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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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marked as duplicate by Ben Crowell, rob, BMS, John Rennie, Brandon Enright 2 hours ago

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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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
    
This question has received some correct answers and some incorrect ones. The one that it's a duplicate of has an accepted answer that is correct. –  Ben Crowell 7 hours ago

5 Answers 5

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
    
Two photons of the same frequency have the same "amplitude", No, their wavefunctions can differ. –  Ben Crowell yesterday
    
@BenCrowell I am responding to the question, where amplitude is connected with energy , that is why I put the quotation marks. I give the link where a quantum mechanical amplitude is of course defined, in case the questioner is more sophisticatedin physics than the question. –  anna v yesterday

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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If you insist on thinking of photons as waves (which is fine as long as you ignore absorption, though you should really think of it as a disturbance in an electromagnetic field), you can more or less think of all of their amplitudes as being equal, and this is why your premise doesn't make sense.

More precisely, the amplitude of a single photon isn't constant in space and time, but in a broad sense the "total amplitude" of all photons is the same, which I explain more below. It is important to remember that they are not some infinitely extending wave (which would imply infinite energy) but rather localized probability density which is a wave packet. This is the wave dual of the particle.

(A longer wave packet would have a smaller average amplitude, while a shorter one would have a greater amplitude, but the integral of the squared amplitude will always be 1)

And the amplitude of this wave does not even correspond to any observable quantity. In fact, according to schroedinger's equation, the amplitude of this wavefunction is not even a real number, but rather complex. However, the square of the amplitude IS something meaningful: probability density. When you are talking about individual photons (or any quantum particle), rather than thinking about energy intensity you should be thinking about the probability intensity of being here or there or being in a certain state. When you aggregate many photons together, the amplitudes of their superimposed probability densities would correspond in a sense to a real wave intensity on average, and that would be your connection to a classical idea of amplitude. But again, speaking of a single photon, an "intensity" doesn't really make sense.

The energy of a wave is proportional to the integral of the square of it's amplitude. (Note that it is not simply proportional to A^2. This would be the power, not the energy) This is why the energy of a wave packet of constant amplitude is proportional to the square of the amplitude. But this is not the case with a quantum wavefunction as it is a localized wave packet. Since this constant (the integral of probability of being anywhere in space) is always 1, the average or total square amplitude is constant for all photons so it does not appear in the equation for energy.

Comment on Ben cromwell's answer (sorry I can't comment):

@ben Crowell, while your answer is correct, I don't think it directly answers the OP's question. I want to point out that your second explanation is clearly the more relevant one and is at the core of what I was said in my answer. OP is starting with the concept that a photon can be thought of as a wave or wavefunction, which is true (although this comes with many caveats, the basic idea is true), and waves have amplitude. The key thing we agree on is that the amplitude is not constant over space or even for a given photon wavefunction as it depends on boundary conditions. This is exactly my point: since it must obey normalization, you can think (very roughly) of it's "total amplitude" as being constant and this is why it doesn't affect total energy.

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If you insist on thinking of photons as waves (which is fine of course), you can think of all of their amplitudes as being equal. Not true. If you think their amplitudes are all the same, what do you think it is? –  Ben Crowell yesterday
    
What do you mean by "it"? What defines different phots is their different frequencies. –  Paul yesterday
    
When I said amplitudes are the same, I meant that in a vague sort of way because an individual doesn't even have a constant amplitude of course, as it is really described by a wave function, whose amplitudes are not even real numbers. But in a broad sense this amplitude (or the square of it) must integrate to 1. My point was more to give intuition about what the "amplitude" of a single photon even means, and show how it has a totally different meaning than the amplitude of a certain radiation, which consists of many different photons. –  Paul 11 hours ago
    
"It" refers to the amplitude. When I said amplitudes are the same, I meant that in a vague sort of way because an individual doesn't even have a constant amplitude of course, as it is really described by a wave function, whose amplitudes are not even real numbers. So what do you think the magnitude of the amplitude is? But in a broad sense this amplitude (or the square of it) must integrate to 1. Right, which is not consistent with the amplitude being a fixed number for all photons of a given frequency. –  Ben Crowell 9 hours ago
    
The amplitude is a complex number, whose magnitude has no real interpretation though it does when you square it and integrate over some spatial domain (representing the probability of being in that spatial domain). What I mean by the A being constant is that the "total" amplitude (or total square amplitude) is constant in the sense of integrating over all space. Namely it is always 1. I am trying to give some justification to the common answer that all photons have the same A, and also say that while it probably isn't what the OP thought it is, saying photons don't have amplitude is incorrect –  Paul 9 hours ago

Some of the other answers have suggested that the amplitude of the photon's wavefunction is well defined, and that it has the same value for any two photons of the same energy. Whatever else we may say, this can't possibly be right. The amplitude of an electromagnetic wave is defined either by its electric field or by its magnetic field. (In a sensible system of units, these two fields have the same units.) If this amplitude is the same for any two photons with frequencies of $10^8$ Hz, then what is this amplitude? There is no possible answer that makes sense on dimensional grounds.

There are two possible answers to this question, depending on how strict one wants to be.

(1) Strict answer: Photons don't have wavefunctions. See What equation describes the wavefunction of a single photon?

(2) Not-so-strict answer: In some contexts, you can get away with considering a photon to have a wavefunction. It is then possible to calculate its amplitude. The amplitude can be different for different photons of the same frequency. See Amplitude of an electromagnetic wave containing a single photon , which this question is basically a duplicate of.

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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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This doesn't answer the question. –  Ben Crowell yesterday

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