A pair of electrons has spin 0 which makes any such system a boson rather than a fermion. The Pauli exclusion principle does not apply therefore to paired electrons and any such two electrons can coexist in the same quantum state. Wouldn't it make electron shells collapse all into the lowest 1s²?

How can paired electrons still be considered fermions if their total spin is not half-integral.

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    $\begingroup$ What mechanism produces electron pairing in an atom? $\endgroup$ Jan 22 at 12:44

1 Answer 1


The abstraction of two fermions as one single particle creates not an ordinary boson, but a hard core boson. Hard core bosons share the Pauli exclusion principle with fermions in the sense that two hard core bosons cannot occupy the same state. This is true regardless of wether the fermions have actually paired through an interaction, or you just arbitrarily chose to describe them in pairs

That way, even if you were to describe the electrons in an atom by pairing them into bosons, these would be hard core bosons and so they could not all collapse to the lowest orbital because that would be already occupied.

The two pictures are thus consistent, as they should be.

EDIT: I have recently come across this answer by @Chiral Anomaly to a related question. Because of that I've come to realize my answer above is incomplete, if not plainly wrong. Consequently, my arguments in the comment section below might be wrong, I will need to review them.

As they discuss in the linked answer, you actually can put two composite bosons in the same state (meaning $(b^\dagger)^2\neq 0$), as long as the bosons are formed by entangled fermions.

However, as far as I know, composite fermions, coming from entangled fermions or not, still have distinct features from pure bosons. Let us take the example from the linked post: \begin{equation} b^\dagger(f)=\sum_{n,m}f(n,m)a_n^\dagger a_m^\dagger, \tag{1} \end{equation} where $a_n^\dagger$ are orthogonal fermionic modes. First, it can be checked that their commutation relations are in general not strictly bosonic, meaning \begin{equation} [b,b^\dagger] \neq 1, \tag{2} \end{equation} for any normalization.

Moreover, if the entanglement between the generating fermions occurs inside a finite subspace (the sum over $n,~m$ has a finite number of elements $N$), then there is a maximum number of composite bosons that can be put in a given state, meaning $(b^\dagger)^{N+1} = 0$.

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    $\begingroup$ On the other hand, can't superconductivity be viewed as Bose condensation of Cooper pairs? $\endgroup$ Jan 22 at 13:24
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    $\begingroup$ @Roger Vadim The pairs condense to not exactly the same state, though. Each pair in the BCS groundstate has a unique quantum number (the relative momentum/wavenumber). (More technically the Bose-Einstein condensation and the BCS condensation are different regimes, but the ideas we discuss are true for both, I believe.) $\endgroup$ Jan 22 at 13:36
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    $\begingroup$ In the BEC regime I think the relevant quantum number might be position, as the pair size is considerably smaller, but I am talking out of my depth now. Still I believe each pair must have a unique quantum number $\endgroup$ Jan 22 at 13:47
  • $\begingroup$ How does this relate to the existence of Bose Einstein condensation in eg rubidium? These atoms are made of fermions but apparently can bose-condense. $\endgroup$ Jan 22 at 17:28
  • $\begingroup$ @Jahan Claes, I don't know much about these systems, but from what I read it appears they measure a peak in the density of states around zero velocity. This is consistent with the hard-core bosons view because this peak has a finite width, such that the states are not bunched up to exactly zero velocity, but are distributed around it. However, I don't think this finite width is enough to verify that the particles that are condensing are hard-core bosons, as opposed to usual bosons, because a finite uncertainty in velocity is also caused by confinement and other experimental factors. $\endgroup$ Jan 29 at 15:26

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