Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
up vote 3 down vote accepted

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.

share|cite|improve this answer
You truly have overestimated my aptitude, yet thanks, however, for the bits that I could understand. – Meow 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? – Meow 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?)? – Meow Sep 15 '12 at 20:38

Yes, the cooper pairs form a BEC. See Kardar's Statistical Physics of Particles for a detailed derivation of this. I think Kittel's Thermal Physics might also have some explanation (on an easier level).

For a BEC to form, the bosons should be non-interacting. To good approximation, this is the case for cooper pairs.

share|cite|improve this answer
How do they actually 'stick together', though? Is it essentially the way that all BEC stick together, and I need to read into it more? – Meow Sep 14 '12 at 18:44
This is not really true--- the field can be considered a BEC of Cooper pairs in a certain sense, it is the expected value of a fermion bilinear, but the pairs wouldn't bind in isolation from the Fermi sea, and the description is entirely weak coupling, you never have to make a binding potential. This is not the correct explanation, although it is what happens at strong coupling and the two are continuously linked by varying the coupling. – Ron Maimon Sep 14 '12 at 18:50
It is also not true that a BEC requires the bosons to be noninteracting. Also interaction (repulsive) bosons form a BEC. In order for the BEC to be superfluid there is even interaction needed! – Fabian Sep 14 '12 at 21:49
Of course there are more interesting BECs. The simplest however is just a bose gas, which is more-or-less a good intuitive model of cooper pairs. – Ryan Thorngren Sep 14 '12 at 22:03
@user404153: It's a good intuitive model, but it's not the real deal. The OP asked what's going on, not the simplest model. I agree that this is the simplest model, it's essentially Landau's. The BCS model though is more interesting because of the nonlocal nature of the bilinear, which allows an arbitrarily weak interaction to cause condensation, when an arbitrarily weak interaction does not cause binding. – Ron Maimon Sep 15 '12 at 5:09

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.

share|cite|improve this answer
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… – 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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.