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BEC and JJ

  1. Nov 18, 2007 #1
    Repulsive interactions are an attribute of Cooper Pairs, which mean that Cooper Pairs don't behave like an ideal gas . Are these the only attributes that prevent Cooper pairs to behave like an "ideal" boson? How could we show that the two superconductors of a JJ behave just like two BEC's?
  2. jcsd
  3. Nov 18, 2007 #2


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    I am not sure I understand the questions. AFAIK you do not need a condensate in order to form a JJ (I presume you mean a Josephson junction). You do not even need a microsopic model for what is going on in the two banks in ordet to derive the Josephson equations (which is fortunate since this means that we can model high-Tc junctions).
    There is a very nice (and simple) derivation of the equations in the Feynman lectures. I like it because it shows that the Josephson effect is very "general" (at least once the equations have been generalized to non-sinusoidal CPRs, Feynman's approach can be easily modified to cover this).
    Hence, it is not surprising that it can be created using so many different systems including two BEC.

    Alexander Golubov wrote a very nice review of the Josephson effects a few years ago (which includes unconventional current-phase relations); you should be able to find it using Google schoolar (I don't remember the title but I do have it on my computer at work).
  4. Nov 30, 2007 #3
    i appreciate the above reply
  5. Dec 28, 2007 #4
    I'm not interested in the derivations of the well know ac and dc josephson equations. However, I am interested in trying to relate the two islands of a josephson junction with two BE condensates of Cooper pairs when the charging energy is sufficiently much larger than the Josephson coupling energy. I'm just having a difficult time in understanding this analogy. I understand that quasiparticle tunneling is minimal, when we have strong repulsive interactions between the two electrons. Any help would be well appreciated.
    Last edited: Dec 28, 2007
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