Is the Singlet state for Helium with 2 electrons symmetric rather than anti-symmetric as is meant to be for fermions?

Question

I'm looking at two-electron Helium atoms where one electron is in the ground state (due to if it were in other states, it's de-excitation would simply lead to the ionization of the electron). The other electron is in an arbitrary excited state. What does this notation mean in terms of the allowed wavefunctions?:

Specifically, for the singlet state, I understand that the both electrons must have opposite spin to give a total overall spin of Zero (correct me if I'm wrong). But why does this mean suddenly the wavefunction, $\psi_{+}$ is allowed to be symmetric. I thought all fermion wavefunctions had to be anti-symmetric havinga $\psi_{-}$ state.

For the triplet state, as I understand it, the electrons have a total spin number of 1 meaning that they have to have the same spin as each other (+1/2 each or -1/2 each). Is this correct? Further to this, I don't specifically understand the notation in the equations above. Specfically, I don't understand the last bit added on the end: What does this actually mean?

I understand this might be a silly question as it could just be me mixing up notation but any help will be appreciated. Thank You! :)

EDIT: See AV23 Comment below. Answer! :) Basically, find the $\hat{S}^{2}$ eigenvalues to deduce S using below:

Answer 1

The $|00\rangle$ and $|1,M_S\rangle$ represent the spin singlet and triplet states. The overall wavefunction must contain both the 'space' part and the 'spin' part. We can schematically express this as follows: $$\psi \sim \psi(\mathbf{r}_1, \mathbf{r}_2) |s\rangle$$

Now, Pauli's exclusion principle demands the antisymmetry of the overall wavefunction. For the singlet state, as $|00\rangle = \frac{1}{\sqrt 2}\left(|+-\rangle - |-+\rangle\right)$ is already antisymmetric, $\psi_+$ can be antisymmetric only if the space part is symmetric.

The triplet states: $$|1, 1\rangle = |++\rangle$$ $$|1, 0\rangle = \frac{1}{\sqrt 2}\left(|+-\rangle + |-+\rangle \right)$$ $$|1, -1\rangle = |--\rangle$$ are all symmetric, and therefore, the space part corresponding to them must be antisymmetric.

Note that the triplet state $|1, 0\rangle$ does not have both individual spins in the same direction. However, it still has a total spin of $1$.

In summary, $\psi_+$ and $\psi_-$ are both antisymmetric; it is just that their space & spin parts are symmetric & antisymmetric and antisymmetric & symmetric respectively.