Possible eigenstates for identical fermions?

Suppose we have two identical $s=1/2$ fermions which can be in the (unperturbed) one-particle states $\psi_a(\mathbf{r})$ and $\psi_b(\mathbf{r})$. My professor claims the possible eigenstates are the following:

Symmetric spatial part

$\psi_1 = \psi_a(\mathbf{r_1})\psi_a(\mathbf{r_2}) \vert 00\rangle$

$\psi_2 = \psi_b(\mathbf{r_1})\psi_b(\mathbf{r_2}) \vert 00\rangle$

$\psi_3 = \dfrac{1}{2}(\psi_a(\mathbf{r_1})\psi_b(\mathbf{r_2}) + \psi_b(\mathbf{r_1})\psi_a(\mathbf{r_2})) \vert 00 \rangle$

Anti-symmetric spatial part

$\psi_4 = \dfrac{1}{2}(\psi_a(\mathbf{r_1})\psi_b(\mathbf{r_2})-\psi_b(\mathbf{r_1})\psi_a(\mathbf{r_2})) \vert 1m \rangle$

where $m=-1,0,1$.

My questions are:

1) How can two identical fermions occupy the same state ($\psi_{1,2}$)?

2) Why is the normalization factor $\dfrac{1}{2}$ rather than $\dfrac{1}{\sqrt{2}}$ for $\psi_{3,4}$?

• The total wave function of the two fermion system must be anti-symmetric...here you have two parts, space and spin and one of them can be antisymmetric to render the total wave function antisymmetric. – drvrm Aug 15 '18 at 17:33
• @drvrm Yes, that is clear to me. However, what is not clear is how the fermions can occupy the same spatial state (which is what happens in $\psi_1$ and $\psi_2$). Does this not violate the Pauli exclusion principle? – Lozansky Aug 15 '18 at 17:39
• @Lozansky-no it should not as their spin states are different.however if their space functions are antisymmetric then they can have symmetric spin functions. – drvrm Aug 15 '18 at 19:02
• Related : Total spin of two spin-1/2 particles. – Frobenius Aug 15 '18 at 21:38
• @Lozansky-the factor half..my guess is it may be due to two normalization constants – drvrm Aug 16 '18 at 19:49

$\Psi = \psi \chi$, where $\psi$ is the wave function and $\chi$ is the spin part. I will not go into the details of the math here, just look at it like: there is a spatial and a spin part. What matters is that $\Psi$ is antisymmetric, so either $\psi$ symmetric and $\chi$ antisymmetric or vice versa.