Some people say it is bound state, some say it is not. Which is more accurate? Problem is that I read in some books, including Ziman, that Cooper pairs are bound states but my teacher says that it is not true and that Bardeen had to explain it many times even to his peers.Now, i know that it has something to do with the resonance in scattering cross section...but oversimplifications with hand-waving about phonons that mediate interaction creating an attractive force and a bound state, like some kind of electron-electron molecule just make me angry. I know it is phonon mediated but it is not that simple, right?
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1What's your definition of a bound state? – ACuriousMind Nov 14 '14 at 19:15
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1Cooper pairs are not bound states! – mcodesmart Nov 14 '14 at 19:50
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1Bound state, a state in which there is some kind of potential, keeping the particle like electron, orbiting a proton, eg... – Žarko Tomičić Nov 15 '14 at 14:02
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So, it is some kind of correlated motion, and, interestingly, time-reversed relatively? – Žarko Tomičić Nov 17 '14 at 10:33
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meaning, momentum k+ and k-, or something? of two electrons... – Žarko Tomičić Mar 31 '15 at 16:02
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@FraSchelle Your comments could probably be turned into an answer – By Symmetry Sep 11 '16 at 10:26
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@BySymmetry Thank for the suggestion. I did it, and erase the comments. – FraSchelle Oct 08 '17 at 13:48
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Yes, Cooper pairs are bound states of electrons held together by an attractive potential mediated by phonons. It's explained pretty clearly in the Wikipedia page. But they're often very loosely bound so the individual electrons remain very far apart, so you shouldn't think of a Cooper pair as a localized particle-like object.
Regarding your comment to your question, the electrons almost always have opposite momenta, but there are exotic "FFLO" states where they do not.
tparker
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As Cooper wrote in his paper (1956), one Cooper pair is a bound-state. The point is : it's a bound-state if you consider 2-electrons on top of the Fermi sea with an attractive potential. In the many-body case, better not to think in term of bound state. A Cooper pair is a fictitious particle which has no clear meaning, although the main characteristics of superconductivity (bosonic electrodynamics of charge 2e) suggests that such a picture might be fruitful.
In field theory, a Cooper pair is just a non-trivial correlation between fermionic creation operators in the ground state : the anomalous Green-Gork'ov functions $F^{\dagger}\sim\left\langle c^{\dagger}c^{\dagger}\right\rangle $.
Following some comments from @tparker : Cooper pair has nothing to do with any kind of motion. "Cooper pair" simply means that the correlation function made from two creation operator (say $\left\langle c^{\dagger}c^{\dagger}\right\rangle $ with the average procedure done on the ground state) is non-vanishing, that's all. For historical/conventional superconductivity, the correlations appeared between two time-reversal electronic states, the so-called $s$-wave superconductivity. It makes the superconducting state robust to some interaction, in particular it gives to the condensate an immunity with respect to disorder (the Anderson's theorem). But this can be seen as extra properties on top of the correlations.
FraSchelle
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The question asked about physics and not formalism. I feel this answer did not clarify any physics, and simply restated the formalism. – ComptonScattering Oct 27 '21 at 14:16