Deriving the London's equation for superconductor


by Trave11er
Tags: deriving, equation, london, superconductor
Trave11er
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#1
Dec16-12, 10:02 AM
P: 70
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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andrien
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#2
Dec17-12, 12:54 AM
P: 972
you may start here
http://en.wikipedia.org/wiki/London_equations
v is written as p/m and usual prescription of minimal coupling is provided
p--->{p-(e/c)A}
DrDu
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#3
Dec17-12, 01:42 AM
Sci Advisor
P: 3,361
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/

Trave11er
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#4
Dec17-12, 10:02 AM
P: 70

Deriving the London's equation for superconductor


Thank you for the gem, DrDu - it is beautiful.
Trave11er
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#5
Dec18-12, 05:46 AM
P: 70
Actually, it is beautiful, but I don't understand much. Can you provide an explanation without calling for field theory?
DrDu
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#6
Dec18-12, 09:17 AM
Sci Advisor
P: 3,361
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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