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Bose-Einstein condensate of photons

  1. Jan 23, 2012 #1
    Does the photon number state of n photons in the same mode (i.e. Fock state |n>) constitute a Bose-Einstein condensate of photons?
     
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  3. Jan 23, 2012 #2

    Bill_K

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    No. For a closed system of bosons with a finite mass the number of particles is fixed, and if it is cooled eventually a critical temperature Tc is reached below which a finite fraction of them are forced to occupy the ground state. But if you cool a photon gas the number of photons simply decreases and Bose-Einstein condensation never occurs.
     
  4. Jan 23, 2012 #3

    Cthugha

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    Is that necessarily so? Last year the group aroun Martin Weitz from the university of Bonn had two high-profile publications on a photon-number conserving way of thermalizing a photon gas and on photon BEC (Nature Phys. 6, 512 (2010) and Nature 468, 545 (2010)).

    I mean, these publications sure are pretty controversial, but just stating it never occurs without detailed discussion is pretty harsh.

    Just to be clear, I would like to stress and support the opinion that it is indeed true that a photon number Fock state of course does not constitute a BEC.
     
  5. Jan 23, 2012 #4
    But is there a physical explanation for this? I thought the low temperature requirement for BEC was only caused by the need to achieve sufficient overlapping of de Broglie waves. In principle, if you take a large photon number state |n>, all photons being in the same mode, it seems that the mere concentration of photons should provide for a substantial overlap? And isn't it true that low temperature of a photon gas just means we imitate the black body spectrum at low T, that is, go to the infrared/RF range?
     
  6. Jan 23, 2012 #5

    Cthugha

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    The necessary temperature for condensation indeed depends on the mass of the particles. For examples there are proposals to create polariton condensates at room temperatures and already demonstrations of that for temperatures way above 100K.

    A photon number state is not a BEC because it will lack coherence. If you take for example Legett's book on quantum fluids, you will see that you also need off-diagonal long-range order, which roughly translates into long-range spatial coherence or a fixed phase relationship over larger distances. This is not necessarily realized for a realistic photon number state (remember photon number-phase uncertainty). It is realized for a completely delocalized photon number state, but these are pretty theoretical constructs, as far as I know.
     
  7. Jan 23, 2012 #6
    For example, in a real laser (whose output is essentially coherent)?
     
  8. Jan 24, 2012 #7

    Cthugha

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    Lasers emit coherent states, which are not Fock states. Coherent states have a Poissonian photon number distribution around some mean photon number as indicated by a g2 of 1. Fock states have a g2 of 1-(1/n) where n is the photon number.
     
  9. Jan 24, 2012 #8

    DrDu

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    The important point with a BEC being a BEC is that it has to be an equilibrium state at some (low) temperature. A laser is not an equilibrium state.
    The Photon BEC described by Martin Weiz is not a condensate of free photons but rather of photons interacting with matter, i.e. polaritons.
     
  10. Jan 24, 2012 #9

    DrDu

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    Off diagonal long range order can also be realized in states with fixed particle number. E.g. any condensate of atoms contains a fixed number of atoms. A description in terms of phase eigenstates merely simplifies the mathematical description. Anyhow a BEC is strictly defined only in the thermodynamical limit when N goes to infinity and the density N/V is constant. Then it makes little difference to distinguish between states with infinitely but fixed number of particles and infinite but unsharp number of particles.
     
  11. Jan 24, 2012 #10
    What's the theoretical justification for this? Is it theoretically impossible for the amount of photons to stay the same but for their frequency to decrease? Maybe because of momentum concerns?
     
    Last edited: Jan 24, 2012
  12. Jan 24, 2012 #11

    Cthugha

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    Weitz himself strongly opposes that point of view. To enter the polariton regime, interaction with matter is not enough, but you need strong coupling (or in the semiclassical regime non-perturbative coupling) between the light field and matter excitations which amounts to reversible spontaneous emission. I am quite sure that is not the case in the Weitz-paper as scattering rates are too small and he operates in the weak coupling regime. That of course does not mean no coupling and it should indeed be stressed that these are not free photons, but cavity photons. The question is rather whether he found a good way to have a "masked" common laser. I personally also do not like that paper too much, but mostly because of the double standards it shows in refereeing. Compared to the amount of data that was required to get the first serious paper on polariton condensation by Kasprzak published, the evidence for the photon BEC provided by Weitz and his coworkers is tiny. No g2 measurements, not much about first-order coherence,...

    The role of equilibrium is also a bit more complicated as has been pointed out in terms of "steady-state" polariton condensation. The most thorough discussion I am aware of comes from the group of Peter Littlewood (M. H. Szymańska, J. Keeling, and P. Littlewood, "Mean-field theory and fluctuation spectrum of a pumped decaying Bose-Fermi system across the quantum condensation transition", Phys. Rev. B 75 195331 (2007)) with the basic result that any distribution function diverging at the chemical potential can cause instability of the normal state and the onset of condensation. I do not watch this branch of theory too closely, but I am quite sure they stepped beyond mean-field in the meantime.

    Sure. However, the number of real systems showing BEC in the strictest sense will amount pretty much to zero.

    Yes, but for particles which vanish upon detection like photons this is somewhere between very hard and impossible to achieve. It is easier for atoms.
     
    Last edited: Jan 24, 2012
  13. Jan 25, 2012 #12

    DrDu

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    Yes, I agree on that. Whether the photons are merely non free or polaritons in sensu stricto is however secondary in my oppinion. The important point is that their dispersion relation is changed so as to render them effectively massive.
     
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