In Cooper pair, is there an essential condition on spin of electrons forming pair? Also explain the reason behind it. And what is the charge on them.


The quantum state of electrons forming Cooper pairs are plane wave (free electrons from the Fermi sea of the conducting band). So the spatial part of the quantum state of the pair is symmetric under exchange of the two particles. The total state shall be anti-symmetric (Fermi rule), so the angular momentum part shall be anti-symmetric. But then since neither electron has an orbital momentum, and assuming the pair has no orbital momentum, we are just adding two spin 1/2. The result is textbook: we have an anti-symmetric state with spin 0 and three symmetric states of spin 1. So the spin shall be 0.

As for the charge, the answer is straightforward: -2e

  • $\begingroup$ What about charge on them? $\endgroup$ – Ka Sikh May 15 '17 at 16:00
  • $\begingroup$ yes, sorry, just added that $\endgroup$ – user154997 May 15 '17 at 16:28
  • $\begingroup$ The projection $m_\ell$ of the orbital angular momentum quantum number $\ell$ applies to the system with angular momentum, not to each particle in the system; it doesn't make sense to talk about the $m$ for "each electron." (This is different from the atomic case, where we sometimes forget that the nucleus is involved.) Also a Cooper pair of electrons with $\ell=1$ must have $S=1$, in order for the entire wavefunction to be antisymmetric, in analogy to the protons in orthohydrogen and parahydrogen. I don't know whether the $\ell=1$ pairs are involved in superconductivity or not. $\endgroup$ – rob May 15 '17 at 16:35
  • $\begingroup$ @rob My prose is quite confusing indeed. I was thinking each electron is bound to a nuclei and those l,m referred to that quantum state. $\endgroup$ – user154997 May 15 '17 at 16:41
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    $\begingroup$ My understanding of superconductivity is not what I'd like, but I thought the Cooper pairs formed in the Fermi gas of free electrons in the conductor and that those electrons are not associated with any of the positive ions that make up the crystal lattice. $\endgroup$ – rob May 15 '17 at 16:44

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