In every crystal there is a set of spatial ground states, which may be occupied by electrons with opposite spins, i.e. by singlet electron pairs. If every singlet electron pair in the ground state is permanent (i.e. non-breaking), then the singlet pair is energetically more favorable than unpaired electrons. Indeed, the spin alignment of unpaired electrons randomly fluctuates (for example, due to thermal pair-breaking), forming either a temporary singlet or triplet. Thus, there is a non-zero probability that the unpaired electrons are a triplet and cannot simultaneously occupy the lowest spatial energy state. One of unpaired electrons must take a higher state, increasing the total system energy. A stable electron pair becomes possible when there is no thermal energy for the electron transfer into the higher state. Then the pairing energy is the difference between two energies: E1. Energy of the permanent singlet in the spatial ground state; E2. Energy of unpaired electrons avoiding the same spatial ground state.

The pairing energy (E1-E2) is not necessarily negligible, so below a certain temperature (say Tc) the electron pairs become stable. Can we assert that the electron pairing in superconductors is a direct consequence of the Exclusion principle? (the key word here is "direct", i.e. pairing without any mediators as phonons, excitons, plasmons, magnons etc.)

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    $\begingroup$ Only having electrons in singlet states is not enough to get superconductivity. $\endgroup$
    – Samuel
    Nov 2, 2022 at 6:42
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    $\begingroup$ Please, correct me if I am wrong, I am not an expert in the field. When you say "singlet electron pairs" it seems that this is not really the same as Cooper pairs. Moreover, even having Cooper pairing is probably not enough to explain the 100% of different existing superconductors.en.wikipedia.org/wiki/Cooper_pair $\endgroup$
    – Quillo
    Nov 2, 2022 at 7:31
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    $\begingroup$ I don't really get what this question is asking, but Bose-Einstein condensation is probably less directly related to Pauli exclusion than you think, see physics.stackexchange.com/a/706581/50583 $\endgroup$
    – ACuriousMind
    Nov 2, 2022 at 8:39
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    $\begingroup$ Yes, two electrons in a singlet state is not the same as a Cooper pair. For the latter, the electrons need to have opposite momenta. BCS theory can explain low $T_C$ superconductors, so called s-wave superconductivity. There are adaptations for d-wave SC, but it does not completely explain high $T_C$ SC. $\endgroup$
    – Samuel
    Nov 2, 2022 at 8:44

1 Answer 1


Thank you, all comments are correct. Indeed, singlet states are not enough to get superconductivity (SC), singlet pairs are not the same as superconducting pairs and Bose-Einstein condensation is directly related to SC. You found the correct answer: for the superconductivity the singlet electron pairs must form Bose-Einstein-condensate (BEC). Then the energy of every pair is minimum and quantized as energy of bosons in the BEC, so the pair dynamics is non-dissipative below a certain pair-breaking temperature. And the pair-breaking energy (kTc) is exactly the energy necessary to destroy the permanent singlet, i.e. to create unpaired electrons from stable singlets. The Pauli Principle provides that the unpaired electrons cannot stay permanently and simultaneously in the lowest energy state, the unpaired electrons must take other (higher) energy states. Thus, the pairing energy follows directly from the Pauli Principle.


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