Microscopic theory of superconductivity

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Discussion Overview

The discussion revolves around the microscopic theory of superconductivity, specifically focusing on the role of Cooper pairs and the underlying quantum mechanics involved. Participants seek to understand the concept without delving into complex quantum mechanics.

Discussion Character

  • Exploratory, Technical explanation, Debate/contested

Main Points Raised

  • Some participants mention that electrons form 'Cooper pairs', which is suggested to result in zero resistance.
  • One participant asserts that the resistance is actually zero, explaining that Cooper pairs are quasiparticles that behave as Bosons, allowing them to occupy the ground state without scattering.
  • Another participant challenges the feasibility of discussing the microscopic theory without sophisticated quantum mechanics, stating that it is fundamentally a quantum mechanical process.
  • A question is raised about whether Cooper pairs form a degenerate gas.
  • A link is provided to external resources discussing the connection between electron pairs and diamagnetism.

Areas of Agreement / Disagreement

Participants express differing views on the necessity of sophisticated quantum mechanics for understanding superconductivity. There is also a lack of consensus on the nature of Cooper pairs and their implications for resistance.

Contextual Notes

Some statements rely on specific definitions of terms like 'zero resistance' and 'degenerate gas', which may not be universally agreed upon. The discussion includes unresolved questions about the nature of Cooper pairs and their behavior.

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?
 

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