I am thinking about something I learnt as an undergraduate. In the section on identical particles (page 212) of Griffiths book on quantum mechanics he speaks of helium and says that:

> 
> The excited states of helium consist of one electron in the hydrogenic
>  ground state, and the other in an excited state
>  $\psi\_{100}\psi\_{nlm}$.
> 
> 
> 

My interest is regarding the quantum numbers. And that he concludes there are precisely three. My feeling is that this is a consequence of the Pauli exclusion principle. But [this thread](https://physics.stackexchange.com/q/518635/97775) explains that the Pauli exclusion principle is the observation that a slater determinant is zero if two columns in the determinant are equal. For the elements of the periodic table, it is not the case — as Hartree-Fock shows. Therefore, I see no reason to believe that the Pauli exclusion principle applies. This makes me wonder how we can know that these are precisely the quantum numbers of helium, for example.

Is there a deeper theorem that still ensures this, or is this experimental?

---

Edit:
I have found out the following: To say that $k \in \mathbb{Z}$ is a quantum number of $\psi$, is to say that there there is a hermitian operator that commutes with $H$ and we denote the indices of the eigenvalues of the commuting operator by $k$. We can have multiple quantum numbers if several operators commute with $H$. Because if $K$ is a hilbert space, $H,L$ are normal on $K$ and $[H,L] = 0$, there is a basis $\{e\_k\}\_{k=1}^\infty$ for $K$ such that $e\_k$ is an eigenvector of $H$ and $K$ for each $k \in \mathbb{N}$. In the case that there are multiple operators that commute, there are multiple quantum numbers as a consequence. 

In the case of hydrogen, both $L\_z,L^2$ commute with $H$ ($H$ always commutes with itself) so there are eigenvalues $\{\varepsilon\_n\}\_{n=1}^\infty$, $\{\lambda\_l\}\_{l=1}^\infty$ and $\{\mu\_m\}\_{m=1}^\infty$ such that
$$
L\_z e\_k = \mu\_m e\_k, \qquad L^2 e\_k = \lambda\_l e\_k, \qquad H e\_k = \varepsilon\_n e\_k.
$$
Therefore the notation $\psi\_{nlm}$ is used to indicate that when one of the operators $H,L^2,L\_z$ act on $\psi\_{nlm}$, we get the values $\varepsilon\_n, \lambda\_l,\mu\_m$. 

The question therefore reduces to: Are there exactly three physically relevant operators that commute with $H$ for helium?