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If you increase the number of photons leaving a light source you increase the brightness/intensity, so it could be said that intensity is directly proportional to number of photons. However intensity is directly proportional to amplitude squared, so this leads me to wonder that if you increase the amplitude of a light wave, are you just creating more photons rather than affecting each individual photon in some way?

Vishnu JK
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Royalpuker
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2 Answers2

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Yeah, pretty much. When you increase the amplitude of a light wave, you are essentially just sending more photons of the same kind. The energy of each photon is $hf$ (though with some allowance for wavepackets, where each photon will come with as a probability distribution over a range of frequencies), so if you increase the energy flux you need to increase the photon flux.

When you actually dig down on it things are not quite as simple, of course, because the exact value of the electric field is a quantum mechanical variable analogous to the position variable of a quantum harmonic oscillator, and you can't really describe it as a variable with a well-defined oscillation. (In particular, for example, if you know precisely the intensity of the light, then you lose all information about the phase of the oscillations, and vice versa.) However, the electric field is essentially confined to a region that grows as the square root of the number of photons in the mode.

Emilio Pisanty
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  • Emilio is right. The intensity of the field is essentially the mean number of photons; or more precisely, the expectation value of the number operator for the state of the field. – Girish Kulkarni Jul 28 '16 at 16:16
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You are exactly right. The intensity increases the number of photons without changing the frequency or distribution of that frequency. It is that simple, even at the deepest level of understanding, without even getting into quantum fields or modes.

kdo2
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