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## Main Question or Discussion Point

Usual way of expressing pauli exclusion principle is, that the two identical fermions must be on different states. This seems a confusing statement, since the meaning of the term "different states" isn't very clear. Suppose I have two electrons, otherwise in the same states, but other one being spin up, and other one spin down. So the spin states are [tex](1,0)[/tex] and [tex](0,1)[/tex]. When I rotate the system about the x-axis with 90 degrees, the spin states become [tex](1,-i)/\sqrt{2}[/tex] and [tex](-i,1)/\sqrt{2}[/tex]. Now they are not in the "different states" in a naive sense, but this certainly cannot be a violation of the pauli exclusion principle. I can see that these vectors are still orthogonal. This was expected, since their orthogonality doesn't depend on the chosen representation. So wouldn't it be more correct to say, that the states of two identical fermions must be orthogonal? When I read and think at the same time, I see the expression "different states" meaning "orthogonal states". Is this correct?

The more fundamental fact is that the two particle wavefunction must satisfy the identity [tex]\psi(x_1,x_2)=-\psi(x_2,x_1)[/tex]. When two fermions are occupying states, whose spatial representations are [tex]\psi_1(x)[/tex] and [tex]\psi_2(x)[/tex], according to all books, the two particle wavefunction must be [tex]\psi(x_1,x_2)=(1/\sqrt{2})(\psi_1(x_1)\psi_2(x_2)-\psi_2(x_1)\psi_1(x_2))[/tex], so that the antisymmetry property is satisfied. The texts then say, that now we see that [tex]\psi_1[/tex] and [tex]\psi_2[/tex] must be different, because if they were the same the wavefunction would vanish. But how do you see that they must be orthogonal? Okey, if these states were eigenstates of some usual variable, everything would be fine, but if [tex]\psi_1[/tex] and [tex]\psi_2[/tex] are not eigenstates of some usual variable (like energy), then what? If antisymmetry of [tex]\psi(x_1,x_2)[/tex] implies orthogonality of [tex]\psi_1[/tex] and [tex]\psi_2[/tex], it can probably be shown with some calculation, but I haven't seen such anywhere.

It doesn't fully make sense to think particles occupying different (or orthogonal) states, since in the antisymmetrical wavefunction none of the particles is in some specific state. Still, in statistical stuff it is usually assumed that fermions simply are on the different states. Is this some kind of statistical approximation? You know, even if it wasn't fully correct, it would give correct statistical results and so on?

The more fundamental fact is that the two particle wavefunction must satisfy the identity [tex]\psi(x_1,x_2)=-\psi(x_2,x_1)[/tex]. When two fermions are occupying states, whose spatial representations are [tex]\psi_1(x)[/tex] and [tex]\psi_2(x)[/tex], according to all books, the two particle wavefunction must be [tex]\psi(x_1,x_2)=(1/\sqrt{2})(\psi_1(x_1)\psi_2(x_2)-\psi_2(x_1)\psi_1(x_2))[/tex], so that the antisymmetry property is satisfied. The texts then say, that now we see that [tex]\psi_1[/tex] and [tex]\psi_2[/tex] must be different, because if they were the same the wavefunction would vanish. But how do you see that they must be orthogonal? Okey, if these states were eigenstates of some usual variable, everything would be fine, but if [tex]\psi_1[/tex] and [tex]\psi_2[/tex] are not eigenstates of some usual variable (like energy), then what? If antisymmetry of [tex]\psi(x_1,x_2)[/tex] implies orthogonality of [tex]\psi_1[/tex] and [tex]\psi_2[/tex], it can probably be shown with some calculation, but I haven't seen such anywhere.

It doesn't fully make sense to think particles occupying different (or orthogonal) states, since in the antisymmetrical wavefunction none of the particles is in some specific state. Still, in statistical stuff it is usually assumed that fermions simply are on the different states. Is this some kind of statistical approximation? You know, even if it wasn't fully correct, it would give correct statistical results and so on?