# In what way do Cooper pairs of electrons bond and stay bonded in superconductors?

I understand how electrons initially move into another's vicinity, but nowhere can I find a fathomable answer to this. Also, does the pairs forming 'a condensate' mean a Bose-Einstein condensate?

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They don't really bind, the interaction can be too weak to bind, but the pairs condense anyway. The theory is interesting precisely because the condensation is happening from an interaction too weak to form actual pairs. When the interaction is strong, the pairs can be considered as effective particles, and the theory of superconductivity is different and simpler--- it's a BEC of the pairs.

The standard description is BCS theory, and I'll explain it Bogoliubov's way, which is found in countless places. I'll consider the electron field interacting with a nonrelativistic instantaneous potential. This is not accurate in traditional superconductors already, because the interaction is phononic and retarded, but whatever, the phenomenon doesn't care about this very much.

you have an electron field $\psi(k)$, and the interaction Hamiltonian is

$$V(k_\mathrm{in}-k_\mathrm{out}) sum_{s} \bar{\psi}^{s_1}_{k_1}\bar{\psi}^{s_2}_{k_2} \psi^{s_3}_{k_3} \psi^{s_4}_{k_4} \delta(\sum k)$$

Where the $\delta$ is multiplied by 3 $2\pi$ factors as always, I assume V is spherical symmetric, and there are two spin fluids which are degenerate in energy. The main assumption, which is justified self-consistently is that the field $\psi^0_{k}\psi^1_{-k}$ has an expectation value. You can see this is self consistent from the fact that if V is negative at $2|k_f|$, this expected value lowers the energy classically.

The exected value of $\psi\psi$ is a type of condensation, but it isn't local condensation because there is no real local bilinear field that it is the exectation value of. In the limit that the coupling is strong, you can make a local field which creates a bound Cooper pair, but this is only linked adiabatically to the BCS picture.

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You truly have overestimated my aptitude, yet thanks, however, for the bits that I could understand. –  Alyosha Sep 14 '12 at 19:35
@Alyosha: Sorry, it's not so hard--- the condensate is of pairs, but they aren't close to each other, they are nonlocally paired, the condensate is in k-space. You can think of it as follows: there is a fermi sea of electrons, and which electron is making the bound state with each other electron is not well defined, as they swap in and out with each other, but the end result is that there is a BEC anyway. –  Ron Maimon Sep 14 '12 at 19:53
... also, I tried to make a much longer answer, but this site is having incredible lag problems regarding typing TeX heavy answers. It used to be that turning off the preview fixed this, but no more. –  Ron Maimon Sep 14 '12 at 19:56
I have, bar the mathematics, understood that far. But what stops the electrons from doing this at relatively high temperatures (i.e. forgetting high temperature superconductors)? It can't just be the disruptive thermal motions, as not all materials superconduct. Or are the electrons which demonstrate this effect in the conduction band beforehand in superconducting materials, even when hot, and just need to have the thermal motion removed to begin condensing? –  Alyosha Sep 15 '12 at 20:34
And, remembering the forgotten, why do some materials' electrons condense at higher temperatures (or: what mechanism DO they use, then, if not a BEC, to superconduct?)? –  Alyosha Sep 15 '12 at 20:38
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The attraction is caused by screening, which means that when two electrons come together, they tend to push away the nearby electrons, which leaves a net positive charge in the area that keeps those two electrons together.

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The attraction is usually by phonons, which is the deformation of the heavy nuclei, not the other electrons. You can tell it's this by the isotope effect--- the critical temperature depends on the nuclear mass. This means the nuclear inertia is important, that the nuclei are moving during the binding interaction. –  Ron Maimon Sep 15 '12 at 0:47
@RonMaimon, I used Rae's Quantum Physics as my source, but it seems there is some controversy about the role of phonons. Some people think they don't play a in high temperature superconductors, for example in news.sciencemag.org/sciencenow/2012/03/… –  jcohen79 Sep 16 '12 at 22:21
Phonons play absolutely no role in HighTc materials, this is certain, because the phonon pairing is no stronger in highTc materials, and phonons always gives S-wave pairing since it is approximately isotropic in the scales where it is stronger than Coulomb repulsion. The pairing in HighTc is due to a mechanism that I think I know, but nobody else agrees with me on. I can tell you if you ask a question on it. The phonons are the source of the force in regular superconductivity for sure, since the critical temperature isotope dependence is correctly predicted from this hypothesis. –  Ron Maimon Sep 16 '12 at 23:45