 Solution

Start from a gauge-invariant Ginzburg-Landau theory of superconductivity gradient energy $|(-i\hslash \nabla -q\mathbf{A})\psi {|}^2/(2m)$, with $\psi$ the coherent pair order parameter and $q$ its charge. Varying the free energy with respect to $\mathbf{A}$ gives the current density displayed in the source. Write $\psi =\sqrt{n_s}{e}^{i\theta }$ to obtainSingle-valuedness around a loop in superconducting material requires $\oint \nabla \theta \cdot d\mathbf{l}=2\pi n$. Integrating the current relation therefore gives the exact fluxoid conditionFor a thick ring choose the contour sufficiently deep in the material that the current term is negligible. Then flux itself is approximately quantized, with positive quantumThe sign of the pair charge is absorbed into the orientation of $n$. In a thin ring with appreciable kinetic inductance, it is the fluxoid, not exactly the geometrical flux alone, that remains quantized. A state with nonzero phase-gradient current is a concrete counterexample to dropping the current integral without justification. The flux-only identity thus uses the ordinary thick-ring or negligible-kinetic-inductance limit in addition to phase single-valuedness.