How Does Antisymmetry Define Electron States in Quantum Chemistry?

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kof9595995
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Let's say a two electron state for Helium atom, I've seen author (Sakurai, Modern QM,section 6.4) wrote [tex]\Phi ({x_1},{x_2})\chi[/tex] as the most general form, where [tex]\Phi ({x_1},{x_2})[/tex] is either a symmetric or antisymmetric wave function, and [tex]\chi[/tex] is the singlet or triplet state respectively. But how about this kinda state:
[tex]{\psi _a}({x_1}){\psi _b}({x_2})| \uparrow \downarrow > - {\psi _a}({x_2}){\psi _b}({x_1})| \downarrow \uparrow >[/tex]
It's also antisymmetric, so isn't it also a possible state?
 

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It's hard to believe but I came up against exactly the same problem just this morning. I wasn't doing the helium atom but I had the same problem.

The state you've written can also be written as the superposition of Sakurai's states, but oddly enough if you've quoted him correctly he doesn't allow that: you quote his as saying the most general state is of type such and such, not superpositions of such states.

I'm pretty sure your example must also be a legal state.
 
I agree, it's just that all books I've read discussed Helium in terms of spin-singlet and spin-triplet state, I just don't see what's really nice about this basis.
 
As long as spin orbit coupling is negligible (as is certainly the case in He), the spin and the hamiltonian have a common basis. So it makes good sense to use an basis of eigenstates of the spin. Half of the problem of finding the eigenstates of the Hamiltonian is then already solved. A more general state will be time dependent and is therefore usually not of the same interest.
 
It's a fair enough reason, thanks.
 
We've managed to describe almost all chemistry in terms of that basis, so I'd say it's pretty useful.
You'd have difficulty finding a chemist who doesn't think in terms of doubly-occupied spatial orbitals.

Quantum chem isn't an exception here either, even when dealing with correlated systems and DFT methods,
the general way of looking at stuff is in terms of how the various Slater determinants contribute in this basis.