Microscopic theory of superconductivity

welatiger
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hello everybody
i want informational about microscopic theory of superconductivity but without sophisticated quantum mechanics
can you help me ?
 
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the electrons form 'cooper pairs'. somehow that results in virtually zero resistance
 


Not virtually zero, but actually zero resistance.

The Cooper pair is a quasiparticle. Unlike the two electrons inside it, the Cooper pair is a Boson, not a Fermion. Therefore the Cooper pair is not limited by the Pauli exclusion principle and all the Cooper pair are in the ground state (lowest quantum numbers).

In the case of a free electron (Fermion) conductivity requires electrons to be in excited states because they are above the Fermi energy so scattering of the excited electrons drops them back into the ground state. For the Boson there is no such scattering process.
 


welatiger said:
hello everybody
i want informational about microscopic theory of superconductivity but without sophisticated quantum mechanics
can you help me ?

No. The microscopic theory requires "sophisticated quantum mechanics". This is fundamentally a quantum mechanical process.
 


do the cooper pairs form a degenerate gas?
 
From the BCS theory of superconductivity is well known that the superfluid density smoothly decreases with increasing temperature. Annihilated superfluid carriers become normal and lose their momenta on lattice atoms. So if we induce a persistent supercurrent in a ring below Tc and after that slowly increase the temperature, we must observe a decrease in the actual supercurrent, because the density of electron pairs and total supercurrent momentum decrease. However, this supercurrent...
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