My motivation for this question comes from the periodic table. There, the many-body ground state electronic configurations of certain atoms like boron or carbon can have a nonzero total orbital angular momentum. This can be interpreted as arising from the Pauli exclusion principle if one takes the approximate picture of electrons filling a Slater determinant of single-body orbitals, some of which have nonzero orbital angular momentum. If I instead imagine that electrons were bosons, it's much more tempting that the total orbital angular momentum would be zero for all atoms regardless of the number of electrons. This would stem from an approximate picture of these bosonic electrons filling the $1s$ state repeatedly, but it's not so clear to me whether there's a nice proof of this in the presence of electron-electron Coulomb repulsion.
Is there a proof that the ground states of atoms featuring spinless, bosonic "electrons" would have total orbital angular momentum zero?
As an aside, I suspect that more generally there's a well-known proof that a Hamiltonian for spinless bosons of the form $H=\sum_i \left( \frac{p_i^2}{2m} + V(|\vec{r}_i|) \right) + \frac{1}{2}\sum_j \sum_{i \neq j} U(|\vec{r}_i - \vec{r}_j|)$ necessarily has a ground state of zero angular momentum.
