Total irreducible representation of a quantum system - Slater determinant?

I'm beginning to study quantum chemistry and I became pretty confused about the total state of the molecule.

When we describe the single orbitals using irreducible representations like in the picture below, then we can get the total quantum state by direct product $$(2a_1)^2(1b_2)^2(3a_1)^2(1b_1)^2=(A_1)^2\otimes (B_2)^2\otimes (A_1)^2\otimes (B_1)^2 =A_1,$$

as it is described in this answer. But when I read about multi-particle wavefunctions, I realized, that the wavefunction of the whole system should be antisymmetric, so I should assemble Slater determinant to get the multi-particle wavefunction with the enforced antisymmetry property.

Questions

1) How is Slater determinant connected with the direct product above?

2) Does $(A_1)^2$ mean $A_1 \otimes A_1$ or is it purely the notation for 2 electrons in the orbital?

• I haven't read through it in detail, but I think this should help: google.com/url?sa=t&source=web&rct=j&url=http://… – Tyberius Mar 1 '18 at 23:59
• (1) It isn't. (2) Yes. – Ivan Neretin Mar 2 '18 at 4:40
• @IvanNeretin Would you mind to respond in more detail? So, why can we find the total irreducible representation using direct product? And by "yes" you mean, that $()^2$ is purely the notation and not power? – Eenoku Mar 2 '18 at 9:32
• The Slater determinant is just a linear combination of these direct products. The symmetry will still be the same. – orthocresol Mar 2 '18 at 10:14
• Yes (extra text to satisfy comment length) – pentavalentcarbon Mar 2 '18 at 13:27