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Bose-einsten condensation

  1. May 21, 2005 #1
    I was reading today about superfluids and the like. They were talking about how at a certain low temperature bosons will all go into their ground state. I'm trying to get a physical feel for what this means. If a mole of helium 4 atoms(boson) are cooled so that condensation occurs. What exactly happens? I can imagine, if there is no degeneracy pressure, then the mole of atoms can become extremely dense. But how dense can they get? Is the nucleus nucleus coulomb repulsion the only thing stopping them from becoming unified?
  2. jcsd
  3. May 22, 2005 #2
    Don't you mean fermions?
  4. May 22, 2005 #3
    I'm not exactly sure, don't take this as gospel, but I'd imagine that when they came closely packed, occupying similar space, that the constituent particles of He4 (neutrons and protons) are actually Fermions, so they would have to satisfy Pauli Exclusion Principle...

    For example, a Cooper Pair in a Superconductor is a Boson, with constituents of 2 electrons, but the two spins are in opposite directions. So if another electron came within range, it would destroy the Cooper Pair (Boson) by means of breaking the binding energy. So I think a similar process may be availible for liquid Helium.

    Superconductors and Superfluids are similar concepts involving Bose-Einstein condensation. Superconductors involve zero resistance, Superfluids involve zero viscosity.

    May I make it clear I have only studied superconductors before, not superfluids, so my superfluid assumptions are based on analogy from superconductors.

    I have often asked these questions myself before, but couldn't be bothered to seek an answer, and these were the conclusions I reached.

    Hopefully a better informed person will be able to shed some real light.

    Also, I believe that neutrons in neutron stars are superfluid - Bose-Einstein condensation again - I think they form neutron pairs, similar to Cooper Pairs. But they are obviously really dense.
    Last edited: May 22, 2005
  5. May 23, 2005 #4
    One can ask this, what 'state' of matter is a single He4 ?..and what state does transition occur, ie..Gas>..Liquid..>Solid..? or Solid>..Liquid>..Gas,
  6. May 23, 2005 #5


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    No, one cannot ask that question. BE condensation, and other phase transition are meaningless for ONE particle, be it He4, etc. BE condensation and phase transition (even quantum phase transition), are COLLECTIVE phenomena, which can only occur when there's a gazillion particle interactions. Take 12 water molecules, and they'll NEVER solidify into a solid.

  7. May 23, 2005 #6
    Exactly?..thats why I had asked the question in the first place!


    The fact that you answered in the way you did, makes it quite simple and plain to see.

    The question has been asked and answered in many PHASE TRANSITION papers, one comes to the question of He3 ?

    For the original poster:http://boojum.hut.fi/research/theory/he3.html

    one can test:http://boojum.hut.fi/research/applied/rotating3he.html

    and understand the original question?

    Thus:The reason for the different behavior of 4He and 3He is quantum mechanics. 4He is a boson. The appearance of the superfluid phase in 4He is related to Bose condensation, where a macroscopic fraction of the atoms is in the lowest-energy one-particle state. 3He is a fermion (like electron) and it is forbidden by the Pauli exclusion principle that more than one fermion is in the same one-particle state. The superfluidity arises from formation of weakly bound pairs of fermions, so called Cooper pairs. The pairs behave as bosons. In the superfluid state there is a macroscopic occupation of a single Cooper pair state.

    Quote taken from here:http://boojum.hut.fi/research/theory/helium.html

    Interest is on the idea of ONE-PARTICLE-STATE?..collectively, the behaviour is that the SUPERFLUID..exhibits 'Single Particle' SOLID? status!
    Last edited: May 23, 2005
  8. May 23, 2005 #7


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    There is one important error in your posting (and the references you cited). Superfludity in He3 is NOT the same as the s-wave Cooper pairings. The Cooper pairings in standard conventional superconductors is known to have a zero net angular momentum. This results in an "s-wave" order parameter and a singlet state. The superfludity in He3 is due to the pairing of two He3 atoms, but NOT in a singlet, s-wave state, but in a TRIPLET, p-wave state! The spins from each He3 allign parallel to each other, not opposite as in the conventional Cooper pairs.

    [Also, take note the the standard name given to "Cooper pairs" are always reserved for paired electrons. Paired anything else are never called "cooper pairs"].

    Please refer to Leggett's work on He3 superfluidity to get the actual picture from the "horse's mouth".

  9. May 23, 2005 #8
    Thanks ZapperZ, I may have extended far beyond my knowledge on CM understanding, I have a very limited knowledge on Leggett, but I am just learning,, slow, but hopefully sure.
  10. May 25, 2005 #9
    So what happens and can this relate to the case of neutrons within neutron stars?
  11. May 25, 2005 #10


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    Off the top of my head, from what I can remember, the pairing in neutron stars is spin-triplet state, so it has the same characteristics as the superfluidity in He3. Both appears to be sufficiently described by the BCS theory.

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