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The most general for of the Landau-Ginzburg free energy is$^1$ \begin{align} F_s(T)&=F_n(t)+\int_{V} d^3\vec{r}\left\{\frac{1}{2m^\ast}|(-i\nabla-e^\ast\vec{A})\psi|^2+a(T)|\psi|^2+\frac{b(T)}{2}|\psi|^4\right\}\\[5pt] &+\frac{1}{2}\int_{\mathbb{R}^3} d^3\vec{r}\ B^2.\tag{1}\label{1} \end{align} The GL order parameter $\psi(\vec{r})=|\psi(\vec{r})|e^{i\phi(\vec{r})}$ is related - as well known - to the BCS gap parameter $\Delta$, which is why I'm only focusing on the GL theory here. Writing down the EoM of \eqref{1}, we find a minimally coupled Gross-Pitaevskii equation for the order parameter $\psi$ and the Ampére-Maxwell law (magnetostatic limit) for the magnetic field, with $$\vec{j}=-\left[\frac{e}{m}\nabla\theta+\frac{2e^2}{m}\vec{A}\right]$$ being the supercurrent. Now, fixing London's gauge $\nabla\cdot\vec{A}=0$, this implies that $\nabla\phi=0$ (cf. this answer$^2$), which means that the phase is constant with this gauge choice. Now, my question is the following: the modulusu of order parameter is constant in the case of homogeneous media - where the gradient term of the free energy is absent. In other words, in the bulk of the material the order parameter is constant, and as we approach the surface, even in absence of a magnetic field it gets away from the bulk value, as it can be seen studying the Gross-Pitaevskii equation$^3$, so it is not an consequence of the gauge choice. On the other hand, $\nabla \phi=0$ seems to be a consequence merely of the gauge choice and to hold everywhere in the medium, not just in the bulk, so is the phase of a superconductor just constant everywhere with this gauge? In my current understanding, the question is relevant even for more practical question, as the Josephson effect is caused by the current due to the phase difference between the superconductors

$^1$The subscripts 's' and 'n' correspond to the superconducting and to the normal phase respectively; the first integral is over the sample, the second over all space. I'm also using natural units.

$^2$Note that they use the notation $\theta$ for what I have called $\phi$. Moreover, without resorting to QM, that $\nabla\cdot\vec{j}=0$ is already evident by the magnetostatic form of the EoM (Ampére-Maxwell).

$^3$Check e.g. Michael Tinkham, Introduction To Superconductivity (1996, McGraw-Hill College), section 4.2.1.

Tobias Fünke
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Under some assumptions, indeed, $\nabla\phi=0$ hence $\phi$ is constant in all path connected components of a superconductor. However, a typical Josephson junction separates two superconducting components with non superconducting ones (e.g. SIS or SNS junctions). Each superconducting components have their respective phases, but since they are not path connected, there is no reason to believe that they share the same phase. In general they do not, and this is precisely described by the Josephson equations.

LPZ
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