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Super fluids and pauli exclusion principle

  1. Feb 3, 2006 #1
    I don't understand how can atoms like heilium-4 which has same amounts of fermions behave like bosons and do not obey pauli's exclusion principle. Do the heilium physically stack up together like laser light, or does it just seem to do that? could someone please explain how superfluids work.
     
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  3. Feb 3, 2006 #2
    I don't understand much about the exclusion principle, but I do know that two fermions can together act like one boson. This is because they both have half integer spins and two half integers add up to a whole integer, like a boson.
     
  4. Feb 3, 2006 #3

    ZapperZ

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    Fermions can form what is known as "composite bosons". This is where their spins line up to form a conglomerate that has a net integer spin.

    He4 has 2 protons, 2 neutrons, and 2 electrons. Below the superfluid transition, the energy state for all of them to be in a particular configuration is lower than if they are all pointing in all random directions. That's why the whole atom becomes an object (they are all entangled with each other) with a net spin of zero, a boson.

    You could go even further. If you have He3, which is the same as He4, but missing a neutron, then the best you can get is an atom with a net spin of 1/2. However, below an even lower temperature, a single He3 atom can actually pair up (with the help of external agents such as the other He3 atoms) with another He3 atom to form a composite boson consisting of paired up He3 atoms. This pairing is similar to the BCS-type pairing of Cooper pairs in superconductivity.

    Zz.
     
  5. Feb 3, 2006 #4
    When ,say heilium, goes in a super fluid state can a batch of it occupy the same amount of space as one atom of heilium? if it can't, why not because the exculsion principle is no longer obeyed so can't it all merge together to only occupy a point in space.
     
  6. Feb 3, 2006 #5

    ZapperZ

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    Here's the problem with your question.

    Fermi-Dirac and Bose-Einstein statistics kick in when the particles' wavefunction make a significant overlap. When this happens, something call indistinguishibility of the particles becomes important. Now this doesn't mean the particles all look the same. We have the same thing in classical statistics. What this means in this case is that you cannot, even in principle, distinguish one particle with the next. You cannot tag a particle and follow it along anymore.

    So the question on whether they all "occupy a point in space" is meaningless because you can't look at one particle, point to where it is, and look at another and point to where it is.

    The BE statistics indicates that the overall wavefunction must be even. Note that this overall wavefunction includes the product of the spin part and the spatial part. What this means is that they can all occupy the SAME state, and be described by the SAME wavefunction.

    There is an added complication here that I am hesitating to get into, so I'll mention it only in passing. While each of the fermions form a composite wavefunction, the fermions themselves STILL MAINTAIN their F-D statistics. In other words, if one electron has a k1up and the other pair has a -k1down configuration, the OTHER pair of electrons can only have k2up,-k2down, etc... I.e. each of the fermions in the whole condensate STILL can only occupy a unique state to maintain the fermionic statistics. so while the composite boson can all condense to the same state, the consituent fermions cannot. This already prevents things from "occuping the same point" in space, which in itself is difficult already due to local coulomb repulsion.

    Zz.
     
  7. Feb 4, 2006 #6

    Gokul43201

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    I'd like to raise one small point which I've been trying to ignore...and Zz can do this better justice. The OP seems to be using the term 'superfluid' interchangeably with 'BEC'.

    While the two phenomena are closely related, they are not the same thing. In fact, it is possible to have a BEC that does not exhibit superfluidity (and for instance, in 2D, the converse is possible).

    Superfluidity is a transport characteristic - it suggests that there is no mechanism for dissipation of flow. On the other hand, macroscopic condensation is an outcome of quantum statistics - it tells you that there is no way to distinguish between (possibly composite) particles.
     
  8. Feb 4, 2006 #7
    Will superfluids always be Bose-Einstein Condenstates's?
     
    Last edited by a moderator: Feb 4, 2006
  9. Feb 4, 2006 #8

    Gokul43201

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    For the most part, yes...but not always. For instance, the cooper pairs of BCS exhibit superfluid flow. Also, in 2 dimensions (eg: on the surface of a liquid, or in a quantum well), a true BEC is not possible; yet it is possible to have what is known as a Kosterlitz-Thouless Superfluid (KTS).

    However, these states (the BCS state, the BEC and the KTS) are all closely related.
     
  10. Mar 3, 2006 #9
    Can anyone tell me why it is only fermions which must obey the Pauli Exclusion Principle? - my physics teacher can't tell me!
     
  11. Mar 3, 2006 #10

    ZapperZ

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    That is like asking "why do charge particles have electric fields?"

    While those questions are still be asked and studied, as it stands now, these are part of the properties of how we define those things. We would not have known about charged particles without its electric field, and we won't have known about fermions if it weren't for the exclusion principle.

    So a fermion could easily ask you "Why do you look like that?" You could answer "well, how I look is part of how people define and recognize me. So my looks is me!" And a fermion would reply "DITTO!"

    Of course, I'm assuming a fermion can talk, something that isn't part of how it is defined.

    Zz.
     
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