Question about second quantization I have a question about second quantization. For two fermions, one in state $\alpha_1$ and the other in state $\alpha_2$, using occupation number representation, one can express them as $|n_{\alpha_1}=1,n_{\alpha_2}=1\rangle$. Now, if I exchange these two fermions, after that there is still one particle in $\alpha_1$ and one particle in $\alpha_2$, and I can still write them as $|n_{\alpha_1}=1,n_{\alpha_2}=1\rangle$. But we know, for fermions, if one exchange them, there is an extra minus sign introduced. So what's wrong in the above statement? Thank you.
 A: The two kets you have written down are NOT equivalent. The first state when written correctly is:
$|\alpha_1\rangle \otimes |\alpha_2 \rangle$ while on exchange, it becomes $|\alpha_2\rangle \otimes |\alpha_1 \rangle$. You can interpret this in the following way: Label particles 1 and 2. $|\alpha_1\rangle \otimes |\alpha_2 \rangle$ means particle one is in state $\alpha_1$ and particle two in state $\alpha_2$. On exchange, particle one sits in $\alpha_2$ and particle two in $\alpha_1$. If these were fermions the following relationship would hold:
$|\alpha_1\rangle \otimes |\alpha_2 \rangle=-|\alpha_2\rangle \otimes |\alpha_1 \rangle$, which is achieved in the standard manner, using canonical anti commutation relationship.
A: I will use creation operators for the state because it is easier to see the solution to the problem. You can build a two particle state from the vacuum $|0\rangle$ in two distinct ways:
$$ a_1^\dagger a_2^\dagger |0\rangle, \qquad a_2^\dagger a_1^\dagger |0\rangle. $$
Since the fermion creation operators anticommute ($\{a_1^\dagger,a_2^\dagger\}=0$), the two kets differ by a minus sign. Does that mean there is an ambiguity? No, because kets are only defined up to a complex phase, and all physical observables, the eigenvalues of a Hermitian operator, are the same.
You only need to be careful when you are calculating a matrix element, because you must be consistent on your choice. For example, if you choose $|\alpha\rangle = a_1^\dagger a_2^\dagger |0\rangle$, then you must choose for the bra
$$ \langle\alpha| = (a_1^\dagger a_2^\dagger |0\rangle)^\dagger = \langle 0|a_2a_1.$$
This is another way to see that the order of the creation operators doesn't matter. If you had picked the second option, the new ket and also the new bra would flip the sign, so that $\langle\alpha|O|\alpha\rangle$ is independent of your choice.
