# Are combined fermion wavefunctions still antisymmetric after wavefunction collapse?

I suppose $$|A,B\rangle = |A\rangle |B\rangle$$ is a "2 non-identical particle state" in your notation, so that your first spin state looks symmetric. This is impossible for identical electrons, unless there is a "spatial part" of the wave function that, in this case, must be antisymmetric (the spatial wavefunction must be antisymmetric if the two electrons are in a spin triplet state but symmetric if they are in a spin singlet state).

How identical particles work: you cannot have a state as $$|A\rangle |B\rangle$$, where the first particle is in state $$A$$, but you have to sum more information, so to hide the possibility to know which is the "first particle". The obvious way is to go for
$$|A,B\rangle_{a} \propto |A\rangle |B\rangle - |B\rangle |A\rangle$$ or $$|A,B\rangle_{s} \propto |A\rangle |B\rangle + |B\rangle |A\rangle$$ The states labelled by $$a$$ or $$s$$ are the physical states of identical particles. There is no notion as "the first particle is in $$A$$", so this is not something you can measure. Hence, no collapse to $$|A\rangle |B\rangle$$ or $$|B\rangle |A\rangle$$, as no device can distinguish identical particles in the first place.

It depends on what the notation $$\left|\uparrow\downarrow\right\rangle$$ means. When talking about Bell states, the left and right positions in the ket normally represent different spatial locations. In this case it doesn't matter whether the particles are of the same type since they can be distinguished by location. Even if both particles are electrons, the states $$(\left|\uparrow\downarrow\right\rangle+\left|\downarrow\uparrow\right\rangle)/\sqrt2$$ and $$\left|\uparrow\downarrow\right\rangle$$ are fine. If you wrote them out in gory detail they would be something like $$(\left|\uparrow_A\downarrow_B\right\rangle-\left|\downarrow_B\uparrow_A\right\rangle+\left|\downarrow_A\uparrow_B\right\rangle-\left|\uparrow_B\downarrow_A\right\rangle)/2$$ $$(\left|\uparrow_A\downarrow_B\right\rangle-\left|\downarrow_B\uparrow_A\right\rangle)/\sqrt2$$

where $$\uparrow_A$$ means an electron with position $$A$$ and spin $$\uparrow$$, and left and right placement in the ket is now an arbitrary label instead of an indicator of spatial position.