Consider the helium atom with two electrons, but ignore coupling of angular momenta, relativistic effects, etc.
The spin state of the system is a combination of the triplet states and the singlet state. I will denote a linear combination of the three triplet states as $\lvert\chi_+\rangle$ (because it's symmetrical under exchange of electrons) and $\lvert\chi_-\rangle$ the singlet state (because it's anti-symmetric).
Then, the orbital state of the electrons. Suppose one electron is in the state $\lvert\phi_a\rangle$; the other in the state $\lvert\phi_b\rangle$. The orbital state of the system is:
$$ \lvert\phi_{\pm}\rangle = \frac{1}{\sqrt 2} \left (\lvert\phi_a\rangle\lvert\phi_b\rangle \pm \lvert\phi_b\rangle\lvert\phi_a\rangle \right )$$
Because the overal state $\lvert\psi\rangle$ of the electrons must be anti-symmetric, is it correct to construct it as following:
$$\lvert\psi\rangle = \lvert\phi_{\pm}\rangle \lvert\chi_{\mp}\rangle \text{ ?}$$
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