Past exam of the condensed matter physics major option course of the University of Oxford / 2016 / 8 / b / Solution

Let $m,q,n_s$ denote pair mass, charge and number density, respectively. A uniform-phase condensate in the stated gauge has zero canonical momentum $m{\mathbf{v}}_s+q\mathbf{A}=0$. ThereforeFor static fields, $\nabla \times \mathbf{B}={\mu }_0{\mathbf{j}}_s$ and $\nabla \cdot \mathbf{B}=0$. Taking a curl and using the vector identity for a double curl givesFor electronic pairs, $m\simeq 2m_e$ and $q=-2e$, while $n_s$ is the pair density, not twice that density. In a general gauge the canonical momentum is $\hslash \nabla \theta$; taking the curl away from vortices gives the same London equation.