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Since any source of light will have a finite duration, the light emited won't have a particular frecuency. It will be a sum of different frequencies (infinite, I think) if we apply Fourier's series (integral).

Would this mean that any photon's frequency will have some uncertainty or something like the photon emitted would be a combination of different photons?

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duplicate of physics.stackexchange.com/questions/68506/… –  Ben Crowell Jun 22 '13 at 19:55

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If you look at it from the quantum mechanical point of view you might invoke the energy-time uncertainty relation, which for a foton might become:

$\Delta\omega\Delta t \geq \frac{1}{2}$.

Which says that for very short times the uncertainty on the frequency can become very large.

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Is this a good interpretation of the energy-time uncertainty? I only know that $\Delta t$ refers to the evolution of the Hamiltonian. –  jinawee Jun 22 '13 at 19:31
    
This mechanism is, in fact, how very short particle lifetimes are measured. You find the width of the line in energy space and apply the HUP in it $\Delta E \Delta t$ form. –  dmckee Jun 22 '13 at 19:38
    
The $\Delta t$ refers to the amount of time it takes for the expectationvalue of an observable to change by one standarddeviation. So it depends on what observable you are looking at, which in your case probably is the amplitude of the light in a given range of frequencies. –  Nick Jun 22 '13 at 19:46

I don't think it is a combination of different photons. I think the uncertainty of the frequency is lower if the duration is longer.

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