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BEC condenstate thermal statistics vs. coherence

  1. Mar 5, 2012 #1
    Hi. I'm reading an introductory section on the Bose-Einstein condensation of a non-interacting, spinless boson gas. I'm confused by the claim that the ground state is in a coherent state with eigenvalue sqrt(N0) exp(i theta), where N0 is the expected number of particles in the ground state. The justification is that the commutator [a0/sqrt(V), a0*/sqrt(V)] = 1/V goes to zero in the thermodynamic limit V = volume goes to infinity (a0 annihilation operator for ground state). Therefore a0 acts like a complex number and so the ground state must be in a coherent state. Huh? Who asked you to divide by V anyway? Totally opaque. And doesn't statistical mechanics say the system is in an ensemble of definite particle eigenstates with probability exp(-beta*mu*N)/Z (i.e. NOT a coherent superposition of definite particle states)?? Does someone understand this better?
     
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  3. Mar 6, 2012 #2

    DrDu

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    The expectation value of a_0^*a_0 is N, so in the thermodynamic limit you have to divide by V so as to obtain something finite, i.e. N/V.
    The vanishing of the commutator (and of the commutators with all other well defined operators) means that by Schur's theorem a/\sqrt{V} is represented as a pure number iff the representation of the operator algebra is irreducible. However, noone forces you to use an irreducible representation. You can also work with a state of the BEC with the number of particles being sharp. However, calculations become somewhat more involved. Nevertheless, this is the correct description for finite systems.
    And finally, no, statistical mechanics does not tell you that particle number has to be sharp as N is an operator in that expression you write down and not a number.
     
  4. Mar 6, 2012 #3
    OK, you've given me quite a bit to chew on there. I might have a follow-up later, but thanks for the head-start.
     
  5. Mar 11, 2012 #4
    I understand that exp(beta*mu*N)/Z could be viewed as an operator and therefore that <N> = Tr[ N exp(beta*mu*N)/Z ] could be calculated in any basis (say coherent states). But this doesn't tell you about the distribution of particle number. In the grand canonical ensemble the density operator rho = Sum[ exp(beta*mu*n)/Z |n><n|, n=0..+inf ], i.e. diagonal only in the basis of definite particle states. If as claimed the ground state is actually in a coherent state with eigenvalue alpha, then shouldn't we have rho = |alpha><alpha| (i.e. in a pure state)? In an experiment I imagine it would be difficult to measure anything but <N>. Is this a case of trying to apply statistical mechanics to rigidly? Or am I interpreting wrong?
     
    Last edited: Mar 11, 2012
  6. Mar 12, 2012 #5

    DrDu

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    One of the problems here is that the density operator generally does not exist in the thermodynamical limit of infinite sample size.
     
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