# Ginzburg-Landau boundary condition in the 1D no fields case

It is commonly seen that in finding the coherence length from Ginzburg-Landau, that the following equation is found: $\frac{\partial^2 f}{\partial \eta^2} + f(1-f^2) = 0$ which is for a superconductor filling the infinite half space of $x$ from 0 to infinity. This is the normalized version where $\eta = x/\xi$. Now usually one boundary condition used is that $f(x=0) = 0$, which seems reasonable, you would expect the order parameter might go to zero at the boundary. The solution given is $f(\eta) = \tanh \left(\frac{\eta}{\sqrt{2}} \right)$ However if one considers that usual boundary condition $(-i \hbar \nabla \psi - \frac{e^*}{c}{\bf A} \psi)|_{n} = 0$ Then this means surely that $\frac{\partial f}{\partial x}|_{x = 0} = 0$ (1) at the boundary. This means that the previously given solution cannot be correct, and the solution must just be $f = 1$.

However I can see that the current across the surface is still zero for the usual tanh solution. I would like to know, why is the boundary condition (1) disregarded here?

Both the $\psi=\psi^*=0$ Dirichlet boundary condition and the Neumann-lke condtion ${\bf n}\cdot (\nabla-2ie{\bf A}/\hbar)\psi=0$, make mathematical sense for the Landau-Ginzburg free energy functional in that either makes the integrated-out boundary term vanish. It's really physics that selects the one you should use. If something (a change of material, say) forces the order parameter to zero at the surface then Dirichlet wins. If the order parameter is non-zero at the surface then Neumann makes sense. As the coherence length is defined by bending the order parameter, I think Dirichlet is Ok.

• Thanks for the answer! I shall have to do some more reading about the Dirichlet boundary condition. Aug 16, 2018 at 19:19