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GaugeInvariance< Superconductivity.Superconductivity | Index | Superconductivity.PauliLimit > (Sparse notes) London discusses superconductivity in terms of momentum conservation and the minimal substitution, deriving in non relativistic limit that {$\mathbf v = - \frac q m \mathbf A$}, hence {$\mathbf j = \frac {n e^2} m \mathbf A$}. This is apparently not gauge invariant, since velocity of electrons is an observable and the vector potential should not be (if it must retain gauge invariance). However Bardeen's Phys Rev B 1 469 (1950) points out that there is a gauge invariant equivalent formulation, by considering the phase of the electron wave function {$\psi=\psi_0e^{i\theta}$} In the proposed gauge invariant formulation {$$\begin{equation}\mathbf A \rightarrow\mathbf A^\prime-\frac \hbar e\boldsymbol \nabla \theta\end{equation}$$} allows alternative gauge choices that must obey to the conditions {$$\boldsymbol \nabla \cdot\left(\mathbf A+\frac \hbar e\boldsymbol \nabla \theta\right) = 0\qquad \left(\mathbf A+\frac \hbar e\boldsymbol \nabla \theta\right)_{\perp}=0$$} in the bulk and on the surface, respectively. There is still a continuous symmetry breaking brought about by superconductivity. It lies is the fact that two originally independent gauge invariances, that of the quadripotential of electromagnetism and that of the phase of the complex quantum wavefunction, both not observables, are now linked by this condition. Thus they remain not observable, but correlated. Notice that the symmetry breaking does not refer to the Hamiltonian, but rather to the ground state wavefunction. This is general, like in the case of ferromagnetism. The massless Goldstone mode of ferromagnetism (acoustic spin wave excitations at q=0) here is replaced by a massive Higg's boson (the Cooper pair?) that generates a gap in the excitations. One can choose gauges where {$\boldsymbol\nabla\cdot\mathbf A\ne 0$} and the gauge invariant version of the London equation is {$$\mathbf J = -\frac {nq^2} m \left( A + \frac \hbar e\boldsymbol \nabla \theta\right)$$} with the steady state requirement that {$\boldsymbol\nabla\cdot\mathbf A= -\frac \hbar e \nabla^2 \theta$} < Superconductivity.Superconductivity | Index | Superconductivity.PauliLimit > |