Deriving the London's equation for superconductor

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SUMMARY

The London equation for superconductors can be derived from the principle that the canonical momentum of the ground state is zero. This derivation is supported by references to Kittel's "Quantum Theory of Solids" and the minimal coupling prescription, where momentum is adjusted by the vector potential A. The concept of broken symmetry, as discussed by Weinberg, provides a broader understanding of this phenomenon. The rigidity of the wavefunction, as noted by the Londons, indicates that the expectation of momentum remains unchanged in the presence of an applied transversal field.

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  • Understanding of superconductivity principles
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  • Knowledge of minimal coupling in quantum mechanics
  • Basic grasp of broken symmetry concepts in physics
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Physicists, materials scientists, and students studying superconductivity and quantum mechanics will benefit from this discussion.

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The equation can be obtained from the fact that the "canonical momentum of the ground grstate of superconductor is zero", but where does this fact follow from.
P.S. Jackson gives a vague reference to Kittel, which I couldn't find in his Introduction_to_solid_state/Quantum_theory_of_Solids.
 
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There are plenty of explanations around. Maybe the most general is due to broken symmetry arguments as expounded by Weinberg:
http://ptp.ipap.jp/link?PTPS/86/43/
 
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Thank you for the gem, DrDu - it is beautiful.
 
Actually, it is beautiful, but I don't understand much. Can you provide an explanation without calling for field theory?
 
In Kittel Quantum theory of solids he discusses how a superconductor reacts to an applied transversal field of long wavelength. He finds that the wavefunction remains unchanged to lowest order. Hence the expectation of the momentum <p>=0 also does not change with A. This has already been called "rigidity of the wavefunction" by the Londons.
 

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