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What do you mean by Cooper pair in real space and k space?

  1. Apr 7, 2015 #1


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    So i came to know that electrons residing across a tiny shell around fermi momentum takes part into the game of forming cooper pair but this phenomenon in real space is actually different. here all the electrons are taking part into this. what does it mean. i am not able to comprehend this physically. Also I have just started to read these so don't know much. But one thing i AM short of worried about why cooper pairs themselves don't scatter and contribute to resistivity. Is it because they are like bosons So that they just can pass through each other effectively(what i mean here is two fermions as a hole can be thought of as bosons I think at least theoretically two fermi operators does commute with other two).. Also why should not they get affected by impurities that is not clear to me as well. if someone physically try to motivate this that would be helpful,since algebra is fairly simple at this level but physical picture is somewhat obscure.
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  3. Apr 8, 2015 #2


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    Also cooper pairs break up in scattering. However unlike normal electrons getting scattered in a current carrying state, the state that they get scattered into is energetically higher than the unscattered state. So sooner or longer, the electrons will recombine into the cooper pair they started off.
    Of course, the energetic stabilisation of the current carrying state is due to the cooper pairs forming kind of a Bose Einstein condensate. The state with no current is energetically lower than the state carrying current, however, to get there, you would have to break all cooper pairs (carrying momentum) and re-pair them into cooper pairs with no momentum. But you can't get this breaking cooper pairs only one by one in scattering.
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