Deriving the London's equation for superconductor

[Trave11er]

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.

 

[andrien]

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]

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]

Thank you for the gem, DrDu - it is beautiful.

 

[Trave11er]

Actually, it is beautiful, but I don't understand much. Can you provide an explanation without calling for field theory?

 

[DrDu]

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.