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Laser vs. photon amplifier

  1. Jan 22, 2012 #1
    What is the difference between a laser and a one-atom photon amplifier? Can we just say that a laser is many one-atom photon amplifiers put together? Is there any difference in the resulting photon statistics?
     
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  3. Jan 22, 2012 #2

    Drakkith

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  4. Jan 22, 2012 #3
    I mean the amplifier in the form of an excited two-level atom. And then, if we bring a billion of such atoms together (atomic gas, or maybe we can even let the atoms form a crystal), my question is, will we then obtain an ordinary gas laser (variants: solid-medium laser, optical parametric amplifier laser), and will the photon statistics remain the same?
     
  5. Jan 23, 2012 #4

    Cthugha

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    Well, that depends strongly on the parameters of your system, e.g. your cavity quality factor, beta factor and so on.

    Generally speaking you will get a transition from bunching to coherent emission with increaing excitation density for many-atom lasers and a transition from antibunching to coherent emission with increasing excitation density for a single-atom laser.
     
  6. Jan 23, 2012 #5
    Could you elaborate on the meaning of the term "excitation density"? Does it have something to do with the wavelength of emitted light?
     
  7. Jan 23, 2012 #6

    Cthugha

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    No, every usual laser has a threshold, as you need to create an inversion. As you drive the system harder (that can be optical excitation, electrical carrier injection or whatever), you invert the system and at some excitation density the system starts lasing. Usually that just means increasing the power of your excitation laser.
     
  8. Jan 23, 2012 #7
    But that's a one-atom laser, the atom is already in the excited state, so how can you increase that?


    Why will there be antibunching for a one-atom laser? If it's one-atom, it can only emit one stimulated photon, in the same state as incident photon, creating a |2> state at the output (unless it emits spontaneously, and we get |1,1> output state). But do Fock states really show untibunching behavior? Since one can always write |2> as the tensor product |1>|1>, shouldn't both photons behave independently and trigger the detector without any mutual correlation?
     
  9. Jan 24, 2012 #8

    Cthugha

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    Lasing needs a system with more than two levels. You need at least an excited state that decays on a fast scale towards another state which decays towards the ground state in the lasing transition. You then pump the upper excited state.


    Yes, Fock states show antibunching. It is their signature to have g2 below 1. However, stimulated emission does not create Fock states. The photon number is not fixed, but varies. You do not have an assured stimulated emission process producing two photons, but you can have such a process. This is a difference.
     
  10. Jan 24, 2012 #9
    Thanks for the answer. But it raises 3 new questions that I don't understand how to answer.

    1. Antibunching in a Fock state |n> would mean that the times of detection of the n photons as represented by points on the time axis would tend to "repel each other". How is this behavior compatible with independence of photons in such a state, as described by the tensor product |n> = |1>|1>|1>......|1> ? Does it have something to do with the indeterminacy of phase when we have a definite amplitude? Or is it just the fact that absorption of one photon decreases the number of photons in the beam, making absorption of the remaining photons less likely? (But even if it becomes less likely now, it should be just as unlikely in all future moments, so there will still be no preferred time for the other photons to get absorbed?)

    2. What kind of state does stimulated emission produce in the general case - superposition of Fock states, mixture of Fock states, or several Fock states entangled?

    3. Why is the number of emitted photons uncertain when the energy difference between the excited and ground levels of an atom is well known? (Since did not restrict the timing of the experiment to a narrow time window, the energy-time uncertainty shouldn't get in the way). One photon in the output beam can in principle give rise to two photons, but such parametric down-conversion would require that we let it pass through a nonlinear crystal?
     
  11. Jan 24, 2012 #10

    Cthugha

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    1) Even if photons are independent of each other each detection process still destroys one of them, thereby reducing the probability to detect further photons. Therefore to have real statistical independency of detection events, one needs to introduce a slight amount of noise that cancels the lowering of the photon number by the detection events. This is realized for a Poissonian photon number distribution.

    2) The kind of state produced by stimulated emission (which is not exactly the same as lasing!) depends on the input state.

    3) Again, are you talking about lasing or a single stimulated emission process? These two are not the same. Of course if you imagined an ideal efficiency stimulated emission process and a Fock input state, you could get a different Fock input state out. But that has nothing to do with what happens in a laser.
     
  12. Jan 24, 2012 #11
    If input state is |n> (n-photon Fock state), and we are talking about the ideal process of stimulated emission by a single, two-level, excited atom, will the output state be the superposition a|n+1> + b|n,1> with the second term due to spontaneous emission?

    In what way is lasing different from a single ideal efficiency stimulated emission process (repeated multiple times according to the number of atoms in the laser)? I understand that sub-Poissonian statistics/anitubunching of output Fock states will give way to Poissonian statistics of coherent output light. But what is the underlying reason for this change?
     
  13. Jan 24, 2012 #12

    Cthugha

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    Well, there will be a certain probability for spontaneous emission, yes. Whether this is a realistic number or negligible depends on the exact choice of two-level system you are using and its surroundings. Usually spontaneous emission can go into many modes of the system which should also be taken into account.

    In lasers you never have any Fock states to start with. You pump your 3- or 4-level system and need spontaneous emission into the cavity to start operation. It is then this emission into the cavity that starts stimulated emission. Spontaneous emission will usually show bunched emission (g2 larger than 1). Of course this is not possible due to blockade effects if you have a single emitter. Now one should distinguish between two different systems termed single-atom lasers. There have been such devices in the strong coupling regime between light and matter as realized by Kimble in 2003. These indeed stay non-classical in the "lasing" regime and are sometimes termed thresholdless lasers. However, they are not really coherent.
    On the other hand you can also consider a single emitter in a classical laser with high quality factor which is kept inverted by pumping. This will get coherent in the end.

    If you want to understand that, you can simply put up the rate equations yourself. A simple birth-death model (Phys. Rev. A 50, 4318–4329 (1994)). You need to include spontaneous emission, losses through the cavity mirrors, terms that couple the inversion of your lasing medium to the photon number inside the cavity and so on and can then calculate the expectation value ,<n^2>/<n>^2.

    From a more intuitive point of view, you will find that some steady state will develop which is governed by the interplay of stimulated emission and losses. As long as you keep the pumping constant, you will find that there is always some "restoring force" driving the system back to steady state. If you lose a photon, the average inversion will increase which in turn will cause the emission range to increase. If you have more photons compared to the steady state photon number the inversion and also the emission rate will drop. This kind of feedback is not found for a simple non-classical single photon emitter for example.
     
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