condensed matter - Why do superconductors conduct electricity without resistance?

Many authors have suggested that persistent currents in superconducting rings arise from the energy gap in the single-particle spectrum. Indeed, the argument has been put forward many times on this site! It is usually suggested that because there is an energy gap, Cooper pairs are prevented from scattering out of the condensate.

However, this cannot be correct. For one, high temperature superconductors have d-wave symmetry, which implies a node (i.e. it takes zero energy to excite an electron along this direction). This seems to suggest that a complete gap is not necessary for persistent currents. Furthermore, it has been shown by Abrikosov and Gorkov that when one introduces magnetic impurities into an s-wave superconductor, the gap closes before persistent currents are destroyed.

Therefore, the single-particle gap is not a necessary condition for superconductivity and any attempt to explain persistent currents by appealing to an energy gap in the single-particle spectrum cannot be correct.

Is there therefore a simple way to understand why persistent currents exist in a superconductor intuitively? What are the necessary requirements?

Answer

A superconductor conducts electricity without resistance because the supercurrent is a collective motion of all the Cooper pairs present.

In a regular metal the electrons more or less move independenly. Each electron carries a current $-e \textbf{v}(\textbf{k})$, where $\textbf{k}$ is its momentum and $\textbf{v}(\textbf{k}) = \partial E(\textbf{k})/\partial \textbf{k}$ is the semiclassical velocity. If an electron gets scattered from momentum $\textbf{k}$ to $\textbf{k}'$ it gives a corresponding change in the current. A sequence of such processes can cause the current to degrade.

In a superconductor, the story is totally different because the Cooper pairs are bosons and are condensed. This means that Cooper pairs self-organize into a non-trivial collective state, which can be characterized by an order parameter $\langle \Psi(\textbf{x}) \rangle = \sqrt{n} e^{i\theta(\textbf{x})}$ (where $\Psi$ is the annihilation operator for Cooper pairs.) which varies smoothly in space. Since the current operator can be written in terms of $\Psi$ it follows that gradients of $\theta$ give rise to currents of the condensate: $\textbf{j} = n(\nabla \theta + \textbf{A})$. All the small-scale physics (such as scattering) gets absorbed into the effective macroscopic dynamics of this order parameter (Landau-Ginzburg theory).

One should think of every single Cooper pair in the system taking part in some kind of delicate quantum dance, with the net effect being a current flow. But this dance is a collective effect and so it's not sensitive to adding or removing a few Cooper pairs. Therefore, scattering processes don't affect the current.