Photons and physical interpretation of Fourier transform

[Jarl]

When we study physics at the faculty we are told that any non-sinusoidal wave can be regarded as a combination of sinusoidal waves of different frecuencies, with the ‘weight’ of the different frecuencies given by the Fourier transform. On the other hand, if we have an electromagnetic wave, we know that it is the combined result of many photons of different energies, and it is generally assumed that the interval of the Fourier space where Fourier transform is not zero gives us the energies of the photons in the beam, through the famous relation E=h·f.

However, if we have a radiofrecuency wave (for example), and we switch it on and off abruptly, we introduce much higher frecuencies in the Fourier transform ¿does it mean that we’ll have photons at these high frecuencies?. It is hard to believe so.

A more clear case, a continuous (cuasi)monocromatic wave source, a detector, and a shutter between them. For the source, the radiation have only one frecuency, but after the shutter, the Fourier transform of the wave has also higher frecuencies. The photons, however shouldn’t have more energy.

So ¿in which cases or under which conditions can we relate the ‘matematical’ frecuencies given by Fourier analysis with photons energies?

 

[berkeman]

Switching the beam on and off does not generate higher frequency components, it generates low-frequency components at the rate that you are switching the beam on and off.

 

[Jarl]

Switching the beam on and off does not generate higher frequency components, it generates low-frequency components at the rate that you are switching the beam on and off.

If we allow the wave to pass only during some short time (i.e, only a pulse) we introduce high frecuencies in the Fourier transform.
Anyway, the Fourier transform is modified in some way, while the photon energies shouldn't change.The question is about it.