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- Newsgroups: sci.physics
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- From: marty@amaterasu.physics.uiuc.edu (Marty Gelfand)
- Subject: Re: the nature of exclusion
- References: <1992Aug14.210429.23650@galois.mit.edu> <1992Aug18.115432.2269@aoa.aoa.utc.com> <18AUG199208291048@zeus.tamu.edu>
- Message-ID: <Bt6qBw.2A5@news.cso.uiuc.edu>
- Sender: usenet@news.cso.uiuc.edu (Net Noise owner)
- Organization: Dept. of Physics, Univ. of Illinois at Urbana-Champaign
- Date: Tue, 18 Aug 1992 15:14:18 GMT
- Lines: 38
-
- >carl@aoa.aoa.utc.com (Carl Witthoft) writes...
- >>An interesting continuation of this: if you pull enough energy
- >>out of the right fermionic system and squish it enough, fermions
- >>occasionally pair off to form boson-like units. These units then
- >>drop into a common energy state. Cold He3 is one example, if I
- >>remember correctly :=).
- And David Ring asks...
- >This one has always bugged the heck out of me. Say A1 and A2 pair up
- >and B1 and B2 pair up. And each pair is in the same quantum state, then
- >either A1 is in the same state as B1, or A1 is in the same state as B2.
- >Both of which are forbidden. What's wrong?
- >
- What's wrong is the picture that Carl has sketched. In fact there are
- some important differences between superfluid He3, or indeed
- superconductivity in electronic systems, and superfluidity in an ideal
- Bose gas. There are even important differences between superfluidity in
- He4 and in an ideal bose gas (which are particularly striking in the
- vicinity of the phase transition).
- Superfluid He3 is actually too complicated for me to discuss (I really
- don't know it well enough to write off the top of my head). Let me say
- only that it is possible to have a superfluid ground state even with
- purely repulsive interactions, so that that conventional notion of
- pairing is quite strained. Superfluidity in He3 is probably due
- primarily to this "Kohn-Luttinger" mechanism.
- Instead, let's look at a simpler matter: s-wave (which is to say,
- conventional) superconductivity. The Bardeen-Cooper-Schrieffer (BCS)
- wavefunction is a good starting point for the description of
- superconductivity in the weak-coupling limit. It looks like
- \prod_k (u_k + v_k a^\dagger_{k,up} a^\dagger_{-k,down}) | vacuum >
- where the u's and v's are c-numbers, and a^\dagger's are creation operators.
- This is manifestly antisymmetric. One can say that the states
- k,up and -k,down are "paired", since there are strong (complete!)
- correlations between the occupancies of those states. One can also that
- "all the pairs are in the same state" since they all have zero total
- momentum. But nonetheless the wavefunction is antisymmetric.
- Perhaps the bottom line is that electrons don't pair, states do.
- I hope someone will correct me if this is wrong...
- --marty gelfand
-