Or must the spin state being measured be in the same direction as $\hat S_x $?
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3$\begingroup$ What you're asking doesn't really make sense. $\hat{S}_x|\uparrow_z\rangle$ is not an operator, it's a vector, so you can't compute its eigenvalue, because it doesn't have one, because it's not an operator. You can compute the eigenvalues and eigenvectors of $\hat{S}_x$, though. They are $\pm\hbar/2$ and $|\uparrow_z\rangle \pm |\downarrow_z\rangle$. Can you clarify your question? Also, "must the spin state being measured be in the same direction as the operator" doesn't make sense either... $\endgroup$– marchMar 15 at 16:03
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1$\begingroup$ ...because the quantity measured corresponds to an operator, so we choose the operator based on what measurement we're making. Can you clarify? $\endgroup$– marchMar 15 at 16:03
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$\begingroup$ My apologies, I've corrected the terminology now. $\endgroup$– TariusMar 15 at 16:10
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$\begingroup$ The question is still incoherent. What do you mean by “the vector $S_x$?” $S_x$ is not a vector. $\endgroup$– SandejoMar 15 at 16:19
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2$\begingroup$ $\hat{S}_x$ is an operator, $|\uparrow_z\rangle$ is a state(-vector) and so is $\hat{S}_x|\uparrow_z\rangle$. It only makes sense to talk about eigenvalues and eigenvectors of an operator, i.e. $\hat{S}_x$. Also: $\hat{S}_x$ does not indicate a direction, since it is an operator, and not a vector in the 3-dimensional coordinate space. $\endgroup$– SamuelMar 15 at 16:31
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Is it possible to calculate the eigenvalue for the spin vector $ S_x |↑_z⟩ $?
If you are asking whether $S_x|\uparrow_z\rangle$ is an eigenvector of some operator, the answer is yes. $$ S_x|\uparrow_z\rangle = \frac{1}{2}|\downarrow_z\rangle\;, $$ which is an eigenvector of $S_z$ with eigenvalue $-1$.
If you are asking whether or not it is possible to find the eigenvectors of $S_x$, the answer is yes.
In terms of eigenvectors of $S_z$ you can write the eigenvectors of $S_x$ as: $$ |+\rangle =\frac{1}{\sqrt{2}}\left(|\uparrow_z\rangle + |\downarrow_z\rangle\right) $$ and $$ |-\rangle =\frac{1}{\sqrt{2}}\left(|\uparrow_z\rangle - |\downarrow_z\rangle\right) $$
