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Some questions about BEC

  1. May 13, 2005 #1
    I just read a review paper on Boson-Einstein condensation for the first time. I have 2 questions.
    1.Can anyone tell me why all experimental realization of BEC are done in alkali metal gas, like Na, Cs? Can the interaction between atoms be totally neglected?
    2.A facinating property of BEC gas is its coherence, i.e. all atoms are precisely in the same state(ground state) with the same phase factor. I don't know why. Suppose as the tempreture is lowered, the atoms go to the ground state one by one, why they will neccessarily be in the same phase?
     
  2. jcsd
  3. May 13, 2005 #2

    ZapperZ

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    In principle, Bose-Einstein condensaton (BEC) can occur with almost any atom, be it due to the spin of the whole atom itself (He4), or the formation of composite boson of 2 or more atoms (such as in He3 or Y). The reason why the first realization of BEC for atoms was achieved for alkali metals is due to the nature of the magnetic trapping being used to cool the gas. Alkali metals have 1 unpaired s-orbital valence shell electron and can be trapped in a magnetic field. It also doesn't have the added complications of p and d-orbital valence shell, but it doesn't mean such atoms cannot achieve BEC.

    Now I'm not sure what you mean by "interaction between atoms" can be totally neglected. After all, they all condense BECAUSE of the "interaction" of the spin symmetry. At such low temperatures, thermal agitation and coulombic repulsion have been significantly lowered so that the spin interactions dominates.

    They don't condensed to the "same phase". It is just that the whole glob can be described via a single wavefunction. The coherence can be either temporal or spatial. This means that one atom differs from another only by a phase, but can still be described by the same wavefunction. That is typically what is meant by coherence.

    Zz.
     
  4. May 13, 2005 #3
    can i say this way? The state of the whole system is the superposition of all single atom wavefunctions. since all atoms are in the same state( on the same ray), but their phase are randomly distributed. we sum up all the wavefunctions, we get zero???the atoms cancel each other because of their randomness in phase?
     
  5. May 13, 2005 #4

    ZapperZ

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    Eh?

    Forget about the atom's individual wavefunction. That is no longer significant. Consider it as simply one entity - a boson (this will work only for non-composite boson case). You now have a particle of an integer spin. This is what is condensing. Everybody that condenses can be described crudely as "sin(kx - phi). The only difference is that particle as phi=12, the other particle phi=23, and the other particle phi=374, etc. But they all have the same wavefunction. They only differ in phase.

    There is also no reason why you have to "sum" this.

    Zz.
     
  6. May 13, 2005 #5
    yeah, i know what i was saying is wrong(it doesn't happen in experiment). but since the principle of superposition is a fundamental rule in QM and we did it in other occasions, like the two slit experiment, why "sum" all the individual wavefunctions leads to trouble here?
     
  7. May 13, 2005 #6

    ZapperZ

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    Superpostion of what?

    The principle of superpostion is a PART of QM, but it isn't QM.

    You need to keep in mind that when we have BEC, we are dealing with a many-body problem and not just one, two, three, twelve, 36, etc particles. You cannot write a wavefunction to account for each individual consitutents of the glob - that will get you nowhere fast.

    I think the problem we have here is the meaning of "coherence", not "superposition".

    Zz.
     
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