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I don't understand how two photons of the same frequency can have different amplitudes, neither how to produce them.

I know that classically the square of the amplitude is proportional to the energy, but photons aren't classical particles.

My understanding is that a photon's energy is $h\nu$ - what does the square of its amplitude represent, then? Are there bounds to the amplitude of an EM wave?

Take two waves of amplitudes $A_1$ and $A_2$ and frequency $f_0$. If $A_2 = 2A_1$, can the wave with amplitude $A_2$ be said to carry/be two $A_1$ photons of frequency $f_0$?

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You are mixing classical and quantum descriptions and that is a messy tar pit where few venture. But the important thing to get here is that the amplitude of the electric field oscillation is a wave property and does not have a simply analogue in photons. The phrase you are looking for is "wave packet", but don't expect it to come with easy math. –  dmckee May 6 '12 at 1:22
    
A recent related question: physics.stackexchange.com/questions/27809/… –  dmckee May 6 '12 at 1:26
    
@dmckee I don't fear the math. Maybe the wording of my question makes me look like a poor bloke, but I'm really just struggling with the concepts - I'm a student of chemistry, and such physics questions are pretty much avoided in the curriculum. –  CHM May 6 '12 at 1:27
    
@dmckee I've read the question. Doesn't answer how such photons are produced :/ –  CHM May 6 '12 at 1:28
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3 Answers 3

I am making this an answer because it is too long for comments and in a sense does answer with a reference.

This is a wrong idea of photons that you have:

I don't understand how two photons of the same frequency can have different amplitudes, neither how to produce them.

Two photons of the same frequency have the same "amplitude", since the only thing they have to differ from each other is frequency. Otherwise they are identical. They are the quantum substructure of classical electromagnetism. The macroscopic electric field of a wave consisting of photons does have an amplitude which is statistically built up from the individual photons.

@Lubos Motl has an extensive article of how classical fields are built up from the quantum substructure, and mathematical ability is necessary to understand it. As @dmckee says wave packets come in.

But he also uses the electromagnetic field in a simple example.

The electromagnetic example starts at this paragraph:

However, in the rest of this section, I want to focus on another way how to see classical physics of fields emerge out of large ensembles of photons, one that mimics the thermodynamic limit of statistical physics (even in the context of classical mechanics).

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Two photons of the same frequency are indistinguishable except for their spin orientation. When they become an ensemble of photons it is the spin orientation of each with respect to the others that has to coherently sum up to the classical electromagnetic wave. Thus the sum of their energy will add up to the energy of the wave they build up, which will be proportional to E**2, E the electric field. In this sense we can count the number of photons in a single frequency wave. –  anna v Aug 12 '13 at 8:47
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Take two waves of amplitudes $A_1$ and $A_2$ and frequency $f_0$. If $A_2 = 2A_1$, can the wave with amplitude $A_2$ be said to carry/be two $A_1$ photons of frequency $f_0$?

Kind of like that. Replace the word "photons" with "quanta" and you'll be pretty close.

As you may know, the EM field can be broken down into components at various frequencies, or modes, each of which can store energy independently. The energy stored in a mode of frequency $f$ can only be a multiple of $hf$. That's all the formula $E=hf$ means. If an EM wave has energy $2hf$, then it has two quanta of energy at that frequency. These quanta are not the same as what we actually call photons, though. Whereas a quantum of energy is localized to a single frequency, a photon (as we usually think of it) is localized in position, and thus consists of excitations at many different frequencies.

The amplitude of the electric field of an EM wave is proportional to the square root of the energy it carries. And since $E = nhf$ for a single mode, a monochromatic EM wave will have an amplitude proportional to $\sqrt{n}$, the square root of the excitation number of that mode. But remember, excitation number is not the same as photon number, so it's not as simple as saying that the amplitude is proportional to the square root of the number of photons. That's just an approximation that works when the number of photons is large enough that you can treat the system classically.

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Here is a simple calculation to think about. Imagine a generic radio transmitter emitting, say, 1 watt. That's 1 joule/second. Suppose the frequency is 100 MHz. Take the Einstein relation E=hf, where h is Planck's constant. Figure out the flux of photons (number emitted per second). To speak of photons in a QM sense the number of photons needs to be relatively small. The sea of photons emitted by our transmitter behave collectively as a wave: in effect a Bose-Einstein condensate.

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