Why is a state with large number of photons not classical?

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In the last paragraph of these notes, https://ocw.mit.edu/courses/physics...g-2016/lecture-notes/MIT8_04S16_LecNotes3.pdf, it says how a state with large number of photons is not classical. Why is that? I thought quantum mechanics' laws were most applicable when we dealt with small scale object, such as one photon or two. But I were to have like a 20W bulb or something which gave the same number of photons, will it still be treated quantum mechanically?
 

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  • #2
Drakkith
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The last paragraph for reference:

Classical behavior is a subtle limit of quantum mechanics: a classical electromagnetic field requires a large number of photons. Any state with an exact, fixed number of photons, even if large, is not classical, however. Classical electromagnetic states are so-called coherent states, in which the number of photons fluctuates.


it says how a state with large number of photons is not classical.
That's not what it says. It says that a state with a fixed number of photons, even if large, is not a classical situation. Classical states do not have fixed numbers of photons.
 
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  • #3
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The last paragraph for reference:

Classical behavior is a subtle limit of quantum mechanics: a classical electromagnetic field requires a large number of photons. Any state with an exact, fixed number of photons, even if large, is not classical, however. Classical electromagnetic states are so-called coherent states, in which the number of photons fluctuates.




That's not what it says. It says that a state with a fixed number of photons, even if large, is not a classical situation. Classical states do not have fixed numbers of photons.
Yeah, sorry I misquoted it. But why is it so? Will a beam of light with a specific intensity not output a fixed number of photons?
 
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Drakkith
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Yeah, sorry I misquoted it. But why is it so? Will a beam of light with a specific intensity not output a fixed number of photons?
It will not. I don't know much more than that though, but I'm sure someone around here does.
 
  • #5
Dale
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@Drakkith hit it on the head earlier when he mentioned coherent states. A coherent state is an eigenstate of the annihilation operator, so you can remove a photon without changing the state. Thus the number of photons is uncertain and there is an uncertainty relation between the photon number and the phase.
 
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@Drakkith hit it on the head earlier when he mentioned coherent states. A coherent state is an eigenstate of the annihilation operator, so you can remove a photon without changing the state. Thus the number of photons is uncertain and there is an uncertainty relation between the photon number and the phase.
But in an earlier post, coherent states were used to refer to classical EM waves, right? But here, you talked about the uncertainty principle which is a quantum mechanical property. How are they consistent?
 
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Ibix
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But in an earlier post, coherent states were used to refer to classical EM waves, right? But here, you talked about the uncertainty principle which is a quantum mechanical property. How are they consistent?
A classical state can be described quantum mechanically. In fact it must be possible (the so called "correspondence principle"), or quantum mechanics couldn't be consistent with classical optics, and it would have to be incorrect.

All Dale and the lecture notes are saying is that simply having a boatload of photons is not enough to make the classical approximation valid. Unless further conditions are satisfied (coherent states) you have a boatload of photons that you need quantum mechanics to describe. If they're in a coherent state you can use quantum mechanics if you want, but classical optics is probably easier.
 
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Dale
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But in an earlier post, coherent states were used to refer to classical EM waves, right? But here, you talked about the uncertainty principle which is a quantum mechanical property. How are they consistent?
Coherent states are quantum mechanical states. They correspond to classical EM waves by the correspondence principle.
 
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  • #9
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A classical state can be described quantum mechanically. In fact it must be possible (the so called "correspondence principle"), or quantum mechanics couldn't be consistent with classical optics, and it would have to be incorrect.

All Dale and the lecture notes are saying is that simply having a boatload of photons is not enough to make the classical approximation valid. Unless further conditions are satisfied (coherent states) you have a boatload of photons that you need quantum mechanics to describe. If they're in a coherent state you can use quantum mechanics if you want, but classical optics is probably easier.
I see, that makes sense. And why do you *need* a coherent state in order to be able to approximate the state using classical E&M? I know in case of position, when you have a whole bunch of particles, their uncertainty in position is smaller than their collective size and you can use classical mechanics to get a definite position, but how does coherent state allow for a classical approximation?
 
  • #10
Cthugha
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And why do you *need* a coherent state in order to be able to approximate the state using classical E&M?
Because noise matters. One of the central features of the classical realm is the assumption that measurements can be noninvasive, so they do not alter the state that is measured. Considering photons in the quantum realm, this is obviously impossible. A detection event will destroy a photon and therefore change the photon number. However, the closest thing you can get is a state, where the mean photon number does not change upon detection of a photon.

Consider two extreme cases of light fields. Consider pulsed light fields for easier intuition. Both light fields contain 3 photons on average. One of them has exactly 3 photons in each pulse. The other has 6 photons in 50% of the pulses and 0 photons in the other 50% of the pulses. Now how do you change the state of the light field, if you detect one photon? For the first light field, you will alway have exactly 2 photons afterwards. The mean photon number has been reduced by one. For the second light field, you will never detect a photon in the cases that there are 0 photons in the pulse. So if you detect a photon, you know that there were 6 photons in the pulse and there are now 5 left. So by destroying a photon, you have effectively even increased the mean photon number of the remaining field to 5, which is pretty counterintuitive, but correct.

Now, if one can decrease or increase the mean photon number of the remaining light field by destroying one photon, it is quite intuitive that light fields with such a photon number distribution must exist, that the mean photon number will not be changed at all when a photon is destroyed. These states of the light field are coherent states and the corresponding photon number distribution is Poissonian.
 
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  • #11
DrDu
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The usual argument is that in states with fixed photon number, the expectation values of the electromagnetic field all vanish, even if the intensity is high and the phase of the electromagnetic field is completely undefined.
 

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