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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

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

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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.

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  • $\begingroup$ Thanks a lot! A last question to strengthen this statement: In my exercise sheet it was statet that: One of the particles is known to be in the state described by $|\phi>$ and the other in the state described by $|\chi>$. The state of the system is then for fermions: $|\Psi>=\frac{1}{2}(|\psi(1)>|\chi(2)>-|\psi(2)>|\chi(1)>)$ and the notation that $|\psi(k)>$ denotes that the ket is a function of all variables of particle k (same for $|\chi(k)>$). But apperenltly this is wrong according to what you said? So the statement on my sheet is wrong? $\endgroup$ – Martin Dec 1 '15 at 14:19
  • $\begingroup$ @Martin: Yes, I would call that notation wrong in the abstract bra-ket notation. When you actually write wavefunctions, it becomes muddier, because the multiplication of functions is commutative: $\psi(\vec x_1)\cdot\chi(\vec x_2) = \chi(\vec x_2)\cdot\psi(\vec x_1)$ $\endgroup$ – ACuriousMind Dec 1 '15 at 14:41
  • $\begingroup$ Thanks a lot again! That was also my problem, in the wavefunction formalism everything worked, but when i used this braket-notation from the sheet it didnt...but now i know where i'm at! Another side question: Is it right if I write : $<a(1)|b(2)>=\int dx_1dx_2 <a(1)|x_1><x_1|x_2><x_2|b(2)>==\int dx a^*(x)b(x)$? because that would also clearifty some of my problems, meaning exactly this wave-function commutative-ness $\endgroup$ – Martin Dec 1 '15 at 14:47

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