What is the correct way of writing the antisymmetrized state of two identical fermions?

Question

I am just confused:

If I have 2 identical fermions, where one of them is in state A and the other one is in state b, and they are normalised and orthogonal, which statement is right:

1) $|\Psi\rangle=\frac{1}{\sqrt{2}}\left(|a_1\rangle|b_2\rangle-|a_2\rangle|b_1\rangle\right)$

or

2) $|\Psi\rangle=\frac{1}{\sqrt{2}}\left(|a_1\rangle|b_2\rangle-|b_1\rangle|a_2\rangle\right)$

when '1' and '2' denote the coordinates. I think it should be 2, but in my script its different...

edit: Okay, I now think 1) should be right, but it arises another question: If I calculate $\langle\Psi|\Psi\rangle$ I get

$\frac{1}{2}(\langle a_1|a_1\rangle\langle b_2|b_2\rangle+\langle a_2|a_2\rangle\langle b_1|b_1\rangle-\langle a_1|a_2\rangle\langle b_2|b_1\rangle-\langle a_2|a_1\rangle\langle b_1|b_2\rangle)=\frac{1}{2}(1+1-1-1)=0$

Answer 1

Your second option is correct.

Writing $$ \lvert a_1\rangle\lvert b_2 \rangle - \lvert a_2 \rangle \lvert b_1\rangle$$ does not make sense. The notation $\lvert \psi_1\rangle\lvert\phi_2\rangle$ is shorthand for the state $\lvert\psi_1\rangle\otimes\lvert\phi_2\rangle$ in the tensor product $\mathcal{H}_1\otimes\mathcal{H}_2 $ of the Hilbert spaces $\mathcal{H}_1$ and $\mathcal{H}_2$ of the individual partices. But $\lvert a_2\rangle\lvert b_1\rangle$ would live in $\mathcal{H}_2\otimes\mathcal{H}_1 $, which is a different (although isomorphic) space, and you cannot add/subtract elements that lie in different vector spaces.