Let's say $ \begin{bmatrix} 1\\ 0\\ \end{bmatrix} $ and $ \begin{bmatrix} 0\\ 1\\ \end{bmatrix} $ are the eigenvector of $\hat S_z$, is the state $ -1\begin{bmatrix} 1\\ 0\\ \end{bmatrix} +0\begin{bmatrix} 1\\ 0\\ \end{bmatrix} = \begin{bmatrix} -1\\ 0\\ \end{bmatrix} $ the same spin state of $ \begin{bmatrix} 1\\ 0\\ \end{bmatrix} $? If yes why do we have two different spinors for indicate the same state?
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2 Answers
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Yes they are, and not only those two but there are infinitely many vectors representing the same physical state. Don't forget that physical states in QM are represented by rays in Hilbert space. So in general, any state $|\psi\rangle$ it represents the same physical situation as $e^{i\phi}|\psi\rangle$. In your case you are just picking on the particular case of $\phi = \pi$
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2$\begingroup$ To substantiate this, notice that all predictions in QM can be written in terms of probabilities, and that probabilities can be written $\langle \psi | P| \psi \rangle$, where $P$ is a projector onto the eigenvectors (if $P$ is unfamiliar to you, replace it with $|\phi \rangle \langle \phi |$, that is the special case with only one eigenvector corresponding to the eigenvalue). Then if you replace $|\psi\rangle$ with $e^{i \alpha} |\psi\rangle$, that expression is unchanged bc $|e^{-i\alpha}|^2=1 $. So multiplying $\psi$ by a complex number of size $|z|=1$ does not affect predictions in QM. $\endgroup$ Commented Apr 24, 2020 at 19:02
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They both represent the same eigenstate. There is a global phase in the last eigenstate, but in the end, the global phases are not crucial since what is important is the square module of your eigenstate, which has the same value.