= Solution
Use SI units and normalize the condensate <wavefunction> so $|\psi|^2=n$. With <mechanical momentum> operator $-i\hbar\nabla-qA$, the charge <current density> is
$$
\boxed{j=\frac{q\hbar}{2mi}(\psi^*\nabla\psi-\psi\nabla\psi^*)-\frac{q^2}{m}|\psi|^2A.}
$$
For uniform $n$ and $\psi=\sqrt n\,e^{i\vartheta}$, this becomes $j=(nq/m)(\hbar\nabla\vartheta-qA)$.
Under the <gauge transformation> $A\mapsto A+\nabla\chi$, the <wavefunction> transforms as $\psi\mapsto e^{iq\chi/\hbar}\psi$, so $\vartheta\mapsto\vartheta+q\chi/\hbar$. The two added gradients cancel in $\hbar\nabla\vartheta-qA$, while $|\psi|^2$ stays fixed. Thus \b[the superconducting current is gauge invariant]. The same cancellation in the full displayed current proves it without requiring a constant amplitude.
For uniform carrier density in a region with smooth single-valued phase, taking a curl gives the <London equation> $\nabla\times j=-nq^2B/m$. In the time-independent setting <Ampère's circuital law> gives $\nabla\times B=\mu_0j$, and <Gauss's law for magnetism> gives $\nabla\cdot B=0$. Therefore
$$
-\nabla^2B=\nabla\times(\nabla\times B)=\mu_0\nabla\times j=-\frac{\mu_0nq^2}{m}B,
$$
so the <Helmholtz equation> and <London penetration depth> are
$$
\boxed{\nabla^2B=\frac{B}{\ell^2},\qquad\ell=\sqrt{\frac{m}{\mu_0nq^2}}.}
$$
The equation has the modified, exponentially decaying Helmholtz sign. For a planar surface at $x=0$ with material in $x>0$, the bounded solution is $B(x)=B(0)e^{-x/\ell}$; the exponentially growing solution is excluded in the bulk. Hence a macroscopic region far from the surface has negligible field: \b[magnetic flux is expelled from the bulk], the <Meissner effect>. The literal assertion of zero flux everywhere is the bulk limit: finite penetration remains within distance $\ell$ of a surface. Vortices or flux trapped through a hole require different phase or topology assumptions and are not ruled out by this smooth-phase derivation.
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