What does it mean that a Cooper pair behaves as a boson but respects the obligations of fermions?

Question

I refer to the fact that it has integer spin, but antisymmetric wavefunction. How is this possible?

Answer 1

Well, $1/2\otimes1/2=0\oplus1$, so a system with two fermions has integer spin. But it is still a two fermion system, and therefore its wavefunction must be antisymmetric, as usual. This is not specific to Cooper pairs, but is basic Quantum Mechanics... [what is specific to Cooper pairs is that their size is $\gg a_0$, which means they are highly de-localised; this in turns means that they have mixed statistics: they have integer spin and yet they are not bosons].

The wavefunction of any system of any number of fermions must be antisymmetric (and this is one of the postulates of QM). The antisymmetry of the wavefunction is about the total wavefunction, that is, space wavefunction and spin wavefunction; you may have symmetric $\psi(r_1,r_2)$ and antisymmetric $\psi(s_1,s_2)$ or vice-versa. If you write $q=(\boldsymbol r,s)$, then the wavefunction of the pair is $$ \psi(q_1,q_2)=\color{red}-\psi(q_2,q_1) $$

If $\psi(q_1,q_2)=\psi_\mathrm{space}(\boldsymbol r_1,\boldsymbol r_2)\psi_\mathrm{spin}(s_1,s_2)$, then there are two possibilities: $$ \begin{cases}\psi_\mathrm{space}(\boldsymbol r_1,\boldsymbol r_2)=+\psi_\mathrm{space}(\boldsymbol r_2,\boldsymbol r_1)\\ \psi_\mathrm{spin}(\boldsymbol s_1,\boldsymbol s_2)=-\psi_\mathrm{spin}(\boldsymbol s_2,\boldsymbol s_1) \end{cases}\qquad\text{singlet} $$ or $$ \begin{cases}\psi_\mathrm{space}(\boldsymbol r_1,\boldsymbol r_2)=-\psi_\mathrm{space}(\boldsymbol r_2,\boldsymbol r_1)\\ \psi_\mathrm{spin}(\boldsymbol s_1,\boldsymbol s_2)=+\psi_\mathrm{spin}(\boldsymbol s_2,\boldsymbol s_1) \end{cases}\qquad\text{triplet} $$

In some cases the ground state is a singlet and in some others it's a triplet. It turns out that the Cooper pairs is a singlet.

For more details, see this SE post or this article (DOI: 10.1002/pssb.200461754).