Is there a frequency eigenstate for light?

In summary: Basically, if you measure the correlation between two particles in a state, you can determine that the state is an eigenstate of the annihilation operator. However, this does not mean that the photons in the state are actually annihilating with each other. In fact, if you measure the correlation between two particles in a state, you can determine that the state is an eigenstate of the annihilation operator, but you can also determine that the photons in the state are not actually annihilating with each other. So while the state is an eigenstate of the annihilation operator, the photons in the state are not actually annihilating with each other.
  • #1
snoopies622
846
28
I thought it was the coherent state, but since that is an eigenstate of the annihilation operator, and the annihilation operator is not hermitian, then it has no corresponding observable, and I'm assuming that one can observe frequency.

Thanks.
 
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  • #2
Considering that ##E=h\nu##, isn't that the same as asking if there is an energy eigenstate.
 
  • #3
Well, inside a cube of light, the total energy depends on both the number of photons and the energy (frequency) of each photon. Since the number of photons is by itself an operator, I thought there would have to be a separate one for frequency.

On the other hand, if frequency is a parameter instead of an operator, then maybe not.

I was really wondering about the mutual compatibility of certain kinds of knowledge of light. Is it possible, for example, to know that I have exactly, say, three photons inside of a cube, and to also know the exact frequency of each one?

(And if so, would this simply mean that I would have no knowledge of their phase as well? I do remember reading in an old Dirac paper that energy and phase were conjugate variables. And what about polarity? Is that simply another parameter and therefore not a problem? And so on . . . )
 
  • #4
I think this is what the Fock space provides. There is a Hilbert space vector

##|n_1,n_2,\ldots\rangle##

where we can in principle know all occupation numbers for all states. So for each state we know the frequency and the energy (as parameters determined by the size of the box) and we know the occupation number.
 
  • #5
Doesn't the uncertainty principle demand that if the spacetime dimensions of the box are finite, then it is impossible to know exactly the energy/momentum content inside, and thus both the number of photons and the energy/momentum/frequency of each one simultaneously?
 
  • #6
Why do you think so?

Suppose you have a state

##|0,\ldots,0,1,0,\ldots\rangle##

with "1" at the i-the position. Then the occupation number is

##N_i = 1##

the frequency is

##\omega = \omega_i##

and the energy is

##E = \hbar\omega_i##
 
  • #7
I'm thinking about the wave/particle dual nature of light. To have light energy contained in a finite region of spacetime, does it not have to be like a pulse, that is, a mix of many wavelengths instead of just one?
 
  • #8
You are right, for a wave packet instead of a plane wave energy and frequency are subject to uncertainty.
 
  • #9
So then, for a plane wave with an exactly defined wavelength, if we take a box of finite dimensions inside of it, can it still contain an exactly known number of photons? Does measuring the photon number change this wavelength certainty?
 
  • #10
Theoretically if the box is perfectly mirrored it allows for exact plane waves and exactly defined frequencies
 
  • #11
Oh I didn't mean a closed box with reflecting sides, I just meant, "inside this particular cube of spacetime," and in particular I was imagining a (non physical) cube with edges much smaller than the wavelength of the light itself.
 
  • #12
Actually this reminds me, I asked a question similar to this one almost three years ago, and it was there that I first learned about coherent light.

https://www.physicsforums.com/showthread.php?t=470413

But if light with a well-defined wavelength is coherent, and coherent light is light with an indefinite photon number, then . . well, you can see why I would think that wavelength and photon number are non-commuting observables.
 
  • #13
You want to look at the relationship between the wave and photon descriptions of light?
The number of photons is proportional to the intensity of the light.

But I think you are mixing models here.
 
  • #14
snoopies622 said:
But if light with a well-defined wavelength is coherent, and coherent light is light with an indefinite photon number, then . . well, you can see why I would think that wavelength and photon number are non-commuting observables.

You are mixing up different meanings of coherence here. First-order coherence is what you measure in a double slit or Michelson interferometer and is related to how well defined the phase of your field is. This is indeed related to the spectral purity of your light field. The power spectral density and the first-order field coherence form a Fourier transform pair. Or simply speaking: A spectrally narrow field will have a long coherence time and a well defined phase over long times and a spectrally broad field will have a short coherence time and the phase will randomize over short time scales. All of these quantities are still classical and on the field level. On this level coherence is just a matter of long or short coherence times, but there is no completely incoherent light.

Now, second-order coherence is the meaning of coherence in the sense of quantum optics and the coherence found in lasers. Here, a coherent state means that a state is an eigenstate of the photon annihilation operator and it also means that joint detection rates factorize. Basically, that means that you do not get any additional information by detecting a photon. For such states the photon number uncertainty is small, but present (it is the uncertainty given by a Poissonian distribution). However, this is not related to wavelength. The uncertainty in the width of the spectral distribution and the uncertainty in photon number are two different and not necessarily related sources for uncertainty of the total energy. For example you could have spectrally broad single photon states (no photon number uncertainty, large spectral uncertainty) or spectrally filtered thermal light (huge photon number uncertainty, but small spectral uncertainty).

The pairs of noncommuting quantities you get are time-energy and photon number-phase. Both are "weak" uncertainty relations because there is neither a strict time, nor a strict phase operator for light.
 
  • #15
Thanks Cthugha, much to chew on here.
 
  • #16
Cthugha said:
The pairs of noncommuting quantities you get are time-energy and photon number-phase. Both are "weak" uncertainty relations because there is neither a strict time, nor a strict phase operator for light.

Just a thought, I don't know if it means anything: The product of time and energy, as well as that of position and momentum, have the dimensions of Planck's constant, while neither photon number nor phase have dimensions at all.
 

Related to Is there a frequency eigenstate for light?

1. What is a frequency eigenstate?

A frequency eigenstate for light refers to a specific state of light where it has a well-defined frequency. In other words, the light waves are all oscillating at the same frequency, resulting in a pure and single-colored light.

2. Is there only one frequency eigenstate for light?

No, there are actually an infinite number of frequency eigenstates for light. This is because light can have an infinite range of frequencies, from radio waves to gamma rays, and each frequency represents a different eigenstate.

3. How are frequency eigenstates related to the color of light?

The color of light is directly related to its frequency eigenstate. For example, red light has a lower frequency than blue light, and thus, a different eigenstate. This is why different colors of light appear different to our eyes.

4. Can light exist in a superposition of frequency eigenstates?

Yes, light can exist in a superposition of frequency eigenstates. This means that it can have a combination of multiple frequencies, resulting in a mixture of colors. This is commonly seen in white light, which is a superposition of all visible frequencies.

5. How are frequency eigenstates important in understanding the behavior of light?

Frequency eigenstates are crucial in understanding the behavior of light because they help us to predict and explain how light interacts with matter. The frequency of light determines its energy and how it will interact with different materials, which is essential in fields such as optics and spectroscopy.

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