a find better one, let me know, I'll be grateful.). 

Dear Mario,

"Now, n is related to the amplitude of that component,"

Before you take my words as dogma - I have no practical physicistic experiences - I am an Informationscientistist (only)...

What you mean is in my knowledge the spin of an electron. In BOHR's Model the different energylevels are known in that way. A photon would be in your understanding a "something" which can transport energy differeneces from one energy level to another.

Not definable would be: all "properties of a photon" - like frequency or amplitude of a wave, polarisation or (even) spin. I can agree that imagination. We have to wait till we know more....

When we mention energy of an electromagnetic wave we always say it's energy in terms of Amplitude square , and say it's frequency independent.

However , when we mention this in quantum mechanics it's enertgy is hf and its frequency dependent. And we say higher frequency means higher energy, now we no longer relates it to it's amplitude???

Is everyone trying to answer the same ?

Subhasish, think of 10 particles at 1 eV in a certain volume of space. Now think of 100 particles at 1 eV in an identical volume of space. The particles in both volumes are all at the energy of 1 eV, but the total amount of energy in each volume is represented by the amplitude as described above.

So yes, the horizontal axis is what the energy (wavelength, frequency, etc) level of the field is at, while the vertical axis is the amount of energy at that level.

That was very crude, but I hope accurate and sensible enough.

Dear Maa'm ,

Higher no. of particles means higher chance of observing a Photon/EVENT ( Amplitude square is high), understood.

But how to think or relate increasing frequency , keeping Amplitude constant means lower chance of observing a Photon/EVENT?

Is there any specific probability distribution explaining this phenomenon?

You do not understand your TV!!!

https://en.wikipedia.org/wiki/Amplitude_modulation#:~:text=Amplitude%20modulation%20(AM)%20is%20a,such%20as%20an%20audio%20signal.