Past exam of the mathematics course of the University of Cambridge / 2010 / ii / Paper 4 / 35B / Solution

Write the superconducting order parameter as $\psi =\sqrt{n_s}{e}^{i\vartheta }$, with carrier mass $m$. Minimal electromagnetic coupling gives mechanical momentum $\hslash \nabla \vartheta -q\mathbf{A}$, so the current density isIt also follows by substituting $\psi$ into $(q\hslash /m)\mathrm{Im}({\psi }^{\ast }\nabla \psi )-(q^2/m)|\psi {|}^2\mathbf{A}$. In a gauge where the phase is constant, this reduces to the London equation $\mathbf{j}=-n_s{q}^2\mathbf{A}/m$.

Under $\mathbf{A}\mapsto \mathbf{A}+\nabla \chi$, the phase changes by $\vartheta \mapsto \vartheta +q\chi /\hslash$. The two gradient changes cancel, proving gauge invariance of the actual current. A formula omitting the phase term is valid only in its chosen gauge.

For a uniform superconductor without vortices, taking a curl gives $\nabla \times \mathbf{j}=-n_s{q}^2\mathbf{B}/m$. Combine this with the stationary Ampère's law $\nabla \times \mathbf{B}={\mu }_0\mathbf{j}$ and $\nabla \cdot \mathbf{B}=0$ to obtainFor a magnetic field parallel to the planar interface, translation symmetry reduces the equation to ${\mathbf{B}}^{\prime \prime }=\mathbf{B}/{\lambda }_L^2$. Boundedness and decay as $z\to \infty$, together with the tangential boundary value, give the Meissner effectHere ${\lambda }_L$ is the London penetration depth.

The parallel-field assumption is necessary for this one-dimensional Meissner solution. If the printed arbitrary constant ${\mathbf{B}}_0$ has a nonzero normal component, $\nabla \cdot \mathbf{B}=0$ forces that component to be constant across a translation-invariant half-space, whereas the London equation forces it to vanish. Thus no decaying planar Meissner state with that boundary value exists; one must change the geometry or allow a different magnetic state.