The usual explanation for superconductivity is that the electrons form Cooper pairs, which are bosons. This effective boson then condenses. E.g., quoting Wikipedia,

Therefore, unlike electrons, multiple Cooper pairs are allowed to be in the same quantum state, which is responsible for the phenomenon of superconductivity.

This doesn't make much sense to me. If the electron is described by a field $\psi_\alpha$, then the pair is described by some composite $\Phi\sim\psi_\alpha\psi^\dagger_\beta$. It is true that $\Phi$ is a boson, but it is not a regular boson: its square vanishes, $\Phi^2\sim\psi^2\psi^{\dagger2}\equiv0$, since $\psi^2=0$. So the pairs are not actually allowed to be in the same state, $\Phi$ cannot have arbitrarily large occupation number.

In some cases one can make this issue explicit. For example in 2d one can bosonize a fermionic degree of freedom, $\phi\sim \bar\psi\psi$. The field $\phi$ is a boson, but it turns out that $\phi^2$ is a singular field (it creates null states), as one would expect from its fermionic version. Similarly in 3d Minwalla and collaborators have shown that the bosonized version of a fermion field still satisfies the (appropriate) version of spin-statistics, as expected from the fermionic presentation.

So, how can I understand Cooper pairs? I find it hard to believe the usual claim that they can be in the same quantum state.


1 Answer 1


Disclaimer: I'm not a condensed matter theorist, and I find most statistical physics utterly impenetrable. However, this question also bothered me long enough to try to figure it out. Here's my unqualified interpretation:

Of course, you are correct that "Cooper pairs are bosons and hence can show Bose-Einstein condensation" doesn't really make a lot of sense. What is actually meant by that pithy phrase is the following:

  1. We're only looking at electrons near the Fermi surface, i.e. those that participate in conduction.

  2. These electrons can form Cooper pairs, which are long-range bound states whose energy is lower than that of the two unbound electrons.

  3. Since the electrons are all near the Fermi surface, they all have about the same energy.

  4. We can now think of the conductor as a "gas of electron pairs", with the ground state being the Cooper pair state and the excited states being unbound states of the electrons. Since the electrons all have about the same energy, the pairs "behave like bosons" - the energy of the Cooper pair of two electrons is about the same as that of any other Cooper pair, and so as long as we just care about energy we don't really case whether the Cooper pairs are actually the same state (in the sense forbidden by Pauli exclusion), we can just treat it like that. The pairs are still distinct quantum states - their momenta still differ, within a small band around the Fermi surface.

  5. Therefore, we have "bosons" (=electron pairs) that can either occupy a ground state - the Cooper pair - or an excited unbound state, even if these aren't actual single quantum mechanical states. By the general logic of Bose-Einstein condensation, we get that there is some critical temperature below which a lot of these bosons will condense into the ground state, i.e. below which the electrons will form Cooper pairs. Note that it is crucial for this that the Cooper pair state is gapped to the unbound state, i.e. that there is an actual energy gap between it and the free pair, just as Bose-Einstein condensation relies on there being a gap between the ground state and the first excited state.

Hence, the magic here isn't that the Cooper pairs somehow magically allow us to circumvent Pauli exclusion, it's that they provide a gapped ground state that allows for condensation of electron pairs from free states into this bound state to happen. Without the pair state, there's just a continuous range of states around the Fermi surface, and so no gap, and no condensation.

  • $\begingroup$ thanks, sounds very reasonable to me. Not sure why someone disagrees with this, it would be nice to hear some explicit complaints, if they are any $\endgroup$ Commented May 13, 2022 at 17:40
  • $\begingroup$ I think this sounds pretty good, I will just comment that in the BCS ground state it is actually the case that all the electrons are paired into Cooper pairs (see my answer linked above). However, this pairing doesn't really have any physical significance when far away from the Fermi surface where they are all locked into their states. $\endgroup$
    – Rococo
    Commented May 28, 2023 at 4:46

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