# A question about Pauli’s exclusion principle and electron orbital

According to Pauli’s exclusion principle, $s$ orbital contains at most two electrons with the opposite spin(up and down). Why can't $s$ orbital contain a third electron whose state is the linear combination of spin up and down?

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Edit: The Pauli exclusion principle is a consequence of the fact that electrons (and other spin-$\frac 12$ particles) are fermions, which means the wavefunction of the system must change sign when we exchange two such identical particles. If you think about the $s$-orbital, where a single electron has only two possible states, $\vert\uparrow\rangle$ and $|\downarrow\rangle$, you'll quickly see that the only possible state of two electrons is $$|\uparrow\downarrow\rangle - |\downarrow\uparrow\rangle$$ (Let me forget about normalisation, for simplicity.) The first arrow gives the spin of the first electron, and the second arrow gives the spin of the second electron. There is no way to get a totally anti-symmetric three-electron state.
What about hlew's comment, about one electron being in state $|\uparrow\rangle + |\downarrow\rangle$, and the other in state $|\uparrow\rangle - |\downarrow\rangle$? If one particle is in state $|a\rangle$, and the other in state $|b\rangle$, you might think that the overall system is in state $|a\rangle|b\rangle = |ab\rangle$, but because the electrons are identical fermions, we don't know 'which one is in which state', so we must take the anti-symmetric combination, i.e. $|ab\rangle - |ba\rangle$; in this case, we get \begin{eqnarray*} &&(|\uparrow\rangle - |\downarrow\rangle)(|\uparrow\rangle + |\downarrow\rangle) - (|\uparrow\rangle + |\downarrow\rangle)(|\uparrow\rangle - |\downarrow\rangle)\\[2ex] &\propto& |\uparrow\downarrow\rangle - |\downarrow\uparrow\rangle \end{eqnarray*} The $|\uparrow\uparrow\rangle$ and $|\downarrow\downarrow\rangle$ pieces cancel out; this is the exclusion principle.