I cannot comprehend how the increasing of amplitude of an electromagnetic wave increases the number of photons. How does this even happen. I am also not able able to make sense of that fact that, when you increase frequency, you increase the energy of each individual photon. How does this happen? Is there a better way to visualize this
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Related: How do you visualize a quantized electromagnetic field? – Ruslan May 25 '21 at 08:25
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The second statement is a direct postulate of Planck's Quantum Theory. Perhaps you can visualise this in the way that a highly energetic particle will be able to make rapid oscillations and hence have a higher frequency. – Ambica Govind May 25 '21 at 11:02
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I very rarely think about changing the energy or count of photons once they exist, only what they are when generating the photons. Is there a situation you are specifically concerned about where you change the energy or count of existing photons? – Matt May 25 '21 at 19:58
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When you increase the amplitude of your light source while keeping the frequency same, you are essentially increasing the power output of the source. However since the energy of individual photons(which depends only on the frequency of the photons) is the same, the number of photons increases correspondingly.
Use:
$\mathrm{P}\!\cdot\!\mathrm{t} = h\nu n$
Where P is the power of the source, n is the no of photons, h is plank's constant and $\nu$ is frequency of incident electromagnetic radiation.
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Strictly speaking, the amplitude of the field (meaning the operators of the electric and magnetic fields, see here) and the photon number do not commute. Thus, the claims that increasing the amplitude increases the number of photons or vice versa are technically not correct.
The state with definite photon number will have uncertain amplitude, whereas a state with well-dfeined amplitude (a coherent state) will have high uncertainty in the photon number. One however can calculate this uncertainty, which is given by a Poissonian distrubution (see here), and for high amplitudes proportional to the field intensity: $$ (\Delta n)^2\propto E^2 $$
Roger V.
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