Deriving the London's equation for superconductorby Trave11er Tags: deriving, equation, london, superconductor 

#1
Dec1612, 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. 



#2
Dec1712, 12:54 AM

P: 986

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} 



#3
Dec1712, 01:42 AM

Sci Advisor
P: 3,375

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/ 



#4
Dec1712, 10:02 AM

P: 70

Deriving the London's equation for superconductor
Thank you for the gem, DrDu  it is beautiful.




#5
Dec1812, 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?




#6
Dec1812, 09:17 AM

Sci Advisor
P: 3,375

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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