Antisymmetry requirement for the total wavefunction My understanding is that if we are dealing with a system of two electrons, the total wavefunction needs to be antisymmetric only when the two electrons have same value of n and l ( i.e. they are equivalent). In the case of non-equivalent electrons, the total wavefunction doesn't necessarily have to be antisymmetric.
Is it correct? If not, can you explain the requirement of antisymmetry in case of equivalent and non-equivalent electrons?
( I was reading about the term symbols and terms like ${^3S}$ had me question my understanding of Pauli exclusion principle, as both spin and space parts are symmetric here.)
 A: The wavefunction is always antisymmetric. It does not matter if they have similar, different, or even identical numbers for $n$ and $l$.
But let's be clear. The wavefunction is a function from the configuration space into the joint spin state. So for a spin 1/2 particle the wavefunction is a function from $\mathbb R^3$ to $\mathbb C^2$ whereas for two spin 1/2 particles the wavefunction is a function from $\mathbb R^6$ to $\mathbb C^2\otimes\mathbb C^2$ and it has to be antisymmetric if they are indistinguishable fermions, symmetric if they are indistinguishable bosons, and no restrictions (could be symmetric, antisymmetric, or neither) if they are distinguishable.
And by the symmetries we mean things like $\Psi(\vec a,\vec b)=-\Psi(\vec b,\vec a).$
So let's do an example. If $\psi_{nlm}:\vec r\mapsto\psi_{nlm}(\vec r)$ is a solution with a particular value of $n,$ $l,$ and $m$ for the scalar (spin zero) Schrödinger equation for a 1/r potential then you could make a spin 1/2 solution by having $$\Psi:\mathbb R^3\rightarrow\mathbb C^2:\vec r\mapsto\psi_{nlm}(\vec r)\left[\begin{matrix}\alpha\\ \beta\end{matrix}\right].$$
For a pair of indistinguishable spin 1/2 particles (e.g. two electrons, two muons, etcetera) a possible wavefunction is
$$\Psi:\mathbb R^6\rightarrow\mathbb C^2\otimes\mathbb C^2,$$ $$\Psi:(\vec a,\vec b)\mapsto\psi_{nlm}(\vec a)\psi_{NLM}(\vec b)\left[\begin{matrix}\alpha\\ \beta\end{matrix}\right] \otimes\left[\begin{matrix}A\\ B\end{matrix}\right]-\psi_{nlm}(\vec b)\psi_{NLM}(\vec a)\left[\begin{matrix}\alpha\\ \beta\end{matrix}\right] \otimes\left[\begin{matrix}A\\ B\end{matrix}\right].$$
In fact you can build more complicated states out of these antisymmetric states. But the note complicated states will remain antisymmetric.
