Two electrons are in a sphere. How does the total spin depend on the size of the sphere? I was looking at Landau's entrance exam problem set from this compilation
http://people.tamu.edu/~abanov/QE/TM-QM.pdf
I stumbled across problem 186 which asks:
Two electrons are inside a sphere. Find (qualitatively) how the total spin depends on the radius of the sphere $R$.
I would think naively that the fact that there are two electrons are important but I fail to see why the spins are affected. $SO(3)$ should still be preserved under scaling of the sphere so the spins should be unaffected.
 A: I have to imagine that the question is referring to the total spin of the ground state.
The ground state of two non-interacting fermions confined to a spherical well is the state in which they have the same (ground state) spatial wavefunction while occupying the antisymmetric spin singlet, so their total spin is $s=0$.  In the first excited state, the spatial wavefunction $\propto \psi_0(x)\psi_1(y) - \psi_0(y)\psi_1(x)$ is antisymmetric and they occupy the symmetric spin triplet with $s=1$.
Once we turn electrostatic interactions on, there will be an extra contribution to the total energy which affects the symmetric spatial wavefunctions more than the antisymmetric spatial wavefunctions - heuristically, when the spatial wavefunction is symmetric, the electrons are "closer together" on average.  This contribution scales like $1/R$ where $R$ is the radius of the sphere.  On the other hand, the difference between the ground state and first excited state of the non-interacting particles scales like $1/R^2$.
As a result, there will be a fight between the "kinetic" term and the potential energy term, with the dominant term depending on the size of the sphere. In turn, this affects whether the ground state wavefunction will be symmetric (which requires the antisymmetric $s=0$ spin singlet) or antisymmetric (which requires the symmetric $s=1$ spin triplet).
