# 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?

• That mechanism is only for low temperature superconductors. It is known it is invalid for high temperature superconductors which operate via a different, as yet not fully understood mechanism. – John Meacham Sep 26 '16 at 23:49
• @JohnMeacham This is not even true for low temperature superconductors. As I said, Abrikosov and Gorkov showed that the energy gap closes before persistent currents disappear when introducing magnetic impurities into a low temperature superconductor. The energy gap is not necessary even for low temperature superconductors. – Xcheckr Sep 26 '16 at 23:52
• Think about the alignment of the moving charges and the self-inductance which self-holds for low temperatures. It is not really imaginable that any current will flow without any loses but indeed there are not (electric) loses. So the current at the beginning one need to build up the magnetic field which than is frozen. Seems to be unexpected? Could be proofed by interrupting the coils wire in one position. – HolgerFiedler Sep 27 '16 at 8:38

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.

• I like this explanation, though one thing about it still eludes me. The current-carrying state is not the ground state of the superconductor. Does your explanation therefore still apply? Or does your argument only work for the BCS wavefunction (which is the ground state wavefunction)? – Xcheckr Sep 28 '16 at 2:44
• I think that the equation $b |\Psi\rangle \propto |\Psi\rangle$ should still hold for the current-carrying state, as this state is still characterized by a condensate of Cooper pairs (albeit one with a spatially varying phase). However, I'm not an expert... – Dominic Else Sep 28 '16 at 3:17
• Interesting argument, but I'm not so sure about this. The condition $b |\Psi\rangle \propto |\Psi \rangle$ is the condition of a coherent state. But it seems to me that one does not see a property analogous to persistent supercurrents in any coherent state- for example, a coherent state of light. – Rococo Sep 28 '16 at 4:25
• @Rococo The important thing is the macroscopic coherence between a large number of particles. I tried to improve my answer above. – Dominic Else Sep 28 '16 at 6:14
• @Xcheckr you said The current-carrying state is not the ground state of the superconductor. Does your explanation therefore still apply? but superconductor has non zero ground state current. – L.K. Apr 12 '17 at 16:11

A superconductor is characterized by two main properties:

1. zero resistivity, and
2. the Meissner effect.

Equivalently, these can be stated more succinctly as

1. $E = 0$ (remember that resistivity is defined as $\frac{E}{j}$), and
2. $B = 0$.

So even more succinctly: superconductors are characterized by no internal electromagnetic fields!

What is the intuitive reason for this? It can be understood from the fundamental/microscopic property of superconductors: superconductors can be described in terms of superpositions of electrons and holes. Note that these two components have different electric charges, hence such a superposition can only be coherent if nothing couples to the charges inside a SC! Indeed, if there were an electromagnetic field inside the SC, it would couple differently to the electron and hole, decohering the superposition and destroying the SC. [Of course this doesn't do full justice to the theory of superconductivity, since this reasoning doesn't explain why we have superpositions of holes and electrons. Rather, my point is that once we start from that, then the aforementioned is hopefully intuitive.]

• I must downvote this for a common but incorrect assertion: "It should be emphasized that (contrary to the impression unfortunately given in some text books) the (meta)stability of current-carrying states is by no means a trivial consequence of the diamagnetic (Meissner) effect. In fact, it is easy to construct models that show the latter feature but not the former,the simplest example being the noninteracting Bose gas." -from courses.physics.illinois.edu/phys598sc1/fa2015/Lectures/… – Rococo Sep 27 '16 at 23:00
• The answer, as given in the same notes, involves the energy cost (absent in Schroedinger evolution but present in Gross-Pitaevskii) of changing the number of nodes in the wavefunction. Maybe I will try to make a proper answer about this later; or anyone else is also welcome to do so. – Rococo Sep 27 '16 at 23:03
• @Rococo Thanks for the correction! To be honest I'm surprised to find out that the Meissner effect doesn't imply supercurrents (and it's also not obvious to me that a charged Bose gas doesn't have supercurrents). Anyway, I edited my answer accordingly. (Luckily it was not crucial to the main point that I was trying to get across.) – Ruben Verresen Sep 28 '16 at 0:22
• Fair enough, downvote retracted. I'll have to think more about the actual argument.. – Rococo Sep 28 '16 at 4:31

This can be explained by the BCS THEORY.

According to this theory, a Cooper pair of two electrons can be formed due to the net attractive force between the two electrons. It would be very surprising that how can two electrons attract each other because we know that like charges repel each other. But it can be made possible because of electron - lattice - electron interaction , i. e, lattice ions acts as a mediator for providing net attractive force between the two electrons.

Since the net spin of Cooper pair is zero, all the electrons that form the cooper pair can be accommodated in the same ground state (Pauli exclusion principle), so electrons condensate to a single state.

We know that resistance occurs due to scattering of electrons by the lattice vibrations. But here since all the electrons are in a single state so the lattice have to scatter all the electrons simultaneously. It's all or none situation.

A large energy would be required to scatter huge number of electrons so lattice ions are not able to scatter the electrons and hence due to this resistivity drops to zero. So Resistivity of superconductors is zero for temperature below $T_c$.

• Thank you for your response, but your explanation does not include one important fact: the phenomenon of persistent currents is a metastable non-equilibrium phenomenon, while your answer describes an equilibrium situation. While I think your answer has some correctness to it, it cannot be the whole answer. – Xcheckr May 8 '17 at 14:32

While I don't disagree with much of what is said, I think it is important to take note of Landau's argument for superfluidity. They key question is why a supercurrent can't decay and lose energy by exciting low energy quasiparticles. This is quite subtle because circulation quantization (quantization of the fluxoid) is not enough to prevent this decay process from occurring. One could maintain circulation quantization and still slowly reduce the current by reducing the magnitude of the order parameter (i.e., generating heat).

If there is a gap it is obvious that one cannot dissipate energy into low energy excitations. However, for gapless superconductivity, it is less clear. For nodal superconductors, as long as the dispersion around the node is linear, Landau's argument shows that below a critical velocity, one cannot have a process that reduces the supercurrent and excites quasiparticles.