I've been introduced to multiple particle systems in quantum mechanics, and in the case of the $2$-electron system, I'm facing this massive confusion.
In the ground state of a $2$-electron system, you have both electrons occupying the same exact energy level, one is spin up, the other is spin down. My question is, why does this wave function have to have a singlet spin part?
I've read many answers here, and it claims that, since the spatial wavefunction for two electrons in the same energy level, is symmetric under exchange, the spin part must be antisymmetric. This is because, the particle is a fermion, and must always have an antisymmetric wavefunction.
However, according to the books that I've read, ( Griffiths, Zettili ), the concept of the symmetric and antisymmetric wave function is important only when the two particles are indistinguishable. However, that is not the case here. Due to the spin projection, the two particles here are distinguishable. In that case, why do we care so much about creating an antisymmetric wave function ?
Is the book wrong or misleading here, or is there a better reason for the singlet state? The book and the answers over here, clearly seem to contradict each other, unless I'm missing something. Any explanation would be highly appreciated.
Best Answer
Yes but which one has its spin up, and which one its spin down?
Surely calling the first electron "one" and the second electron "the other" is arbitrary so the trick we have to deal with this arbitrariness and make the state independent of this arbitrary labelling is to make the wavefunction either symmetric (for bosons) or antisymmetric (for fermions). This way, there is no such thing as electron "one" and electron "other". That's why we make a state symmetric or antisymmetric.