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.