Coulomb gauge or London gauge in superconductivity?

Question

The expression for the supercurrent is given by $${\vec j}_s=\alpha\nabla\theta+\beta{\vec A}$$ where $\alpha$ and $\beta$ are two constants, irrelevant for the question here. Here, $\theta$ stands for the phase of the macroscopic wavefunction and ${\vec A}$ stands for the vector potential. In my answer to this question, I have shown that in the Coulomb gauge ($\nabla\cdot{\vec A}=0$), the expression for ${\vec j}_s$ simplifies to $${\vec j}_s=\beta{\vec A}.$$ I seem to be able to derive this using the Coulomb gauge only.

But this Wikipedia reference tells that Coulomb gauge is not enough. We need to impose the London gauge conditions to derive ${\vec j}_s=\beta{\vec A}$ which seemingly renders my answer wrong. But I cannot figure out what is incorrect in my answer.

Answer 1

The Coulomb gauge condition $\nabla\cdot A$ does not uniquely specify a gauge. For any given field configuration, there are still infinitely many choices of $A$ which have that property. One says that there is still *residual gauge freedom * after imposing the Coulomb gauge condition. Normally this freedom goes away if you demand that the fields vanish at spatial infinity, so the Coulomb gauge is said to be complete, but that is not the case here as I’ll explain momentarily.

In your answer, you argue that in any gauge which satisfies this condition, $\nabla^2\theta \propto \nabla\cdot j$, which vanishes if the condensate wavefunction is time-independent, which is true. However, you then argue that $\nabla\theta$ must vanish due to the Helmholtz theorem, and this is wrong for two reasons.

First, Helmholtz requires that the field in question be twice differentiable on the region in question. Since the phase of the condensate is not defined outside of the superconductor, that means you must restrict your attention to the half-space which is bounded by the surface of the superconductor. This means that boundary conditions need to be supplied for spatial infinity on one side and the superconductor surface on the other.

Secondly, you assume that $\nabla\theta$ must vanish in the bulk, but there’s no reason for that to be true. Even if the current vanishes in the bulk, $\nabla\theta$ need not, as long as it is canceled by $A$ there.

In summary, you supplemented the condition that $\nabla \cdot A$ with two additional tacit assumptions on $\nabla\theta$ - namely that it vanishes in the bulk, and that its normal component at the surface vanishes too. Only with these additional constraints can you can conclude that the phase is constant, but these constraints constitute constraints on $A$. The combination of these constraints - plus the Coulomb gauge condition - uniquely defines the London gauge.

Answer 2

$\textbf{div} A=0$ does not fix the gauge. We need boundary conditions.

1. The London gauge says $A.n=0$ on $\partial\Omega$ where $\Omega$ is the domain of the material under analysis. Check "Superconductivity of metals and alloys" DeGenne eqn. 2-14. This also referred to as the "Coulomb Gauge" when people consider finite domains; check "Vortices in the Magnetic Ginzburg-Landau Model"-E. Sandier, S. Serfaty.
2. The Coulomb Gauge says $A=0$ as $|x|\to\infty$.
3. When solving the Ginzburg Landau equations for calculating $H_{c2}$, one usually works with the periodic system where $A.n=0$ on $\partial\Omega$ does not make sense. There one picks a gauge at random. Books usually say $A=H x \textbf{e}_y$ to satisfy $\textbf{curl} A=H \textbf{e}_z$ and $\textbf{div} A=0$.

To finish, while $\textbf{div} A=0$ might help you simplify the expression for the current, it is not sufficient to fix the gauge.

In the following, I shall explain a bit more about the London Gauge as specified in the Wikipedia article.

The Wikipedia article says that $A=0$ in the bulk Superconductor (SC), I'm not sure if that can be established. If you work with the London gauge given in the Wikipedia article, it is possible to establish $A.n=0$

Proof that $A=0$ in the SuperConducting bulk, based on Wikipedia article. A bit more rigor is needed, this is just a formal argument.

The necessary condition for the Meissner effect in the SC bulk is $h=0\implies \textbf{curl } A=0\implies A=\nabla\chi$ for some $\chi$; here $h$ is the microscopic magnetic field. At the minimum, the phase of the wave function satisfies $\nabla(\theta-\chi)=0$. If we impose the stricter version of the London gauge $A.n=0$ on the Surface of the bulk SC phase, we can show that $A=0$.

We denote the Superconducting bulk as $\Omega_S=\Omega\cap\{x\vert u(x)=u_{\infty}\}$ where the SC bulk is identified by the condition $u=u_{\infty}$. If we set $A.n=0$ on $\partial\Omega_S$, then we get the following differential equations, $$\textbf{div} A=\Delta \chi=0~~\text{in}~\Omega_S$$ and $$A.n=\dfrac{\partial\chi}{\partial n}=0~~\text{in}~\partial\Omega_S$$ The solution to this is $\chi$ is constant. Now this means $A=\nabla\chi=0$. \textbf{end of Proof} (Could find the qed symbol).

While this gauge gives a great condition that $A=0$ in $\Omega_S$, this makes the Ginzburg Landau (GL) minimization problem highly nonlinear and very hard to solve. This is saying that in order to solve the GL minimization problem numerically, you have to solve for $u$ (which depends on $A$) and then identify $u=u_{\infty}$ and then solve for $A$ with the condition $A.n=0$ on $\partial\Omega_S$. Instead, it's much easier to impose the simpler condition that $A.n=0$ on $\partial\Omega$, the London gauge as stated by De-Genne. Given that the problem is "Gauge Invariant" and any choice of gauge is going to give you the same answer, it seems like an unnecessary choice to make to impose the conditions put forward in the Wikipedia article.