I'm trying to improve my conceptual understanding of Cooper pairs in a superconductor. Not how they come about such as discussed in this question, but more in the sense of what kind of picture I should have of the pair inside the SC itself.

The way I understand it is as follows: Cooper pairs are in fact bound states of two electrons. However, they are not bound states like we know them from molecules where the two guys are flying together all the time, like in an oxygen molecule. They are bound, but weakly bound. So I would suppose that for both electrons, if you follow them individually, they kind of fly around randomly. But if you study the correlations between them, you will see that there are non-zero correlations between them.

Is this an agreeable way of looking at them, or can someone offer an alternative/more refined/different picture? And how does one 'evolve' this picture of weakly-bound/correlated pairs into a single wavefunction with just a phase and an amplitude? Is that (as the above answer described) due to the fact that the pairs are (due to their weak binding) in general quite far away from each other, and thus their wavefunctions overlap with essentially all other pairs in the SC, forming the condensate?

I'm putting this in italics as it kind of distracts from the question, but does motivate my thought process a little. The reason I started thinking about this is motivated by Andreev bound states in SNS junctions (which result in the DC Josephson effect). To understand that one can think of an electron in the normal metal (N), which when incident upon the S part nucleates a Cooper pair in the S part while turning into a hole moving in the other direction in the N part. The hole then arrives at the other S part and destroys a Cooper pair there, before being reflected back as an electron again and repeating the process, leading to a supercurrent. Of course one can instead say that you have two pieces of superconductors with overlapping wavefunctions, resulting in the same physics. I understand that multiple pictures can be used to explain the same physics, but I'm trying to motivative how the two are connected.

  • $\begingroup$ This might help. physics.stackexchange.com/questions/146821/… $\endgroup$ Oct 6, 2017 at 15:41
  • $\begingroup$ That's useful indeed, emphasizing that in a many-body system it is better to think about the correlated nature. I'm still trying to make it fit together though, as in some scenario's the pair picture works extremely well, such as for understanding the DC Josephson effect/Andreev bound states in SNS junctions. Using the correlation picture there is much less intuitive than Cooper pairs, but one should probably still be able to think of it in that way, somehow. $\endgroup$
    – user129412
    Oct 7, 2017 at 11:21

1 Answer 1


There are many many ways to think of the superconducting condensate. I listed a few of them in a related question. Depending on the context, to use one picture or an other might be more or less illuminating.

What is nevertheless clear is that Cooper pairs are not some kind of bound state, except for the really stringent Cooper problem (more about that in this answer). By the way, if it was a bound state, the two-electrons should stay together in the flow of the Josephson current, the notion of Andreev reflexion should not exist, ... as it is the custom for a molecule to not disintegrate at zero-temperature. (Note : surely you may weaken the notion of bound state and continue to think of Cooper pairs in that way, here I use the (strict) equivalence bound state = molecule).

As for the wave-function and the phase problem, it has been discussed in some other places many times. The crucial aspect I'd like to mention is the fact that the correlations between two electrons behave as bosonic excitations. Hence you might see a superconductor as a Bose-Einstein-like condensate (with initial particle of charge 2e). A Bose-Einstein condensate has a global phase and is well-known to be described by a single wave-function (having a global phase and behaving as a unique wave-function mean the same thing).

I did the following remark many times on this website. Usually it remains unclear for readers. I think it's because many users of this website have background in high-energy physics. Anyways, here it goes. Superconductivity is not the process of making correlations between two electrons mediated by a coherent phonon at low temperature. It's the process of making correlated bosonic excitations (usually called the Cooper pairs) on top of a Landau liquid (usually thought off as constituted by bare electrons) via bosonic virtual exchange (usually named phonon). Nevertheless, an electron in a Landau liquid is an emergent concept and it is not an elementary particle (for historical reasons it's unfortunately called an electron), a Cooper pair is an emergent concept and it is not a bound state, and a phonon is just one description of a cristal in terms of a bosonic bath. Because everything in solids are in fact correlations described in a quantum field fashion, Cooper pairs is just one more possibility of correlations at low temperatures.

  • $\begingroup$ Thanks for linking the two answers, and especially for the last paragraph. I myself am not a high-energy physics guy but am instead just beginning to operate in the realm of condensed matter (with an emphasis on quantum information processing applications), so that is a perspective that is useful for me to understand and embrace. The electron being emergent will take some more thinking from my side, but overall I am happy with this answer, also thanks to the other parts such as the connection to the BEC. $\endgroup$
    – user129412
    Oct 9, 2017 at 7:24
  • $\begingroup$ Be careful, when I say electrons are emergent I just mean that they are quasiparticle of the Landau-Fermi liquid : the bare electron surrounded by the other electrons and the phonons behave like a free electron. That's the meaning of the Landau liquid theory. But clearly, a superconductor is nothing but the gas of these quasiparticles which condense according to the Cooper mechanism. $\endgroup$
    – FraSchelle
    Oct 9, 2017 at 7:45

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