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Regarding BEC: Bose distribution and probabilities

  1. Feb 26, 2012 #1
    Hello PF. I won't just lurk around today, I will pose a question.

    I was looking at a dilute ultracold bosonic gas and was trying to see how one can predict the existance of a BEC and got stuck on this:
    I was comparing probabilities between finding the system in the lowest state (energetically) and the first excited state through the use of the Boltzmann (and also Gibbs) factor.

    Let E0 be the energy of the lowest state and also let me set this as my zero on the energy scale (E0 = 0), and let E1 be the energy of the first excited state. Now let's check the ratio of probability

    P(E0)/P(E1) = exp(-E0/kT)/exp(-E1/kT) = exp(E1/kT), ok!

    However, energy between these levels can be (for example) of the order of E1/k = 10-14 (temperature units) and thus exp(E1/kT) ≈ 1.
    This gives (from above) the relation P(E0) ≈ P(E1). So the system is as likely to be in the lowest state as the first excited one.

    Now I turn my head and look at the Bose distribution function
    fBE = 1/(exp[(Ex-μ)/kT] - 1)
    which describes the mean occupation of bosons in a state of energy Ex at temperature T, k and μ are Boltzmanns constant and the chemical potential (negative here!) respectivley.
    Now I get something completely different from before (at least that's how it appears to me)! Let's say I have a system consisting of around N = 1022 bosons, and as above the energy difference (in temperature units) is of the order 10-14. I set the temperature to about T = 1 mK and let Ex = E1.

    When I now check the ratio fBE/N ≈ 5*10-12.
    So there's not a substantial fraction of particles at all in the first excited state! They must (nearly) all be in the lowest state. Since at this temperature the chemical potential is very close to E0 and thus fBE is really large. E0 dominates!

    Now how do I resolve this? From looking at fBE I can see that there's an onset of condensation at low temperatures. But when looking at probabilities the lowest state is as good as the first excited one. What differs between these viewpoints!?
  2. jcsd
  3. Feb 27, 2012 #2


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    Yes, true. The system is nearly as likely to be in a state with N_0 particles in the lowest level as to be in a state with N_0-1 particles in the lowest level and 1 particle in the excited level.
  4. Feb 27, 2012 #3
    I cannot see how that solves my issue, sorry.
    All the bosons should have that same probability since it's assumed here they don't interact. And since they don't obey the Pauli principle the can cram either levels without restriction (only restriction is that the number of atoms N is conserved). Therefore the system in it's entirety can have the state of energy N*E0 or N*E1 equally probable. Right?
  5. Feb 27, 2012 #4


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    No, the Boltzmann factor for all particles to be in the first excited state is exp(-e N E_1/kT) as compared to all particles being in the ground state. This is infinitesimally small.
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